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Learning Objectives
• To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems:
1. those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and
2. those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
The equilibrium constant for the decomposition of $CaCO_{3(s)}$ to $CaO_{(s)}$ and $CO_{2(g)}$ is $K = [CO_2]$. At 800°C, the concentration of $CO_2$ in equilibrium with solid $CaCO_3$ and $CaO$ is $2.5 \times 10^{-3}\; M$. Thus $K$ at 800°C is $2.5 \times 10^{-3}$ (remember that equilibrium constants are unitless). A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane).
This reaction can be written as follows:
$\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1}$
and the equilibrium constant $K = [\text{isobutane}]/[\text{n-butane}]$. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,
\begin{align*} K&=\dfrac{[\text{isobutane}]}{[\texT{n-butane}]} \[4pt] &=\dfrac{0.041\; M}{0.016\,M} \[4pt] &= 2.6 \label{Eq2} \end{align*}
Thus the equilibrium constant for the reaction as written is 2.6.
Example $1$
The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:
$\ce{2SO2(g) + O2(g) \rightleftharpoons 2SO3(g)} \nonumber$
A mixture of $SO_2$ and $O_2$ was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained
• $5.0 \times 10^{-2}\; M\; SO_3$,
• $3.5 \times 10^{-3}\; M\; O_2$, and
• $3.0 \times 10^{-3}\; M\; SO_2$.
Calculate $K$ and $K_p$ at this temperature.
Given: balanced equilibrium equation and composition of equilibrium mixture
Asked for: equilibrium constant
Strategy
Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain $K$.
Solution
Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,
$K=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})}=7.9 \times 10^4 \nonumber$
To solve for $K_p$, we need to identify $\Delta n$ where $Δn = 2 − 3 = −1$ and then
\begin{align*}K_p &=K(RT)^{Δn} \[4pt] K_p &=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1} \[4pt] K_p &=1.2 \times 10^3\end{align*}
Exercise $1$
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI (g) } \nonumber$
A mixture of $H_2$ and $I_2$ was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained
• $1.37\times 10^{−2}\; M\; HI$,
• $6.47 \times 10^{−3}\; M\; H_2$, and
• $5.94 \times 10^{-4}\; M\; I_2$.
Calculate $K$ and $K_p$ for this reaction.
Answer ($K$)
$K = 48.8$
Answer ($K_p$)
$K_p = 48.8$
Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example $2$ shows one way to do this.
Example $2$
A 1.00 mol sample of $NOCl$ was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of $Cl_2$. Calculate $K$ at this temperature. The equation for the decomposition of $NOCl$ to $NO$ and $Cl_2$ is as follows:
$\ce{2 NOCl (g) \rightleftharpoons 2NO(g) + Cl2(g)} \nonumber$
Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium
Asked for: $K$
Strategy:
1. Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).
2. Calculate all possible initial concentrations from the data given and insert them in the table.
3. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.
4. Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.
Solution
A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2} \nonumber$
To obtain the concentrations of $NOCl$, $NO$, and $Cl_2$ at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial
Change
Final
B Initially, the system contains 1.00 mol of $NOCl$ in a 2.00 L container. Thus $[NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M$. The initial concentrations of $NO$ and $Cl_2$ are $0\; M$ because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of $Cl_2$ in a 2.00 L container, so $[Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M$. We insert these values into the following table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change
Final 0.028
C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of $Cl_2$, the substance for which initial and final concentrations are known:
$Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M \nonumber$
According to the coefficients in the balanced chemical equation, 2 mol of $NO$ are produced for every 1 mol of $Cl_2$, so the change in the $NO$ concentration is as follows:
$Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M \nonumber$
Similarly, 2 mol of $NOCl$ are consumed for every 1 mol of $Cl_2$ produced, so the change in the $NOCl$ concentration is as follows:
$Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M \nonumber$
We insert these values into our table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change −0.056 +0.056 +0.028
Final 0.028
D We sum the numbers in the $[NOCl]$ and $[NO]$ columns to obtain the final concentrations of $NO$ and $NOCl$:
$[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M \nonumber$
$[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M \nonumber$
We can now complete the table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)} \nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.444 0.056 0.028
We can now calculate the equilibrium constant for the reaction:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4} \nonumber$
Exercise $2$
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia ($NH_3$) by reacting $0.1248\; M \;H_2$ and $0.0416\; M \;N_2$ at about 500°C. At equilibrium, the mixture contained 0.00272 M $NH_3$. What is $K$ for the reaction
$N_2+3H_2 \rightleftharpoons 2NH_3 \nonumber$
at this temperature? What is $K_p$?
Answer ($K$)
$K = 0.105$
Answer ($K_p$)
$K_p = 2.61 \times 10^{-5}$
Calculating Equilibrium Concentrations from the Equilibrium Constant
To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation $\ref{Eq1}$), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example $2$.
$\text{n-butane} (g) \rightleftharpoons \text{isobutane} (g)$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial
Change
Final
The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as +x, then the change in the concentration of n-butane is Δ[n-butane] = −x. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.
$\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)}$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial 1.00 0
Change −x +x
Final (1.00 − x) (0 + x) = x
Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\dfrac{x}{1.00−x}=2.6$
Rearranging and solving for $x$,
\begin{align*} x &=2.6(1.00−x) = 2.6−2.6x \[4pt] x+2.6x &= 2.6 \[4pt] x&=0.72 \end{align*}
We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n-butane and isobutane listed in the table:
$[\text{n-butane}]_f = (1.00 − x) M = (1.00 − 0.72) M = 0.28\; M$
$[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M$
We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same $K$ that we used in the calculation:
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) =2.6$
This is the same $K$ we were given, so we can be confident of our results.
Example $3$ illustrates a common type of equilibrium problem that you are likely to encounter.
Example $3$
The water–gas shift reaction is important in several chemical processes, such as the production of H2 for fuel cells. Thisreaction can be written as follows:
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber$
$K = 0.106$ at 700 K. If a mixture of gases that initially contains 0.0150 M $H_2$ and 0.0150 M $CO_2$ is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?
Given: balanced equilibrium equation, $K$, and initial concentrations
Asked for: final concentrations
Strategy:
1. Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x. From the values in the table, calculate the final concentrations.
2. Write the equilibrium equation for the reaction. Substitute appropriate values from the ICE table to obtain x.
3. Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K.
Solution
A The initial concentrations of the reactants are $[H_2]_i = [CO_2]_i = 0.0150\; M$. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of $H_2O$ as x, then $Δ[H_2O] = +x$. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x. For example, 1 mol of $CO$ is produced for every 1 mol of $H_2O$, so the change in the $CO$ concentration can be expressed as $Δ[CO] = +x$. Similarly, for every 1 mol of $H_2O$ produced, 1 mol each of $H_2$ and $CO_2$ are consumed, so the change in the concentration of the reactants is $Δ[H_2] = Δ[CO_2] = −x$. We enter the values in the following table and calculate the final concentrations.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber$
ICE $[H_2]$ $[CO_2]$ $[H_2O]$ $[CO]$
Initial 0.0150 0.0150 0 0
Change −x −x +x +x
Final (0.0150 − x) (0.0150 − x) x x
B We can now use the equilibrium equation and the given $K$ to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x)}=\dfrac{x^2}{(0.0150−x)^2}=0.106 \nonumber$
We could solve this equation with the quadratic formula, but it is far easier to solve for $x$ by recognizing that the left side of the equation is a perfect square; that is,
$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \nonumber$
Taking the square root of the middle and right terms,
$\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \nonumber$
$x =(0.326)(0.0150)−0.326x \nonumber$
$1.326x=0.00489 \nonumber$
$x =0.00369=3.69 \times 10^{−3} \nonumber$
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M$
• $[H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M$
• $[CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M$
We can check our work by inserting the calculated values back into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107 \nonumber$
To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.
Exercise $3$
Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)} \nonumber$
$K = 54$ at 425°C. If 0.172 M $H_2$ and $I_2$ are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?
Answer
$[HI]_f = 0.270 \;M$ $[H_2]_f = [I_2]_f = 0.037\; M$
In Example $3$, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example $4$.
Example $4$
In the water–gas shift reaction shown in Example $3$, a sample containing 0.632 M $\ce{CO2}$ and 0.570 M $\ce{H_2}$ is allowed to equilibrate at 700 K. At this temperature, $K = 0.106$. What is the composition of the reaction mixture at equilibrium?
Given: balanced equilibrium equation, concentrations of reactants, and $K$
Asked for: composition of reaction mixture at equilibrium
Strategy:
1. Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations (x) and the final concentrations.
2. Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x.
3. Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K.
Solution
A $[CO_2]_i = 0.632\; M$ and $[H_2]_i = 0.570\; M$. Again, x is defined as the change in the concentration of $H_2O$: $Δ[H_2O] = +x$. Because 1 mol of $CO$ is produced for every 1 mol of $H_2O$, the change in the concentration of $CO$ is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of $H_2$ and $CO_2$ are consumed for every 1 mol of $H_2O$ produced, $Δ[H_2] = Δ[CO_2] = −x$. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)} \nonumber$
ICE $H_{2(g)}$ $CO_{2(g)}$ $H_2O_{(g)}$ $CO_{(g)}$
Initial 0.570 0.632 0 0
Change −x −x +x +x
Final (0.570 − x) (0.632 − x) x x
B We can now use the equilibrium equation and the known $K$ value to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \nonumber$
In contrast to Example $3$, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:
$x^2 = 0.106(0.360 − 1.20x + x^2) \nonumber$
Collecting terms on one side of the equation,
$0.894x^2 + 0.127x − 0.0382 = 0 \nonumber$
This equation can be solved using the quadratic formula:
$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \nonumber$
$x =0.148 \text{ and } −0.290 \nonumber$
Only the answer with the positive value has any physical significance, so $Δ[H_2O] = Δ[CO] = +0.148 M$, and $Δ[H_2] = Δ[CO_2] = −0.148\; M$.
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M$
• $[H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M$
• $[CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M$
We can check our work by substituting these values into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \nonumber$
Because K is essentially the same as the value given in the problem, our calculations are confirmed.
Exercise $4$
The exercise in Example $1$ showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which $K = 54$ at 425°C. If a sample containing 0.200 M $H_2$ and 0.0450 M $I_2$ is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?
Answer
$[H_I]_f = 0.0882\; M$
$[H_2]_f = 0.156\; M$
$[I_2]_f = 9.2 \times 10^{−4} M$
In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ($K ≤ 10^{−3}$) or very large ($K ≥ 10^3$), which means that the change in the concentration (defined as x) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example $5$.
Example $5$
Atmospheric nitrogen and oxygen react to form nitric oxide:
$\ce{N2 (g) + O2 (g) \rightleftharpoons 2NO(g)}$
with $K_p = 2.0 \times 10^{−31}$ at 25°C.
What is the partial pressure of $\ce{NO}$ in equilibrium with $N_2$ and $O_2$ in the atmosphere (at 1 atm) $P_{N_2} = 0.78\; atm$ and $P_{O_2} = 0.21\; atm$?
Given: balanced equilibrium equation and values of $K_p$, $P_{O_2}$, and $P_{N_2}$
Asked for: partial pressure of NO
Strategy:
1. Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.
2. Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration (x).
3. Calculate the partial pressure of $NO$. Check your answer by substituting values into the equilibrium equation and solving for $K$.
Solution
A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of $O_2$ is 0.21 atm and that of $N_2$ is 0.78 atm. If we define the change in the partial pressure of $NO$ as 2x, then the change in the partial pressure of $O_2$ and of $N_2$ is −x because 1 mol each of $N_2$ and of $O_2$ is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
$\ce{N2 (g) + O2 (g) \rightleftharpoons 2NO(g)}$
ICE $P_{N_2}$ $P_{O_2}$ $P_{NO}$
Initial 0.78 0.21 0
Change −x −x +2x
Final (0.78 − x) (0.21 − x) 2x
B Substituting these values into the equation for the equilibrium constant,
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \nonumber$
In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation,
$\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31 \nonumber}$
$\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \nonumber$
$x^2=\dfrac{0.33 \times 10^{−31}}{4} \nonumber$
$x^=9.1 \times 10^{−17} \nonumber$
C Substituting this value of x into our expressions for the final partial pressures of the substances,
• $P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm$
• $P_{N_2}=(0.78−x) \;atm=0.78 \;atm$
• $P_{O_2}=(0.21−x) \;atm=0.21\; atm$
From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, $2.0 \times 10^{−16}$ is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or $10^{−3} > K > 10^3$, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic $NO$, an ingredient of smog, does not form from atmospheric concentrations of $N_2$ and $O_2$ to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 \times 10^{−31} \nonumber$
The final $K_p$ agrees with the value given at the beginning of this example.
Exercise $5$
Under certain conditions, oxygen will react to form ozone, as shown in the following equation:
$3O_{2(g)} \rightleftharpoons 2O_{3(g)} \nonumber$
with $K_p = 2.5 \times 10^{−59}$ at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere ($P_{O_2}=0.21\; atm$)?
Answer
$4.8 \times 10^{−31} \;atm$
Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (K ≥ 10^3). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example $6$.
Example $6$
The chemical equation for the reaction of hydrogen with ethylene ($C_2H_4$) to give ethane ($C_2H_6$) is as follows:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \nonumber$
with $K = 9.6 \times 10^{18}$ at 25°C. If a mixture of 0.200 M $H_2$ and 0.155 M $C_2H_4$ is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?
Given: balanced chemical equation, $K$, and initial concentrations of reactants
Asked for: equilibrium concentrations
Strategy:
1. Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.
2. Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).
3. Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.
Solution:
A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example $5$. If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \nonumber$
IACE $[H_{2(g)}]$ $[C_2H_{4(g)}]$ $[C_2H_{6(g)}]$
Initial 0.200 0.155 0
Assuming 100% reaction 0.045 0 0.155
Change +x +x −x
Final (0.045 + x) (0 + x) (0.155 − x)
B Substituting values into the equilibrium constant expression,
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \nonumber$
Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows:
$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \nonumber$
$x=3.6 \times 10^{−19} \nonumber$
C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:
• $[C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M$
• $[C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M$
• $[H_2]_f = (0.045 + x) \;M = 0.045 \;M$
We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \nonumber$
This $K$ value agrees with our initial value at the beginning of the example.
Exercise $6$
Hydrogen reacts with chlorine gas to form hydrogen chloride:
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \nonumber$
with $K_p = 4.0 \times 10^{31}$ at 47°C. If a mixture of 0.257 M $H_2$ and 0.392 M $Cl_2$ is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?
Answer
$[H_2]_f = 4.8 \times 10^{−32}\; M$
$[Cl_2]_f = 0.135\; M$
$[HCl]_f = 0.514\; M$
Summary
Various methods can be used to solve the two fundamental types of equilibrium problems:
1. those in which we calculate the concentrations of reactants and products at equilibrium and
2. those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.
When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.5%3A_Equilibrium_Calculations_for_Gas-Phase_and_Heterogenous_Reactions.txt |
Learning Objectives
• To predict in which direction a reaction will proceed.
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination.
The Reaction Quotient
To determine whether a system has reached equilibrium, chemists use a quantity called the Reaction Quotient ($Q$). The expression for the Reaction Quotient has precisely the same form as the equilibrium constant expression from the Law of Mass Action, except that $Q$ may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction:
$aA+bB \rightarrow cC+dD$
the Reaction Quotient is defined as follows:
$Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq1}$
To understand how information is obtained using a Reaction Quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide,
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}. \label{Rx}$
For this reaction, $K = 4.65 \times 10^{−3}$ at 298 K. We can write $Q$ for this reaction as follows:
$Q=\dfrac{[NO_2]^2}{[N_2O_4]} \label{Eq2}$
The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of $Q$ were calculated for each. Each experiment begins with different proportions of product and reactant:
Table $1$: Three Conditions of the dissociation of dinitrogen tetroxide to nitrogen dioxide (Equation $1$)
Experiment $[NO_2]\; (M)$ $[N_2O_4]\; (M)$ $Q = \dfrac{[NO^2]^2}{[N^2O^4]}$
1 0 0.0400 $\dfrac{0^2}{0.0400}=0$
2 0.0600 0 $\dfrac{(0.0600)^2}{0}=\text{undefined}$
3 0.0200 0.0600 $\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{−3}$
As these calculations demonstrate, $Q$ can have any numerical value between 0 and infinity (undefined); that is, $Q$ can be greater than, less than, or equal to K.
Comparing the magnitudes of $Q$ and K enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes $Q$ approach K. If $Q = K$, for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed. If $Q < K$, then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants. Conversely, if $Q > K$, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products. These points are illustrated graphically in Figure $1$.
If $Q < K$, the reaction will proceed to the right as written. If $Q > K$, the reaction will proceed to the left as written. If $Q = K$, then the system is at equilibrium.
Example $1$: steam-reforming
At elevated temperatures, methane ($CH_4$) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction:
$\ce{CH4(g) + H2O (g) \rightleftharpoons CO (g) + 3H2(g)} \nonumber$
$K = 2.4 \times 10^{−4}$ at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If $1.2 × 10^{−2}$ mol of $CH_4$, $8.0 × 10^{−3}$ mol of $H_2O$, $1.6 \times 10^{−2}$ mol of $CO$, and $6.0 \times 10^{−3}$ mol of $H_2$ are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce $CO$ and $H_2$ or to the left to form $CH_4$ and $H_2O$?
Given: balanced chemical equation, $K$, amounts of reactants and products, and volume
Asked for: direction of reaction
Strategy:
1. Calculate the molar concentrations of the reactants and the products.
2. Use Equation $\ref{Eq1}$ to determine $Q$. Compare $Q$ and $K$ to determine in which direction the reaction will proceed.
Solution:
A: We must first find the initial concentrations of the substances present. For example, we have $1.2 \times 10^{−2} mol$ of $CH_4$ in a 2.0 L container, so
\begin{align*} [CH_4]&=\dfrac{1.2\times 10^{−2} mol}{2.0\; L} \[4pt] &=6.0 \times 10^{−3} M \end{align*}
We can calculate the other concentrations in a similar way:
• $[H_2O] = 4.0 \times 10^{−3} M$,
• $[CO] = 8.0 \times 10^{−3} M$, and
• $[H_2] = 3.0 \times 10^{−3} M$.
B: We now compute $Q$ and compare it with $K$:
\begin{align*} Q&=\dfrac{[CO][H_2]^3}{[CH_4][H_2O]} \[4pt] &=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})} \[4pt] &=9.0 \times 10^{−6} \end{align*}
Because $K = 2.4 × 10^{−4}$, we see that $Q < K$. Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming $\ce{H2}$ and $\ce{CO}$ at the expense of $H_2O$ and $CH_4$.
Exercise $1$
In the water–gas shift reaction introduced in Example $1$, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen:
$\ce{ CO (g) + H2O (g) \rightleftharpoons CO2(g) +H2(g)} \nonumber$
$K = 0.64$ at 900 K. If 0.010 mol of both $\ce{CO}$ and $\ce{H_2O}$, 0.0080 mol of $\ce{CO_2}$, and 0.012 mol of $\ce{H_2}$ are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written?
Answer
$Q = 0.96$ (Q > K), so the reaction will proceed to the left, and $CO$ and $H_2O$ will form.
Predicting the Direction of a Reaction with a Graph
By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which $Q = K$. Such a graph allows us to predict what will happen to a reaction when conditions change so that $Q$ no longer equals $K$, such as when a reactant concentration or a product concentration is increased or decreased.
Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:
$\ce{PbCO3(s) \rightleftharpoons PbO(s) + CO2(g)} \label{Eq3}$
Because $PbCO_3$ and $PbO$ are solids, the equilibrium constant is simply $K = [CO_2]$. At a given temperature, therefore, any system that contains solid $PbCO_3$ and solid $PbO$ will have exactly the same concentration of $CO_2$ at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure $2$, which shows a plot of $[CO_2]$ versus the amount of $PbCO_3$ added. Initially, the added $PbCO_3$ decomposes completely to $CO_2$ because the amount of $PbCO_3$ is not sufficient to give a $CO_2$ concentration equal to $K$. Thus the left portion of the graph represents a system that is not at equilibrium because it contains only CO2(g) and PbO(s). In contrast, when just enough $PbCO_3$ has been added to give $[CO_2] = K$, the system has reached equilibrium, and adding more $PbCO_3$ has no effect on the $CO_2$ concentration: the graph is a horizontal line. Thus any $CO_2$ concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough $PbCO_3$ and $PbO$ are present. For example, the point labeled A in Figure $2$ lies above the horizontal line, so it corresponds to a $[CO_2]$ that is greater than the equilibrium concentration of $CO_2$ (Q > K). To reach equilibrium, the system must decrease $[CO_2]$, which it can do only by reacting $CO_2$ with solid $PbO$ to form solid $PbCO_3$. Thus the reaction in Equation $\ref{Eq3}$ will proceed to the left as written, until $[CO_2] = K$. Conversely, the point labeled B in Figure $2$ lies below the horizontal line, so it corresponds to a $[CO_2]$ that is less than the equilibrium concentration of $CO_2$ (Q < K). To reach equilibrium, the system must increase $[CO_2]$, which it can do only by decomposing solid $PbCO_3$ to form $CO_2$ and solid $PbO$. The reaction in Equation $\ref{Eq3}$ will therefore proceed to the right as written, until $[CO_2] = K$.
In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:
$\ce{CdO (s) + H2(g) \rightleftharpoons Cd (s) + H2O(g)} \label{Eq4}$
and the equilibrium constant K is $[H_2O]/[H_2]$. If $[H_2O]$ is doubled at equilibrium, then [H2] must also be doubled for the system to remain at equilibrium. A plot of $[H_2O]$ versus $[H_2]$ at equilibrium is a straight line with a slope of K (Figure $3$). Again, only those pairs of concentrations of $H_2O$ and $H_2$ that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation $\ref{Eq4}$ will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in Figure $3$ lies below the line, indicating that the $[H_2O]/[H_2]$ ratio is less than the ratio of an equilibrium mixture (Q < K). Thus the reaction in Equation $\ref{Eq4}$ will proceed to the right as written, consuming $H_2$ and producing $H_2O$, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in Figure $3$ lies above the line, indicating that the $[H_2O]/[H_2]$ ratio is greater than the ratio of an equilibrium mixture (Q > K). Thus the reaction in Equation $\ref{Eq4}$ will proceed to the left as written, consuming $H_2O$ and producing $H_2$, which causes the concentration ratio to move down and to the right toward the equilibrium line.
In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:
$NH_4I_{(s)} \rightleftharpoons NH_{3(g)}+HI_{(g)} \label{Eq5}$
For this system, $K$ is equal to the product of the concentrations of the two products: $[NH_3][HI]$. If we double the concentration of NH3, the concentration of HI must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure $4$. As a result, for a given concentration of either $HI$ or $NH_3$, only a single equilibrium composition that contains equal concentrations of both $NH_3$ and HI is possible, for which $[NH_3] = [HI] = K^{1/2}$. Any point that lies below and to the left of the equilibrium curve (such as point A in Figure $4$) corresponds to $Q < K$, and the reaction in Equation $\ref{Eq5}$ will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure $4$) corresponds to $Q > K$, and the reaction in Equation $\ref{Eq5}$ will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium.
Effect of Change in Pressure on Equilibrium
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for $K_c$) or partial pressure (for $K_p$). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which $\ce{NO}$, $\ce{O_2}$, and $\ce{NO_2}$ are at equilibrium:
$\ce{2NO}_{(g)}+\ce{O}_{2(g)} \rightleftharpoons \ce{2NO}_{2(g)} \label{Eq8}$
The formation of additional amounts of $\ce{NO2}$ decreases the total number of molecules in the system because each time two molecules of $\ce{NO_2}$ form, a total of three molecules of $\ce{NO}$ and $\ce{O_2}$ are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of $\ce{NO_2}$ into $\ce{NO}$ and $\ce{O_2}$, which tends to restore the pressure.
Now consider this reaction:
$\ce{N2}(g)+\ce{O2}(g)\rightleftharpoons\ce{2NO}(g) \label{Eq9}$
Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.
Effect of Change in Temperature on Equilibrium
Changing concentration or pressure perturbs an equilibrium because the Reaction Quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle.
When hydrogen reacts with gaseous iodine, heat is evolved.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI (g)} \;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{Eq10}$
Because this reaction is exothermic, we can write it with heat as a product.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI (g) + heat} \label{Eq11}$
Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of $\ce{H2}$ and $\ce{I2}$ and a reduction in the concentration of $\ce{HI}$. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.
When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the $\ce{HI}$ system increases the equilibrium constant: At the new equilibrium the concentration of $\ce{HI}$ has increased and the concentrations of $\ce{H2}$ and $\ce{I2}$ decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.
Temperature affects the equilibrium between $\ce{NO_2}$ and $\ce{N_2O_4}$ in this reaction
$\ce{N_2O}_{4(g)} \rightleftharpoons \ce{2NO}_{2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{Eq12}$
The positive $ΔH$ value tells us that the reaction is endothermic and could be written
$\ce{heat}+\ce{N_2O}_{4(g)} \rightleftharpoons \ce{2NO}_{2(g)} \label{Eq13}$
At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown $\ce{NO_2}$ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless $\ce{N_2O_4}$ increases, and the concentration of brown $\ce{NO_2}$ decreases, causing the brown color to fade.
Temperature is Neither a Reactant nor Product
It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" ($\Delta{H} > 0$), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" ($\Delta{H} < 0$), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below.
Example $2$: Steam Reforming of Methane
The commercial production of hydrogen is carried out by treating natural gas with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”):
$\ce{CH_4 + H_2O <=> CH_3OH + H_2} \nonumber$
Given the following boiling points: CH4 (methane) = –161°C, H2O = 100°C, CH3OH = 65°, H2 = –253°C, predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C.
Solution
To identify the influence of changing pressure on the three reaction conditions, we need to identify the correct reaction including the phase of each reactant and product. Calculate the change in the moles of gas for each process:
Temperature
Equation
$Δn_g$
shift
50° $[\ce{CH4(g) + H2O(l) → CH3OH(l) + H2(g)}$ 0 none 75° $\ce{CH4(g) + H2O(l) → CH3OH(g) + H2(g)}$ +1 to left 120° $\ce{CH4(g) + H2O(g) → CH3OH(g) + H2(g)}$ 0 none
Exercise $2$
What will happen to the equilibrium when the volume of the system is decreased?
$\ce{2SO2 (g) + O2 (g) \rightleftharpoons 2SO3 (g)} \nonumber$
Answer
Decreasing the volume leads to an increase in pressure which will cause the equilibrium to shift towards the side with fewer moles. In this reaciton, there are three moles on the reactant side and two moles on the product side, so the new equilibrium will shift towards the products (to the right).
Summary
The Reaction Quotient (Q) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. Reaction Quotient: $Q =\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber$ The Reaction Quotient ($Q$ or $Q_p$) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, $Q = K$. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium.
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.
Effects of Disturbances of Equilibrium and K
Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K
reactant added added reactant is partially consumed toward products none
product added added product is partially consumed toward reactants none
decrease in volume/increase in gas pressure pressure decreases toward side with fewer moles of gas none
increase in volume/decrease in gas pressure pressure increases toward side with more moles of gas none
temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes
temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes
Footnotes
1. 1 Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.
Glossary
Le Chatelier's principle
when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium
concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress
change to a reaction's conditions that may cause a shift in the equilibrium | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.6%3A_Reaction_Directions_%28Empirical_Explanation%29.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously:
• $ΔS_{univ} > 0$ (2nd law of thermodynamics),
• $ΔG_{sys} < 0$ (Gibbs Energy version), and
• the relative magnitude of the reaction quotient $Q$ versus the equilibrium constant $K$.
Recall that if $Q < K$, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if $Q > K$, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If $Q = K$, then the system is at equilibrium, and no net reaction occurs. Table $1$ summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes.
Table $1$: Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
*Spontaneous in the reverse direction.
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
Because all three criteria assess the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction ($ΔG°$) and the equilibrium constant ($K$).
Free Energy and the Equilibrium Constant
Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. "Free Energy", ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating $ΔH$ from the equation for $ΔG$. The general relationship can be shown as follow (derivation not shown):
$\Delta G = V \Delta P − S \Delta T \label{18.29}$
If a reaction is carried out at constant temperature ($ΔT = 0$), then Equation $\ref{18.29}$ simplifies to
$\Delta{G} = V\Delta{P} \label{18.30}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the $V$ in Equation $\ref{18.30}$ by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express $\Delta{G}$ in terms of the initial and final pressures ($P_i$ and $P_f$, respectively):
\begin{align} \Delta G &=\left(\dfrac{nRT}{P}\right)\Delta P \[4pt] &=nRT\dfrac{\Delta P}{P}=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31} \end{align}
If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
$G − G^° = nRT\ln{P}$
This can be rearranged as follows:
$G = G^° + nRT\ln {P} \label{18.32}$
As you will soon discover, Equation $\ref{18.32}$ allows us to relate $ΔG^o$ and $K_p$. Any relationship that is true for $K_p$ must also be true for $K$ because $K_p$ and $K$ are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA+bB \rightleftharpoons cC+dD \label{18.33}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for $ΔG$:
\begin{align} \Delta{G}&=\sum_m G_{products}−\sum_n G_{reactants} \[4pt] &=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34} \end{align}
Substituting Equation $\ref{18.32}$ for each term into Equation $\ref{18.34}$,
$ΔG=[(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)]$
Combining terms gives the following relationship between $ΔG$ and the reaction quotient $Q$:
\begin{align} \Delta G &=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right) \[4pt] &=\Delta G^\circ+RT\ln Q \label{18.35} \end{align}
where $ΔG°$ indicates that all reactants and products are in their standard states. For gases at equilibrium ($Q = K_p$,), and as you’ve learned in this chapter, $ΔG = 0$ for a system at equilibrium. Therefore, we can describe the relationship between $ΔG^o$ and $K_p$ for gases as follows:
\begin{align} 0 &= ΔG° + RT\ln K_p \label{18.36a} \[4pt] ΔG° &= −RT\ln K_p \label{18.36b} \end{align}
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then $K_p = 1$, and neither reactants nor products are favored: the system is at equilibrium.
For a spontaneous process under standard conditions, $K_{eq}$ and $K_p$ are greater than 1.
To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure $3$). If a system is present with reactants and products present in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.
Example $2$
$ΔG^o$ is −32.7 kJ/mol of N2 for the reaction
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}\nonumber$
This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate $ΔG$ for the same reaction under the following nonstandard conditions:
• $P_{\textrm N_2}$ = 2.00 atm,
• $P_{\textrm H_2}$ = 7.00 atm,
• $P_{\textrm{NH}_3}$ = 0.021 atm,
• and T = 100°C.
Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
1. Using the values given and Equation $\ref{18.35}$, calculate $Q$.
2. Substitute the values of ΔG° and Q into Equation $\ref{18.35}$ to obtain $ΔG$ for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation $\ref{18.35}$. Substituting the partial pressures given, we can calculate $Q$:
$Q=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}}=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4\times10^{-7}\nonumber$
B Substituting the values of ΔG° and Q into Equation $\ref{18.35}$,
\begin{align*} \Delta G &=\Delta G^\circ+RT\ln Q \[4pt] &=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right] \[4pt] &=-32.7\textrm{ kJ}+(-44\textrm{ kJ}) \[4pt] &=-77\textrm{ kJ/mol of N}_2 \end{align*}
Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise $2$
Calculate $ΔG$ for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, $P_{\mathrm{O_2}}$ = 0.200 atm, and $P_{\mathrm{NO_2}}$ = 1.00 × 10−4 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer
−92.9 kJ/mol of O2; the reaction is spontaneous to the right as written, so products are favored.
Example $3$
Calculate Kp for the reaction of H2 with N2 to give NH3 at 25°C. $ΔG^o$ for this reaction is −32.7 kJ/mol of N2.
Given: balanced chemical equation from Example $2$, ΔG°, and temperature
Asked for: Kp
Strategy:
Substitute values for ΔG° and T (in kelvin) into Equation $\ref{18.36b}$ to calculate $K_p$, the equilibrium constant for the formation of ammonia.
Solution
In Example 10, we used tabulated values of ΔGf to calculate ΔG° for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation $\ref{18.36b}$,
\begin{align*} \Delta G^\circ &=-RT\ln K_\textrm p \[4pt] \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \end{align*}
Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
\begin{align*}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \[4pt] K_\textrm p &=5.4\times10^5\end{align*}
Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. However, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise $3$
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. ΔG° for this reaction is −70.5 kJ/mol of O2.
Answer
2.2 × 1012
Although $K_p$ is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant $K$ is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of $K_p$ and $K$ previously and showed that they are related:
$K_p = K(RT)^{Δn} \label{18.37}$
where $Δn$ is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, $Δn = 0$, so $K_p = K$. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation $\ref{18.36b}$ can be written in a more general form:
$ΔG° = −RT \ln K \label{18.38}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation $\ref{18.38}$ for the difference between $K_p$ and $K$. Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equations $\ref{18.38}$ with $ΔG^o = ΔH^o − TΔS^o$ provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
$ΔG° = ΔH° − TΔS° = −RT \ln K \label{18.39}$
Notice that $K$ becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible.
Temperature Dependence of the Equilibrium Constant
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation $\ref{18.39}$, which can be rearranged as follows:
$\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}$
Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation $\ref{18.40}$ agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation $\ref{18.40}$ also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K1 and K2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively. Applying Equation $\ref{18.40}$ gives the following relationship at each temperature:
\begin{align}\ln K_1&=\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \[4pt] \ln K_2 &=\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R}\end{align}
Subtracting $\ln K_1$ from $\ln K_2$,
$\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}$
Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K1) allow us to calculate the value of the equilibrium constant at any other temperature (K2), assuming that ΔH° and ΔS° are independent of temperature. Equation $\ref{18.41}$ is often referred to as the van 't Hoff equation after Dutch chemist Jacobus Henricus van 't Hoff in 1884 in his book Études de Dynamique chimique (Studies in dynamic chemistry).
Example $4$
The equilibrium constant for the formation of NH3 from H2 and N2 at 25°C was calculated to be Kp = 5.4 × 105 in Example $2$. What is Kp at 500°C? (Use the data from Example $1$)
Given: balanced chemical equation, ΔH°, initial and final T, and Kp at 25°C
Asked for: Kp at 500°C
Strategy:
Convert the initial and final temperatures to kelvin. Then substitute appropriate values into Equation $\ref{18.41}$ to obtain $K_2$, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set $T_1$ = 25°C = 298 K and $T_2$ = 500°C = 773 K, then from Equation $\ref{18.41}$ we obtain the following:
\begin{align*}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \[4pt] \dfrac{K_2}{K_1}&=1.3\times10^{-10} \[4pt] K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align*}
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise $4$
The equilibrium constant for the reaction of $\ce{NO}$ with $\ce{O2}$ to give $\ce{NO2}$ at 25°C is $K_p = 2.2 \times 10{12}$. Use the $ΔH^o_f$ values in the exercise in Example $4$ to calculate $K_p$ for this reaction at 1000°C.
Answer
5.6 × 10−4
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express $ΔG$ in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or Kp > 1, and products are favored over reactants. If ΔG° > 0, then K or Kp < 1, and reactants are favored over products. If ΔG° = 0, then K or Kp = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.7%3A_Reaction_Directions_%28Thermodynamic_Explanation%29.txt |
A partitioning of a compound exist between a mixture of two immiscible phases at equilibrium, which is a measure of the difference in solubility of the compound in these two phases. If one of the solvents is a gas and the other a liquid, the "gas/liquid partition coefficient" is the same as the dimensionless form of the Henry's law constant. A solute can partition when one or both solvents is a solid (e.g., solid solution).
Liquid-Liquid Extractions
We call the process of moving a species from one phase to another phase an extraction. Simple extractions are particularly useful for separations where only one component has a favorable partition coefficient. Several important separation techniques are based on a simple extraction, including liquid–liquid, liquid–solid, solid–liquid, and gas–solid extractions. The most important group of separation techniques uses a selective partitioning of the solute between two immiscible phases. If we bring a phase containing a solute, S, into contact with a second phase, the solute partitions itself between the two phases, as shown by the following equilibrium reaction.
$\textrm S_\textrm{phase 1}\rightleftharpoons \textrm S_\textrm{phase 2}\label{7.20}$
The equilibrium constant for this equilbrium is
$K_\textrm D=\mathrm{\dfrac{[S_{phase\;2}]}{[S_{phase\;1}]}}$
is called the distribution constant or partition coefficient. If $K_D$ is sufficiently large, then the solute moves from phase 1 to phase 2. The solute remains in phase 1 if the partition coefficient is sufficiently small. When we bring a phase containing two solutes into contact with a second phase, if KD is favorable for only one of the solutes a separation of the solutes is possible. The physical states of the phases are identified when describing the separation process, with the phase containing the sample listed first. For example, if the sample is in a liquid phase and the second phase is a solid, then the separation involves liquid–solid partitioning.
It often happens that two immiscible liquid phases are in contact, one of which contains a solute. How will the solute tend to distribute itself between the two phases? One’s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases, since this would correspond to the maximum dispersion (randomness) of the solute. This, however, does not take into the account the differing solubilities the solute might have in the two liquids; if such a difference does exist, the solute will preferentially migrate into the phase in which it is more soluble (Figure $1$ ).
Figure $1$: Scheme for a simple liquid–liquid extraction in which the solute’s partitioning depends only on the KD equilibrium.
Biomagnification
The transport of substances between different phases is of immense importance in such diverse fields as pharmacology and environmental science. For example, if a drug is to pass from the aqueous phase with the stomach into the bloodstream, it must pass through the lipid (oil-like) phase of the epithelial cells that line the digestive tract. Similarly, a pollutant such as a pesticide residue that is more soluble in oil than in water will be preferentially taken up and retained by marine organism, especially fish, whose bodies contain more oil-like substances; this is basically the mechanism whereby such residues as DDT can undergo biomagnification as they become more concentrated at higher levels within the food chain. For this reason, environmental regulations now require that oil-water distribution ratios be established for any new chemical likely to find its way into natural waters. The standard “oil” phase that is almost universally used is octanol, C8H17OH.
In preparative chemistry it is frequently necessary to recover a desired product present in a reaction mixture by extracting it into another liquid in which it is more soluble than the unwanted substances. On the laboratory scale this operation is carried out in a separatory funnel as shown below. The two immiscible liquids are poured into the funnel through the opening at the top. The funnel is then shaken to bring the two phases into intimate contact, and then set aside to allow the two liquids to separate into layers, which are then separated by allowing the more dense liquid to exit through the stopcock at the bottom.
If the distribution ratio is too low to achieve efficient separation in a single step, it can be repeated; there are automated devices that can carry out hundreds of successive extractions, each yielding a product of higher purity. In these applications our goal is to exploit the Le Chatelier principle by repeatedly upsetting the phase distribution equilibrium that would result if two phases were to remain in permanent contact.
Example $1$
The distribution ratio for iodine between water and carbon disulfide is 650. Calculate the concentration of I2 remaining in the aqueous phase after 50.0 mL of 0.10M I2 in water is shaken with 10.0 mL of CS2.
Solution
The equilibrium constant is
$K_d = \dfrac{C_{CS_2}}{C_{H_2O}} = 650 \nonumber$
Let m1 and m2 represent the numbers of millimoles of solute in the water and CS2 layers, respectively. Kd can then be written as (m2/10 mL) ÷ (m1/50 mL) = 650. The number of moles of solute is (50 mL) × (0.10 mmol mL–1) = 5.00 mmol, and mass conservation requires that m1 + m2 = 5.00 mmol, so m2 = (5.00 – m1) mmol and we now have only the single unknown m1. The equilibrium constant then becomes
$\dfrac{(5.00 – m_1)\, mmol / 10\, mL}{ m_1\, mmol / 50\, mL} = 650 \nonumber$
Simplifying and solving for $m_1$ yields
$\dfrac{(0.50 – 0.1)m_1}{0.02\, m_1} = 650 \nonumber$
with m1 = 0.0382 mmol.
The concentration of solute in the water layer is (0.0382 mmol) / (50 mL) = 0.000763 M, showing that almost all of the iodine has moved into the CS2 layer.
Chromatographic Separations
In an extraction, the sample is one phase and we extract the analyte or the interferent into a second phase. We also can separate the analyte and interferents by continuously passing one sample-free phase, called the mobile phase, over a second sample-free phase that remains fixed or stationary. The sample is injected into the mobile phase and the sample’s components partition themselves between the mobile phase and the stationary phase. Those components with larger partition coefficients are more likely to move into the stationary phase, taking a longer time to pass through the system. This is the basis of all chromatographic separations. Chromatography provides both a separation of analytes and interferents, and a means for performing a qualitative or quantitative analysis for the analyte.
Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure $3$ provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced at the top of the column as a narrow band. Ideally, the solute’s initial concentration profile is rectangular (Figure $\PageIndex{1a}$). As the sample moves down the column the solutes begin to separate (Figures $\PageIndex{1b,c}$), and the individual solute bands begin to broaden and develop a Gaussian profile. If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure $\PageIndex{1d}$).
We can follow the progress of the separation either by collecting fractions as they elute from the column (Figure $4$), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a chromatogram (Figure $5$), and consists of a peak for each solute. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.8%3A_Distribution_of_a_Single_Species_between_Immiscible_Phases_-_Extraction.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q1
Write an expression for the equilibrium constant $K_c$ for each reaction below.
1. $\ce{2 NO(g) + O2(g) <=> 2 NO2(g)}$
2. $\ce{N2H4(g) + 3 O2(g) <=> 2 NO2(g) + H2O(g)}$
3. $\ce{O3(g) + H2O(l) <=> H2(g) + 2 O2(g)}$
Solution
For a general reaction $\ce{aA + bB <=> cC + dD}$ then Law of Mass Action argues we can construct an equilibrium constant thusly
$K_c = \ce{\dfrac{[C]^c[D]^d}{[A]^a[B]^b}} \nonumber$
This is the general equations to substitute the reactants and product concentrations (with coefficients as exponents).
$K_c = \ce{\dfrac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}} \nonumber$
1. $K_c = \ce{\dfrac{[NO2]^2}{[NO]^2 [O2]}}$
2. $K_c = \ce{\dfrac{[H2O][NO2]^2}{[N2H4] [O2]^3}}$
3. $K_c = \ce{\dfrac{[O2]^2 [H2]}{[O3]}}$
Note that for part c, $\ce{H2O (l)}$ is not included because pure solids and liquids are not included in the equilibrium constant.
Q3
Gaseous chlorine reacts with water vapor to form hydrogen chloride and oxygen gas. Write down the equilibrium expression ($K_p$) for the reaction.
Solution
First, a balanced reaction is constructed to ensure the equilibrium expression will be correctly states.
$\ce{2Cl_2 + 2H_2O <=> 4HCl + O2} \nonumber$
Equilibrium expression $K_p$ is as follows:
$K_p = \dfrac{p^4_{\ce{HCl}} p_{\ce{O2}}}{p^2_{\ce{Cl2}}p^2_{\ce{H2O}}}\nonumber$
Q11
For each thermodynamics equilibrium expressions outlined below, determine one possible chemical reaction related to each.
1. $K=\frac{a\{\ce{CO}\}^{2}}{a\{\ce{CO_{2}}\}}\nonumber$
2. $K=\frac{a\{\ce{H_{2}O}\}^{2}}{a\{\ce{H_{2}}\}^{2} a\{\ce{O_{2} }\} }\nonumber$
3. $K=\frac{a\{\ce{Cl_{2}O}\}^{2}}{a\{\ce{Cl_{2}}\}^{2} a\{\ce{O_{2}}\} }\nonumber$
4. $K=\frac{a\{\ce{COCl_{2}}\}}{a\{\ce{CO}\}\cdot a\{\ce{Cl_{2}}\}}\nonumber$
Solution
1. $\ce{CO2(g) + C(s) <=> 2 CO(g)}\nonumber$
2. $\ce{2 H2(g) + O2(g) <=> 2 H2O(g)}\nonumber$
3. $\ce{2 Cl2(g) + O2(g) <=> 2 Cl2O(g)}\nonumber$
4. $\ce{CO(g) + Cl2(g) <=> COCl2(g)}\nonumber$
Q15A
Based on the following data, calculate the equilibrium constant and the value of ΔG at 273 K.
a. $\ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}$ with
• $[\ce{CO2}]_{\text{eq}} = 0.0954 \,\text{M}$
• $[\ce{H2}]_{\text{eq}} = 0.0454 \,\text{M}$
• $[\ce{CO}]_{\text{eq}} = 0.0046 \,\text{M}$
• $[\ce{H2O}]_{\text{eq}} = 0.0046 \,\text{M}$
b. $\ce{2NO(g) + 2H2(g) <=> N2(g) + 2H2O(g)}$ with
• $[\ce{NO}]_{\text{eq}} = 0.45\, \text{M}$
• $[\ce{H2}]_{\text{eq}} = 0.63 \,\text{M}$
• $[\ce{N2}]_{\text{eq}} = 0.95 \,\text{M}$
• $[\ce{H2O}]_{\text{eq}} = 1.3 \, \text{M}$
Solution
a) \begin{align*} K &= \dfrac{\ce{[CO]}\ce{[H2O]}}{\ce{[CO2]}\ce{[H2]}} \[4pt] &= \dfrac{(0.0046)(0.0046)}{(0.0954)(0.0454)} \[4pt] &= 4.9 \times 10^{-3} \end{align*}
$\Delta G = R T \ln(K) = (8.3145 \, \text{J/(K mol)})(273 \, \text{K}) \ln(4.9 \times 10^{-3}) = -12072.288 \, \text{J/mol} \nonumber$
b) $K = \ce{\frac{[N2][H2O]^2}{[NO]^2[H2]^2}} = \frac{[0.95][1.3]^2}{[0.45]^2[0.63]^2} = 19.98 \nonumber$
$\Delta G = R T \ln(K) = (8.3145 \, \text{J/K mol})(273 \, \text{K}) \ln(19.98) = 6797.62 \, \text{J/mol} \nonumber$
Q15B
Calculate the appropriate K value for the following reactions using the information provided. Assume that they are preformed under standard conditions (and 298.15 K).
1. $\ce{2 NO2(g) <=> 2 NO(g) + O2(g) }\nonumber$
2. $\ce{NH4+ (aq) + H2O(l) <=> H3O+ (aq) + NH3 (aq)}\nonumber$
3. Jim’s brother, Txeltoqlztop, goes to the bathroom only to find out that his sibling did not refill the toilet paper roll! To get revenge, Txeltoqlztop decides to fill his brother’s room with carbon monoxide, via the following reaction: $\ce{2 CH4 (g) + 3 O2 (g) <=> 2 CO(g) + 4 H2O(g)} \nonumber$ Calculate the K value for this equation under standard conditions, interpret what it means in regards to the direction this equation will go, and compare it to the other two.
Solution
The core of this problem is to find the K value and understanding what information it can describe about the reaction. Since the problem provides many, many $\Delta{G_{f}}$, it is likely that to calculate the K value, a relationship needs to be drawn from $\Delta{G}$ to K. This relationship is as follows:
$\Delta{G_{\text{rxn}}^{\circ}}= -RT \ln\left(K_{p}\right)\nonumber$
Relevant Information obtained from the thermodynamic tables
$\Delta{G_f^{\circ}}\; \ce{NO2(g)} = 51.30 \mathrm{\dfrac{kJ}{mol}}$ $\Delta{G_f^{\circ}}\; \ce{H3O^+(aq)} = -103.45 \mathrm{\dfrac{kJ}{mol}}$
$\Delta{G_f^{\circ}}\; \ce{NO(g)} = 86.57 \mathrm{\dfrac{kJ}{mol}}$ $\Delta{G_f^{\circ}}\; \ce{NH3(aq)} = -26.5 \mathrm{\dfrac{kJ}{mol}}$
$\Delta{G_f^{\circ}}\; \ce{O2(g)} = 0 \mathrm{\dfrac{kJ}{mol}}$ $\Delta{G_f^{\circ}}\; \ce{CH4(g)} = -50.84 \mathrm{\dfrac{kJ}{mol}}$
$\Delta{G_f^{\circ}}\; \ce{NH4^+ (aq)} = -79.37 \mathrm{\dfrac{kJ}{mol}}$ $\Delta{G_f^{\circ}}\; \ce{CO(g)} = -137.28 \mathrm{\dfrac{kJ}{mol}}$
$\Delta{G_f^{\circ}}\; \ce{H2O(l)} = -237.14 \mathrm{\dfrac{kJ}{mol}}$ $\Delta{G_f^{\circ}}\; \ce{H2O(g)} = -228.61 \mathrm{\dfrac{kJ}{mol}}$
Note that the relationship here is between $\Delta{G}$ and $K_p$, which is:
$K_p \approx \dfrac{P_{\text{products}}}{P_{\text{reactants}}} \nonumber$
Where $P$ is partial pressure of the products or reactants. The reason this is an approximation is because the equilibrium actually measures the activities of the species, but by using just the partial pressures a good estimate is procured. Since the equation relating $K_p$ and $\Delta{G}$ only has these two variables as unknown, once $\Delta{G}$ is found, algebra can find $K_p$. The process for utilizing Hess's Law to find state variables is outlined extensively. The values below were found using the exact same process, now for $\Delta{G}$ (Just in case, the Hess law formula is provided).
$\Delta{G^{\circ}} = \sum{n_{\text{products}} \times {\Delta{G_{f_{\text{products}}}^{\circ}}}} - \sum{n_{\text{reactants}}\times{\Delta{G_{f_{\text{reactants}}}^{\circ}}}}$
$\Delta G^{\circ} (\ce{2 NO2(g)<=> 2 NO(g) + O2 (g)}) = 70.42 \, \mathrm{\dfrac{kJ}{mol}}$
$\Delta G^{\circ} (\ce{NH4^+ (aq) + H2O(l) <=> H3O^+(aq) + NH3 (aq)}) = -437.93 \mathrm{\dfrac{kJ}{mol}}$
$\Delta G^{\circ} (\ce{2 CH4 (g) + 3 O2 (g) <=> 2 CO(g) + 4 H2O (l)}) = -1121.44 \mathrm{\dfrac{kJ}{mol}}$
To find K, plug in the $\Delta{G}$ into the equation above. (Note that the solution does NOT show the conversion of the gas constant from J to kJ!!)
$\Delta{G_{\text{rxn}}^{\circ}}= -\left(8.314 \, \mathrm{\dfrac{J}{mol \, K}} \right)\left(298.15 \, \mathrm{K}\right)\times{\ln\left(K_{p}\right)}$
$70.54 \mathrm{\dfrac{kJ}{mol}} = -\left(8.314 \mathrm{\dfrac{J}{mol \, K}}\right)\left(298.15 \, \mathrm{K}\right)\times{\ln\left(K_{p}\right)}$
$-\dfrac{70.54 \mathrm{\dfrac{kJ}{mol}}}{\left(8.314 \mathrm{\dfrac{J}{molK}}\right)\left(298.15 \, \mathrm{K}\right)}= \ln\left(K_{p}\right)$
$e^{-\dfrac{70.54 \mathrm{\dfrac{kJ}{mol}}}{\left(8.314 \mathrm{\dfrac{J}{mol \, K}}\right)\left(298.15 \, \mathrm{K}\right)}} = K_{p}$
$K_{p} \approx e^{-28.457} \approx 4.3776\times10^{-13}$
Thus, the $K_{p}$ value for $\ce{2 NO2 (g) <=> 2 NO (g) + O2 (g)}$ is a very small number!. Since the equilibrium value depicts the ratio of products to reactants at the equilibrium point, this very tiny $K_{p}$ value means that the reaction will heavily favor the reactants. The third reaction where Txeltoqlztop wants to poison his brother with $\ce{CO}$ also follows the same process, and the answer is such:
$K_{p} \approx e^{452}$
$e^{452}$ is a really big number! That means, while the previous reaction will stay mostly on the reactant sides, this reaction heavily favors the products side, meaning that pretty much all of the methane and oxygen that Txeltoqlztop uses will be transformed into carbon monoxide, which is all the better for him!
The same method can be applied to the aqueous reaction, but the $K$ value calculated is $K_{c}$ rather than $K_{p}$.
$K_{c} \approx e^{-176.65} \approx 1.898 \times 10^{-77}$*
*These equilibrium constants are based solely off the calculated Gibbs energy, some of the reactions, therefore, may have actual K values much different from the ones calculated.
Abstract: use of $\Delta G_{\text{rxn}}^{\circ} = -RT \times \ln\left(K_p\right)$
Q15C
Use the thermodynamic data in Tables T1 and T2 to calculate the equilibrium constants for the following equations at 25°C.
1. $\ce{H2(g) + 2F(g) <=> 2HF(g)}$
2. $\ce{FeCl3(s) <=> FeCl2(s) + Cl2(g)}$
3. $\ce{CaSO4(s) <=> Ca^{2+}(aq) + SO4^{2-} (aq)}$
Solution
For each of the chemical reactions, the main equation needed is
$\Delta G_{\text{rxn}} = -R \times T \times \ln K \nonumber$
Where R is 8.134 J K-1 mol-1 and T is equal to 298K. In order to determine the value of Grxn, the following the reaction is needed
$\Delta G_{\text{rxn}} = \sum n G_{f}^{\circ} (\text{products}) - \sum n G_{f}^{\circ} (\text{reactants}) \nonumber$
a. By looking in the tables, the $\Delta G$ of formation can be found and they are listed for each compound below
H2(g) = 203.3 J mol-1, F2(g) = 62.3 J mol-1, HF(g)=-275.4 J mol-1
Next, these values are plugged into the second equation to the the change in Gibbs energy for the reaction
$\Delta G_{\text{rxn}} = -275.4\; \mathrm{J \, mol^{-1}} - (203.3\; \mathrm{J \, mol^{-1}} + 62.3\; \mathrm{J \, mol^{-1}}) = -514\; \mathrm{J \, mol^{-1}} \nonumber$
Using the the value for the Gibbs Energy, the value of K can be determined using the first equation
$-541\; \mathrm{J \, mol^{-1}} = -8.314\; \mathrm{J\, K^{-1} mol^{-1}} \times 298 \, \text{K} \times \ln K \nonumber$
$\dfrac{-541\; \mathrm{J \, mol^{-1}}}{-8.314\; \mathrm{J\, K^{-1}\, mol^{-1}} \times 298 \, \text{K}} = \ln K \nonumber$
$K = 1.25 \nonumber$
b. The $\Delta G$ of formation for the compounds are
FeCl3(s) = -334.0 J mol-1, FeCl2(s) = -302.3 J mol-1, Cl(g)=105.3 J mol-1
The calculations are
$\Delta G_{\text{rxn}} = (-302.3\; \mathrm{J\, mol^{-1}} + 105.3\; \mathrm{J\, mol^{-1}}) + 334.0\; \mathrm{J\, mol^{-1}} =137\; \mathrm{J\, mol^{-1}} \nonumber$
$137\; \mathrm{J \, mol^{-1}} = -8.314\; \mathrm{J \, K^{-1} \, mol^{-1}} \times 298 \, \mathrm{K} \times \ln K \nonumber$
$\dfrac{137\; \mathrm{J \, mol^{-1}}}{-8.314\; \mathrm{J \, K^{-1} \, mol^{-1}} \times 298 \, \mathrm{K}} = \ln K \nonumber$
$K = 0.945 \nonumber$
c. The $\Delta G$ of formation for the compounds are
CaSO4(s)= -1322.0 J/mol, Ca2+(aq) = -553.58 J/mol, SO42-(aq) = -744.53 J/mol
The calculations are
$\Delta G_{\text{rxn}} = (-553.58\; \mathrm{J \, mol^{-1}} - 744.53\; \mathrm{J\, mol^{-1}}) + 1322.0\; \mathrm{J\, mol^{-1}} = 23.89\; \mathrm{J\, mol^{-1}} \nonumber$
$23.89\; \mathrm{J\, mol^{-1}} = -8.314\; \mathrm{J\, K^{-1}\, mol^{-1}} \times 298 \, \text{K} \times \ln K \nonumber$
$\dfrac{23.89\; \mathrm{J\, mol^{-1}}}{-8.314\; \mathrm{J\, K^{-1}\, mol^{-1}} \times 298 \, \text{K}} = \ln K \nonumber$
$K = 0.99 \nonumber$
Q17
To ignite this house on fire, you dowsed the house in some ethanol and used a combustion reaction:
$\ce{C2H5OH(l) + O2(g) <=> H2O(l) + CO2 (g)}\nonumber$
This reaction has an equilibrium constant of K1. If you multiply this equation by 5, will the equilibrium constant change? If so, write this new equilibrium constant, K2, in terms of K1.
Solution
Because:
$K = \ce{\dfrac{[A]^{a}[B]^{b}}{[C]^{c}[D]^{d}}}$
If we multiply a chemical equation by a constant, the exponents of K will be multiplied by that multiple as well.
For this question, first balance this chemical equation, $\ce{C2H5OH(l) + 3 O2(g) <=> 3 H2O(l) + 2 CO2 (g)}\nonumber$ the corresponding equilibrium constant is $K_1 = \ce{\frac{[CO2]^2}{[O2]^3}} \nonumber$ since liquid is ignored.
When we multiply this equation by 5, the chemical equation becomes, $\ce{5 C2H5OH(l) + 15 O2(g) <=> 15 H2O(l) + 10 CO2 (g)}\nonumber$ the corresponding equilibrium constant is $K_2 = \ce{\frac{[CO2]^10}{[O2]^15}} = \left(\ce{\frac{[CO2]^2}{[O2]^3}}\right)^5 \nonumber$
Therefore, for this reaction, the equilibrium constant changes and $K_2 = K_1^5$.
Q21
An unknown gas B5D10 reacts at room temperature and becomes B10D20, the reaction is illustrated with the equation
$\ce{2 B5D10(g) <=> B10D20(g)} \nonumber$
At room temperature the equilibrium partial pressure of B5D10 is $1.23 \times 10^{-4} \, \text{atm}$ and that of B10D20 is $3.14 \times 10^{-16} \, \text{atm}$. Given these constants, what is $K_p$ of this reaction at room temperature.
Solution
Given that we already know the partial pressures and this reaction is at equilibrium, we simply substitute the values into the equilibrium equation and solve for Kp.
$K_p=\dfrac{[\text{products}]}{[\text{reactants}]}\nonumber$
$K_p=\dfrac{[3.14 \times 10^{-16}]}{[1.23 \times 10^{-4}]^2}=2.08 \times 10^{-8}\nonumber$
$K_p=2.08 \times 10^{-8}\nonumber$
Q23A
Suppose there exists a compound that can be found in two forms, red and blue, such that
$\ce{X_{red} <=> X_{blue}}\nonumber$
During equilibrium at 25.0°C, 17% of the substance is determined to be in blue form. What is the equilibrium constant for this reaction?
Solution
The equilibrium constant, $\ce{K_{eq}}$, for the chemical equation
$\ce{X_{red} <=> X_{blue}}\nonumber$
can be calculated by constructing the following equation:
$\ce{K_{eq}} = \frac{ \left[ \ce{X_{blue}} \right] }{ \left[ \ce{X_{red}} \right] } \nonumber$
Given that 17% of the concentration is $\ce{X_{blue}}$, that means the remaining 83% is $\ce{X_{red}}$, which can be plugged into the equation to yield
$\ce{K_{eq}} = \frac{ \left[ 0.17 \right] }{ \left[0.83 \right] } = 0.205 \nonumber$
which can be further reinforced by viewing it as the fact that since $\ce{K_{eq}} \lt 1$, the concentration of products present is expected to be smaller than the concentration of reactants.
Q23B
1,2-dimethylcyclohexane can exist in both boat and chair conformation. There is equilibrium between the two forms. What is the equilibrium constant for the reaction of boat conformation to chair conformation if the molecules in the boat form are 68.7%?
Solution
$\ce{[Boat] -> [Chair]} \nonumber$
Let B stand for Boat and C stand for Chair, then the equilibrium constant for this reaction is
$K=\ce{\frac{[C]}{[B]}} \nonumber$
For [C] and [B] here, we can use the molar concentration $c = n / V$, then $\ce{[C]} = n(\ce{C}) / V \nonumber$ $\ce{[B]} = n(\ce{B}) / V \nonumber$ So $K = \frac{n(\ce{C}) / V}{n(\ce{B}) / V} = \frac{n(\ce{C})}{n(\ce{B})} = \frac{n(\ce{C}) / n_{\text{tot}}}{n(\ce{B}) / n_{\text{tot}}} \nonumber$
From the question, it consists only with boat and chair conformation, so $n(\ce{C}) / n_{\text{tot}} = 0.687$, and the remaining is for boat, so $n(\ce{B}) / n_{\text{tot}} = 1 - 0.687 = 0.313$.
Therefore, the equilibrium constant is $K = \frac{n(\ce{C}) / n_{\text{tot}}}{n(\ce{B}) / n_{\text{tot}}} = \frac{0.687}{0.313} = 2.19 \nonumber$
Q28
The $K_p$ for the following reaction is 3.0 at 1000 K.
$\ce{CO2(g) + C(s) <=> 2 CO(g)} \nonumber$
1. If the pressure in the container where the reaction occurs was initially 0.48 bar, calculate the partial pressures of $\ce{CO2}$ and $\ce{CO}$ at equilibrium.
2. Calculate the percentage of $\ce{CO2}$ that has been dissociated.
Solution
a)
$\ce{CO2 (g) + C(s) <=> 2CO (g)} \nonumber$
$\ce{CO2 (g)}$ $\ce{CO (g)}$
Initial 0.48 0
Change $-x$ $+2x$
Equilibrium $0.48 - x$ $2x$
$K_p = \frac{p(\ce{CO})^2}{p(\ce{CO2})} \nonumber$
⇒ $3 = \frac{(2x)^2}{(0.48-x)} \nonumber$
⇒ $4x^2 = 3(0.48-x) \nonumber$
⇒ $4x^2 = 1.44 - 3x \nonumber$
⇒ $4x^2 + 3x - 1.44 = 0 \nonumber$
By solving the quadratic equation, the positive value we get is
$x = 0.33 \nonumber$
Hence at equilibrium,
$p(\ce{CO}) = 2x = 0.66 \nonumber$
$p(\ce{CO2}) = 0.48 - x = 0.48 - 0.33 = 0.15 \, \text{bar}\nonumber$
b) For fraction that has not reacted,
$\frac{p(\ce{CO2(final)})}{p(\ce{CO2(initial)})} \times 100 \% \nonumber$
The unreacted percentage is
$= \frac{0.15}{0.48} \times 100 \% = 31.25 \% \nonumber$
Hence, 68.75% of the CO2 has reacted.
Q29
A 100.0 mL glass bulb was filled which a weighed sample of solid $\ce{XO3}$, where $\ce{X}$ is an unknown element. The bulb was then attached to a pressure gauge and heated to 325 K, at which, the pressure was read to be 0.891 atm. Given that all of the $\ce{XO3}$ in the bulb at 325 K was in the gas phase, and it also partially dissociated into $\ce{O2 (g)}$ and $\ce{XO (g)}$:
$\ce{XO3 (g) <=> O2 (g) + XO (g)} \nonumber$
At 325 K, Kp = 3.14 for this reaction. Calculate the partial pressures of all three species in the bulb at equilibrium.
Solution
$\ce{XO3 (g) <=> O2 (g) + XO (g)} \nonumber$
Let Y be the initial partial pressure of X2O5, and x is the change in partial pressure of O2:
ICE $\ce{XO3}$ $\ce{O2}$ $\ce{XO}$
Initial $y$ 0 0
Change $-x$ $+x$ $+x$
Equilibrium $y-x$ $x$ $x$
The total pressure is 0.891 atm, therefore:
$(y - x) + x + x = 0.891 \, \text{atm} \nonumber$
$y + x = 0.891 \, \text{atm} \nonumber$
$y = 0.891 - x \nonumber$
So the ICE table now would be:
ICE $\ce{XO3}$ $\ce{O2}$ $\ce{XO}$
Initial $0.891 - x$ 0 0
Change $- x$ $+ x$ $+ x$
Equilibrium $0.891 - 2x$ $x$ $x$
$K_{p} = 3.14 = \dfrac{P_{\ce{XO}} \cdot P_{\ce{O2}}}{P_{\ce{XO3}}} = \dfrac{x^2}{0.891 - 2x} \nonumber$
$x^2 + 6.28x - 2.80 = 0 \nonumber$
$x = 0.418 \nonumber$
$P_{\ce{O2}} = P_{\ce{XO}} = 0.418 \, \text{atm} \nonumber$
$P_{\ce{XO3}} = 0.891 - 2(0.418) = 0.055 \, \text{atm} \nonumber$
Q31
The equilibrium constant $K_c$ for the reaction
$\ce{F2(g) + H2(g) <=> 2 HF(g)}\nonumber$
at 298 K is $5.07 \times 10^4$. Hydrogen with a partial pressure of 0.03500 atm is mixed with fluorine with a partial pressure of 0.06800 atm, and allowed to reach the equilibrium. What is the partial pressure of each of the gasses at the equilibrium?
Solution
We are looking for partial pressures of the reactants and we have starting partial pressures and a $K_c$ value. First we need to convert $K_c$ to $K_p$
$K_p = K_c (RT)^{\Delta{n_{\text{gas}}}}\nonumber$
$K_p = (5.07 \times 10^4) \times (0.0821 \times 298 \, \text{K})^{0} = 5.07 \times 10^4 \nonumber$
The equilibrium constant can be constructed from the equation given
$K_p = \ce{\dfrac{[HF]^2}{[H2][F2]}} \nonumber$
Use an ICE table to find the equilibrium partial pressures of the equilibrium
Reaction $\ce{H2 (g)}$ $\ce{F2 (g)}$ $\ce{2HF (g)}$
I $0.035$ $0.068$ $0$
C $-x$ $-x$ $+2x$
E $0.035-x$ $0.068-x$ $2x$
$5.07 \times 10^4 = \dfrac{[2x]^2}{[0.035 - x]^2[0.068 - x]^2} \nonumber$
Simplify the equation and solve as you wish. You can't use the quadratic formula because you end up with a degree three equation but you can put the equation into a graphing calculator and solve for $x$ using the zero function.
$x = 0.0286 \, \text{atm} \nonumber$
$\ce{[HF]} = 2 x = 0.0571 \, \mathrm{atm} \nonumber$
$\ce{[H2]} = 0.035 - x = 0.00643 \, \mathrm{atm} \nonumber$
$\ce{[F2]} = 0.068 - x = 0.0394 \, \mathrm{atm} \nonumber$
Q33
At 25 °C, the equilibrium constant for the reaction below has the value of $1.7 \times 10^{-13}$:
$\ce{ N2O(g) + 1/2 O2(g) <=> 2 NO(g)} \nonumber$
In a container where $\ce{N2O}$ has an initial partial pressure of 0.62 atm, $\ce{O2}$ has a pressure of 0.24 atm, and $\ce{NO}$ has an initial pressure of 0.08 atm, what will the partial pressure of the three gases be after reaching equilibrium at the same temperature?
Q35
The reaction:
$\ce{2HI(g) <=> H2(g) + I2(g)} \nonumber$
has an equilibrium constant $K_c = 1.82 \times 10^{-2}$ at 698 K. If the reaction took place in a tank at 698 K and started off with HI having a partial pressure of 1 atm, what would the partial pressures of all the gases be at equilibrium?
Solution
First, we need to convert $K_c$ into $K_p$ with the equation
$K_p = K_c (RT)^{\Delta n} \nonumber$
$K_p = 1.82 \times 10^{-2} (0.0821 \, \text{L atm / K mol} \times 698 \, \text{K})^{2-2} \nonumber$
$K_p = 1.82 \times 10^{-2} \nonumber$
An ICE table can also be constructed for the reaction:
$\ce{2HI (g) <=> H2 (g) + I2 (g)} \nonumber$
ICE Table $\ce{2HI (g)}$ $\ce{H2(g)}$ $\ce{I2(g)}$
Initial
1
0
0
Change
$-2x$
$+x$
$+x$
Equilibrium
$1-2x$
$x$
$x$
$0.82 \times 10^{-2} = \dfrac{x^2}{1-2x} \nonumber$
$x^2 = 1.82 \times 10^{-2} - 0.0368 x \nonumber$
$x = 0.118488 \nonumber$
$\text{partial pressure of } \ce{HI} = 0.763 \, \text{atm} \nonumber$
$\text{partial pressure of } \ce{H2} = \text{partial pressure of } \ce{I2} = 0.1189 \, \text{atm} \nonumber$
Q37
What is the concentration of a $\ce{XY}$ that will be found in equilibrium at room temperature, given that $\ce{X2}$ has a concentration of $3.14 \times 10^{-15} \, \text{M}$, $\ce{Y2}$ has a concentration of $1.23 \times 10^{-4}\, \text{M}$, and the $K_c$ of the reaction is $2.22 \times 10^{-7} \, \text{M}$.
Solution
First we must state what the formulas for both the reaction and Kc and then we can substitute the known values and solve for the concentration of XY as an unknown variable.
$\ce{2XY(g) <=> X2 (g) + Y2(g)} \nonumber$
$K_c = \mathrm{\dfrac{[Products]}{[Reactants]}} \nonumber$
$K_{c}= \ce{\dfrac{[X2] [Y2]}{[XY]^2}} = \dfrac{[3.14\times10^{-15}][1.23\times10^{-4}]}{\ce{[XY]}^{2}}=2.22\times10^{-7} \nonumber$
$K_{c}=\dfrac{[3.86\times10^{-19}]}{\ce{[XY]}^{2}}=2.22\times10^{-7} \nonumber$
$K_{c}=3.86\times10^{-19}=2.22\times10^{-7}\ce{[XY]}^{2} \nonumber$
$\ce{[XY]}=1.32 \times 10^{-6} \nonumber$
So the concentration of $\ce{XY}$ is $1.32 \times 10^{-6} \, \text{M}$ in equilibrium at 25ºC.
Q41
Hydrogen and oxygen gas react with each other to form gaseous water with an equilibrium constant for the reaction is $K_c = 1.33 \times 10^{20}$ at 1000 K.
1. Consider a system at 1000.0 K in which 4.00 atm of oxygen is mixed with 0.500 atm of hydrogen and no water is initially present. What is the concentration of hydrogen gas after equilibration.
2. Consider a system, also at 1000.0 K, where 0.250 atm of oxygen is mixed with 0.500 atm of hydrogen and 2.000 atm of water. What is the concentration of hydrogen and oxygen gas after equilibration.
Solution
a)$\ce{2 H2 (g) + O2 (g) <=> 2 H2O(l)} \nonumber$
$K=3.4 = \frac{1}{P_{\ce{H2}}^2 P_{\ce{O2}}} \nonumber$
$P_{\ce{H2}}=1.2\, \text{atm} \nonumber$
$3.4 = \frac{1}{(1.2)^2 P_{\ce{O2}}} \nonumber$
$P_{\ce{O2}}=\frac{1}{(1.2)^2 (3.4)}= 0.204\, \text{atm} \nonumber$
b) Excess $\ce{H2O(l)}$ is added drive the reaction to the left towards the reactant to reach the equilibrium; therefore, pressure of H2 and O2 increased.
$\ce{2H2(g)}$ $\ce{O2(g)}$ $\ce{2H2O(l)}$
Initial 0 0.3 N/A
Change $+2x$ $+x$ N/A
Equilibrium $2x$ $0.3+x$ N/A
$K=\frac{1}{(2x)^{2}(0.3+x)}=3.4 \nonumber$
$3.4(0.3+x)(2x)^{2}=1 \nonumber$
$x=0.339\, \text{atm} \nonumber$
$P_{\ce{O2}}=0.3+0.339=0.639\, \text{atm} \nonumber$
$P_{\ce{H2}}=2(0.339)=0.678\, \text{atm} \nonumber$
Q43
If you decompose ammonium nitrate into water and nitrous oxide, which further decomposes into oxygen and nitrogen gas, making the final reaction:
$\ce{ 2 NH4NO3(s) <=> 2 N2(g) + O2(g) + 4 H2O(g)} \nonumber$
This reaction takes place only in the molten salt, above its melting point of 169.6°C (for the anhydrous product). So, you get your 170°C blow torch and begin decomposing the ammonium nitrate under a sealed hood. Upon decomposition, all of the ammonium nitrate is gone, i.e. the reaction goes to completion.
1. What can be said of the equilibrium constant of this reaction at 170°C?
2. Now, in another universe, the partial pressure of N2(g) is 0.3 atm initially. Will the reaction still go to, essentially, completion?
Solution
(a) If there is no ammonium nitrate left, then the reaction was heavily products favoring, meaning K is extremely high at 170°C.
Note: It's more accurate to say that Q is a big number when there is no reactants left after reaction occurred. Reviewer's comment: Q stands for the current state. K is the equilibrium constant. When there is no reactant left after reaction occurred, the reaction is at the equilibrium state, so Q = K. Since Q is a big number, then K is also a big number.
(b) There would have to be an enormous amount of Nitrogen to force the reaction backwards. Because 0.3 atm is likely not enough to push back an essentially irreversible reaction, it has no noticeable affect on the amount of end product. (Note: If the reactants and products are essentially gas, we calculate the equilibrium constant not by its concentration, but by its pressure.)
Q45
Al2Cl6 at a partial pressure of 0.600 atm is placed in a closed container at 454 K. Al3Cl9 (partial pressure $1.98 \times 10^{-3} \, \text{atm}$) is also placed in it as well. Argon is added to raise the total pressure up to 1.00 atm.
Find whether if there is going to be net production or consumption of Al3Cl9 given $K_p = 1.04 \times 10^{-4}$. Then find the final pressure of Al3Cl9
Solution
The reaction is $\ce{ 3 Al2Cl6 (g) <=> 2 Al3Cl9 (g)} \nonumber$ Because argon does not react, it can be ignored.
$Q=\frac{(\ce{[Al3Cl9]})^{2}}{(\ce{[Al2Cl6]})^{3}}= \frac{(1.98 \cdot 10^{-3})^{2}}{(0.600)^{3}}=1.82 \cdot 10^{-5} < K_p = 1.04 \cdot 10^{-4} \nonumber$
There will be production of Al3Cl9.
ICE Table Al2Cl6 Al3Cl9
Initial 0.600 $1.98 \times 10^{-3}$
Change $-3y$ $+2y$
Equilibrium $0.6 - 3y$ $1.98 \times 10^{-3} + 2y$
$K= 1.04 \cdot 10^{-4} \frac{(\ce{[Al3Cl9]})^{2}}{(\ce{[Al2Cl6]})^{3}}= \frac{(1.98 \times 10^{-3} + 2y)^{2}}{(1 - (1.98 \times 10^{-3} + 2y))^{3}} \nonumber$ Solving out $y = 0.0041 \, \mathrm{atm} \nonumber$
Al3Cl9 final pressure = 0.0102 atm.
Q46
The thermal decomposition of NH4Cl solid proceeds as follows:
$\ce{NH4Cl(s) <=> NH3(g) + HCl(g)} \nonumber$
the equilibrium constant at 275°C is $1.04 \times 10^{-2}$. If the partial pressures of $\ce{NH3(g)}$ and $\ce{HCl(g)}$ are equal, and the total partial pressure of the system is 0.200 atm, in what direction does the reaction proceed? What will be formed?
Solution
First, we make two equations to solve out the partial pressures. To get the first equation, we sum over the partial pressures to the total pressure. $\ce{NH4Cl}$ is a solid, so it doesn't contribute to the partial pressure
$P_{\ce{NH3}} + P_{\ce{HCl}} = 0.200 \, \mathrm{atm} \nonumber$
The second equation is from the question that partial pressures of $\ce{NH3(g)}$ and $\ce{HCl(g)}$ are equal
$P_{\ce{NH3}} = P_{\ce{HCl}} \nonumber$
Hence, $P_{\ce{NH3}} = P_{\ce{HCl}} = 0.100 \, \mathrm{atm} \nonumber$
$Q(\text{reaction quotient}) = P_{\ce{NH3}}P_{\ce{HCl}} = {(0.100)^2}\nonumber$
$10^{-2} < 1.04 \times 10^{-2}\nonumber$
$Q < K\nonumber$
Hence, the reaction will slightly proceed more in the forward direction.
NH3 and HCl will be formed.
Q47
A tube contains a mixture of NO2 and N2O4 gas is set at 298K, in which the initial partial pressure NO2 is 0.38 atm, and the initial partial pressure of N2O4 is 0.59 atm. NO2 is a brownish gas, while N2O4 is a colorless gas.
$\ce{2NO2(g) \rightleftharpoons N2O4(g)} \nonumber$
1. What is the reaction quotient at the start of this reaction?
2. As the reaction shown above reaches equilibrium, the mixture becomes more brown. With this in mind, is the equilibrium constant, Kp, greater than or less than the Q calculated in part a?
Solution
a) $\ce{2 NO2 (g) \rightleftharpoons N2O4(g)}$
$P_{N_{2}O_{4}} = 0.59 \; atm \; and \; P_{NO_{2}} = 0.38 \; atm$
$Q_{p} = \frac{P_{N_{2}O_{4}}}{(P_{NO_{2}})^{2}}$
$Q_{p} = \frac{0.59}{0.38^{2}} = 4.1$
b) Since the mixture becomes more brown as the reaction reaches equilibrium, more NO gas is being produced, therefore the reverse reaction is favored, so Qp > kp, in conclusion, Kp < 4.1.
Q49
At 500 K the equilibrium constant for the reaction:
$\ce{ PCl3 (g) + Cl2 (g) <=> PCl5 (g)} \nonumber$
is $K_{eq}=2.1$. Suppose the equilibrium is disturbed and shifts, while temperature remains constant. Calculate the reaction quotient for each one of the possible changes in equilibrium concentration given below and evaluate in what direction the reaction will shift to reestablish equilibrium.
1. [PCl3]= 0.0563, [Cl2]= 0.0784, [PCl5]= 0.8934.
2. [PCl3]=0.4390 , [Cl2]= 0.3547, [PCl5]= 0.1048.
3. [PCl3]= 0.4018, [Cl2]= 0.8690, [PCl5]= 0.7205.
Q48
NO2 has a brownish color. At elevated temperatures, NO2 reacts with CO
$\ce{NO2 (g) + CO (g) <=> NO(g) + CO2 (g)}\nonumber$
The other gases in this equation are colorless. When a gas mixture is prepared at 600K, in which 9.6 atm is the initial partial pressure of both NO2 and CO, and 5.4 atm is the partial pressure of both NO and CO2, the brown color of the mixture observed begins to get stronger as the reaction progresses toward equilibrium. Give a condition that must be satisfied by the equilibrium constant K.
Q51
The equilibrium constant ($K_p$) for the reaction
$\ce{As4(g) <=> 2 As2(g)} \nonumber$
is $5.5837 \times 10^{-4}$ at 1090 K.
1. The initial molarity of $\ce{[As4]}$ is 2.3 M and the initial molarity of $\ce{[As2]}$ is 0.001 M. Calculate the reaction quotient $Q$ and determine which way the reaction proceeds.
2. Calculate the molarity of each compound at the equilibrium.
3. Which direction will the reaction proceed if the pressure on the reaction decreases? Explain your answer.
Solution
(a) First, because you are given the initial molarities of your products and reactants and a Kp instead of a Kc value, you have to convert Kp to Kc using the formula:
$K_p = K_c(RT)^{\Delta{n_{\text{gas}}}}\nonumber$
Plug in your given values
5.5837 x 10-4=Kc(.0821*1090)2-1
5.5837 x 10-4=Kc(89.489)
Kc=6.24 x 10-6
Set up a reaction quotient for the equation given
Q = $\ce{\frac{ [As2]^2}{[As4]}}$
Q = $\frac{ [.001]^2}{[2.3]}$
Q= 4.35 x 10-7
Reaction will progress towards the right to reach equilibrium
(b) Set up an ice chart to find the equilibrium molarities
Reaction P4 2P2
I 2.3 .001
C -x +2x
E 2.3-x .001+2x
6.24 x 10-6 = $\frac{ [.001+2x]^2}{[2.3-x]}$
4x2+.00400624x-1.3352x10-5=0
Use the quadratic equation to find x
x= -.0024, .00139
.001-.0024 is negative, which can't happen in real life so we know x actually equals .00139
[P4]= 2.29861 M
[P2]= .00378 M
(c) The reaction will progress towards the right because as pressure decreases, the reaction moves towards the side with the most moles of gas, which is the products
Q55
Consider a reaction involving only gaseous compounds. The product yield at equilibrium decreases when the temperature and volume increase.
1. Is this reaction endothermic or exothermic?
2. Does the number of gas molecules in this reaction increase or decrease?
Solution
Exothermic
Decreases
In exothermic reactions, product yield is indirectly related to temperature. In endothermic reactions, product yield is directly related to temperature. These properties can be further explained when heat is thought of as a reactant or a product. In exothermic reactions, heat is a product, and in endothermic reactions, heat is a reactant. When heat is added to a reaction, the equilibrium shifts to the opposite side to relieve stress on the system.
The number of gas molecules is directly related to the volume of the container in which the reaction is happening. For example, if the volume of the container is decreased, the reaction will shift to whichever side has fewer moles of gas.
Q57
the Haber process, also called the Haber–Bosch process, is an artificial nitrogen fixation process and is the main industrial procedure for the production of ammonia today.
$\ce{N3 + 3H2 <=> 2NH3} \quad (\Delta H=-92.4\,kJ/mol) \nonumber$
Determine the best condition of temperature and pressure to yield the most ammonia.
Solution
Because Haber process is an exothermic reaction, a lower temperature will maximize the yield of product. Also, according to the ideal gas law, because the moles of gas are proportional to the volume, and the volume is inverse proportional to the pressure; a higher pressure is required to maximize the yield of product.
Q59
At room temperature, a vessel is filled with gaseous carbon dioxide. Some water is added at 2.00 atm. It is well-shaken, integrating carbon dioxide gas into the water. 0.5 kg of the solution is taken out and boiled to extract 2.50 L of carbon dioxide. The system is at 10 degree Celsius and 1.00 atm. What is the Henry’s Law constant for carbon dioxide in water?
Solution
given that the extracted solution:
$PV=nRT$
$n=\frac{PV}{RT}=\frac{1atm\times 2.5L}{0.082\frac{L\ atm}{mol\ K} \times 283.15L}=0.1077\ mol$
$mass_{CO_{2}}=n_{CO_{2}}\times molar\ mass = 0.1077\ mol \times 44\ g/mol=4.7388\ g$
$mass_{H_{2}O}=500\ g - 4.7388\ g = 495.29\ g$
$n_{H_{2}O}=\frac{495.29\ g}{18\ g/mol} = 27.5 mol H2O \[Solubility\ of\ CO_{2}\ in\ water=\frac{0.1077\ mol}{0.5\ kg}=0.2154\ mol/kg$
Due to the Ideal gas law, the ratio of moles equals to the ratio of pressure
pressure: pressure = mol:mol = 0.1077: 27.51
because the total pressure is 2 atm, partial pressure of CO2 = 0.0724 atm
Due to Henry's law:
$k=\frac{c}{P}=\frac{0.2154mol/kg}{0.0724\ atm}=2.9751\frac{mol}{kg\ atm}$
Q61
The equilibrium constant for a reaction increases from $\ce{K= 4.5 \times 10^{-2}}$ to $\ce{K= 4.5}$ as the temperature increases from $0.00\text{°C}$ to $300.0\text{°C}$. Assuming the change in enthalpy ($\ce{\Delta}\text{H}$) and change in entropy ($\ce{\Delta}\text{S}$) are constant over this range of temperatures, what is $\ce{\Delta}\text{S}$ for this reaction? Is this reaction endothermic or exothermic?
Solution
Assuming $\ce{\Delta}\text{H}$ and $\ce{\Delta}\text{S}$ are constant, we can set up a system of equations as follows:
$\{\begin{array}{ll} \ce{\Delta}{G_{0.00\text{°C}}} = {\Delta}\text{H} - \text{T}{\Delta}\text{S} \ {\Delta}{G_{300.0\text{°C}}} = {\Delta}\text{H} - \text{T}{\Delta}\text{S} \ \end{array}$
We can determine $\ce{\Delta}{G_{0.00\text{°C}}}$ and $\ce{\Delta}{G_{300.0\text{°C}}}$ by utilizing the equation below that relates $\ce{\Delta}{G}$ to $\ce{K_{eq}}$:
$\ce{\Delta}{G} \ = \ -\text{RT ln(K)}$ Easily remembered using the mnemonic: "${\Delta}\text{G}$et in the $-\text{R}$igh$\text{T}$ $\text{l}$a$\text{n}$e, $\text{K}$?"
$\ce{\Delta}{G_{0.00\text{°C}}} \ = \ - \left(8.314 \frac{\text{J}}{\text{K} \times \text{mol}} \right) \left( 273.15 \text{K} \right) \left( \text{ln}( 4.5 \times 10^{-2}) \right)\ = \ 7042.5 \frac{\text{J}}{ \text{mol}}$
$\ce{\Delta}{G_{300.0\text{°C}}} \ = \ - \left(8.314 \frac{\text{J}}{\text{K} \times \text{mol}} \right) \left( 573.15 \text{K} \right) \left( \text{ln}( 4.5) \right)\ = \ -7167.2 \frac{\text{J}}{ \text{mol}}$
These values can then be plugged back into the original system of equations to obtain
$\{\begin{array}{ll} \ce{\Delta}{G_{0.00\text{°C}}} = 7042.5 \frac{\text{J}}{\text{mol}} = {\Delta}\text{H} - \left( 273.15 \text{K} \right) {\Delta}\text{S} \ {\Delta}{G_{300.0\text{°C}}} = -7167.2 \frac{\text{J}}{ \text{mol}} = {\Delta}\text{H} - \left( 573.15 \text{K} \right) {\Delta}\text{S} \ \end{array}$
To calculate the values of the remaining two unknown variables, a substitution can be made that expresses $\ce{\Delta}\text{H}$ in terms of $\ce{\Delta}\text{S}$.
$-7167.2 \frac{\text{J}}{\text{mol}} + \left( 573.15 \text{K} \right) {\Delta}\text{S} = {\Delta}\text{H}$
This substitution can be plugged into the first equation to calculate the value for $\ce{\Delta}\text{S}$.
$7042.5 \frac{\text{J}}{\text{mol}} = -7167.2 \frac{\text{J}}{\text{mol}} + \left( 573.15 \text{K} \right) {\Delta}\text{S} - \left( 273.15 \text{K} \right) {\Delta}\text{S}$
${\Delta} \text{S} = 47.37 \frac{\text{J}}{\text{K} \times \text{mol}}$
$\ce{\Delta}\text{H}$ can be calculated now by plugging in the value for ${\Delta} \text{S}$ into either one of the equations in the original set.
$7042.5 \frac{\text{J}}{\text{mol}} = {\Delta}\text{H} - \left( 273.15 \text{K} \right) \left( 47.37 \frac{\text{J}}{\text{K} \times \text{mol}} \right)$
$\ce{\Delta}\text{H} = 19980 \frac{\text{J}}{\text{mol}} = 19.98 \frac{\text{kJ}}{\text{mol}}$
Because $\ce{\Delta}\text{H}$ is positive, we know the reaction is endothermic.
Q63A
The equilibrium constant for the reaction $\ce{N2O4(g) <=> 2 NO2(g)}$ is measured to be $5.9\times10^{-3}$ at 298\;K and $1.3\times10^{-6}$ at $398\;K$
1. Calculate $\Delta G^{\circ}$ at 298 K for the reaction.
2. Calculate $\Delta H$ and $\Delta S^{\circ}$, assuming the enthalpy and entropy changes are independent of temperature between 298 K and 398 K.
Solution
a.Use the equation $\Delta G^{\circ}=-RT\ln K$
$\Delta G^{\circ}=-RT \ln K=(-8.3145\; \mathrm{J \, K \, mol^{-1}}) \times (298\; \mathrm{K})(\ln 5.9 \times 10^{-3})=-1.29\times10^{4}\; \mathrm{J\, mol^{-1}}$
For one mole of the reaction as written, $\Delta G^{\circ}$ is -1290 kJ.
b. Use van't Hoff equation and the values of K at 298 K and at 398 K to obtain $\Delta H^{\circ}$. Then get $\Delta S^{\circ}$ from $\Delta G^{\circ}$ and the equation $\Delta G^{\circ}=-RT \ln K$
$\ln(\frac{1.3 \times 10^{-6}}{5.9 \times 10^{-3}})=\frac{\Delta H^{\circ}}{8.3145\; \mathrm{J\,K^{-1} \, mol^{-1}}}(\frac{1}{398\;K}-\frac{1}{298\;K})$
$\Delta\;H^{\circ}=83.0\;kJ$
The answer equals both $\Delta\;H^{\circ}$ at 298 K and 398 K because it is assumed in the derivation of the van't Hoff equation that $\Delta\;H^{\circ}$ is independent of temperature. Next,
$\Delta\;S^{\circ}=\frac{\Delta\;H^{\circ}-\Delta\;G^{\circ}}{T}=\frac{83.0\times10^{3}\;J-(-1.29\times10^{4}\;J)}{298\;K}=323\;J\;K^{-1}$
Q63B
For the reaction: $\ce{ Br2(g) <=> 2Br(g)}$, the equilibrium constant is $4.64 \times 10^{-29}$ at 25oC and $1.1 \times 10^{-3}$ at 1280 K.
1. What is ∆G of the reaction at 298 K?
2. Assuming that $∆H^o$ and $∆S^o$ are independent of temperature, what are the values of enthalpy and entropy when the temperature changes to 1280 K?
Solution
a. ∆G = RTln(k)
= (8.3145JK-1mol-1)(298K)ln(4.64x10-29)
= -161.65kJ
a) $ln \left(\dfrac{1.1x10^{-3}}{4.64x10^{-29}}\right) = \dfrac{-∆ H}{R} \left( \dfrac{1}{1280K} - \dfrac{1}{298K} \right) \nonumber$
Solve for ∆ H.
∆ H = 188.70 KJ
$ln \left({k}\right) = \dfrac{-∆ H}{RT} + \dfrac{ ∆S}{R} \nonumber$
$ln \left({1.1x10^{-3}}\right) = \dfrac{-188699.30}{(8.3245;\ {JK^{-1}mol^{-1})(1280:\K} + \dfrac{ ∆S}{R} \nonumber$
∆S =90.78 JK-1mol-1
Q65
Calculate the equilibrium constant for the reaction,
2 NO2 (g) $\rightleftharpoons$ N2O4 (g) at 500 K,
given the equilibrium constant K = 4.65 x 10-3 at 298 K, and ΔH° = -57.2 kJ mol-1. Assume that ΔH° does not change between 298 K and 500 K
Solution
Use the van 't Hoff equation to calculate K500 from K298 and the standard enthalpy of the reaction ΔH°:
$ln \frac{K_{500}}{K_{298}} = - \dfrac{\Delta H°}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}} ) \nonumber$
=> K500 = 3.366 x 10-7
Solve for the unknown variable, in this case it is K500. T1 is equal to 298K while T2 is equal to 500K.
Q69
The equilibrium constant for the following reaction was experientially calculated at different temperatures.
$\ce{N2 (g) + 3H2 (g) \rightleftharpoons 2NH3 (g)} \nonumber$
Temperature (°C) 300 400 450 500 550 600
Equilibrium Constant (Keq) 4.34 x 10-3 1.64 x 10-4 4.51 x 10-5 1.45 x 10-5 5.38 x 10-6 2.25 x 10-6
1. Plot $\ln K$ as a function of $1/T$ (K-1)
2. What is $\Delta Hº$ of the equilibrium reaction with ammonia? HINT: use the slope to find the $\Delta Hº$
Solution
The graph bellow shows what the graph should look like:
First we must write the van't Hoff equation:
$\ln K = - \dfrac{\Delta{H^o}}{R} \dfrac{1}{T} + \dfrac{\Delta{S^o}}{R}\label{20}\nonumber$
Writing the equation like this how it can be compared with y = mx + b. With this information we then know that the slope of the graphe is equal to -$\Delta{H^o}$/R. So:
$\Delta{H^o}$ = -mR were R = 8.3145 J K-1mol-1
$\Delta{H^o}$ = -104.9 kJ mol-1
Q71
At a certain temperature, $0.5\ \text{mol}$ $\ce{I_{2\, (aq)}}$ is dissolved in $2\ \text{L}$ $\ce{H_{2}O}$ and shaken vigorously in a container with $2\ \text{L}$ of $\ce{CCl_4}$, an immiscible liquid. The container is then left to equilibrate during which time the iodine redistributes itself between the layers of $\ce{H_{2}O}$ and $\ce{CCl_4}$ as indicated by the following reaction:
$\ce{I_{2\, (aq)} \rightleftharpoons I_{2} (CCl_4)}\nonumber$
The final concentration of $\ce{I_{2\, (aq)}}$ is determined to be $0.0710\ \text{M}$. What is the partition coefficient $\left(\text{K}\right)$ for this process?
Solution
The word "equilibrate" is a dead giveaway for "ICE Table" so we can establish the following ICE Table based on the information provided for the reaction provided.
$\ce{I_{2\, (aq)} \rightleftharpoons I_{2} (CCl_4)}$
ICE Table $\ce{I_{2\, (aq)}}$ $\ce{I_{2} (CCl_4)}$
Initial $0.250 \text{M}$ $0 \text{M}$
Change $- \text{x}$ $+ \text{x}$
Equilibrium $0.0710 \text{M}$ $+ \text{x}$
$\text{x}$ can be calculated by solving the following equation:
$0.250 \text{M} - \text{x} = 0.0710 \text{M}$
$\text{x} = 0.179 \text{M}$
This value can then be plugged into the ICE Table to obtain equilibrium values for $\ce{I_{2\, (aq)}}$ and $\ce{I_{2} (CCl_4)}$. (Remember that the two liquids are immiscible so the molarity isn't $0.125 \text{M}$ as $\ce{I_{2} (CCl_4)}$ does not contribute to the initial molarity of the solution.)
ICE Table $\ce{I_{2\, (aq)}}$ $\ce{I_{2} (CCl_4)}$
Initial $0.250 \text{M}$ $0 \text{M}$
Change $- 0.179 \text{M}$ $+ 0.179 \text{M}$
Equilibrium $0.0710 \text{M}$ $+ 0.179 \text{M}$
The equilibrium concentrations of $\ce{I_{2\, (aq)}}$ and $\ce{I_{2} (CCl_4)}$ can then be plugged into the equilibrium constant expression and solve for the partition coefficient.
$\text{K} = \frac{ 0.179 \text{M} }{ 0.0710 \text{M} }$
$\text{K} = 2.52$
Q73
At 25°C, 0.130 g/L of chloroacetic acid (C2H3ClO2) dissolve in of water, and at the same temperature 0.2834 g/L of chloroacetic acid dissolve in 1-octanol.
a. Calculate the following equilibrium constants
$\ce{C2H2ClO2H(s)} \rightleftharpoons \ce{C2H2ClO2H}_{(aq)} \nonumber$
$\ce{C2H2ClO2H(s)} \rightleftharpoons \ce{C2H2ClO2H}_{(octanol)} \nonumber$
b. Calculate the participation coefficient, $K$, of the combined reactions.
$\ce{C2H2ClO2H(aq)} \rightleftharpoons \ce{C2H2ClO2H}_{(octanol)} \nonumber$
Solution
First, we will need g/L to M so that we can calculate the equilibrium constant.
$\left(\dfrac{0.130\; \cancel{g\; C_{2}H_{3}ClO_{2}} }{1\;L}\right) \left(\dfrac{1\: mol}{94.494\: \cancel{ g\; C_{2}H_{3}ClO_{2}}} \right) = 0.0013757\; M\nonumber$
$\left(\dfrac{0.2834\; \cancel{g\; C_{2}H_{3}ClO_{2}} }{1\;L}\right) \left(\dfrac{1\: mol}{94.494\: \cancel{ g\; C_{2}H_{3}ClO_{2}}} \right) = 3.00x10^{-3}\; M\nonumber$
$K = \dfrac{K_{forminde}}{K_{aq}}=\dfrac{3.00 \times 10^{-3}}{0.001376} =2.18\nonumber$
$\log_{10} K_{oct/wat} =0.338\nonumber$
Q75
At 298 K, the two gases NOBr and Br2 have the number of moles of 1.25 and 8.5 x 10-5 at equilibrium with a sufficient amount of solid NO. Determine the equilibrium constant of the reaction:
$2NOBr_{(g)} \rightleftharpoons 2NO_{(s)} + Br_{2(g)} \nonumber$
Solution
Use: Pgas = ngas(RT/V)
K = PBr2 / P2NOBr = nBr2 / n2NOBr = 5.44 x 10-5
The volume is needed to solve this problem. It is necessary to find the value of K regardless of it is Kp or Kc.
Q79
At 300 K and 1 atm, N2O4 is partly dissociated into NO2. The density of the equilibrium mixture was 2.076 g/L. What is the degree of dissociation of N2O4 under these circumstances?
Solution
$N_2O_4(g) \rightarrow 2NO_2(g) \nonumber$
moles of N2O4 at equilibrium = 1-x moles of NO2 at equilibrium = 2x
Therefore, the degree of dissociation of SO3 is 0.8.
(Multiplying number of moles by molar mass to get the total mass)
92*(1- x) + [46*(2x)/ (1+x) ]= [ 2.076 g/L * 0.08206 (L*atm)/(mol*K) ] / 1 atm
On solving the equation, we get the equation, x= 0.8
Q85
The decompressions of calcium carbonate ($\ce{CaCO3}$) occurs according to the following reaction:
$\ce{ CaCO3 (s) \rightleftharpoons CaO(s) + CO2 (g)}\nonumber$
100 g of solid calcium carbonate ($\ce{CaCO3}$) and 50 g of calcium oxide ($\ce{CaO}$) are placed in an evacuated round bottom flask that is heated up to 35°C. At this temperature the pressure in the round bottom flask is found to be 0.5 atm.
1. What is the equilibrium constant of the decompressions of calcium carbonate ($\ce{CaCO3}$) at 35°C?
2. If 100 g of calcium oxide (CaO) at 35°C was added to the flask what would be the new total pressure of the flask and why?
3. If the pressure in the round bottom flask would to increase to 1.0 atm, then what would would be the new equilibrium constant?
Solution
1. First we need to write the equilibrium expression for the reaction: $K_{eq} = p\ce{CO2}]$ Now we just need to plug in the pressure of CO2: Keq = [0.5] = 0.5
2. The total pressure would still be 0.5 atm because solids have an activity of 1 so they don't effect the equilibrium pressure.
3. The new equilibrium constant would be 1.0 atm.
Q87
Let's say you have the reaction: $A(s) + B(g) \leftrightharpoons C(s) + D(g)\nonumber$
At 25°C, this reaction has an equilibrium constant of 157.2. Find the partial pressures of $B(g)$ and $D(g)$ when the total pressure of the system is equal to 4.3 atm.
Solution
First, the equilibrium expression of this reaction is
$K_c = \frac {D(g)}{B(g)} = 157.2\nonumber$
Note, the A(s) and C(s) are not included in this expression because they are solids. Solids have an activity equal to 1 which basically means they do not affect the equilibrium of the reaction.
To calculate the partial pressure of our unknowns, we can use Dalton's Law of partial pressures: $P_A = X_A P_{Total}\nonumber$
Since we are looking for the partial pressure of each of our gases, and we already have the total pressure of the system, we need to find the mole fraction of each of our gases. To find it, we can manipulate the above equilibrium expression to get: $D(g) = 157.2 \; B(g)\nonumber$
Here, we see that at equilibrium, there are 157.2 molecules of $D(g)$ for every 1 molecule of $B(g)$
Now that we have the ratio of molecules in this reaction, we can construct workable mole fractions.
$X_D = \frac {157.2}{1 + 157.2}$
$X_B = \frac {1}{1 + 157.2}$
Finally, we can plug everything in to find our partial pressures.
$P_D = \frac {157.2}{158.2} \times (4.3 \; atm) = 4.27 \; atm$
$P_B = \frac {1}{157.8} \times (4.3 \; atm) = 0.027 \; atm$
Q91
Given the reaction
$C_3H_5N(g) + 3 \; H_2(g) \leftrightharpoons C_5H_{11}N(g)\nonumber$
whose equilibrium constant is solved by this equation $log_{10}(K) = -20.28 + \frac {10.56 \; K}{T}\nonumber$
1. Solve for $K$ when $T = 200 \; K$
2. What is partial pressure of $C_3H_5N(g)$ when $T = 200 \; K$ and the partial pressure of hydrogen is 1.0 atm and the partial pressure of $C_5H_{11}N(g)$ is $2.0 \times 10^{-5}$ atm.
Solution
Part a)
$log_{10}(K) = -20.28 + \frac {10.56 \; K}{200 \; K}\nonumber$
$K = 10^{-20.23} = 5.89 \times 10^{-21}$
part b)
Set up the equilibrium constant of the reaction, plug in your partial pressures, and solve for $C_3H_5N(g)$
$K = \frac {[C_5H_{11}N]}{[C_5H_5N] \; [H_2]^3}$
$K = \frac {2.0 \times 10^{-5}}{[C_5H_5N] \; [1]}$
$[C_5H_5N] = 3.40 \times 10^{15}$
Q93
Two equal volume solutions of 2M HCOO- and 2M HCOOH are mixed together. Assume that the volumes are additive. The value of the equilibrium constant Ka for HCOOH is 1.77 x 10-4. What is the ratio between the amounts of HCOOH and HCOO- in the solution at equilibrium?
Solution
The chemical equation for this reaction is
$HCOOH_{(aq)}+H_{2}O_{l}\leftrightarrow H_{3}O^{+}_{(aq)}+HCOO^{-}_{(aq)}\nonumber$
This indicates that
$K_{a}=\frac{[HCOO^{-}][H_{3}O^{+}]}{[HCOOH]}\nonumber$
Because the volumes are equal and additive, the concentrations of the two solutions become half of what they were initially.
At equilibrium,
$[HCOO^{-}]=1+x\nonumber$
$[H_{3}O^{+}]=x\nonumber$
$[HCOOH]=1-x\nonumber$
Plugging in these values into the equilibrium expression gives
$1.77\bullet 10^{-4}=\frac{[1+x][x]}{[1-x]}\nonumber$
Solving the equation for x gives $x= 1.77\bullet 10^{-4}M\nonumber$
The ratio of HCOOH and HCOO- is $\frac{1+x}{1-x}\nonumber$ and due to x being such a small value, the ratio between the amounts of HCOOH and HCOO- is essentially 1.
Q101
The $∆G_f^o$ of $\ce{PbO(s)}$ is -188.95 kJ mol-1. Calculate the equilibrium pressure of O2(g) over a sample of pure PbO(s) in contact with pure Pb(s) at 25°C. PbO(s) decomposes according to the equation
$\ce{PbO_{(s)} <=> Pb (s) + 1/2 O2(g)}\nonumber$
Solution
Since Gf° for O2(g) and Pbs(s) is 0, for the reaction
$\Delta G^{\circ}=(0+0)-(-188.95kJ mol^{-1})=188.95kJ mol^{-1}\nonumber$
The equilibrium constant of the reaction can be computed by
$ln K= -\frac{\Delta G^{\circ}}{RT}=-76.22\nonumber$
$K=7.898\bullet 10^{-34}\nonumber$
From the chemical equation, we can tell
$K=(P_{O_{2}})^{1/2}\nonumber$
Therefore the equilibrium pressure of O2 at 25°C
$P_{O_{2}}=6.2\bullet 10^{-67}atm\nonumber$
Q105
If that weren’t enough, Jim’s sister, ., knocked his grass juice (a mixture of 1-Hexanol, $Cl_{2}$, $CaCl_{2}$, and cyclohexane) into the family’s benzene-ethanol vat! Assume that the two solvents in the vat are immiscible with each other but are miscible with the grass juice components. Jim separates the benzene and ethanol, and drinks the one with the greater amount of 1-Hexanol. Which one does he drink? Which of the two solvents have the highest concentration of the other components of the grass juice?
Solution
There are two immiscible solvents in the family vat. What this means is that they do not mix/ dissolve well in each other. Consequently, when . poured the grass juice into this vat, the solution technically contains two solvents. Evidently, what this means is that each of the solutes (which are the grass juice constituents mentioned in the question) dissolves to a different extent in each solvent. When the question asks which solvent has the greater amount of 1-Hexanol, it is the same as asking does 1-Hexanol dissolve “more” in either benzene or ethanol (the two solvents). The concept of “like dissolves like” is very important here, and to solve this problem first benzene must be identified as non-polar and driven by dispersion forces, while ethanol is polar. Since polar solvents dissolve polar solutes, then ethanol should be able to dissolve $CaCl_{2}$ and 1-Hexanol, both polar solutes, much better, while benzene can dissolve $Cl_{2}$ and cyclohexane, which are nonpolar, to a greater extent. The aptitude that each solvent has in dissolving the solutes directly correlate to the concentrations of solutes that are present in the solvent.
Thus, Jim would drink the ethanol, since it has the highest concentration of the polar 1-Hexanol. Also, the benzene would have the highest concentration of $Cl_{2}$ and cyclohexane, and the ethanol would also have the highest concentration of $CaCl_{2}$, which is ionic and therefore “very polar”.
Abstract: “Like dissolves Like”: polar solvents dissolve polar solutes, and nonpolar solvents dissolve polar solutes
Q107
Calculate the pH of a 0.5 M $H_3PO_4$ acid solution given that $K_{a1}=7.1 \times 10^{-3}$, $K_{a2}=6.3 \times 10^{-8}$ and $K_{a3}=4.2 \times 10^{-13}$.
Solution
A14.107V2
1.25
H14.107V2
Review on how to calculate the pH of a polyprotic acids here.
S14.107V2
$7.1 \times 10^{-3}=\dfrac{x^2}{(0.5-x)}\nonumber$
$x=0.0561\;M=[H^+]\nonumber$
$pH=-\log[H^+]=1.25\nonumber$
We assume that all of the $H^+$ in solution comes from the dissociation of the first H+ because the second and the third is too small to have a significant effect in the pH of the solution.
Q107
The Haber-Bosch process is a method invented in the 20th century as a method for hydrogen fixation. It is widely used in industry today, and involves reacting nitrogen and hydrogen in the following manner:
$\ce{N2 (g) + 3H2 (g) \rightleftharpoons 2 NH3 (g)}\nonumber$
Calculate $\Delta{G_{rxn}^{\circ}}$. If the pressure of the products and reaction are all kept constant at the same value, in which direction will the reaction go?
Relevant information:
$\Delta{G_f^{\circ}}\; NH_{2\, (g)} = 199.83\dfrac{kJ}{mol}$
Solution
The process of calculating $\Delta{G}$ should be very familiar after reading Q15, so this solution will not elaborate much on this method. With that being said, the $\Delta{G_{rxn}^{\circ}}$ is as follows:
$\Delta{G_{rxn}^{\circ}} = 399.66\dfrac{kJ}{mol}$
Both $N_{2}$ and $H_{2}$ have respective $\Delta{G_{f}}$ of zero. Now, remember that if $\Delta{G_{rxn}^{\circ}}$ is negative ($\Delta{G_{rxn}^{\circ}}$<0), then the reaction is spontaneous in the direction mentioned. However, in this case the $\Delta{G_{rxn}^{\circ}}$ is definitely NOT less than zero, so it is not spontaneous in the direction $N_{2\, (g)} + 3H_{2\, (g)} \rightarrow 2 NH_{3\, (g)}$. However, if the chemical formula is "flipped" so that it goes from $N_{2\, (g)} + 3H_{2\, (g)} \leftarrow 2 NH_{3\, (g)}$, Hess's law states that the corresponding $\Delta{G_{rxn}^{\circ}}$ is also "flipped" from positive to negative (or vice versa, depending on the value. For this situation, however, the $\Delta{G_{rxn}^{\circ}}$ does turn negative). What information can be drawn from this is that, since a negative $\Delta{G_{rxn}^{\circ}}$ means that the reaction will be spontaneous, then it can be inferred that the reaction will favor going from $N_{2\, (g)} + 3H_{2\, (g)} \leftarrow 2 NH_{3\, (g)}$, rather than $N_{2\, (g)} + 3H_{2\, (g)} \rightarrow 2 NH_{3\, (g)}$. Without even having to calculate the equilibrium constant, from this information alone, it can be inferred that, at equilibrium, the reactants are favored over the products. Therefore, if both gasses are kept at constant pressure, the reaction will tend to increase concentration of reactants. Thus, the reaction will go left.
Abstract: Find the $\Delta{G_{rxn}^{\circ}}$, understand what implications it has in regards to equilibrium, and apply this knowledge to infer that the reaction will favor the reactants.
Q109
At 273 K, the equation: H2(g) + I2(g) → 2HI(g) has an equilibrium constant of 60.2
1. Solve for ΔGº at 273 K
2. How does G change when the pressure of HI is 2 atm, I2 is 4 atm and H2 is 6 atm?
3. Is this reaction spontaneous?
Solution
a) ΔG° = -RT ln(K)
ΔG° = -(8.3145 J/K mol)(273 K) ln(60.2)
ΔG° = -9301.1 J/mol
b) ΔG = ΔG° + RT ln(Q)
ΔG = -9301.1 J/mol + (8.3145 J/K mol)(273 K) ln(PHI / PH2PI2)
ΔG = -9301.1 J/mol + (8.3145 J/K mol)(273 K) ln(2 / (6x4))
ΔG = -14941.49 J/mol.
c) This reaction proceeds spontaneously in the forward direction.
Q109
At 773 K the equilibrium constant for the Haber-Bosch process is 1.45x10-5. The chemical reaction involved in the Haber-Bosch process is:
$N_{2\, (g)} + 3H_{2\, (g)} \rightleftharpoons 2NH_{3\: (g)} \nonumber$
1. What is $\Delta G°(773)$ for this reaction?
2. What is $\Delta G$ at 773 K for transforming 1 mole of N2 and 3 moles H2 held at 30 atm to 2 moles of NH3 held at 1 atm?
3. In which direction would the previous reaction run spontaneously?
Solution
a. The value of $\Delta G°(773)$ can be found using the equation $\Delta G° = -RT \ln K$ where T and K are already known to be 773 K and 1.45x10-5 respectively while R=8.134 J K-1 mol-1.
$\Delta G°= -8.314\; J\; K^{-1}\; mol^{-1} \times 773K \times ln(1.45 \times 10^{-5}) = 70052\; J \nonumber$
$\Delta G°= 70052\; J \nonumber$
b. The equation needed to determine $\Delta G$ is $\Delta G = RTln(Q/K)$. The values of R, T, and K are known while Q is
$Q = \dfrac{P_{NH_{3}}^{2}/P_{ref}} {(P_{N_{2}}/P_{ref})(P_{H_{2}}^{3}/P_{ref})} \nonumber$
Pref =1 atm and the other pressures are stated in the question. So to find Q, the pressures have to be plugged into the equation
$Q = \dfrac{(1)^{2}} {(20)(20)^{3}} = \dfrac{1}{160000} = 6.25 \times 10^{-6}\nonumber$
Now that the value of Q is known, $\Delta G$ can be found
$\Delta G= -8.314\; J\; K^{-1}\; mol^{-1} \times 773K \times ln(\frac{6.25 \times 10^{-6}}{1.45 \times 10^{-5}}) = -5291\; J \nonumber$
$\Delta G = -5291\; J \nonumber$
c. The previous reaction would run towards the products spontaneously because $\Delta G < 0$.
Q111
The following reaction is exothermic
$\ce{SnO2(s) + 2H2(g) → Sn(g) + 2H2O(l)} \nonumber$
1. What is the equilibrium expression based on this chemical equation?
2. What happens to the reaction if more H2 is added? SnO2 is added into the system?
3. What effect would an increase in temperature have on the system?
4. What affect would an increase in pressure have on the system?
Solution
1. Add texts here. Do not delete this text first.K = [Sn] / [H2] This is because solids are not included in the equilibrium expression
2. If H2 were added, the reaction would shift right, towards the products, more Sn is produced. If SnO2 were added, there would be no effect on the equilibrium because SnO2 is not involve in the equilibrium process (K).
3. The reaction would shift towards the reactants. The reaction is exothermic, so heat is treated like a product.
4. An increase in pressure would shift the reaction towards the side that has less mole of gas. Therefore, in this reaction, an increased in pressure would shift the reaction to the right, more products are produced. Q111
Q112
Given the chemical reaction
$\ce{ SnO2(s) \rightleftharpoons Sn (s) + O2 (g)} \nonumber$
1. Formulate the equilibrium expression using activities
2. State the effect on $p\ce{O2}$ at equilibrium if more $\ce{SnCl4(l)}$ is added to the system
3. State the effect on $p\ce{O2}$ at equilibrium if more $\ce{O2(g)}$ is added to the system
Solution
a. The activities of solids and liquids are always equal to 1. The activities of gases are equal to the partial pressure of the gas in bars. Therefore the equilibrium expression would be
$K = P_{O_{2}} \nonumber$
b. $P_{O_{2}}$ would not change because the activity of a liquid is always 1 meaning it has no effect on the equilibrium constant
c. $P_{O_{2}}$ would decrease because the extra gas would push the system out of equilibrium and in order for the system to return to equilibrium, some of the O2 gas would have to react with some of the Sn(s). This would decrease the amount of O2(g) present in the system and therefore decrease the the pressure of O2(g).
Q113
Arrange the following solutions in order of most acidic to basic: NaOH, NaCN, H2SO4, NH4NO3, NaCl
Solution
$\ce{H2SO4 > NH4NO3 > NaCl > NaCN > NaOH }$
Review acids bases here.
• H2SO4 is a strong acid.
• NH4NO3 is formed by a weak base (NH4OH) and a strong acid (HNO3). Hence, NH4NO3 is acidic.
• NaCl is formed by a strong base (NaOH) and a strong acid (HCl). Hence, it is a neutral salt.
• NaCN is formed by a strong base (NaOH) and a weak acid (HCN). Hence, NaCN is basic.
• NaOH is a strong base.
Q115
Given that the $K_b$ values for $\ce{NH_3}$ and $\ce{NH_2NH_2}$ are $1.8 \times 10^{-5}$ and $8.5 \times 10^{-7}$, respectively. Which is the stronger acid?
Solution
$\ce{NH2NH2}$
Calculate the $K_a$ values first. Review on how to identify the strength of an acid from its $K_a$ value here.
For NH3,
$K_a= \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\nonumber$
For NH2NH2,
$K_a= \dfrac{(1.0 \times 10^{-14}}{8.5 \times 10^{-7}}= 1.18 \times 10^{-8}\nonumber$
Since the Ka value of NH2NH2 is larger, NH2NH2 is the stronger acid.
Q119
Calculate the pH of the following solutions.
1. 0.5 M $\ce{NaCN}$ ($K_a \text{ of } \ce{HCN}=6.2 \times 10^{-10}$)
2. 0.01 M $\ce{NH4Cl}$ ($K_b \text{ of } \ce{NH3} = 1.8 \times 10^{-5}$)
Solution
11.5 5.63
Ka=(1*10-14)/(1.8*10-5)=5.56*10-10
5.56*10-10=x2/(0.01-x)
x=2.36*10-6
pH= -log(x)= 5.63
H14.119V2
Review on the properties of salts here.
$\ce{CN- (aq) + H2O (l) <=> HCN (aq) + OH- (aq)}$
Kb=(1*10-14)/(6.2*10-10)=1.62*10-5
$1.62 \times 10^{-5} =\dfrac{x^2}{0.5-x}$
x=0.00283
pOH= -log(x)= 2.55
pH= 14-pOH= 11.5
$\ce{NH4^{+}(aq) + H2O (l) <=>NH3 (aq) + H3O^{+} (aq)}$
Q131
Identify the stronger acid in each of the following combination. Explain.
1. HF and HBr
2. H2SO4 and H2SO3
3. HOCl and HOBr
Solution
14.131V2
HBr H2SO4 HOCl
Study the effect of structure on acids and bases.
S14.131V2
1. HBr is the stronger acid because Br less electronegative and therefore less polar as compared to HF.
2. H2SO4 is the stronger acid because as the number of attached oxygen on the O-H bonds increases, the less polar the molecule is.
3. For oxyacids, the less polar oxyacids are the ones that are more electronegative. Since Cl is more electronegative than Br, HOCl is the stronger acid.
Q137
Identify the Lewis acid and Lewis base for each of the following reaction.
1. $\ce{H^{+} (aq) + NH3 (aq) <=> NH4^{+} (aq)}$
2. $\ce{Al^{3+} (aq) + 6H2O (l) <=> Al(H2O)6^{3+} (aq)}$
3. $\ce{CO2 (g) + H2O (l) <=> H2CO3 (aq)}$
Solution
Lewis acid
Lewis base
1.
H+
NH3
2.
Al3+
H2O
3.
CO2
H2O
H14.137V2
Study the concept of electron pairs here.
S14.137V2
H+ is the electron-pair acceptor while NH3 is the electron-pair donor. Al3+ is the electron-pair acceptor while H2O is the electron-pair donor. CO2 is the electron-pair acceptor while H2O is the electron-pair donor.
Q139
Beryllium hydroxide is an amphoteric compound. It can react as an acid as well as a base. Please write the reactions of Be(OH)2 and describe its role base on Brønsted-Lowry and Lewis theories.
Solution
Be(OH)2(s)+2H+(aq)-> Be2+(aq)+2H2O(l) (Brønsted-Lowry base, 2H+ acceptor)
Be(OH)2(s)+2OH-(aq)-> Be(OH)42-(aq) (Lewis acid, electron pair acceptor)
H14.139.2P
Consider Brønsted-Lowry and Lewis theories.
http://chemwiki.ucdavis.edu/Textbook...cids_and_Bases
| textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.E%3A_Chemical_Equilibria_%28Exercises%29.txt |
Acids and bases have been defined differently by three sets of theories. One is the Arrhenius definition, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H+) ions while bases produce hydroxide (OH-) ions in solution. On the other hand, the Bronsted-Lowry definition defines acids as substances that donate protons (H+) whereas bases are substances that accept protons. Also, the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors. Acids and bases can be defined by their physical and chemical observations.
• 15.1: Classifications of Acids and Bases
In chemistry, acids and bases have been defined differently by three sets of theories: One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H+) ions while bases produce hydroxide (OH-) ions in solution. The other two definitions are discussed in detail alter in the chapter and include the Brønsted-Lowry definition and the Lewis theory.
• 15.2: Properties of Acids and Bases in Aqueous Solutions
We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibria in terms of the Brønsted–Lowry model and then proceed to a quantitative description the following sections.
• 15.3: Acid and Base Strength
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons.
• 15.4: Equilibria Involving Weak Acids and Bases
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons.
• 15.5: Buffer Solutions
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (\(HA\)) and its conjugate weak base (\(A^−\)). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. Buffers are characterized by their pH range and buffer capacity.
• 15.6: Acid-Base Titration Curves
The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component.
• 15.7: Polyprotic Acids
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. It is often (but not always) appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.
• 15.8: Organic Acids and Bases - Structure and Reactivity
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an H+ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of H+, making the conjugate acid a stronger acid.
• 15.9: A Deeper Look - Exact Treatment of Acid-Base Equilibria
The methods for dealing with acid-base equilibria that we developed in the earlier units of this series are widely used in ordinary practice. Although many of these involve approximations of various kinds, the results are usually good enough for most purposes. In this unit, we look at exact, or "comprehensive" treatment of some of the more common kinds of acid-base equilibria problems.
• 15.E: Acid-Base Equilibria (Exercises)
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
15: AcidBase Equilibria
Learning Objectives
• To understand the differences between the three definitions of Acids and Bases
• Identify acids, bases, and conjugate acid-base pairs according to the three definitions of Acids and Bases
• To understand the concept of conjugate acid–base pairs in acid/base reactions
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Carl Axel Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Acids and bases are common solutions that exist everywhere. Almost every liquid that we encounter in our daily lives consists of acidic and basic properties, with the exception of water. They have completely different properties and are able to neutralize to form H2O, which will be discussed later in a subsection. Acids and bases can be defined by their physical and chemical observations (Table $1$).
Table $1$: General Properties of Acids and Bases
ACIDS BASES
produce a piercing pain in a wound. give a slippery feel.
taste sour. taste bitter.
are colorless when placed in phenolphthalein (an indicator). are pink when placed in phenolphthalein (an indicator).
are red on blue litmus paper (a pH indicator). are blue on red litmus paper (a pH indicator).
have a pH<7. have a pH>7.
produce hydrogen gas when reacted with metals.
produce carbon dioxide when reacted with carbonates.
Common examples: Lemons, oranges, vinegar, urine, sulfuric acid, hydrochloric acid
Common Examples: Soap, toothpaste, bleach, cleaning agents, limewater, ammonia water, sodium hydroxide.
Acids and bases in aqueous solutions will conduct electricity because they contain dissolved ions. Therefore, acids and bases are electrolytes. Strong acids and bases will be strong electrolytes. Weak acids and bases will be weak electrolytes. This affects the amount of conductivity.
In chemistry, acids and bases have been defined differently by three sets of theories: One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H+) ions while bases produce hydroxide (OH-) ions in solution. The other two definitions are discussed in detail include the Brønsted-Lowry definition the defines acids as substances that donate protons (H+) whereas bases are substances that accept protons and the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors.
Arrhenius Acids and Bases
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. The Arrhenius definition of acid-base reactions is a development of the "hydrogen theory of acids". It was used to provide a modern definition of acids and bases, and followed from Arrhenius's work with Friedrich Wilhelm Ostwald in establishing the presence of ions in aqueous solution in 1884. This led to Arrhenius receiving the Nobel Prize in Chemistry in 1903.
An Arrhenius acid is a compound that increases the concentration of $\ce{H^{+}}$ ions that are present when added to water.
$\ce{HCl(g) \rightarrow H^{+}(aq) + Cl^{-}(aq)} \label{eq1}$
In this reaction, hydrochloric acid ($\ce{HCl}$) gas dissociates into hydrogen ($\ce{H^{+}}$) and chloride ($\ce{Cl^{-}}$) ions when dissolved in water, thereby releasing $\ce{H^{+}}$ ions into solution. These $\ce{H^{+}}$ ions form the hydronium ion ($\ce{H_3O^{+}}$) when they combine with water molecules. Both processes can be represented in a chemical equation by adding $\ce{H2O}$ to the reactants side of Equation \ref{eq1} and switching hydronium ions for free protons.
$\ce{ HCl(g) + H2O(l) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)} \label{eq2}$
An Arrhenius base is a compound that dissociates to yield hydroxide ions ($\ce{OH^{−}}$) in aqueous solution. Common examples of Arrhenus bases include Sodium hydroxide ($\ce{NaOH}$), Potassium hydroxide ($\ce{KOH}$), Magnesium hydroxide $\ce{Mg(OH)2}$, and Calcium hydroxide ($\ce{Ca(OH)2}$. All of these bases are solids at room temperature and when dissolved in water, will generate a metal cation and the hydroxide ion ($\ce(OH^{-}}$), for example, Sodium hydroxide
$\ce{NaOH(s) -> Na^{+}(aq) + OH^{-}(aq)} \label{eq3}$
All Arrhenius acids have easily detachable hydrogen that leave to form hydronium ions in solution and all Arrhenius bases have easily detachable OH groups that form hydroxide ions in solution.
Limitation of the Arrhenius Definition of Acids and Bases
The Arrhenius definitions of acidity and alkalinity are restricted to aqueous solutions and refer to the concentration of the solvated ions. Under this definition, pure $\ce{H2SO4}$ or $\ce{HCl}$ dissolved in toluene are not acidic, despite the fact that both of these acids will donate a proton to toluene. In addition, under the Arrhenius definition, a solution of sodium amide ($\ce{NaNH2}$) in liquid ammonia is not alkaline, despite the fact that the amide ion ($\ce{NH2^{-}}$) will readily deprotonate ammonia. Thus, the Arrhenius definition can only describe acids and bases in an aqueous environment. The Arrhenius definition can only describe acids and bases in protic solvents and environments (e.g., water, alcohols, within proteins etc.).
Brønsted-Lowry Acids and Bases
The Arrhenius definitions identified an acid as a compound that dissolves in water to yield hydronium ions (Equation \ref{eq2}\) and a base as a compound that dissolves in water to yield hydroxide ions (Equation \ref{eq3}). As mentioned above, this definition limited. We extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, $\ce{H^+}$. A proton is what remains when a normal hydrogen atom, $\ce{^1_1H}$, loses an electron. A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base).
Acids may be compounds such as $\ce{HCl}$ or $\ce{H2SO4}$, organic acids like acetic acid ($\ce{CH_3COOH}$) or ascorbic acid (vitamin C), or $\ce{H2O}$. Anions (such as $\ce{HSO_4^-}$, $\ce{H_2PO_4^-}$, $\ce{HS^-}$, and $\ce{HCO_3^-}$) and cations (such as $\ce{H_3O^+}$, $\ce{NH_4^+}$, and $\ce{[Al(H_2O)_6]^{3+}}$) may also act as acids.
Bases fall into the same three categories and may be neutral molecules (such as $\ce{H_2O}$, $\ce{NH_3}$, and $\ce{CH_3NH_2}$), anions (such as $\ce{OH^-}$, $\ce{HS^-}$, $\ce{HCO_3^-}$, $\ce{CO_3^{2−}}$, $\ce{F^-}$, and $\ce{PO_4^{3−}}$), or cations (such as $\ce{[Al(H_2O)_5OH]^{2+}}$). The most familiar bases are ionic compounds such as $\ce{NaOH}$ and $\ce{Ca(OH)_2}$, which contain the hydroxide ion, $\ce{OH^-}$. The hydroxide ion in these compounds accepts a proton from acids to form water:
$\ce{H^+ + OH^- \rightarrow H_2O} \label{15.1.1}$
We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):
\begin{align*} \text{acid} &\rightleftharpoons \text{proton} + \text{conjugate base} \[4pt] \ce{HF} &\rightleftharpoons \ce{H^+ + F^-} \[4pt] \ce{H_2SO_4} &\rightleftharpoons \ce{H^+ + HSO_4^{−}} \[4pt] \ce{H_2O} &\rightleftharpoons \ce{H^+ + OH^-} \[4pt] \ce{HSO_4^-} &\rightleftharpoons \ce{H^+ + SO_4^{2−}} \[4pt] \ce{NH_4^+} &\rightleftharpoons \ce{H^+ + NH_3} \end{align*}
We call the product that results when a base accepts a proton the base’s conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base):
\begin{align*} \text{base} + \text{proton} &\rightleftharpoons \text{conjugate acid} \[4pt] \ce{OH^- + H^+} &\rightleftharpoons \ce{H2O} \[4pt] \ce{H_2O + H^+} &\rightleftharpoons \ce{H3O+} \[4pt] \ce{NH_3 + H^+} &\rightleftharpoons \ce{NH4+} \[4pt] \ce{S^{2-} + H^+} &\rightleftharpoons \ce{HS-} \[4pt] \ce{CO_3^{2-} + H^+} &\rightleftharpoons \ce{HCO3-} \[4pt] \ce{F^- + H^+} &\rightleftharpoons \ce{HF} \end{align*}
In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
with
$K =\dfrac{[\ce{H3O^{+}}][\ce{F^{-}}]}{[\ce{HF}]}\nonumber$
When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding ammonia to water yields hydroxide ions and ammonium ions:
with
$K =\dfrac{[\ce{C5NH6^{+}}][\ce{OH^{-}}]}{[\ce{C5NH5}]} \nonumber$
Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization. Pure water undergoes autoionization to a very slight extent. Only about two out of every $10^9$ molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water ($K_w$):
$\ce{H_2O(l) + H2O(l) <=> H_3O^{+}(aq) + OH^{-}(aq)}$
with
$K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{15.1.4}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, $K_w$ has a value of $1.0 \times 10^{−14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for $K_\ce{w}$ is approximately $5.1 \times 10^{−13}$, roughly 100-times larger than the value at 25 °C.
Example $1$: Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\ce{[H_3O^+]} = \ce{[OH^- ]}$. At 25 °C:
\begin{align*} K_\ce{w} &=\ce{[H_3O^+][OH^- ]} \[4pt] &= [\ce{H3O^{+}}]^2 \[4pt] &= [\ce{OH^{-}} ]^2 \[4pt] &=1.0 \times 10^{−14} \end{align*}
So
$\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal $1.0 \times 10^{−7}\; M$.
Exercise $1$
The ion product of water at 80 °C is $2.4 \times 10^{−13}$. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer
$\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7} \; M\nonumber$
It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. Example $2$ demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
Example $2$: The Inverse Proportionality of $\ce{[H_3O^+]}$ and $\ce{[OH^- ]}$
The Inverse Proportionality of [H3O+] and [OH-] A solution of carbon dioxide in water has a hydronium ion concentration of $2.0 \times 10^{−6}\; M$. What is the concentration of hydroxide ion at 25 °C?
Solution
We know the value of the ion-product constant for water at 25 °C:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber$
$K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14} \nonumber$
Thus, we can calculate the missing equilibrium concentration.
Rearrangement of the Kw expression yields that $[\ce{OH^- }]$ is directly proportional to the inverse of [H3O+]:
$[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber$
The hydroxide ion concentration in water is reduced to $5.0 \times 10^{−9}\: M$ as the hydrogen ion concentration increases to $2.0 \times 10^{−6}\; M$. This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the $\ce{[OH^- ]}$ is reduced relative to that in pure water.
A check of these concentrations confirms that our arithmetic is correct:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber$
Exercise $2$
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
$\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber$
Amphoteric Species
Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:
$\ce{HCO^-3(aq) + H2O(l) <=> CO^{2-}3(aq) + H3O^+(aq)} \nonumber$
$\ce{HCO^{-}3(aq) + H2O(l) <=> H2CO3(aq) + OH^{-}(aq)} \nonumber$
Example $3$: The Acid-Base Behavior of an Amphoteric Substance
Write separate equations representing the reaction of $\ce{HSO3-}$
1. as an acid with $\ce{OH^-}$
2. as a base with HI
Solution
1. $\ce{HSO3-}(aq)+ \ce{OH^-}(aq)\rightleftharpoons \ce{SO3^2-}(aq)+ \ce{H_2O}_{(l)}$
2. $\ce{HSO3-}(aq)+\ce{HI}(aq)\rightleftharpoons \ce{H2SO3}(aq)+\ce{I-}(aq)$
Exercise $3$
Write separate equations representing the reaction of $\ce{H2PO4-}$
1. as a base with HBr
2. as an acid with $\ce{OH^-}$
Answer a
$\ce{H2PO4-}(aq)+\ce{HBr}(aq)\rightleftharpoons \ce{H3PO4}(aq)+\ce{Br-}(aq)$
Answer b
$\ce{H2PO4-}(aq)+\ce{OH^-} (aq)\rightleftharpoons \ce{HPO4^2-}(aq)+ \ce{H_2O}_{(l)}$
Lewis Acids and Bases
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. The Brønsted–Lowry concept of acids and bases defines a base as any species that can accept a proton, and an acid as any substance that can donate a proton. Lewis proposed an alternative definition that focuses on pairs of electrons instead. According to Lewis:
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Lewis Acid-Base Neutralization Involving Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
Note
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur.
The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" $\ce{H2O}$. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate acid is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, $\ce{F^{–}}$ accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
All Brønsted–Lowry bases (proton acceptors), such as $\ce{OH^{−}}$, $\ce{H2O}$, and $\ce{NH3}$, are also electron-pair donors. Thus the Lewis definition of acids and bases does not contradict the Brønsted–Lowry definition. Rather, it expands the definition of acids to include substances other than the $\ce{H^{+}}$ ion.
Lewis Acid-Base Neutralization without Transferring Protons
Electron-deficient molecules, such as BCl3, contain less than an octet of electrons around one atom and have a strong tendency to gain an additional pair of electrons by reacting with substances that possess a lone pair of electrons. Lewis’s definition, which is less restrictive than either the Brønsted–Lowry or the Arrhenius definition, grew out of his observation of this tendency. A general Brønsted–Lowry acid–base reaction can be depicted in Lewis electron symbols as follows:
The proton (H+), which has no valence electrons, is a Lewis acid because it accepts a lone pair of electrons on the base to form a bond. The proton, however, is just one of many electron-deficient species that are known to react with bases. For example, neutral compounds of boron, aluminum, and the other Group 13 elements, which possess only six valence electrons, have a very strong tendency to gain an additional electron pair. Such compounds are therefore potent Lewis acids that react with an electron-pair donor such as ammonia to form an acid–base adduct, a new covalent bond, as shown here for boron trifluoride (BF3):
The bond formed between a Lewis acid and a Lewis base is a coordinate covalent bond because both electrons are provided by only one of the atoms (N, in the case of F3B:NH3). After it is formed, however, a coordinate covalent bond behaves like any other covalent single bond.
Species that are very weak Brønsted–Lowry bases can be relatively strong Lewis bases. For example, many of the group 13 trihalides are highly soluble in ethers (R–O–R′) because the oxygen atom in the ether contains two lone pairs of electrons, just as in H2O. Hence the predominant species in solutions of electron-deficient trihalides in ether solvents is a Lewis acid–base adduct. A reaction of this type is shown in Figure $1$ for boron trichloride and diethyl ether:
Figure $1$: Lewis Acid/Base reaction of boron trichloride and diethyl ether reaction
Note
• Electron-deficient molecules (those with less than an octet of electrons) are Lewis acids.
• The acid-base behavior of many compounds can be explained by their Lewis electron structures.
Many molecules with multiple bonds can act as Lewis acids. In these cases, the Lewis base typically donates a pair of electrons to form a bond to the central atom of the molecule, while a pair of electrons displaced from the multiple bond becomes a lone pair on a terminal atom.
Figure $2$: Lewis Acid/Base reaction of the hydroxide ion with carbon dioxide
Example $4$
Identify the acid and the base in each Lewis acid–base reaction.
1. BH3 + (CH3)2S → H3B:S(CH3)2
2. CaO + CO2 → CaCO3
3. BeCl2 + 2 Cl → BeCl42−
Given: reactants and products
Asked for: identity of Lewis acid and Lewis base
Strategy:
In each equation, identify the reactant that is electron deficient and the reactant that is an electron-pair donor. The electron-deficient compound is the Lewis acid, whereas the other is the Lewis base.
Solution:
1. In BH3, boron has only six valence electrons. It is therefore electron deficient and can accept a lone pair. Like oxygen, the sulfur atom in (CH3)2S has two lone pairs. Thus (CH3)2S donates an electron pair on sulfur to the boron atom of BH3. The Lewis base is (CH3)2S, and the Lewis acid is BH3.
2. As in the reaction shown in Equation 8.21, CO2 accepts a pair of electrons from the O2− ion in CaO to form the carbonate ion. The oxygen in CaO is an electron-pair donor, so CaO is the Lewis base. Carbon accepts a pair of electrons, so CO2 is the Lewis acid.
3. The chloride ion contains four lone pairs. In this reaction, each chloride ion donates one lone pair to BeCl2, which has only four electrons around Be. Thus the chloride ions are Lewis bases, and BeCl2 is the Lewis acid.
Exercise $\PageIndex{4A}$
Identify the acid and the base in each Lewis acid–base reaction.
1. (CH3)2O + BF3 → (CH3)2O:BF3
2. H2O + SO3 → H2SO4
Answer a
Lewis base: (CH3)2O; Lewis acid: BF3
Answer b
Lewis base: H2O; Lewis acid: SO3
Exercise $\PageIndex{4B}$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $Al(OH)_3 + OH^– \rightarrow Al(OH)_4^–$
2. $SnS_2 + S^{2–} \rightarrow SnS_3^{2–}$
3. $Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}$
4. $AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–$
5. $Fe^{2+} + NO \rightarrow Fe(NO)^{2+}$
6. $[Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}$
Summary
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, $\ce{OH^-}$ when it undergoes autoionization:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber$
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
$K_\ce{w}=\mathrm{[H_3O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber$
Glossary
acid ionization
reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
amphiprotic
species that may either gain or lose a proton in a reaction
amphoteric
species that can act as either an acid or a base
autoionization
reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionization
reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
Brønsted-Lowry acid
proton donor
Brønsted-Lowry base
proton acceptor
conjugate acid
substance formed when a base gains a proton
conjugate base
substance formed when an acid loses a proton
ion-product constant for water (Kw)
equilibrium constant for the autoionization of water | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.1%3A_Classifications_of_Acids_and_Bases.txt |
Learning Objectives
• Explain the characterization of aqueous solutions as acidic, basic, or neutral
• To know the relationship between acid or base strength and the magnitude of $K_a$, $K_b$, $pK_a$, and $pK_b$.
• Express hydronium and hydroxide ion concentrations on the pH and pOH scales
• Perform calculations relating pH and pOH
We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibria in terms of the Brønsted–Lowry model and then proceed to a quantitative description the following sections.
Free Hydrogen Ions do not Exist in Water
Owing to the overwhelming excess of $H_2O$ molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water. The hydrogen ion in aqueous solution is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high charge density of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a hydronium ion, $H_3O^+$. In a sense, $H_2O$ is acting as a base here, and the product $H_3O^+$ is the conjugate acid of water:
Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H+ and $H_2O$ is so strong that writing “H+(aq)” hardly does it justice, although it is formally correct. The formula $H_3O^+$ more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water. However, the equation
$\ce{HA → H^{+} + A^{–}} \nonumber$
is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus it is permissible to talk about “hydrogen ions” and use the formula H+ in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions.
Interestingly, experiments indicate that the proton does not stick to a single $H_2O$ molecule, but changes partners many times per second. This molecular promiscuity, a consequence of the uniquely small size and mass the proton, allows it to move through the solution by rapidly hopping from one $H_2O$ molecule to the next, creating a new $H_3O^+$ ion as it goes. The overall effect is the same as if the $H_3O^+$ ion itself were moving. Similarly, a hydroxide ion, which can be considered to be a “proton hole” in the water, serves as a landing point for a proton from another $H_2O$ molecule, so that the OH ion hops about in the same way.
The hydronium ion is an important factor when dealing with chemical reactions that occur in aqueous solutions. Because hydronium and hydroxide ions can “move without actually moving” and thus without having to plow their way through the solution by shoving aside water molecules as do other ions, solutions which are acidic or alkaline have extraordinarily high electrical conductivities. The hydronium ion has a trigonal pyramidal geometry and is composed of three hydrogen atoms and one oxygen atom. There is a lone pair of electrons on the oxygen giving it this shape. The bond angle between the atoms is 113 degrees.
As H+ ions are formed, they bond with $H_2O$ molecules in the solution to form $H_3O^+$ (the hydronium ion). This is because hydrogen ions do not exist in aqueous solutions, but take the form the hydronium ion, $H_3O^+$. A reversible reaction is one in which the reaction goes both ways. In other words, the water molecules dissociate while the OH- ions combine with the H+ ions to form water. Water has the ability to attract H+ ions because it is a polar molecule. This means that it has a partial charge, in this case the charge is negative. The partial charge is caused by the fact that oxygen is more electronegative than hydrogen. This means that in the bond between hydrogen and oxygen, oxygen "pulls" harder on the shared electrons thus causing a partial negative charge on the molecule and causing it to be attracted to the positive charge of H+ to form hydronium. Another way to describe why the water molecule is considered polar is through the concept of dipole moment. The electron geometry of water is tetrahedral and the molecular geometry is bent. This bent geometry is asymmetrical, which causes the molecule to be polar and have a dipole moment, resulting in a partial charge.
The pH Scale
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm:
$\mathrm{pX=−\log X} \label{$1$}$
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
$\mathrm{pH=-\log[H_3O^+]}\label{$2$}$
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
$\mathrm{[H_3O^+]=10^{−pH}}\label{$3$}$
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
$\mathrm{pOH=-\log [OH^−]}\label{$4$}$
or
$\mathrm{[OH^-]=10^{−pOH}} \label{$5$}$
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_w$ expression:
\begin{align*} K_\ce{w} &=\ce{[H_3O^+][OH^- ]} \[4pt] -\log K_\ce{w} &= -\log([H_3O^+][OH^−]) \[4pt] &=-\log[H_3O^+] + -\log[OH^-] \[4pt] p \mathit{K}_w &=pH + pOH \end{align*}
At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$ and so:
$\mathrm{14.00=pH + pOH} \label{$9$}$
The hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times 10^{-7}\; M$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{$1$0}$
$\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{$1$1}$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities less than $1.0 \times 10^{-7}\; M$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities greater than $1.0 \times 10^{-7}\; M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Since the autoionization constant $K_w$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:
\begin{align*} \mathrm{pH} &= -\log[H_3O^+] \[4pt] &= -\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{$1$2} \end{align*}
\begin{align*} \mathrm{pOH} &= -\log[OH^-] \[4pt] &= -\log(4.9\times 10^{−7}) \[4pt] &=6.31 \label{$1$3} \end{align*}
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table $1$).
Table $1$: Summary of Relations for Acidic, Basic and Neutral Solutions
Classification Relative Ion Concentrations pH at 25 °C pH at 80 °C
acidic [H3O+] > [OH] pH < 7 pH < 6.31
neutral [H3O+] = [OH] pH = 7 pH = 6.31
basic [H3O+] < [OH] pH > 7 pH > 6.31
Figure $2$ shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.
Example $1$: Calculation of pH from $\ce{[H_3O^+]}$
What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \times 10^{−3}\; M$?
Solution
\begin{align*} \mathrm{pH} &= -\log [H_3O^+] \[4pt] &= -\log(1.2 \times 10^{−3}) \[4pt] &=−(−2.92) =2.92 \end{align*}
Exercise $1$
Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water:
$\ce{CO}_{2(aq)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)}\nonumber$
Air-saturated water has a hydronium ion concentration caused by the dissolved $\ce{CO_2}$ of $2.0 \times 10^{−6}\; M$, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
Answer
5.70
Example $2$: Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3.
Solution
\begin{align*} \mathrm{pH} =-\log [\ce{H_3O^+}] &=7.3 \[4pt] [\ce{H_3O^+}] &=10^{−7.3} \[4pt] [\ce{H_3O^+}] &=5\times 10^{−8}\;M \end{align*}
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.)
Exercise $\PageIndex{2A}$
Calculate the hydronium ion concentration of a solution with a pH of −1.07.
Answer
12 M
This uses the definition of pH that we commonly use:
$pH=-\log[H_3O^+] \nonumber$
So for this solution:
\begin{align*} \mathrm{pH} =-\log [\ce{H_3O^+}] &= -1.07 \[4pt] [\ce{H_3O^+}] & = 10^{+1.07} \[4pt] &= 12\, (\text{with significant figure}) \end{align*}
However, at this high concentration, the solution will be non-ideal and we have to use the proper definition in terms of hydronium activities
$pH=-\log a\{H_3O^+\} \nonumber$
See this module for more details.
Exercise $\PageIndex{2B}$
The ionization constant of water $\ce{K_w}$ at 37 °C is $2.42 \times 10^{-14}$. What is the pH for a neutral solution at this human physiological temperature? Is the water acidic, basic or neutral?
Answer
$\ce{K_w} = \ce{[H3O^{+}] [OH^{-}]} = 2.42 \times 10^{-14}\nonumber$
and
\begin{align*} \ce{[H3O^{+}]} = \ce{[OH^{-}]} &= \sqrt{2.42 \times 10^{-14} } \[4pt] &= 1.55 \times 10^{-7} \end{align*}
$pH=-\log[H_3O^+] = -\log 1.55 \times 10^{-7} = 6.81$
If we use the definition of acidic systems like in Figure $2$, then we would (incorrectly) argue the solution is acidic. However, since $\ce{[OH^{-}]} = \ce{[H3O^{+}]}$, the solution is still neutral. This is only a strange idea, if one ignores the temperature dependence of $\ce{K_w}$.
Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid:
$\ce{H_2O}_{(l)}+\ce{CO}_{2(g)}⟶\ce{H_2CO}_{3(aq)} \label{$1$4}$
$\ce{H_2CO}_{3(aq)} \rightleftharpoons \ce{H^+}_{(aq)} + \ce{HCO^-}_{3(aq)} \label{$1$5}$
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
$\ce{H_2O}_{(l)} + \ce{SO}_{3(g)} ⟶\ce{H_2SO}_{4(aq)} \label{$1$6}$
$\ce{H_2SO}_{4(aq)} ⟶ \ce{H^+}_{(aq)} + \ce{HSO}_{4(aq)}^- \label{$1$7}$
Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Figure $3$: (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $3$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure $4$).
Figure $4$: (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of \$1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther).
Example $3$: Calculation of pOH
What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH at 25 °C?
Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:
$\mathrm{pOH=-\log[OH^− ]=-\log 0.0125}$
$=−(−1.903)=1.903$
The pH can be found from the $\ce{pOH}$:
$\mathrm{pH+pOH=14.00}$
$\mathrm{pH=14.00−pOH=14.00−1.903=12.10}$
Exercise $3$
The hydronium ion concentration of vinegar is approximately $4 \times 10^{−3}\; M$ at 25 °C. What are the corresponding values of pOH and pH?
Answer
pOH = 11.6, pH = 2.4
The pH of a solution may also be visually estimated using colored indicators (Figure $5$).
Figure $5$: (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-M solutions of progressively weaker acids: HCl (pH = l), CH3CO2H (pH = 3), and NH4Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C6H5NH2 (pH = 9), NH3 (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa).
Summary
The concentration of hydronium ion in a solution of an acid in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7.00 and pOH = 7.00
Key Equations
• $\ce{pH}=-\log[\ce{H3O+}]$
• $\ce{pOH} = -\log[\ce{OH-}]$
• [H3O+] = 10−pH
• [OH] = 10−pOH
• pH + pOH = pKw = 14.00 at 25 °C
Glossary
acidic
describes a solution in which [H3O+] > [OH]
basic
describes a solution in which [H3O+] < [OH]
neutral
describes a solution in which [H3O+] = [OH]
pH
logarithmic measure of the concentration of hydronium ions in a solution
pOH
logarithmic measure of the concentration of hydroxide ions in a solution | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.2%3A_Properties_of_Acids_and_Bases_in_Aqueous_Solutions.txt |
Learning Objectives
• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)$
Water is the base that reacts with the acid $\ce{HA}$, $\ce{A^{−}}$ is the conjugate base of the acid $\ce{HA}$, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of $\ce{H3O+}$ and $\ce{A^{−}}$ when the acid ionizes in water; Figure $1$ lists several strong acids. A weak acid gives small amounts of $\ce{H3O+}$ and $\ce{A^{−}}$.
Figure $1$: Some of the common strong acids and bases are listed here.
Six Strong Acids Six Strong Bases
$\ce{HClO4}$ perchloric acid $\ce{LiOH}$ lithium hydroxide
$\ce{HCl}$ hydrochloric acid $\ce{NaOH}$ sodium hydroxide
$\ce{HBr}$ hydrobromic acid $\ce{KOH}$ potassium hydroxide
$\ce{HI}$ hydroiodic acid $\ce{Ca(OH)2}$ calcium hydroxide
$\ce{HNO3}$ nitric acid $\ce{Sr(OH)2}$ strontium hydroxide
$\ce{H2SO4}$ sulfuric acid $\ce{Ba(OH)2}$ barium hydroxide
The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid $\ce{HA}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)$
we write the equation for the ionization constant as:
$K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is also the solvent. If the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The larger the $K_a$ of an acid, the larger the concentration of $\ce{H3O+}$ and $\ce{A^{−}}$ relative to the concentration of the nonionized acid, $\ce{HA}$. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase.
The following data on acid-ionization constants indicate the order of acid strength: $\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}$
\begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \[4pt] \ce{HSO4-}(aq)+\ce{H2O}(aq) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned}
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:
$\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon}$
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Example $1$: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
$\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber$
The chemical equation for the dissociation of the nitrous acid is:
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber$
Since $10^{−pH} = \ce{[H3O+]}$, we find that $10^{−2.09} = 8.1 \times 10^{−3}\, M$, so that percent ionization (Equation \ref{PercentIon}) is:
$\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber$
Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.
Exercise $1$
Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89.
Answer
1.3% ionized
We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq)$
Water is the acid that reacts with the base, $\ce{HB^{+}}$ is the conjugate acid of the base $\ce{B}$, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Figure $1$ lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.
As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, $\ce{B}$:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq),$
we write the equation for the ionization constant as:
$K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}$
where the concentrations are those at equilibrium. Again, we do not include [H2O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:
\begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned}
A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.
Consider the ionization reactions for a conjugate acid-base pair, $\ce{HA − A^{−}}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)$
with $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$.
$\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq)$
with $K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}$.
Adding these two chemical equations yields the equation for the autoionization for water:
\begin{align*} \cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l) &⇌ \ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)} \[4pt] \ce{2H2O}(l) &⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \end{align*}
As shown in the previous chapter on equilibrium, the $K$ expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ $K$ expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:
$K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}×\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w}$
For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion ($\ce{CH3COO-}$), is 5.6 × 10−10. The product of these two constants is indeed equal to $K_w$:
$K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w}$
The extent to which an acid, $\ce{HA}$, donates protons to water molecules depends on the strength of the conjugate base, $\ce{A^{−}}$, of the acid. If $\ce{A^{−}}$ is a strong base, any protons that are donated to water molecules are recaptured by $\ce{A^{−}}$. Thus there is relatively little $\ce{A^{−}}$ and $\ce{H3O+}$ in solution, and the acid, $\ce{HA}$, is weak. If $\ce{A^{−}}$ is a weak base, water binds the protons more strongly, and the solution contains primarily $\ce{A^{−}}$ and $\ce{H3O^{+}}$—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure $2$).
Figure $3$ lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.
The first six acids in Figure $3$ are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.
Those acids that lie between the hydronium ion and water in Figure $3$ form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure $3$ exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.
The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure $3$. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.
Example $2$: The Product Ka × Kb = Kw
Use the $K_b$ for the nitrite ion, $\ce{NO2-}$, to calculate the $K_a$ for its conjugate acid.
Solution
Kb for $\ce{NO2-}$ is given in this section as 2.17 × 10−11. The conjugate acid of $\ce{NO2-}$ is HNO2; Ka for HNO2 can be calculated using the relationship:
$K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \nonumber$
Solving for Ka, we get:
\begin{align*} K_\ce{a} &=\dfrac{K_\ce{w}}{K_\ce{b}} \[4pt] &=\dfrac{1.0×10^{−14}}{2.17×10^{−11}} \[4pt] &=4.6×10^{−4} \end{align*}
This answer can be verified by finding the Ka for HNO2 in Table E1
Exercise $2$
We can determine the relative acid strengths of $\ce{NH4+}$ and $\ce{HCN}$ by comparing their ionization constants. The ionization constant of $\ce{HCN}$ is given in Table E1 as 4.9 × 10−10. The ionization constant of $\ce{NH4+}$ is not listed, but the ionization constant of its conjugate base, $\ce{NH3}$, is listed as 1.8 × 10−5. Determine the ionization constant of $\ce{NH4+}$, and decide which is the stronger acid, $\ce{HCN}$ or $\ce{NH4+}$.
Answer
$\ce{NH4+}$ is the slightly stronger acid (Ka for $\ce{NH4+}$ = 5.6 × 10−10).
Summary
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion.
Key Equations
• $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$
• $K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}$
• $K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})$
Glossary
acid ionization constant (Ka)
equilibrium constant for the ionization of a weak acid
base ionization constant (Kb)
equilibrium constant for the ionization of a weak base
leveling effect of water
any acid stronger than $\ce{H3O+}$, or any base stronger than OH will react with water to form $\ce{H3O+}$, or OH, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong
percent ionization
ratio of the concentration of the ionized acid to the initial acid concentration, times 100 | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.3%3A_Acid_and_Base_Strength.txt |
Learning Objectives
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).
Acetic acid ($\ce{CH3CO2H}$) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure $1$). The remaining weak acid is present in the nonionized form.
For acetic acid, at equilibrium:
$K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber$
Table $1$: Ionization Constants of Some Weak Acids
Ionization Reaction Ka at 25 °C
$\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}$ 1.2 × 10−2
$\ce{HF + H2O ⇌ H3O+ + F-}$ 3.5 × 10−4
$\ce{HNO2 + H2O ⇌ H3O+ + NO2-}$ 4.6 × 10−4
$\ce{HNCO + H2O ⇌ H3O+ + NCO-}$ 2 × 10−4
$\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}$ 1.8 × 10−4
$\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}$ 1.8 × 10−5
$\ce{HCIO + H2O ⇌ H3O+ + CIO-}$ 2.9 × 10−8
$\ce{HBrO + H2O ⇌ H3O+ + BrO-}$ 2.8 × 10−9
$\ce{HCN + H2O ⇌ H3O+ + CN-}$ 4.9 × 10−10
Table $1$ gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1.
At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).
For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber$
This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.
We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure $2$). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, $K_b$, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber$
The ionization constants of several weak bases are given in Table $2$ and Table E2.
Table $2$: Ionization Constants of Some Weak Bases
Ionization Reaction Kb at 25 °C
$\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}$ 5.9 × 10−4
$\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}$ 4.4 × 10−4
$\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}$ 6.3 × 10−5
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$ 1.8 × 10−5
$\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}$ 4.3 × 10−10
Example $3$: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and $\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M$. What is the value of $K_a$ for acetic acid?
Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \[4pt] &=1.77×10^{−5} \end{align*}
Exercise $3$
What is the equilibrium constant for the ionization of the $\ce{HSO4-}$ ion, the weak acid used in some household cleansers:
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber$
In one mixture of NaHSO4 and Na2SO4 at equilibrium, $\ce{[H3O+]}$ = 0.027 M; $\ce{[HSO4- ]}=0.29\:M$; and $\ce{[SO4^2- ]}=0.13\:M$.
Answer
$K_a$ for $\ce{HSO_4^-}= 1.2 ×\times 10^{−2}$
Example $4$: Determination of Kb from Equilibrium Concentrations
Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, $\ce{[C8H10N4O2H+]}$ = 5.0 × 10−3 M, and [OH] = 2.5 × 10−3 M?
Solution
At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber$
so
$K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber$
Exercise $4$
What is the equilibrium constant for the ionization of the $\ce{HPO4^2-}$ ion, a weak base:
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber$
In a solution containing a mixture of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ at equilibrium with:
• $[\ce{OH^{−}}] = 1.3 × 10^{−6} M$
• $\ce{[H2PO4^{-}]=0.042\:M}$ and
• $\ce{[HPO4^{2-}]=0.341\:M}$.
Answer
Kb for $\ce{HPO4^2-}=1.6×10^{−7}$
Example $5$: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, $\ce{HNO2}$, is 2.34. What is its $K_a$?
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber$
Solution
We determine an equilibrium constant starting with the initial concentrations of HNO2, $\ce{H3O+}$, and $\ce{NO2-}$ as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)
We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:
We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):
To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate $\ce{[H3O+]}$, the equilibrium concentration of $\ce{H3O+}$, from the pH:
$\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber$
The change in concentration of $\ce{H3O+}$, $x_{\ce{[H3O+]}}$, is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, $\mathrm{[H_3O^+]_i}$. The initial concentration of $\ce{H3O+}$ is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).
The change in concentration of $\ce{NO2-}$ is equal to the change in concentration of $\ce{[H3O+]}$. For each 1 mol of $\ce{H3O+}$ that forms, 1 mol of $\ce{NO2-}$ forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.
Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:
Finally, we calculate the value of the equilibrium constant using the data in the table:
$K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber$
Exercise $5$
The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.
Answer
$K_b = 1.8 × 10^{−5}$
Example $6$: Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.
What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber$
Solution
1. Determine x and equilibrium concentrations. The equilibrium expression is:
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber$
The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.
The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}
Now solve for $x$. Because the initial concentration of acid is reasonably large and $K_a$ is very small, we assume that $x << 0.534$, which permits us to simplify the denominator term as $(0.534 − x) = 0.534$. This gives:
$K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber$
Solve for $x$ as follows:
\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \[4pt] &=9.6×10^{−5} \[4pt] x &=\sqrt{9.6×10^{−5}} \[4pt] &=9.8×10^{−3} \end{align*}
To check the assumption that $x$ is small compared to 0.534, we calculate:
\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}
$x$ is less than 5% of the initial concentration; the assumption is valid.
We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:
\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \[4pt] &=9.8×10^{−3}\:M \end{align*}
The pH of the solution can be found by taking the negative log of the $\ce{[H3O+]}$, so:
$pH = −\log(9.8×10^{−3})=2.01 \nonumber$
Exercise $6$: acetic acid
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber$
Hint
Determine $\ce{[CH3CO2- ]}$ at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or $\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100$.
Answer
percent ionization = 1.3%
The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.
Example $7$: Equilibrium Concentrations in a Solution of a Weak Base
Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber$
Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example $6$. The reactants and products will be different and the numbers will be different, but the logic will be the same:
1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber$
If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:
$x=4.0×10^{−3} \nonumber$
This change is less than 5% of the initial concentration (0.25), so the assumption is justified.
Recall that, for this computation, $x$ is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):
\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \[4pt] &=4.0×10^{−3}\:M \end{align*}
Then calculate pOH as follows:
$\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber$
Using the relation introduced in the previous section of this chapter:
$\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber$
permits the computation of pH:
$\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber$
Check the work. A check of our arithmetic shows that $K_b = 6.3 \times 10^{−5}$.
Exercise $7$
1. Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 6.3 × 10−5.
2. Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or $\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%$
Answer a
$7.56 × 10^{−4}\, M$, 2.33%
Answer b
2.33%
Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.
Example $8$: Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the $\ce{HSO4-}$ ion, a weak acid. What is the pH of a 0.50-M solution of $\ce{HSO4-}$?
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber$
Solution
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of $\ce{HSO4-}$ so that we can use $\ce{[H3O+]}$ to determine the pH. As in the previous examples, we can approach the solution by the following steps:
1. Determine $x$ and equilibrium concentrations. This table shows the changes and concentrations:
2. Solve for $x$ and the concentrations.
As we begin solving for $x$, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of $x$.
At equilibrium:
$K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber$
If we assume that x is small and approximate (0.50 − x) as 0.50, we find:
$x=7.7×10^{−2} \nonumber$
When we check the assumption, we confirm:
$\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber$
which for this system is
$\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber$
The value of $x$ is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find $x$.
The equation:
$K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber$
gives
$6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber$
or
$x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber$
This equation can be solved using the quadratic formula. For an equation of the form
$ax^{2+} + bx + c=0, \nonumber$
$x$ is given by the quadratic equation:
$x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber$
In this problem, $a = 1$, $b = 1.2 × 10^{−3}$, and $c = −6.0 × 10^{−3}$.
Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:
$x=7.2×10^{−2} \nonumber$
Now determine the hydronium ion concentration and the pH:
\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \[4pt] &=7.2×10^{−2}\:M \end{align*}
The pH of this solution is:
$\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber$
Exercise $8$
1. Show that the quadratic formula gives $x = 7.2 × 10^{−2}$.
2. Calculate the pH in a 0.010-M solution of caffeine, a weak base:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber$
Hint
It will be necessary to convert [OH] to $\ce{[H3O+]}$ or pOH to pH toward the end of the calculation.
Answer
pH 11.16
The Relative Strengths of Strong Acids and Bases
Strong acids, such as $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order $\ce{HCl < HBr < HI}$, and so $\ce{HI}$ is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.
Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, $\ce{NH2-}$, are such strong bases that they react completely with water:
$\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq)$
$\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq)$
Thus, O2− and $\ce{NH2-}$ appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is $\ce{HF < HCl < HBr < HI}$. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\ce{CH4 < NH3 < H2O < HF}$; across the third row, it is $\ce{SiH4 < PH3 < H2S < HCl}$ (see Figure $3$).
Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, $\ce{O2S(OH)2}$, sulfurous acid, $\ce{OS(OH)2}$, nitric acid, $\ce{O2NOH}$, perchloric acid, $\ce{O3ClOH}$, aluminum hydroxide, $\ce{Al(OH)3}$, calcium hydroxide, $\ce{Ca(OH)2}$, and potassium hydroxide, $\ce{KOH}$:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure $4$).
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\ce{Al(H2O)3(OH)3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\ce{Al(H2O)3(OH)3}$, is converted into the soluble ion, $\ce{[Al(H2O)2(OH)4]-}$, by reaction with hydroxide ion:
$[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber$
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The $\ce{Al(H2O)3(OH)3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\ce{[Al(H2O)6]^3+}$ by reaction with hydronium ion:
$\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber$
In this case, protons are transferred from hydronium ions in solution to $\ce{Al(H2O)3(OH)3}$, and the compound functions as a base.
Summary
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
Key Equations
• $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$
• $K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}$
• $K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})$
• $\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100$
Glossary
acid ionization constant (Ka)
equilibrium constant for the ionization of a weak acid
base ionization constant (Kb)
equilibrium constant for the ionization of a weak base
leveling effect of water
any acid stronger than $\ce{H3O+}$, or any base stronger than OH will react with water to form $\ce{H3O+}$, or OH, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong
oxyacid
compound containing a nonmetal and one or more hydroxyl groups
percent ionization
ratio of the concentration of the ionized acid to the initial acid concentration, times 100 | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.4%3A_Equilibria_Involving_Weak_Acids_and_Bases.txt |
Learning Objectives
• To understand how adding a common ion affects the position of an acid–base equilibrium.
• To know how to use the Henderson-Hasselbalch approximation to calculate the pH of a buffer.
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid ($HA$) and its conjugate base $(A^−$) or a weak base ($B$) and its conjugate acid ($BH^+$), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect: Weak Acids Combined with Conjugate Bases
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of $\ce{H^{+}}$). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows:
$\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[\ce{H^{+}}][\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2}$
Sodium acetate ($\ce{CH_3CO_2Na}$) is a strong electrolyte that ionizes completely in aqueous solution to produce $\ce{Na^{+}}$ and $\ce{CH3CO2^{−}}$ ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added $\ce{CH_3COO^{−}}$ and some of the $\ce{H^{+}}$ ions originally present in solution.
Because $\ce{Na^{+}}$ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which $[\ce{H^{+}}]$ is less than the initial value. Because $[\ce{H^{+}}]$ has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as $\ce{HCl}$ to the system, $[\ce{H^{+}}]$ increases. Once again the equilibrium is temporarily disturbed, but the excess $\ce{H^{+}}$ ions react with the conjugate base ($\ce{CH_3CO_2^{−}}$), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [$\ce{CH_3CO_2^{−}}$] than before. In both cases, only the equilibrium composition has changed; the ionization constant $K_a$ for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case $\ce{CH3CO2^{−}}$, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Example $1$
A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given: solution concentration and pH, $pK_a$, and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for: pH and percent ionization of formic acid
Strategy:
1. Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
2. Substitute the expressions for the final concentrations into the expression for Ka. Calculate $[\ce{H^{+}}]$ and the pH of the solution.
3. Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution:
A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The $\ce{Na^{+}}$ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:
$\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber$
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
Final Concentration
ICE $[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
Initial 0.150 $1.00 \times 10^{−7}$ 0.100
Change −x +x +x
Equilibrium (0.150 − x) x (0.100 + x)
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
\begin{align*} K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \[4pt] &\approx \dfrac{x(0.100)}{0.150} \[4pt] &\approx 10^{−3.75} \[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber
Rearranging and solving for $x$,
\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \[4pt] &=2.7 \times 10^{−4}\[4pt] &=[H^+] \end{align*} \nonumber
The value of $x$ is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover,
$K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber$
which is greater than $1.0 \times 10^{−6}$, so again, our assumption is justified. The final pH is:
$pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber$
compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of $\ce{H^{+}}$ ions, driving the equilibrium to the left.
C Because $HCl$ is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
$HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber$
initial concentrations, changes in concentration, and final concentrations
$[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
initial 0.150 0.200 0
change −x +x +x
final (0.150 − x) (0.200 + x) x
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final $[\ce{HCO2^{-}}]$. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber$
Rearranging and solving for $x$,
\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
$\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber$
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding $\ce{H^{+}}$ ions drives the dissociation equilibrium to the left.
Exercise $1$
A 0.225 M solution of ethylamine ($\ce{CH3CH2NH2}$ with $pK_b = 3.19$) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
1. the pH of the solution if enough solid ethylamine hydrochloride ($\ce{EtNH3Cl}$) is added to make the solution 0.100 M in $\ce{EtNH3^{+}}$
2. the percentage of ethylamine that is ionized if enough solid $\ce{NaOH}$ is added to the original solution to give a final concentration of 0.050 M $\ce{NaOH}$
Answer a
11.16
Answer b
1.3%
A Video Discussing the Common Ion Effect: The Common Ion Effecr(opens in new window) [youtu.be]
The Common Ion Effect: Weak Bases Combined with Conjugate Acids
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base ($B$) and its conjugate acid ($BH^+$). The general equation for the ionization of a weak base is as follows:
$B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3}$
If the equilibrium constant for the reaction as written in Equation $\ref{Eq3}$ is small, for example $K_b = 10^{−5}$, then the equilibrium constant for the reverse reaction is very large: $K = \dfrac{1}{K_b} = 10^5$. Adding a strong base such as $OH^-$ to the solution therefore causes the equilibrium in Equation $\ref{Eq3}$ to shift to the left, consuming the added $OH^-$. As a result, the $OH^-$ ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the $OH^-$ ion concentration, the reaction will proceed to the left to counteract the stress.
If the $pK_b$ of the base is 5.0, the $pK_a$ of its conjugate acid is
$pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber$
Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
$BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4}$
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation $\ref{Eq4}$ shifts to the left. As a result, the $H^+$ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb $H^+$ and $OH^-$ ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on $K$), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure $1$, when $NaOH$ is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M $NaOH$ to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of $NaOH$ solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the $NaOH$ solution results in only a relatively small change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the $pK_a$ or $pK_b$ of the weak acid or weak base. The procedure is analogous to that used in Example $1$ to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is $HA \leftrightharpoons H^+ + A^−$, for which the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5}$
This equation can be rearranged as follows:
$[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6}$
Taking the logarithm of both sides and multiplying both sides by −1,
\begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align}
Replacing the negative logarithms in Equation $\ref{Eq7}$,
$pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8}$
or, more generally,
$pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9}$
Equation $\ref{Eq8}$ and Equation $\ref{Eq9}$ are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their $K_a$ values.
There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations:
• $[base] = [acid]$: Under these conditions, $\dfrac{[base]}{[acid]} = 1 \nonumber$ in Equation \ref{Eq9}. Because $\log 1 = 0$, $pH = pK_a \nonumber$ regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base.
• $[base]/[acid] = 10$: In Equation $\ref{Eq9}$, because $\log 10 = 1$, $pH = pK_a + 1. \nonumber$
• $[base]/[acid] = 100$: In Equation $\ref{Eq9}$, because $\log 100 = 2$, $pH = pK_a + 2. \nonumber$
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = $pK_a$ − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
If [base] = [acid] for a buffer, then pH = $pK_a$. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example $2$
What is the pH of a solution that contains
1. 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$? (The $pK_a$ of formic acid is 3.75.)
2. 0.0135 M $\ce{HCO2H}$ and 0.0215 M $\ce{HCO2Na}$?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The $pK_b$ of pyridine is 8.77.)
Given: concentration of acid, conjugate base, and $pK_a$; concentration of base, conjugate acid, and $pK_b$
Asked for: pH
Strategy:
Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH.
Solution:
According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is
$pH = pK_a + \log([A−]/[HA]). \nonumber$
A
Inserting the given values into the equation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result makes sense because the $[A^−]/[HA]$ ratio is between 1 and 10, so the pH of the buffer must be between the $pK_a$ (3.75) and $pK_a + 1$, or 4.75.
B
This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
C
In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion ($HPy^+$). We will therefore use Equation $\ref{Eq9}$, the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and $[acid] = [HPy^{+}] = 0.234\, M$. We also are given $pK_b = 8.77$ for pyridine, but we need $pK_a$ for the pyridinium ion. Recall from Equation 16.23 that the $pK_b$ of a weak base and the $pK_a$ of its conjugate acid are related:
$pK_a + pK_b = pK_w. \nonumber$
Thus $pK_a$ for the pyridinium ion is $pK_w − pK_b = 14.00 − 8.77 = 5.23$. Substituting this $pK_a$ value into the Henderson-Hasselbalch approximation,
\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \[4pt] & =5.23 −0.294 \[4pt] &=4.94 \end{align*} \nonumber
Once again, this result makes sense: the $[B]/[BH^+]$ ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the $pK_a$ (5.23) and $pK_a − 1$, or 4.23.
Exercise $2$
What is the pH of a solution that contains
1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The $pK_a$ of benzoic acid is 4.20, and the $pK_b$ of trimethylamine is also 4.20.
Answer a
4.08
Answer b
9.68
A Video Discussing Using the Henderson Hasselbalch Equation: Using the Henderson Hasselbalch Equation(opens in new window) [youtu.be] (opens in new window)
The Henderson-Hasselbalch approximation ((Equation $\ref{Eq8}$) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example $3$.
Example $3$
The buffer solution in Example $2$ contained 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$ and had a pH of 3.95.
1. What is the final pH if 5.00 mL of 1.00 M $HCl$ are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M $NaOH$ are added?
Given: composition and pH of buffer; concentration and volume of added acid or base
Asked for: final pH
Strategy:
1. Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example $1$. Then calculate the amount of acid or base added.
2. Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch approximation (Equation \ref{Eq9}) to obtain the pH.
Solution:
The added $\ce{HCl}$ (a strong acid) or $\ce{NaOH}$ (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:
$100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber$
$100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber$
The millimoles of $\ce{H^{+}}$ in 5.00 mL of 1.00 M $\ce{HCl}$ is as follows:
$5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber$
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
$\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber$
initial amounts, changes in amounts, and final amounts:
$HCO^{2−} (aq)$ $H^+ (aq)$ $HCO_2H (aq)$
Initial 21.5 mmol 5.00 mmol 13.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 16.5 mmol ∼0 mmol 18.5 mmol
The final amount of $H^+$ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final $[H^+]$ and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example $1$ or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of $HCO_2^−$ and $HCO_2H$, which can be calculated using the number of millimoles ($n$) of each and the total volume ($VT$). Substituting these values into the Henderson-Hasselbalch approximation (Equation $\ref{Eq9}$):
\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber
Once again, this result makes sense on two levels. First, the addition of $HCl$has decreased the pH from 3.95, as expected. Second, the ratio of $HCO_2^−$ to $HCO_2H$ is slightly less than 1, so the pH should be between the $pK_a$ and $pK_a$ − 1.
A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of $HCO_2H$ and 21.5 mmol of $HCO_2^−$. The number of millimoles of $OH^-$ in 5.00 mL of 1.00 M $NaOH$ is as follows:
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
$\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber$
initial amounts, changes in amounts, and final amounts
$HCO_2H (aq)$ $OH^−$ $HCO^−_2 (aq)$
Initial 13.5 mmol 5.00 mmol 21.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 8.5 mmol ∼0 mmol 26.5 mmol
The final amount of $OH^-$ in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both $HCO_2^−$ and $HCO_2H$ into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the $pK_a$ and $pK_a$ + 1, as expected for a solution with a $HCO_2^−/HCO_2H$ ratio between 1 and 10.
Exercise $3$
The buffer solution from Example $2$ contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
1. What is the final pH if 12.0 mL of 1.5 M $\ce{NaOH}$ are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M $\ce{HCl}$ are added?
Answer a
5.30
Answer b
4.42
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations.
A Video Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Acid to a Buffer(opens in new window) [youtu.be]
The Change in pH with the Addition of a Strong Base to a Buffer:
The results obtained in Example $3$ and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of $HCl$ or $NaOH$ solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to $1.1 \times 10^{−4}$ M HCl). In this case, adding 5.00 mL of 1.00 M $HCl$ would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M $NaOH$ would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
The most effective buffers contain equal concentrations of an acid and its conjugate base.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure $2$ for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of $CH_3CO_2^−$ to $CH_3CO_2H$ from 1:1 reduces the buffer capacity of the solution.
A Video Discussing The Buffer Region: The Buffer Region (opens in new window) [youtu.be]
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure $3$. As indicated by the labels, the region around $pK_a$ corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the $pK_a$ to approximately a pH value of 1 unit greater than the $pK_a$, which is why buffer solutions usually have a pH that is within ±1 pH units of the $pK_a$ of the acid component of the buffer.
This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to $K_a$. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to $K_b$. However, we can calculate either $K_a$ or $K_b$ from the other because they are related by $K_w$.
Blood: A Most Important Buffer
Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $\ce{CO2}/\ce{HCO3^{−}}$ system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is usually written as follows:
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10}$
with $K_a = 4.5 \times 10^{−7}$ and $pK_a = 6.35$ at 25°C. In fact, Equation $\ref{Eq10}$ is a grossly oversimplified version of the $\ce{CO2}/\ce{HCO3^{-}}$ system because a solution of $\ce{CO2}$ in water contains only rather small amounts of $H_2CO_3$. Thus Equation $\ref{Eq10}$ does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C.
As shown in Equation $\ref{Eq11}$, $\ce{CO2}$ is in equilibrium with $\ce{H2CO3}$, but the equilibrium lies far to the left, with an $\ce{H2CO3}/\ce{CO2}$ ratio less than 0.01 under most conditions:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11}$
with $K′ = 4.0 \times 10^{−3}$ at 37°C. The true $pK_a$ of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a $K_a$ of $2.0 \times 10^{−4}$, which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling $\ce{H2CO3}$ from both sides give the following overall equation for the reaction of $\ce{CO2}$ with water to give a proton and the bicarbonate ion:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a}$
with $K'=4.0 \times 10^{−3} (37°C)$
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b}$
with $K_a=2.0 \times 10^{−4} (37°C)$
$\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c}$
with $K=8.0 \times 10^{−7} (37°C)$
The $K$ value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid ($K_a$) and the equilibrium constant (K) for the reaction of $\ce{CO2 (aq)}$ with water to give carbonic acid. The equilibrium equation for the reaction of $\ce{CO2}$ with water to give bicarbonate and a proton is therefore
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13}$
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law,
$[\ce{CO2}]=k P_{\ce{CO2}} \nonumber$
where $k$ is the Henry’s law constant for $\ce{CO2}$, which is $3.0 \times 10^{−5} \;M/mmHg$ at 37°C. Substituting this expression for $[\ce{CO2}]$ in Equation \ref{eq13},
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber$
where $P_{\ce{CO2}}$ is in mmHg. Taking the negative logarithm of both sides and rearranging,
$pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15}$
Thus the pH of the solution depends on both the $\ce{CO2}$ pressure over the solution and $[\ce{HCO3^{−}}]$. Figure $4$ plots the relationship between pH and $[\ce{HCO3^{−}}]$ under physiological conditions for several different values of $P_{\ce{CO2}}$, with normal pH and $[\ce{HCO3^{−}}]$ values indicated by the dashed lines.
According to Equation \ref{Eq15}, adding a strong acid to the $\ce{CO2}/\ce{HCO3^{−}}$ system causes $[\ce{HCO3^{−}}]$ to decrease as $\ce{HCO3^{−}}$ is converted to $\ce{CO2}$. Excess $\ce{CO2}$ is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in $P_{\ce{CO2}}$. Because the change in $[\ce{HCO3^{−}}]/P_{CO_2}$ is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the $\ce{OH^{-}}$ reacts with $\ce{CO2}$ to form $\ce{HCO3^{−}}$, but $\ce{CO2}$ is replenished by the body, again limiting the change in both $[\ce{HCO3^{−}}]/P_{\ce{CO2}}$ and pH. The $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower $P_{\ce{CO2}}$ at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and $[\ce{HCO3^{-}}]$. The increase in pH and decrease in $[\ce{HCO3^{−}}]$ in response to the decrease in $P_{\ce{CO2}}$ are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
A Video Summary of the pH Curve for a Strong Acid/Strong Base Titration:
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid ($HA$) and its conjugate weak base ($A^−$). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the $K_a$ or $K_b$), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their $K_a$ values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $CO_2/HCO_3^−$ system, which dominates the buffering action of blood plasma. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.5%3A_Buffer_Solutions.txt |
Learning Objectives
• To calculate the pH at any point in an acid–base titration.
In an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration.
Titrations of Strong Acids and Bases
Figure $\PageIndex{1a}$ shows a plot of the pH as 0.20 M $\ce{HCl}$ is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as $\ce{HCl}$ is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of $\ce{HCl}$ (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M $\ce{NaOH}$ is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of $\ce{NaOH}$ as shown in Figure $\PageIndex{1b}$. As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.
Suppose that we now add 0.20 M $\ce{NaOH}$ to 50.0 mL of a 0.10 M solution of $\ce{HCl}$. Because $\ce{HCl}$ is a strong acid that is completely ionized in water, the initial $[H^+]$ is 0.10 M, and the initial pH is 1.00. Adding $\ce{NaOH}$ decreases the concentration of H+ because of the neutralization reaction (Figure $\PageIndex{2a}$):
$\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber$
Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the $\ce{H^{+}}$ ions originally present have been consumed. For the titration of a monoprotic strong acid ($\ce{HCl}$) with a monobasic strong base ($\ce{NaOH}$), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
$moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1}$
If 0.20 M $\ce{NaOH}$ is added to 50.0 mL of a 0.10 M solution of $\ce{HCl}$, we solve for $V_b$:
$V_b(0.20 Me)=0.025 L=25 mL \nonumber$
At the equivalence point (when 25.0 mL of $\ce{NaOH}$ solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more $\ce{NaOH}$ produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M $NaOH$.
As shown in Figure $\PageIndex{2b}$, the titration of 50.0 mL of a 0.10 M solution of $\ce{NaOH}$ with 0.20 M $\ce{HCl}$ produces a titration curve that is nearly the mirror image of the titration curve in Figure $\PageIndex{2a}$. The pH is initially 13.00, and it slowly decreases as $\ce{HCl}$ is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M $\ce{HCl}$.
The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.
Example $1$: Hydrochloric Acid
Calculate the pH of the solution after 24.90 mL of 0.200 M $\ce{NaOH}$ has been added to 50.00 mL of 0.100 M $\ce{HCl}$.
Given: volumes and concentrations of strong base and acid
Asked for: pH
Strategy:
1. Calculate the number of millimoles of $\ce{H^{+}}$ and $\ce{OH^{-}}$ to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.
2. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.
Solution
A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of $\ce{H^{+}}$ in 50.00 mL of 0.100 M $\ce{HCl}$ can be calculated as follows:
$50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber$
The number of millimoles of $\ce{NaOH}$ added is as follows:
$24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber$
Thus $\ce{H^{+}}$ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of $\ce{OH^{-}}$ to the $\ce{HCl}$ solution. Because only 4.98 mmol of $OH^-$ has been added, the amount of excess $\ce{H^{+}}$ is 5.00 mmol − 4.98 mmol = 0.02 mmol of $H^+$.
B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of $\ce{H^{+}}$ is as follows:
$\left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber$
Hence,
$pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber$
This is significantly less than the pH of 7.00 for a neutral solution.
Exercise $1$
Calculate the pH of a solution prepared by adding $40.00\; mL$ of $0.237\; M$ $HCl$ to $75.00\; mL$ of a $0.133 M$ solution of $NaOH$.
Answer
11.6
pH after the addition of 10 ml of Strong Base to a Strong Acid:
https://youtu.be/_cM1_-kdJ20 (opens in new window)
pH at the Equivalence Point in a Strong Acid/Strong Base Titration:
https://youtu.be/7POGDA5Ql2M
Titrations of Weak Acids and Bases
In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding $K_a$ or $K_b$. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, $[\ce{H^{+}}]$ of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its $pK_a$ and its concentration. Because only a fraction of a weak acid dissociates, $[\(\ce{H^{+}}]$ is less than $[\ce{HA}]$. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration.
Figure $\PageIndex{3a}$ shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M $\ce{NaOH}$ superimposed on the curve for the titration of 0.100 M $\ce{HCl}$ shown in part (a) in Figure $2$. Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the $\ce{HCl}$ solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the $\ce{HCl}$ titration; the magnitude of the pH change at the equivalence point depends on the $pK_a$ of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess $\ce{NaOH}$ present, regardless of whether the acid is weak or strong.
The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the $K_a$ or $K_b$.
The titration curve in Figure $\PageIndex{3a}$ was created by calculating the starting pH of the acetic acid solution before any $\ce{NaOH}$ is added and then calculating the pH of the solution after adding increasing volumes of $NaOH$. The procedure is illustrated in the following subsection and Example $2$ for three points on the titration curve, using the $pK_a$ of acetic acid (4.76 at 25°C; $K_a = 1.7 \times 10^{-5}$.
Calculating the pH of a Solution of a Weak Acid or a Weak Base
As explained discussed, if we know $K_a$ or $K_b$ and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define $x$ as $[\ce{H^{+}}]$ due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:
$\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber$
table of concentrations for the ionization of 0.100 M acetic acid
ICE $[CH_3CO_2H]$ $[H^+]$ $[CH_3CO_2^−]$
initial 0.100 $1.00 \times 10^{−7}$ 0
change −x +x +x
final 0.100 − x x x
In this and all subsequent examples, we will ignore $[H^+]$ and $[OH^-]$ due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified.
Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),
\begin{align*} K_a &=\dfrac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \[4pt] &=\dfrac{(x)(x)}{0.100 - x} \[4pt] &\approx \dfrac{x^2}{0.100} \[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber
Solving this equation gives $x = [H^+] = 1.32 \times 10^{-3}\; M$. Thus the pH of a 0.100 M solution of acetic acid is as follows:
$pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber$
pH at the Start of a Weak Acid/Strong Base Titration: https://youtu.be/AtdBKfrfJNg
Calculating the pH during the Titration of a Weak Acid or a Weak Base
Now consider what happens when we add 5.00 mL of 0.200 M $\ce{NaOH}$ to 50.00 mL of 0.100 M $CH_3CO_2H$ (part (a) in Figure $3$). Because the neutralization reaction proceeds to completion, all of the $OH^-$ ions added will react with the acetic acid to generate acetate ion and water:
$CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2}$
All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine $[\ce{H^{+}}]$ of the resulting solution.
Step 1
To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of $\ce{CH_3CO_2H}$ in the original solution and the amount of $\ce{OH^{-}}$ in the $\ce{NaOH}$ solution that was added. The acetic acid solution contained
$50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber$
The $\ce{NaOH}$ solution contained
5.00 mL=1.00 mmol $NaOH$
Comparing the amounts shows that $CH_3CO_2H$ is in excess. Because $OH^-$ reacts with $CH_3CO_2H$ in a 1:1 stoichiometry, the amount of excess $CH_3CO_2H$ is as follows:
5.00 mmol $CH_3CO_2H$ − 1.00 mmol $OH^-$ = 4.00 mmol $CH_3CO_2H$
Each 1 mmol of $OH^-$ reacts to produce 1 mmol of acetate ion, so the final amount of $CH_3CO_2^−$ is 1.00 mmol.
The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations.
$\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber$
ICE table
ICE $[\ce{CH_3CO_2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH_3CO_2^{−}}]$
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol
This ICE table gives the initial amount of acetate and the final amount of $OH^-$ ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of $CH_3CO_2^−$ in equilibrium is insignificant compared to the amount of $OH^-$ added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of $OH^-$, but the amount of $OH^-$ due to the autoionization of water is insignificant compared to the amount of $OH^-$ added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.
Step 2
To calculate $[\ce{H^{+}}]$ at equilibrium following the addition of $NaOH$, we must first calculate [$\ce{CH_3CO_2H}$] and $[\ce{CH3CO2^{−}}]$ using the number of millimoles of each and the total volume of the solution at this point in the titration:
$final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber$ $\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber$ $\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber$
Knowing the concentrations of acetic acid and acetate ion at equilibrium and $K_a$ for acetic acid ($1.74 \times 10^{-5}$), we can calculate $[H^+]$ at equilibrium:
$K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber$
$\left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber$
Calculating $−\log[\ce{H^{+}}]$ gives
$pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber$
Comparing the titration curves for $\ce{HCl}$ and acetic acid in Figure $\PageIndex{3a}$, we see that adding the same amount (5.00 mL) of 0.200 M $\ce{NaOH}$ to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for $\ce{HCl}$ (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example $2$, we calculate another point for constructing the titration curve of acetic acid.
pH Before the Equivalence Point of a Weak Acid/Strong Base Titration:
https://youtu.be/znpwGCsefXc
Example $2$
What is the pH of the solution after 25.00 mL of 0.200 M $\ce{NaOH}$ is added to 50.00 mL of 0.100 M acetic acid?
Given: volume and molarity of base and acid
Asked for: pH
Strategy:
1. Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$. Determine which species, if either, is present in excess.
2. Tabulate the results showing initial numbers, changes, and final numbers of millimoles.
3. If excess acetate is present after the reaction with $\ce{OH^{-}}$, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.
4. Calculate $K_b$ using the relationship $K_w = K_aK_b$. Calculate [OH−] and use this to calculate the pH of the solution.
Solution
A Ignoring the spectator ion ($Na^+$), the equation for this reaction is as follows:
$CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber$
The initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$ are as follows:
25.00 mL(0.200 mmol OH−mL=5.00 mmol $OH-$
$50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber$
The number of millimoles of $OH^-$ equals the number of millimoles of $CH_3CO_2H$, so neither species is present in excess.
B Because the number of millimoles of $OH^-$ added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.
$CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber$
results of the neutralization reaction
ICE $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH3CO2^{−}}]$
initial 5.00 mmol 5.00 mmol 0 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 0 mmol 0 mmol 5.00 mmol
C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:
$[\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber$
The equilibrium reaction of acetate with water is as follows:
$\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber$
The equilibrium constant for this reaction is
$K_b = \dfrac{K_w}{K_a} \label{16.18}$
where $K_a$ is the acid ionization constant of acetic acid. We therefore define x as $[\ce{OH^{−}}]$ produced by the reaction of acetate with water. Here is the completed table of concentrations:
$H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber$
completed table of concentrations
$[\ce{CH3CO2^{−}}]$ $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$
initial 0.0667 0 1.00 × 10−7
change −x +x +x
final (0.0667 − x) x x
D We can obtain $K_b$ by substituting the known values into Equation \ref{16.18}:
$K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23}$
Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for $x$:
\begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber
Thus $[OH^{−}] = 6.22 \times 10^{−6}\, M$ and the pH of the final solution is 8.794 (Figure $\PageIndex{3a}$). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce $\ce{OH^{-}}$.
Exercise $2$
Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M $\ce{HCl}$ solution to 125.0 mL of a 0.150 M solution of ammonia. The $pK_b$ of ammonia is 4.75 at 25°C.
Answer
9.23
As shown in part (b) in Figure $3$, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.
The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure $4$ illustrates the shape of titration curves as a function of the $pK_a$ or the $pK_b$. As the acid or the base being titrated becomes weaker (its $pK_a$ or $pK_b$ becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.
One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures $\PageIndex{4a}$ and $\PageIndex{4b}$ for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows:
$K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
If $[HA] = [A^−]$, this reduces to $K_a = [H_3O^+]$. Taking the negative logarithm of both sides,
$−\log K_a = −\log[H_3O+] \nonumber$
From the definitions of $pK_a$ and pH, we see that this is identical to
$pK_a = pH \label{16.52}$
Thus the pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid, as indicated in part (a) in Figure $4$ for the weakest acid where we see that the midpoint for $pK_a$ = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base).
The pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid.
Titrations of Polyprotic Acids or Bases
When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the $pK_a$ values are separated by at least three $pK_a$ units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid $H_3PO_4$ with $\ce{NaOH}$ is illustrated in Figure $5$ and shows two well-defined steps: the first midpoint corresponds to $pK_a$1, and the second midpoint corresponds to $pK_a$2. Because HPO42 is such a weak acid, $pK_a$3 has such a high value that the third step cannot be resolved using 0.100 M $\ce{NaOH}$ as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure $5$. The initial pH is high, but as acid is added, the pH decreases in steps if the successive $pK_b$ values are well separated. Table E1 lists the ionization constants and $pK_a$ values for some common polyprotic acids and bases.
Example $3$
Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M $\ce{NaOH}$ solution to 100.0 mL of a 0.0510 M solution of oxalic acid ($\ce{HO_2CCO_2H}$), a diprotic acid (abbreviated as $\ce{H2ox}$). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion ($\ce{O2CCO2^{2−}}$, abbreviated $\ce{ox^{2-}}$).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as $\ce{Ca^{2+}}$ are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.
Given: volume and concentration of acid and base
Asked for: pH
Strategy:
1. Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.
2. Calculate the concentrations of all the species in the final solution. Determine $\ce{[H{+}]}$ and convert this value to pH.
Solution:
A Table E5 gives the $pK_a$ values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:
$100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber$
$55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber$
The strongest acid ($H_2ox$) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of $OH^-$ to react with Hox−, forming ox2 and H2O. The reactions can be written as follows:
$\underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber$
$\underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber$
In tabular form,
Solutions to Example 17.3.3
$\ce{H2ox}$ $\ce{OH^{-}}$ $\ce{Hox^{−}}$ $\ce{ox^{2−}}$
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol
B The equilibrium between the weak acid ($\ce{Hox^{-}}$) and its conjugate base ($\ce{ox^{2-}}$) in the final solution is determined by the magnitude of the second ionization constant, $K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}$. To calculate the pH of the solution, we need to know $\ce{[H^{+}]}$, which is determined using exactly the same method as in the acetic acid titration in Example $2$:
$\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber$
Thus the concentrations of $\ce{Hox^{-}}$ and $\ce{ox^{2-}}$ are as follows:
$\left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber$
$\left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber$
We can now calculate [H+] at equilibrium using the following equation:
$K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber$
Rearranging this equation and substituting the values for the concentrations of $\ce{Hox^{−}}$ and $\ce{ox^{2−}}$,
$\left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber$
So
$pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber$
This answer makes chemical sense because the pH is between the first and second $pK_a$ values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than $pK_{a1}$), but we added only enough to titrate less than half of the second, less acidic proton, with $pK_{a2}$. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to $pK_{a2}$.
Exercise $3$: Piperazine
Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine ($pK_{b1}$ = 4.27, $pK_{b2}$ = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M $\ce{HCl}$ (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.
Answer
pH=4.9
Indicators
In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.
We can describe the chemistry of indicators by the following general equation:
$\ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber$
where the protonated form is designated by $\ce{HIn}$ and the conjugate base by $\ce{In^{−}}$. The ionization constant for the deprotonation of indicator $\ce{HIn}$ is as follows:
$K_{In} =\dfrac{ [\ce{H^{+}} ][ \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3}$
The $pK_{in}$ (its $pK_a$) determines the pH at which the indicator changes color.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, Hydrangea macrophylla flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure $6$). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers.
Irrespective of the origins, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a $pK_{in}$ that is within one pH unit of the expected pH at the equivalence point of the titration.
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure $7$ shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure $8$. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. The pH ranges over which two common indicators (methyl red, $pK_{in} = 5.0$, and phenolphthalein, $pK_{in} = 9.5$) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the $\ce{HCl}$ titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of $\ce{NaOH}$ has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the $\ce{HCl}$ titration, the phenolphthalein indicator will turn pink when about 50 mL of $\ce{NaOH}$ has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of $\ce{NaOH}$ will therefore cause the methyl red indicator to change color, resulting in a huge error.
The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the $pK_a$ of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure $9$).
pH Indicators: pH Indicators(opens in new window) [youtu.be]
Summary and Takeaway
Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the $pK_a$, and the $pK_b$ of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the $pK_a$ of the weak acid or the $pK_b$ of the weak base. Thus titration methods can be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.6%3A_Acid-Base_Titration_Curves.txt |
Learning Objectives
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as $\ce{HCl}$, $\ce{HNO3}$, and $\ce{HCN}$ that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:
$\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq) \nonumber$
$\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq) \nonumber$
$\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq) \nonumber$
Even though it contains four hydrogen atoms, acetic acid, $\ce{CH3CO2H}$, is also monoprotic because only the hydrogen atom from the carboxyl group ($\ce{-COOH}$) reacts with bases:
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic Acids
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
• The first ionization is
$\ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq) \nonumber$
with $K_{\ce a1} > 10^2;\: {complete\: dissociation}$.
• The second ionization is
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq) \nonumber$
with $K_{\ce a2}=1.2×10^{−2}$.
This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, $\ce{H2CO3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.
• First Ionization
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \nonumber$
with
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
• Second Ionization
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
with
$K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11} \nonumber$
$K_{\ce{H2CO3}}$ is larger than $K_{\ce{HCO3-}}$ by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the $\ce{HCO3-}$ formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and $\ce{HCO3-}$ are practically equal in a pure aqueous solution of H2CO3.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H3O+ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.
Example $1$: Ionization of a Diprotic Acid
When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are $\ce{[H3O+]}$, $\ce{[HCO3- ]}$, and $\ce{[CO3^2- ]}$ in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}$
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}$
Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than $\ce{HCO3-}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem:
1. Using the customary four steps, we determine the concentration of H3O+ and $\ce{HCO3-}$ produced by ionization of H2CO3.
2. Then we determine the concentration of $\ce{CO3^2-}$ in a solution with the concentration of H3O+ and $\ce{HCO3-}$ determined in (1).
To summarize:
1. First Ionization: Determine the concentrations of $\ce{H3O+}$ and $\ce{HCO3-}$.
Since \ref{step1} is has a much bigger $K_{a1}=4.3×10^{−7}$ than $K_{a2}=4.7×10^{−11}$ for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber$
As for the ionization of any other weak acid:
An abbreviated table of changes and concentrations shows:
Abbreviated table of changes and concentrations
ICE Table $\ce{H2CO3}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{HCO3-}(aq)$
Initial (M) $0.033 \:M$ - $0$ $0$
Change (M) $- x$ - $+x$ $+x$
Equilibrium (M) $0.033 \:M - x$ - $x$ $x$
Substituting the equilibrium concentrations into the equilibrium constant gives us:
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber$
Solving the preceding equation making our standard assumptions gives:
$x=1.2×10^{−4} \nonumber$
Thus:
$\ce{[H2CO3]}=0.033\:M \nonumber$
$\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber$
2. Second Ionization: Determine the concentration of $CO_3^{2-}$ in a solution at equilibrium.
Since the \ref{step1} is has a much bigger $K_a$ than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the \ref{step2}:
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
ICER Table for the \ref{step2}:
ICE Table $\ce{HCO3-}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{CO3^2-}(aq)$
Initial (M) $1.2×10^{−4}\:M$ - $1.2×10^{−4}\:M$ $0$
Change (M) $- y$ - $+y$ $+y$
Equilibrium (M) $1.2×10^{−4}\:M - y$ - $1.2×10^{−4}\:M + y$ $y$
\begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*} \nonumber
To avoid solving a quadratic equation, we can assume $y \ll 1.2×10^{−4}\:M$ so
$K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber$
Rearranging to solve for $y$
$y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber$
$[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber$
To summarize:
In part 1 of this example, we found that the $\ce{H2CO3}$ in a 0.033-M solution ionizes slightly and at equilibrium $[\ce{H2CO3}] = 0.033\, M$, $[\ce{H3O^{+}}] = 1.2 × 10^{−4}$, and $\ce{[HCO3- ]}=1.2×10^{−4}\:M$. In part 2, we determined that $\ce{[CO3^2- ]}=5.6×10^{−11}\:M$.
Exercise $2$: Hydrogen Sulfide
The concentration of $H_2S$ in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate $\ce{[H3O+]}$, $\ce{[HS^{−}]}$, and $\ce{[S^{2−}]}$ in the solution:
$\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber$
$\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber$
Answer
$[\ce{H2S}] = 0.1 M$, $\ce{[H3O+]} = [HS^{−}] = 0.0001\, M$, $[S^{2−}] = 1 × 10^{−19}\, M$
We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).
Triprotic Acids
A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:
• The first ionization is
$\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \nonumber$
with $K_{\ce a1}=7.5×10^{−3}$.
• The second ionization is
$\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \nonumber$
with $K_{\ce a2}=6.2×10^{−8}$.
• The third ionization is
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \nonumber$
with $K_{\ce a3}=4.2×10^{−13}$.
As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.
Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:
$\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber$
and
$\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq) \nonumber$
Summary
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.
Glossary
diprotic acid
acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps
diprotic base
base capable of accepting two protons. The protons are accepted in two steps
monoprotic acid
acid containing one ionizable hydrogen atom per molecule
stepwise ionization
process in which an acid is ionized by losing protons sequentially
triprotic acid
acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.7%3A_Polyprotic_Acids.txt |
Learning Objectives
• To understand how molecular structure affects the strength of an acid or base.
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the $\ce{A–H}$ or $\ce{B–H^+}$ bond, the less likely the bond is to break to form $H^+$ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with $pK_a$ values in parentheses:
$H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}$
Stability of the Conjugate Base
Whether we write an acid–base reaction as $AH \rightleftharpoons A^−+H^+$ or as $BH^+ \rightleftharpoons B+H^+$, the conjugate base ($A^−$ or $B$) contains one more lone pair of electrons than the parent acid ($AH$ or $BH^+$). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of $H^+$ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses:
$CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}$
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of $CH_4$ is $CH_3^−$, and the conjugate base of $HF$ is $F^−$. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the $F^−$ ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, $\ce{HF}$ has a greater tendency to dissociate to form $H^+$ and $F^−$ than does methane to form $H^+$ and $CH_3^−$, making HF a much stronger acid than $CH_4$.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula $HE$, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form $E^−$ and $H^+$. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.
the strongest acid Known: The hydrohelium Cation
The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, $\ce{HeH^{+}}$, which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol.
Ball and stick model of the hydrohelium ion. (CC BY-SA 3.0; CCoil).
$\ce{HeH^{+}}$ cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a hypothetical aqueous acidity using Hess's law:
HHe+(g) H+(g) + He(g) +178 kJ/mol
HHe+(aq) HHe+(g) +973 kJ/mol
H+(g) H+(aq) −1530 kJ/mol
He(g) He(aq) +19 kJ/mol
HHe+(aq) H+(aq) + He(aq) −360 kJ/mol
A free energy change of dissociation of −360 kJ/mol is equivalent to a pKa of −63.
It has been suggested that $\ce{HeH^{+}}$ should occur naturally in the interstellar medium, but it has not yet been detected.
Inductive Effects
Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
$HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}$
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5
As the electronegativity of $X$ increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as $H^+$.
The acidity of oxoacids, with the general formula $HOXO_n$ (with $n$ = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom $X$. As shown in Figure $1$, the $K_a$ values of the oxoacids of chlorine increase by a factor of about $10^4$ to $10^6$ with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure $1$ show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure $1$ and Figure $2$, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from $HClO$ to $HClO_4$ (also written as $HOClO_3$, while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as $H^+$ ions, thereby increasing the strength of the acid.
At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure $2$, the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms.
Electron delocalization in the conjugate base increases acid strength.
The electrostatic potential plots in Figure $2$ demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in $ClO_4^+$, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion ($ClO_4^−$), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion ($OCl^−$), the negative charge is largely localized on a single oxygen atom (Figure $2$). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.
Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, $H_3PO_4$ is a weak acid, $H_2SO_4$ is a strong acid, and $HClO_4$ is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to $Cl$, which causes electrons to be drawn from oxygen to the central atom, weakening the $\ce{O–H}$ bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table $1$ shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid $(H_2CO_3$) were a discrete molecule with the structure $\ce{(HO)_2C=O}$, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid ($H_3PO_4$), for which pKa1 = 2.16. Instead, the tabulated value of $pK_{a1}$ for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, $H_2CO_3$ is only a minor component of the aqueous solutions of $CO_2$ that are referred to as carbonic acid. Similarly, if phosphorous acid ($H_3PO_3$) actually had the structure $(HO)_3P$, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as $HOCl$ (pKa = 7.40). In fact, the $pK_{a1}$ for phosphorous acid is 1.30, and the structure of phosphorous acid is $\ce{(HO)_2P(=O)H}$ with one H atom directly bonded to P and one $\ce{P=O}$ bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as $H_3PO_4$. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table $1$: Values of pKa for Selected Polyprotic Acids and Bases
*$H_2CO_3$ and $H_2SO_3$ are at best minor components of aqueous solutions of $CO_{2(g)}$ and $SO_{2(g)}$, respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.
Polyprotic Acids Formula $pK_{a1}$ $pK_{a2}$ $pK_{a3}$
carbonic acid* “$H_2CO_3$” 6.35 10.33
citric acid $HO_2CCH-2C(OH)(CO_2H)CH_2CO_2H$ 3.13 4.76 6.40
malonic acid $HO-2CCH_2CO_2H$ 2.85 5.70
oxalic acid $HO_2CCO_2H$ 1.25 3.81
phosphoric acid $H_3PO_4$ 2.16 7.21 12.32
phosphorous acid $H_3PO_3$ 1.3 6.70
succinic acid $HO_2CCH_2CH_2CO_2H$ 4.21 5.64
sulfuric acid $H_2SO_4$ −2.0 1.99
sulfurous acid* “$H_2SO_3$” 1.85 7.21
Polyprotic Bases Formula $pK_{b1}$ $pK_{b2}$
ethylenediamine $H_2N(CH_2)_2NH_2$ 4.08 7.14
piperazine $HN(CH_2CH_2)_2NH$ 4.27 8.67
propylenediamine $H_2N(CH_2)_3NH_2$ 3.45 5.12
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
$pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber$
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the $\ce{–CH_3}$ group of acetic acid by a $\ce{–CF_3}$ group results in about a 10,000-fold increase in acidity!
Example $1$
Arrange the compounds of each series in order of increasing acid or base strength.
1. sulfuric acid [$H_2SO_4$, or $(HO)_2SO_2$], fluorosulfonic acid ($FSO_3H$, or $FSO_2OH$), and sulfurous acid [$H_2SO_3$, or $(HO)_2SO$]
2. ammonia ($NH_3$), trifluoramine ($NF_3$), and hydroxylamine ($NH_2OH$)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or base strengths
Strategy:
Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.
Solution:
Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, $FSO_3H$ is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
$pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber$
The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an $H^+$ ion. Thus $NF_3$ is predicted to be a much weaker base than $NH_3$. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in $NH_3$ by $OH$ will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured $pK_b$ values:
$pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber$
Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured.
Exercise $1$
Arrange the compounds of each series in order of
1. decreasing acid strength: $H_3PO_4$, $CH_3PO_3H_2$, and $HClO_3$.
2. increasing base strength: $CH_3S^−$, $OH^−$, and $CF_3S^−$.
Answer a
$HClO-3 > CH_3PO_3H_2 > H_3PO_4$
Answer a
$CF_3S^− < CH_3S^− < OH^−$
Summary
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an $H^+$ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of $H^+$, making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an $\ce{O–H}$ bond and allow hydrogen to be more easily lost as $H^+$ ions.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.8%3A_Organic_Acids_and_Bases_-_Structure_and_Reactivity.txt |
Learning Objectives
• Calculate the pH when two weak acids are present in a solution.
• Calculate the pH when the concentration of the acid is very dilute.
• Calculate the pH by including the autoionization of water.
A common paradigm in solving for pHs in weak acids and bases is that the equilibria of solutions containing one weak acid or one weak base. In most cases, the amount of $\ce{H+}$ from the autoionization of water is negligible. For very dilute solutions, the amount of $\ce{H+}$ ions from the autoionization of water must also be taken into account. Thus, a strategy is given here to deal with these systems.
When two or more acids are present in a solution, the concentration of $\ce{H+}$ (or pH) of the solution depends on the concentrations of the acids and their acidic constants Ka. The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, Ka's. Similarly, the concentration of $\ce{OH-}$ ions in a solution containing two or more weak bases depends on the concentrations and Kb values of the bases. For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases.
Dissociation of Acids and Bases in Water Couple Two Equilibria with a Common Ion H+
If the pH is between 6 and 8, the contribution due to autoionization of water to $\ce{[H+]}$ should also be considered. When autoionization of water is considered, the method is called the exact pH calculation or the exact treatment. This method is illustrated below. When the contribution of pH due to self-ionization of water cannot be neglected, there are two coupled equilibria to consider:
$\ce{HA \rightleftharpoons H+ + A-}$
ICE Table $\ce{HA}$ $\ce{H+}$ $\ce{A-}$
Initial $C$ 0 0
Change $- x$ ${\color{Red} x}$ $x$
Equilibrium $C-x$ ${\color{Red} x}$ $x$
and
$\ce{H2O \rightleftharpoons H+ + OH-}$
ICE Table $\ce{H2O}$ $\ce{H+}$ $\ce{OH-}$
Equilibrium - ${\color{Red} y}$ $y$
Thus,
\begin{align} \ce{[H+]} &= ({\color{Red} x+y})\ \ce{[A- ]} &= x\ \ce{[OH- ]} &= y \end{align}
and the two equilibrium constants are
$K_{\large\textrm{a}} = \dfrac{ ({\color{Red} x + y})\, x}{C - x} \label{1}$
and
$K_{\large\textrm{w}} = ({\color{Red} x + y})\, y \label{2}$
Although you may use the method of successive approximation, the formula to calculate the pH can be derived directly from Equations $\ref{1}$ and $\ref{2}$. Solving for $\color{ref} x$ from Equation $\ref{2}$ gives
$x = \dfrac{K_{\large\textrm{w}}}{y} - y$
and substituting this expression into $\ref{1}$ results in
$K_{\large\textrm{a}} = \dfrac{({\color{Red} x+y}) \left(\dfrac{K_{\large\textrm{w}}}{y} - y\right)}{C - \dfrac{K_{\large\textrm{w}}}{y} + y}$
Rearrange this equation to give:
\begin{align} \ce{[H+]} &= ({\color{Red} x+y})\ &= \dfrac{C - \dfrac{K_{\large\textrm{w}}}{y} + y}{\dfrac{K_{\large\textrm{w}}}{y} - y}\, K_{\large\textrm{a}} \end{align}
Note that
$\dfrac{K_{\large\textrm w}}{y} = \ce{[H+]}$
so
$y = \dfrac{K_{\large\textrm w}}{\ce{[H+]}}.$
Thus, we get:
$\ce{[H+]} = \dfrac{ C - \ce{[H+]} + \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}\, K_{\large\textrm{a}} \label{Exact}$
As written, Equation $\ref{Exact}$ is complicated, but can be put into a polynomial form
$\ce{[H+]^3} + K_{\large\textrm{a}} \ce{[H+]^2} - \left( K_{\large\textrm{w}} + C K_{\large\textrm{a}} \right) \ce{[H+]} - K_{\large\textrm{w}} K_{\large\textrm{a}} =0 \label{Exact2}$
Solving for the exact hydronium concentration requires solving a third-order polynomial. While this is analytically feasible, it is an awkward equation to handle. Instead, we often consider two approximations to Equation $\ref{Exact2}$ that can made under limiting conditions.
Case 1: High Concentration Approximation
If $[H^+] > 1 \times 10^{-6}$, then
$\dfrac{K_w}{[H^+]} < 1 \times 10^{-8}.$
This is small indeed compared to $[H^+]$ and $C$ in Equation $\ref{Exact}$. Thus,
$[H^+] \approx \dfrac{C - [H^+]}{[H^+]} K_{\large\textrm{a}}$
$[H^+]^2 + K_{\large\textrm{a}} [H^+] - C K_{\large\textrm{a}} \approx 0 \label{Quad}$
Equation $\ref{Quad}$ is a quadratic equation with two solutions. However, only one will be positive and real:
$[H^+] \approx \dfrac{-K_{\large\textrm{a}} + \sqrt{K_{\large\textrm{a}}^2 + 4 C K_{\large\textrm{a}}}}{2}$
Case 2: Low Concentration Approximation
If $[H^+] \ll C$, then
$C - [H^+] \approx C$
Equation $\ref{Exact}$ can be simplified
\begin{align} [H^+] &\approx \dfrac{C}{[H^+]}\: K_{\large\textrm{a}}\ [H^+] &\approx \sqrt{C K_{\large\textrm{a}}} \end{align}
The treatment presented in deriving Equation $\ref{Exact}$ is more general, and may be applied to problems involving two or more weak acids in one solution.
Example $1$
Calculate the $\ce{[H+]}$, $\ce{[Ac- ]}$, and $\ce{[Cc- ]}$ when the solution contains 0.200 M $\ce{HAc}$ ($K_a = 1.8 \times 10^{-5}$), and 0.100 M $\ce{HCc}$ (the acidity constant $K_c = 1.4 \times 10^{-3}$). ($\ce{HAc}$ is acetic acid whereas $\ce{HCc}$ is chloroacetic acid).
Solution
Assume x and y to be the concentrations of $\ce{Ac-}$ and $\ce{Cc-}$, respectively, and write the concentrations below the equations:
$\begin{array}{ccccc} \ce{HAc &\rightleftharpoons &H+ &+ &Ac-}\ 0.200-x &&x &&x\ \ \ce{HCc &\rightleftharpoons &H+ &+ &Cc-}\ 0.100-y &&y &&y \end{array}$
$\ce{[H+]} = (x + y) \nonumber$
Thus, you have
$\dfrac{(x + y)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{Ex1.1}$
$\dfrac{(x + y)\, y}{0.100 - y} = 1.4\times 10^{-3} \label{Ex1.2}$
Solving for x and y from Equations $\ref{Ex1.1}$ and $\ref{Ex1.2}$ may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since $\ce{HAc}$ is a weaker acid than is $\ce{HCc}$, you expect x << y. Further, y << 0.100. Therefore, $x + y \approx y$ and 0.100 - y => 0.100. Equation $\ref{Ex1.2}$ becomes:
$\dfrac{ ( y)\, y}{0.100} = 1.4 \times 10^{-3} \label{Ex1.2a}$
which leads to
\begin{align*} y &= (1.4 \times 10^{-3} \times 0.100)^{1/2}\ &= 0.012 \end{align*}
Substituting $y$ in Equation $\ref{Ex1.1}$ results in
$\dfrac{(x + 0.012)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{1'}$
This equation is easily solved, but you may further assume that $0.200 - x \approx 0.200$, since $x << 0.200$. Thus,
\begin{align*} x &= \dfrac{-0.012 + (1.44\times 10^{-4} + 1.44\times 10^{-5})^{1/2}}{2}\ &= 2.9\times 10^{-4}\:\: \longleftarrow \textrm{Small indeed compared to 0.200} \end{align*}
You had a value of 0.012 for y by neglecting the value of x in Equation $\ref{Ex1.2}$. You can now recalculate the value for y by substituting values for x and y in Equation $\ref{Ex1.2}$.
$\dfrac{(2.9\times 10^{-4} + y)\, y}{0.100 - 0.012} = 1.4\times 10^{-3} \label{2"}$
Solving for y in the above equation gives
$y = 0.011 \nonumber$
You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from $2.9 \times 10^{-4}$ to a new value of $3.2 \times 10^{-4}$. Use your calculator to obtain these values. Further refinement does not lead to any significant changes for x or y.
Discussion
You should write down these calculations on your note pad, since reading alone does not lead to thorough understanding.
Example $2$
A weak acid $\ce{HA}$ has a $K_a$ value of $4.0 \times 10^{-11}$. What are the pH and the equilibrium concentration of $\ce{A-}$ in a solution of 0.0010 M $\ce{HA}$?
Solution
For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a).
Method (a)
The two equilibrium equations are:
$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-};\ 0.0010-x &&x &&x\ \ \ce{H2O &\rightleftharpoons &H+ &+ &OH-}\ &&y &&y \end{array}$
$\ce{[H+]} = (x+y)$
\begin{align} \dfrac{(x+y)\, x}{0.0010-x} &= 4.0\times 10^{-}11 \label{3}\ \ (x+y)\, y &= 1\times 10^{-}14 \label{4} \end{align}
Assume y << x, and x << 0.0010, then you have
\begin{align} \dfrac{(x )\, x}{0.0010} &= 4.0\times 10^{-11} \label{3'} \ x &= (0.0010 \times 4.0e^{-11})^{1/2}\ &= 2.0\times 10^{-7} \end{align}
Substituting $2.0 \times ^{-7}$ for x in 4 and solving the quadratic equation for y gives,
$(2.0\times 10^{-}7+y)\, y = 1\times 10^{-14} \nonumber$
$y = 4.1\times 10^{-8} \nonumber$
Substituting $4.1 \times 10^{-8}$ in Equation $\ref{3}$, but still approximating 0.0010-x by 0.0010:
$\dfrac{(x+4.1\times 10^{-8})\, x}{0.0010} = 4.0\times 10^{-11} \label{3''}$
Solving this quadratic equation for a positive root results in
$x = 1.8 \times 10^{-7} \;\text{M} \longleftarrow \textrm{Recall }x = \ce{[A- ]} \nonumber$
\begin{align*} \ce{[H+]} &= x + y\ &= (1.8 + 0.41)\,1\times 10^{-7}\ &= 2.2\times 10^{-7}\ \ce{pH} &= 6.65 \end{align*}
The next method uses the formula derived earlier.
Method (b)
Using the formula from the exact treatment, and using $2 \times 10^{-7}$ for all the $\ce{[H+]}$ values on the right hand side, you obtain a new value of $\ce{[H+]}$ on the left hand side,
\begin{align*} \ce{[H+]} &= \dfrac{C - \ce{[H+]} + \dfrac{K_{\large\textrm w}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm w}}{\ce{[H+]}}} K_{\large\textrm a}\ &= 2.24\times 10^{-7}\ \ce{pH} &= 6.65 \end{align*}
The new $\ce{[H+]}$ enables you to recalculate $\ce{[A- ]}$ from the formula:
\begin{align*} (2.24\times 10^{-7}) \ce{[A- ]} &= C K_{\large\textrm a} \nonumber\ \ce{[A- ]} &= \dfrac{(0.0010) (4.0\times 10^{-11})}{2.24\times 10^{-7}} \nonumber\ &= 1.8\times 10^{-7} \nonumber \end{align*}
DISCUSSION
You may have attempted to use the approximation method:
\begin{align*} x &= (C K_{\large\textrm a})^{1/2} \nonumber\ &= 2.0\times 10^{-7}\: \mathrm{M\: A^-,\: or\: H^+;\: pH = 6.70} \nonumber \end{align*}
and obtained a pH of 6.70, which is greater than 6.65 by less than 1%. However, when an approximation is made, you have no confidence in the calculated pH of 6.70.
Summary
Water is both an acid and a base due to the autoionization,
$\ce{H2O \rightleftharpoons H+ + OH-} \nonumber$
However, the amount of $\ce{H+}$ ions from water may be very small compared to the amount from an acid if the concentration of the acid is high. When calculating $\ce{[H+]}$ in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.9%3A_A_Deeper_Look_-_Exact_Treatment_of_Acid-Base_Equilibria.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q3
Vinegar contains acetic acid $\mathrm{CH_3COOH}$. What species serves as a base when vinegar is mixed with baking soda, sodium bicarbonate, during the preparation of bread?
Solution
$NaHCO_3 + CH_3COOH \rightarrow NaCH_3COO + CO_2 + H_2O\nonumber$
$\ce{NaHCO_3}$ serves as the base.
Q7
1. Thinking of acid-base reaction in terms of oxide donors and oxide acceptors, is the base oxide donors or oxide acceptors?
2. Identify the acid and base in the reaction:
$\ce{CaO + CO2 \rightleftharpoons CaCO3 }\nonumber$
Solution
1. The base is the oxide donor. We can distinctly see this in the autoionization of H2O, where OH- counts as the basic part
2. CaO is the base, and CO2 is the acid.
Q5
Baking soda known as $\mathrm{NaHCO_3}$ is formed by adding water and carbon dioxide to sodium carbonate.
1. Write a balanced equation for this chemical reaction
2. Is this a Brønsted-Lowry acid-base reaction? What is a Brønsted-Lowry acid? What is a Brønsted-Lowry base?
Solution
1. $\mathrm{Na_2CO_3 + H_2O + CO_2 \rightarrow 2 NaHCO_3}$
2. This is not a Brønsted-Lowry acid-base reaction because it does not involve a transfer of an H+ ion. A Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor.
Q9
Identify each of the following oxides as an acid or base anhydride:
1. $\mathrm{CaO}$
2. $\mathrm{P_2O_5}$
Solution
1. $\mathrm{CaO}$ is the base anhydride of calcium hydroxide $\mathrm{Ca(OH)_2}$.
2. $\mathrm{P_2O_5}$ is the acid anhydride of phosphoric acid $\mathrm{H_3PO_4}$.
Q11
Al(III) oxide is amphorteric. What is the balanced chemical equation of Al(III) oxide react with aqueous $\ce{H_2SO_4}$? What is the balanced equation of it reacts with $\ce{KOH}$?
Solution
$\ce{Al2O3(s) + 3H2SO4(aq) \rightleftharpoons Al2(SO4)3(aq) + 3H2O(l)} \nonumber$
$\ce{Al2O3 + 2KOH + 3H2O \rightleftharpoons 2KAl(OH)4 \nonumber$
Q13
The $\ce{[H_3O^+]}$ concentration in a glass of orange juice is 3.96 x 10-5 M. What is the juice's pH?
Solution
$pH = -\log[H_3O+] \nonumber$
Since the concentration of hydronium ions is given, the pH calculation is as follows:
$pH = -\log[3.96 \times 10^{-5}] = 4.4023\nonumber$
pH is a measure of hydrogen ions in a solution. This concentration defines the acidity or alkalinity of a solution.
Q17
The pKw of an unknown salty water at 25 °C is 13.665. This differs from the usual Kw of 14.00 at this temperature because dissolved salts make this unknown salty water a non- ideal solution. If the pH in the salty water is 7.8, what are the concentrations of H3O+ and OH- in the salty water at 25 °C?
Solution
$\mathrm{pH = -log[H_3O^+]}$, hence,
$\mathrm{[H_3O^+] = 10^{-7.8} = 1.5849 \times 10^-8 \: M}$
Since $\mathrm{pK_w = [OH^-][H_3O^+]}$,
$\mathrm{[OH^-] = \frac{10^{-13.665}}{1.5849 \times 10^{-8} \: M} = 1.3645 \times 10^{-6} M }$
Q19
When rubidium (Rb) solid is added to water, there is an instantaneous and vigorous reaction (i.e., an explosion) as this video demonstrates. Based on this information, which of these two equations is a more accurate representation the reaction?
$\ce{ 2Rb(s) + 2H2O(l) \rightarrow 2RbOH(aq) + H2(aq)} \nonumber$
$\ce{ 2Rb(s) + 2H3O^{+}(aq) \rightarrow 2Rb^{+}(aq) + H2(aq) +2H2O(l)} \nonumber$
Solution
The equation:
$\ce{2Rb(s) + 2H2O(l) \rightarrow 2RbOH(aq) + H2 (aq)}\nonumber$
Is a better representation of Rb being placed into water. It represents the direct interaction between the alkali metal and water. We know from the problem that the reaction is fast, vigorous, and forceful. The second equation represents a reaction in equilibrium, one that we would expect to be slow as there are only 1x10-7mol/L of H3O+ in a 1L of water. Even if the reaction were vigorous at the low concentration of hydronium ions, there would not be enough of them to keep up with the speed of the reaction (they would become a limiting reactant), hence slowing or stopping the reaction from proceeding.
Q23
In the following chemical equation determine which species is the strongest acid and which is the strongest base, using the Brønsted–Lowry definition. At equilibrium, is there a greater concentration of reactants or products present?
1. $\ce{HIO_{3\, (aq)} + HCOO^{-}_{(aq)} \rightleftharpoons HCOOH_{(aq)} + IO^{-}_{3\, (aq)}}\nonumber$
2. $\left(\ce{HIO_{3\, (aq)} \rightleftharpoons IO^{-}_{3\, (aq)}} \qquad \ce{K_{a} = 1.6 \times 10^{-1}} \right) \nonumber$
3. $\left(\ce{HCOOH_{(aq)} \rightleftharpoons HCOO^{-}_{(aq)}} \qquad \ce{K_{a} = 1.8 \times 10^{-4}} \right) \nonumber$
Solution
A Brønsted–Lowry acid is the species that donates protons in a solution. When comparing two different weak acids in solution, like:
$\ce{HIO_{3\, (aq)} + HCOO^{-}_{(aq)} \rightleftharpoons HCOOH_{(aq)} + IO^{-}_{3\, (aq)}}\nonumber$
We can compare their abilities to donate protons to see which one is the stronger of the two weak acids.
$\ce{HIO_{3\, (aq)} + H_{2}O_{(aq)} \rightleftharpoons IO^{-}_{3\, (aq)} + H_{3}O^{+}_{(aq)}} \qquad \ce{K_{a} = 1.6 \times 10^{-1}} \nonumber$
$\ce{HCOOH_{(aq)} + H_{2}O_{(aq)} \rightleftharpoons HCOO^{-}_{(aq)} + H_{3}O^{+}_{(aq)}} \qquad \ce{K_{a} = 1.8 \times 10^{-4}} \nonumber$
Seeing that $\ce{K_{a} = 1.6 \times 10^{-1}} \gt \ce{K_{a} = 1.8 \times 10^{-4}}$, we know $\ce{HIO_{3}}$ is the stronger acid.
A Brønsted–Lowry base is the species that accepts protons. So in this case, we must examine which of the two weak acids has a stronger conjugate base, which means we must find the $\ce{K_{b}}$ for the reactions of the conjugate bases.
We know that:
$\ce{K_{w} = \ K_{a} \times K_{b}} \nonumber$
and
$\ce{K_{w}} = 1.0 \times 10^{-7} \nonumber$
So we can find $\ce{K_{b}}$ by dividing $\ce{K_{w}}$ by $\ce{K_{a}}$ which ultimately gives us:
$\ce{IO^{-}_{3\, (aq)} + H_{2}O_{(aq)} \rightleftharpoons HIO_{3\, (aq)} + OH^{-}_{(aq)}} \qquad \ce{K_{b} = 6.3 \times 10^{-14}} \nonumber$
$\ce{HCOO^{-}_{(aq)} + H_{2}O_{(aq)} \rightleftharpoons HCOOH_{(aq)} + OH^{-}_{(aq)}} \qquad \ce{K_{b} = 5.6 \times 10^{-11}} \nonumber$
Seeing that $\ce{K_{b} = 5.6 \times 10^{-11}} \gt \ce{K_{a} = 6.3 \times 10^{-14}}$, we know $\ce{HCOO^{-}}$ is the stronger base.
To determine whether there is a greater concentration of reactants or products present, the K value for the overall reaction must be determined. The overall reaction is the product of the first given reaction and the reverse of the second given reaction. Dividing the first value for Ka by the second gives
$K=888\nonumber$
$K>1\nonumber$ which indicates that at equilibrium, there is a greater concentration of products than reactants.
Q27
Acetic acid gives vinegar a sour taste and strong aroma. Its $K_a$ value is 1.75 x 10-5. What is the pH of the solution if 0.59 grams of acetic acid is dissolved in 40 mL of water?
Solution
First, convert grams of acetic acid to moles.
$(0.59\; g\; CH_{3}COOH) \left(\dfrac{1\: mol}{60.05\: g\; CH_{3}COOH} \right) = 0.0098\; mol\nonumber$
Then, find the molarity of acetic acid by dividing the number of moles of acetic acid by the number of liters of water.
$\dfrac{0.0098\: mol}{0.04\: {L\; water}} = 0.246\; M\nonumber$
Using the molarity and Ka, construct and solve an ICE table to find out how much the acetic acid dissociates.
$\ce{CH_3COOH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COO^{-}_{(aq)} + H_3O+_{(aq)}}$
$CH_3COOH$ $H_2O$ $CH_3COO^-$ $H_3O^+$
I 0.246 --- 0 0
C -x --- +x +x
E 0.246-x --- x x
The acid dissociation constant works in the below equation:
$K_a = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]} \nonumber$
Plug in the final concentration values from the ICE table and solve for x.
$1.75 \times 10^{-5} = \dfrac{x^2}{0.246-x} \nonumber$
$x = 0.0021 \nonumber$
Use the calculated value of x to calculate the concentration of H3O+, which in this case is equal to x. Then, plug this concentration into the equation for pH.
$pH = -log[0.0021] \nonumber$
$pH = 2.6848 \nonumber$
Q29
1. A student prepares a solution of 0.60M of formic acid carefully in a water bath that remains constant at 25oC, determine the pH of the solution. (Ka of formic acid: 1.8 x 10-4)
2. How many grams of trichloroacetic acid should be dissolved per liter of deionized water so that the solution of trichloroacetic acid would have the same pH as that of the formic acid solution in a)? (Ka of trichloroacetic acid is 2.2 x 10-1)
Solution
Construct an ICE table based on the equation:
$\mathrm{HCOOH + H_2O \rightleftharpoons HCO_2^- + H_3O^+} \nonumber$
$HCOOH$ $H_2O$ $HCO_2^-$ $H_3O^+$
I
0.6
-
0
0
C
-x
-
+x
+x
E
0.6-x
-
x
x
Ka can then be equated to an algebraic expression:
$\mathrm{Ka= \dfrac{x^2 }{0.6-x}}\nonumber$
$\mathrm{1.8 \times 10^{-4}= \dfrac{x^2 }{0.6-x}}\nonumber$
$\mathrm{x=0.0103026}\nonumber$
or
$\mathrm{x=-0.0104}\nonumber$
x must be postive. Thus, x = 0.0103026.
$\mathrm{[H_3O^+]}\nonumber$
$\mathrm{pH = -log(0.0103026) = 1.987}\nonumber$
Because we know the pH we want to attain (1.987), we have start by first finding $\mathrm{[H_3O^+]}$:
$\mathrm{-log[H_3O^+] = 1.987}\nonumber$
$\mathrm{[H_3O^+] = 10^{-1.987}}\nonumber$
$\mathrm{[H_3O^+] = 0.0103026}\nonumber$
An ICE table can also be constructed for the reaction:
$CCl_3CO_2H + H_2O \rightleftharpoons CCl_3CO_2^- + H_3O^+\nonumber$
$CCl_3CO_2H$ $H_2O$ $CCl_3CO_2^-$ $H_3O^+$
I
x
-
0
0
C
-0.010303
-
+0.010303
+0.010303
E
x-0.010303
-
0.010303
0.010303
$\mathrm{K_a = \dfrac{(0.010303)^2}{(x-0.010303)}}\nonumber$
$\mathrm{2.2 \times 10^{-1} = \dfrac{(0.010303)^2}{(x-0.010303)}}\nonumber$
$\mathrm{x = 0.0107855 M}\nonumber$
To calculate the mass of trichloroacetic acid, we can calculate the molarity of trichloroacetic acid by its molar mass.
$\mathrm{Mass \: of \: trichloroacetic \: acid= 0.0107855 \: M \times (35.5 \dfrac{g}{mol} \times 3 + 12 \dfrac{g}{mol} \times 2 + 16 \dfrac{g}{mol} \times 2 + 1 \dfrac{g}{mol})= 1.76 \: g}\nonumber$
Q41
Rank each of the 0.2 M solution below in an order of increasing pH: $NH_4I$, $KF$, $HCl$, $KCl$, $KOH$.
Solution
$HCl< NH_{4}I<KF<KCl<KOH\nonumber$
Q43
0.040 mol of Diethylamine ($\ce{C_4H_11N}$, pKb =11.09) is titrated with 0.015 mol of $\ce{HCl}$ in a 1.00L wash bottle, calculate the pH value of the solution.
Solution
Because equivalence point is not reached yet, we can employ Henderson-Hasselbalch equation.
\begin{align*} pOH &\approx pK_b + \log \frac{[BH^+]}{[B]} \[5pt] pOH &\approx (11.09) + \log \frac{[C_4H_{12}N^+]}{[C_4H_{11}N]} \[5pt] pOH &\approx (11.09) + \log \frac{0.015}{0.040-15} \[5pt] pOH &\approx 10.87 \end{align*}
At room temperature $K_w = 14$ and
\begin{align*} pH &= pK_w - pOH \[5pt] &= 14.00 - 10.87 \[5pt] &= 3.13 \end{align*}
Q45
Prepare a Hypochlorous acid/Hypochlorite buffer at pH 7.
1. Suppose you only have 0.5 mol of lithium hypochlorite, but an infinite supply of hypochlorous acid (pKa = 7.53) and water. How many moles of hypochlorous acid should you use, assuming you use all 0.5 mol of lithium hypochlorite, and dilute the buffer to 100 mL?
2. Suppose Professor Güntherfœrd's arm was doused in about 0.12 moles total $HCl$. Will this buffer be enough to bring her arm up to a pH over 6.5?
Solution
Using the Henderson–Hasselbalch approximation:
$\mathrm{pH \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber$
It's easy to re-arrange the equation to solve for $\mathrm{[HClO]_\circ}$. Multiplying by 100 mL yields the moles of acid added.
$\mathrm{pH \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber$
$\mathrm{7 \approx pK_a + \log{\dfrac{[ClO^-]_o}{[HClO]_o}}}\nonumber$
$\mathrm{7 \approx 7.53 + \log{\dfrac{5}{[HClO]_o}}}\nonumber$
$\mathrm{[HClO]_\circ \approx 16.94}\nonumber$
$\mathrm{Moles \: HClO \approx 1.694}\nonumber$
Q49
A student is given 500 mL of a 0.500 M acetic acid solution and wants to create a pH 5.0 buffer. How many mL of 1 M $\ce{NaOH}$ must be added to the original solution? Acetic acid has a pKa of 4.756.
Solution
We first use the Henderson-Hasselbach approximation to determine the required ratio of base to acid in the solution:
\begin{align*} pH&=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \[5pt] 5.00 &=4.756+\log\left(\dfrac{[base]}{[acid]}\right) \[5pt] 0.244 &=\log\left(\dfrac{[base]}{[acid]}\right) \[5pt] 1.75 &=\dfrac{[base]}{[acid]} \end{align*}\nonumber
Next, we will determine the molar amount of base required to get a pH of 5.00. In order to simplify things we will first solve for the molar quantities as of each using the simple ratio above in relation to 1.75:
\begin{align*} \text{Acetic Acid in 500 mL solution} &= {(0.5 \; M) \times (0.5 \; L)} \[5pt] &= 0.25 \; mol \end{align*}\nonumber
$NaOH\; in\; 500\; mL\; solution={(0.25 \; mol) \times (1.75)} +0.25=0.6875+ \; mol \nonumber$
Now we'll determine the volume of 1M NaOH needed to raise the pH:
$\left(\dfrac{1 \; mol}{1 \; L}\right)=\left(\dfrac{0.6875 \; mol}{x}\right)\nonumber$
$s=0.6875\; L\; NaOH\;\nonumber$
Now we'll check to make sure everything is right:
$Acetic\; Acid\;=\left(\dfrac{0.25}{0.9375}\right)=0.2667M\nonumber$
$NaOH\;=\left(\dfrac{0.6875 - 0.25}{0.9375}\right)=0.4667M\nonumber$
$5.00=4.756+\log\left(\dfrac{[base]}{[acid]}\right)\nonumber$
$pH=4.756+\log\left(\dfrac{[0.4667]}{[0.2667]}\right)\nonumber$
$pH=4.756+0.243=4.999\nonumber$
We get a pH of 4.999 which is about 5.00 and isn't exactly 5.00 due to rounding early in the problem, so the problem was done correctly.
Q51
0.15M of HBr is added into 50mL of 0.1M Ca(OH)2 for the titration.
1. What is the pH of the solution before HBr is added?
2. What is the pH of the solution at the point when it needs 1 mL of HBr to neutralize the solution?
3. What is the pH of the solution when it is titrated 1 mL past neutralization?
Solution
$2HBr+Ca(OH)_{2}\rightleftharpoons 2 H_{2}O+CaBr_{2}$
a) $Ca(OH)_2$ is strong base. They dissociate completely.
$\mathrm{[OH^{-}]=2[Ca(OH)_{2}]=0.2M}\nonumber$
$\mathrm{pH = pK_w - pOH}\nonumber$
$\mathrm{pH = 14 + log(0.2 \; M)}\nonumber$
$\mathrm{pH = 13.3}\nonumber$
b) Because Ca(OH)2 and HBr are both strong base and strong acid, at equilibrium, the pH is 7.00.
$\mathrm{mole_{Ca(OH)_{2}}=(Volume)(Molarity)}$
$\mathrm{mole_{Ca(OH)_{2}}=(0.05L)(0.1M)}$
$\mathrm{mole_{Ca(OH)_{2}}=0.005\ mol}$
@ equilibrium;
$\mathrm{mole_{HBr}=2\ mole_{Ca(OH)_{2}}}$
$\mathrm{mole_{HBr}=2(0.005\ mol)=0.01\ mol}$
$\mathrm{Volume_{HBr} = \frac{mole_{HBr}}{molarity_{HBr}}=\frac{0.01\ mol}{0.15M}=0.0667\ L=66.7\ mL}$
$\mathrm{Total\ Volume\ at\ equilibrium=66.7mL + 50mL=116.7mL}$
$\mathrm{Total\ Volume\ 1mL\ short\ of\ equilibrium=116.7mL-1mL=115.7mL}$
$\mathrm{Volume_{HBr}\ 1mL\ short\ of\ equilibrium=66.7mL-1mL=65.7\ mL}$
$\mathrm{mols \; of \; OH^- \; not \; neutralized \; by \; HBr = 0.01 \; mol - 0.0657 \; L \times (0.15 \;M) =1.45 \times 10^{-4} \; mol }$
$\mathrm{pH = 14 + log(\dfrac{1.45^{-4} \; mol}{0.1157 \; L}) = 11.1}$
c)
At 1mL after equilibrium, Ca(OH)2 has been neutralized by HBr. Only HBr exists in the solution.
$\mathrm{ mole_{HBr}=(0.001L)(0.15M)=1.5\times 10^{-4}\ mol}$
$\mathrm{Volume=Total\ Volume+1mL=117.7mL=0.1177\ L}$
$\mathrm{M_{HBr}\ in\ the\ solution=\frac{1.5\times 10^{-4}\ mol}{0.1177\ L}=1.274\times 10^{-3}M}$
$\mathrm{M_{HBr}\ in\ the\ solution=M_{H^{+}}\ in\ the\ solution}$
$\mathrm{pH=-log[H^{+}]=-log(1.274\times 10^{-3}M)=2.89}$
Q55
Kb at 25°C for Diethylamine ($\ce{(C_2H_5)_2NH}$) is $1.3 \times 10^{-3}$. Consider the titration of 50.00 mL of a 0.1000 M solution of Diethylamine with 0.100 M $\ce{HCl}$ added with the following volumes: 0, 10.00, 50.00 mL. Calculate pH for each solutions. At an unknown volume beyond 50.00 mL, the pH is 3.90. Find the corresponding amount volume of $\ce{HCl}$ needed to obtain that pH.
Solution
When HCl volume = 0.00 mL.
$DiEtNH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons DiEtNH_{2(aq)}^{+}+OH_{aq}^{-}$
$\frac{[DiEtNH_{2}^{+}][OH^{-}]}{[DiEtNH]}= 1.3\times 10^{-3}$
$[OH^-] = [DiEtNH_2^+] = y\nonumber$
$[DiEtNH] = 0.1000 - y \nonumber$
$\frac{y^{2}}{0.1000-y}= 1.3\times 10^{-3}$
$\mathrm{y = 0.01077M= [OH^-] }\nonumber$
$\mathrm{pOH = 1.97}\nonumber$
$\mathrm{pH = 12.03}\nonumber$
When HCl volume = 10.00 mL.
$\mathrm{[DiEtNH_{2}^{+}]=\frac{(0.1000M)(0.01L)}{(0.050+0.010)L}= 0.0167 \; M}$
$\mathrm{[DiEtNH]=\frac{(0.1000M(0.050L)-(0.1000M)(0.01L)}{(0.050+0.010)L}= 0.0667 \; M}$
Plug it back to Henerson Hasselbalch equation
$\mathrm{pOH = pK_b + \log(\dfrac{[BH^+]}{[B]})}$
$\mathrm{pOH = -\log(1.3 \times 10^{-3}) +\log(\dfrac{[0.0167 \; M]}{[0.0667 \; M]})} \[\mathrm{pH = pH - pOH = 14.00 - 2.28 = 11.72}$
When HCl volume = 50 mL.
The titration is at the equivalence point. At equivalence, the reaction consists of 100 mL of 0.050 mol $\ce{DiEtNH_2^+}$
$\mathrm{DiEtNH_{2(aq)}^{+}+H_{2}O_{(l)}\rightleftharpoons DiEtNH_{(aq)}+H_{3}O_{(aq)}^{+}}$
$\mathrm{K_a= \dfrac{1.00 \times 10^{-14}}{1.3 \times 10^{-3}} = 7.69 \times 10^{-12}}\nonumber$
$\mathrm{K_{a}= 7.69\times 10^{-12}=\frac{x^{2}}{(0.5000-x)};x=[H_{3}O^{+}]}$
$\mathrm{x = 1.96 \times 10^{-6}; pH = 5.71}\nonumber$
Beyond the equivalence point
Beyond the equivalence point, solution behaves like $HCl$.
Given pH = 3.90.
$\mathrm{10^{-3.90}=\frac{(z-0.050 \; L)}{0.100 \; L+z} \times 0.1000 \; M}$
$\mathrm{z= 0.0502 \: L}$
Q57
Sodium Bicarbonate ($\ce{NaHCO_3}$) is a very weak base when dissolved in water. Some amount of sodium bicarbonate is dissolved in 125 mL of a 0.25 M solution of HNO3. The 168 mL of 0.15 M NaOH was used to titrate the solution. How many grams of sodium bicarbonate were added?
Solution
We are titrating an acid with two bases so solve for the amount of acid the $\ce{NaOH}$ neutralizes and the remaining moles of acid will be the number of moles of sodium bicarbonate.
$\mathrm{Moles \: HNO_3= \frac{0.25 \; moles}{1 \: L} \times 0.125 \: L = 0.03125 \: moles}\nonumber$
$\mathrm{Moles \: NaOH= \frac{0.15 \: moles}{1 \: L} \times 0.168 \: L =0.0252 \: moles} \nonumber$
$\mathrm{0.03125-0.0252= 0.00605 \; moles \; sodium \; bicarbonate}\nonumber$
Now just multiply by sodium bicarbonate's molar mass (84.007 $\frac{g}{mol}$ ) to find the mass of sodium bicarbonate added
$\mathrm{0.00605 \times 84.007 = 0.51 g}\nonumber$
Q59
What is the molarity of a $\ce{HNO3}$ water solution if it requires 31.80 mL of such solution to titrate 0.0662 g Aniline in 100 mL aqueous solution to equivalence point? What will the pH value be at the equivalence point if $\mathrm{K_b(Aniline) = 3.8 \times 10^{-10}}$?
Solution
Aniline and $\ce{HNO3}$ react with a one-to-one stoichiometry
$\ce{C6H5NH2(aq) + HNO3(aq) \rightleftharpoons C6H5NH3^{+}(aq) + NO3^{-} (aq) } \nonumber$
Therefore
$M(HNO_{3})=\dfrac{0.0662 \; g}{93.13 \; g/mol \;} \times \dfrac{1}{0.03180 \; L}=0.0224 \; M\nonumber$
Suppose at the equivalence point all Aniline is converted to its conjugate acid, then its concentration equals
$[\ce{C6H5NH3^{+}}]= \dfrac {0.0662\,g} {(93.13\,g/mol ) \; 0.1318\,L}=0.00539\, M \nonumber$
Also as some Aniline's conjugate acid reacts with water,
$K_a=\dfrac{K_w}{K_b}=\dfrac{1.0 \times 10^{-14}}{3.8\times 10^{-10}}=2.63 \times 10^{-5}=\dfrac{[H_3O^+][C_6H_5NH_2]}{[C_6H_5NH_3^+]} = \dfrac{x^2}{0.00539-x}\nonumber$
Therefore,
$x =[H_{3}O^{+}]=3.636\times 10^{-4}M\nonumber$
so $\mathrm{pH=3.44}$. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.E%3A_Acid-Base_Equilibria_%28Exercises%29.txt |
Solubility equilibrium is a type of dynamic equilibrium. It exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The solid may dissolve unchanged, with dissociation or with chemical reaction with another constituent of the solvent, such as acid or alkali. Each type of equilibrium is characterized by a temperature-dependent equilibrium constant. Solubility equilibria are important in pharmaceutical, environmental and many other scenarios.
• 16.1: The Nature of Solubility Equilibria
Dissolution of a salt in water is a chemical process that is governed by the same laws of chemical equilibrium that apply to any other reaction. There are, however, a number of special aspects of of these equilibria that set them somewhat apart from the more general ones that are covered in the lesson set devoted specifically to chemical equilibrium. These include such topics as the common ion effect, the influence of pH on solubility, and supersaturation.
• 16.2: Ionic Equilibria between Solids and Solutions
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
• 16.3: Precipitation and the Solubility Product
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation.
• 16.4: The Effects of pH on Solubility
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH.
• 16.5: Complex Ions and Solubility
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf).
• 16.6: A Deeper Look: Selective Precipitation of Ions
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis, a procedure for discovering the identity of metal ions present in the mixture. The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective.
• 16.E: Solubility and Precipitation (Exercises)
16: Solubility and Precipitation Equilibria
Learning Objectives
• Understand the qualitative nature of dissolving a salt in water
Dissolution of a salt in water is a chemical process that is governed by the same laws of chemical equilibrium that apply to any other reaction. There are, however, a number of special aspects of of these equilibria that set them somewhat apart from the more general ones that are covered in the lesson set devoted specifically to chemical equilibrium. These include such topics as the common ion effect, the influence of pH on solubility, supersaturation, and some special characteristics of particularly important solubility systems... all explained in what follows.
The Dissolution of Salts in Water
Drop some ordinary table salt into a glass of water, and watch it "disappear". We refer to this as dissolution, and we explain it as a process in which the sodium and chlorine units break away from the crystal surface, get surrounded by H2O molecules, and become hydrated ions.
$\ce{NaCl_(s) \rightarrow Na^{+}(aq) + Cl^{–}(aq)} \nonumber$
The designation (aq) means "aqueous" and comes from aqua, the Latin word for water. It is used whenever we want to emphasize that the ions are hydrated — that H2O molecules are attached to them.
Remember that solubility equilibrium and the calculations that relate to it are only meaningful when both sides (solids and dissolved ions) are simultaneously present. But if you keep adding salt, there will come a point at which it no longer seems to dissolve. If this condition persists, we say that the salt has reached its solubility limit, and the solution is saturated in NaCl. The situation is now described by
$\ce{NaCl_(s) <=>Na^{+}(aq) + Cl^{–}(aq)} \nonumber$
in which the solid and its ions are in equilibrium.
Salt solutions that have reached or exceeded their solubility limits (usually 36-39 g per 100 mL of water) are responsible for prominent features of the earth's geochemistry. They typically form when NaCl leaches from soils into waters that flow into salt lakes in arid regions that have no natural outlets; subsequent evaporation of these brines force the above equilibrium to the left, forming natural salt deposits. These are often admixed with other salts, but in some cases are almost pure NaCl. Many parts of the world contain buried deposits of NaCl (known as halite) that formed from the evaporation of ancient seas, and which are now mined.
Expressing solubilities
Solubilities are most fundamentally expressed in molar (mol L–1 of solution) or molal (mol kg–1 of water) units. But for practical use in preparing stock solutions, chemistry handbooks usually express solubilities in terms of grams-per-100 ml of water at a given temperature, frequently noting the latter in a superscript. Thus 6.9 20 means 6.9 g of solute will dissolve in 100 mL of water at 20° C.
When quantitative data are lacking, the designations "soluble", "insoluble", "slightly soluble", and "highly soluble" are used. There is no agreed-on standard for these classifications, but a useful guideline might be that shown below.
What determines solubility?
The solubilities of salts in water span a remarkably large range of values, from almost completely insoluble to highly soluble. Moreover, there is no simple way of predicting these values, or even of explaining the trends that are observed for the solubilities of different anions within a given group of the periodic table.
Ultimately, the driving force for dissolution (and for all chemical processes) is determined by the Gibbs free energy change. Dissolution of a salt is conceptually understood as a sequence of the two processes depicted above:
1. breakup of the ionic lattice of the solid (i.e., lattice energy)
2. followed by attachment of water molecules to the released ions (Solvation or Hydration Energy).
The first step consumes a large quantity of energy, something that by itself would strongly discourage solubility. But the second step releases a large amount of energy and thus has the opposite effect. Thus the net energy change depends on the sum of two large energy terms (often approaching 1000 kJ/mol) having opposite signs. Each of these terms will to some extent be influenced by the size, charge, and polarizability of the particular ions involved, and on the lattice structure of the solid. This large number of variables makes it impossible to predict the solubility of a given salt.
Nevertheless, there are some clear trends for how the solubilities of a series of salts of a given anion (such as hydroxides, sulfates, etc.) change with a periodic table group. And of course, there are a number of general solubility rules.
Solubility Rules
1. Salts containing Group I elements are soluble (Li+, Na+, K+, Cs+, Rb+). Exceptions to this rule are rare. Salts containing the ammonium ion (NH4+) are also soluble.
2. Salts containing nitrate ion (NO3-) are generally soluble.
3. Salts containing Cl -, Br -, I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are all insoluble.
4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually anything else is insoluble.
5. Most sulfate salts are soluble. Important exceptions to this rule include BaSO4, PbSO4, Ag2SO4 and SrSO4 .
6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble.
7. Most sulfides of transition metals are highly insoluble. Thus, CdS, FeS, ZnS, Ag2S are all insoluble. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
8. Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. Some other insoluble carbonates include FeCO3 and PbCO3.
9. Chromates are frequently insoluble. Examples: PbCrO4, BaCrO4
10. Phosphates are frequently insoluble. Examples: Ca3(PO4)2, Ag3PO4
11. Fluorides are frequently insoluble. Examples: BaF2, MgF2 PbF2.
A solution must be saturated to be in equilibrium with the solid. This is a necessary condition for solubility equilibrium, but it is not by itself sufficient. True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems.
Undersaturated and supersaturated solutions
If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be undersaturated. A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.
Solubility and Temperature
Solubility usually increases with temperature - but not always. This is very apparent from the solubility-vs.-temperature plots shown in Figure $1$. (Some of the plots are colored differently in order to make it easier to distinguish them where they crowd together.) The temperature dependence of any process depends on its entropy change — that is, on the degree to which thermal kinetic energy can spread throughout the system. When a solid dissolves, its component molecules or ions diffuse into the much greater volume of the solution, carrying their thermal energy along with them. So we would normally expect the entropy to increase — something that makes any process take place to a greater extent at a higher temperature.
So why does the solubility of cerium sulfate (green plot) diminish with temperature? Dispersal of the Ce3+ and SO42– ions themselves is still associated with an entropy increase, but in this case the entropy of the waterdecreases even more owing to the ordering of the H2O molecules that attach to the Ce3+ ions as they become hydrated. It's difficult to predict these effects, or explain why they occur in individual cases — but they do happen. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.1%3A_The_Nature_of_Solubility_Equilibria.txt |
Learning Objectives
• To calculate the solubility of an ionic compound from its Ksp
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
The Solubility Product
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}$
As you will discover in more advanced chemistry courses, basic anions, such as S2−, PO43−, and CO32−, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}$
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43− ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table $1$, which shows that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equations $\ref{Eq2a}$ and $\ref{Eq2b}$, changes in pH can affect the solubility of a compound as discussed later.
As with K, the concentration of a pure solid does not appear explicitly in Ksp.
Table $1$: Solubility Products for Selected Ionic Substances at 25°C
Solid Color Ksp Solid Color Ksp
*These contain the Hg22+ ion.
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, $K_{sp}$, like $K$, is defined in terms of the molar concentrations of the component ions.
A color photograph of a kidney stone, 8 mm in length. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O2CCO2)·H2O and Ca3(PO4)2. from Wikipedia.
Example $1$: Kidney Stones
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: Ksp
Strategy:
1. Write the balanced dissolution equilibrium and the corresponding solubility product expression.
2. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2−) are as follows:
$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2−] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2− ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \nonumber$
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise $1$
One crystalline form of calcium carbonate (CaCO3) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer 4.5 × 10−9
The reaction of weakly basic anions with $H_2O$ tends to make the actual solubility of many salts higher than predicted.
A crystal of calcite (CaCO3), illustrating the phenomenon of double refraction. When a transparent crystal of calcite is placed over a page, we see two images of the letters. Image used with permisison from Wikipedia
Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example $1$. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations (ICE Tables), remembering that the concentration of the pure solid is essentially constant.
Example $2$
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C
Given: Ksp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
1. Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
2. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
1. A The dissolution equilibrium for Ca3(PO4)2 (Equation $\ref{Eq2a}$) is shown in the following ICE table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43− ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43−] will be +2x. We can insert these values into the table.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq)
$\ce{Ca3(PO4)2}$ $[\ce{Ca^{2+}}]$ $\ce{[PO4^{3−}}]$
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):
\begin{align*}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \2.07\times10^{-33}&=108x^5 \1.92\times10^{-35}&=x^5 \1.14\times10^{-7}\textrm{ M}&=x\end{align*}
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43−] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
1. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
$\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}$
Exercise $2$
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature
Answer a
1.28 × 10−4 M
Answer a
3.54 mg
Relating Solubilities to Solubility Constants
The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration.
For example, let us denote the solubility of Ag2CrO4 as S mol L–1. Then for a saturated solution, we have
• $[Ag^+] = 2S$
• $[CrO_4^{2–}] = S$
Substituting this into Eq 5b above,
$(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}$
$S= \left( dfrac{K_{sp}}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}$
thus the solubility is $8.8 \times 10^{–5}\; M$.
Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A2B and AB2, say, on the basis of their Ks values.
It is meaningless to compare the solubilities of two salts having different formulas on the basis of their Ks values.
Example $2$
The solubility of CaF2 (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of Ks under these conditions.
Solution
moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = $2.05 \times 10^{-5}$ mol
\begin{align*} S &= \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} \[4pt] &= 2.05 \times 10^{-4} M \end{align*}
\begin{align*}K_{sp} &= [Ca^{2+}][F^–]^2 \[4pt] &= (S)(2S)^2 \[4pt] &= 4 × (2.05 \times 10^{–4})^3 \[4pt] &= 3.44 \times 10^{–11} \end{align*}
Example $3$
Estimate the solubility of La(IO3)3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 10–12.
Solution
The equation for the dissolution is
$\ce{La(IO_3)_3 <=> La^{3+ } + 3 IO3^{–}} \nonumber$
If the solubility is S, then the equilibrium concentrations of the ions will be
[La3+] = S and [IO3] = 3S. Then Ks = [La3+][IO3]3 = S(3S)3 = 27S4
27S4 = 6.2 × 10–12, S = ( ( 6.2 ÷ 27) × 10–12 )¼ = 6.92 × 10–4 M
[IO3] = 3S = 2.08 × 10–5 (M)
Example $4$: Cadmium
Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH)2 (Ks = 2.5E–14). If 1000 L of a certain wastewater contains Cd2+ at a concentration of 1.6E–5 M, what concentration of Cd2+ would remain after addition of 10 L of 4 M NaOH solution?
Solution
As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.
Volume of treated water: 1000 L + 10 L = 1010 L
Concentration of OH on addition to 1000 L of pure water:
(4 M) × (10 L)/(1010 L) = 0.040 M
Initial concentration of Cd2+ in 1010 L of water:
$(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M$
The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)2 is formed — that is, all of the Cd2+ gets precipitated.
Concentrations [Cd2+], M [OH], M
initial 1.6E–5 0.04
change –1.6E–5 –3.2E–5
final: 0 0.04 – 3.2E–5 ≈ .04
Now "turn on the equilibrium" — find the concentration of Cd2+ that can exist in a 0.04M OH solution:
Concentrations [Cd2+], M [OH], M
initial o 0.04
change +x +2x
at equilibrium x .04 + 2x.04
Substitute these values into the solubility product expression:
Cd(OH)2(s) = [Cd2+] [OH]2 = 2.5E–14
[Cd2+] = (2.5E–14) / (16E–4) = 1.6E–13 M
Note that the effluent will now be very alkaline:
$pH = 14 + log 0.04 = 12.6$
so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.
Summary
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.2%3A_Ionic_Equilibria_between_Solids_and_Solutions.txt |
Learning Objectives
• Define Ksp, the solubility product.
• Explain solid/solution equilibria using $K_{sp}$ and Qsp.
• Calculate molarity of saturated solution from Ksp.
• Calculate $K_{sp}$ from molarity of saturated solution.
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation. The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions.
Properties of Precipitates
Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid (or just the supernate). The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting.
Figure $1$: Above is a diagram of the formation of a precipitate in solution. (Public Domain; ZabMilenko)
The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as a double displacement reaction where the partners "switching; that is, the two reactants each "lose" their partner and form a bond with a different partner:
A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:
$\ce{CdSO4(aq) + K2S (aq) \rightarrow CdS (s) + K2SO4(aq)}$
Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble. However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution.
Solubility Rules
Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule.
1. Salts formed with group 1 cations and $\ce{NH_4^{+}}$ cations are soluble. There are some exceptions for certain $Li^+$ salts.
2. Acetates ($\ce{C2H3O2^{-}}$), nitrates ($\ce{NO3^{-}}$), and perchlorates ($\ce{ClO4^{-}}$) are soluble.
3. Bromides, chlorides, and iodides are soluble.
4. Sulfates ($\ce{SO4^{2-}}$) are soluble with the exception of sulfates formed with $\ce{Ca^{2+}}$, $\ce{Sr^{2+}}$, and $\ce{Ba^{2+}}$.
5. Salts containing silver, lead, and mercury (I) are insoluble.
6. Carbonates ($\ce{CO3^{2-}}$), phosphates ($\ce{PO4^{3-}}$), sulfides, oxides, and hydroxides ($\ce{OH^{-}}$) are insoluble. Sulfides formed with group 2 cations and hydroxides formed with calcium, strontium, and barium are exceptions.
If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.
Net Ionic Equations
To understand the definition of a net ionic equation, recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair:
$\color{blue}{A}\color{red}{B}\color{black} (aq) + \color{blue}{C}\color{red}{D}\color{black} (aq) → \color{blue}{A}\color{red}{D}\color{black} (aq) \color{black}+ \color{blue}{C}\color{red}{B}\color{black} (s)$
The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below:
$\color{blue}{A}^+ \color{black} (aq) + \color{red}{B}\color{black}^- (aq) + \color{blue}{C}\color{black}^+ (aq) + \color{red}{D}^-\color{black} (aq) → \color{blue}{A}^+\color{black} (aq) + \color{red}{D}^-\color{black} (aq) + \color{blue}{C}\color{red}{B}\color{black} (s)$
In the equation above, A+ and D- ions are present on both sides of the equation. These are called spectator ions because they remain unchanged throughout the reaction. Since they go through the equation unchanged, they can be eliminated to show the net ionic equation:
$\color{red}{B}\color{black}^- (aq) + \color{blue}{C}\color{black}^+ (aq) → + \color{blue}{C}\color{red}{B}\color{black} (s)$
The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs
Equilibrium and non-Equilibrium Conditions
The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibria. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations. An ion product can in principle have any positive value, depending on the concentrations of the ions involved. Only in the special case when its value is identical with Ks does it become the solubility product. A solution in which this is the case is said to be saturated. Thus when
$[\ce{Ag^{+}}]^2 [\ce{CrO4^{2-}}] = 2.76 \times 10^{-12}$
at the temperature and pressure at which this value $K_{sp}$ of applies, we say that the "solution is saturated in silver chromate".
The ion product $Q$ is analogous to the reaction quotient $Q$ for gaseous equilibria.
As summarized in Figure $3$, there are three possible conditions for an aqueous solution of an ionic solid:
• $Q < K_{sp}$. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
• $Q = K_{sp}$. The solution is saturated and at equilibrium.
• $Q > K_{sp}$. The solution is supersaturated, and ionic solid will precipitate.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Example $3$: barium milkshakes
Barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is $1.08 \times 10^{−10}$ at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that $\ce{NaCl}$ is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
1. Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
2. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
3. Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution
A The only slightly soluble salt that can be formed when these two solutions are mixed is $\ce{BaSO4}$ because $\ce{NaCl}$ is highly soluble. The equation for the precipitation of $\ce{BaSO4}$ is as follows:
$\ce{BaSO4(s) <=> Ba^{2+} (aq) + SO^{2−}4(aq)} \nonumber$
The solubility product expression is as follows:
$K_{sp} = [\ce{Ba^{2+}}][\ce{SO4^{2−}}] = 1.08 \times 10^{−10} \nonumber$
B To solve this problem, we must first calculate the ion product:
$Q = [\ce{Ba^{2+}}][\ce{SO4^{2−}}] \nonumber$
using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of $\ce{BaCl2}$ solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
\begin{align*} \textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+} \[4pt] [\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+} \end{align*}
Similarly, the concentration of SO42− after mixing is the total number of moles of SO42− in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
\begin{align*} \textrm{moles SO}_4^{2-} &=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-} \[4pt] [\mathrm{SO_4^{2-}}] &=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-} \end{align*}
We can now calculate $Q$:
$Q = [\ce{Ba^{2+}}][\ce{SO4^{2−}}] = (2.9 \times 10^{−4})(1.8 \times 10^{−4}) = 5.2 \times 10^{−8} \nonumber$
C We now compare $Q$ with the $K_{sp}$. If Q > Ksp, then $\ce{BaSO4}$ will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that $\ce{BaSO4}$ will precipitate when the two solutions are mixed. In fact, $\ce{BaSO4}$ will continue to precipitate until the system reaches equilibrium, which occurs when
$[\ce{Ba^{2+}}][\ce{SO4^{2−}}] = K_{sp} = 1.08 \times 10^{−10}. \nonumber$
Exercise $3$
The solubility product of calcium fluoride ($\ce{CaF2}$) is $3.45 \times 10^{−11}$. If 2.0 mL of a 0.10 M solution of $\ce{NaF}$ is added to 128 mL of a $2.0 \times 10^{−5}\,M$ solution of $\ce{Ca(NO3)2}$, will $\ce{CaF2}$ precipitate?
Answer
Yes, since $Q_{sp} = 4.7 \times 10^{−11} > K_{sp}$.
A solution must be saturated to be in equilibrium with the solid. This is a necessary condition for solubility equilibrium, but it is not by itself sufficient. True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems.
Undersaturated and supersaturated solutions
If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be undersaturated. A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.
How to know the saturation status of a solution? Just comparing the ion product Qs with the solubility product Ksp. as shown in Table $1$.
Table $1$: relationship among $Q_{sp}$, $K_{sp}$ and saturation
$Q_{sp}/K_{sp}$ Status
> 1 Product concentration too high for equilibrium; net reaction proceeds to left.
= 1 System is at equilibrium; no net change will occur.
< 1 Product concentration too low for equilibrium; net reaction proceeds to right.
For example, for the system
$\ce{Ag2CrO4(s) <=> 2 Ag^{+} + CrO_4^{2–}} \label{4ba}$
a solution in which Qs < Ks (i.e., Ks /Qs > 1) is undersaturated (blue shading) and the no solid will be present. The combinations of [Ag+] and [CrO42–] that correspond to a saturated solution (and thus to equilibrium) are limited to those described by the curved line. The pink area to the right of this curve represents a supersaturated solution.
For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, Qsp is greater than Ksp. When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring $Q_{sp}$ to approach $K_{sp}$. For example Sodium acetate trihydrate, $\ce{NaCH3COO\cdot 3H2O}$, when heated to 370 K will become a liquid and stays as a liquid when cooled to room temperature or even below 273 K (Video $1$). As soon as a seed crystal is present, crystallization occurs rapidly. In such a process, heat is released since this is an exothermic process $\Delta H < 0$.
Video $1$: "hot ice" (sodium acetate) crystallized from a non-equilibrium supersaturated ($Q_{sp} > K_{sp}$) solution
Example $1$
A sample of groundwater that has percolated through a layer of gypsum ($\ce{CaSO4}$) with $K_{sp} = 4.9 \times 10^{–5} = 10^{–4.3}$) is found to have be $8.4 \times 10^{–5}\; M$ in Ca2+ and $7.2 \times 10^{–5}\; M$ in SO42–. What is the equilibrium state of this solution with respect to gypsum?
Solution
The ion product
$Q_s = (8.4 \times 10^{–5})(7.2 \times 10^{-5}) = 6.0 \times 10^{–4} \nonumber$
exceeds $K_{sp}$, so the ratio $K_{sp} /Q_{sp} > 1$ and the solution is supersaturated in $\ce{CaSO_4}$.
Relating Solubilities to Solubility Constants
The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration.
For example, let us denote the solubility of Ag2CrO4 as S mol L–1. Then for a saturated solution, we have
• $[Ag^+] = 2S$
• $[CrO_4^{2–}] = S$
Substituting this into Eq 5b above,
$(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}$
$S= \left( \dfrac{K_{sp}}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}$
thus the solubility is $8.8 \times 10^{–5}\; M$.
Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A2B and AB2, say, on the basis of their Ks values.
It is meaningless to compare the solubilities of two salts having different formulas on the basis of their Ks values.
Example $2$
The solubility of CaF2 (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of Ks under these conditions.
Solution
moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = $2.05 \times 10^{-5}$ mol
$S = \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} = 2.05 \times 10^{-4} M$
$K_{sp}= [Ca^{2+}][F^–]^2 = (S)(2S)^2 = 4 × (2.05 \times 10^{–4})^3 = 3.44 \times 10^{–11}$
Example $3$
Estimate the solubility of La(IO3)3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 10–12.
Solution
The equation for the dissolution is
$La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–$
If the solubility is S, then the equilibrium concentrations of the ions will be
[La3+] = S and [IO3] = 3S. Then Ks = [La3+][IO3]3 = S(3S)3 = 27S4
27S4 = 6.2 × 10–12, S = ( ( 6.2 ÷ 27) × 10–12 )¼ = 6.92 × 10–4 M
[IO3] = 3S = 2.08 × 10–5 (M)
Example $4$: Cadmium
Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH)2 (Ks = 2.5E–14). If 1000 L of a certain wastewater contains Cd2+ at a concentration of 1.6E–5 M, what concentration of Cd2+ would remain after addition of 10 L of 4 M NaOH solution?
Solution
As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.
Volume of treated water: 1000 L + 10 L = 1010 L
Concentration of OH on addition to 1000 L of pure water:
(4 M) × (10 L)/(1010 L) = 0.040 M
Initial concentration of Cd2+ in 1010 L of water:
$(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M$
The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)2 is formed — that is, all of the Cd2+ gets precipitated.
Concentrations $[\ce{Cd^{2+}}],\, M$ $[\ce{OH^{–}}],\, M$
initial 1.6E–5 0.04
change –1.6E–5 –3.2E–5
final: 0 0.04 – 3.2E–5 ≈ .04
Now "turn on the equilibrium" — find the concentration of Cd2+ that can exist in a 0.04M OH solution:
Concentrations $[\ce{Cd^{2+}}],\, M$ $[\ce{OH^{–}}],\, M$
initial o 0.04
change +x +2x
at equilibrium x .04 + 2x.04
Substitute these values into the solubility product expression:
Cd(OH)2(s) = [Cd2+] [OH]2 = 2.5E–14
[Cd2+] = (2.5E–14) / (16E–4) = 1.6E–13 M
Note that the effluent will now be very alkaline:
$pH = 14 + \log 0.04 = 12.6$
so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.
The Common Ion Effect
It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. This is just what would be expected on the basis of the Le Chatelier Principle; whenever the process
$CaF_{2(s)} \rightleftharpoons Ca^{2+} + 2 F^– \label{7}$
is in equilibrium, addition of more fluoride ion (in the form of highly soluble NaF) will shift the composition to the left, reducing the concentration of Ca2+, and thus effectively reducing the solubility of the solid. We can express this quantitatively by noting that the solubility product expression
$[Ca^{2+}][F^–]^2 = 1.7 \times 10^{–10} \label{8}$
must always hold, even if some of the ionic species involved come from sources other than CaF2(s). For example, if some quantity x of fluoride ion is added to a solution initially in equilibrium with solid CaF2, we have
• $[Ca^{2+}] = S$
• $[F^–] = 2S + x$
so that
$K_{sp} = [Ca^{2+}][ F^–]^2 = S (2S + x)^2 . \label{9a}$
University-level students should be able to derive these relations for ion-derived solids of any stoichiometry. In most practical cases x will be large compared to S so that the 2S term can be dropped and the relation becomes
$K_{sp} ≈ S x^2$
$S = \dfrac{K_{sp}}{x^2} \label{9b}$
The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as Na2CrO4.
What's different about the plot on the right? If you look carefully at the scales, you will see that this one is plotted logarithmically (that is, in powers of 10.) Notice how a much wider a range of values can display on a logarithmic plot. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations (such as that made in Equation $\ref{9b}$) can yield straight-lines within the range of values for which the approximation is valid.
Example $5$: strontium sulfate
Calculate the solubility of strontium sulfate (Ks = 2.8 × 10–7) in
1. pure water and
2. in a 0.10 mol L–1 solution of Na2SO4.
Solution:
(a) In pure water, Ks = [Sr2+][SO42–] = S2
S = √Ks = (2.8 × 10–7)½ = 5.3 × 10–4
(b) In 0.10 mol L–1 Na2SO4, we have
Ks = [Sr2+][SO42–] = S × (0.10 + S) = 2.8 × 10–7
Because S is negligible compared to 0.10 M, we make the approximation
Ks = [Sr2+][SO42–] ≈ S × (0.10 M) = 2.8 × 10–7
so S ≈ (2.8 × 10–7) / 0.10M = 2.8 × 10–6 M — which is roughly 100 times smaller than the result from (a).
Note
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example $6$
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
1. Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
2. Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43−] is +2x. We can insert these values into the ICE table.
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+} (aq) + 2PO^{3−}_{4(aq)}$
$\ce{Ca3(PO4)2}$ $\ce{[Ca2+]}$ $\ce{[PO43−]}$
initial pure solid 0.20 0
change +3x +2x
final pure solid 0.20 + 3x 2x
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43−]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\begin{align}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33} \x^2&=6.5\times10^{-32} \x&=2.5\times10^{-16}\textrm{ M}\end{align}
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example $2$—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise $6$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
Applications and Examples
Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied.
Example $7$
Complete the double replacement reaction and then reduce it to the net ionic equation.
$NaOH (aq) + MgCl_{2 \;(aq)} \rightarrow \nonumber$
First, predict the products of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”).
$2NaOH (aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2 \nonumber$
Second, consult the solubility rules to determine if the products are soluble. Group 1 cations ($Na^+$) and chlorides are soluble from rules 1 and 3 respectively, so $NaCl$ will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus $Mg(OH)_2$ will form a precipitate. The resulting equation is the following:
$2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl (aq) + Mg(OH)_{2\;(s)} \nonumber$
Third, separate the reactants into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms:
$2Na^+ (aq) + 2OH^- (aq) + Mg^{2+} (aq) + 2Cl^- (aq) \rightarrow Mg(OH)_{2\;(s)} + 2Na^+ (aq) + 2Cl^- (aq) \nonumber$
Lastly, eliminate the spectator ions (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final net ionic equation is:
$Mg^{2+} (aq) + 2OH^- (aq) \rightarrow Mg(OH)_{2(s)} \nonumber$
Example $8$
Complete the double replacement reaction and then reduce it to the net ionic equation.
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow \nonumber$
Solution
The predicted products of this reaction are $CoSO_4$ and $NaCl$. From the solubility rules, $CoSO_4$ is soluble because rule 4 states that sulfates ($SO_4^{2-}$) are soluble. Similarly, we find that $NaCl$ is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows:
$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl (aq) \nonumber$
Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.):
Co2- (aq) + 2Cl-(aq) + 2Na+ (aq) + SO42-(aq) Co2- (aq) + SO42-(aq) + 2Na+ (aq) + 2Cl-(aq)
No precipitation reaction
This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs.
Example $9$
Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations.
1. $Fe(NO_3)_{3\;(aq)} + NaOH (aq) \rightarrow$
2. $Al_2(SO_4)_{3\;(aq)} + BaCl_{2\;(aq)} \rightarrow$
3. $HI (aq) + Zn(NO_3)_{2\;(aq)} \rightarrow$
4. $CaCl_{2\;(aq)} + Na_3PO_{4\;(aq)} \rightarrow$
5. $Pb(NO_3)_{2\;(aq)} + K_2SO_{4 \;(aq)} \rightarrow$
Solutions
a. Regardless of physical state, the products of this reaction are $Fe(OH)_3$ and $NaNO_3$. The solubility rules predict that $NaNO_3$ is soluble because all nitrates are soluble (rule 2). However, $Fe(OH)_3$ is insoluble, because hydroxides are insoluble (rule 6) and $Fe$ is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows:
$Fe^{3+} (aq) + NO^-_{3\;(aq)} + Na^+ (aq) + 3OH^- (aq) \rightarrow Fe(OH)_{3\;(s)} + Na^+ (aq) + NO^-_{3\;(aq)} \nonumber$
Canceling out spectator ions leaves the net ionic equation:
$Fe^{3+} (aq) + OH^- (aq) \rightarrow Fe(OH)_{\;3(s)} \nonumber$
b. From the double replacement reaction, the products are $AlCl_3$ and $BaSO_4$. $AlCl_3$ is soluble because it contains a chloride (rule 3); however, $BaSO_4$ is insoluble: it contains a sulfate, but the $Ba^{2+}$ ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing):
$2Al^{3+} (aq) + 6Cl^- (aq) + 3Ba^{2+} (aq) + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+} (aq) +6Cl^- (aq) + 3BaSO_{4\;(s)} \nonumber$
Canceling out spectator ions leaves the following net ionic equation:
$Ba^{2+} (aq) + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)} \nonumber$
c. From the double replacement reaction, the products $HNO_3$ and $ZnI_2$ are formed. Looking at the solubility rules, $HNO_3$ is soluble because it contains nitrate (rule 2), and $ZnI_2$ is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs.
d. The products of this double replacement reaction are $Ca_3(PO_4)_2$ and $NaCl$. Rule 1 states that $NaCl$ is soluble, and according to solubility rule 6, $Ca_3(PO_4)_2$ is insoluble. The ionic equation is:
$Ca^{2+} (aq) + Cl^- (aq) + Na^+ (aq) + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+ (aq) + Cl^- (aq) \nonumber$
After canceling out spectator ions, the net ionic equation is given below:
$Ca^{2+} (aq) + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} \nonumber$
e. The first product of this reaction, $PbSO_4$, is soluble according to rule 4 because it is a sulfate. The second product, $KNO_3$, is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs.
Summary
In contrast to $K_{sp}$, the ion product ($Q_{sp}$) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.3%3A_Precipitation_and_the_Solubility_Product.txt |
Learning Objectives
• To understand why the solubility of many compounds depends on pH.
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH.
The Effect of Acid–Base Equilibria the Solubility of Salts
We begin our discussion by examining the effect of pH on the solubility of a representative salt, $\ce{M^{+}A^{−}}$, where $\ce{A^{−}}$ is the conjugate base of the weak acid $\ce{HA}$. When the salt dissolves in water, the following reaction occurs:
$\ce{MA (s) \rightleftharpoons M^{+} (aq) + A^{-} (aq)} \label{17.13a} \nonumber$
with
$K_{sp} = [\ce{M^{+}}][\ce{A^{−}}] \label{17.13b} \nonumber$
The anion can also react with water in a hydrolysis reaction:
$\ce{A^{-} (aq) + H2O (l) \rightleftharpoons OH^{-} (aq) + HA (aq)} \label{17.14}$
Because of the reaction described in Equation $\ref{17.14}$, the predicted solubility of a sparingly soluble salt that has a basic anion such as S2−, PO43, or CO32 is increased. If instead a strong acid is added to the solution, the added H+ will react essentially completely with A to form HA. This reaction decreases [A], which decreases the magnitude of the ion product
$Q = [\ce{M^{+}}][\ce{A^{-}}]$
According to Le Chatelier’s principle, more MA will dissolve until $Q = K_{sp}$. Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)2 is relatively insoluble in water:
$Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+} (aq) + 2OH^− (aq) \label{17.15a}$
with
$K_{sp} = 5.61 \times 10^{−12} \label{17.15b}$
When acid is added to a saturated solution that contains excess solid Mg(OH)2, the following reaction occurs, removing OH from solution:
$H^+ (aq) + OH^− (aq) \rightarrow H_2O (l) \label{17.16}$
The overall equation for the reaction of Mg(OH)2 with acid is thus
$Mg(OH)_{2(s)} + 2H^+ (aq) \rightleftharpoons Mg^{2+} (aq) + 2H_2O (l) \label{17.17}$
As more acid is added to a suspension of Mg(OH)2, the equilibrium shown in Equation $\ref{17.17}$ is driven to the right, so more Mg(OH)2 dissolves.
Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF2 is a sparingly soluble salt:
$CaF_{2(s)} \rightleftharpoons Ca^{2+} (aq) + 2F^− (aq) \label{17.18a}$
with
$K_{sp} = 3.45 \times 10^{−11} \label{17.18b}$
When strong acid is added to a saturated solution of CaF2, the following reaction occurs:
$H^+ (aq) + F^− (aq) \rightleftharpoons HF (aq) \label{17.19}$
Because the forward reaction decreases the fluoride ion concentration, more CaF2 dissolves to relieve the stress on the system. The net reaction of CaF2 with strong acid is thus
$CaF_{2(s)} + 2H^+ (aq) \rightarrow Ca^{2+} (aq) + 2HF (aq) \label{17.20}$
Example $1$ shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt.
Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution.
Example $1$
Lead oxalate (PbC2O4), lead iodide (PbI2), and lead sulfate (PbSO4) are all rather insoluble, with Ksp values of 4.8 × 10−10, 9.8 × 10−9, and 2.53 × 10−8, respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities?
Given: Ksp values for three compounds
Asked for: relative solubilities in acid solution
Strategy:
Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions.
Solution
The solubility Equilibria for the three salts are as follows:
$PbC_2O_{4(s)} \rightleftharpoons Pb^{2+} (aq) + C_2O^{2−}_{4(aq)} \nonumber$
$PbI_{2(s)} \rightleftharpoons Pb^{2+} (aq) + 2I^− (aq) \nonumber$
$PbSO_{4(s)} \rightleftharpoons Pb^{2+} (aq) + SO^{2−}_{4(aq)} \nonumber$
The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI2 will not greatly affect its solubility; the acid will simply dissociate to form H+(aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO2CCO2H), which is a weak diprotic acid (pKa1 = 1.23 and pKa2 = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions:
$C_2O^{2−}_{4(aq)} + H^+ (aq) \rightarrow HO_2CCO^−_{2(aq)} \nonumber$
$HO_2CCO^−_{2(aq)} + H^+ (aq) \rightarrow HO_2CCO_2H (aq) \nonumber$
These reactions will decrease [C2O42], causing more lead oxalate to dissolve to relieve the stress on the system. The pKa of HSO4 (1.99) is similar in magnitude to the pKa1 of oxalic acid, so adding a strong acid to a saturated solution of PbSO4 will result in the following reaction:
$SO^{2-}_{4(aq)} + H^+ (aq) \rightleftharpoons HSO^-_{4(aq)} \nonumber$
Because HSO4 has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO4 will be significantly less than for PbC2O4.
Exercise $1$
Which of the following insoluble salts—AgCl, Ag2CO3, Ag3PO4, and/or AgBr—will be substantially more soluble in 1.0 M HNO3 than in pure water?
Answer
Ag2CO3 and Ag3PO4
Solubility Products and pH: https://youtu.be/XJ0s5SATZgQ
Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility Equilbria(part (a) in Figure $1$:). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows:
$\ce{CO2(aq) + H2O (l) \rightleftharpoons H^{+} (aq) + HCO^{−}3(aq)} \label{17.21}$
$\ce{HCO^{−}3(aq) \rightleftharpoons H^{+} (aq) + CO^{2-}3(aq)} \label{17.22}$
$\ce{Ca^{2+} (aq) + CO^{2−}3(aq) \rightleftharpoons CaCO3(s)} \label{17.23}$
Limestone deposits that form caves consist primarily of CaCO3 from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO3 in CO2-rich water rises toward Earth’s surface or is otherwise heated, CO2 gas is released as the water warms. CaCO3 then precipitates from the solution according to the following equation (part (b) in Figure $1$:):
$Ca^{2+} (aq) + 2HCO^−_{3(aq)} \rightleftharpoons CaCO_{3(s)} + CO_{2(g)} + H_2O (l) \label{17.24}$
The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated.
When groundwater-containing atmospheric CO2 (Equations $\ref{17.21}$ and $\ref{17.22}$) finds its way into microscopic cracks in the limestone deposits, CaCO3 dissolves in the acidic solution in the reverse direction of Equation $\ref{17.24}$. The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in Equation $\ref{17.24}$ to shift to the right. A circular layer of solid CaCO3 is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China (Figure $2$).
Acidic, Basic, and Amphoteric Oxides and Hydroxides
One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxides and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxides or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. There is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:
$Cs_2O (s) + H_2O (l) \rightarrow 2Cs^+ (aq) + 2OH^− (aq) \label{17.25}$
$SO_{3(g)} + H_2O (l) \rightarrow H_2SO_{4(aq)} \label{17.26}$
Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed
Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions.
The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete Mn+ cations and O2− anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: Eδ+–Oδ−. The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO3, which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO42):
$MoO_{3(s)} + 2OH^− (aq) \rightarrow MoO^{2−}_{4(aq)} + H_2O (l) \label{17.27}$
As shown in Figure $3$, there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called amphoteric oxides (from the Greek ampho, meaning “both,” as in amphiprotic). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in Video $1$, for example, mixing the amphoteric oxide Cr(OH)3 (also written as Cr2O3•3H2O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH)3 to dissolve to give a bright violet solution of Cr3+(aq), which contains the [Cr(H2O)6]3+ ion, whereas adding strong base gives a green solution of the [Cr(OH)4] ion. The chemical equations for the reactions are as follows:
$\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \label{17.28}$
$\mathrm{Cr(OH)_3(s)}+\mathrm{OH^-(aq)}\rightarrow\underset{\textrm{green}}{\mathrm{[Cr(OH)_4]^-}}\mathrm{(aq)}\label{17.29}$
Video $1$: Chromium(III) Hydroxide [Cr(OH)3 or Cr2O3•3H2O] is an Example of an Amphoteric Oxide. All three beakers originally contained a suspension of brownish purple Cr(OH)3(s) (center). When concentrated acid (6 M H2SO4) was added to the beaker on the left, Cr(OH)3 dissolved to produce violet [Cr(H2O)6]3+ ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH)3 to dissolve, producing green [Cr(OH)4]ions. For a more complete description, see https://www.youtube.com/watch?v=IQNcLH6OZK0
Example $2$
Aluminum hydroxide, written as either Al(OH)3 or Al2O3•3H2O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base.
Given: amphoteric compound
Asked for: dissolution reactions in acid and base
Strategy:
Using Equations $\ref{17.28}$ and $\ref{17.29}$ as a guide, write the dissolution reactions in acid and base solutions.
Solution
1. An acid donates protons to hydroxide to give water and the hydrated metal ion, so aluminum hydroxide, which contains three OH ions per Al, needs three H+ ions:
$Al(OH)_{3(s)} + 3H^+ (aq) \rightarrow Al^{3+} (aq) + 3H_2O (l) \nonumber$
In aqueous solution, Al3+ forms the complex ion [Al(H2O)6]3+.
1. In basic solution, OH is added to the compound to produce a soluble and stable poly(hydroxo) complex:
$Al(OH)_{3(s)} + OH^− (aq) \rightarrow [Al(OH)_4]^− (aq) \nonumber$
Exercise $2$
Copper(II) hydroxide, written as either Cu(OH)2 or CuO•H2O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base.
Answer
$Cu(OH)_{2(s)} + 2H^+ (aq) \rightarrow Cu^{2+} (aq) + 2H_2O (l) \nonumber$
$Cu(OH)_{2(s)} + 2OH^− (aq) \rightarrow [Cu(OH)_4]^2_{−(aq)} \nonumber$
Selective Precipitation Using pH
Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate.
The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations.
Suppose, for example, we have a solution that contains 1.0 mM Zn2+ and 1.0 mM Cd2+ and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilbria can be written as follows:
$ZnS (s) \rightleftharpoons Zn^{2+} (aq) + S^{2−} (aq) \label{17.30a}$
with
$K_{sp}= 1.6 \times 10^{−24} \label{17.30b}$
and
$CdS (s) \rightleftharpoons Cd^{2+} (aq) + S^{2−} (aq) \label{17.31a}$
with
$K_{sp} = 8.0 \times 10^{−27} \label{17.31b}$
Because the S2− ion is quite basic and reacts extensively with water to give HS and OH, the solubility equilbria are more accurately written as $MS (s) \rightleftharpoons M^{2+} (aq) + HS^− (aq) + OH^−$ rather than $MS (s) \rightleftharpoons M^{2+} (aq) + S^{2−} (aq)$. Here we use the simpler form involving S2−, which is justified because we take the reaction of S2− with water into account later in the solution, arriving at the same answer using either equilibrium equation.
The sulfide concentrations needed to cause $ZnS$ and $CdS$ to precipitate are as follows:
$K_{sp} = [Zn^{2+}][S^{2−}] \label{17.32a}$
$1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\label{17.32b}$
$1.6 \times 10^{−21}\; M = [S^{2−}]\label{17.32c}$
and
$K_{sp} = [Cd^{2+}][S^{2−}] \label{17.33a}$
$8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\label{17.33b}$
$8.0 \times 10^{−24}\; M = [S^{2−}] \label{17.33c}$
Thus sulfide concentrations between 1.6 × 10−21 M and 8.0 × 10−24 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H2S contains 0.10 M H2S at 20°C. The pKa1for H2S is 6.97, and pKa2 corresponding to the formation of [S2−] is 12.90. The equations for these reactions are as follows:
$H_2S (aq) \rightleftharpoons H^+ (aq) + HS^− (aq) \label{17.34a}$
with
$pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \label{17.34b}$
$HS^− (aq) \rightleftharpoons H^+ (aq) + S^{2−} (aq) \label{17.34c}$
with
$pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \label{17.34d}$
We can show that the concentration of S2− is 1.3 × 10−13 by comparing Ka1 and Ka2 and recognizing that the contribution to [H+] from the dissociation of HS is negligible compared with [H+] from the dissociation of H2S. Thus substituting 0.10 M in the equation for Ka1 for the concentration of H2S, which is essentially constant regardless of the pH, gives the following:
$K_{\textrm{a1}}=1.1\times10^{-7}=\dfrac{[\mathrm{H^+}][\mathrm{HS^-}]}{[\mathrm{H_2S}]}=\dfrac{x^2}{0.10\textrm{ M}} \x=1.1\times10^{-4}\textrm{ M}=[\mathrm{H^+}]=[\mathrm{HS^-}] \label{17.35}$
Substituting this value for [H+] and [HS] into the equation for Ka2,
$K_{\textrm{a2}}=1.3\times10^{-13}=\dfrac{[\mathrm{H^+}][\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}=\dfrac{(1.1\times10^{-4}\textrm{ M})x}{1.1\times10^{-4}\textrm{ M}}=x=[\mathrm{S^{2-}}]$
Although [S2−] in an H2S solution is very low (1.3 × 10−13 M), bubbling H2S through the solution until it is saturated would precipitate both metal ions because the concentration of S2− would then be much greater than 1.6 × 10−21 M. Thus we must adjust [S2−] to stay within the desired range. The most direct way to do this is to adjust [H+] by adding acid to the H2S solution (recall Le Chatelier's principle), thereby driving the equilibrium in Equation $\ref{17.34d}$ to the left. The overall equation for the dissociation of H2S is as follows:
$H_2S (aq) \rightleftharpoons 2H^+ (aq) + S^{2−} (aq) \label{17.36}$
Now we can use the equilibrium constant K for the overall reaction, which is the product of Ka1 and Ka2, and the concentration of H2S in a saturated solution to calculate the H+ concentration needed to produce [S2−] of 1.6 × 10−21 M:
$K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \label{17.37}$
\begin{align}\mathrm{[H^+]^2}=\dfrac{K[\mathrm{H_2S}]}{[\mathrm{S^{2-}}]}=\dfrac{(1.4\times10^{-20})(\textrm{0.10 M})}{1.6\times10^{-21}\textrm{ M}}&=0.88 \ [\mathrm{H^+}]&=0.94\end{align} \label{17.38}
Thus adding a strong acid such as HCl to make the solution 0.94 M in H+ will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H2S.
Example $3$
A solution contains 0.010 M Ca2+ and 0.010 M La3+. What concentration of HCl is needed to precipitate La2(C2O4)3•9H2O but not Ca(C2O4)•H2O if the concentration of oxalic acid is 1.0 M? Ksp values are 2.32 × 10−9 for Ca(C2O4) and 2.5 × 10−27 for La2(C2O4)3; pKa1 = 1.25 and pKa2 = 3.81 for oxalic acid.
Given: concentrations of cations, Ksp values, and concentration and pKa values for oxalic acid
Asked for: concentration of HCl needed for selective precipitation of La2(C2O4)3
Strategy:
1. Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La2(C2O4)3•9H2O.
2. Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox2] needed to precipitate La2(C2O4)3•9H2O into the overall equation for the dissociation of oxalic acid to calculate the required [H+].
Solution
A Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox2 for oxalate, we write the solubility product expression for calcium oxalate as follows:
$K_{sp} = [Ca^{2+}][ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \nonumber$
$[ox^{2−}] = 2.32 \times 10^{−7}\; M \nonumber$
The expression for lanthanum oxalate is as follows:
$K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \nonumber$
$[ox^{2−}] = 2.9 \times 10^{−8}\; M \nonumber$
Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between $2.9 \times 10^{−8} M$ and $2.32 \times 10^{−7} M$.
B To prevent Ca2+ from precipitating as calcium oxalate, we must add enough H+ to give a maximum oxalate concentration of 2.32 × 10−7 M. We can calculate the required [H+] by using the overall equation for the dissociation of oxalic acid to oxalate:
$HO_2CCO_2H (aq) \rightleftharpoons 2H^+ (aq) + C_2O^{2−}_{4(aq)}$
K = Ka1Ka2 = (10−1.25)(10−3.81) = 10−5.06 = 8.7×10−6
Substituting the desired oxalate concentration into the equilibrium constant expression,
\begin{align}8.7\times10^{-6}=\dfrac{[\mathrm{H^+}]^2[\mathrm{ox^{2-}}]}{[\mathrm{HO_2CCO_2H}]} &= \dfrac{[\mathrm{H^+}]^2(2.32\times10^{-7})}{1.0} \ [\mathrm{H^+}] &=\textrm{6.1 M}\end{align} \nonumber
Thus adding enough HCl to give [H+] = 6.1 M will cause only La2(C2O4)3•9H2O to precipitate from the solution.
Exercise $3$
A solution contains 0.015 M Fe2+ and 0.015 M Pb2+. What concentration of acid is needed to ensure that Pb2+ precipitates as PbS in a saturated solution of H2S, but Fe2+ does not precipitate as FeS? Ksp values are 6.3 × 10−18 for FeS and 8.0 × 10−28 for PbS.
Answer
0.018 M H+
Summary
The anion in sparingly soluble salts is often the conjugate base of a weak acid that may become protonated in solution, so the solubility of simple oxides and sulfides, both strong bases, often depends on pH. The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.4%3A_The_Effects_of_pH_on_Solubility.txt |
Learning Objectives
• To be introduced to complex ions, including ligands.
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
Figure $1$: The Formation of Complex Ions. An aqueous solution of $\ce{CuSO4}$ consists of hydrated Cu2+ ions in the form of pale blue [Cu(H2O)6]2+ (left). The addition of aqueous ammonia to the solution results in the formation of the intensely blue-violet [Cu(NH3)4(H2O)2]2+ ions, usually written as [Cu(NH3)4]2+ ion (right) because ammonia, a stronger base than H2O, replaces water molecules from the hydrated Cu2+ ion. For a more complete description, see www.youtube.com/watch?v=IQNcLH6OZK0.
The Formation Constant
The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows:
\begin{align}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\hspace{5mm}K_1 \ \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}\hspace{3mm}K_2 \ \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}\hspace{3mm}K_3 \ \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}}\hspace{3mm}K_4 \end{align} \label{17.3.1}
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $\ce{NH_3}$ ligands, not six.
$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Common Complex Ions
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32− ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42− ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of $\ce{Cu(NO3)2•6H2O}$ is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{17.3.2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{17.3.2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
$\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber$
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,
\begin{align*}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times10^{-14}\end{align*}
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise $1$
The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6].
Answer
2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}$
with
$K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}$
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}$
with
$K_f = 2.9 \times 10^{13} \label{17.3.5b}$
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide (HgI2) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer a
1.9 × 10−10 M
Answer a
1.4 M
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$\ce{Ca^{2+} (aq) + O3POPO^{4−}4(aq) \rightleftharpoons [Ca(O3POPO3)]^{2−} (aq)} \label{17.3.7a}$
with
$K_f = 4\times 10^4\label{17.3.7b}$
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Figure $2$: An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $2$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Summary
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.5%3A_Complex_Ions_and_Solubility.txt |
Learning Objectives
• To know how to separate metal ions by selective precipitation.
• To understand how several common metal cations can be identified in a solution using selective precipitation.
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis, a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure $1$.
Group 1: Insoluble Chlorides
Most metal chloride salts are soluble in water; only $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M $\ce{HCl}$, thereby causing $\ce{AgCl}$, $\ce{PbCl2}$, and/or $\ce{Hg2Cl2}$ to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation.
Group 2: Acid-Insoluble Sulfides
Next, the acidic solution is saturated with $\ce{H2S}$ gas. Only those metal ions that form very insoluble sulfides, such as $\ce{As^{3+}}$, $\ce{Bi^{3+}}$, $\ce{Cd^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Hg^{2+}}$, $\ce{Sb^{3+}}$, and $\ce{Sn^{2+}}$, precipitate as their sulfide salts under these acidic conditions. All others, such as $\ce{Fe^{2+}}$ and $\ce{Zn^{2+}}$, remain in solution. Once again, the precipitates are collected by filtration or centrifugation.
Group 3: Base-Insoluble Sulfides (and Hydroxides)
Ammonia or $\ce{NaOH}$ is now added to the solution until it is basic, and then $\ce{(NH4)2S}$ is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions $\ce{Co^{2+}}$, $\ce{Fe^{2+}}$, $\ce{Mn^{2+}}$, $\ce{Ni^{2+}}$, and $\ce{Zn^{2+}}$ precipitate as their sulfides, and the trivalent metal ions $\ce{Al^{3+}}$ and $\ce{Cr^{3+}}$ precipitate as their hydroxides: $\ce{Al(OH)3}$ and $\ce{Cr(OH)3}$. If the mixture contains $\ce{Fe^{3+}}$, sulfide reduces the cation to $\ce{Fe^{2+}}$, which precipitates as $\ce{FeS}$.
Group 4: Insoluble Carbonates or Phosphates
The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When $\ce{Na2CO3}$ is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding $\ce{(NH4)2HPO4}$ causes the same metal ions to precipitate as insoluble phosphates.
Group 5: Alkali Metals
At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals ($\ce{Li^{+}}$, $\ce{Na^{+}}$, $\ce{K^{+}}$, $\ce{Rb^{+}}$, and $\ce{Cs^{+}}$) and ammonium ($\ce{NH4^{+}}$). We now take a second sample from the original solution and add a small amount of $\ce{NaOH}$ to neutralize the ammonium ion and produce $\ce{NH3}$. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present.
Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$, are all quite insoluble in water. Because $\ce{PbCl2}$ is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any $\ce{PbCl2}$ present. Isolating the solution and adding a small amount of $\ce{Na2CrO4}$ solution to it will produce a bright yellow precipitate of $\ce{PbCrO4}$ if $\ce{Pb^{2+}}$ were in the original sample (Figure $2$).
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any $\ce{AgCl}$ because $\ce{Ag^{+}}$ forms a stable complex with ammonia: $\ce{[Ag(NH3)2]^{+}}$. In addition, $\ce{Hg2Cl2}$ disproportionates in ammonia.
$\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber$
to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
$\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber$
Any silver ion in the solution is then detected by adding $\ce{HCl}$, which reverses the reaction and gives a precipitate of white $\ce{AgCl}$ that slowly darkens when exposed to light:
$\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber$
Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups.
Summary
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.6%3A_A_Deeper_Look%3A_Selective_Precipitation_of_Ions.txt |
These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.
Q5
State the balanced equation representing the dissolution of potassium chromate and write its solubility product expression.
Answer
$\mathrm{K_2CrO_4 \rightleftharpoons 2K^+ + CrO_4^{2-} }\nonumber$
$\mathrm{K_{sp} = [K^+]^2 [CrO_4^{2-}] }\nonumber$
Q11
Consider Mercury(I) bromide. Estimate the concentration of $Hg_2^{2+}$ and $Br^-$ given the $K_{sp}$ as 5.6 x 10-23.
Answer
$\ce{Hg2Br2 (s) <=> Hg2^{2+} (aq) + 2Br^- (aq)}$
$K_{sp}=5.6\cdot 10^{-23}=[Hg_{2}^{2+}][Br^{-}]^{2}=x(2x)^{2}$
$\mathrm{x = 2.41 \times 10^{-8} = [Hg_2^{2+}] } \nonumber$
$\mathrm{[Br^-] = 4.82 \times 10^{-8}}\nonumber$
Q13
The solubility product constant of Calcium chlorate ($Ca(ClO_3)_2$) of water is 7.1 × 10-7 at 25oC. How many grams of $Ca(ClO_3)_2$ were dissolved in 750 mL?
Answer
First, write out the solubility product constant expression for $\ce{Ca(ClO3)2}$
$\ce{ Ca(ClO3)2 (s) <=> Ca^{2+} (aq) + 2 ClO3^- (aq)} \nonumber$
$\mathrm{K_{sp} = [Ca^{2+}][ClO_3^-]^2}\nonumber$
ICE Table $Ca(ClO_3)_2$ $Ca^{2+}$ $ClO_3^-$
Initial - 0 0
Change - +x +2x
Equilibrium - x 2x
$Ca(ClO_3)_2$ does not matter because it is a solid
$\mathrm{K_{sp}= [x] \cdot [2x]^2}\nonumber$
$\mathrm{7.1 \times 10^{-7} = 4x^3}\nonumber$
$\mathrm{x = 0.0056 \; M}\nonumber$
The number of moles of $Ca^{2-}$ is the same as the number of moles of $Ca(ClO_3)_2$ because matter cannot be created or destroyed.
$\mathrm{\dfrac{0.0056 \: moles}{L} \times 0.750 \: L = 0.0042 \; moles \; Ca(ClO_3)_2}\nonumber$
Finally, the mass can be calculated by multiplying the number of moles with the molar mass:
$\mathrm{0.0042 \: moles Ca(ClO_3)_2 \times (\dfrac{206.98 \: g}{1 \; mole} ) = 0.872 \: g }\nonumber$
Q15
At 25°C, water dissolves 0.8108g of PbCl2 per liter, calculate the Ksp of PbCl2 at 25°C.
Answer
$PbCl_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^{-}(aq)\nonumber$
$K_{sp}=[Pb^{2+}]\cdot [Cl^{-}]^{2}\nonumber$
$n(PbCl_{2})=\dfrac{0.8108g}{278.1g/mol}=2.9155 \times 10^{-3}mol=n(Pb^{2+})=0.5n(Cl^{-})\nonumber$
$K_{sp}=\dfrac{2.9155 \times 10^{-3}mol}{1L}\cdot (\dfrac{5.831 \times 10^{-3}mol}{1L})^{2}=9.91 \times 10^{-8}\nonumber$
Q17
0.986635g of $\ce{BaCO_3}$ is dissolved in 1.00dm3 water at 80oC. the solution was then cooled down to 25oC. Given that the Ksp of $\ce{BaCO_3}$ in water is 2.58x10-9 at 25oC, determine with an explanation whether or not a precipitate will form at 25oC.
Answer
x moles of BaCO3 dissolves into x moles of Ba2+ and x moles CO32- due to equal stoichiometric coefficients (since solution is 1.00L, the moles also equal to molarities). Calculate the moles of BaCO3, then square the result to find Qsp, then compare Qsp to Ksp at 250C.
$\mathrm{moles \: of \; BaCO_3 = 0.986635 \; g \times (137.327 \dfrac{g}{mol} + 12 \dfrac{g}{mol} + 3 \times 16 \dfrac{g}{mol}) = 5.0 \times 10^{-3} \; mol }\nonumber$
$\mathrm{Q_{sp} = (5.0 \times 10^{-3} \; M)^2 = 2.5 \times 10^{-5} }\nonumber$
$\mathrm{2.5 \times 10^{-5} > 2.58 \times 10^{-9}}\nonumber$
$\mathrm{Q_{sp} > K_{sp} }\nonumber$
Therefore, precipitate forms.
Q19
Suppose that you take a solution of 500.0-mL of 0.001234 M $FeF_2$ and mix it well with a 500.0-mL solution of 0.003142 M $KOH$ at 25ºC. Given that $\mathrm{K_{sp} = 4.87 \times 10^{-17} }$ is for $Fe(OH)_2$, determine whether or not a precipitate will be formed from the mixture.
Answer
Before starting any calculations it is important to write down the equation for precipitation of the Fe(OH)2.
$\ce{Fe(OH)2}(s) \rightleftharpoons \ce{Fe^2+}(aq)+\ce{2OH-}(aq)\nonumber$
State the solubility product expression and set it equal to the solubility product constant:
$K_{\ce sp}={[Fe^2+][OH-]^2}=4.87\times10^{-17}\nonumber$
Now that the essential basic steps are out of the way, find the new concentrations of both $\ce{Fe^{2+} }$ Fe2+ and OH- by first finding the number of moles in the original unmixed solutions and then dividing that number by the total new volume in liters.
$\textrm{moles Fe}^{2+}=\textrm{500} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.001234\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=6.17\times10^{-4}\textrm{ mol Fe}^{2+}\nonumber$
$\textrm[{Fe}^{2+}]=\left(\dfrac{6.17\times10^{-4}\textrm{ mol Fe}^{2+}}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{\textrm{1000} \cancel{ mL}\;}{\textrm{1 L}} \right )=6.17\times10^{-4}\textrm{ M Fe}^{2+}\nonumber$
$\textrm{moles OH-}=\textrm{500} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.003142\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=1.57\times10^{-3}\textrm{ mol OH-}\nonumber$
$\textrm[{OH-}]=\left(\dfrac{1.57\times10^{-3}\textrm{ mol OH-}}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{\textrm{1000} \cancel{ mL}\;}{\textrm{1 L}} \right )=1.57\times10^{-3}\textrm{ M OH-}\nonumber$
Now that the new concentrations have been found they can be substituted in the Qsp equation and solved for.
$Q_{\ce sp}={[6.17\times10^{-4}][1.57\times10^{-3}]^2}=1.52\times10^{-9}\nonumber$
As shown above the $\mathrm{Q_{sp} > K_{sp}}$, therefore a precipitate does in fact form.
Q23
The solubility product constant of $\ce{Ag_2CO_3}$ is $\mathrm{8.1 \times 10^{-12}}$ at 25oC. Calculate $\ce{[Ag^+]}$ and $\ce{[CO_3^{2-}]}$ in the solution at equilibrium when 6 mL of 0.1M $\ce{Na_2CO_3}$ are mixed with 3 mL of 0.5M $\ce{AgNO_3}$.
Answer
$2AgNO_{3(aq)}+Na_2CO_{3(aq)} \rightleftharpoons Ag_{2}CO_{3(s)}+2NaNO_{3(aq)}$
$mole_{AgNO_{3}}=mole_{Ag^{+}}=(0.003 \; L)(0.5 \; M)=1.5\times 10^{-3}\ mol$
$mole_{Na_2CO_{3}}=mole_{CO_{3}^{2-}}=(0.006 \; L)(0.1 \; M)=6\times 10^{-4}\ mol$
$Total\ Volume = 3 \; mL + 6 \; mL = 9 \; mL = 0.009 \; L$
$M_{CO_{3}^{2-}}=\dfrac{6\times 10^{-4}mol}{0.009 \; L}=0.0667 \; M$
$M_{Ag^{+}}=\dfrac{1.5\times 10^{-3}\ mol}{0.009 \; L}=0.1667 \; M$
$2Ag^{+} +CO_{3}^{2-} \rightarrow Ag_{2}CO_{3(s)}$
$\ce{2Ag^+}$ $\ce{CO_3^{2-}}$ $\ce{Ag_2CO_3}$
Initial 0.1667M 0.0667M X
Change -2x -x X
Equilibrium 0.1667-2x 0.0667-x X
$K_{sp}=8.1\times 10^{-12}=[CO_{3}^{-}][Ag^{+}]^{2}$
$8.1\times 10^{-12}=[0.0667-x ][0.1667-2x]^{2}$
$x=0.06669999M$
$[Ag^{+}]=0.1667-2x=0.03336M$
$[CO_{3}^{2-}]=0.0667-x=7.30\times 10^{-9}$
Q25
Your nemesis has prepared a perfectly saturated solution of CaF2. To sabotage their work, you add 10 M NaF to their solution. How many grams of CaF2 fall out of solution if they had prepared 500 mL.
$\mathrm{CaF_2 :K_{sp} = 4.0 \times 10^{-11}}$
Answer
First, calculate the molar solubility of CaF2 from it's Ksp with the expression:
$K_{sp} = (s)(2s)^2= 4s^{3}\nonumber$
$4.0 \times 10^{-11} = 4s^3\nonumber$
$s = 2.15 \times 10^{-4}\nonumber$
Then, create an ICE table for the addition of F- as NaF:
Products $\ce{F^-}$ $\ce{Ca^{2+}}$
Initial s+10 s
Change -2x -x
Equilibrium (s+10) -2x s-x
Now, it's reasonable to assume that 10>>x and s, so our expression simplifies to:
$K_{sp} = 10^{2}(s-x)$
Solving for x yields the moles/L which fall out, so multiplying x by 0.5 L (500 mL) yields the correct answer.
$K_{sp} = 4.0 \times 10^{-11} = 10^2(s-x)\nonumber$
$x = 0.000215 \; M\nonumber$
$0.5 \; L (0.000215 \; M) (78.07 \; \dfrac{g}{mol}) = 0.00839 \; g\nonumber$
Q27
Given copper(II) hydroxide, $\ce{Cu(OH)_2}$. The concentration of $\ce{Cu^{2+}}$ and $\ce{OH^{-}}$ at equilibrium in 25°C water is 1.765 x 10-7M and 3.530 x 10-7M respectively.
1. Find the Ksp.
2. Find the molar solubility of $\ce{Cu(OH)_2}$ in 0.100M $\ce{NaOH}$.
Answer
$Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^{-}$
1)
Since the solubilities given are already at equilibrium, Ksp is found as such:
$K_{sp}= [Cu^{2+}][OH^{-}]^{2}=(1.765 \times 10^{-7}M)(3.530 \times 10^{-7}M)^{2}=2.2 \times 10^{-20}$
2)
ICE Table $\mathrm{Cu(OH)_{2(s)}}$ $\mathrm{Cu^{2+}_{(aq)}}$ $\mathrm{OH^-_{(aq)}}$
Initial - 0 0.100
Change - +Z +2Z
Equilibrium - Z 0.100+2Z
Apply the ice table result to the Ksp found in 1), we get
$K_{sp}=2.2\times 10^{-20} =[Cu^{2+}][OH^{-}]^{2}=(z)(0.100+2z)^{2}$
Z can be found either by plugging in quadratic formula or finding intersection in a graphing calculator.
$K_{sp}=2.2\times 10^{-20} =(z)(0.100+2z)^{2}$
$\mathrm{z = 2.2 \times 10^{-18} \; M}\nonumber$
Q28
Barium Carbonate (BaCO3) has a solubility product of Ksp = 8.1 x 10-9 at 25°C for the equilibrium
$BaCO_3 \rightleftharpoons Ba^{2+} + CO_3^{2-} \nonumber$
1. Calculate the molar solubility of Barium Carbonate in the water at 25°C.
2. Calculate the molar solubility of Barium Carbonate in 0.1M PbCO3.
Answer
a) The expression for solubility product is
$[Ba^{2+}][{CO_{3}}^{2-}] = K_{sp}$
Let $[Ba^{2+}] = S$
$S^{2} = K_p = 8.1 \times 10^{-9}$
Hence, S = 9 x 10-5
Molar Solubility of Barium Carbonate = 9 x 10-5 M
b) due to the common ion effect, $PbCO_3$ reduces the molar solubility of Barium Carbonate.
$BaCO_{3} \rightleftharpoons Ba^{2+}\:\:\:\:+\:\:\:\: CO^{2-}_3$
$\ce{Ba^{2+}}$ $\ce{CO^{2-}_3}$
Initial 0 0.1
Change +S +S
Final +S 0.1+S
$[Ba^{2+}][{CO_3}^{2-}] = K_{sp}$
$S(0.1+S) = K_{sp}$
$S+0.1 \approx 0.1$
⇒ $S(0.1) = K_{sp}$
⇒ $S = \dfrac{8.1\times10^{-9}}{0.1} = 8.1\times10^{-8}$
⇒ $S = 8.1 \times 10^{-8} << 0.1$
Hence, Molar Solubility = 8.1 x 10-8 M
Q33
Determine how will the solubility of the following compounds change (increase, decrease, or unchanged) if the neutral solution is made to be more acidic. Explain your answer in one sentence.
1. $\ce{NO_3^-}$
2. $\ce{NH_3}$
3. $\ce{CO_3^{2-}}$
Answer
As the acidity of a solution increases, the solubility of salts consist of conjugate bases of weak acids will increases.
1. $\ce{NO_3^-}$ is the conjugate base of a strong acid so the solubility will not change
2. $\ce{NH_3}$ is the conjugate base of a weak acid so the solubility will increase
3. $\ce{CO_3^{2-}}$ is also a conjugate base of a weak acid so the solubility will increase
Q37
18-crown-6, or $\ce{C_12H_24O_6}$, is an organic compound that can bind to group 1 metal ions by wrapping around them when in aqueous solutions. Due to its particular size, it can fit the $K^+$ ion better than any other.
Ball-and-stick model of the 18-crown-6 potassium complex in crystalline (18-crown-6)potassium chlorochromate, [K(C12H24O6)][CrClO3]. X-ray diffraction data from S. A. Kotlyar, R. I. Zubatyuk, O. V. Shishkin, G. N. Chuprin, A. V. Kiriyak and G. L. Kamalov (February 2005). "(18-Crown-6)potassium chlorochromate". Acta. Cryst. E61 (2): m293-m295. DOI:10.1107/S1600536805000085. (Public domain; Ben Mills).
This physical property can be seen in the equilibrium constants for 18-crown-6 and various alkali metals:
$\ce{Na+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow Na-crown+_{(aq)}}$
K = 6.6
$\ce{K+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow K-crown+_{(aq)}}$
K = 111.6
$\ce{Rb+_{(aq)} + C_12H_24_O_{6(aq)} \rightarrow Rb-crown+_{(aq)}}$
K = 36
If an aqueous solution is made that is 0.006 M in both 18-crown-6 and Rb+, what is the concentration of unbound Rb+ at equilibrium? If the same solution is made, but with K+ instead of Rb+ ions, what is the concentration of unbound K+ ions at equilibrium?
Answer
First, construct an ICE table.
$\ce{Rb+_{aq} + C_12H_24O_{6(aq)} \rightleftharpoons Rb-crown+_{aq}}\nonumber$
$\ce{Rb^+}$ $\ce{C_{12}H_{24}O_6}$ Rb-crown+
I 0.006 0.006 0
C -x -x +x
E 0.006-x 0.006-x x
Then, use the K value of the reaction to solve for x. Use x to solve for the concentration of the desired ion.
$K = \dfrac{[Rb-crown+]}{[Rb+][C_{12}H_{24}O_6]}$
$36 = \dfrac{[x]}{[0.006-x][0.006-x]}$
$x = 9.2954 \times 10^{-4}$
$[Rb^+] = 0.0507 M$
Repeat this process for the other reaction.
$\ce{K+_{aq} + C_{12}H_{24}O_{6(aq)} \rightleftharpoons K-crown^+_{aq}}$
$\ce{K^+}$ $\ce{C_{12}H_{24}O_6}$ K-crown+
I 0.006 0.006 0
C -x -x +x
E 0.006-x 0.006-x x
$K = \dfrac{[K-crown^+]}{[K^+][C_{12}H_{24}O_6]}$
$111.6 = \dfrac{[x]}{[0.006-x][0.006-x]}$
$x = 0.0019$
$[K^+] = 0.0041 M$
Q39
Is it easier to dissolve CuBr into 1M of NaBr solution than to dissolve it into pure water? Does the answer change if when the concentration of NaBr is 0.5M?
Answer
1. When CuBr is added into NaBr solution, the common ion effect occurs. Common ion effect explains that the solvent is responsible for the decrease in the solubility of the precipitate when the soluble compound contains one of the ions of the precipitate that is added to the solution in equilibrium. According to Le Chatelier's principle, some of the ions in excess should be removed from the solution. So it is harder to dissolve CuBr into NaBr solution than dissolve it into pure water.
2. If the concentration is changed to 0.5M, it is still harder to dissolve CuBr in NaBr solution, but less difficult.
Q41
Identify the salt $\ce{NH_{4}F}$ as either acidic, basic, or neutral. Then explain the mechanism behind why the salt is acidic, basic, or neutral using chemical equations and their $\ce{K_{a}}$ or $\ce{K_{b}}$ values provided here and here.
Answer
• For a salt to have an acidic character, it must be able to accept $\text{OH}^{-}$ ions when it dissociates in an aqueous solution and leave an excess of $\text{H}^{+}$ in the solution to have an acidic effect.
• For a salt to have a basic character, it must be able to either accept an $\text{H}^{+}$ when it dissociates in an aqueous solution and leave an excess of $\text{OH}^{-}$ in the solution to have an basic effect.
$\ce{ NH_{4}F}$ can be classified as either acidic or basic depending upon which species is present in a greater concentration ( $\text{H}^{+}$ or $\text{OH}^{-}$ ) in the solution when $\ce{NH_{4}F}$ is added.
$\ce{ NH^{+}_{4(aq)} + F^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + HF_{(aq)} + H_{3}O^{+}_{(aq)} + OH^{-}_{(aq)}}$
$\ce{ K_a = 5.6 \times 10^{-10}}$ for $\ce{NH^{+}_{4(aq)} + H_{2}O_{(1)} \rightleftharpoons NH_{3(aq)} + H_{3}O^{+}_{(aq)}}$
$\ce{ K_b = 1.6 \times 10^{-11}}$ for $\ce{ F^{-}_{(aq)} + H_{2}O_{(1)} \rightleftharpoons HF_{(aq)} + OH^{-}_{(aq)}}$
Since $\ce{ {K}_{a} \gt {K}_{b} }$ , $\ce{ NH_{4}^{+}}$ has a greater affect on the solution and therefore the salt is acidic.
** See key for problem 15.23 on how to calculate $\ce{K_{b}}$ or $\ce{K_{a}}$ given $\ce{K_{a}}$ or $\ce{K_{b}}$
Q43
For the complex ion, $Fe(H_2O)_{6(aq)}^{3+}$, the acid ionization constant for 298K is $\mathrm{7.7 \times 10^{-3}}$. What is the pH of a 50 mL solution that contains 0.02 mols of $Fe(NO_3)_2$?
Answer
Complex ion problems are solved identically to other equilibrium problems.
First, write the balanced equation:
$[Fe(H_2O)_{6}]^{3+}_{(aq)} + H_2O_{(aq)} \rightleftharpoons [Fe(H_2O)_5OH]^{2+}_{(aq)} + H_3O^{+}_{(aq)} \nonumber$
Then, create and solve an ICE table.
$[Fe(H_2O)_6]^{3+}$ $H_2O$ $H_3O^{3+}$ $[Fe(H_2O)_5OH]^{2+}$
0.40 - 0 0
-x - +x +x
0.40-x - x x
From the ICE table:
$K_a = \dfrac{[H_3O^{3+}][Fe(H_2O)_5OH^{2+}]}{[Fe(H_2O)_6^{3+}]}\nonumber$
$7.7 \times 10^{-3} = K_{a} = \dfrac{\left[x\right]\left[x\right]}{\left[0.4-x\right]}\nonumber$
By rearranging the equation:
$\mathrm{x^{2}+0.0077x-0.00308=0}\nonumber$
and using the quadratic,
$\mathrm{\dfrac{-0.0077\pm{\sqrt {0.0077^{2}-4\left(1\right)\left(-0.00308\right)}}}{2}}\nonumber$
$\mathrm{x=0.05178}\nonumber$
$\mathrm{pH = -log[H_3O^+] = 1.286}\nonumber$
Q45
Calculate the concentration of $\mathrm{H_3O^+}$ in a 0.3 M solution of oxalic acid ( $\mathrm{pK_{a1} = 1.25 ; pK_{a2} = 3.81}$).
Answer
$\mathrm{K_a = 10^{-1.25} =0.0562}$
Balanced equation: $\mathrm{C_2H_2O_4 + H_2O \rightleftharpoons C_2HO_4^- + H_3O^+}\nonumber$
$\ce{C_2H_2O_4}$ $\ce{C_2HO_4^{-}}$ $\ce{H_3O^+}$
I 0.3 0 0
C -x +x +x
E 0.3-x x x
At equilibrium:
$\mathrm{\frac{(x)^2}{(0.3-x)} = K_{a1}}\nonumber$
$\mathrm{x = 0.1048}\nonumber$
Balanced equation: $\mathrm{C_2HO_4^- + H_2O \rightleftharpoons C_2O_4^{2-} + H_3O^+}\nonumber$
$\ce{C_2HO_4^-}$ $\ce{C_2O_4^{2-}}$ $\ce{H_3O^+}$
I 0.1048 0 0.1048
C -x +x +x
E 0.1048-x x x+0.1048
At equilibrium:
$\mathrm{K_{a2} = 10^{-3.81} }\nonumber$
$\mathrm{\frac{x(x+0.1048)}{(0.1048-x)} = K_{a2}}\nonumber$
$\mathrm{x = 0.000154}\nonumber$
$\mathrm{[H_3O^+] = 0.1048 + 0.000154 \; M = 0.105 \; M}\nonumber$
Q49
An aqueous solution at 25 °C is 0.10 M in both Ba2+ and Ca2+ ions. One wants to separate the two ions by taking advantage of the different solubility of BaCO3 and CaCO3.
$BaCO_3 \; K_{sp} = 2.58 \times 10^{-9} \; M$
$CaCO_3 \; K_{sp} = 3.36 \times 10^{-9} \; M$
What is the highest possible CO32- concentration that allows only one salt to present at equilibrium? Which ion is present in the solid, Ba2+ or Ca2+?
Answer
$\mathrm{BaCO_{3(s)} \leftrightharpoons Ba^{2+}_{(aq)} + CO_{3(aq)}^{2-}}\nonumber$
$\mathrm{[CO_3^{2-}] < \dfrac{K_{sp}}{[Ba^{2+}]} = \dfrac{2.58 \times 10^{-9}}{0.1 \; M} = 2.58 \times 10^{-8} \; M \; (Using \: the \; formula \; for \; K_{sp})}\nonumber$
$\mathrm{CaCO_{3(s)} \rightleftharpoons Ca_{(aq)}^{2+} + CO_{3(aq)}^{2-} }\nonumber$
$\mathrm{[CO_3^{2-}] < \dfrac{K_{sp}}{[Ca^{2+}]} = \dfrac{3.36 \times 10^{-9}}{0.1 \; M} = 3.36 \times 10^{-8} \; M}\nonumber$
Therefore, the highest possible $CO_3^{2-}$ concentration that allows only one salt to present at equilibrium is $\mathrm{2.58 \times 10^{-8} \; M}$. $Ba^{2+}$ is present in the solid.
Q51
A solution was created by dissolving 1.31 kg of Ba(NO3)2 into 1 L of 0.50 M Ca(NO3)2. Determine which molecule is more soluble. Then determine the concentration of SO4 ion is needed to precipitate all of one molecule but leave the other molecule completely unreacted, use Table E3.
Answer
First we must determine the mass action equation for dissolving the salts:
$\mathrm{K_{sp} = [Ba^+][SO_4^-] = 1.08 \times 10^{-10} } \nonumber$
$\mathrm{K_{sp} = [Ca^+][SO_4^-] = 4.93 \times 10^{-5} } \nonumber$
We will use CaSO4 as the salt because it is the most soluble.
For Ca+ to to remain in solution, its reaction quotient must remain smaller than then its Ksp:
$\mathrm{Q = [Ca^+][SO_4^-] < 4.93 \times 10^{-5}}\nonumber$
Now we just need to impute the concentration of Ca+ and solve for the concentration of SO4 ion:
$\mathrm{[0.5 \; M][SO_4^-] < 4.93 \times 10^{-5}}\nonumber$
$\mathrm{[SO_4^-] < 9.86 \times 10^{-5} \; M}\nonumber$
So as long as the concentration of SO4 ion is smaller than 9.86 x 10-5 M, no CaSO4 will precipitate
Q55
Let's say you have a solution that is saturated with $HF$ at a concentration of $[HF] = 0.20 M$.
1. If you want some element $X$ whose concentration is equal to 0.20 M to exist entirely in this solution, what will be the highest pH possible for this mixture? ($K_a \; of \; HF \; is \; 6.6 \times 10^{-4}$ and $K \; of \; XF(s) = 2.5 \times 10^{-16})$. Hint: the reaction is given as $XF(s) + H_2O(l) \leftrightharpoons X(aq) + F^-(aq) + OH^-(aq)$
2. What would be the concentration of some element $G$ in equilibrium with solid $GF$ in this solution at the pH value found in part a) given that $K \; of \; GF(s) = 4.9 \times 10^{-22})$.
Answer
Part a)
Since you have the equation $XF(s) + H_2O(l) \leftrightharpoons X(aq) + F^-(aq) + OH^-(aq)$, the equilibrium constant K will be equal to
$K = [X] \; [F^-] \; [OH^-] = 2.5 \times 10^{-16}$
We now replace these variables above with known derived equations
$K = [X] \left(\dfrac {[K_a] [HF]}{[H_3O^+]} \right) \left(\dfrac {[K_w]}{[H_3O^+]}\right) = 2.5 \times 10^{-16}$
Plug in known values and solve for $[H_3O^+]$
$[H_3O^+] = \sqrt{\dfrac {(0.20) \; (6.6 \times 10^{-4}) \; (0.20) \; (1.0 \times 10^{-14})}{2.5 \times 10^{-16}}}$
$[H_3O^+] = 0.0325 \; M$
Now we solve for pH and see that $pH = -log(0.0325) = 1.50\nonumber$
Part b)
For part b), we solve the question in a very similar manner as used in part a) with the only differences being a few different variables we plug in. Now, we are solving for our concentration of some element $G$ using the value of $H_3O^+$ we found above.
Given the reaction $GF(s) + H_2O(l) \leftrightharpoons G(aq) + F^-(aq) + OH^-(aq)$, the equilibirum constant K will be equal to
$K = [G] \; [F^-] \; [OH^-] = 4.90 \times 10^{-22}$
We now replace these variables above with known derived equations and this time, solve for $[G]$ using the $[H_3O^+]$ value found above
$K = [G] \; (\dfrac {[K_a] [HF]}{[H_3O^+]}) \; (\dfrac {[K_w]}{[H_3O^+]}) = 4.9 \times 10^{-22}$
$[G] = 3.92 \times 10^{-7} \; M$
Q63
200 mL of 0.5M NaCl and 800 mL of 0.10M AgNO3 are mixed together. Calculate the mass of NaNO3 precipitated. Assume:
• that the volumes are additive,
• that AgCl is completely insoluble, and
• any other substances that may form are completely soluble.
Answer
Because NaCl and AgNO3 are strong electrolytes, they dissolve in water by dissociation.
NaNO3 precipitates according to the net ionic equation
$Na^{+}_{(aq)}+NO^{-}_{3(aq)}\rightarrow NaNO_{3(s)}\nonumber$
First calculate the moles of each ion present.
$n_{Na^{+}}=0.2L\times 0.5M=0.1 mol Na^{+}\nonumber$
$n_{NO_{3}^{-}}=0.8L\times 0.1M= 0.08 mol NO_{3}^{-}\nonumber$
Assuming that NaNO3 is completely insoluble, the reaction continues until the entire 0.08 mol of NO3- is consumed, which is the limiting reactant, creating 0.08 mol NaNO3.
$0.08 \; mol \; NaNO_{3}(84.99\dfrac{g}{mol})=6.80 \; g \; NaNO_{3}\nonumber$
Q69
Siobhan has a $1L$ solution with containing $0.050M$ of $CO_{3\, (aq)}^{2-}$ and $0.100M$ of $F_{(aq)}^{-}$. She titrates the solution with a 0.100 M titrand of Iron Iodide, $FeI_{2}$. The $K_{sp}$ expressions are:
$FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}$ $K_{sp}$= $3.13\times{10}^{-11}$
$FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}$ $K_{sp}$= $2.36\times{10}^{-6}$
At what volume of iron iodide will the precipitate $FeCO_{3\, (s)}$ appear? How about $FeF_{2\, (s)}$?
Answer
This problem primarily focuses on expanding on how the solubility constant $K_{sp}$ can predict which solution will titrate first. When Iron Iodide is dripped into the solution, it disassociates into Iron(II) and Iodide Ions.
$FeI_{2} \rightarrow Fe^{2+}_{(aq)} + 2I^-_{aq}$
This facilitates a "common ion effect", where the ions from the dissolution of the titrand Iron Iodide contribute to shifting the concentration of the other two reactions:
$FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}$
$FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}$
As seen here, the introduction of $Fe^{2+}$ ions will mean there is a larger iron ion concentration, which shifts both of the above reactions to the left. This concept that explains this phenomena is colloquially known as Le Chatlier's principle. However, in order to precisely determine at what point precipitate will form, the $K_{sp}$ values need to be taken into consideration. Recall that the K constant is defined in terms of either concentrations or partial pressures. As the species in this problem are only aqueous solutes and not gaseous compounds, the resulting $K_{sp}$ will be defined by concentrations as depicted below:
$FeCO_{3\, (s)} \rightleftharpoons CO_{3\, (aq)}^{2-} + Fe_{(aq)}^{2+}$ $K_{sp}= \dfrac{\left[CO_{3}^{2-}\right]\left[Fe^{2+}\right]}{1} = 3.13\times{10}^{-11}$
$FeF_{2\, (s)} \rightleftharpoons 2 F_{(aq)}^{-} + Fe_{(aq)}^{2+}$ $K_{sp}= \dfrac{\left[F^{-}\right]^{2}\left[Fe^{2+}\right]}{1} = 2.36\times{10}^{-6}$
Notice that each $K_{sp}$ is divided by 1. This is to represent the solid reactants. Also, based on the information above, recognize that, at equilibrium, the product of the concentrations of the two ions should equal to thier $K_{sp}$, and since the problem provided both $\left[CO^{2+}_{3}\right]$ and $\left[F^{-}\right]$. By substituting these concentrations into the respective $K_{sp}$ formulas, the equations can be algebraically manipulated to provide the $\left[Fe^{2+}\right]$ at equilibrium. This process is seen below for $FeF_{2\, s}$. The process is basically the same to finding $\left[Fe^{2+}\right]$ for $\left[CO^{2+}_{3}\right]$, but there is no need to square root the value since there is no stoichiometric coefficients for that particular equation. In contrast, $F^{-}$ has a stoichiometric coefficient of 2). Since only initial concentrations were given, we can determine at which point $Q_{sp} > K_{sp}$
$Q_{sp}= \dfrac{\left[F^{-}\right]^{2}\left[Fe^{2+}\right]}{1} = 2.36\times{10}^{-6}$
$Q_{sp}= \left(0.05M\right)^{2}\left[Fe^{2+}\right] = 2.36\times{10}^{-6}$
$\left[Fe^{2+}\right] = \dfrac{2.36\times{10}^{-6}}{\left(0.100M^{2}\right)}= 2.36\times10^{-4}M$
Therefore, $\left[Fe^{2+}\right]$ for $FeF_{2}= 2.36\times10^{-4}M$
And $\left[Fe^{2+}\right]$ for $FeCO_{3}=6.26\times10^{-10} M$, using the same method.
What this information tells us is the calculated $\left[Fe^{2+}\right]$ is the concentration of $Fe^{2+}$ that one would expect to be in equilibrium at the given concentration of the other ion. Therefore, if $\left[Fe^{2+}\right]$ is higher than this value, then the resulting reaction quotient will be larger than $K_{sp}$, causing the reaction to shift towards the product and consequently causing the precipitate to form. From the values, it is clear that $FeCO_{3\, (s)}$ will end up precipitating first, and then $FeF_{2\, s}$.
Finding the exact volume that the precipitate for each reaction will form is relatively straightforward. Based on the information overviewed before, after the $\left[Fe^{2+}\right]$ reaches the calculated point, precipitate will form. Therefore, to find the volume of titrant needed, just set up the following equation:
$molarity_{titrant}\times{V_{titrant}} = mol_{titrant}= mol_{Fe^{2+} to precipitate}$
$0.100M\times{V_{titrant}}= \left(6.26\times10^{-10}M\right)\left(1L\right)$
$0.100M\times{V_{titrant}}= 6.26\times10^{-10}mol$
$V_{titrant}= \dfrac{6.26\times10^{10}mol}{0.100M}$
$V_{titrant}=6.26\times10^{-9}L$ for $FeCO_{3\, (s)}$ precipitate to form.
Via the same process, $V_{titrant}=0.00236L$ for $FeF_{2\, (s)}$ to form.
Abstract: Use $K_{sp}$ to determine $Q_{sp}$ concentration $\left[Fe^{2+}\right]$. Divide by molarity of titrant to get desired volume for precipitate to form.
Q73
Complex ion plays an important role in the level of solubility. For example:
$Hg^{2+}_{(aq)} + 2I^-_{(aq)} \rightleftharpoons HgI_{2 (s)}\nonumber$
This reaction results in the formation of solid HgI2. However, when solid HgI2 reacts in the same process:
$HgI_{2(s)} + I^-_{(aq)} \rightleftharpoons HgI^-_{3(aq)}\nonumber$
The solubility increases greatly and the reaction results in an aqueous product. Explain this phenomenon.
Answer
The first reaction produces a solid precipitate and the second reaction is soluble because complex ions are created in the reaction. $HgI_3^-$ is a complex ion, which means it has Hg ion in the center that is surrounded by iodine molecules that are able to act as Lewis bases and attract the protons of the Lewis acid in the reaction. These properties are brought out as a result of the reaction and result in an aqueous product.
Q73
Why does the inclusion of $NH_3$ increase the solubility of $CuCl_2$ in an aqueous solution?
Answer
The solubility of Cu2+ increases when NH3 present because NH3 forms a complex ion with Cu2+. The corresponding the chemical reaction for is as follows,
$\mathrm{CuCl_{2(s)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Cl_{(aq)}^- + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}}\nonumber$
The formation of the complex ion pulls equilibrium towards the right which allows more CuCl2 to dissolve. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/16%3A_Solubility_and_Precipitation_Equilibria/16.E%3A_Solubility_and_Precipitation_%28Exercises%29.txt |
Electrochemistry is the study of electricity and how it relates to chemical reactions. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as redox reaction, or oxidation-reduction reaction.
17: Electrochemistry
Learning Objectives
• To understand the basics of voltaic cells
• To connect voltage from a voltaic cell to underlying redox chemistry
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$\ce{Zn(s) + Br2(aq) \rightarrow Zn^{2+} (aq) + 2Br^{−} (aq)} \nonumber$
The half-reactions are as follows:
reduction half-reaction:
$\ce{Br2 (aq) + 2e^{−} \rightarrow 2Br^{−} (aq)} \nonumber$
oxidation half-reaction:
$\ce{Zn (s) \rightarrow Zn^{2+} (aq) + 2e^{−} }\nonumber$
Each half-reaction is written to show what is actually occurring in the system; $\ce{Zn}$ is the reductant in this reaction (it loses electrons), and $\ce{Br2}$ is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation $1$). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
In any redox reaction, the number of electrons lost by the oxidation reaction(s) equals the number of electrons gained by the reduction reaction(s).
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.
There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction ($ΔG < 0$) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur ($ΔG > 0$). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $1$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells.
Voltaic (Galvanic) Cells
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
$\ce{Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)} \label{20.3.4}$
We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
This same reaction can be carried out using the galvanic cell illustrated in Figure $\PageIndex{3a}$. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of $\ce{Cu^{2+}}$ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of $\ce{Zn^{2+}}$ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge, a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are $\ce{Na^{+}}$ or $\ce{K^{+}}$ and $\ce{NO3^{−}}$ or $\ce{SO4^{2−}}$, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to $\ce{Zn^{2+}}$ ions at the zinc electrode (the anode), and $\ce{Cu^{2+}}$ ions are reduced to $\ce{Cu}$ metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of $\ce{Zn^{2+}}$ ions in the solution increases; simultaneously, the copper strip gains mass, and the concentration of $\ce{Cu^{2+}}$ ions in the solution decreases (Figure $\PageIndex{3b}$). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the $\ce{Zn^{2+}}$ solution would increase as the zinc metal dissolves, and the total positive charge in the $\ce{Cu^{2+}}$ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the $\ce{Zn^{2+}}$ solution and a flow of cations into the $\ce{Cu^{2+}}$ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential ($E_{cell}$) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (Figure $\PageIndex{3a}$). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example $1$.
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
Example $1$
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
$\ce{3Sn(s) + 2NO3^{-}(aq) + 8H^{+} (aq) \rightarrow 3Sn^{2+} (aq) + 2NO (g) + 4H2O (l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
1. Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
2. From the direction of electron flow, assign each electrode as either positive or negative.
Solution
A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:
reduction: $\ce{NO3^{−} (aq) + 4H^{+}(aq) + 3e^{−} → NO(g) + 2H2O(l)} \nonumber$
oxidation: $\ce{Sn(s) → Sn^{2+}(aq) + 2e^{−}} \nonumber$
Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.
Exercise $1$
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of $\ce{MnO_4^{−}}$ in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of $\ce{Sn^{2+}}$ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
$\ce{2MnO^{−}4(aq) + 5Sn^{2+}(aq) + 16H^{+}(aq) \rightarrow 2Mn^{2+}(aq) + 5Sn^{4+}(aq) + 8H2O(l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.
Answer a
\begin{align*} \ce{MnO4^{−}(aq) + 8H^{+}(aq) + 5e^{−}} &→ \ce{Mn^{2+}(aq) + 4H2O(l)} \[4pt] \ce{Sn^{2+}(aq)} &→ \ce{Sn^{4+}(aq) + 2e^{−}} \end{align*} \nonumber
Answer b
The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
Answer c
The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.
Electrochemical Cells: Electrochemical Cells(opens in new window) [youtu.be]
Constructing Cell Diagrams (Cell Notation)
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the $\ce{Zn/Cu}$ cell shown in Figure $\PageIndex{3a}$ is written as follows:
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
$\ce{Pt(s)\, | \, H2(g) | HCl(aq, \, 1\,M)\,|\, AgCl(s) \,Ag(s)} \nonumber$
This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
cathode reaction:
$\ce{AgCl (s) + e^{−} \rightarrow Ag(s) + Cl^{−}(aq)} \nonumber$
anode reaction:
$\ce{ 1/2 H2(g) -> H^{+}(aq) + e^{-}} \nonumber$
overall:
$\ce{ AgCl(s) + 1/2H2(g) -> Ag(s) + Cl^{-} + H^{+}(aq)} \nonumber$
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Example $2$
Draw a cell diagram for the galvanic cell described in Example $1$. The balanced chemical reaction is as follows:
$\ce{3Sn(s) + 2NO^{−}3(aq) + 8H^{+}(aq) \rightarrow 3Sn^{2+}(aq) + 2NO(g) + 4H2O(l)} \nonumber$
Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution
The anode is the tin strip, and the cathode is the $\ce{Pt}$ electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus $\ce{Sn(s)∣Sn^{2+}(aq)}$. We could include $\ce{H2SO4(aq)}$ with the contents of the anode compartment, but the sulfate ion (as $\ce{HSO4^{−}}$) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction ($\ce{NO}$) and the $\ce{Pt}$ electrode. These are written as $\ce{HNO3(aq)∣NO(g)∣Pt(s)}$, with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
$\ce{Sn(s)\,|\,Sn^{2+}(aq)\,||\,HNO3(aq)\,|\,NO(g)\,|\,Pt_(s)} \nonumber$
The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise $2$
Draw the cell diagram for the following reaction, assuming the concentration of $\ce{Ag^{+}}$ and $\ce{Mg^{2+}}$ are each 1 M:
$\ce{Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)} \nonumber$
Answer
$\ce{ Mg(s) \,|\,Mg^{2+}(aq, \;1 \,M )\,||\,Ag^+(aq, \;1\, M)\,|\,Ag(s)} \nonumber$
Cell Diagrams: Cell Diagrams(opens in new window) [youtu.be]
Summary
A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur. Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.1%3A_Electrochemical_Cells.txt |
Learning Objectives
• To understand the relationship between cell potential and the equilibrium constant.
• To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of $\ce{Co(s)}$ with $\ce{Ni^{2+}(aq)}$ to form $\ce{Ni(s)}$ and $\ce{Co^{2+}(aq)}$ occurs spontaneously, but if we reduce the concentration of $\ce{Ni^{2+}}$ by a factor of 100, so that $\ce{[Ni^{2+}]}$ is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential & Gibbs Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1}$
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2}
The total charge transferred from the reductant to the oxidant is therefore $nF$, where $n$ is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell ($w_{max}$) is equal to the product of the cell potential ($E^°_{cell}$) and the total charge transferred during the reaction ($nF$):
$w_{max} = nFE_{cell} \label{20.5.3}$
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
The change in free energy ($\Delta{G}$) is also a measure of the maximum amount of work that can be performed during a chemical process ($ΔG = w_{max}$). Consequently, there must be a relationship between the potential of an electrochemical cell and $\Delta{G}$; this relationship is as follows:
$\Delta{G} = −nFE_{cell} \label{20.5.4}$
A spontaneous redox reaction is therefore characterized by a negative value of $\Delta{G}$ and a positive value of $E^°_{cell}$, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and $E^°_{cell}$ is as follows:
$\Delta{G^°} = −nFE^°_{cell} \label{20.5.5}$
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example $1$
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: $ΔG^o$ for the reaction and spontaneity
Strategy:
1. From the relevant half-reactions and the corresponding values of $E^o$, write the overall reaction and calculate $E^°_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate $ΔG^o$. If $ΔG^o$ is negative, then the reaction is spontaneous.
A
As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of $E^o$. From Table P2, we can find the reduction and oxidation half-reactions and corresponding $E^o$ values:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of $E^o$ is not affected:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber
B
We can now calculate ΔG° using Equation $\ref{20.5.5}$. Because six electrons are transferred in the overall reaction, the value of $n$ is 6:
\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \& =-15.6 \times 10^4\textrm{ J} \ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber
Thus $ΔG^o$ is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise $1$
Use the data in Table P2 to calculate $ΔG^o$ for the reduction of ferric ion by iodide:
$\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of $\ce{Fe^{3+}(aq)}$ to $\ce{Fe(s)}$ is not listed in the table, but two related reductions are given:
$\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6}$
$\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7}$
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because $E^o$ is not a state function. However, because $ΔG^o$ is a state function, the sum of the $ΔG^o$ values for the individual reactions gives us $ΔG^o$ for the overall reaction, which is proportional to both the potential and the number of electrons ($n$) transferred. To obtain the value of $E^o$ for the overall half-reaction, we first must add the values of $ΔG^o (= −nFE^o)$ for each individual half-reaction to obtain $ΔG^o$ for the overall half-reaction:
\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9}$
Three electrons ($n = 3$) are transferred in the overall reaction, so substituting into Equation $\ref{20.5.5}$ and solving for $E^o$ gives the following:
\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber
This value of $E^o$ is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of $E^o$ for half-reactions cannot be added to give $E^o$ for the sum of the half-reactions; only values of $ΔG^o = −nFE^°_{cell}$ for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between $\Delta{G^°}$ and the equilibrium constant $K$, to obtain a relationship between $E^°_{cell}$ and $K$. Recall that for a general reaction of the type $aA + bB \rightarrow cC + dD$, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta{G°} = −RT \ln K \label{20.5.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation $\ref{20.5.5}$), we can write
$−nFE^°_{cell} = −RT \ln K \label{20.5.12}$
Rearranging this equation,
$E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}$
For $T = 298\, K$, Equation $\ref{20.5.12B}$ can be simplified as follows:
\begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align}
Thus $E^°_{cell}$ is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of $E^°_{cell}$ and vice versa.
Example $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: $K$
Strategy:
1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and $E^o_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Use Equation $\ref{20.5.13}$ to solve for $\log K$ and then $K$.
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber
B Two electrons are transferred in the overall reaction, so $n = 2$. Solving Equation $\ref{20.5.13}$ for log K and inserting the values of $n$ and $E^o$,
\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \[4pt] K & =2.3\times10^{69}\end{align*} \nonumber
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of $\ce{Sn^{2+}(aq)}$ with oxygen to produce $\ce{Sn^{4+}(aq)}$ and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber$
Answer
$5.7 \times 10^{72}$
Figure $1$ summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between $\Delta{G}$ and the reaction quotient $Q$.
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.2%3A_The_Gibbs_Free_Energy_and_Cell_Voltage.txt |
Learning Objectives
• Relate cell potentials to Gibbs energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, $\Delta{G}$, is given as follows:
$\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1}$
We also know that $ΔG = −nFE_{cell}$ (under non-standard conditions) and $ΔG^o = −nFE^o_{cell}$ (under standard conditions). Substituting these expressions into Equation $\ref{Eq1}$, we obtain
$−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2}$
Dividing both sides of this equation by $−nF$,
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}$
Equation $\ref{Eq3}$ is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium ($ΔG = 0$), then Equation $\ref{Eq3}$ reduces to Equation $\ref{Eq31}$ and $\ref{Eq32}$ because $Q = K$, and there is no net transfer of electrons (i.e., Ecell = 0).
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31}$
since
$E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32}$
Substituting the values of the constants into Equation $\ref{Eq3}$ with $T = 298\, K$ and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in $Q$):
$E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4}$
The Power of the Nernst Equation
The Nernst Equation ($\ref{Eq3}$) can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation $\ref{Eq4}$ allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation $\ref{Eq4}$ that the cell potential changes by 0.0591/n V for each 10-fold change in the value of $Q$ because log 10 = 1.
Example $1$
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
$\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber$
Calculate $E_{cell}$ for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl] = 0.0030 M, $P_\mathrm{Cl_2}$ = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation $\ref{Eq4}$ instead of Equation $\ref{Eq3}$. The overall reaction involves the net transfer of two electrons:
$2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber$
$2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber$
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation $\ref{Eq4}$,
\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ & =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \ & =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture
Exercise $1$
Molecular oxygen will not oxidize $MnO_2$ to permanganate via the reaction
$\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber$
Calculate $E_{cell}$ for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, $P_\mathrm{O_2}$= 0.20 atm, [MNO4] = 1.0 × 10−4 M, and T = 25°C.
Answer
Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5}$
The reaction quotient is therefore $Q = [Zn^{2+}]/[Cu^{2+}]$. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation $\ref{Eq4}$:
\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\ & =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6}
Thus the initial voltage is greater than E° because $Q<1$. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with $\log{Q}$ over this range is linear with a slope of −0.0591/n, as illustrated in Figure $1$. As the reaction proceeds still further, $Q$ continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of $Q$ when Ecell = 0 is calculated as follows:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \ E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ \log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \ Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7}
Recall that at equilibrium, $Q = K$. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8}$
cathode:
$\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9}$
anode:
$\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10}$
Overall
$\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11}$
As the reaction progresses, the concentration of $Ag^+$ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the $Ag^+$ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of $Ag(s)$ in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align*} \nonumber
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example $2$
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
$\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
\begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align*} \nonumber
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise $2$
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber$
What will be the potential when the circuit is closed?
Answer
0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products ($K_{sp}$) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure $1$, which is designed to measure the solubility product of silver chloride:
$K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber$
In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl] and substituting:
\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber
The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align}
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation $\ref{Eq122}$ for $K_{sp}$,
\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt] K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example $3$: Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure $1$, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: Ksp
Strategy:
1. From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
2. Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation $\ref{Eq12}$ and solve for Ksp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of $\ce{Pb^{2+}}$ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
\begin{align*}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \ [\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)
so
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \ \textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \ -7.77 & =\log K_\textrm{sp} \ 1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber
Exercise $3$
A concentration cell similar to the one described in Example $3$ contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer
5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example $3$, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated below.
Example $4$: Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
$\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber$
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
1. Write the overall cell reaction.
2. Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table P2):
Zn(s) + 2H2+(aq) → Zn2+(aq) + H2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation $\ref{Eq4}$), we can calculate [H+] under nonstandard conditions:
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \ \textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \ 16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \ 8.46 &=-\log[\mathrm{H^+}] \ 8.5 &=\mathrm{pH}\end{align*} \nonumber
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise $4$
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows:
$Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber$
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb2+ in the groundwater.
Answer
$1.2 \times 10^{−9}\; M$
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.3%3A_Concentration_Effects_and_the_Nernst_Equation.txt |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
Batteries
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure $1$). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry.
Leclanché Dry Cell
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cell is actually a “wet cell”: the electrolyte is an acidic water-based paste containing $MnO_2$, $NH_4Cl$, $ZnCl_2$, graphite, and starch (part (a) in Figure $1$). The half-reactions at the anode and the cathode can be summarized as follows:
• cathode (reduction):
$\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber$
The $\ce{Zn^{2+}}$ ions formed by the oxidation of $\ce{Zn(s)}$ at the anode react with $\ce{NH_3}$ formed at the cathode and $\ce{Cl^{−}}$ ions present in solution, so the overall cell reaction is as follows:
• overall reaction:
$\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3}$
The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the $\ce{MnO2}$ that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the $\ce{Zn}$ anode reacts spontaneously with $\ce{NH4Cl}$ in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline battery is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
• cathode (reduction)
$\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber$
• overall reaction:
$\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber$
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Button Batteries
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either $\ce{HgO}$ or $\ce{Ag2O}$ as the oxidant rather than $\ce{MnO2}$ in Figure $\PageIndex{1b}$).
The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown):
• cathode (mercury battery): $\ce{HgO(s) + H2O(l) + 2e^{−} -> Hg(l) + 2OH^{−}(aq)} \nonumber$
• Anode (mercury battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• overall reaction (mercury battery): $\ce{Zn(s) + 2HgO(s) -> 2Hg(l) + ZnO(s)} \nonumber$ with $E_{cell} = 1.35 \,V$.
• cathode reaction (silver battery): $\ce{Ag2O(s) + H2O(l) + 2e^{−} -> 2Ag(s) + 2OH^{−}(aq)} \nonumber$
• anode (silver battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• Overall reaction (silver battery): $\ce{Zn(s) + 2Ag2O(s) -> 2Ag(s) + ZnO(s)} \nonumber$ with $E_{cell} = 1.6 \,V$.
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as $\ce{Hg}$ and $\ce{Ag}$.
Lithium–Iodine Battery
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the lithium–iodine battery. The anode is lithium metal, and the cathode is a solid complex of $I_2$. Separating them is a layer of solid $LiI$, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
• cathode (reduction):
$I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11}$
• anode (oxidation):
$2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12}$
• overall:
$2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a}$
with $E_{cell} = 3.5 \, V$
As shown in part (c) in Figure $1$, a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using $TiS_2$, for example, for the cathode.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
Nickel–Cadmium (NiCad) Battery
The nickel–cadmium, or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure $2$, the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
The electrode reactions during the discharge of a $NiCad$ battery are as follows:
• cathode (reduction):
$2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13}$
• anode (oxidation):
$Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14}$
• overall:
$Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15}$
$E_{cell} = 1.4 V$
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
$NiO(OH)_{(s)} + MH \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16}$
The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators"
Lead–Acid (Lead Storage) Battery
The lead–acid battery is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells.
As shown in Figure $3$, the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide ($PbO_2$). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M $H_2SO_4$). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
• cathode (reduction):
$PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17}$
with $E^°_{cathode} = 1.685 \; V$
• anode (oxidation):
$Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18}$
with $E^°_{anode} = −0.356 \; V$
• overall:
$Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19}$
and $E^°_{cell} = 2.041 \; V$
As the cell is discharged, a powder of $PbSO_4$ forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer.
A hydrometer can be used to test the specific gravity of each cell as a measure of its state of charge (www.youtube.com/watch?v=SRcOqfL6GqQ).
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and $PbSO_4$ is converted back to metallic lead and $PbO_2$. If the battery is recharged too vigorously, however, electrolysis of water can occur:
$2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX}$
This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the $PbSO_4$ or $PbO_2$ particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
Fuel Cells
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in Figure $4$. The electrode reactions are as follows:
• cathode (reduction):
$O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20}$
• anode (oxidation):
$2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21}$
• overall:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22}$
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of $O_2 (g)$ at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of $O_2 (g)$.
Summary
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.4%3A_Batteries_and_Fuel_Cells.txt |
Learning Objectives
• To understand the process of corrosion.
Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Corrosion is a REDOX process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide ($\ce{Fe2O3 \cdot xH2O}$), commonly known as rust, that does not provide a tight protective film (Figure $1$). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
• at cathode: $\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber$ with $E^o_{SRP}=1.23\; V$.
• at anode: $\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber$ with $E^o_{SRP} = −0.45\; V$.
• overall: $\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3}$ with $E^o_{cell} = 1.68\; V$.
The $\ce{Fe^{2+}}$ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing $\ce{Fe^{3+}}$, as represented in the following equation:
$\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4}$
The sign and magnitude of $E^o_{cell}$ for the corrosion process (Equation $\ref{Eq3}$) indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3 provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure $2$).
Prophylactic Protection
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy).
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^{2+}}$ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in Table P2 show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure $3$). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
Cathodic Protection
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as $\ce{Zn}$ (E° = −0.76 V for $\ce{Zn^{2+} + 2e^{−} -> Zn}$) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
$\underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5}$
$\underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6}$
$\underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7}$
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans.
In a similar strategy, sacrificial electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure $4$). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example $1$
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
1. If the boat is immersed in seawater, what corrosion reaction will occur? What is $E^o°_{cell}$?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, $E^o°_{cell}$, and preventive measures
Strategy:
1. Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate $E^o°_{cell}$.
2. Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution
1. A According to Table P2, both copper and tin are less active metals than iron (i.e., they have higher positive values of $E^o°_{cell}$ than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which $\ce{O2}$ is reduced, and the iron screws will act as anodes at which iron dissolves:
\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber
Over time, the iron screws will dissolve, and the boat will fall apart.
1. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation $\ref{Eq7}$).
Exercise $1$
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
1. Do you accept his proposal?
2. What else should you have the plumber do while at your home?
Answer a
Not unless you plan to sell the house very soon because the $\ce{Cu/Fe}$ pipe joints will lead to rapid corrosion.
Answer b
Any existing $\ce{Pb/Fe}$ joints should be examined carefully for corrosion of the iron pipes due to the $\ce{Pb–Fe}$ junction; the less active $\ce{Pb}$ will have served as the cathode for the reduction of $\ce{O2}$, promoting oxidation of the more active $\ce{Fe}$ nearby.
Summary
Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.5%3A_Corrosion_and_Its_Prevention.txt |
Learning Objectives
• To understand electrolysis and describe it quantitatively.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
Electrolytic Cells
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a $\,1\; M\, Cd^{2+}$ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (Figure $\PageIndex{1a}$).
The overall reaction is as follows:
$\ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber$
with $E°_{cell} = 0.74\; V$
This reaction is thermodynamically spontaneous as written ($ΔG^o < 0$):
\begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber
In this direction, the system is acting as a galvanic cell.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (Figure $\PageIndex{1b}$). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
• half-reaction at the cathode:
$\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}$
with $E^°_{cathode} = −0.40 \, V$
• half-reaction at the anode:
$\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}$
with $E^°_{anode} = 0.34 \, V$
• Overall Reaction:
$\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}$
with $E^°_{cell} = −0.74 \: V$
Because $E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ by $Cu$—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table $1$.
Table $1$: Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
Ecell > 0 < 0
Electrode Process
anode oxidation oxidation
cathode reduction reduction
Sign of Electrode
anode +
cathode +
Electrolytic Reactions
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten $\ce{NaCl}$, for example, and an electrical potential is applied, $\ce{Cl^{-}}$ is oxidized at the anode, and $\ce{Na^{+}}$ is reduced at the cathode. The overall reaction is as follows:
$\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}$
This is the reverse of the formation of $\ce{NaCl}$ from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of $\ce{NaCl}$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $\ce{NaCl}$ in a Downs cell (Figure $2$). In this specialized cell, $\ce{CaCl2}$ (melting point = 772°C) is first added to the $\ce{NaCl}$ to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
$\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7}$
Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
1. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table P2, which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions.
2. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl2 is electrolyzed, Cl is oxidized because it is the only anion present, but either Na+ or Ca2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al3+), but there are three species that can be oxidized: C, O2−, and F.
In the Hall–Heroult process, C is oxidized instead of O2− or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
Example $1$
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
1. List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 7.5, determine which species will be reduced and which species will be oxidized.
2. Identify the products that will form at each electrode.
Solution
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise $1$
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer
Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (Figure $3$).
The reactions that occur are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}$
• anode: $2H_2O_{(l)} → O_{2(g)} + 4H^+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}$
• overall: $2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}$
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH] = 1.0 × 10−7. Assuming that $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage, represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3, SO42, PO43, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
Electroplating
In a process called electroplating, a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure $4$.
The half-reactions in electroplating a fork, for example, with silver are as follows:
• cathode (fork): $\ce{Ag^{+}(aq) + e^{−} -> Ag(s)} \quad E°_{cathode} = 0.80 V\ \nonumber$
• anode (silver bar): $\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \quad E°_{anode} = 0.80 V \nonumber$
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because $E^o_{cell} = 0\, V$, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
Quantitative Considerations
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction
$\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber$
1 mol of electrons reduces 1 mol of $\ce{Ag^{+}}$ to $\ce{Ag}$ metal. In contrast, in the reaction
$\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber$
1 mol of electrons reduces only 0.5 mol of $\ce{Cu^{2+}}$ to $\ce{Cu}$ metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge ($q$ in coulombs) transferred is the product of the current ($I$ in amperes) and the time ($t$, in seconds):
$q = I \times t \label{20.9.14}$
The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of $\ce{CuSO4}$ for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \[4pt] &=\mathrm{220\;A\cdot s} \[4pt] &=\textrm{220 C} \end{align*} \nonumber
The number of moles of electrons transferred to $\ce{Cu^{2+}}$ is therefore
\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
Example $2$
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$
B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
coulombs = (1.85 × 102mol e)(96,485 C/mol e) = 1.78 × 103 C / mole
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise $2$
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer
5.8 h
Electroplating: Electroplating(opens in new window) [youtu.be]
Summary
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time:
$q = I \times t \nonumber$
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/17%3A_Electrochemistry/17.7%3A_A_Deeper_Look%3A_Electrolysis_of_Water_and_Aqueous_Solutions.txt |
This chapter will present a quantitative description of when the chemical composition of a system is not constant with time. Chemical kinetics is the study of reaction rates, the changes in the concentrations of reactants and products with time. With a discussion of chemical kinetics, the reaction rates or the changes in the concentrations of reactants and products with time are studied. The techniques you are about to learn will enable you to describe the speed of many such changes and predict how the composition of each system will change in response to changing conditions. As you learn about the factors that affect reaction rates, the methods chemists use for reporting and calculating those rates, and the clues that reaction rates provide about events at the molecular level.
18: Chemical Kinetics
Learning Objectives
• To determine the reaction rate of a reaction.
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.
The progress of a simple reaction (A → B) is shown in Figure $1$; the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure $1$. The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
$\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}$
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation $\ref{Eq1}$ is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
Reaction rates generally decrease with time as reactant concentrations decrease.
A Video Discussing Average Reaction Rates. Video Link: Introduction to Chemical Reaction Kinetics(opens in new window) [youtu.be] (opens in new window)
Determining the Reaction Rate of Hydrolysis of Aspirin
We can use Equation $\ref{Eq1}$ to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure $2$.
Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table $1$ and are shown in the graph in Figure $3$.
Table $1$: Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
The data in Table $1$ were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).
The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align*} \nonumber
The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \&=2\times10^{-5}\textrm{ M/h}\end{align*} \nonumber
If the reaction rate is calculated during the last interval given in Table $1$(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
\begin{align*}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \&=8.2\times10^{-6}\textrm{ M/h}\end{align*} \nonumber
Calculating the Reaction Rate of Fermentation of Sucrose
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide:
$\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2}$
The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in Chapter 10, the volume of CO2 gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
$\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3}$
The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation $\ref{Eq3}$ so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with time, so its rate of change is automatically expressed as a positive value.
Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation $\ref{Eq2}$) corresponds to sucrose, so the reaction rate is generally defined as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4}$
Example $1$: Decomposition Reaction I
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)} \nonumber$
Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.
Given: balanced chemical equation
Asked for: reaction rate expressions
Strategy:
1. Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
2. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.
Solution
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B The balanced chemical equation shows that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that the rate of change of [N2O5] and [NO2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Exercise $1$: Contact Process I
The contact process is used in the manufacture of sulfuric acid. A key step in this process is the reaction of $SO_2$ with $O_2$ to produce $SO_3$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}$
Instantaneous Rates of Reaction
The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.
The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.
Example $2$: Decomposition Reaction II
Using the reaction shown in Example $1$, calculate the reaction rate from the following data taken at 56°C:
$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber$
calculate the reaction rate from the following data taken at 56°C:
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175
Given: balanced chemical equation and concentrations at specific times
Asked for: reaction rate
Strategy:
1. Using the equations in Example $1$, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.
2. Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example $1$, the reaction rate can be evaluated using any of three expressions:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t} \nonumber$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})} \nonumber$
B Substituting actual values into the expression,
$\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}$
Similarly, NO2 can be used to calculate the reaction rate:
$\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s} \nonumber$
Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used:
$\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s} \nonumber$
Again, this is the same value obtained from the N2O5 and NO2 data. Thus, the reaction rate does not depend on which reactant or product is used to measure it.
Exercise $2$: Contact Process II
Using the data in the following table, calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148
Answer:
9.0 × 10−6 M/s
Summary
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• General definition of rate for A → B: $\textrm{rate}=\frac{\Delta [\textrm B]}{\Delta t}=-\frac{\Delta [\textrm A]}{\Delta t} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.1%3A_Rates_of_Chemical_Reactions.txt |
Learning Objectives
• To apply rate laws to zeroth, first and second order reactions.
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is
$\text{rate} = k. \nonumber$
We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of $−k$. The value of $k$ is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of $k$, a positive value.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$[A] = [A]_0 − kt \label{14.4.2}$
where $[A]_0$ is the initial concentration of reactant $A$. Equation \ref{14.4.2} has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = [A]$, $mx = −kt$, and $b = [A]_0$.)
Units
In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate. At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure $2$, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where \ce{NAD^{+}}\) (nicotinamide adenine dinucleotide) and $\ce{NADH}$ (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (Figure $\PageIndex{3a}$). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure $3$).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$[A] = [A]_0e^{−kt} \label{14.4.6}$
where $[A]_0$ is the initial concentration of reactant $A$ at $t = 0$; $k$ is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation $\ref{14.4.6}$ predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation $\ref{14.4.6}$ and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of $A$ and $t$:
$\ln[A] = \ln[A]_0 − kt \label{14.4.7}$
Because Equation $\ref{14.4.7}$ has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = \ln[A]$ and $b = \ln[A]_0$, a plot of $\ln[A]$ versus $t$ for a first-order reaction should give a straight line with a slope of $−k$ and an intercept of $\ln[A]_0$. Either the differential rate law (Equation $\ref{14.4.5}$) or the integrated rate law (Equation $\ref{14.4.7}$) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure $5$ is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure $4$ have been studied extensively to find ways of maximizing the concentration of the active species.
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table $1$. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.
Table $1$: Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table $1$ shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table $1$. For example, substituting the values for Experiment 3 into Equation $\ref{14.4.5}$,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example $1$
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
data for the reaction at 650°C
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
1. Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
2. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M)
1.6 × 10−6 s−1 = k
Exercise $1$
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber$
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Data for the reaction at 320°C
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer
first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure $\PageIndex{6a}$ shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.
The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure $6$. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure $6$ for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}} \[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}} \[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure $6$ are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Video Example Using the First-Order Integrated Rate Law Equation:
Example $2$
If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M (k = 1.6 × 10−6 s−1) ?
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
1. Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
2. Given a concentration [A], solve the integrated rate law for time t.
Solution
The exponential form of the integrated rate law for a first-order reaction (Equation $\ref{14.4.6}$) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having the rate constant of k = 1.6 × 10−6 s−1, we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt} \[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
We could also have used the logarithmic form of the integrated rate law (Equation $\ref{14.4.7}$):
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \[4pt] &=-3.912-0.0576=-3.970 \nonumber \[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for $t$. Equation $\ref{14.4.7}$ gives the following:
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right) \[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber
Exercise $2$
In the exercise in Example $1$, you found that the decomposition of sulfuryl chloride ($\ce{SO2Cl2}$) is first order, and you calculated the rate constant at 320°C.
1. Use the form(s) of the integrated rate law to find the amount of $\ce{SO2Cl2}$ that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
2. How long would it take for 90% of the SO2Cl2 to decompose?
Answer a
0.0252 M
Answer b
29 h
Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form
$\ce{2A → products.}\nonumber$
A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}$
Because Equation $\ref{14.4.9}$ has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Second-order reactions generally have the form 2A → products or A + B → products.
Video Discussing the Second-Order Integrated Rate Law Equation: Second-Order Integrated Rate Law Equation(opens in new window) [youtu.be]
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure $7$
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation $\ref{14.4.8}$) or the integrated rate law (Equation $\ref{14.4.9}$).
Table $2$: Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table $2$. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber$
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer]2
This means that the reaction is second order in the monomer. Using Equation $\ref{14.4.8}$ and the data from any row in Table $2$, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\begin{align}\textrm{rate}&=k[\textrm A]^2 \8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure $\PageIndex{8a}$. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in Figure $\PageIndex{8b}$. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.
Example $3$
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experimental data for the reaction at 300°C and four initial concentrations of NO2
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2 \5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber
Exercise $3$
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Some initial rate data at 25°C
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
Answer
second order in HO2; k = 1.4 × 109 M−1·s−1
If a plot of reactant concentration versus time is not linear, but a plot of 1/(reactant concentration) versus time is linear, then the reaction is second order.
Example $4$
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation \ref{14.4.9}) and the rate constant calculated above.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for $t$.
Solution
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9},
\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation $\ref{14.4.9}$ for t, using the concentrations given.
\begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise $4$
In the previous exercise, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation $\ref{14.4.9}$) and the rate constant calculated in the exercise in Example $3$.
Answer
2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $A + B \rightarrow products$, in which the reaction is first order in $A$ and first order in $B$. The differential rate law for this reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \nonumber$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• zeroth-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber$ $[A] = [A]_0 − kt \nonumber$
• first-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber$ $[A] = [A]_0e^{−kt} \nonumber$ $\ln[A] = \ln[A]_0 − kt \nonumber$
• second-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber$ $\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.2%3A_Rate_Laws.txt |
Learning Objectives
• To determine the individual steps of a simple reaction.
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.
In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
$\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1}$
For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.
The overall sequence of elementary reactions is the mechanism of the reaction.
Molecularity and the Rate-Determining Step
To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
$\ce{NO2(g) + CO(g) -> NO(g) + CO2 (g)} \label{14.6.2}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of $\ce{NO2}$ with a molecule of $\ce{CO}$ that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
$rate = k[\ce{NO2}]^2 \label{14.6.3}$
The fact that the reaction is second order in $[\ce{NO2}]$ and independent of $[\ce{CO}]$ tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be
$rate = k[\ce{NO2}][\ce{CO}]. \nonumber$
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
two-step mechanism
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\textrm{slow}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{elementary reaction}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\rightarrow\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{elementary reaction}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$ $\textrm{overall reaction}$
According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The $\ce{NO3}$ molecule is an intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.
The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.
Using Molecularity to Describe a Rate Law
The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)
Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table $1$). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is
$rate = k[A]. \nonumber$
For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure $1$. For a bimolecular elementary reaction of the form A + B → products, the general rate law is
$rate = k[A][B]. \nonumber$
Table $1$: Common Types of Elementary Reactions and Their Rate Laws
Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third
For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).
Identifying the Rate-Determining Step
Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
rate laws for each elementary reaction in our example as well as for the overall reaction.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\mathrm{k_1}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{rate}=k_1[\mathrm{NO_2}]^2\textrm{ (predicted)}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{rate}=k_2[\mathrm{NO_3}][\mathrm{CO}]\textrm{ (predicted)}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\xrightarrow{k}\mathrm{NO}+\mathrm{CO_2}$ $\textrm{rate}=k[\mathrm{NO_2}]^2\textrm{ (observed)}$
The experimentally determined rate law for the reaction of $NO_2$ with $CO$ is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so $k$ for the overall reaction must equal $k_1$. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.
Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.
Example $1$: A Reaction with an Intermediate
In an alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{k_1}\mathrm{N_2O_4}$
$\textrm{step 2}$ $\underline{\mathrm{N_2O_4}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO}+\mathrm{NO_2}+\mathrm{CO_2}}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$
Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?
Given: elementary reactions
Asked for: rate law for each elementary reaction and overall rate law
Strategy:
1. Determine the rate law for each elementary reaction in the reaction.
2. Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.
Solution
A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].
B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.
Exercise $1$
Iodine monochloride ($\ce{ICl}$) reacts with $\ce{H2}$ as follows:
$\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber$
The experimentally determined rate law is $rate = k[\ce{ICl}][\ce{H2}]$. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: $\ce{HI}$ is an intermediate.)
Answer
Solutions to Exercise 14.6.1
$\textrm{step 1}$ $\mathrm{ICl}+\mathrm{H_2}\xrightarrow{k_1}\mathrm{HCl}+\mathrm{HI}$ $\mathrm{rate}=k_1[\mathrm{ICl}][\mathrm{H_2}]\,(\textrm{slow})$
$\textrm{step 2}$ $\underline{\mathrm{HI}+\mathrm{ICl}\xrightarrow{k_2}\mathrm{HCl}+\mathrm{I_2}}$ $\mathrm{rate}=k_2[\mathrm{HI}][\mathrm{ICl}]\,(\textrm{fast})$
$\textrm{sum}$ $\mathrm{2ICl}+\mathrm{H_2}\rightarrow\mathrm{2HCl}+\mathrm{I_2}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Example $2$ : Nitrogen Oxide Reacting with Molecular Hydrogen
Assume the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process:
the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process
$\textrm{step 1}$ $\mathrm{NO}+\mathrm{NO}\xrightarrow{k_1}\mathrm{N_2O_2}$ $\textrm{(fast)}$
$\textrm{step 2}$ $\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}$ $\textrm{(slow)}$
$\textrm{step 3}$ $\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}$ $\textrm{(fast)}$
Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction:
$\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed}$
Answer
• Step 1: $rate = k_1[\ce{NO}]^2$
• Step 2: $rate = k_2[\ce{N_2O_2}][\ce{H_2}]$
• Step 3: $rate = k_3[\ce{N_2O}][\ce{H_2}]$
The overall reaction is then
$\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber$
• Rate Determining Step : #2
• Yes, because the rate of formation of $[\ce{N_2O_2}] = k_1[\ce{NO}]^2$. Substituting $k_1[\ce{NO}]^2$ for $[\ce{N_2O_2}]$ in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where $k = k_1k_2$.
Reaction Mechanism (Slow step followed by fast step): Reaction Mechanism (Slow step Followed by Fast Step)(opens in new window) [youtu.be] (opens in new window)
Summary
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step.
18.4: Reaction Mechanisms and Rate
note sure without book | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.3%3A_Reaction_Mechanisms.txt |
Learning Objectives
• To understand why and how chemical reactions occur.
It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates.
Microscopic Factor 1: Collisional Frequency
Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The collisional frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency ($Z_{AB}$) between two species in a gas is straightforward, it is beyond the scope of this text and the equation for collisional frequency of $A$ and $B$ is the following:
$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq}$
with
• $N_A$ and $N_B$ are the numbers of $A$ and $B$ molecules in the system, respectively
• $r_a$ and $r_b$ are the radii of molecule $A$ and $B$, respectively
• $k_B$ is the Boltzmann constant $k_B$ =1.380 x 10-23 Joules Kelvin
• $T$ is the temperature in Kelvin
• $\mu_{AB}$ is calculated via $\mu_{AB} = \frac{m_Am_B}{m_A + m_B}$
The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that $Z_{AB}$ increases with increasing density (i.e., increasing $N_A$ and $N_B$), with increasing reactant size ($r_a$ and $r_b$), with increasing velocities (predicted via Kinetic Molecular Theory), and with increasing temperature (although weakly because of the square root function).
A Video Discussing Collision Theory of Kinetics: Collusion Theory of Kinetics (opens in new window) [youtu.be]
Microscopic Factor 2: Activation Energy
Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time.
The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy ($E_a$). We will define this concept using the reaction of $\ce{NO}$ with ozone, which plays an important role in the depletion of ozone in the ozone layer:
$\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber$
Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate.
Experimental rate law for this reaction is
$\text{rate} = k [\ce{NO}][\ce{O3}] \nonumber$
and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. Figure $1$ shows a plot of the rate constant of the reaction of $\ce{NO}$ with $\ce{O3}$ at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the Clausius-Claperyon equation). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier.
In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complex or the transition state of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily.
Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence.
We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. Figure $2$ shows a plot for the NO–O3 system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction ($ΔE$) is negative, which means that the reaction releases energy. (In this case, $ΔE$ is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction ($E_a$ is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur.
Figure $\PageIndex{3a}$ illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, Figure $\PageIndex{3b}$ illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and $ΔE > 0$. Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere.
For similar reactions under comparable conditions, the one with the smallest Ea will occur most rapidly.
Whereas $ΔE$ is related to the tendency of a reaction to occur spontaneously, $E_a$ gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest $E_a$ will occur more rapidly.
Figure $4$ shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than Ea; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than Ea. Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier.
Video Discussing Transition State Theory: Transition State Theory(opens in new window) [youtu.be]
Microscopic Factor 3: Sterics
Even when the energy of collisions between two reactant species is greater than $E_a$, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For $\ce{NO}$ and $\ce{O3}$ to produce $\ce{NO2}$ and $\ce{O2}$, a terminal oxygen atom of $\ce{O3}$ must collide with the nitrogen atom of $\ce{NO}$ at an angle that allows $\ce{O3}$ to transfer an oxygen atom to $\ce{NO}$ to produce $\ce{NO2}$ (Figure $4$). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of $\ce{NO}$ and $\ce{O3}$ result in a reaction at kinetic energies greater than $E_a$, most collisions of $\ce{NO}$ and $\ce{O3}$ are unproductive. The fraction of orientations that result in a reaction is called the steric factor ($\rho$) and its value can range from $\rho=0$ (no orientations of molecules result in reaction) to $\rho=1$ (all orientations result in reaction).
Macroscopic Behavior: The Arrhenius Equation
The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 1030 times per second. If every collision produced two molecules of $\ce{NO}$, the atmosphere would have been converted to $\ce{NO}$ and then $\ce{NO2}$ a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide.
For an $A + B$ elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship:
$\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber$
where
$\text{rate} = k[A][B] \label{14.5.2}$
Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, $A$, called the frequency factor:
$k=Ae^{-E_{\Large a}/RT} \label{14.5.3}$
The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, $A$ is actually not constant (Equation \ref{freq}). Instead, $A$ increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time.
Equation $\ref{14.5.3}$ is known as the Arrhenius equation and summarizes the collision model of chemical kinetics, where $T$ is the absolute temperature (in K) and R is the ideal gas constant [8.314 J/(K·mol)]. $E_a$ indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large $E_a$ increases rapidly with increasing temperature, whereas the reaction rate with a smaller $E_a$ increases much more slowly with increasing temperature.
If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of Equation $\ref{14.5.3}$,
\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align}
Equation $\ref{14.5.4}$ is the equation of a straight line,
$y = mx + b \nonumber$
where $y = \ln k$ and $x = 1/T$. This means that a plot of $\ln k$ versus $1/T$ is a straight line with a slope of $−E_a/R$ and an intercept of $\ln A$. In fact, we need to measure the reaction rate at only two temperatures to estimate $E_a$.
Knowing the $E_a$ at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining $E_a$ from reaction rates measured at several temperatures is illustrated in Example $1$.
A Video Discussing The Arrhenius Equation: The Arrhenius Equation(opens in new window) [youtu.be]
Example $1$: Chirping Tree Crickets
Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping ($f$) as a function of temperature ($T$). Use the data in the following table, along with the graph of ln[chirping rate] versus $1/T$ to calculate $E_a$ for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F).
Chirping Tree Crickets Frequency Table
Frequency (f; chirps/min) ln f T (K) 1/T (K)
200 5.30 299 3.34 × 10−3
179 5.19 298 3.36 × 10−3
158 5.06 296 3.38 × 10−3
141 4.95 294 3.40 × 10−3
126 4.84 293 3.41 × 10−3
112 4.72 292 3.42 × 10−3
100 4.61 290 3.45 × 10−3
89 4.49 289 3.46 × 10−3
79 4.37 287 3.48 × 10−3
Given: chirping rate at various temperatures
Asked for: activation energy and chirping rate at specified temperature
Strategy:
1. From the plot of $\ln f$ versus $1/T$, calculate the slope of the line (−Ea/R) and then solve for the activation energy.
2. Express Equation \ref{14.5.4} in terms of k1 and T1 and then in terms of k2 and T2.
3. Subtract the two equations; rearrange the result to describe k2/k1 in terms of T2 and T1.
4. Using measured data from the table, solve the equation to obtain the ratio k2/k1. Using the value listed in the table for k1, solve for k2.
Solution
A If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of $\ln f$ versus $1/T$ should give a straight line (Figure $6$).
Also, the slope of the plot of $\ln f$ versus $1/T$ should be equal to $−E_a/R$. We can use the two endpoints in Figure $6$ to estimate the slope:
\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber
A computer best-fit line through all the points has a slope of −6.67 × 103 K, so our estimate is very close. We now use it to solve for the activation energy:
\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber
B If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use Equation \ref{14.5.4} to express the known rate constant ($k_1$) at the first temperature ($T_1$) as follows:
$\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber$
Similarly, we can express the unknown rate constant ($k_2$) at the second temperature ($T_2$) as follows:
$\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber$
C These two equations contain four known quantities (Ea, T1, T2, and k1) and two unknowns (A and k2). We can eliminate A by subtracting the first equation from the second:
\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber
Then
$\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber$
D To obtain the best prediction of chirping rate at 308 K (T2), we try to choose for T1 and k1 the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for T1 = 296 K, where f = 158, and using the $E_a$ calculated previously,
\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \[4pt] &=0.87 \end{align*} \nonumber
Thus k308/k296 = 2.4 and k308 = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute.
Exercise $\PageIndex{1A}$
The equation for the decomposition of $\ce{NO2}$ to $\ce{NO}$ and $\ce{O2}$ is second order in $\ce{NO2}$:
$\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber$
Data for the reaction rate as a function of temperature are listed in the following table. Calculate $E_a$ for the reaction and the rate constant at 700 K.
Data for the reaction rate as a function of temperature
T (K) k (M−1·s−1)
592 522
603 755
627 1700
652 4020
656 5030
Answer
$E_a$ = 114 kJ/mol; k700= 18,600 M−1·s−1 = 1.86 × 104 M−1·s−1.
Exercise $\PageIndex{1B}$
What $E_a$ results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C?
Answer
about 51 kJ/mol
A Video Discussing Graphing Using the Arrhenius Equation: Graphing Using the Arrhenius Equation (opens in new window) [youtu.be] (opens in new window)
Summary
For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is $k=Ae^{-E_{\Large a}/RT}$. A minimum energy (activation energy,v$E_a$) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: $k=Ae^{-E_{\Large a}/RT}$. A plot of the natural logarithm of k versus 1/T is a straight line with a slope of −Ea/R. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.5%3A_Effect_of_Temperature_on_Reaction_Rates.txt |
under construction
18.7: Kinetics of Catalysis
Learning Objectives
• To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.
Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower Ea, but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst (Figure $1$). Nevertheless, because of its lower Ea, the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes.
A catalyst affects Ea, not ΔE.
Heterogeneous Catalysis
In heterogeneous catalysis, the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.
An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure $2$, the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.
Figure $2$ shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.
Several important examples of industrial heterogeneous catalytic reactions are in Table $1$. Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.
Table $1$: Some Commercially Important Reactions that Employ Heterogeneous Catalysts
Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate $\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}$ $\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}$
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
Homogeneous Catalysis
In homogeneous catalysis, the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds (Table $2$), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis.
Table $2$: Some Commercially Important Reactions that Employ Homogeneous Catalysts
Commercial Process Catalyst Reactants Final Product
Union Carbide [Rh(CO)2I2] CO + CH3OH CH3CO2H
hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H
hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO
adiponitrile process Ni/PR3complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene
Enzymes
Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrate.
Because enzymes can increase reaction rates by enormous factors (up to 1017 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water (Figure $3$).
Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research.
Summary
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.6%3A_A_Deeper_Look%3A_Reaction_Dynamics.txt |
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei. Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst.
We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe.
19: Nuclear Chemistry
Learning Objectives
• To understand the factors that affect nuclear stability.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
The Atomic Nucleus
Each element can be represented by the notation \(^A_Z \textrm X\), where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons, and an atom with a particular number of protons and neutrons is called a nuclide. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
stable isotopes of oxygen represented in different ways
\(^A_Z \textrm X\) \(\ce{^{16}_8 O}\) \(\ce{^{17}_8 O}\) \(\ce{^{18}_8 O}\)
\(^A \textrm X\) \(\ce{^{16} O}\) \(\ce{^{17} O}\) \(\ce{^{18} O}\)
\(\textrm{element-A:}\) \(\textrm{oxygen-16}\) \(\textrm{oxygen-17}\) \(\textrm{oxygen-18}\)
Because the number of neutrons is equal to AZ, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive, emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopes.
Nuclear Stability
The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force, an extremely powerful but very short-range attractive force between nucleons (Figure \(1\)). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(2\). The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure \(3\), more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Most stable nuclei contain even numbers of both neutrons and protons
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(2\), the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Origin of the Magic Numbers
Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the Nuclear Shell Model and the Liquid Drop Model. Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{15}^{30} P}\)
2. \(\ce{_{43}^{98} Tc}\)
3. tin-118
4. \(\ce{_{94}^{239} Pu}\)
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(2\), its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is.
b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is.
c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable.
d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive.
Exercise \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{90}^{232} Th}\)
2. \(\ce{_{20}^{40} Ca}\)
3. \(\ce{_8^{15} O}\)
4. \(\ce{_{57}^{139} La}\)
Answer a
radioactive
Answer b
stable
Answer c
radioactive
Answer d
stable
Superheavy Elements
In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements, with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.1%3A_Mass-Energy_Relationships_in_Nuclei.txt |
Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
Relative Abundances of the Elements on Earth and in the Known Universe
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure $1$. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure $1$ illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table $1$ for some common elements.
Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Table $1$: Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Synthesis of the Elements in Stars
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure $2$). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
$_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta \_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma \_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}$
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two $\gamma$ rays, and a great deal of energy:
$4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2}$
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
$2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3}$
Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
$_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4}$
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
$_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5}$
$_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6}$
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
$_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7}$
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
The Formation of Heavier Elements in Supernovas
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure $2$). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure $3$). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
$_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}$
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
Example $1$: Carbon Burning Stars
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution
1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{12}^{24}\textrm{Mg}$.
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{10}^{20}\textrm{Ne}+\,_2^4\alpha$.
Exercise $1$
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer
19 neutrons; $_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta$
Summary
Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.2%3A_Nuclear_Decay_Processes.txt |
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation:
$\ln\dfrac{[\textrm A]_0}{[\textrm A]}=kt \label{21.4.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation $\ref{21.4.1}$ gives
$\ln\dfrac{[\textrm A]_0}{[\textrm A]_0/2}=\ln 2=kt_{1/2}$
Substituting $\ln{2} \approx 0.693$ into the equation results in the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure $1$, and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100\%}{2}=50\%$ $\dfrac{1}{2}(100\%)=50\%$
2 $\dfrac{50\%}{2}=25\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right)(100\%)=25\%$
3 $\dfrac{25\%}{2}=12.5\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right )\left (\dfrac{1}{2}\right)(100\%)=12.5\%$
n $\dfrac{100\%}{2^n}$ $\left(\dfrac{1}{2}\right)^n(100\%)=\left(\dfrac{1}{2}\right)^n\%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example $1$
The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
1. Use Equation $\ref{21.4.2}$ to calculate the half-life of the reaction.
2. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
3. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution
A We can calculate the half-life of the reaction using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}$
Thus it takes almost 8 h for half of the cis-platin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^5}=\dfrac{0.053\textrm{ M}}{32}=0.0017\textrm{ M}$
After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^{10}}=\dfrac{0.053\textrm{ M}}{1024}=5.2\times10^{-5}\textrm{ M}$
C The percent completion after 5 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-0.0017\textrm{ M})(100)}{0.053}=97\%$
The percent completion after 10 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-5.2\times10^{-5}\textrm{ M})(100)}{0.053\textrm{ M}}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise $1$
Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C.
1. What is the half-life for the reaction under these conditions?
2. If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer a
4.3 × 105 s = 120 h = 5.0 days;
Answer b
4.8 × 10−3 M
Radioactive Decay Rates
Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity (A) as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \label{21.4.3}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A = kN \label{21.4.4}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation $\ref{21.4.3}$ and Equation $\ref{21.4.4}$, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \label{21.4.5}$
Equation $\ref{21.4.5}$ is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation $\ref{21.4.5}$) or the integrated rate law:
$N = N_0e^{−kt}$
$\ln \dfrac{N}{N_0}=-kt \label{21.4.6}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.6, along with some of their applications.
Table $2$: Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
Note
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$\ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure $2$). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example $2$
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation $\ref{21.4.4}$ to calculate N0/N. Then substitute the value for the half-life of 14C into Equation $\ref{21.4.2}$ to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation $\ref{21.4.6}$ to obtain the elapsed time.
Solution
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation $\ref{21.4.6}$) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\begin{align}\ln\dfrac{N}{N_0}&=-kt \ \dfrac{\ln(N/N_0)}{k}&=t\end{align}
A From Equation $\ref{21.4.4}$, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_0}{A}=\dfrac{kN_0}{kN}=\dfrac{N_0}{N}=\dfrac{15}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\textrm{ yr}}=1.22\times10^{-4}\textrm{ yr}^{-1}$
B Substituting into the equation for t,
$t=\dfrac{\ln(N_0/N)}{k}=\dfrac{\ln(15/8.0)}{1.22\times10^{-4}\textrm{ yr}^{-1}}=5.2\times10^3\textrm{ yr}$
From our calculations, the man died 5200 yr ago.
Exercise $2$
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer
30,000 yr
Summary
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.3%3A_Kinetics_of_Radioactive_Decay.txt |
Learning Objectives
• To know the differences between ionizing and nonionizing radiation and their effects on matter.
• To identify natural and artificial sources of radiation.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
Ionizing versus Nonionizing Radiation
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
$\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}}$
Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure $1$). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle:
$\text{1 MeV/particle} = \text{96 billion J/mol}. \nonumber$
The Effects of Ionizing Radiation on Matter
The effects of ionizing radiation depend on four factors:
1. The type of radiation, which dictates how far it can penetrate into matter
2. The energy of the individual particles or photons
3. The number of particles or photons that strike a given area per unit time
4. The chemical nature of the substance exposed to the radiation
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in Figure $2$. Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table $1$.
Table $1$: Some Properties of Ionizing Radiation
Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air*
*Distance at which half of the radiation has been absorbed.
α particles 3–9 < 0.05 mm < 10 cm
β particles ≤ 3 < 4 mm 1 m
x-rays <10−2 < 1 cm < 3 m
γ rays 10−2–101 < 20 cm > 3 m
There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
$\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}}$
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man) was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
Wilhelm Röntgen
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
Natural Sources of Radiation
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (Figure $3$). One component of background radiation is cosmic rays, high-energy particles and $\gamma$ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as $\ce{^{14}C}$:
$\ce{^{14}_7 N + ^1_0 n \rightarrow ^{14}_6 C + ^1_1p }\label{Eq3}$
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as $\ce{^{232}Th}$ and $\ce{^{238}U}$ as well as radioactive daughter isotopes, such as$\ce{^{226}Ra}$. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the $K^+$ ion. Naturally occurring potassium contains 0.0117% $\ce{^{40}K}$, which decays by emitting both a β particle and a (\gamma\) ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 $\ce{^{40}K}$ nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
$\ce{^{222}_{86} Rn \rightarrow ^4_2\alpha + ^{218}_{84} Po + ^4_2\alpha + ^{214}_{82} Pb } \label{Eq4}$
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of $\ce{^{218}Po}$ releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The $\ce{^{218}Po}$ isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
Artificial Sources of Radiation
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Example $1$
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
1. Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
2. Determine the number of decays per year for this amount of 40K.
3. Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
$\textrm{moles }^{40}\textrm K= 140\textrm{ g K} \times \dfrac{0.0117\textrm{ mol }^{40}\textrm K}{100\textrm{ mol K}}\times\dfrac{1\textrm{ mol K}}{40.0\textrm{ g K}}=4.10\times10^{-4}\mathrm{\,mol\,^{40}K} \nonumber$
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
$\dfrac{\textrm{decays}}{\textrm{year}}=4.10\times10^{-4}\mathrm{\,mol^{40}\,K}\times\dfrac{1.05\times10^7\textrm{ decays/s}}{\mathrm{1.00\,mol\,^{40}K}}\times\dfrac{60\textrm{ s}}{1\textrm{ min}}\times\dfrac{60\textrm{ min}}{1\textrm{ h}}\times\dfrac{24\textrm{ h}}{1\textrm{ day}}\times\dfrac{365\textrm{ days}}{1\textrm{ yr}}$
C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event:
\begin{align*}\textrm{total energy per year}&=\dfrac{1.36\times10^{11}\textrm{ decays}}{\textrm{yr}}\times\dfrac{1.32\textrm{ MeV}}{\textrm{decays}}\times\dfrac{10^6\textrm{ eV}}{\textrm{MeV}}\times\dfrac{1.602\times10^{-19}\textrm{ J}}{\textrm{eV}}\&=2.87\times10^{-2}\textrm{ J/yr}\end{align*} \nonumber
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
\begin{align*}\textrm{radiation dose per year}&=\dfrac{2.87\times10^{-2}\textrm{ J/yr}}{\textrm{70.0 kg}}\times\dfrac{1\textrm{ rad}}{1\times10^{-2}\textrm{ J/kg}}\&=4.10\times10^{-2}\textrm{ rad/yr}=41\textrm{ mrad/yr}\end{align*} \nonumber
This corresponds to almost half of the normal background radiation most people experience.
Exercise $1$
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer
5.7 × 103 rad/yr (which is 10 times the fatal dose)
Assessing the Impact of Radiation Exposure
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table $2$. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table $2$: The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) Symptoms/Effects
< 5 no observable effect
5–20 possible chromosomal damage
20–100 temporary reduction in white blood cell count
50–100 temporary sterility in men (up to a year)
100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded
> 300 permanent sterility in women
> 500 fatal to 50% within 30 days; destruction of bone marrow and intestine
> 3000 fatal within hours
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in Figure $4$, but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Summary
Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.4%3A_Radiation_in_Biology_and_Medicine.txt |
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n}\label{5.2.16}$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $1$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$:
$^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows:
$^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows:
$^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus
$^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore
$^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows:
$^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta$
Exercise $1$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer
1. $^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
2. $^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
3. $^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $2$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $2$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer
1. beta decay
2. positron emission or electron capture
3. alpha decay | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.5%3A_Nuclear_Fission.txt |
Learning Objectives
• Describe the nuclear reactions in a nuclear fusion reaction
• Quantify the energy released or absorbed in a fusion reaction
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1H$ and a triton, $^3_1H$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $2$).
Example $1$
Calculate the energy released in each of the following hypothetical processes.
1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$
2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$
3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$
Solution
1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$
2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$
3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$
Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.
Nuclear Reactors
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $3$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.6%3A_Nuclear_Fusion_and_Nucleosynthesis.txt |
Spectroscopy is the use of the absorption, emission, or scattering of electromagnetic radiation by atoms or molecules (or atomic or molecular ions) to qualitatively or quantitatively study the atoms or molecules, or to study physical processes. The interaction of radiation with matter can cause redirection of the radiation and/or transitions between the energy levels of the atoms or molecules. A transition from a lower level to a higher level with transfer of energy from the radiation field to the atom or molecule is called absorption. A transition from a higher level to a lower level is called emission if energy is transferred to the radiation field, or nonradiative decay if no radiation is emitted. Redirection of light due to its interaction with matter is called scattering, and may or may not occur with transfer of energy, i.e., the scattered radiation has a slightly different or the same wavelength.
20: Molecular Spectroscopy and Photochemistry
The electromagnetic spectrum
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio.
Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: shorter wavelengths correspond to higher energy.
High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10-16 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10-9 m), while radio waves can be several hundred meters in length.
The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of particles, called photons, rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as:
$E = \dfrac{hc}{\lambda} \tag{4.1.1}$
where E is energy in kJ/mol, λ (the Greek letter lambda) is wavelength in meters, c is 3.00 x 108 m/s (the speed of light), and h is 3.99 x 10-13 kJ·s·mol-1, a number known as Planck’s constant.
Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s-1.
When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression:
$\lambda \nu = c \tag{4.1.2}$
where ν (the Greek letter ‘nu’) is frequency in s-1. Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 1014 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
(Image from Wikipedia commons)
Notice that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest.
Exercise 4.4: Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kJ/mol of photons?
Solutions
Overview of a molecular spectroscopy experiment
In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed.
Here is the key to molecular spectroscopy: a given molecule will specifically absorb only those wavelengths which have energies that correspond to the energy difference of the transition that is occurring. Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of ΔE, the molecule will specifically absorb radiation with wavelength that corresponds to ΔE, while allowing other wavelengths to pass through unabsorbed.
By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure.
These generalized ideas may all sound quite confusing at this point, but things will become much clearer as we begin to discuss specific examples.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/20%3A_Molecular_Spectroscopy_and_Photochemistry/20.1%3A_General_Aspects_of_Molecular_Spectroscopy.txt |
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring that is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the stretching mode of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 1013 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 1013 Hz.
If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
Exercise 4.5: : Express the wavenumber value of 3000 cm-1 in terms of wavelength (in meter units) frequency (in Hz), and associated energy (in kJ/mol).
The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm-1 (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 1013 Hz, and a ΔE value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: a strong peak in the 1650-1750 cm-1 region is a dead giveaway for the presence of a carbonyl group. Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm-1).
The jagged peak at approximately 2900-3000 cm-1 is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum.
You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm-1 region. This part of the spectrum is called the fingerprint region. While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. It was the IR fingerprint region of the suspicious yellow paint that allowed for its identification as a pigment that could not possibly have been used by the purported artist, William Aiken Walker.
Now, let’s take a look at the IR spectrum for 1-hexanol.
As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1.
We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
You can see many more examples of IR spectra in the Spectral Database for Organic Compounds
Exercise 4.6: Explain how you could use the C-C and C-H stretching frequencies in IR spectra to distinguish between four constitutional isomers: 1,2-dimethylcyclohexene, 1,3-octadiene, 3-octyne, and 1-octyne.
Exercise 4.7: Using the online Spectral Database for Organic Compounds, look up IR spectra for the following compounds, and identify absorbance bands corresponding to those listed in the table above. List actual frequencies for each signal to the nearest cm-1 unit, using the information in tables provided on the site.
a) 1-methylcyclohexanol
b) 4-methylcyclohexene
c) 1-hexyne
d) 2-hexyne
e) 3-hexyne-2,5-diol
Exercise 4.8: A carbon-carbon single bond absorbs in the fingerprint region, and we have already seen the characteristic absorption wavelengths of carbon-carbon double and triple bonds. Rationalize the trend in wavelengths. (Hint - remember, we are thinking of bonds as springs, and looking at the frequency at which they 'bounce').
Solutions
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in table 1 in the tables section at the end of the text.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol (this type of reaction is discussed in detail in chapter 16).
Kahn Academy video tutorials on infrared spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/20%3A_Molecular_Spectroscopy_and_Photochemistry/20.2%3A_Vibrations_and_Rotations_of_Molecules%3A_In.txt |
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated $\pi$ systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n \rightarrow \pi^*$ transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise 4.9: What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the lmax of b-carotene?
Exercise 4.10: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solutions
Protecting yourself from sunburn
Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.
Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (para-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.
Looking at UV-vis spectra
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
Schematic for a UV-Vis spectrophotometer
(Image from Wikipedia Commons)
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is $\lambda_{max}$, which is the wavelength at maximal light absorbance. As you can see, NAD+ has $\lambda_{max} = 260\;nm$. We also want to record how much light is absorbed at $\lambda_{max}$. Here we use a unitless number called absorbance, abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \dfrac{I_0}{I} \tag{4.3.1}$
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Exercise 4.11: Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solutions
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.
If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
$\epsilon = \dfrac{A}{c} \tag{4.3.2}$
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L* mol-1*cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 L* mol-1*cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Exercise 4.12: 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Kahn Academy video tutorials on UV-Vis spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/20%3A_Molecular_Spectroscopy_and_Photochemistry/20.3%3A_Excited_Electronic_States%3A_Electronic_Spe.txt |
Discussion Questions
• What are crystal defects and how are they classified?
• How do impurities affect the structure and properties of a solid?
• What are color centers and how do they affect electric conductivity of solids?
Few, if any, crystals are perfect in that all unit cells consist of the ideal arrangement of atoms or molecules and all cells line up in a three dimensional space with no distortion. Some cells may have one or more atoms less whereas others may have one or more atoms than the ideal unit cell. The imperfection of crystals are called crystal defects. Crystal defects are results of thermodynamic equilibrium contributed also by the increase in entropy TS term of the Gibb's free energy:
$\Delta G = \Delta H - T \Delta S \nonumber$
Only at the unattainable absolute zero K will a crystal be perfect, in other words, no crystals are absolutely perfect. However, the degree of imperfection vary from compound to compound. On the other hand, some solid-like structure called flickering clusters also exist in a liquid. For example, the density of water is the highest at 277 K. The flickering clusters increase as temperature drops below 277 K, and the water density decreases as a result. The missing and lacking of atoms or ions in an ideal or imaginary crystal structure or lattice and the misalignment of unit cells in real crystals are called crystal defects or solid defects. Crystal defects occur as points, along lines, or in the form of a surface, and they are called point, line, or plane defects respectively.
Point Defects
Point defects can be divided into Frenkel defects and Schottky defects, and these often occur in ionic crystals. The former are due to misplacement of ions and vacancies. Charges are balanced in the whole crystal despite the presence of interstitial or extra ions and vacancies. On the other hand, when only vacancies of cation and anions are present with no interstitial or misplaced ions, the defects are called Schottky defects.
Point defects are common in crystals with large anions such as AgBr, AgI, RbAgI4. Due to the defects, the ions have some freedom to move about in crystals, making them relatively good conductors. These are called ionic conductors, unlike metals in which electrons are responsible for electric conductivity. Recently, ionic conductors have attracted a lot of attention because the fuel cell and battery technologies require conducting solids to separate the electrodes.
Line Defects
Line defects are mostly due to misalignment of ions or presence of vacancies along a line. When lines of ions are missing in an otherwise perfect array of ions, an edge dislocation appeared. Edge dislocation is responsible for the ductility and malleability. In fact the hammering and stretching of materials often involve the movement of edge dislocation. Movements of dislocations give rise to their plastic behavior. Line dislocations usually do not end inside the crystal, and they either form loops or end at the surface of a single crystal.
A dislocation is characterized by its Burgers vector: If you imagine going around the dislocation line, and exactly going back as many atoms in each direction as you have gone forward, you will not come back to the same atom where you have started. The Burgers vector points from start atom to the end atom of your journey (This "journey" is called Burgers circuit in dislocation theory).
In this electron microscope image of the surface of a crystal, you see point defects and a Burger journey around an edge dislocation. The dislocation line is in the crystal, and the image shows its ending at the surface. A Burger vector is approximately perpendicular to the dislocation line, and the missing line of atoms is somewhere within the block of the Buerger journey.
If the misalignment shifts a block of ions gradually downwards or upwards causing the formation of a screw like deformation, a screw dislocation is formed. The diagram here shows the idealized screw dislocation.
Line defects weakens the structure along a one-dimensional space, and the defects type and density affects the mechanical properties of the solids. Thus, formation and study of dislocations are particularly important for structural materials such as metals. This link gives some impressive images of dislocations. Chemical etching often reveal pits which are visible under small magnifications.
Example $1$
The Table of X-ray Crystallographic Data of Minerals (The CRC Handbook of Chemistry and Physics) list the following for bunsenite (NiO): Crystal system: cubic, structure type: rock salt, a = 4.177*10-8 cm. In the table of Physical Constant of Inorganic Compounds, the density of bunsenite (NiO) is 6.67g / cm3. From these values, evaluate the cell volume (volume of the unit cell), sum of Ni and O radii (rNi + rO), 2 Molar volumes, (X-ray) density, and the Schottky defect vacancy rate.
Solution
Actually, most of the required values have been listed in the table, but their evaluations illustrate the methods. These values are evaluated below:
Cell volume = a3
= (4.177*10-8)3
= 72.88*10-24 cm3
$r_{Ni}+r_O = \dfrac{a}{ 2} \nonumber$
$= 2.088 \times 10^{-8} \nonumber$
X-ray density = 4*(58.69+16.00) / (6.023*1023*72.88*10-24)
= 6.806 g / cm3,
Compared to the observed density = 6.67 g / cm3. The molar volumes (58.69+16.00) / density are thus
74.69 / 6.806 = 10.97 cm3; and 74.69 / 6.67 = 11.20 cm3
The vacancy rate = 6.806 - 6.67 / 6.806
= 0.02 (or 2%)
DISCUSSION
These methods are hints to assignments.
From the given conditions, we cannot calculate individual radii of Ni and O, but their sum is calculable.
Plane Defects
Plane defects occur along a 2-dimensional surface. The surface of a crystal is an obvious imperfection, because these surface atoms are different from those deep in the crystals. When a solid is used as a catalyst, the catalytic activity depends very much on the surface area per unit mass of the sample. For these powdery material, methods have been developed for the determination of unit areas per unit mass.
Another surface defects are along the grain boundaries. A grain is a single crystal. If many seeds are formed when a sample starts to crystallize, each seed grow until they meet at the boundaries. Properties along these boundaries are different from the grains. A third plane defects are the stacking faults. For example, in the close packing arrangement, the adjancent layers always have the AB relationship. In a ccp (fcc) close packing sequence, ...ABCABC..., one of the layer may suddenly be out of sequence, and become ..ABABCABC.... Similarly, in the hcp sequence, there is a possibility that one of the layer accidentally startes in the C location and resulting in the formation of a grain boundary.
How do impurities affect the structure and properties of a solid?
You already know that to obtain a perfectly pure substance is almost impossible. Purification is a costly process. In general, analytical reagent-grade chemicals are of high purity, and yet few of them are better than 99.9% pure. This means that a foreign atom or molecule is present for every 1000 host atoms or molecules in the crystal. Perhaps the most demanding of purity is in the electronic industry. Silicon crystals of 99.999 (called 5 nines) or better are required for IC chips productions. These crystal are doped with nitrogen group elements of P and As or boron group elements B, Al etc to form n- aand p-type semiconductors. In these crystals, the impurity atom substitute atoms of the host crystals.
Presence minute foreign atoms with one electron more or less than the valence four silicon and germanium host atoms is the key of making n- and p-type semiconductors. Having many semiconductors connected in a single chip makes the integrated circuit a very efficient information processor. The electronic properties change dramatically due to these impurities. This is further described in Inorganic Chemistry by Swaddle.
In other bulk materials, the presence of impurity usually leads to a lowering of melting point. For example, Hall and Heroult tried to electrolyze natural aluminum compounds. They discovered that using a 5% mixture of Al2O3 (melting point 273 K) in cryolite Na3AlF6 (melting point 1273 K) reduced the melting point to 1223 K, and that enabled the production of aluminum in bulk. Recent modifications lowered melting temperatures below 933 K. Some types of glass are made by mixing silica (SiO2), alumina (Al2O3), calcium oxide (CaO), and sodium oxide (Na2O). They are softer, but due to lower melting points, they are cheaper to produce.
Color centers and how do they affect electric conductivity of solids?
Color centers are imperfections in crystals that cause color (defects that cause color by absorption of light). Due to defects, metal oxides may also act as semiconductors, because there are many different types of electron traps. Electrons in defect region only absorb light at certain range of wavelength. The color seen are due to lights not absorbed. For example, a diamond with C vacancies (missing carbon atoms) absorbs light, and these centers give green color as shown here. Replacement of Al3+ for Si4+ in quartz give rise to the color of smoky quartz.
A high temperature phase of ZnOx, (x < 1), has electrons in place of the O2- vacancies. These electrons are color centers, often referred to as F-centers (from the German word farben meaning color). Similarly, heating of ZnS to 773 K causes a loss of sulfur, and these material fluoresces strongly in ultraviolet light. Some non-stoichiometric solids are engineered to be n-type or p-type semiconductors. Nickel oxide NiO gain oxygen on heating in air, resulting in having Ni3+ sites acting as electron trap, a p-type semiconductor. On the other hand, ZnO lose oxygen on heating, and the excess Zn metal atoms in the sample are ready to give electrons. The solid is an n-type semiconductor.
Questions
1. Why does the density of water decreases when the temperature decreases from 277 to 273 K?
2. What type of defects is due to misplaced atoms or ions and vacancies of the same in a crystalline material.
3. What defects reduce the density of a solid?
4. What type of material do the fuel cell and battery technologies need to separate the electrodes?
Solutions
1. Hint: Due to the formation of flickering clusters.
Skill -
Correlate properties of a material to its structure.
2. Hint: Frenkel defects.
Skill -
Explain Frendel and Schottky defects.
3. Hint: Vacancies of Schottky defects reduce the density.
Skill -
Correlate properties of a material to its structure.
4. Hint: Ionic conductors.
Skill -
Specify the desirable properties of a material. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.4%3A_Defects_and_Amorphous_Solids.txt |
Learning Objectives
• To understand the relationship between the lattice energy and physical properties of an ionic compound.
• To use the Born–Haber cycle to calculate lattice energies.
Recall that the reaction of a metal with a nonmetal usually produces an ionic compound; that is, electrons are transferred from the metal (the reductant) to the nonmetal (the oxidant). Ionic compounds are usually rigid, brittle, crystalline substances with flat surfaces that intersect at characteristic angles. They are not easily deformed, and they melt at relatively high temperatures. $\ce{NaCl}$, for example, melts at 801°C. These properties result from the regular arrangement of the ions in the crystalline lattice and from the strong electrostatic attractive forces between ions with opposite charges.
While formation of ion pairs from isolated ions releases large amounts of energy, even more energy is released when these ion pairs condense to form an ordered three-dimensional array. In such an arrangement each cation in the lattice is surrounded by more than one anion (typically four, six, or eight) and vice versa, so it is more stable than a system consisting of separate pairs of ions, in which there is only one cation–anion interaction in each pair. Note that $r_0$ may differ between the gas-phase dimer and the lattice.
An ionic lattice is more stable than a system consisting of separate ion pairs.
Calculating (Ionic) Lattice Energies
The lattice energy of nearly any ionic solid can be calculated rather accurately using a modified form of Coulomb's law:
$U=−\dfrac{k′Q_1Q_2}{r_0} \label{21.5.1}$
where $U$, which is always a positive number, represents the amount of energy required to dissociate 1 mol of an ionic solid into the gaseous ions. The proportionality constant in Equation $\ref{21.5.1}$ is expanded below, but it is worthwhile to discuss its general features first.
If we assume that $ΔV = 0$, then the lattice energy, $U$, is approximately equal to the change in enthalpy, $ΔH$:
$\ce{MX(s) \rightarrow M^{+n} (g) + X^{−n} (g)} \;\;\; ΔH \approx U \label{21.5.2}$
As before, $Q_1$ and $Q_2$ are the charges on the ions and $r_0$ is the internuclear distance. We see from Equation $\ref{21.5.1}$ that lattice energy is directly related to the product of the ion charges and inversely related to the internuclear distance. The value of the constant $k′$ depends on the specific arrangement of ions in the solid lattice and their valence electron configurations. Representative values for calculated lattice energies, which range from about 600 to 10,000 kJ/mol, are listed in Table $1$. Energies of this magnitude can be decisive in determining the chemistry of the elements.
Table $1$: Representative Calculated Lattice Energies
Substance U (kJ/mol)
NaI 682
CaI2 1971
MgI2 2293
NaOH 887
Na2O 2481
NaNO3 755
Ca3(PO4)2 10,602
CaCO3 2804
Because the lattice energy depends on the product of the charges of the ions, a salt having a metal cation with a +2 charge (M2+) and a nonmetal anion with a −2 charge (X2−) will have a lattice energy four times greater than one with $\ce{M^{+}}$ and $\ce{X^{−}}$, assuming the ions are of comparable size (and have similar internuclear distances). For example, the calculated value of $U$ for $\ce{NaF}$ is 910 kJ/mol, whereas $U$ for $\ce{MgO}$ (containing $\ce{Mg^{2+}}$ and $\ce{O^{2−}}$ ions) is 3795 kJ/mol.
Because lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. This effect is illustrated in Figure $1$, which shows that lattice energy decreases for the series $\ce{LiX}$, $\ce{NaX}$, and $\ce{KX}$ as the radius of $\ce{X^{−}}$ increases. Because $r_0$ in Equation $\ref{21.5.1}$ is the sum of the ionic radii of the cation and the anion (r0 = r+ + r), r0 increases as the cation becomes larger in the series, so the magnitude of U decreases. A similar effect is seen when the anion becomes larger in a series of compounds with the same cation.
Because the ionic radii of the cations decrease in the order $\ce{K^{+} > Na^{+} > Li^{+}}$ for a given halide ion, the lattice energy decreases smoothly from $\ce{Li^{+}}$ to $\ce{K^{+}}$. Conversely, for a given alkali metal ion, the fluoride salt always has the highest lattice energy and the iodide salt the lowest.
Lattice energies are highest for substances with small, highly charged ions.
Example $1$
Arrange GaP, BaS, CaO, and RbCl in order of increasing lattice energy.
Given: four compounds
Asked for: order of increasing lattice energy
Strategy:
Using Equation $\ref{21.5.1}$, predict the order of the lattice energies based on the charges on the ions. For compounds with ions with the same charge, use the relative sizes of the ions to make this prediction.
Solution:
The compound GaP, which is used in semiconductor electronics, contains Ga3+ and P3− ions; the compound BaS contains Ba2+ and S2− ions; the compound CaO contains Ca2+ and O2− ions; and the compound RbCl has Rb+ and Cl ions. We know from Equation $\ref{21.5.1}$ that lattice energy is directly proportional to the product of the ionic charges. Consequently, we expect RbCl, with a (−1)(+1) term in the numerator, to have the lowest lattice energy, and GaP, with a (+3)(−3) term, the highest. To decide whether BaS or CaO has the greater lattice energy, we need to consider the relative sizes of the ions because both compounds contain a +2 metal ion and a −2 chalcogenide ion. Because Ba2+ lies below Ca2+ in the periodic table, Ba2+ is larger than Ca2+. Similarly, S2− is larger than O2−. Because the cation and the anion in BaS are both larger than the corresponding ions in CaO, the internuclear distance is greater in BaS and its lattice energy will be lower than that of CaO. The order of increasing lattice energy is RbCl < BaS < CaO < GaP.
Exercise $1$
Arrange InAs, KBr, LiCl, SrSe, and ZnS in order of decreasing lattice energy.
Answer
InAs > ZnS > SrSe > LiCl > KBr
Lattice Energy also Depends on Crystal Structure
There are many other factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulsion from ions of opposite charge and ions of the same charge.
As an example, let us consider the the $\ce{NaCl}$ crystal. In the following discussion, assume r be the distance between Na+ and Cl- ions. The nearest neighbors of Na+ are 6 Cl- ions at a distance 1r, 12 Na+ ions at a distance 2r, 8 Cl- at 3r, 6 Na+ at 4r, 24 Na+ at 5r, and so on. Thus, the electrostatic potential of a single ion in a crystal by approximating the ions by point charges of the surrounding ions:
$E_{ion-lattice} = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{12.5.4}$
The Madelung constant $M$ - named after Erwin Medelung - is a geometrical factor that depends on the arrangement of ions in the solid. For example, $M$ for $\ce{NaCl}$ is a poorly converging series of interaction energies:
$M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{21.5.5}$
with
• $Z$ is the number of charges of the ions, (e.g., 1 for $\ce{NaCl}$ ),
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}\; C$),
• $4\pi \epsilon_o$ is 1.11265x10-10 C2/(J m).
The Madelung constant depends on the structure type and Equation $\ref{21.5.5}$ is applicable only for the sodium chloride (e.g., rock salt) lattice geometry. Other values for other structural types are given in Table $2$.
Table $2$: Madelung Constants. A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
Compound
Crystal Lattice
M
A : C Type
NaCl NaCl 1.74756 6 : 6 Rock salt
CsCl CsCl 1.76267 6 : 6 CsCl type
CaF2 Cubic 2.51939 8 : 4 Fluorite
CdCl2 Hexagonal 2.244
MgF2 Tetragonal 2.381
ZnS (wurtzite) Hexagonal 1.64132
TiO2 (rutile) Tetragonal 2.408 6 : 3 Rutile
bSiO2 Hexagonal 2.2197
Al2O3 Rhombohedral 4.1719 6 : 4 Corundum
There are other factors to consider for the evaluation of lattice energy and the treatment by Max Born and Alfred Landé led to the formula for the evaluation of lattice energy for a mole of crystalline solid. The Born–Landé equation (Equation $\ref{21.5.6}$) is a means of calculating the lattice energy of a crystalline ionic compound and derived from the electrostatic potential of the ionic lattice and a repulsive potential energy term
$U= \dfrac{N_A M z^{+}z^{-} e^2}{4\pi \epsilon_o r_o} \left( 1 - \dfrac{1}{n} \right) \label{21.5.6}$
where
• $N_A$ is Avogadro constant
• $M$ is the Madelung constant for the lattice
• $z^{+}$ is the charge number of cation
• $z^{−}$ is the charge number of anion
• $e$ is elementary charge ($1.6022×10^{−19} C$)
• $ε_0$ is the permittivity of free space ($8.854 \times 10^{-12} C^2/m$)
• $r_0$ is the distance to closest ion
• $n$ is the Born exponent that is typically between 5 and 12 and is determined experimentally. $n$ is a number related to the electronic configurations of the ions involved (Table $3$).
Table $3$: $n$ values for select solids
Atom/Molecule n
He 5
Ne 7
Ar 9
Kr 10
Xe 12
LiF 5.9
LiCl 8.0
LiBr 8.7
NaCl 9.1
NaBr 9.5
Example $2$
Estimate the lattice energy for $\ce{NaCl.}$
Solution
Using the values giving in the discussion above, the estimation is given by Equation \ref{21.5.6}
\begin{align*}U_{NaCl} &= \dfrac{(6.022 \times 10^{23} /mol) (1.74756) (1)(1) (1.6022 \times 10 ^{-19}\,C)^2}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m)} \left( 1 - \dfrac{1}{9.1} \right) \[4pt] &= - 756\, kJ/mol \end{align*}
Discussion
Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Hable cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment.
The Relationship between Lattice Energies and Physical Properties
The magnitude of the forces that hold an ionic substance together has a dramatic effect on many of its properties. The melting point, for example, is the temperature at which the individual ions have enough kinetic energy to overcome the attractive forces that hold them in place. At the melting point, the ions can move freely, and the substance becomes a liquid. Thus melting points vary with lattice energies for ionic substances that have similar structures. The melting points of the sodium halides (Figure $2$), for example, decrease smoothly from NaF to NaI, following the same trend as seen for their lattice energies (Figure $1$). Similarly, the melting point of MgO is 2825°C, compared with 996°C for NaF, reflecting the higher lattice energies associated with higher charges on the ions. In fact, because of its high melting point, MgO is used as an electrical insulator in heating elements for electric stoves.
The hardness of ionic materials—that is, their resistance to scratching or abrasion—is also related to their lattice energies. Hardness is directly related to how tightly the ions are held together electrostatically, which, as we saw, is also reflected in the lattice energy. As an example, MgO is harder than NaF, which is consistent with its higher lattice energy.
In addition to determining melting point and hardness, lattice energies affect the solubilities of ionic substances in water. In general, the higher the lattice energy, the less soluble a compound is in water. For example, the solubility of NaF in water at 25°C is 4.13 g/100 mL, but under the same conditions, the solubility of MgO is only 0.65 mg/100 mL, meaning that it is essentially insoluble.
High lattice energies lead to hard, insoluble compounds with high melting points.
The Born–Haber Cycle
In principle, lattice energies could be measured by combining gaseous cations and anions to form an ionic solid and then measuring the heat evolved. Unfortunately, measurable quantities of gaseous ions have never been obtained under conditions where heat flow can be measured. Instead, lattice energies are found using the experimentally determined enthalpy changes for other chemical processes, Hess’s law, and a thermochemical cycle called the Born–Haber cycle. Developed by Max Born and Fritz Haber in 1919, the Born–Haber cycle describes a process in which an ionic solid is conceptually formed from its component elements in a stepwise manner.
Let’s use the Born–Haber cycle to determine the lattice energy of $\ce{CsF(s)}$. $\ce{CsF}$ is a nearly ideal ionic compound because $\ce{Cs}$ is the least electronegative element that is not radioactive and F is the most electronegative element. To construct a thermochemical cycle for the formation of $\ce{CsF}$, we need to know its enthalpy of formation, ΔHf, which is defined by the following chemical reaction:
$2Cs_{(s)}+F_{2(g)} \rightarrow 2CsF_{(s)} \label{21.5.7}$
Because enthalpy is a state function, the overall $ΔH$ for a series of reactions is the sum of the values of $ΔH$ for the individual reactions. We can therefore use a thermochemical cycle to determine the enthalpy change that accompanies the formation of solid CsF from the parent elements (not ions). The Born–Haber cycle for calculating the lattice energy of cesium fluoride is shown in Figure $1$. This particular cycle consists of six reactions, Equation $\ref{21.5.7}$ plus the following five reactions:
Reaction 1
$Cs_{(s)} \rightarrow Cs_{(g)}\;\;\; ΔH_1=ΔH_{sub}=76.5\; kJ/mol \label{21.5.8a}$
This equation describes the sublimation of elemental cesium, the conversion of the solid directly to a gas. The accompanying enthalpy change is called the enthalpy of sublimation (ΔHsub) (Table $4$) and is always positive because energy is required to sublime a solid.
Table $4$: Selected Enthalpies of Sublimation at 298 K
Substance ΔHsub (kJ/mol)
Li 159.3
Na 107.5
K 89.0
Rb 80.9
Cs 76.5
Be 324.0
Mg 147.1
Ca 177.8
Sr 164.4
Ba 180.0
Reaction 2:
$\ce{Cs(g) -> Cs^{+}(g) + e^{-}} \;\;\; ΔH_2=I_1=375.7\; kJ/mol \label{21.5.8b}$
This equation describes the ionization of cesium, so the enthalpy change is the first ionization energy of cesium. Recall that energy is needed to ionize any neutral atom. Hence, regardless of the compound, the enthalpy change for this portion of the Born–Haber cycle is always positive.
Reaction 3:
$\ce{1/2 F2(g) -> F(g)} \;\;\; ΔH_3=\frac{1}{2}D=79.4\; kJ/mol \label{21.5.8c}$
This equation describes the dissociation of fluorine molecules into fluorine atoms, where $D$ is the energy required for dissociation to occur (Table $5$). We need to dissociate only $\ce{1/2}$ mol of $F_{2(g)}$ molecules to obtain 1 mol of $F_{(g)}$ atoms. The ΔH for this reaction, too, is always positive because energy is required to dissociate any stable diatomic molecule into the component atoms.
Table $5$: Selected Bond Dissociation Enthalpies at 298 K
Substance D (kJ/mol)
H2(g) 436.0
N2(g) 945.3
O2(g) 498.4
F2(g) 158.8
Cl2(g) 242.6
Br2(g) 192.8
I2(g) 151.1
Reaction 4:
$F_{(g)}+ e^- \rightarrow F^-_{(g)} \;\;\; ΔH_4= EA = –328.2\; kJ/mol \label{21.5.8d}$
This equation describes the formation of a gaseous fluoride ion from a fluorine atom; the enthalpy change is the electron affinity of fluorine. Recall that electron affinities can be positive, negative, or zero. In this case, ΔH is negative because of the highly negative electron affinity of fluorine.
Reaction 5:
$Cs^+_{(g)} + F^–_{(g)}→CsF_{(s)} \;\;\; ΔH_5=–U \label{21.5.8e}$
This equation describes the formation of the ionic solid from the gaseous ions. Because Reaction 5 is the reverse of the equation used to define lattice energy and U is defined to be a positive number, ΔH5 is always negative, as it should be in a step that forms bonds.
If the enthalpy of formation of CsF from the elements is known (ΔHf = −553.5 kJ/mol at 298 K), then the thermochemical cycle shown in Figure $3$ has only one unknown, the quantity ΔH5 = −U. From Hess’s law, we can write
$ΔH_f = ΔH_1 + ΔH_2 + ΔH_3 + ΔH_4 + ΔH_5 \label{21.5.9}$
We can rearrange Equation $\ref{21.5.9}$ to give
$−ΔH_5 = ΔH_1 + ΔH_2 + ΔH_3 + ΔH_4 − ΔH_f \label{21.5.10}$
Substituting for the individual ΔHs, we obtain
Substituting the appropriate values into this equation gives
$U = 76.5\; kJ/mol + 375.7 \;kJ/mol + 79.4\; kJ/mol + (−328.2\; kJ/mole) − (−553.5\; kJ/mol) = 756.9\; kJ/mol \label{21.5.11}$
U is larger in magnitude than any of the other quantities in Equation $\ref{21.5.1}$1. The process we have used to arrive at this value is summarized in Table $6$.
Table $6$: Summary of Reactions in the Born–Haber Cycle for the Formation of CsF(s)
Reaction Enthalpy Change (kJ/mol)
(1) Cs(s) → Cs(g) ΔHsub = 76.5
(2) Cs(g) → Cs + (g) + e I1 = 375.7
(3) ½F2(g) → F(g) ½D = 79.4
(4) F(g) + e → F(g) EA = −328.2
(5) Cs + (g) + F(g) → CsF(s) U = −756.9
total Cs(s) + ½F2(g) → CsF(s) ΔHf = −553.5
Predicting the Stability of Ionic Compounds
Equation $\ref{21.5.9}$ may be used as a tool for predicting which ionic compounds are likely to form from particular elements. As we have noted, ΔH1Hsub), ΔH2 (I), and ΔH3 (D) are always positive numbers, and ΔH2 can be quite large. In contrast, ΔH4 (EA) is comparatively small and can be positive, negative, or zero. Thus the first three terms in Equation $\ref{21.5.9}$ make the formation of an ionic substance energetically unfavorable, and the fourth term contributes little either way. The formation of an ionic compound will be exothermic (ΔHf < 0) if and only if ΔH5 (−U) is a large negative number. This means that lattice energy is the most important factor in determining the stability of an ionic compound. Another example is the formation of BaO:
$Ba_{(s)}+\frac{1}{2}O_{2(g)} \rightarrow BaO_{(s)} \label{21.5.11a}$
The enthalpy change is just the enthalpy of formation (e.g, $ΔH=ΔH_f$) with a Born–Haber cycle is compared with that for the formation of $\ce{CsF}$ in Figure $4$.
The lattice energy of BaO, with a dipositive cation and a dinegative anion, dominates the Born–Haber cycle.
• Reaction 1: $Ba_{(s)}→Ba_{(g)} \;\;\; ΔH_1=ΔH_{sub}=180.0\; kJ/mol \label{ 21.5.12}$ More than twice as much energy is required to sublime barium metal (180.0 kJ/mol) as is required to sublime cesium (76.5 kJ/mol).
• Reaction 2: $Ba_{(s)}→Ba^{2+}_{(g)}+2e^– \;\;\; ΔH_2=I_1+I_2=1468.1\; kJ/mol \label{ 21.5.13}$ Nearly four times the energy is needed to form Ba2+ ions (I1 = 502.9 kJ/mol, I2 = 965.2 kJ/mol, I1 + I2 = 1468.1 kJ/mol) as Cs+ ions (I1 = 375.7 kJ/mol).
• Reaction 3: $\dfrac{1}{2}O_{2(g)}→O{(g)} \;\;\; ΔH_3=\frac{1}{2}D=249.2\; kJ/mol \label{21.5.14}$ Because the bond energy of O2(g) is 498.4 kJ/mol compared with 158.8 kJ/mol for F2(g), more than three times the energy is needed to form oxygen atoms from O2 molecules as is required to form fluorine atoms from F2.
• Reaction 4: $O_{(g)} + 2e^– →O^{2–}_{(g)} \;\;\; ΔH_4=EA_1+EA_2=603\; kJ/mol \label{21.5.15}$ Forming gaseous oxide (O2−) ions is energetically unfavorable. Even though adding one electron to an oxygen atom is exothermic (EA1 = −141 kJ/mol), adding a second electron to an O(g) ion is energetically unfavorable (EA2 = +744 kJ/mol)—so much so that the overall cost of forming O2−(g) from O(g) is energetically prohibitive (EA1 + EA2 = +603 kJ/mol). If the first four terms in the Born–Haber cycle are all substantially more positive for BaO than for CsF, why does BaO even form? The answer is the formation of the ionic solid from the gaseous ions (Reaction 5):
• Reaction 5: $Ba^{2+}_{(g)}+O^{2–}_{(g)} → BaO_{(s)} \;\;\; ΔH_5=–U \label{21.5.16}$
Remember from Equations $\ref{21.5.1}$ and $\ref{21.5.6}$ that lattice energies are directly proportional to the product of the charges on the ions and inversely proportional to the internuclear distance. Although the internuclear distances are not significantly different for BaO and CsF (275 and 300 pm, respectively), the larger ionic charges in $\ce{BaO}$ produce a much higher lattice energy. Substituting values for $\ce{BaO}$ (ΔHf = −548.0 kJ/mol) into the equation and solving for U gives:
\begin{align*} U&=ΔH_{sub}(Ba)+[I_1(Ba)+I_2(Ba)]+\frac{1}{2}D(O_2)+[EA_1(O)+EA_2(O)]−ΔH_f(BaO)\;\;\; \label{21.5.17} \[4pt] &=180.0\; kJ/mol + 1468.1 \; kJ/mol + 249.2\; kJ/mol + 603\; kJ/mol−(−548.0\; kJ/mol) \[4pt] &= 3048\; kJ/mol \end{align*}
If the formation of ionic lattices containing multiple charged ions is so energetically favorable, why does CsF contain Cs+ and F ions rather than Cs2+ and F2− ions? If we assume that U for a Cs2+F2− salt would be approximately the same as U for BaO, the formation of a lattice containing Cs2+ and F2− ions would release 2291 kJ/mol (3048 kJ/mol − 756.9 kJ/mol) more energy than one containing Cs+ and F ions. To form the Cs2+ ion from Cs+, however, would require removing a 5p electron from a filled inner shell, which calls for a great deal of energy: I2 = 2234.4 kJ/mol for Cs. Furthermore, forming an F2− ion is expected to be even more energetically unfavorable than forming an O2− ion. Not only is an electron being added to an already negatively charged ion, but because the F ion has a filled 2p subshell, the added electron would have to occupy an empty high-energy 3s orbital. Cesium fluoride, therefore, is not Cs2+F2− because the energy cost of forming the doubly charged ions would be greater than the additional lattice energy that would be gained.
Lattice energy is usually the most important energy factor in determining the stability of an ionic compound.
Example $3$
Use the thermodynamics data in the reference tables to calculate the lattice energy of $\ce{MgH2}$.
Given: chemical compound and data from figures and tables
Asked for: lattice energy
Strategy:
A Write a series of stepwise reactions for forming $\ce{MgH2}$ from its elements via the gaseous ions.
B Use Hess’s law and data from the specified figures above and tables to calculate the lattice energy.
Solution:
A Hess’s law allows us to use a thermochemical cycle (the Born–Haber cycle) to calculate the lattice energy for a given compound. We begin by writing reactions in which we form the component ions from the elements in a stepwise manner and then assemble the ionic solid:
(1) $Mg_{(s)} → Mg_{(g)}$ $ΔH_1=$ $ΔH=ΔH_{sub}(Mg)$
(2) $Mg_{(g)} → Mg^{2+}_{(g)} + 2e^-$ $ΔH_2=$ $I_1(Mg)+I_2(Mg$
(3) $H_{2(g)} → 2H_{(g)}$ $ΔH_3=$ $D(H_2)$
(4) $2H_(g)+2e^− →2H^-_{(g)}$ $ΔH_4=$ $2EA(H)$
(5) $Mg2+(g) +2H^−_{(g)} → MgH_{2(s)}$ $ΔH_5=$ $−U$
total $Mg_{(s)}+H_{2(g)} \rightarrow MgH_{2(s)}$ $ΔH=$ $ΔH_f$
B Table A6 lists the first and second ionization energies for the period 3 elements [I1(Mg) = 737.7 kJ/mol, I2(Mg) = 1450.7 kJ/mol]. First electron affinities for all elements are given in Figure $1$ [EA(H) = −72.8 kJ/mol]. Table $4$ lists selected enthalpies of sublimation [ΔHsub(Mg) = 147.1 kJ/mol]. Table $5$ lists selected bond dissociation energies [D(H2) = 436.0 kJ/mol]. Enthalpies of formation (ΔHf = −75.3 kJ/mol for MgH2) are listed in Table T2. From Hess’s law, ΔHf is equal to the sum of the enthalpy changes for Reactions 1–5:
For MgH2, U = 2701.2 kJ/mol. Once again, lattice energy provides the driving force for forming this compound because ΔH1, ΔH2, ΔH3 > 0. When solving this type of problem, be sure to write the chemical equation for each step and double-check that the enthalpy value used for each step has the correct sign for the reaction in the direction it is written.
Exercise $3$
Use data from the reference tables to calculate the lattice energy of Li2O. Remember that the second electron affinity for oxygen [O(g) + e → O2−(g)] is positive (+744 kJ/mol).
Answer
2809 kJ/mol
Summary
Ionic compounds have strong electrostatic attractions between oppositely charged ions in a regular array. The lattice energy ($U$) of an ionic substance is defined as the energy required to dissociate the solid into gaseous ions; $U$ can be calculated from the charges on the ions, the arrangement of the ions in the solid, and the internuclear distance. Because U depends on the product of the ionic charges, substances with di- or tripositive cations and/or di- or trinegative anions tend to have higher lattice energies than their singly charged counterparts. Higher lattice energies typically result in higher melting points and increased hardness because more thermal energy is needed to overcome the forces that hold the ions together. Lattice energies cannot be measured directly but are obtained from a thermochemical cycle called the Born–Haber cycle, in which Hess’s law is used to calculate the lattice energy from the measured enthalpy of formation of the ionic compound, along with other thermochemical data. The Born–Haber cycle can be used to predict which ionic compounds are likely to form. Sublimation, the conversion of a solid directly to a gas, has an accompanying enthalpy change called the enthalpy of sublimation.
Key Takeaway
• The lattice energy is usually the most important energy factor in determining the stability of an ionic compound. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.5%3A_Lattice_Energies_of_Crystals.txt |
21.5: A Deeper Look: Lattice Energies of Crystals
Conceptual Problems
1. If a great deal of energy is required to form gaseous ions, why do ionic compounds form at all?
2. What are the general physical characteristics of ionic compounds?
3. Ionic compounds consist of crystalline lattices rather than discrete ion pairs. Why?
4. What factors affect the magnitude of the lattice energy of an ionic compound? What is the relationship between ionic size and lattice energy?
5. Which would have the larger lattice energy—an ionic compound consisting of a large cation and a large anion or one consisting of a large anion and a small cation? Explain your answer and any assumptions you made.
6. How would the lattice energy of an ionic compound consisting of a monovalent cation and a divalent anion compare with the lattice energy of an ionic compound containing a monovalent cation and a monovalent anion, if the internuclear distance was the same in both compounds? Explain your answer.
7. Which would have the larger lattice energy—CrCl2 or CrCl3—assuming similar arrangements of ions in the lattice? Explain your answer.
8. Which cation in each pair would be expected to form a chloride salt with the larger lattice energy, assuming similar arrangements of ions in the lattice? Explain your reasoning.
1. Na+, Mg2+
2. Li+, Cs+
3. Cu+, Cu2+
9. Which cation in each pair would be expected to form an oxide with the higher melting point, assuming similar arrangements of ions in the lattice? Explain your reasoning.
1. Mg2+, Sr2+
2. Cs+, Ba2+
3. Fe2+, Fe3+
10. How can a thermochemical cycle be used to determine lattice energies? Which steps in such a cycle require an input of energy?
11. Although NaOH and CH3OH have similar formulas and molecular masses, the compounds have radically different properties. One has a high melting point, and the other is a liquid at room temperature. Which compound is which and why?
Numerical Problems
1. Arrange SrO, PbS, and PrI3 in order of decreasing lattice energy.
2. Compare BaO and MgO with respect to each of the following properties.
1. enthalpy of sublimation
2. ionization energy of the metal
3. lattice energy
4. enthalpy of formation
3. Use a thermochemical cycle and data from Figure 7.13, Table 7.5, Table 8.2, Table 8.3 to calculate the lattice energy (U) of magnesium chloride (MgCl2).
4. Would you expect the formation of SrO from its component elements to be exothermic or endothermic? Why or why not? How does the valence electron configuration of the component elements help you determine this?
5. Using the information in Problem 4 and Problem 5, predict whether CaO or MgCl2 will have the higher melting point.
6. Use a thermochemical cycle and data from Table 8.2, Table 8.3, and Chapter 25 to calculate the lattice energy of calcium oxide. The first and second ionization energies of calcium are 589.8 kJ/mol and 1145.4 kJ/mol.
Answers
1. Lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the internuclear distance. Therefore, PrI3 > SrO > PbS.
2. U = 2522.2 kJ/mol
3. Despite the fact that Mg2+ is smaller than Ca2+, the higher charge of O2− versus Cl gives CaO a larger lattice energy than MgCl2. Consequently, we expect CaO to have the higher melting point. | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.E%3A_Structure_and_Bonding_in_Solids_%28Exercises%29.txt |
A polymer is a large molecule, or macromolecule, composed of many repeated subunits. Because of their broad range of properties, both synthetic and natural polymers play essential and ubiquitous roles in everyday life. Polymers range from familiar synthetic plastics such as polystyrene to natural biopolymers such as DNA and proteins that are fundamental to biological structure and function.
23: Polymeric Materials and Soft Condensed Matter
Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by Hermann Staudinger, a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of macromolecules composed of 10,000 or more atoms. He formulated a polymeric structure for rubber, based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms polymer and monomer were derived from the Greek roots poly (many), mono (one) and meros (part).
Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections.
23.4: Natural Polymers
Polymers are long chain, giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains, sometimes with branching or cross-linking between the chains.
• Addition Polymers
An addition polymer is a polymer which is formed by an addition reaction, where many monomers bond together via rearrangement of bonds without the loss of any atom or molecule under specific conditions of heat, pressure, and/or the presence of a catalyst.
• Condensation Polymers
Condensation polymers are any kind of polymers formed through a condensation reaction—where molecules join together—losing small molecules as byproducts such as water or methanol, as opposed to addition polymers which involve the reaction of unsaturated monomers.
• Introduction to Polymers
Polymers are substances containing a large number of structural units joined by the same type of linkage. These substances often form into a chain-like structure. Polymers in the natural world have been around since the beginning of time. Starch, cellulose, and rubber all possess polymeric properties. Man-made polymers have been studied since 1832. Today, the polymer industry has grown to be larger than the aluminum, copper and steel industries combined.
• Molecular Weights of Polymers
Most polymers are not composed of identical molecules. The HDPE molecules, for example, are all long carbon chains, but the lengths may vary by thousands of monomer units. Because of this, polymer molecular weights are usually given as averages.
• Polyethylene
Polyethylene is the most popular plastic in the world. This is the polymer that makes grocery bags, shampoo bottles, children's toys, and even bullet proof vests. For such a versatile material, it has a very simple structure, the simplest of all commercial polymers. A molecule of polyethylene is nothing more than a long chain of carbon atoms, with two hydrogen atoms attached to each carbon atom.
• Rubber Polymers
Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers.
Thumbnail: Space-filling model of a section of the polyethylene terephthalate polymer, also known as PET and PETE, a polyester used in most plastic bottles. Color code: Carbon, C (black), Hydrogen, H (white), and Oxygen, O (red). (Public Domain; Jynto). | textbooks/chem/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/23%3A_Polymeric_Materials_and_Soft_Condensed_Matter/23.1%3A_Polymerization_Reactions_for_Synthetic_Polymers.txt |
• 1.1: A Particulate View of the World - Structure Determines Properties
The chemistry-centered reformulation of this principle is that Structure Determines Properties, which is a useful concept in chemistry and in fields that chemistry is important including biology, environmental science, biochemistry, polymer science, medicine, engineering, and nutrition among many others.
• 1.2: Classifying Matter- A Particulate View
Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. Solids are relatively rigid and have fixed shapes and volumes. In contrast, liquids have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. Gases, such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers.
• 1.3: The Scientific Approach to Knowledge
The particulate model of matter is not a clear conclusion to the casual observer of nature. While all modern scientists accept the concept of the atom, when the concept of the atom was first proposed about 2,500 years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of the existence of things that are too small to see. We will spend some time considering the evidence (observations) that convince scientists of the existence of atoms.
• 1.4: Early Ideas about the Building Blocks of Matter
Greek philosophers were "all thought and no action" and did not feel the need to test their theories with reality via experiments and the scientific method. Dalton's atomic theory were based on experimentation and testing ideas against reality.
• 1.5: Modern Atomic Theory and the Laws That Led to It
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. While John Dalton is credited for proposing modern atomic theory. Dalton built his atomic theory upon these laws.
• 1.6: The Discovery of the Electron
The atom was not indivisible as Dalton's theory had postulated. J. J. Thomson proved that atoms were not the most basic form of matter. He demonstrated that cathode rays consist of charged particles, later identified as electrons. He measures the ratio of charge over mass of an electron, Robert Millikan carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron.
• 1.7: The Structure of The Atom
The results of Thomson’s and other experiments implied that electrons were constituents of all matter and hence of all atoms. Since macroscopic samples of the elements are found to be electrically neutral, this meant that each atom probably contained a positively charged portion to balance the negative charge of its electrons.
• 1.8: Subatomic Particles - Protons, Neutrons, and Electrons
To date, about 118 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles.
• 1.9: Atomic Mass- The Average Mass of an Element’s Atoms
There are 21 elements with only one isotope, so all their atoms have identical masses. All other elements have two or more isotopes, so their atoms have at least two different masses. However, all elements obey the law of definite proportions when they combine with other elements, so they behave as if they had just one kind of atom with a definite mass. To solve this dilemma, we define the atomic mass as the weighted average mass of all naturally occurring isotopes of the element.
• 1.10: The Origins of Atoms and Elements
The Earth is composed of 90 chemical elements, of which 81 have at least one stable isotope. Most of these elements have also been detected in stars. Where did these elements come from? The accepted scenario is that the first major element to condense out of the primordial soup was helium, which still comprises about one-quarter of the mass of the known universe. Stellar nucleosynthesis then is responsible for the generation of new elements by nuclear reaction within stars.
Thumbnail: Schematic of Rutherford's famous gold foil experiment.
01: Atoms
Matter is anything that occupies space and has mass. Matter has bulk properties that we can directly observe and investigate. For example, why is water a liquid at room temperature and at atmospheric pressure? What makes a drug effective or toxic? Why is steel a strong solid and not a liquid like mercury? Why can we ignite a hdryogen gas, but not helium (Figure \(1\))?
To address these questions, we must first understand two fundamentals concepts of matter that underlie all of chemistry.
1. All matter is consist of small particles, which is the basis of modern atomic theory, and
2. The structure of these particles determines the properties that the matter.
A common refrain taught in biology classes is that Structure Determines Function. The chemistry-centered reformulation of this concept is that Structure Determines Properties, which is a useful theme in chemistry and in fields that chemistry plays an important role including biology, environmental science, biochemistry, polymer science, medicine, engineering, and nutrition among many others. However, what is meant by properties and structure?
Properties are generally separated into either chemical properties (e.g., will a reaction happen and under what conditions will it occur) or physical properties (e.g., what is a substance' melting points, boiling points, solubility). But what we we mean about structure? This is a catch all phrase for the size, geometry and nature of the atoms of the particulates that the matter is composed of. As you progress through the topics of this chemistry text, we will remind you of this central concept and its applications. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.01%3A_A_Particulate_View_of_the_World_-_Structure_Determines_Properties.txt |
Learning Objectives
• To classify matter into the three primary states of matter
• Separate homogeneous from inhomogeneous mixtures
Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. Solids are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, liquids have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. Gases, such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and pressure (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a physical change. For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance.
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure \(2\)). Gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
Elements, Compounds, and Mixtures
A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is homogeneous. Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys.
If the composition of a material is not completely uniform, then it is heterogeneous (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water. The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods.
Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.
Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. (This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States!)
Most mixtures can be separated into pure substances, which may be either elements or compounds. An element, such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a compound, such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a chemical change. The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 118 elements are known, but millions of chemical compounds have been prepared from these 118 elements. The known elements are listed in the periodic table.
In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process called electrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements (Figure \(3\)). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound.
The overall organization of matter and the methods used to separate mixtures are summarized in Figure \(4\).
Example \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. filtered tea
2. freshly squeezed orange juice
3. a compact disc
4. aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms
5. selenium
Given: a chemical substance
Asked for: its classification
Strategy:
1. Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.
2. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Solution
1. A Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration. B Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. A Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. A A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence a compact disc is not chemically pure. B The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. A Aluminum oxide is a single, chemically pure compound.
5. A Selenium is one of the known elements.
Exercise \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. white wine
2. mercury
3. ranch-style salad dressing
4. table sugar (sucrose)
Answer A
solution
Answer B
element
Answer C
heterogeneous mixture
Answer D
compound
Different Definitions of Matter: https://youtu.be/qi_qLHc8wLk
Summary
The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.02%3A_Classifying_Matter-_A_Particulate_View.txt |
Learning Objectives
• Give a short history of the concept of the atom.
• Describe the contributions of Democritus and Dalton to atomic theory.
• Summarize Dalton's atomic theory and explain its historical development.
We discussed the particulate model of matter, but this is not a clear conclusion to the casual observer of nature. While all modern scientists accept the concept of the atom, when the concept of the atom was first proposed about 2,500 years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of the existence of things that are too small to see. We will spend some time considering the evidence (observations) that convince scientists of the existence of atoms.
About 2,500 years ago, early Greek philosophers believed the entire universe was a single, huge, entity. In other words, "everything was one." They believed that all objects, all matter, and all substances were connected as a single, big, unchangeable "thing." One of the first people to propose "atoms" was a man known as Democritus. As an alternative to the beliefs of the Greek philosophers, he suggested that atomos, or atomon - tiny, indivisible, solid objects - make up all matter in the universe.
Democritus then reasoned that changes occur when the many atomos in an object were reconnected or recombined in different ways. Democritus even extended this theory, suggesting that there were different varieties of atomos with different shapes, sizes, and masses. He thought, however, that shape, size, and mass were the only properties differentiating the different types of atomos. According to Democritus, other characteristics, like color and taste, did not reflect properties of the atomos themselves, but rather, resulted from the different ways in which the atomos were combined and connected to one another.
The early Greek philosophers tried to understand the nature of the world through reason and logic, but not through experiment and observation. As a result, they had some very interesting ideas, but they felt no need to justify their ideas based on life experiences. In a lot of ways, you can think of the Greek philosophers as being "all thought and no action." It's truly amazing how much they achieved using their minds, but because they never performed any experiments, they missed or rejected a lot of discoveries that they could have made otherwise. Greek philosophers dismissed Democritus' theory entirely. Sadly, it took over two millennia before the theory of atomos (or "atoms," as they're known today) was fully appreciated.
Greeks: "All thought and No Action"
Greek philosophers were "all thought and no action" and did not feel the need to test their theories with reality. In contrast, Dalton's efforts were based on experimentation and testing ideas against reality.
While it must be assumed that many more scientists, philosophers, and others studied composition of matter after Democritus, a major leap forward in our understanding of the composition of matter took place in the 1800's with the work of the British scientists John Dalton. His atomic theory is a fundamental concept that states that all elements are composed of atoms. Dalton formulated his theory by focusing on experimental results (in contrast to the ancient Greek philosophers) by studied the weights of various elements and compounds. From his experiments and observations, as well as the work from peers of his time, Dalton proposed a new theory of the atom.The general tenets of this theory were as follows:
• All matter is composed of extremely small particles called atoms.
• Atoms of a given element are identical in size, mass, and other properties. Atoms of different elements differ in size, mass, and other properties.
• Atoms cannot be subdivided, created, or destroyed.
• Atoms of different elements can combine in simple whole number ratios to form chemical compounds.
• In chemical reactions, atoms are combined, separated, or rearranged.
Dalton's atomic theory has been largely accepted by the scientific community, with the exception of three changes. We know now that (1) an atom can be further subdivided, (2) all atoms of an element are not identical in mass, and (3) using nuclear fission and fusion techniques, we can create or destroy atoms by changing them into other atoms.
The evidence for atoms is so great that few doubt their existence. In fact, individual atoms are now routinely observed with state-of-the art technologies.
The Scientific Method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(2\)).
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution acid forms a blue solution and a brown gas. After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in the dinosaur extinction scenario, iridium concentrations in the dinosaur extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why.
It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline.
Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw
Measurementsunderlie the Scientific Method
The results of a scientific experiment must be communicated to be of value. This affords an opportunity for other scientists to check them. It also allows the scientific community, and sometimes the general public, to share new knowledge. Communication, however, is not always as straightforward as it might seem. Ambiguous terminology can often turn a seemingly clear statement into a morass of misunderstanding.
As an example, consider someone tells you that the forecast is a high of 25°. Do you put on a winter jacket, or summer wear? If you are thinking winter, then you interpreted the temperature as 25°F. However if 25°C was meant, which is equal to 77°F, a winter jacket would be far too warm. Or consider filling up a car with gasoline. If you are in the US, you will be dealing in dollars per gallon, whereas, if you were in continental Europe, you will be dealing in Euros per liter. Given that the exchange rate from USD to Euros fluctuates and that there are roughly 3.79 liters in 1 gallon, it is difficult to simply compare numbers between gas prices in the USA, and say, France, if you don't know what units you are using. As a final example, consider the speed 24 meters/sec. Do you interpret this as close to highway speed limits in the US? It is the same speed as 53.7 miles per hour, likely a more familiar set of units for speed for people in the US.
Scientists are not all that different from other people—they too have favorite units which are especially suited to certain areas of research. Nevertheless, scientists have constantly pressed for improvement and uniformity in systems of measurement. The first such action occurred nearly 200 years ago when, in the aftermath of the French Revolution, the metric system spread over most of continental Europe and was adopted by scientists everywhere. The United States nearly followed suit, but in 1799 Thomas Jefferson was unsuccessful in persuading Congress that a system based on powers of 10 was far more convenient and would eventually become the standard of the world.
The metric system has undergone continuous evolution and improvement since its original adoption by France. Beginning in 1899, a series of international conferences have been held for the purpose of redefining and regularizing the system of units. In 1960 the Eleventh Conference on Weights and Measures proposed major changes in the metric system and suggested a new name — the International System of Units — for the revised metric system. (The abbreviation SI, from the French Système International, is commonly used.) Scientific bodies such as the U.S. National Bureau of Standards and the International Union of Pure and Applied Chemistry have endorsed the SI.
Summary
• Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.
• 2,500 years ago, Democritus suggested that all matter in the universe was made up of tiny, indivisible, solid objects he called "atomos." However, other Greek philosophers disliked Democritus' "atomos" theory because they felt it was illogical. Dalton's Atomic Theory is the first scientific theory to relate chemical changes to the structure, properties, and behavior of the atom.
• The natural sciences begin with observation, and this usually involves numerical measurements of quantities such as length, volume, density, and temperature. Most of these quantities have units of some kind associated with them, and these units must be retained when you use them in calculations. Measuring units can be defined in terms of a very small number of fundamental ones that, through "dimensional analysis", provide insight into their derivation and meaning, and must be understood when converting between different unit systems.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.03%3A_The_Scientific_Approach_to_Knowledge.txt |
The ancient Greek philosophers did a lot of discussing, with part of their conversations concerning the physical world and its composition. There were different opinions about what made up matter. Some felt one thing was true while others believed another set of ideas. Since these scholars did not have laboratories and had not developed the idea of the experiment, they were left to debate. Whoever could offer the best argument was considered right. However, often the best argument had little to do with reality. One of their ongoing debates had to do with sand. The question posed was: into how small of pieces can you divide a grain of sand? The prevailing thought at the time, pushed by Aristotle, was that the grain of sand could be divided indefinitely, that you could always get a smaller particle by dividing a larger one and there was no limit to how small the resulting particle could be.
Since Aristotle was such an influential philosopher, very few people disagreed with him. However, there were some philosophers who believed that there was a limit to how small a grain of sand could be divided. One of these philosophers was Democritus (~460 - ~370 B.C.E.), often referred to as the "laughing philosopher" because of his emphasis on cheerfulness (Figure \(1\)). He suggested that atomos, or atomon - tiny, indivisible, solid objects - make up all matter in the universe.
Democritus then reasoned that changes occur when the many atomos in an object were reconnected or recombined in different ways. Democritus even extended this theory, suggesting that there were different varieties of atomos with different shapes, sizes, and masses. He thought, however, that shape, size, and mass were the only properties differentiating the different types of atomos. According to Democritus, other characteristics, like color and taste, did not reflect properties of the atomos themselves, but rather, resulted from the different ways in which the atomos were combined and connected to one another.
Aristotle disagreed with Democritus and offered his own idea of the composition of matter. According to Aristotle, everything was composed of four elements: earth, air, fire, and water. The theory of Democritus explained things better, but Aristotle was more influential, so his ideas prevailed. We had to wait almost two thousand years before scientists came around to seeing the atom as Democritus did.
It is very interesting that Democritus had the basic idea of atoms, even though the had no experimental evidence to support his thinking. We now know more about how atoms hold together in "clusters" (compounds), but the basic concept existed over two thousand years ago. We also know that atoms can be further subdivided, but there is still a lower limit to how small we can break up that grain of sand.
Greeks Philosophers were not Scientists
Greek philosophers were "all thought and no action" and did not feel the need to test their theories with reality via experiments and the scientific method. Dalton's atomic theory were based on experimentation and testing ideas against reality.
Summary
The early Greek philosophers tried to understand the nature of the world through reason and logic, but not through experiment and observation. As a result, they had some very interesting ideas, but they felt no need to justify their ideas based on life experiences. In a lot of ways, you can think of the Greek philosophers as being "all thought and no action." It's truly amazing how much they achieved using their minds, but because they never performed any experiments, they missed or rejected a lot of discoveries that they could have made otherwise. Greek philosophers dismissed Democritus' theory entirely. Sadly, it took over two millennia before the theory of atomos (or "atoms," as they're known today) was fully appreciated.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.04%3A_Early_Ideas_about_the_Building_Blocks_of_Matter.txt |
Learning Objectives
• Correctly define a law as it pertains to science.
• Understand the application of the application of the law of conservation of matter.
• Understand the application of the law of definite proportions
• Understand the application of the law of multiple proportions
• Understand the basis of Dalton's Atomic Theory
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. While John Dalton is credited for proposing modern atomic theory. Dalton built his theory upon laws previously identified by Lavoisier and Proust as a basis for his atomic theory:
1. Law of Conservation of Mass,
2. Law of Definite Proportions, and
3. Law of Multiple Proportions.
Law of Conservation of Mass
"Nothing comes from nothing" is an important idea in ancient Greek philosophy that argues that what exists now has always existed, since no new matter can come into existence where there was none before. Antoine Lavoisier (1743-1794) restated this principle for chemistry. This law, which is central is the law of conservation of matter (also known as the "law of indestructibility of matter") and states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved. According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
Law of Conservation Conservation of Mass states that in a chemical reaction, matter is neither created nor destroyed.
It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before (Figure $1$). When wood burns, it combines with oxygen and changes not only to ashes, but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose we had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also suppose we had been able to measure the oxygen used by the fire and the gases produced by the fire. What would we find? The total mass of matter after the fire would be the same as the total mass of matter before the fire.
Example $1$
If heating 10 grams of $\ce{CaCO3}$ produces 4.4 g of $\ce{CO2}$ and 5.6 g of $\ce{CaO}$, show that these observations are in agreement with the law of conservation of mass.
Solution
• Mass of the reactants, $\ce{CaCO3}$ : $10 \,g$
• Mass of the products, $\ce{CO2}$ and $\ce{CaO}$: $4.4 \,g+ 5.6\, g = 10\, g$.
Because the mass of the reactants = the mass of the products, the observations are in agreement with the law of conservation of mass.
Exercise $1$
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
Answer a
The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
Answer b
The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
The Law of Definite Proportions
Joseph Proust (1754-1826) formulated the law of definite proportions (also called the Law of Constant Composition or Proust's Law), which states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table $1$.
Table $1$: Constant Composition of Isooctane
Sample Carbon Hydrogen Mass Ratio
A 14.82 g 2.78 g $\mathrm{\dfrac{14.82\: g\: carbon}{2.78\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
B 22.33 g 4.19 g $\mathrm{\dfrac{22.33\: g\: carbon}{4.19\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
C 19.40 g 3.64 g $\mathrm{\dfrac{19.40\: g\: carbon}{3.63\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.
Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass.
The Law of Definite Proportions applies when elements are reacted together to form the same product. Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen and oxygen react to form water, the Law of Definite Proportions can not be used to compare one experiment in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another material that can be made from hydrogen and oxygen).
Example $2$: Water
Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow, by condensing steam, from river, sea, pond, etc. It can be from different places: USA, UK, Australia, or anywhere. It can be made by chemical reactions like burning hydrogen in oxygen.
However, if the water is pure, it will always consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation.
The Law of Multiple Proportions
The law of multiple proportions (sometime call Dalton's Law) states that if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers. Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here.
• Lineshows the ratio of the relative weights of the two elements in each compound. These ratios were calculated by simply taking the molar mass of each element, and multiplying by the number of atoms of that element per mole of the compound. Thus for NO2, we have (1 × 14) : (2 × 16) = 13:32. (These numbers were not known in the early days of Chemistry because atomic weights (i.e., molar masses) of most elements were not reliably known.)
• The numbers in Lineare just the mass ratios of O:N, found by dividing the corresponding ratios in line 1. But someone who depends solely on experiment would work these out by finding the mass of O that combines with unit mass (1 g) of nitrogen.
• Line is obtained by dividing the figures the previous line by the smallest O:N ratio in the line above, which is the one for N2O. Note that just as the law of multiple proportions says, the weight of oxygen that combines with unit weight of nitrogen work out to small integers.
• Of course we just as easily could have illustrated the law by considering the mass of nitrogen that combines with one gram of oxygen; it works both ways!
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers.
Example $3$: Oxides of Carbon
Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistent with the law of multiple proportions?
Solution
The Law of Multiple Proportions states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio.
Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is:
$\dfrac{0.571\; \text{g oxygen}}{0.429 \;\text{g carbon}} = 1.33\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is
$\dfrac{0.727\; \text{g oxygen}}{0.273 \;\text{g carbon}} = 2.66\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber$
Dividing the mass of oxygen per g of carbon of the second compound:
$\dfrac{2.66}{1.33} = 2\nonumber$
Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers.
John Dalton and the Atomic Theory
The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure $4$), is a fundamental concept that states that all elements are composed of atoms. Previously, an atom was defined as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of the human pinkie (about 1 cm).
Dalton’s ideas are called the modern atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word atom comes from the Greek word atomos, which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science, and the scientific method was being used to study nature. When Dalton announced a modern atomic theory, he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion.
Dalton's Theory was a powerful development as it explained the three laws of chemical combination (above) and recognized a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Six postulates are involved in Dalton's Atomic Theory:
1. All matter consists of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass, but differ from the atoms of other elements.
3. Atoms cannot be created or destroyed.
4. Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form compound atoms.
5. Atoms of same element can combine in more than one ratio to form two or more compounds.
6. The atom is the smallest unit of matter that can take part in a chemical reaction.
Deficiencies of Dalton's Theory
In light of the current state of knowledge in the field of Chemistry, Dalton’s theory had a few drawbacks. According to Dalton’s postulates,
1. The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions.
2. According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. For example, chlorine has two isotopes with mass numbers 35 and 37.
3. Dalton also claimed that atoms of different elements are different in all respects. This has been proven wrong in certain cases: argon and calcium atoms each have an same atomic mass (40 amu).
4. According to Dalton, atoms of different elements combine in simple whole number ratios to form compounds. This is not observed in complex organic compounds like sugar ($C_{12}H_{22}O_{11}$).
5. The theory fails to explain the existence of allotropes (different forms of pure elements); it does not account for differences in properties of charcoal, graphite, diamond.
The importance of Dalton’s theory should not be underestimated. He displayed exceptional insight into the nature of matter and his ideas provided a framework that was later modified and expanded by other. Consequentiually, John Dalton is often considered to be the father of modern atomic theory.
Summary
This section explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.05%3A_Modern_Atomic_Theory_and_the_Laws_That_Led_to_It.txt |
Learning Objectives
• To become familiar with the feature of an electron.
• Summarize and interpret the results of the experiments of Thomson and Millikan.
Long before the end of the 19th century, it was well known that applying a high voltage to a gas contained at low pressure in a sealed tube (called a gas discharge tube) caused electricity to flow through the gas, which then emitted light (Figure $1$). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; this form of energy was called a "cathode ray". In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles.
More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Aside: Electrostatic Forces
If two objects each have electric charge, then they exert an exert an electric force on each other. The magnitude of the force is linearly proportional the charge on each object and inversely proportional to square distance between each other. The magnitude of this electrostatic force is linearly proportional the distance between them and involves the existence of two types of charge, the observation that like charges repel, unlike charges attract and the decrease of force with distance.
The SI unit of electric charge is the coulomb (C) named after French physicist Charles-Augustin de Coulomb.
The video below shows JJ Thompson used such tube to measure the ratio of charge over mass of an electron.
Millikan’s Oil Drop Experiment: Measuring the Charge of the Electron
The American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure $2$).
Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, $1.6 \times 10^{−19}\, C$. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron ($1 \times (1.6 \times 10^{−19}\, C)$), two electrons ($2 \times (1.6 \times 10^{−19}\, C)$), three electrons ($3 \times (1.6 \times 10^{−19}\, C)$), and so on, on a given oil droplet.
Defintion: Elementary Charge
The charge of an electron is sometimes referred to as the elementary charge and usually denoted by $e$. The elementary charge is a fundamental physical constant and as of May 2019, its value is defined to be exactly $1.602176634 \times 10^{−19}\, C$.
Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research ($1.759 \times 10^{11}\, C/kg$), it only required a simple calculation to determine the mass of the electron as well.
\begin{align*} \mathrm{Mass\: of\: electron} &= \mathrm{1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}} \[4pt] &= \mathrm{9.107\times 10^{-31}\:kg} \end{align*}
Scientists had now established that the atom was not indivisible as Dalton's theory had postulated, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood.
Exercise $1$
In a Millikan's oil drop experiment done in alternate universe, the measured charges on drops are found to be $8 \times 10^{-19}\, C$, $12 \times 10^{-19}\, C$ and $20 \times 10^{-19}\, C$. What is the elementary charge in this universe?
Answer
Millikan experiment involves confirm that the charges of the drops were all small integer multiples of the elementary charge. The charges on the drop are found to be multiple of 4. Hence the small charge are $4 \times 10^{-19} C$. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.06%3A_The_Discovery_of_the_Electron.txt |
Learning Objectives
• Outline milestones in the discovery of the nucleus
• Summarize and interpret the results of the Rutherford's gold-foil scatter experiments
Thompson's Non-nuclear Model
Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure \(1\)). A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons.
Rutherford's Nuclear Model
The next major development in understanding the atom came from Ernest Rutherford (1871 to 1937), a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of \(α\) particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the \(α\) particles using a luminescent screen that glowed briefly where hit by an \(α\) particle.
This experiment showed unambiguously that Thomson’s model of the atom (Figure \(1\)) was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (Figure \(\PageIndex{2a}\)) and examined how the \(α\) particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (Figure \(\PageIndex{2b}\)). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (Figure \(\PageIndex{2c}\)). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source.
Intranuclear Structure
Rutherford and other scientists decided that a positively charged nuclear consists of ‘positive electrons’ to balance the charge of the surrounding electrons; this term proton was formally assigned to this particle by 1920. However, Rutherford could not explain why repulsions between the protons in the nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that electrostatic repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume. Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the nucleus together.
The historical development of the different models of the atom’s structure is summarized in Figure \(3\). Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an \(α\) particle contains two protons and neutrons, and is therefore the nucleus of a helium atom.
Summary
Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.07%3A_The_Structure_of_The_Atom.txt |
Learning Objectives
• To know the meaning of isotopes and atomic masses.
To date, about 118 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$, which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
Location
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
Outside nucleus
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
In nucleus
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
In nucleus
Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge and almost all of the volume of an atom consists of empty space in which electrons reside (Figure $1$). The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The Number of Protons Define the Nature of the Elements
The number of protons in the nucleus of an atom is its atomic number ($Z$). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number ($A$)). The number of neutrons is therefore the difference between the mass number and the atomic number:
\begin{align*} \ce{atomic\: number\:(Z)\: &= \:number\: of\: protons\ mass\: number\:(A)\: &= \:number\: of\: protons + number\: of\: neutrons\ A-Z\: &= \:number\: of\: neutrons} \end{align*}
The identity of an element is defined by $Z$, the number of protons in the nucleus of an atom of the element, which is different for each element. The known elements are arranged in order of increasing Z in the periodic table (Figure $2$). The rationale for the peculiar format of the periodic table is explained later. Each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons.
The symbols for several common elements and their atoms are listed in Table $2$. Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Symbols have one or two letters, for example, H for hydrogen and $\ce{Cl}$ for chlorine. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, $\ce{Co}$ is the symbol for the element cobalt, but $\ce{CO}$ is the notation for the compound carbon monoxide, which contains atoms of the elements carbon ($\ce{C}$) and oxygen ($\ce{O}$). All known elements and their symbols are in the periodic table.
Table $2$: Some Common Elements and Their Symbols
Element Symbol Element Symbol
aluminum Al iron Fe (from ferrum)
bromine Br lead Pb (from plumbum)
calcium Ca magnesium Mg
carbon C mercury Hg (from hydrargyrum)
chlorine Cl nitrogen N
chromium Cr oxygen O
cobalt Co potassium K (from kalium)
copper Cu (from cuprum) silicon Si
fluorine F silver Ag (from argentum)
gold Au (from aurum) sodium Na (from natrium)
helium He sulfur S
hydrogen H tin Sn (from stannum)
iodine I zinc Zn
Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists or locations; for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements
Isotopes: Differing Numbers of Neutrons
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number ($A$), the sum of the numbers of protons and neutrons.
The element carbon ($C$) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
For example, naturally occurring hydrogen has two stable nuclides, $\ce{^{1}_{1}H}$ and $\ce{^{2}_{1}H}$, which also are isotopes of one another. More than 99.98 percent is “light” hydrogen, $\ce{^{1}_{1}H}$. This consists of atoms each of which has one proton, one electron, and zero neutrons (Figure $\PageIndex{1; left}$). The rest is “heavy” hydrogen or deuterium, $\ce{^{2}_{1}H}$, which consists of atoms which contain one electron, one proton, and one neutron (Figure $\PageIndex{1; center}$. Hence the nuclidic mass of deuterium is almost exactly twice as great as for light hydrogen. It is also possible to obtain a third isotope, tritium, $\ce{^{3}_{1}H}$. that consists of atoms whose nuclei contain two neutrons and one proton (Figure $\PageIndex{1; right}$). Its mass is about 3 times that of light hydrogen.
In addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table $2$. Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized 2H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized 3H, is also called tritium and sometimes symbolized T.
Table $3$: Nuclear Compositions of Atoms of the Very Light Elements
Element Symbol Atomic Number Number of Protons Number of Neutrons Mass (amu) % Natural Abundance
hydrogen $\ce{^1_1H}$
(protium)
1 1 0 1.0078 99.989
$\ce{^2_1H}$
(deuterium)
1 1 1 2.0141 0.0115
$\ce{^3_1H}$
(tritium)
1 1 2 3.01605 — (trace)
helium $\ce{^3_2He}$ 2 2 1 3.01603 0.00013
$\ce{^4_2He}$ 2 2 2 4.0026 100
lithium $\ce{^6_3Li}$ 3 3 3 6.0151 7.59
$\ce{^7_3Li}$ 3 3 4 7.0160 92.41
beryllium $\ce{^9_4Be}$ 4 4 5 9.0122 100
boron $\ce{^{10}_5B}$ 5 5 5 10.0129 19.9
$\ce{^{11}_5B}$ 5 5 6 11.0093 80.1
carbon $\ce{^{12}_6C}$ 6 6 6 12.0000 98.89
$\ce{^{13}_6C}$ 6 6 7 13.0034 1.11
$\ce{^{14}_6C}$ 6 6 8 14.0032 — (trace)
nitrogen $\ce{^{14}_7N}$ 7 7 7 14.0031 99.63
$\ce{^{15}_7N}$ 7 7 8 15.0001 0.37
oxygen $\ce{^{16}_8O}$ 8 8 8 15.9949 99.757
$\ce{^{17}_8O}$ 8 8 9 16.9991 0.038
$\ce{^{18}_8O}$ 8 8 10 17.9992 0.205
fluorine $\ce{^{19}_9F}$ 9 9 10 18.9984 100
neon $\ce{^{20}_{10}Ne}$ 10 10 10 19.9924 90.48
$\ce{^{21}_{10}Ne}$ 10 10 11 20.9938 0.27
$\ce{^{22}_{10}Ne}$ 10 10 12 21.9914 9.2
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
Example $1$: Composition of an Atom
Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland. The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.
Solution
The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].
Exercise $1$
An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?
Answer
78 protons; 117 neutrons; charge is 4+
Example $2$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $2$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer
$\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Ions: Charged
The protons and neutrons in the nucleus of an atom are held very tightly by strong nuclear forces. It is very difficult either to separate the nuclear particles or to add extra ones. The electrons, on the other hand, are held to the atom by their electrostatic attraction for the positively charged protons in the nucleus. This force is strong, but not so strong that an atom cannot lose or gain electrons. Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:
$\text{Atomic charge} = \text{number of protons} − \text{number of electrons}$
As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+).
$\ce{Na + e^{-} -> Na^{+}}$
Similarly, a neutral fluorine atom (F = 9) has nine electrons, and if it gains a electron it will become an anion with a 1− charge (9 − 10 = 1−).
$\ce{F - 2e^{-} -> F^{-2}}$
The charge of the species has a profound affect on the properties of the species. For example, neutral sodium atoms are unstable and reacting violently when combined with most substnces. However, sodium cations are quite inert; in fact, we eat them all the time as part of sodium chloride (table salt). As we will discussed later, cations and anions almost always occur together to ensure that matter is neutral.
Summary
The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.08%3A_Subatomic_Particles_-_Protons_Neutrons_and_Electrons.txt |
Learning Objectives
• to know the meaning of isotopes and atomic masses.
There are 21 elements with only one isotope, so all their atoms have identical masses. All other elements have two or more isotopes, so their atoms have at least two different masses. However, all elements obey the law of definite proportions when they combine with other elements, so they behave as if they had just one kind of atom with a definite mass. To solve this dilemma, we define the atomic mass as the weighted average mass of all naturally occurring isotopes of the element.
A atomic mass is defined as
$\text{Atomic mass} = \left(\dfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right) + \left(\dfrac{\%\text{ abundance isotope 2}}{100}\right)\times \left(\text{mass of isotope 2}\right)~ ~ ~ + ~ ~ ... \label{amass}$
Similar terms would be added for all the isotopes that would be found in a bulk sample from nature.
GPAs
The weighted average is analogous to the method used to calculate grade point averages in most colleges:
$\text{GPA} = \left(\dfrac{\text{Credit Hours Course 1}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 1}\right) + \left(\dfrac{\text{Credit Hours Course 2}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 2}\right)~ + ~... \nonumber$
The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes Figure $1$. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows:
$\rm(0.9889 \times 12 \;amu) + (0.0111 \times 13.003355 \;amu) = 12.01 \;amu \label{Eq5}$
Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation.
The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.
Example $1$: Lead
Naturally occurring lead is found to consist of four isotopes:
• 1.40% ${}_{\text{82}}^{\text{204}}\text{Pb}$ whose isotopic mass is 203.973.
• 24.10% ${}_{\text{82}}^{\text{206}}\text{Pb}$ whose isotopic mass is 205.974.
• 22.10% ${}_{\text{82}}^{\text{207}}\text{Pb}$ whose isotopic mass is 206.976.
• 52.40% ${}_{\text{82}}^{\text{208}}\text{Pb}$ whose isotopic mass is 207.977.
Calculate the atomic mass of an average naturally occurring sample of lead.
Solution
This is a direct application of Equation \ref{amass} and is best calculated term by term.
Suppose that you had 1 mol lead. This would contain 1.40% ($\dfrac{1.40}{100}$ × 1 mol) ${}_{\text{82}}^{\text{204}}\text{Pb}$ whose molar mass is 203.973 g mol–1. The mass of 20482Pb would be
\begin{align*} \text{m}_{\text{204}} &=n_{\text{204}}\times \text{ }M_{\text{204}} \[4pt] &=\left( \frac{\text{1}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (203}\text{.973 g mol}^{\text{-1}}\text{)} \[4pt] &=\text{2}\text{0.86 g} \end{align*}
Similarly for the other isotopes
\begin{align*}\text{m}_{\text{206}}&=n_{\text{206}}\times \text{ }M_{\text{206}}\[4pt] &=\left( \frac{\text{24}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (205}\text{.974 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{49}\text{0.64 g} \[6pt]\text{m}_{\text{207}}&=n_{\text{207}}\times \text{ }M_{\text{207}}\[4pt] &=\left( \frac{\text{22}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (206}\text{.976 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{45}\text{0.74 g} \[6pt] \text{m}_{\text{208}}&=n_{\text{208}}\times \text{ }M_{\text{208}}\[4pt] &=\left( \frac{\text{52}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (207}\text{.977 g mol}^{\text{-1}}\text{)}\[4pt] &=\text{108}\text{0.98 g} \end{align*}
Upon summing all four results, the mass of 1 mol of the mixture of isotopes is to be found
$2.86\, g + 49.64\, g + 45.74\, g + 108.98\, g = 207.22\, g\nonumber$
The mass of an average lead atom, and thus lead's atomic mass, is 207.2 g/mol. This should be confirmed by consulting the Periodic Table of the Elements.
Exercise $1$: Boron
Boron has two naturally occurring isotopes. In a sample of boron, $20\%$ of the atoms are $\ce{B}-10$, which is an isotope of boron with 5 neutrons and mass of $10 \: \text{amu}$. The other $80\%$ of the atoms are $\ce{B}-11$, which is an isotope of boron with 6 neutrons and a mass of $11 \: \text{amu}$. What is the atomic mass of boron?
Answer
The mass of an average boron atom, and thus boron's atomic mass, is $10.8 \: \text{amu}$. This should be confirmed by consulting the Periodic Table of the Elements.
But which Natural Abundance should be used?
An important corollary to the existence of isotopes should be emphasized at this point. When highly accurate results are obtained, atomic weights may vary slightly depending on where a sample of an element was obtained. For this reason, the Commission on Isotopic Abundance and Atomic Weights of IUPAC (IUPAC/CIAAWhas redefined the atomic masses of 10 elements having two or more isotopes. The percentages of different isotopes often depends on the source of the element.
For example, oxygen in Antarctic precipitation has an atomic weight of 15.99903, but oxygen in marine $\ce{N2O}$ has an atomic mass of 15.9997. "Fractionation" of the isotopes results from slightly different rates of chemical and physical processes caused by small differences in their masses. The difference can be more dramatic when an isotope is derived from nuclear reactors.
Mass Spectrometry: Measuring the Mass of Atoms and Molecules
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision, the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions.
Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure $2$). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.
The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.
Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is
$\rm{\text{mass of }^2H \over \text{mass of }^{12}C} \times \text{mass of }^{12}C = 0.167842 \times 12 \;amu = 2.104104\; amu \label{Eq4}$
The masses of the other elements are determined in a similar way.
Example $2$: Bromine
Naturally occurring bromine consists of the two isotopes listed in the following table:
Isotope Exact Mass (amu) Percent Abundance (%)
79Br 78.9183 50.69
81Br 80.9163 49.31
Calculate the atomic mass of bromine.
Given: exact mass and percent abundance
Asked for: atomic mass
Strategy:
1. Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.
2. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.
3. Add together the weighted masses to obtain the atomic mass of the element.
4. Check to make sure that your answer makes sense.
Solution:
A The atomic mass is the weighted average of the masses of the isotopes (Equation \ref{amass}. In general, we can write
Bromine has only two isotopes. Converting the percent abundances to mass fractions gives
$\ce{^{79}Br}: {50.69 \over 100} = 0.5069 \nonumber$
$\ce{^{81}Br}: {49.31 \over 100} = 0.4931 \nonumber$
B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:
$\ce{^{79}Br}: 79.9183 \;amu \times 0.5069 = 40.00\; amu$
$\ce{^{81}Br}: 80.9163 \;amu \times 0.4931 = 39.90 \;amu$
C The sum of the weighted masses is the atomic mass of bromine is
40.00 amu + 39.90 amu = 79.90 amu
D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.
Exercise $2$
Magnesium has the three isotopes listed in the following table:
Isotope Exact Mass (amu) Percent Abundance (%)
24Mg 23.98504 78.70
25Mg 24.98584 10.13
26Mg 25.98259 11.17
Use these data to calculate the atomic mass of magnesium.
Answer
24.31 amu
Summary
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol.
1.10: The Origins of Atoms and Elements
The Earth is composed of 90 chemical elements, of which 81 have at least one stable isotope. Most of these elements have also been detected in stars. Where did these elements come from? The accepted scenario is that the first major element to condense out of the primordial soup was helium, which still comprises about one-quarter of the mass of the known universe.
Stellar nucleosynthesis is the generation of new elements by nuclear reaction within stars. According to the Big bang theory for which there is now overwhelming evidence, the universe as we know it had its origin in a point source or singularity that began an explosive expansion about 12-15 billion years ago, and which is still continuing. Following a brief period of extremely rapid expansion. Helium and hydrogen became stable during the first few minutes, along with some of the very lightest nuclides up to lithium, which were formed through various nuclear reaction. Formation of most heavier elements was delayed for about million years until nucleosynthesis commenced in the first stars. Hydrogen still accounts for about 93% of the atoms in the universe.
All elements beyond hydrogen were formed in regions where the concentration of matter was large, and the temperature was high; in other words, in stars. The formation of a star begins when the gravitational forces due to a large local concentration of hydrogen bring about a contraction and compression to densities of around 105 g cm–3. This is a highly exothermic process in which the gravitational potential energy is released as heat, about 1200 kJ per gram, raising the temperature to about 107 K. Under these conditions, hydrogen nuclei possess sufficient kinetic energy to overcome their electrostatic repulsion and undergo nuclear fusion to generate helium (Figure \(1\)). This is known as “hydrogen burning”.
As hydrogen burning proceeds, the helium collects in the core of the star, raising the density to 108 g cm–3 and the temperature to 108 K. This temperature is high enough to initiate helium burning. The size of a star depends on the balance between the kinetic energy of its matter and the gravitational attraction of its mass. As the helium burning runs its course, the temperature drops and the star begins to contract. The course of further nucleosynthesis event and the subsequent fate of the star itself depends on the star’s mass. Fusion into heavier species than iron is also precluded by the electrostatic repulsion of the highly charged nuclei. However,further nuclear processes are responsible for these. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/01%3A_Atoms/1.09%3A_Atomic_Mass-_The_Average_Mass_of_an_Elements_Atoms.txt |
• 2.1: The Metric Mix-up - A \$327 Million Unit Error
Small errors in these unit systems can sometimes harbor massive ramifications. Although NASA declared the metric system as its official unit system in the 1980s, conversion factors remain an issue. The Mars Climate Orbiter, meant to help relay information back to Earth, is one notable example of the unit system struggle.
• 2.2: The Reliability of a Measurement
Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly).
• 2.3: Density
Density ( ρ ) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary.
• 2.4: Energy and Its Units
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
• 2.5: Converting between Units
• 2.6: Problem-Solving Strategies
• 2.7: Solving Problems Involving Equations
Many problems in chemistry involve manipulating equations and require the use of multiple conversion steps. Such problems easy to solve as numerical problems once you understand how to approach them. The four simple steps in problem solving are READ-PLAN-SOLVE-CHECK approach.
• 2.8: Atoms and the Mole - How Many Particles?
02: Matter Measurement and Problem Solving
Small errors in these unit systems can sometimes harbor massive ramifications. Although NASA declared the metric system as its official unit system in the 1980s, conversion factors remain an issue. The Mars Climate Orbiter, meant to help relay information back to Earth, is one notable example of the unit system struggle. The orbiter was part of the Mars Surveyor ’98 program, which aimed to better understand the climate of Mars.
As the spacecraft journeyed into space on September 1998, it should have entered orbit at an altitude of 140-150 km above Mars, but instead went as close as 57 km. This navigation error occurred because the software that controlled the rotation of the craft’s thrusters was not calibrated in SI units. The spacecraft expected newtons, while the computer, which was inadequately tested, worked in pound-forces; one pound force is equal to about 4.45 newtons. Unfortunately, friction and other atmospheric forces destroyed the Mars Climate Orbiter.
Clearly, the 4.45-fold difference between the newtons and foot-pound units is catastrophic for space exploration. The project cost \$327.6 million in total. Tom Gavin, an administrator for NASA's Jet Propulsion Laboratory in Pasadena, stated, "This is an end-to-end process problem. A single error like this should not have caused the loss of Climate Orbiter. Something went wrong in our system processes in checks and balances that we have that should have caught this and fixed it." As most advanced chemistry students can attest, properly using units are similarly critical in chemistry; failure to address them properly can have dire consequences. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.01%3A_The_Metric_Mix-up_-_A_327_Million_Unit_Error.txt |
Learning Objectives
• Define accuracy and precision
• Distinguish exact and uncertain numbers
• Correctly represent uncertainty in quantities using significant figures
• Apply proper rounding rules to computed quantities
Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.
Reporting Measurements to Reflect Certainty
Scientific measurements are of no value (or at least, they're not really scientific) unless they are given with some statement of the errors they contain. If a poll reports that one candidate leads another by 5%, that may be politically useful for the winning candidate to point out. But all respectable polls are scientific, and report errors. If the error in measurement is plus or minus 10%, which indicates anything from the candidate leading by 15% to trailing by 5%, the poll really does not reliably tell who is in the lead. If the poll had an error of 1%, the leading candidate could make a more scientific case that for being in the lead (by a 4% to 6% margin, or 5% +/-1%).
The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. Scientific measurements are reported so that every digit is certain except the last, which is estimated. All digits of a measured quantity, including the certain one, are called significant figures.
To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.
Refer to the illustration in Figure \(1\). The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.
Example \(1\)
How much water (with yellow dye) does the graduated buret in Figure \(2\) contain?
Solution
The amount of water is somewhere between 19 ml and 20 ml according to the marked lines. By checking to see where the bottom of the meniscus lies, referencing the ten smaller lines, the amount of water lies between 19.8 ml and 20 ml. The next step is to estimate the uncertainty between 19.8 ml and 20 ml. Making an approximate guess, the level is less than 20 ml, but greater than 19.8 ml. We then report that the measured amount is approximately 19.9 ml. The graduated cylinder itself may be distorted such that the graduation marks contain inaccuracies providing readings slightly different from the actual volume of liquid present.
This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more sensitive balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.
Accuracy and Precision
Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure \(2\)).
Figure \(1\) help to understand the difference between precision (small expected difference between multiple measurements) and accuracy (difference between the result and a known value).
Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table \(2\).
Table \(2\): Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers
Dispenser #1 Dispenser #2 Dispenser #3
283.3 298.3 296.1
284.1 294.2 295.9
283.9 296.0 296.1
284.0 297.8 296.0
284.1 293.9 296.1
Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).
Summary
Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly).
Glossary
uncertainty
estimate of amount by which measurement differs from true value
significant figures
(also, significant digits) all of the measured digits in a determination, including the uncertain last digit
rounding
procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation
precision
how closely a measurement matches the same measurement when repeated
exact number
number derived by counting or by definition
accuracy
how closely a measurement aligns with a correct value | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.02%3A_The_Reliability_of_a_Measurement.txt |
Learning Objectives
• Define density
• Use density as a conversion factor
The terms heavy and light are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. On the other hand, something else is alluded to when we say that oak is heavier than balsa wood. A small shaving of oak would obviously weigh less than a roomful of balsa wood, but oak is heavier in the sense that a piece of given size weighs more than the same-size piece of balsa.
What we are actually comparing is the mass per unit volume, that is, the density. In order to determine these densities, we might weigh a cubic centimeter of each type of wood. If the oak sample weighed 0.71 g and the balsa 0.15 g, we could describe the density of oak as 0.71 g cm–3 and that of balsa as 0.15 g cm–3. (Note that the negative exponent in the units cubic centimeters indicates a reciprocal. Thus 1 cm–3 = 1/cm3 and the units for our densities could be written as $\frac{\text{g}}{\text{cm}^\text{3}}$, g/cm3, or g cm–3. In each case the units are read as grams per cubic centimeter, the per indicating division.) We often abbreviate "cm3" as "cc", and 1 cm3 = 1 mL exactly by definition.
In general it is not necessary to weigh exactly 1 cm3 of a material in order to determine its density. We simply measure mass and volume and divide volume into mass:
$\text{Density} = \dfrac{\text{mass}}{\text{volume}}$
or
$\rho = \dfrac{m}{V} \quad \label{1}$
where $ρ$ is density, $m$ is mass, and $V$ is the volume.
Example $1$: Density of Aluminum
Calculate the density of:
1. a piece of aluminum whose mass is 37.42 g and which, when submerged, increases the water level in a graduated cylinder by 13.9 ml;
2. an aluminum cylinder of mass 25.07 g, radius 0.750 cm, and height 5.25 cm.
Solution
a) Since the submerged metal displaces its own volume,
\begin{align*} \text{Density} &= \rho =\dfrac{m}{V} \[4pt] &= \dfrac{\text{37.42 g}}{\text{13.9 ml}} \[4pt] &= \text{2.69 g}/\text{ml or 2.69 g ml}^{-\text{1}} \end{align*}
b) The volume of the cylinder must be calculated first, using the formula
\begin{align*} \text{V} &= \pi r^\text{2} h\[4pt] &= \text{3.142}\times\text{(0.750 cm)}^\text{2}\times\text{5.25 cm} \[4pt] &= \text{9.278 cm}^\text{3} \end{align*}
Then
$\rho = \dfrac{m}{V} = \dfrac{\text{25.07 g}}{\text{9.278 cm}^\text{3}} = \begin{cases} 2.70 \dfrac{\text{g}}{\text{cm}^\text{3}} \ \text{2.70 g cm}^{-\text{3}} \ \text{2.70 g}/ \text{cm}^\text{3} \end{cases} \nonumber$
which are all acceptable alternatives.
Note that unlike mass or volume (extensive properties), the density of a substance is independent of the size of the sample (intensive property). Thus density is a property by which one substance can be distinguished from another. A sample of pure aluminum can be trimmed to any desired volume or adjusted to have any mass we choose, but its density will always be 2.70 g/cm3 at 20°C. The densities of some common pure substances are listed below.
Tables and graphs are designed to provide a maximum of information in a minimum of space. When a physical quantity (number × units) is involved, it is wasteful to keep repeating the same units. Therefore it is conventional to use pure numbers in a table or along the axes of a graph. A pure number can be obtained from a quantity if we divide by appropriate units. For example, when divided by the units gram per cubic centimeter, the density of aluminum becomes a pure number 2.70:
$\dfrac{\text{Density of aluminum}}{\text{1 g cm}^{-\text{3}}} = \dfrac{\text{2.70 g cm}^{-\text{3}}}{\text{1 g cm}^{-\text{3}}} = \text{2.70} \nonumber$
Table $1$: Density of Several Substances at 20°C.Anchor
Substance Density / g cm-3
Helium gas 0.000 16
Dry air 0.001 185
Gasoline 0.66-0.69 (varies)
Kerosene 0.82
Benzene 0.880
Water 1.000
Carbon tetrachloride 1.595
Magnesium 1.74
Salt 2.16
Aluminum 2.70
Iron 7.87
Copper 8.96
Silver 10.5
Lead 11.34
Uranium 19.05
Gold 19.32
Therefore, a column in a table or the axis of a graph is conveniently labeled in the following form:
Quantity/units
This indicates the units that must be divided into the quantity to yield the pure number in the table or on the axis. This has been done in the second column of the table.
Converting Densities
In our exploration of density, notice that chemists may express densities differently depending on the subject. The density of pure substances may be expressed in kg/m3 in some journals which insist on strict compliance with SI units; densities of soils may be expressed in lb/ft3 in some agricultural or geological tables; the density of a cell may be expressed in mg/µL; and other units are in common use. It is easy to transform densities from one set of units to another, by multiplying the original quantity by one or more unity factors:
Example $2$: Density of Water
Convert the density of water, 1 g/cm3 to (a) lb/cm3 and (b) lb/ft3
Solution
a. The equality $\text{454 g} = \text{1 lb}$ can be used to write two unity factors,
$\dfrac{\text{454 g}}{\text{1 lb}}\nonumber$
or
$\dfrac{\text{1 lb}}{\text{454 g}} \nonumber$
The given density can be multiplied by one of the unity factors to get the desired result. The correct conversion factor is chosen so that the units cancel:
$\text{1} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{1 lb}}{\text{454 g}} = \text{0.002203} \dfrac{\text{lb}}{\text{cm}^\text{3}}\nonumber$
b. Similarly, the equalities $\text{2.54 cm} = \text{1 inch}$, and $\text{12 inches} = \text{1 ft}$ can be use to write the unity factors:
$\dfrac{\text{2.54 cm}}{\text{1 in}} \text{, } \dfrac{\text{1 in}}{\text{2.54 cm}} \text{, } \dfrac{\text{12 in}}{\text{1 ft}} \text{ and } \dfrac{\text{1 ft}}{\text{12 in}} \nonumber$
In order to convert the cm3 in the denominator of 0.002203 to in3, we need to multiply by the appropriate unity factor three times, or by the cube of the unity factor:
$\text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}}\nonumber$
or
$\text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \big(\dfrac{\text{2.54 cm}}{\text{1 in}}\big)^\text{3} = \text{0.0361 lb}/ \text{in}^\text{3}\nonumber$
This can then be converted to lb/ft3:
$\text{0.0361 lb}/ \text{in}^\text{3}\times \big(\dfrac{\text{12 in}}{\text{1 ft}}\big)^\text{3} = \text{62.4 lb}/\text{ft}^\text{3}\nonumber$
It is important to notice that we have used conversion factors to convert from one unit to another unit of the same parameter | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.03%3A_Density.txt |
Learning Objectives
• To understand the concept of energy and its various forms.
• To know the relationship between energy, work, and heat.
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
Forms of Energy
The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure $1$). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms.
Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.
Energy can be converted from one form to another (Figure $2$) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction.
Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of energy: Energy cannot be created or destroyed.
Energy, Heat, and Work
One definition of energy is the capacity to do work. The easiest form of work to visualize is mechanical work (Figure $3$), which is the energy required to move an object a distance d when opposed by a force F, such as gravity:
work = force x distance
$w=F\,d \label{5.1.1}$
Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write Equation $\ref{5.1.1}$ as follows:
work= mass x acceleration x distance
$w = m\,a\,d \label{5.1.2}$
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth.
Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the force of gravity. The amount of work done (w) and thus the energy required depends on three things:
1. the height of the second floor (the distance d);
2. your mass, which must be raised that distance against the downward acceleration due to gravity; and
3. your path.
In contrast, heat (q) is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature.
Energy is an extensive property of matter—for example, the amount of thermal energy in an object is proportional to both its mass and its temperature. A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy.
Kinetic and Potential Energy
The kinetic energy of an object is related to its mass $m$ and velocity $v$:
$KE=\dfrac{1}{2}mv^2 \label{5.1.4}$
For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is
$KE=\dfrac{1}{2}(1360 kg)(26.8 ms)^2= 4.88 \times 10^5 g \cdot m^2 \label{5.1.5}$
Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is named after the British physicist James Joule (1818–1889), an early worker in the field of energy. is defined as 1 kilogram·meter2/second2 (kg·m2/s2). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 103 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 105 J or 4.88 × 102 kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.
To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is w = Fd. According to Equation $\ref{5.1.2}$, the force (F) exerted by gravity on any object is equal to its mass (m, in this case, 1360 kg) times the acceleration (a) due to gravity (g, 9.81 m/s2 at Earth’s surface). The distance (d) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:
$PE= F\;d = m\,a\;d = m\,g\,h \label{5.1.6a}$
$PE=(1360, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m) = 4.88 \times 10^5\; \frac{Kg \cdot m}{s^2} \label{5.1.6b}$
$=4.88 \times 10^5 J = 488\; kJ \label{5.1.6c}$
The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h.
If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero.
Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.
Units of Energy
The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1°C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0°C–100°C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
$1 \;cal = 4.184 \;J \;\text{exactly} \label{5.1.7a}$
$1 \;J = 0.2390\; cal \label{5.1.7b}$
In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information.
Example $1$
1. If the mass of a baseball is 149 g, what is the kinetic energy of a fastball clocked at 100 mi/h?
2. A batter hits a pop fly, and the baseball (with a mass of 149 g) reaches an altitude of 250 ft. If we assume that the ball was 3 ft above home plate when hit by the batter, what is the increase in its potential energy?
Given: mass and velocity or height
Asked for: kinetic and potential energy
Strategy:
Use Equation 5.1.4 to calculate the kinetic energy and Equation 5.1.6 to calculate the potential energy, as appropriate.
Solution:
1. The kinetic energy of an object is given by $\frac{1}{2} mv^2$ In this case, we know both the mass and the velocity, but we must convert the velocity to SI units: $v= \left(\dfrac{100\; \cancel{mi}}{1\;\cancel{h}} \right) \left(\dfrac{1 \;\cancel{h}}{60 \;\cancel{min}} \right) \left(\dfrac{1 \; \cancel{min}}{60 \;s} \right)\left(\dfrac{1.61\; \cancel{km}}{1 \;\cancel{mi}} \right) (\dfrac{1000\; m}{1\; \cancel{km}})= 44.7 \;m/s \nonumber$
The kinetic energy of the baseball is therefore $KE= 1492 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.49 \times10^2\; J \nonumber$
2. The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus $PE= 149\;\cancel{g} \left(\dfrac{1\; kg}{1000\; \cancel{g}} \right)\left(\dfrac{9.81\; m}{s^2} \right) \left(247\; \cancel{ft} \right) \left(\dfrac{0.3048\; m}{1 \;\cancel{ft}} \right)= 1.10 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.10 \times 10^2\; J \nonumber$
Exercise $1$
1. In a bowling alley, the distance from the foul line to the head pin is 59 ft, 10 13/16 in. (18.26 m). If a 16 lb (7.3 kg) bowling ball takes 2.0 s to reach the head pin, what is its kinetic energy at impact? (Assume its speed is constant.)
2. What is the potential energy of a 16 lb bowling ball held 3.0 ft above your foot?
Answer a
3.10 × 102 J
Answer b
65 J
Summary
All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work. Thermochemistry is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. Energy is the capacity to do work. Mechanical work is the amount of energy required to move an object a given distance when opposed by a force. Thermal energy is due to the random motions of atoms, molecules, or ions in a substance. The temperature of an object is a measure of the amount of thermal energy it contains. Heat (q) is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of potential energy (PE), energy caused by the relative position or orientation of an object. Kinetic energy (KE) is the energy an object possesses due to its motion. The most common units of energy are the joule (J), defined as 1 (kg·m2)/s2, and the calorie, defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.04%3A_Energy_and_Its_Units.txt |
Learning Objectives
• To convert a value reported in one unit to a corresponding value in a different unit using conversion factors.
Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will see how conversion factors representing mathematical functions, like $\rho = m / V$, can be used to transform quantities into different parameters. For example, what is the volume of a given mass of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.
Conversion Factors
A conversion factor is a factor used to convert one unit of measurement into another. A simple conversion factor can be used to convert meters into centimeters, or a more complex one can be used to convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. What always must be remembered is that a conversion factor has to represent a fact; this fact can either be simple or much more complex. For instance, you already know that 12 eggs equal 1 dozen. A more complex fact is that the speed of light is $1.86 \times 10^5$ miles/$\text{sec}$. Either one of these can be used as a conversion factor depending on what type of calculation you might be working with (Table $1$).
Table $1$: Conversion Factors from SI units to English Units
English Units Metric Units Quantity
1 ounce (oz) 28.35 grams (g) *mass
1 fluid once (oz) 29.6 mL volume
2.205 pounds (lb) 1 kilogram (kg) *mass
1 inch (in) 2.54 centimeters (cm) length
0.6214 miles (mi) 1 kilometer (km) length
1 quarter (qt) 0.95 liters (L) volume
*pounds and ounces are technically units of force, not mass.
Of course, there are other ratios which are not listed in Table $1$. They may include:
• Ratios embedded in the text of the problem (using words such as per or in each, or using symbols such as / or %).
• Conversions in the metric system, as covered earlier in this chapter.
• Common knowledge ratios (such as 60 seconds $=$ 1 minute).
If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter.
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} = 10^{-2}\rm{m} \nonumber$
or
$100\; \rm{cm} = 1\; \rm{m} \nonumber$
Suppose we divide both sides of the equation by $1 \text{m}$ (both the number and the unit):
$\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber$
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:
$\dfrac{ \text{100 cm}}{\text{1 m}} = \dfrac{ \text{1000 mm}}{\text{1 m}}= \dfrac{ 1\times 10^6 \mu \text{m}}{\text{1 m}}= 1 \nonumber$
We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units.
Dimensional Analysis
Dimensional analysis is amongst the most valuable tools physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis.
Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as $\mathrm{\frac{100\:cm}{1\:m}}$ and multiply:
$3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber$
The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber$
The final step is to perform the calculation that remains once the units have been canceled:
$\dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \nonumber$
In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:
$\text{quantity}\, \cancel{(\text{old units})} \times \underbrace{\left( \dfrac{\text{new units}}{\cancel{\text{old units}}}\right) }_{\text{conversion factor}=1} = \text{quantity (new units)}$
You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you will encounter will not always be so simple. If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example, we used the fraction $\frac{100 \; \rm{cm}}{1 \; \rm{m}}$ as a conversion factor. Does the conversion factor $\frac{1 \; \rm m}{100 \; \rm{cm}}$ also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:
$3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber$
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out. Figure $1$ shows a concept map for constructing a proper conversion.
Exercise $1$
Perform each conversion.
1. 101,000. ns to seconds
2. 32.08 kg to grams
3. 1.53 grams to cg
Answer a
$1.01000 \times 10^{-4} s$
Answer b
$3.208 \times 10^{4} g$
Answer c
$1.53 \times 10^{2} g$
Summary
• Conversion factors are used to convert one unit of measurement into another.
• Dimensional analysis (unit conversions) involves the use of conversion factors that will cancel units you don't want and produce units you do want.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.05%3A__Converting_between_Units.txt |
We know the conversion factor is correct when units cancel appropriately, but a conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship, not because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.
A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:
$\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume or }m\overset{\rho }{\longleftrightarrow}V\text{ }$
This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written
$\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}$
General Steps in Performing Dimensional Analysis
1. Identify the "given" information in the problem. Look for a number with units to start this problem with.
2. What is the problem asking you to "find"? In other words, what unit will your answer have?
3. Use ratios and conversion factors to cancel out the units that aren't part of your answer, and leave you with units that are part of your answer.
4. When your units cancel out correctly, you are ready to do the math. You are multiplying fractions, so you multiply the top numbers and divide by the bottom numbers in the fractions.
As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.
Example $1$: Volume to Mass Conversion
Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).
Solution
The road map
$V\xrightarrow{\rho }m\text{ } \nonumber$
tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:
$\text{Mass} = m = 47.3 \text{cm}^{3} \times \dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}} \nonumber$
Since the volume units are different, we need a unity factor to get them to cancel:
$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}} \nonumber$
We now have the mass in pounds, but we want it in grams, so another unity factor is needed:
$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{0.9 g} \nonumber$
In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.
2.07: Solving Problems Involving Equations
Many problems in chemistry involve manipulating equations and require the use of multiple conversion steps. Such problems easy to solve as numerical problems once you understand how to approach them. The four simple steps in problem solving are READ-PLAN-SOLVE-CHECK approach.
Steps to Solve Chemistry Problems
1. READ the question: Before you start calculating and manipulating equations, read the complete problem thoroughly to ensure you understand what being asked.
2. PLAN your approach:
• Write down all of the information you have been given. It is not uncommon that problems will give more facts than are required to solve - this is a bit of reality in the problem.
• Identify the equation(s) that are required to use to solve the problem, this often requires manipulating one or more equations to give you the desired answer.
3. SOLVE the problem
• Before calculating results, confirm the correct units required for the equations. You may often are required perform one or more unit conversions before directly using the equations.
• Insert the relevant parameters into the equation(s) and get your answer. Do not forget that most answers will involve units.
4. CHECK your answer: Confirm that you have answered all that is requested in the problem and that the answer seems reasonable. For example, if you are calculating the volume of a sample and your calculated results in cubic kilometers, you probably made an error in a conversion or calculation (unless you are working in an astronomy class).
Example $1$
Example $1$ Example $2$
Steps for Problem Solving The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters? A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
Identify the "given"information and what the problem is asking you to "find." Given: 4.7 L
Find: mL
Given: 18 ms
Find: s
List other known quantities $1\, mL = 10^{-3} L$ $1 \,ms = 10^{-3} s$
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{mL}}{10^{-3}\; \cancel{\rm{L}}} = 4,700\; \rm{mL}$
or
$4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL}$
$18 \; \cancel{\rm{ms}} \times \dfrac{10^{-3}\; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
or
$18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
Think about your result. The amount in mL should be 1000 times larger than the given amount in L. The amount in s should be 1/1000 the given amount in ms.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
• Henry Agnew (UC Davis) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.06%3A_Problem-Solving_Strategies.txt |
Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. It also describes the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10−23 g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, it is therefore essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample.
The Mole: A Chemistry "Dozen"
Because atoms and molecules are extremely small, there are a great many of them in any macroscopic sample. For example a 1 cm3 of mercury would contain $4.080 \times 10^{22}$ mercury atoms. The very large numbers involved in counting microscopic particles are inconvenient to think about or to write down. Chemists have chosen to count atoms and molecules using a unit called the mole (mol), from the Latin moles, meaning “pile” or “heap.” One mole is 6.022 x 1023 of the microscopic particles which make up the substance in question. Thus 6.022 x 1023 Br atoms is referred to as 1 mol Br.
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.
The number of entities composing a mole has been experimentally determined to be $6.02214179 \times 10^{23}$. This is a fundamental constant named Avogadro’s number ($N_A$) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being $6.022 \times 10^{23}/\ce{mol}$.
Using Units of Specific Number of Items is Common
Many familiar items are sold in numerical quantities with distinct names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them.
There is a difference in degree, however, because the mole of anything (6.022 x 1023) is so large. A stack of paper containing a mole of sheets would extend more than a million times the distance from the earth to the sun, and 6.022 x 1023 grains of sand would cover all the land in the world to a depth of nearly 2 ft. Obviously there are a great many particles in a mole of anything.
A mole is formally defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12C weighing exactly 12 g. Consistent with this definition, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound as we will demonstrate) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure $2$).
Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12C, the molar mass of any substance is numerically equivalent to its atomic mass in amu. Per the amu definition, a single 12C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12C contains 1 mole of 12C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12C (Table $1$).
Table $1$: Mass of one mole of elements
Element Average Atomic Mass (amu) Atomic Mass (g/mol) Atoms/Mole
C 12.01 12.01 $6.022 \times 10^{23}$
H 1.008 1.008 $6.022 \times 10^{23}$
O 16.00 16.00 $6.022 \times 10^{23}$
Na 22.99 22.99 $6.022 \times 10^{23}$
Cl 33.45 35.45 $6.022 \times 10^{23}$
Overview of Origin and Properties of the Mole
The word "mole" suggests a small, furry burrowing animal to many. But in this lesson, we look at the concept of the mole in chemistry. Learn the incredible magnitude of the mole--and how something so big can help us calculate the tiniest particles in the world.
Converting between Number of Moles and Number of Atoms
Although chemists usually work with moles as units, occasionally it is helpful to refer to the actual number of atoms or molecules involved. When this is done, the symbol $N$ is used for the number of species and $n$ is used for the number of moles. For example, in referring to 1 mol of helium atoms, we could write
$n_{\ce{He}} = 1 \text{mol} \nonumber$
and
$N_{\ce{He}} = 6.022\times 10^{23} \nonumber$
Obtaining $N$ requires the use of a conversion factor to obtain, which is just $N_A$. This which is defined by the equation
$N_{\text{A}} = \dfrac{N}{n} \label{eq1}$
Since for any substance there are 6.022 × 1023 particles per mole, Equation \ref{eq1} can be expanded:
$\textit{N}_\text{A}=\dfrac{6.022\cdot10^{23}} {1\text{ mol}}=6.022\times 10^{23}\text{ mol}^{\text{–1}} \label{eq2}$
Equation \ref{eq2} is the most important conversion factor in general chemistry. Figure $3$ is a flowchart for converting between the number of moles and the number of atoms. The use of these conversions is illustrated in Example $1$ and Exercise $1$.
Example $1$: Copper Atoms
How many atoms are present in 2.76 mol of copper atoms?
Solution
The definition of a mole is an equality that can be used to construct a conversion factor.
$2.76\, \cancel{mol\, \ce{Cu}}\times \frac{6.022\times 10^{23}atoms\, \ce{Cu}}{\cancel{mol\, \ce{Cu}}}=1.66\times 10^{24} \text{molecules} \, \ce{Cu} \nonumber$
Exercise $1$
How many molecules are present in $4.61 \times 10^{-2}$ mol of helium atoms?
Answer
2.78 × 1022 molecules
Converting between Mass and Number of Moles
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol.
Reminder: Molar Masses (Atomic Masses) are Weighted Averages of Isotopic Masses
As discussed in Section 1.9, The molar mass of naturally-occurring carbon is different from that of carbon-12 isotope because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. Similarly, the molar mass of uranium is 238.03 g/mol.
The mole is the basis of quantitative chemistry and provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms. For example, to convert moles of a substance to mass, the following relationship is used:
$(\text{moles}) \times (\text{molar mass}) \rightarrow \text{mass} \label{3.4.1}$
or, more specifically,
$\cancel{\text{moles}} \times \left ( {\text{grams} \over \cancel{\text{mole}} } \right ) = \text{grams} \nonumber$
Conversely, to convert the mass of a substance to moles:
$\left ( {\text{grams} \over \text{molar mass} } \right ) \rightarrow \text{moles} \label{3.4.2A}$
$\left ( { \text{grams} \over \text{grams/mole}} \right ) = \cancel{\text{grams}} \left ( {\text{mole} \over \cancel{\text{grams}} } \right ) = \text{moles} \label{3.4.2B}$
If know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.
Example $2$: Deriving Moles from Grams for an Element
According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?
Solution
The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.
The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):
The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”
$\mathrm{4.7\; \cancel{g} K \left ( \dfrac{mol\; K}{39.10\;\cancel{g}}\right)=0.12\;mol\; K} \nonumber$
The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.
Exercise $2$: Beryllium
Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?
Answer
0.360 mol
Example $3$: Deriving Grams from Moles for an Element
A liter of air contains $9.2 \times 10^{−4}$ mol argon. What is the mass of Ar in a liter of air?
Solution
The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):
In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):
$\mathrm{9.2 \times10^{-4}\; \cancel{mol} \; Ar \left( \dfrac{39.95\;g}{\cancel{mol}\;Ar} \right)=0.037\;g\; Ar} \nonumber$
The result is in agreement with our expectations, around 0.04 g Ar.
Exercise $3$
What is the mass of 2.561 mol of gold?
Answer
504.4 g
Example $4$: Deriving Number of Atoms from Mass for an Element
Copper is commonly used to fabricate electrical wire (Figure $6$). How many copper atoms are in 5.00 g of copper wire?
Solution
The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (NA) to convert this molar amount to number of Cu atoms:
Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth NA, or approximately 1022 Cu atoms. Carrying out the two-step computation yields:
$\mathrm{5.00\:\cancel{g}\:Cu\left(\dfrac{\cancel{mol}\:Cu}{63.55\:\cancel{g}}\right)\left(\dfrac{6.022\times10^{23}\:atoms}{\cancel{mol}}\right)=4.74\times10^{22}\:atoms\: of\: copper}$
The factor-label method yields the desired cancellation of units, and the computed result is on the order of 1022 as expected.
Exercise $4$
A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?
Answer
$4.586 \times 10^{22}\; Au$ atoms | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/02%3A_Matter_Measurement_and_Problem_Solving/2.08%3A_Atoms_and_the_Mole_-_How_Many_Particles.txt |
In this chapter, we describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not. After reading this chapter, you will know enough about the theory of the electronic structure of atoms to explain what causes the characteristic colors of neon signs, how laser beams are created, and why gemstones and fireworks have such brilliant colors. In later chapters, we will develop the concepts introduced here to explain why the only compound formed by sodium and chlorine is NaCl, an ionic compound, whereas neon and argon do not form any stable compounds, and why carbon and hydrogen combine to form an almost endless array of covalent compounds, such as CH4, C2H2, C2H4, and C2H6.
• 3.1: Schrödinger's Cat
The field of chemistry deals with the structures, bonding, reactivity, and physical properties of atoms, molecules, radicals, and ions all of whose sizes range from ca. 1 Å for atoms and small molecules to a few hundred Å for polymers and biological molecules such as DNA and proteins. Their structures, energies, and other properties have only been successfully described within the framework of quantum mechanics. This is why quantum mechanics has to be mastered as part of learning chemistry.
• 3.2: The Nature of Light
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light, the infrared radiation, the ultraviolet light, and the x-rays. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation.
• 3.3: Atomic Spectroscopy and The Bohr Model
The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. However, more direct evidence was needed to verify the quantized nature of energy in all matter. In this section, we describe how observation of the interaction of atoms with visible light provided this evidence.
• 3.4: The Wavelength Nature of Matter
• 3.5: Quantum Mechanics and The Atom
• 3.6: The Shape of Atomic Orbitals
Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex.
Thumbnail: The \(d_{z^2}\) Orbital.
03: The Quantum-Mechanical Model of the Atom
The field of chemistry deals with the structures, bonding, reactivity, and physical properties of atoms, molecules, radicals, and ions all of whose sizes range from ca. 1 Å for atoms and small molecules to a few hundred Å for polymers and biological molecules such as DNA and proteins. The description of the motions and properties of the particles comprising such small systems has been found to not be amenable to treatment using classical mechanics. Their structures, energies, and other properties have only been successfully described within the framework of quantum mechanics. This is why quantum mechanics has to be mastered as part of learning chemistry.
The concepts of quantum mechanics were invented to explain experimental observations that otherwise were totally inexplicable. This period of invention extended from 1900 when Max Planck introduced the revolutionary concept of quantization to 1925 when Erwin Schrödinger and Werner Heisenberg independently introduced two mathematically different but equivalent formulations of a general quantum mechanical theory. In the early 1930's Schrödinger published a way of thinking about the circumstance of radioactive decay that is still useful. We imagine an apparatus containing just one Nitrogen-13 atom and a detector that will respond when the atom decays. Connected to the detector is a relay connected to a hammer, and when the atom decays the relay releases the hammer which then falls on a glass vial containing poison gas. We take the entire apparatus and put it in a box. We also place a cat in the box, close the lid, and wait 10 minutes.
We then ask: Is the cat alive or dead? The answer according to quantum mechanics is that it is 50% dead and 50% alive. Quantum mechanics often gives odd results from a classical perspective.
In this chapter, we will use quantum mechanismcs to describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.01%3A_Schrodinger%27s_Cat.txt |
Learning Objectives
• Explain the basic behavior of waves, including traveling waves and standing waves
• Describe the wave nature of light
• Use appropriate equations to calculate related light-wave properties such as period, frequency, wavelength, and energy
• Describe the particle nature of light
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation.
Visible light and other forms of electromagnetic radiation play important roles in chemistry, since they can be used to infer the energies of electrons within atoms and molecules. Much of modern technology is based on electromagnetic radiation. For example, radio waves from a mobile phone, X-rays used by dentists, the energy used to cook food in your microwave, the radiant heat from red-hot objects, and the light from your television screen are forms of electromagnetic radiation that all exhibit wavelike behavior.
Waves Nature of Light
A wave is a periodic oscillation that transmits energy through space. Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure $1$). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down.
Waves have characteristic properties (Figure $2$). As you may have noticed in Figure $1$, waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength ($λ$, lowercase Greek lambda). Wavelengths are described by a unit of distance, typically meters. The frequency ($\nu$, lowercase Greek nu) of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations per second (1/s = s−1), which in the SI system is called the hertz (Hz). It is named after German physicist Heinrich Hertz (1857–1894), a pioneer in the field of electromagnetic radiation.
The amplitude, or vertical height, of a wave is defined as half the peak-to-trough height; as the amplitude of a wave with a given frequency increases, so does its energy. As you can see in Figure $2$, two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a wave per unit time is its speed ($v$), which is typically measured in meters per second (m/s). The speed of a wave is equal to the product of its wavelength and frequency:
\begin{align} (\text{wavelength})(\text{frequency}) &= \text{speed} \nonumber \[4pt] \lambda \nu &=v \label{6.1.1a} \[4pt] \left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{\text{wave}}}{\text{second}} \right ) &=\dfrac{\text{meters}}{\text{second}} \label{6.1.1b} \end{align}
Different types of waves may have vastly different possible speeds and frequencies. Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs.
Electromagnetic Radiation
Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation. (Figure $3$). Some forms of electromagnetic radiation are shown in Figure $4$. In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c), which turns out to be a fundamental physical constant with a value of 2.99792458 × 108 m/s (about 3.00 ×108 m/s or 1.86 × 105 mi/s). This is about a million times faster than the speed of sound.
Because the various kinds of electromagnetic radiation all have the same speed ($c$), they differ in only wavelength and frequency. As shown in Figure $4$ and Table $1$, the wavelengths of familiar electromagnetic radiation range from 101 m for radio waves to 10−12 m for gamma rays, which are emitted by nuclear reactions. By replacing $v$ with $c$ in Equation $\ref{6.1.1a}$, we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength:
\begin{align} c&=\lambda \nu \[4pt] \nu &=\dfrac{c}{\lambda } \label{6.1.2} \end{align}
For example, the frequency of radio waves is about 108 Hz, whereas the frequency of gamma rays is about 1020 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10−7 m (700 nm, or 4.3 × 1014 Hz) and 4 × 10−7 m (400 nm, or 7.5 × 1014 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly, when frequency decreases, the wavelength increases.
Within the visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism (Figure $\PageIndex{4a}$). A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms.
Table $1$: Common Wavelength Units for Electromagnetic Radiation
Unit Symbol Wavelength (m) Type of Radiation
picometer pm 10−12 gamma ray
angstrom Å 10−10 x-ray
nanometer nm 10−9 UV, visible
micrometer μm 10−6 infrared
millimeter mm 10−3 infrared
centimeter cm 10−2 microwave
meter m 100 radio
As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength:
\begin{align} E\; &\propto\; \nu \label{6.1.3} \[4pt] & \propto\; \dfrac{1}{\lambda } \label{6.1.4} \end{align}
Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer of the atmosphere absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation.
The energy of electromagnetic radiation increases with increasing frequency and decreasing wavelength.
Example $1$: Wavelength of Radiowaves
Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation?
Given: frequency
Asked for: wavelength
Strategy:
Substitute the value for the speed of light in meters per second into Equation $\ref{6.1.2}$ to calculate the wavelength in meters.
Solution:
From Equation \ref{6.1.2}, we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 108 m/s:
\begin{align*} λ\nu &= c \[4pt] &= 2.998 \times 10^8 m/s \end{align*}
Thus the wavelength $λ$ is given by
\begin{align*} \lambda &=\dfrac{c}{ \nu } \[4pt] &=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right ) \[4pt] &=2.965\; m \end{align*}
Exercise $1$
As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun?
Answer
8.45 mm
Interference and Diffraction
When two or more waves arrive at the same point, they superimpose themselves on one another. More specifically, the disturbances of waves are superimposed when they come together—a phenomenon called superposition. Each disturbance corresponds to a force, and forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves—that is, their amplitudes add. Figures $5$ and $6$ illustrate superposition in two special cases, both of which produce simple results.
Figure $5$ shows two identical waves that arrive at the same point exactly in phase. The crests of the two waves are precisely aligned, as are the troughs. This superposition produces pure constructive interference. Because the disturbances add, pure constructive interference produces a wave that has twice the amplitude of the individual waves, but has the same wavelength.
Figure $6$ shows two identical waves that arrive exactly out of phase—that is, precisely aligned crest to trough—producing pure destructive interference. Because the disturbances are in the opposite direction for this superposition, the resulting amplitude is zero for pure destructive interference—the waves completely cancel.
While pure constructive and pure destructive interference do occur, they require precisely aligned identical waves. The superposition of most waves produces a combination of constructive and destructive interference and can vary from place to place and time to time. Sound from a stereo, for example, can be loud in one spot and quiet in another. Varying loudness means the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers creating sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of airplane jets heard by a stationary passenger. The combined sound can fluctuate up and down in volume as the sound from the two engines varies in time from constructive to destructive.
The Particle Nature of Light
When certain metals are exposed to light, electrons are ejected from their surface (Figure $7$). Classical physics predicted that the number of electrons emitted and their kinetic energy should depend on only the intensity of the light, not its frequency. In fact, however, each metal was found to have a characteristic threshold frequency of light; below that frequency, no electrons are emitted regardless of the light’s intensity. Above the threshold frequency, the number of electrons emitted was found to be proportional to the intensity of the light, and their kinetic energy was proportional to the frequency. This phenomenon was called the photoelectric effect (A phenomenon in which electrons are ejected from the surface of a metal that has been exposed to light).
In 1900, the German physicist Max Planck (1858–1947) proposing that the energy of electromagnetic waves is quantized rather than continuous. Planck postulated that the energy of a particular quantum of radiant energy could be described by the equation
$E=h u \label{6.2.1}$
where the proportionality constant h is called Planck’s constant, one of the most accurately known fundamental constants in science. For our purposes, its value to four significant figures is generally sufficient:
$h = 6.626 \times 10^{−34}\, J \cdot s\, (\text{joule-seconds}) \nonumber$
As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. Although quantization may seem to be an unfamiliar concept, we encounter it frequently. For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. Even electrical charge is quantized: an ion may have a charge of −1 or −2 but not −1.33 electron charges.
Albert Einstein (1879–1955; Nobel Prize in Physics, 1921) quickly realized that Planck’s hypothesis about the quantization of radiant energy could also explain the photoelectric effect. The key feature of Einstein’s hypothesis was the assumption that radiant energy arrives at the metal surface in particles that we now call photons (a quantum of radiant energy, each of which possesses a particular energy energy $E$ given by Equation $\ref{6.2.1}$ Einstein postulated that each metal has a particular electrostatic attraction for its electrons that must be overcome before an electron can be emitted from its surface ($E_o= u_o$). If photons of light with energy less than Eo strike a metal surface, no single photon has enough energy to eject an electron, so no electrons are emitted regardless of the intensity of the light. If a photon with energy greater than Eo strikes the metal, then part of its energy is used to overcome the forces that hold the electron to the metal surface, and the excess energy appears as the kinetic energy of the ejected electron:
\begin{align} \text{ kinetic energy of ejected electron} &=E-E_{o} \nonumber \[4pt] &=h \nu -h \nu _{o} \nonumber \[4pt] &=h\left ( \nu - \nu_{o} \right ) \label{6.2.2} \end{align}
When a metal is struck by light with energy above the threshold energy Eo, the number of emitted electrons is proportional to the intensity of the light beam, which corresponds to the number of photons per square centimeter, but the kinetic energy of the emitted electrons is proportional to the frequency of the light. Thus Einstein showed that the energy of the emitted electrons depended on the frequency of the light, contrary to the prediction of classical physics. Moreover, the idea that light could behave not only as a wave but as a particle in the form of photons suggested that matter and energy might not be such unrelated phenomena after all.
Albert Einstein (1879–1955)
In 1900, Einstein was working in the Swiss patent office in Bern. He was born in Germany and throughout his childhood his parents and teachers had worried that he might be developmentally disabled. The patent office job was a low-level civil service position that was not very demanding, but it did allow Einstein to spend a great deal of time reading and thinking about physics.
In 1905, his "miracle year" he published four papers that revolutionized physics. One was on the special theory of relativity, a second on the equivalence of mass and energy, a third on Brownian motion, and the fourth on the photoelectric effect, for which he received the Nobel Prize in 1921, the theory of relativity and energy-matter equivalence being still controversial at the time
Planck’s and Einstein’s postulate that energy is quantized is in many ways similar to Dalton’s description of atoms. Both theories are based on the existence of simple building blocks, atoms in one case and quanta of energy in the other. The work of Planck and Einstein thus suggested a connection between the quantized nature of energy and the properties of individual atoms.
Example $2$
A ruby laser, a device that produces light in a narrow range of wavelengths emits red light at a wavelength of 694.3 nm (Figure $4$). What is the energy in joules of a single photon?
Given: wavelength
Asked for: energy of single photon.
Strategy:
1. Use Equation $\ref{6.2.1}$ and the relationship between wavelength and frequency to calculate the energy in joules.
Solution:
The energy of a single photon is given by $E = h\nu = \dfrac{hc}{λ}.$
Exercise $2$
An x-ray generator, such as those used in hospitals, emits radiation with a wavelength of 1.544 Å.
1. What is the energy in joules of a single photon?
2. How many times more energetic is a single x-ray photon of this wavelength than a photon emitted by a ruby laser?
Answer a
$1.287 \times 10^{-15}\; J/photon$
Answer a
4497 times
Example $3$: Photoelectric Effect
Identify which of the following statements are false and, where necessary, change the italicized word or phrase to make them true, consistent with Einstein's explanation of the photoelectric effect.
1. Increasing the brightness of incoming light increases the kinetic energy of the ejected electrons.
2. Increasing the wavelength of incoming light increases the kinetic energy of the ejected electrons.
3. Increasing the brightness of incoming light increases the number of ejected electrons.
4. Increasing the frequency of incoming light can increase the number of ejected electrons.
Solution
1. False. Increasing the brightness of incoming light has no effect on the kinetic energy of the ejected electrons. Only energy, not the number or amplitude, of the photons influences the kinetic energy of the electrons.
2. False. Increasing the frequency of incoming light increases the kinetic energy of the ejected electrons. Frequency is proportional to energy and inversely proportional to wavelength. Frequencies above the threshold value transfer the excess energy into the kinetic energy of the electrons.
3. True. Because the number of collisions with photons increases with brighter light, the number of ejected electrons increases.
4. True with regard to the threshold energy binding the electrons to the metal. Below this threshold, electrons are not emitted and above it they are. Once over the threshold value, further increasing the frequency does not increase the number of ejected electrons
Exercise $3$
Calculate the threshold energy in kJ/mol of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is $9.87 \times 10^{14}\; Hz$.
Answer
$3.94 \: kJ/mol$
Summary
Light and other forms of electromagnetic radiation move through a vacuum with a constant speed, c, of 2.998 × 108 m s−1. This radiation shows wavelike behavior, which can be characterized by a frequency, ν, and a wavelength, λ, such that c = λν. Light is an example of a travelling wave. Other important wave phenomena include standing waves, periodic oscillations, and vibrations. Standing waves exhibit quantization, since their wavelengths are limited to discrete integer multiples of some characteristic lengths. Electromagnetic radiation that passes through two closely spaced narrow slits having dimensions roughly similar to the wavelength will show an interference pattern that is a result of constructive and destructive interference of the waves. Electromagnetic radiation also demonstrates properties of particles called photons. The energy of a photon is related to the frequency (or alternatively, the wavelength) of the radiation as E = (or $E=\dfrac{hc}{λ}$), where h is Planck's constant. That light demonstrates both wavelike and particle-like behavior is known as wave-particle duality. All forms of electromagnetic radiation share these properties, although various forms including X-rays, visible light, microwaves, and radio waves interact differently with matter and have very different practical applications. Electromagnetic radiation can be generated by exciting matter to higher energies, such as by heating it.
Key Equations
• c = λν
• $E=hν=\dfrac{hc}{λ}$, where h = 6.626 × 10−34 J s
Summary
Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength ($λ$), the distance between successive waves; frequency ($u$), the number of waves that pass a fixed point per unit time; speed ($v$), the rate at which the wave propagates through space; and amplitude, the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light ($c$). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths.
Glossary
amplitude
extent of the displacement caused by a wave (for sinusoidal waves, it is one-half the difference from the peak height to the trough depth, and the intensity is proportional to the square of the amplitude)
continuous spectrum
electromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun)
electromagnetic radiation
energy transmitted by waves that have an electric-field component and a magnetic-field component
electromagnetic spectrum
range of energies that electromagnetic radiation can comprise, including radio, microwaves, infrared, visible, ultraviolet, X-rays, and gamma rays; since electromagnetic radiation energy is proportional to the frequency and inversely proportional to the wavelength, the spectrum can also be specified by ranges of frequencies or wavelengths
frequency ($\nu$)
number of wave cycles (peaks or troughs) that pass a specified point in space per unit time
hertz (Hz)
the unit of frequency, which is the number of cycles per second, s−1
intensity
property of wave-propagated energy related to the amplitude of the wave, such as brightness of light or loudness of sound
interference pattern
pattern typically consisting of alternating bright and dark fringes; it results from constructive and destructive interference of waves
line spectrum
electromagnetic radiation emitted at discrete wavelengths by a specific atom (or atoms) in an excited state
node
any point of a standing wave with zero amplitude
photon
smallest possible packet of electromagnetic radiation, a particle of light
quantization
occurring only in specific discrete values, not continuous
standing wave
(also, stationary wave) localized wave phenomenon characterized by discrete wavelengths determined by the boundary conditions used to generate the waves; standing waves are inherently quantized
wave
oscillation that can transport energy from one point to another in space
wavelength (λ)
distance between two consecutive peaks or troughs in a wave
wave-particle duality
term used to describe the fact that elementary particles including matter exhibit properties of both particles (including localized position, momentum) and waves (including nonlocalization, wavelength, frequency) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.02%3A_The_Nature_of_Light.txt |
Learning Objectives
• To know the relationship between atomic spectra and the electronic structure of atoms.
The concept of the photon emerged from experimentation with thermal radiation, electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies. The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. However, more direct evidence was needed to verify the quantized nature of energy in all matter. In this section, we describe how observation of the interaction of atoms with visible light provided this evidence.
Line Spectra
Although objects at high temperature emit a continuous spectrum of electromagnetic radiation, a different kind of spectrum is observed when pure samples of individual elements are heated. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. When the emitted light is passed through a prism, only a few narrow lines of particular wavelengths, called a line spectrum, are observed rather than a continuous range of wavelengths (Figure $1$). The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. With sodium, however, we observe a yellow color because the most intense lines in its spectrum are in the yellow portion of the spectrum, at about 589 nm.
Such emission spectra were observed for many other elements in the late 19th century, which presented a major challenge because classical physics was unable to explain them. Part of the explanation is provided by Planck’s equation: the observation of only a few values of λ (or $u$) in the line spectrum meant that only a few values of E were possible. Thus the energy levels of a hydrogen atom had to be quantized; in other words, only states that had certain values of energy were possible, or allowed. If a hydrogen atom could have any value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation.
In 1885, a Swiss mathematics teacher, Johann Balmer (1825–1898), showed that the frequencies of the lines observed in the visible region of the spectrum of hydrogen fit a simple equation that can be expressed as follows:
$u=constant\; \left ( \dfrac{1}{2^{2}}-\dfrac{1}{n^{^{2}}} \right ) \label{6.3.1}$
where n = 3, 4, 5, 6. As a result, these lines are known as the Balmer series. The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the Rydberg equation:
$\dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \label{6.3.2}$
where $n_1$ and $n_2$ are positive integers, $n_2 > n_1$, and $\Re$ the Rydberg constant, has a value of 1.09737 × 107 m−1.
Johann Balmer (1825–1898)
A mathematics teacher at a secondary school for girls in Switzerland, Balmer was 60 years old when he wrote the paper on the spectral lines of hydrogen that made him famous.
Balmer published only one other paper on the topic, which appeared when he was 72 years old.
Like Balmer’s equation, Rydberg’s simple equation described the wavelengths of the visible lines in the emission spectrum of hydrogen (with n1 = 2, n2 = 3, 4, 5,…). More important, Rydberg’s equation also predicted the wavelengths of other series of lines that would be observed in the emission spectrum of hydrogen: one in the ultraviolet (n1 = 1, n2 = 2, 3, 4,…) and one in the infrared (n1 = 3, n2 = 4, 5, 6). Unfortunately, scientists had not yet developed any theoretical justification for an equation of this form.
Bohr's Model
In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. Bohr’s model required only one assumption: The electron moves around the nucleus in circular orbits that can have only certain allowed radii. Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that the electron could occupy only certain regions of space.
Using classical physics, Niels Bohr showed that the energy of an electron in a particular orbit is given by
$E_{n}=\dfrac{-\Re hc}{n^{2}} \label{6.3.3}$
where $\Re$ is the Rydberg constant, h is Planck’s constant, c is the speed of light, and n is a positive integer corresponding to the number assigned to the orbit, with n = 1 corresponding to the orbit closest to the nucleus. In this model n = ∞ corresponds to the level where the energy holding the electron and the nucleus together is zero. In that level, the electron is unbound from the nucleus and the atom has been separated into a negatively charged (the electron) and a positively charged (the nucleus) ion. In this state the radius of the orbit is also infinite. The atom has been ionized.
Niels Bohr (1885–1962)
During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project.
In his final years, he devoted himself to the peaceful application of atomic physics and to resolving political problems arising from the development of atomic weapons.
As n decreases, the energy holding the electron and the nucleus together becomes increasingly negative, the radius of the orbit shrinks and more energy is needed to ionize the atom. The orbit with n = 1 is the lowest lying and most tightly bound. The negative sign in Equation $\ref{6.3.3}$ indicates that the electron-nucleus pair is more tightly bound (i.e. at a lower potential energy) when they are near each other than when they are far apart. Because a hydrogen atom with its one electron in this orbit has the lowest possible energy, this is the ground state (the most stable arrangement of electrons for an element or a compound) for a hydrogen atom. As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure $\PageIndex{2a}$). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited state, defined as any arrangement of electrons that is higher in energy than the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states (Figure $1$).
So the difference in energy ($ΔE$) between any two orbits or energy levels is given by $\Delta E=E_{n_{1}}-E_{n_{2}}$ where n1 is the final orbit and n2 the initial orbit. Substituting from Bohr’s equation (Equation \ref{6.3.3}) for each energy value gives
\begin{align*} \Delta E &=E_{final}-E_{initial} \[4pt] &=-\dfrac{\Re hc}{n_{2}^{2}}-\left ( -\dfrac{\Re hc}{n_{1}^{2}} \right ) \[4pt] &=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.4} \end{align*}
If $n_2 > n_1$, the transition is from a higher energy state (larger-radius orbit) to a lower energy state (smaller-radius orbit), as shown by the dashed arrow in part (a) in Figure $3$. Substituting $hc/λ$ for $ΔE$ gives
$\Delta E = \dfrac{hc}{\lambda }=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.5}$
Canceling $hc$ on both sides gives
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.6}$
Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equations $\ref{6.3.5}$ and $\ref{6.3.6}$ indicates that energy is released as the electron moves from orbit $n_2$ to orbit $n_1$ because orbit $n_2$ is at a higher energy than orbit $n_1$. Bohr calculated the value of $\Re$ from fundamental constants such as the charge and mass of the electron and Planck's constant and obtained a value of 1.0974 × 107 m−1, the same number Rydberg had obtained by analyzing the emission spectra.
We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen ($\PageIndex{3b}$); the lines in this series correspond to transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (Figure $\PageIndex{3a}$). The n = 3 to n = 2 transition gives rise to the line at 656 nm (red), the n = 4 to n = 2 transition to the line at 486 nm (green), the n = 5 to n = 2 transition to the line at 434 nm (blue), and the n = 6 to n = 2 transition to the line at 410 nm (violet). Because a sample of hydrogen contains a large number of atoms, the intensity of the various lines in a line spectrum depends on the number of atoms in each excited state. At the temperature in the gas discharge tube, more atoms are in the n = 3 than the n ≥ 4 levels. Consequently, the n = 3 to n = 2 transition is the most intense line, producing the characteristic red color of a hydrogen discharge (Figure $\PageIndex{1a}$). Other families of lines are produced by transitions from excited states with n > 1 to the orbit with n = 1 or to orbits with n ≥ 3. These transitions are shown schematically in Figure $4$
Using Atoms to Time
In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Telecommunications systems, such as cell phones, depend on timing signals that are accurate to within a millionth of a second per day, as are the devices that control the US power grid. Global positioning system (GPS) signals must be accurate to within a billionth of a second per day, which is equivalent to gaining or losing no more than one second in 1,400,000 years. Quantifying time requires finding an event with an interval that repeats on a regular basis.
To achieve the accuracy required for modern purposes, physicists have turned to the atom. The current standard used to calibrate clocks is the cesium atom. Supercooled cesium atoms are placed in a vacuum chamber and bombarded with microwaves whose frequencies are carefully controlled. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Decay to a lower-energy state emits radiation. The microwave frequency is continually adjusted, serving as the clock’s pendulum.
In 1967, the second was defined as the duration of 9,192,631,770 oscillations of the resonant frequency of a cesium atom, called the cesium clock. Research is currently under way to develop the next generation of atomic clocks that promise to be even more accurate. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals.
Example $1$: The Lyman Series
The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur?
Given: lowest-energy orbit in the Lyman series
Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum
Strategy:
1. Substitute the appropriate values into Equation \ref{6.3.2} (the Rydberg equation) and solve for $\lambda$.
2. Use Figure 2.2.1 to locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \nonumber$
A For the Lyman series, n1 = 1. The lowest-energy line is due to a transition from the n = 2 to n = 1 orbit because they are the closest in energy.
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1} \nonumber$
It turns out that spectroscopists (the people who study spectroscopy) use cm-1 rather than m-1 as a common unit. Wavelength is inversely proportional to energy but frequency is directly proportional as shown by Planck's formula, $E=h u$.
Spectroscopists often talk about energy and frequency as equivalent. The cm-1 unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm-1, the visible from 11,000 to 25.000 cm-1 and the UV between 25,000 and 100,000 cm-1. The units of cm-1 are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm-1.
$\widetilde{ u} =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1} \nonumber$
and
$\lambda = 1.215 \times 10^{−7}\; m = 122\; nm \nonumber$
This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets, producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O2 molecules to O atoms which react with other O2 molecules to form stratospheric ozone.
B This wavelength is in the ultraviolet region of the spectrum.
Exercise $1$: The Pfund Series
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n = 5 orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?
Answer
4.65 × 103 nm; infrared
Bohr’s model of the hydrogen atom gave an exact explanation for its observed emission spectrum. The following are his key contributions to our understanding of atomic structure:
Unfortunately, Bohr could not explain why the electron should be restricted to particular orbits. Also, despite a great deal of tinkering, such as assuming that orbits could be ellipses rather than circles, his model could not quantitatively explain the emission spectra of any element other than hydrogen (Figure $5$). In fact, Bohr’s model worked only for species that contained just one electron: H, He+, Li2+, and so forth. Scientists needed a fundamental change in their way of thinking about the electronic structure of atoms to advance beyond the Bohr model.
Atomic Spectroscopy and the Identification of Elements
Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrum (a spectrum produced by the emission of light by atoms in excited states). The converse, absorption of light by ground-state atoms to produce an excited state, can also occur, producing an absorption spectrum (a spectrum produced by the absorption of light by ground-state atoms).
When an atom emits light, it decays to a lower energy state; when an atom absorbs light, it is excited to a higher energy state.
If white light is passed through a sample of hydrogen, hydrogen atoms absorb energy as an electron is excited to higher energy levels (orbits with n ≥ 2). If the light that emerges is passed through a prism, it forms a continuous spectrum with black lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. These wavelengths correspond to the n = 2 to n = 3, n = 2 to n = 4, n = 2 to n = 5, and n = 2 to n = 6 transitions. Any given element therefore has both a characteristic emission spectrum and a characteristic absorption spectrum, which are essentially complementary images.
Emission and absorption spectra form the basis of spectroscopy, which uses spectra to provide information about the structure and the composition of a substance or an object. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. As an example, consider the spectrum of sunlight shown in Figure $7$ Because the sun is very hot, the light it emits is in the form of a continuous emission spectrum. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. By comparing these lines with the spectra of elements measured on Earth, we now know that the sun contains large amounts of hydrogen, iron, and carbon, along with smaller amounts of other elements. During the solar eclipse of 1868, the French astronomer Pierre Janssen (1824–1907) observed a set of lines that did not match those of any known element. He suggested that they were due to the presence of a new element, which he named helium, from the Greek helios, meaning “sun.” Helium was finally discovered in uranium ores on Earth in 1895. Alpha particles are helium nuclei. Alpha particles emitted by the radioactive uranium pick up electrons from the rocks to form helium atoms.
The familiar red color of neon signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure $5$. Similarly, the blue and yellow colors of certain street lights are caused, respectively, by mercury and sodium discharges. In all these cases, an electrical discharge excites neutral atoms to a higher energy state, and light is emitted when the atoms decay to the ground state. In the case of mercury, most of the emission lines are below 450 nm, which produces a blue light (part (c) in Figure $5$). In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light.
Summary
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy excited state to a higher-energy excited state. This produces an absorption spectrum, which has dark lines in the same position as the bright lines in the emission spectrum of an element. Bohr’s model revolutionized the understanding of the atom but could not explain the spectra of atoms heavier than hydrogen.
Key Concepts
• Electrons can occupy only certain regions of space, called orbits.
• Orbits closer to the nucleus are lower in energy.
• Electrons can move from one orbit to another by absorbing or emitting energy, giving rise to characteristic spectra. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.03%3A_Atomic_Spectroscopy_and_The_Bohr_Model.txt |
Learning Objectives
• To understand the wave–particle duality of matter.
Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a photoelectron from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. Light exhibits a wave-particle duality and may described as either a particle or a wave depending on the experiment.
Diffraction is a Wave Phenomenon
It is well-known that light has the ability to diffract around objects in its path, leading to an interference pattern that is particular to the object. This is, in fact, how holography works (the interference pattern is created by allowing the diffracted light to interfere with the original beam so that the hologram can be viewed by shining the original beam on the image). A simple illustration of light diffraction is the Young double slit experiment (Figure $1$).
Here, we use light pictured as waves in a plane parallel to the double slit apparatus and observe what happens when they impinge on the slits. Each slit becomes a point source for spherical waves that subsequently interfere with each other, giving rise to the light and dark fringes on the screen at the right. Interference is a wave phenomenon in which two waves superimpose to form a resultant wave of greater or lower amplitude. It is a primary property used to identify wave behavior in nature.
According to classical physics, electrons should behave like particles - they travel in straight lines and do not curve in flight unless acted on by an external agent, like a magnetic field. In this model, if we fire a beam of electrons through a double slit onto a detector, we should get two bands of "hits", much as you would get if you fired a machine gun at the side of a house with two windows - you would get two areas of bullet-marked wall inside, and the rest would be intact Figure $\PageIndex{2; left}$. However, if the slits are made small enough and close enough together, experimentalist actually observe the electrons diffracting through the slits and interfering with each other just like light waves (Figure $\PageIndex{2; right}$). This means that the electrons exhibit a similar wave-particle duality that light exhibits. In this case, they must have properties like wavelength and frequency. We can deduce the properties from the behavior of the electrons as they pass through our diffraction grating.
The observation that particles may exhibit wavelength phenomena was a pivotal result in the development of quantum mechanics. For physicists this idea was important because it meant that not only could any particle exhibit wave characteristics, but that one could use wave equations to describe phenomena in matter. In his PhD dissertation submitted to the Sorbonne in 1924, Louis de Broglie (1892–1972) proposed that a particle such as an electron could be described by a wave whose wavelength is given by
$\lambda =\dfrac{h}{mv} \label{6.4.3}$
where
• $\lambda$ is the de Broglie wavelength,
• $h$ is Planck’s constant,
• $m$ is the mass of the particle, and
• $v$ is the velocity of the particle.
It turned out that de Broglie hypothesis was able to accurately describe a range of diffraction phenomena of matter. For his work, de Broglie received the Nobel Prize in Physics in 1929.
If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation (Equation \ref{6.4.3}, which is an extremely small number. As you will calculate in Example $1$, Planck’s constant (6.63 × 10−34 J•s) is so small that the wavelength of a particle with a large mass is simply too short (less than the diameter of an atomic nucleus) to be noticeable.
Example $1$: Wavelength of a Baseball in Motion
Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.
Given: mass and speed of object
Asked for: wavelength
Strategy:
1. Convert the speed of the baseball to the appropriate SI units: meters per second.
2. Substitute values into Equation $\ref{6.4.3}$ and solve for the wavelength.
Solution:
The wavelength of a particle is given by $λ = h/mv$. We know that m = 0.149 kg, so all we need to find is the speed of the baseball:
$v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )$
B Recall that the joule is a derived unit, whose units are (kg•m2)/s2. Thus the wavelength of the baseball is
\begin{align*} \lambda &=\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )} \[4pt] &= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )} \[4pt] &=9.95\times 10^{-35}\; m \end{align*}
(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10−14 m, the wavelength of the baseball is almost unimaginably small.
Exercise $1$: Wavelength of a Neutron in Motion
Calculate the wavelength of a neutron that is moving at 3.00 × 103 m/s.
Answer
1.32 Å, or 132 pm
As you calculated in Example $1$, objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, however, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope, which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure $3$).
The Heisenberg Uncertainty Principle
Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a particle that exhibits wavelike behavior. Hence, an "uncertainty principle" for light is merely a conclusion about the nature of electromagnetic waves and nothing new. De Broglie's idea of wave-particle duality means that particles such as electrons which exhibit an uncertainty principle. This was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principle states that the uncertainty in the position of a particle ($Δx$) multiplied by the uncertainty in its momentum [$Δ(mv)$] is greater than or equal to Planck’s constant divided by $4π$:
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\ge \dfrac{h}{4\pi } \label{6.4.7}$
Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths.
If the precise position $x$ of a particle is known absolutely ($Δx = 0$), then the uncertainty in its momentum must be infinite:
$\left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{6.4.8}$
Because the mass of the electron at rest ($m$) is both constant and accurately known, the uncertainty in $Δ(mv)$ must be due to the $Δv$ term, which would have to be infinitely large for $Δ(mv)$ to equal infinity. That is, according to Equation $\ref{6.4.8}$, the more accurately we know the exact position of the electron (as $Δx → 0$), the less accurately we know the speed and the kinetic energy of the electron (1/2 mv2) because $Δ(mv) → ∞$. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as $Δ(mv) → 0$], then $Δx → ∞$ and we have no idea where the electron is.
Example $2$: The Uncertainty Principle Large and Small
Determine the minimum uncertainties in the positions of the following objects if their speeds are known with a precision of $1.0 \times 10^{-3} m/s$:
1. an electron and
2. a bowling ball of mass 6.0 kg.
Strategy
Given the uncertainty in speed $\Delta u = 1.0 \times 10^{-3} m/s$, we have to first determine the uncertainty in momentum $\Delta p = m\Delta u$ and then invert Equation \ref{Heisen} to find the uncertainty in position
$\Delta x = \dfrac{\hbar}{2\Delta p}. \nonumber$
Solution
1. For the electron: \begin{align*} \Delta p &= m\Delta u \[4pt] &= (9.1 \times 10^{-31} kg)(1.0 \times 10^{-3}m/s) \[4pt] &= 9.1 \times 10^{-34} kg \cdot m/s,\end{align*} \begin{align*} \Delta x &= \frac{\hbar}{2\Delta p} \[4pt] &= 5.8 \, cm. \end{align*}
2. For the bowling ball: \begin{align*} \Delta p &= m\Delta u \[4pt] &= (6.0 \, kg)(1.0 \times 10^{-3}m/s) \[4pt] &= 6.0 \times 10^{-3} kg \cdot m/s, \end{align*} \begin{align*} \Delta x &= \frac{\hbar}{2\Delta p} \[4pt] &= 8.8 \times 10^{-33}m. \end{align*}
Significance
Unlike the position uncertainty for the electron, the position uncertainty for the bowling ball is immeasurably small. Planck’s constant is very small, so the limitations imposed by the uncertainty principle are not noticeable in macroscopic systems such as a bowling ball.
Indeterminacy and Probability Distribution Maps
Before either quantum mechanics, physicists described the world using "Classical Mechanics," which is like what you've probably studied before in physics class: Newtonian mechanics (forces, accelerations, etc), electricity and magnetism using Maxwell's equations. All these approaches work well for big things that are not moving too fast. There are two qualities of classical mechanics that quantum mechanics altered. First, in classical mechanics energy and velocity and such quantities can have any value. If you drop a ball, it accelerates smoothly from 0 to a final velocity, rather than moving jerkily from step to step. The Quantum Mechanics got rid of the assumptions that energy and velocity should be "continuous."
A trajectory is the path that an object with mass in motion follows through space as a function of time. In classical mechanics, a complete trajectory is defined by knowledge of position and momentum, simultaneously. If you launch a missile, a space shuttle or kick a ball (Figure $1$), you can calculate almost exactly the path it will follow; if you have perfect knowledge of the forces acting on it, you can calculate its path perfectly. This is because Newton's laws are completely deterministic - that is because they imply that anything that happens at any future time is completed determined by what happens now, and moreover that everything now was completely determined by what happened at any time in the past.
However, as discussed above, due to Heisenberg uncertainty principle, position and momentum cannot be measured simultaneously. This means the concept of a trajectory in quantum mechanics is not defined and we actually do not know where they are or what path they follow (Figure $4$). Quantum mechanics is a probabilitic theory, that discards the concept of trajectory for small particles, and uses statistics and probability to describe evolution.
Summary
An electron possesses both particle and wave properties. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.
$\lambda =\dfrac{h}{mv} \nonumber$
The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.04%3A_The_Wavelength_Nature_of_Matter.txt |
Learning Objectives
• To apply the results of quantum mechanics to electrons in atoms
The paradox described by Heisenberg’s uncertainty principle and the wavelike nature of subatomic particles such as the electron made it impossible to use the equations of classical physics to describe the motion of electrons in atoms. Scientists needed a new approach that took the wave behavior of the electron into account. In 1926, an Austrian physicist, Erwin Schrödinger (1887–1961; Nobel Prize in Physics, 1933), developed wave mechanics, a mathematical technique that describes the relationship between the motion of a particle that exhibits wavelike properties (such as an electron) and its allowed energies.
Schrödinger started with the simple requirement that the total energy of the electron is the sum of its kinetic and potential energies:
$E = \underbrace{\dfrac{mv^2}{2}}_{\text{kinetic energy}} + \underbrace{\dfrac{-e^2}{r}}_{\text{potential energy}} \label{5.6.1}$
The second term represents the potential energy of an electron (whose charge is denoted by e) at a distance r from a proton (the nucleus of the hydrogen atom). In quantum mechanics it is generally easier to deal with equations that use momentum ($p = mv$) rather than velocity, so the next step is to make this substitution:
$E = \dfrac{p^2}{2m} - \dfrac{e^2}{r} \label{5.6.2}$
This is still an entirely classical relation, as valid for the waves on a guitar string as for those of the electron in a hydrogen atom. The third step takes into account the wavelike character of the electron in the atom, a mathematical expression that describes the position and momentum of the electron at all points in space is applied to both sides of the equation. The function, denoted by $\psi$, "modulates" the equation of motion of the electron so as to reflect the fact that the electron manifests itself with greater probability in some locations that at others. This yields the celebrated Schrödinger equation
$\left( \dfrac{mv^2}{2} - \dfrac{e^2}{r} \right) \psi = E\psi \label{5.6.3}$
which is often written as
$H\psi =E \psi$
where $H$ is the Hamiltonian operator, $E$ is the total energy of the electron, and $\psi$ is the wavefunction of the electron. The Hamiltonian is a combination of operations that are used to extract the total energy of the system (e.g., the sum of kinetic an potential energies) and the wavefunction is a mathematical description of the quantum state of an isolated quantum system. The wavefunction is a complex-valued probability amplitude, and the probabilities for the possible results of measurements made on the system can be derived from it. Although quantum mechanics uses sophisticated mathematics, you do not need to understand the mathematical details to follow our discussion of its general conclusions. We focus on the properties of the wavefunctions that are the solutions of Schrödinger’s equations.
Erwin Schrödinger (1887–1961)
Schrödinger’s unconventional approach to atomic theory was typical of his unconventional approach to life. He was notorious for his intense dislike of memorizing data and learning from books. When Hitler came to power in Germany, Schrödinger escaped to Italy. He then worked at Princeton University in the United States but eventually moved to the Institute for Advanced Studies in Dublin, Ireland, where he remained until his retirement in 1955.
Solutions to the Schrödinger Equation for the Hydrogen Atom
A wavefunction is a mathematical function that relates the location of an electron at a given point in space (identified by x, y, and z coordinates) to the amplitude of its wave, which corresponds to its energy. Thus each wavefunction is associated with a particular energy $E$. The properties of wavefunctions derived from quantum mechanics are summarized here:
A wavefunction uses three variables to describe the position of an electron. A fourth variable is usually required to fully describe the location of objects in motion. Three specify the position in space (as with the Cartesian coordinates x, y, and z), and one specifies the time at which the object is at the specified location. For electrons, we can ignore the time dependence because we will be using standing waves, which by definition do not change with time, to describe the position of an electron.
The magnitude of the wavefunction at a particular point in space is proportional to the amplitude of the wave at that point. Many wavefunctions are complex functions, which is a mathematical term indicating that they contain $\sqrt{-1}$, represented as $i$. Hence the amplitude of the wave has no real physical significance. In contrast, the sign of the wavefunction (either positive or negative) corresponds to the phase of the wave, which will be important in our discussion of chemical bonding. The sign of the wavefunction should not be confused with a positive or negative electrical charge.
The square of the wavefunction at a given point is proportional to the probability of finding an electron at that point, which leads to a distribution of probabilities in space. The square of the wavefunction ($\psi^2$) is always a real quantity [recall that that $\sqrt{-1}^2=-1$] and is proportional to the probability of finding an electron at a given point. More accurately, the probability is given by the product of the wavefunction Ψ and its complex conjugate Ψ*, in which all terms that contain i are replaced by $−i$. We use probabilities because, according to Heisenberg’s uncertainty principle, we cannot precisely specify the position of an electron. The probability of finding an electron at any point in space depends on several factors, including the distance from the nucleus and, in many cases, the atomic equivalent of latitude and longitude. As one way of graphically representing the probability distribution, the probability of finding an electron is indicated by the density of colored dots, as shown for the ground state of the hydrogen atom in Figure $2$.
Describing the electron distribution as a standing wave leads to sets of quantum numbers that are characteristic of each wavefunction. From the patterns of one- and two-dimensional standing waves shown previously, you might expect (correctly) that the patterns of three-dimensional standing waves would be complex. Fortunately, however, in the 18th century, a French mathematician, Adrien Legendre (1752–1783), developed a set of equations to describe the motion of tidal waves on the surface of a flooded planet. Schrödinger incorporated Legendre’s equations into his wavefunctions. The requirement that the waves must be in phase with one another to avoid cancellation and produce a standing wave results in a limited number of solutions (wavefunctions), each of which is specified by a set of numbers called quantum numbers.
Each wavefunction is associated with a particular energy. As in Bohr’s model, the energy of an electron in an atom is quantized; it can have only certain allowed values. The major difference between Bohr’s model and Schrödinger’s approach is that Bohr had to impose the idea of quantization arbitrarily, whereas in Schrödinger’s approach, quantization is a natural consequence of describing an electron as a standing wave.
Schrödinger’s approach uses three quantum numbers (n, l, and ml) to specify any wavefunction. The quantum numbers provide information about the spatial distribution of an electron. Although n can be any positive integer, only certain values of l and ml are allowed for a given value of n.
The Principal Quantum Number
The principal quantum number (n) tells the average relative distance of an electron from the nucleus:
$n = 1, 2, 3, 4,… \label{6.5.1}$
As n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wavefunctions that have the same value of n are said to constitute a principal shell because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels.
The Azimuthal Quantum Number
The second quantum number is often called the azimuthal quantum number (l). The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1:
$l = 0, 1, 2,…, n − 1 \label{6.5.2}$
For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wavefunctions that have the same values of both n and l form a subshell. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space.
The Magnetic Quantum Number
The third quantum number is the magnetic quantum number ($m_l$). The value of $m_l$ describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of $m_l$ depend on the value of l: ml can range from −l to l in integral steps:
$m_l = −l, −l + 1,…, 0,…, l − 1, l \label{6.5.3}$
For example, if $l = 0$, $m_l$ can be only 0; if l = 1, ml can be −1, 0, or +1; and if l = 2, ml can be −2, −1, 0, +1, or +2.
Each wavefunction with an allowed combination of n, l, and ml values describes an atomic orbital, a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals.
The Spin Quantum Number
Example $1$: n=4 Shell Structure
How many subshells and orbitals are contained within the principal shell with n = 4?
Given: value of n
Asked for: number of subshells and orbitals in the principal shell
Strategy:
1. Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells.
2. For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell.
Solution:
A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml.
B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2 = 16
Exercise $1$: n=3 Shell Structure
How many subshells and orbitals are in the principal shell with n = 3?
Answer
three subshells; nine orbitals
Rather than specifying all the values of n and l every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital:
l = 0 1 2 3
Designation s p d f
The principal quantum number is named first, followed by the letter s, p, d, or f as appropriate. (These orbital designations are derived from historical terms for corresponding spectroscopic characteristics: sharp, principle, diffuse, and fundamental.) A 1s orbital has n = 1 and l = 0; a 2p subshell has n = 2 and l = 1 (and has three 2p orbitals, corresponding to ml = −1, 0, and +1); a 3d subshell has n = 3 and l = 2 (and has five 3d orbitals, corresponding to ml = −2, −1, 0, +1, and +2); and so forth.
We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table 6.5.1):
• Each principal shell has n subshells. For n = 1, only a single subshell is possible (1s); for n = 2, there are two subshells (2s and 2p); for n = 3, there are three subshells (3s, 3p, and 3d); and so forth. Every shell has an ns subshell, any shell with n ≥ 2 also has an np subshell, and any shell with n ≥ 3 also has an nd subshell. Because a 2d subshell would require both n = 2 and l = 2, which is not an allowed value of l for n = 2, a 2d subshell does not exist.
• Each subshell has 2l + 1 orbitals. This means that all ns subshells contain a single s orbital, all np subshells contain three p orbitals, all nd subshells contain five d orbitals, and all nf subshells contain seven f orbitals.
Each principal shell has n subshells, and each subshell has 2l + 1 orbitals.
Table $1$: Values of n, l, and ml through n = 4
n l Subshell Designation $m_l$ Number of Orbitals in Subshell Number of Orbitals in Shell
1 0 1s 0 1 1
2 0 2s 0 1 4
1 2p −1, 0, 1 3
3 0 3s 0 1 9
1 3p −1, 0, 1 3
2 3d −2, −1, 0, 1, 2 5
4 0 4s 0 1 16
1 4p −1, 0, 1 3
2 4d −2, −1, 0, 1, 2 5
3 4f −3, −2, −1, 0, 1, 2, 3 7
Summary
There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions (Ψ) to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wavefunctions have five important properties:
1. the wavefunction uses three variables (Cartesian axes x, y, and z) to describe the position of an electron;
2. the magnitude of the wavefunction is proportional to the intensity of the wave;
3. the probability of finding an electron at a given point is proportional to the square of the wavefunction at that point, leading to a distribution of probabilities in space that is often portrayed as an electron density plot;
4. describing electron distributions as standing waves leads naturally to the existence of sets of quantum numbers characteristic of each wavefunction; and
5. each spatial distribution of the electron described by a wavefunction with a given set of quantum numbers has a particular energy.
Quantum numbers provide important information about the energy and spatial distribution of an electron. The principal quantum number n can be any positive integer; as n increases for an atom, the average distance of the electron from the nucleus also increases. All wavefunctions with the same value of n constitute a principal shell in which the electrons have similar average distances from the nucleus. The azimuthal quantum number l can have integral values between 0 and n − 1; it describes the shape of the electron distribution. wavefunctions that have the same values of both n and l constitute a subshell, corresponding to electron distributions that usually differ in orientation rather than in shape or average distance from the nucleus. The magnetic quantum number ml can have 2l + 1 integral values, ranging from −l to +l, and describes the orientation of the electron distribution. Each wavefunction with a given set of values of n, l, and ml describes a particular spatial distribution of an electron in an atom, an atomic orbital. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.05%3A_Quantum_Mechanics_and_The_Atom.txt |
Learning Objectives
• To understand the 3D representation of electronic orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of containing an electron.
One way of representing electron probability distributions was illustrated previously for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at $r = 0$ (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (Figure $\PageIndex{1a}$), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (Figure $\PageIndex{1b}$), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure $1$. In contrast, the surface area of each spherical shell is equal to $4πr^2$, which increases very rapidly with increasing r (Figure $\PageIndex{1c}$). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (Figure $\PageIndex{1d}$). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (Figure $\PageIndex{1d}$).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure $2$ compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (Figure $\PageIndex{2c}$). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability. The existence of these nodes is a consequence of changes of wave phase in the wavefunction Ψ.
s Orbitals (l=0)
Three things happen to s orbitals as n increases (Figure $2$):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure $2$. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals (l=1)
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure $3$. Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a $2p_z$ orbital. As shown in Figure $4$, the other two 2p orbitals have identical shapes, but they lie along the x axis ($2p_x$) and y axis ($2p_y$), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative. Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure $4$.
d Orbitals (l=2)
Subshells with l = 2 have five d orbitals; the first principal shell to have a d subshell corresponds to n = 3. The five d orbitals have ml values of −2, −1, 0, +1, and +2.
The hydrogen 3d orbitals, shown in Figure $5$, have more complex shapes than the 2p orbitals. All five 3d orbitals contain two nodal surfaces, as compared to one for each p orbital and zero for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the $3d_{xy}$, \)3d_{xz}\), and $3d_{yz}$ orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the $3d_{x^2−y^2}$ orbital. The fifth 3d orbital, called the $3d_{z^2}$ orbital, has a unique shape: it looks like a $2p_z$ orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the $3d_{z^2}$ orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in Figure $5$, the phase of the wave function is positive for the two lobes of the $dz^2$ orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in Figure $5$.
f Orbitals (l=3)
Principal shells with n = 4 can have subshells with l = 3 and ml values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven f orbitals. Each f orbital has three nodal surfaces, so their shapes are complex. Because f orbitals are not particularly important for our purposes, we do not discuss them further, and orbitals with higher values of l are not discussed at all.
Orbital Energies
Although we have discussed the shapes of orbitals, we have said little about their comparative energies. We begin our discussion of orbital energies by considering atoms or ions with only a single electron (such as H or He+).
The relative energies of the atomic orbitals with n ≤ 4 for a hydrogen atom are plotted in Figure $6$; note that the orbital energies depend on only the principal quantum number n. Consequently, the energies of the 2s and 2p orbitals of hydrogen are the same; the energies of the 3s, 3p, and 3d orbitals are the same; and so forth. Quantum mechanics predicts that in the hydrogen atom, all orbitals with the same value of n (e.g., the three 2p orbitals) are degenerate, meaning that they have the same energy. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are 4 orbitals with different electron density distributions in the n = 2 principal shell (one 2s and three 2p orbitals), 9 in the n = 3 principal shell, and 16 in the n = 4 principal shell.The different values of l and ml for the individual orbitals within a given principal shell are not important for understanding the emission or absorption spectra of the hydrogen atom under most conditions, but they do explain the splittings of the main lines that are observed when hydrogen atoms are placed in a magnetic field. Figure $6$ shows that the energy levels become closer and closer together as the value of n increases, as expected because of the 1/n2 dependence of orbital energies.
The energies of the orbitals in any species with only one electron can be calculated by a minor variation of Bohr’s equation, which can be extended to other single-electron species by incorporating the nuclear charge $Z$ (the number of protons in the nucleus):
$E=−\dfrac{Z^2}{n^2}Rhc \label{6.6.1}$
In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the $p_x$, $p_y$, and $p_z$ orbitals, are degenerate.
Summary
The four chemically important types of atomic orbital correspond to values of $\ell = 0$, $1$, $2$, and $3$. Orbitals with $\ell = 0$ are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. All orbitals with values of $n > 1$ and $ell = 0$ contain one or more nodes. Orbitals with $\ell = 1$ are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with $\ell = 2$ are d orbitals and have more complex shapes with at least two nodal surfaces. Orbitals with $\ell = 3$ are f orbitals, which are still more complex.
Because its average distance from the nucleus determines the energy of an electron, each atomic orbital with a given set of quantum numbers has a particular energy associated with it, the orbital energy.
$E=−\dfrac{Z^2}{n^2}Rhc \nonumber$
In atoms or ions with only a single electron, all orbitals with the same value of $n$ have the same energy (they are degenerate), and the energies of the principal shells increase smoothly as $n$ increases. An atom or ion with the electron(s) in the lowest-energy orbital(s) is said to be in its ground state, whereas an atom or ion in which one or more electrons occupy higher-energy orbitals is said to be in an excited state. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/03%3A_The_Quantum-Mechanical_Model_of_the_Atom/3.06%3A_The_Shape_of_Atomic_Orbitals.txt |
• 4.1: Aluminum- Low Density Atoms Result in Low Density Metal
• 4.2: The Periodic Law and The Periodic Table
The periodic table is used as a predictive tool that arranges of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals.
• 4.3: Electron Configurations- How Electrons Occupy Orbitals
The relative energy of the subshells determine the order in which atomic orbitals are filled. Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals). Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements.
• 4.4: Electron Configurations, Valence Electrons, and the Periodic Table
Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller.
• 4.5: How Electron Configuration of an Element Relates to this Properties
• 4.6: Periodic Trends in the Size of Atoms and Effective Nuclear Charge
Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of electron shielding and an effective nuclear charge are introduced.
• 4.7: Ions- Configurations, Magnetic Properties, Radii, and Ionization Energy
Generally, the first ionization energy and electronegativity values increase diagonally from the lower left of the periodic table to the upper right, and electron affinities become more negative across a row. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell.
• 4.8: Electron Affinities and Metallic Character
The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table.
04: Periodic Properties of the Elements
Learning Objectives
• To become familiar with the organization of the periodic table.
Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its atomic number (Z), the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the periodic table (Figure \(1\)). The rationale for the peculiar format of the periodic table is explained later. Each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons.
The elements are arranged in a periodic table, which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that facilitates the prediction of many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periods, and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groups, numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elements, listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elements, listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3.
Metals, Nonmetals, and Semimetals
The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in Figure \(1\) divides the elements into metals (in blue, below and to the left of the line) and nonmetals (in bronze, above and to the right of the line). Gold-colored lements that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetals.
The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductile; they can be hammered or pressed into thin sheets or foils because they are malleable; and most have a shiny appearance, so they are lustrous. The vast majority of the known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids.
Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals.
Example \(1\): Classifying Elements
Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal?
Given: element
Asked for: classification
Strategy:
Find selenium in the periodic table shown in Figure \(1\) and then classify the element according to its location.
Solution:
The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure \(1\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties.
Exercise \(1\)
Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal?
Answer
metal
As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of Group 1 are known as the alkali metals, Group 2 are the alkaline earth metals, Group 17 are the halogens, and Group 18 are the noble gases.
Group 1: The Alkali Metals
The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal. The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
Group 2: The Alkaline Earth Metals
The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
Group 17: The Halogens
The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).
Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
Group 18: The Noble Gases
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
The noble gases are unreactive at room temperature and pressure.
Summary
The periodic table is used as a predictive tool. It arranges of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.02%3A_The_Periodic_Law_and_The_Periodic_Table.txt |
Learning Objectives
• Derive the predicted ground-state electron configurations of atoms
• Identify and explain exceptions to predicted electron configurations for atoms and ions
• Relate electron configurations to element classifications in the periodic table
Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom.
Orbital Energies and Atomic Structure
The energy of atomic orbitals increases as the principal quantum number, $n$, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of $l$ differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure $1$ depicts how these two trends in increasing energy relate. The 1s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2s and then 2p, 3s, and 3p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3d orbital is higher in energy than the 4s orbital. Such overlaps continue to occur frequently as we move up the chart.
Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5p orbitals fill immediately after the 4d, and immediately before the 6s. The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n, increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f. Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have +Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1s through 3p), the increase in energy due to n is more significant than the increase due to l; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order.
The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure $2$):
1. The number of the principal quantum shell, n,
2. The letter that designates the orbital type (the subshell, l), and
3. A superscript number that designates the number of electrons in that particular subshell.
For example, the notation 2p4 (read "two–p–four") indicates four electrons in a p subshell (l = 1) with a principal quantum number (n) of 2. The notation 3d8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., l = 2) of the principal shell for which n = 3.
The Aufbau Principle
To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure $3$), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure $3$ illustrates the traditional way to remember the filling order for atomic orbitals.
Since the arrangement of the periodic table is based on the electron configurations, Figure $4$ provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3d orbitals.
We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure $3$ or $4$, we would expect to find the electron in the 1s orbital. By convention, the $m_s=+\dfrac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are:
Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron (n = 1, l = 0, ml = 0, $m_s=+\dfrac{1}{2}$). The second electron also goes into the 1s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, $m_s=−\dfrac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are:
The n = 1 shell is completely filled in a helium atom.
The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure $3$ or $4$). Thus, the electron configuration and orbital diagram of lithium are:
An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital.
An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals (ml = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.
Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals have identical n, l, and ms quantum numbers and differ in their ml quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are:
Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are:
The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s22s22p63s1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called valence electrons, and those occupying the inner shell orbitals are called core electrons ( Figure \PageIndex5\PageIndex5). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s22s22p6) and our abbreviated or condensed configuration is [Ne]3s1.
Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.
$\ce{Li:[He]}\,2s^1\ \ce{Na:[Ne]}\,3s^1 \nonumber$
The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s2 configuration, is analogous to its family member beryllium, [He]2s2. Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s23p1, is analogous to its family member boron, [He]2s22p1.
The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure $6$ shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements.
When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3d level but is, instead, added to the 4s level (Figure $3$ or $4$). As discussed previously, the 3d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4s, which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4s1. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4s subshell and calcium has an electron configuration of [Ar]4s2. This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium.
Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [d orbitals], there are 2l + 1 = 5 values of ml, meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons (l = 3, 2l + 1 = 7 ml values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.
Example $1$: Quantum Numbers and Electron Configurations
What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?
Solution
The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons:
The last electron added is a 3p electron. Therefore, n = 3 and, for a p-type orbital, l = 1. The ml value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these ml values is correct. For unpaired electrons, convention assigns the value of $+\dfrac{1}{2}$ for the spin quantum number; thus, $m_s=+\dfrac{1}{2}$.
Exercise $1$
Identify the atoms from the electron configurations given:
1. [Ar]4s23d5
2. [Kr]5s24d105p6
Answer a
Mn
Answer b
Xe
The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure $3$ or $4$. For instance, the electron configurations of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.
In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4s into the 3d orbital to gain the extra stability of a half-filled 3d subshell (in Cr) or a filled 3d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5s24d3. Experimentally, we observe that its ground-state electron configuration is actually [Kr]5s14d4. We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5s orbital are larger than the gap in energy between the 5s and 4d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.
Electron Configurations and the Periodic Table
As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure $6$), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements.
Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react.
It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure $6$, which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure $6$ show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom.
1. Main group elements (sometimes called representative elements) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure $6$. This category includes all the nonmetallic elements, as well as many metals and the intermediate semimetallic elements. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar]4s23d104p1, which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons.
2. Transition elements or transition metals. These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and (n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure $6$) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure $6$), and we will adopt this usage in this textbook.
3. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure $6$. The valence shells of the inner transition elements consist of the (n – 2)f, the (n – 1)d, and the ns subshells. There are two inner transition series:
1. The lanthanide series: lanthanide (La) through lutetium (Lu)
2. The actinide series: actinide (Ac) through lawrencium (Lr)
Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons.
Electron Configurations of Ions
We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the (n – 1)d or (n – 2)f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle.
Example $2$: Predicting Electron Configurations of Ions
What is the electron configuration and orbital diagram of:
1. Na+
2. P3–
3. Al2+
4. Fe2+
5. Sm3+
Solution
First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.
Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.
1. Na: 1s22s22p63s1. Sodium cation loses one electron, so Na+: 1s22s22p63s1 = Na+: 1s22s22p6.
2. P: 1s22s22p63s23p3. Phosphorus trianion gains three electrons, so P3−: 1s22s22p63s23p6.
3. Al: 1s22s22p63s23p1. Aluminum dication loses two electrons Al2+: 1s22s22p63s23p1 = Al2+: 1s22s22p63s1.
4. Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 = 1s22s22p63s23p63d6.
5. Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6. Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm3+: 1s22s22p63s23p64s23d104p65s24d105p66s24f6 = 1s22s22p63s23p64s23d104p65s24d105p64f5.
Exercise $2$
1. Which ion with a +2 charge has the electron configuration 1s22s22p63s23p63d104s24p64d5?
2. Which ion with a +3 charge has this configuration?
Answer a
Tc2+
Answer b
Ru3+
Summary
The relative energy of the subshells determine the order in which atomic orbitals are filled (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals).
Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements (s and p orbitals), transition elements (d orbitals), and inner transition elements (f orbitals).
Glossary
Aufbau principle
procedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time
core electron
electron in an atom that occupies the orbitals of the inner shells
electron configuration
electronic structure of an atom in its ground state given as a listing of the orbitals occupied by the electrons
Hund’s rule
every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin
orbital diagram
pictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow
valence electrons
electrons in the outermost or valence shell (highest value of n) of a ground-state atom; determine how an element reacts
valence shell
outermost shell of electrons in a ground-state atom; for main group elements, the orbitals with the highest n level (s and p subshells) are in the valence shell, while for transition metals, the highest energy s and d subshells make up the valence shell and for inner transition elements, the highest s, d, and f subshells are included | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.03%3A_Electron_Configurations-_How_Electrons_Occupy_Orbitals.txt |
Learning Objectives
• Describe and explain the observed trends in atomic size, ionization energy, and electron affinity of the elements
The elements in groups (vertical columns) of the periodic table exhibit similar chemical behavior. This similarity occurs because the members of a group have the same number and distribution of electrons in their valence shells. However, there are also other patterns in chemical properties on the periodic table. For example, as we move down a group, the metallic character of the atoms increases. Oxygen, at the top of Group 16 (6A), is a colorless gas; in the middle of the group, selenium is a semiconducting solid; and, toward the bottom, polonium is a silver-grey solid that conducts electricity.
As we go across a period from left to right, we add a proton to the nucleus and an electron to the valence shell with each successive element. As we go down the elements in a group, the number of electrons in the valence shell remains constant, but the principal quantum number increases by one each time. An understanding of the electronic structure of the elements allows us to examine some of the properties that govern their chemical behavior. These properties vary periodically as the electronic structure of the elements changes. They are (1) size (radius) of atoms and ions, (2) ionization energies, and (3) electron affinities.
Variation in Covalent Radius
The quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, to determine their relative sizes that give roughly similar values. We will use the covalent radius (Figure $1$), which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity).
We know that as we scan down a group, the principal quantum number, n, increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its covalent radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the covalent radii of the halogens in Table $1$ and Figure $1$. The trends for the entire periodic table can be seen in Figure $2$.
Table $1$: Covalent Radii of the Halogen Group Elements
Atom Covalent radius (pm) Nuclear charge
F 64 +9
Cl 99 +17
Br 114 +35
I 133 +53
At 148 +85
As shown in Figure $2$, as we move across a period from left to right, we generally find that each element has a smaller covalent radius than the element preceding it. This might seem counterintuitive because it implies that atoms with more electrons have a smaller atomic radius. This can be explained with the concept of effective nuclear charge, $Z_{eff}$. This is the pull exerted on a specific electron by the nucleus, taking into account any electron–electron repulsions. For hydrogen, there is only one electron and so the nuclear charge (Z) and the effective nuclear charge (Zeff) are equal. For all other atoms, the inner electrons partially shield the outer electrons from the pull of the nucleus, and thus:
$Z_\ce{eff}=Z−shielding \nonumber$
Shielding is determined by the probability of another electron being between the electron of interest and the nucleus, as well as by the electron–electron repulsions the electron of interest encounters. Core electrons are adept at shielding, while electrons in the same valence shell do not block the nuclear attraction experienced by each other as efficiently. Thus, each time we move from one element to the next across a period, Z increases by one, but the shielding increases only slightly. Thus, Zeff increases as we move from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by electrons on the right side of the periodic table draws them closer to the nucleus, making the covalent radii smaller.
Thus, as we would expect, the outermost or valence electrons are easiest to remove because they have the highest energies, are shielded more, and are farthest from the nucleus. As a general rule, when the representative elements form cations, they do so by the loss of the ns or np electrons that were added last in the Aufbau process. The transition elements, on the other hand, lose the ns electrons before they begin to lose the (n – 1)d electrons, even though the ns electrons are added first, according to the Aufbau principle.
Example $1$: Sorting Atomic Radii
Predict the order of increasing covalent radius for Ge, Fl, Br, Kr.
Solution
Radius increases as we move down a group, so Ge < Fl (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so Kr < Br < Ge. Putting the trends together, we obtain Kr < Br < Ge < Fl.
Exercise $1$
Give an example of an atom whose size is smaller than fluorine.
Answer
Ne or He
Variation in Ionic Radii
Ionic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; it is smaller than the atom from which it is derived (Figure $3$). For example, the covalent radius of an aluminum atom (1s22s22p63s23p1) is 118 pm, whereas the ionic radius of an Al3+ (1s22s22p6) is 68 pm. As electrons are removed from the outer valence shell, the remaining core electrons occupying smaller shells experience a greater effective nuclear charge Zeff (as discussed) and are drawn even closer to the nucleus.
Cations with larger charges are smaller than cations with smaller charges (e.g., V2+ has an ionic radius of 79 pm, while that of V3+ is 64 pm). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, n.
An anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in $Z_{eff}$ per electron. Both effects (the increased number of electrons and the decreased Zeff) cause the radius of an anion to be larger than that of the parent atom ( Figure $3$). For example, a sulfur atom ([Ne]3s23p4) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3s23p6) is 170 pm. For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii.
Atoms and ions that have the same electron configuration are said to be isoelectronic. Examples of isoelectronic species are N3–, O2–, F, Ne, Na+, Mg2+, and Al3+ (1s22s22p6). Another isoelectronic series is P3–, S2–, Cl, Ar, K+, Ca2+, and Sc3+ ([Ne]3s23p6). For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms.
Variation in Ionization Energies
The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy (IE1). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge:
$\ce{X}(g)⟶\ce{X+}(g)+\ce{e-}\hspace{20px}\ce{IE_1} \nonumber$
The energy required to remove the second most loosely bound electron is called the second ionization energy (IE2).
$\ce{X+}(g)⟶\ce{X^2+}(g)+\ce{e-}\hspace{20px}\ce{IE_2} \nonumber$
The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period.
Figure $4$ graphs the relationship between the first ionization energy and the atomic number of several elements. Within a period, the values of first ionization energy for the elements (IE1) generally increases with increasing Z. Down a group, the IE1 value generally decreases with increasing Z. There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2s2) is an s electron, whereas the electron removed during the ionization of boron ([He]2s22p1) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins.
Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure $4$.
Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table $2$, there is a large increase in the ionization energies (color change) for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.
Table $2$: Successive Ionization Energies for Selected Elements (kJ/mol)
Element IE1 IE2 IE3 IE4 IE5 IE6 IE7
K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343
Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9
Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0
Ga 578.8 1979.4 2964.6 6180 8298.7 10873.9 13594.8
Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available
As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available
Example $2$: Ranking Ionization Energies
Predict the order of increasing energy for the following processes: IE1 for Al, IE1 for Tl, IE2 for Na, IE3 for Al.
Solution
Removing the 6p1 electron from Tl is easier than removing the 3p1 electron from Al because the higher n orbital is farther from the nucleus, so IE1(Tl) < IE1(Al). Ionizing the third electron from
$\ce{Al}\hspace{20px}\ce{(Al^2+⟶Al^3+ + e- )} \nonumber$
requires more energy because the cation Al2+ exerts a stronger pull on the electron than the neutral Al atom, so IE1(Al) < IE3(Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain:
IE1(Tl) < IE1(Al) < IE3(Al) < IE2(Na).
Exercise $2$
Which has the lowest value for IE1: O, Po, Pb, or Ba?
Answer
Ba
Variation in Electron Affinities
The electron affinity [EA] is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion).
$\ce{X}(g)+\ce{e-}⟶\ce{X-}(g)\hspace{20px}\ce{EA_1} \nonumber$
This process can be either endothermic or exothermic, depending on the element. The EA of some of the elements is given in Figure $6$. You can see that many of these elements have negative values of EA, which means that energy is released when the gaseous atom accepts an electron. However, for some elements, energy is required for the atom to become negatively charged and the value of their EA is positive. Just as with ionization energy, subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a –2 ion, and so on.
As we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 (5A), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher n level, which is more difficult to do. Group 2 (2A) has a filled ns subshell, and so the next electron added goes into the higher energy np, so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled np subshell and the next electron must be paired with an existing np electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA.
We also might expect the atom at the top of each group to have the largest EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the greatest EA. The reduction of the EA of the first member can be attributed to the small size of the n = 2 shell and the resulting large electron–electron repulsions. For example, chlorine, with an EA value of –348 kJ/mol, has the highest value of any element in the periodic table. The EA of fluorine is –322 kJ/mol. When we add an electron to a fluorine atom to form a fluoride anion (F), we add an electron to the n = 2 shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the n = 3 shell, it occupies a considerably larger region of space and the electron–electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily.
The properties discussed in this section (size of atoms and ions, effective nuclear charge, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus.
Summary
Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller, because the number of valence electrons has changed while the nuclear charge has remained constant. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Electron affinity (the energy associated with forming an anion) is more favorable (exothermic) when electrons are placed into lower energy orbitals, closer to the nucleus. Therefore, electron affinity becomes increasingly negative as we move left to right across the periodic table and decreases as we move down a group. For both IE and electron affinity data, there are exceptions to the trends when dealing with completely filled or half-filled subshells.
Glossary
covalent radius
one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond
effective nuclear charge
charge that leads to the Coulomb force exerted by the nucleus on an electron, calculated as the nuclear charge minus shielding
electron affinity
energy required to add an electron to a gaseous atom to form an anion
ionization energy
energy required to remove an electron from a gaseous atom or ion. The associated number (e.g., second ionization energy) corresponds to the charge of the ion produced (X2+)
isoelectronic
group of ions or atoms that have identical electron configurations | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.04%3A_Electron_Configurations_Valence_Electrons_and_the_Periodic_Table.txt |
Learning Objectives
• To understand periodic trends in atomic radii.
• To predict relative ionic sizes within an isoelectronic series.
Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes; however, some atoms are larger or smaller than others, and this influences their chemistry. In this section, we discuss how atomic and ion “sizes” are defined and obtained.
Atomic Radii
Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure $1$ which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom.
Figure $1$ also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell (n = 1), it shows only a single peak. In contrast, neon, with filled n = 1 and 2 principal shells, has two peaks. Argon, with filled n = 1, 2, and 3 principal shells, has three peaks. The peak for the filled n = 1 shell occurs at successively shorter distances for neon (Z = 10) and argon (Z = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1s2 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of n. Consequently, the two electrons in the n = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the n = 1 shell also decreases tremendously (the filled 1s orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled n = 2 shell in argon is located closer to the nucleus and has a lower energy than the n = 2 shell in neon.
Figure $1$ illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl2 molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius ($r_{cov}$), which is half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule, of 99 pm or 0.99 Å (Figure $\PageIndex{2a}$). Atomic radii are often measured in angstroms (Å), a non-SI unit: 1 Å = 1 × 10−10 m = 100 pm.
In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. A similar approach for measuring the size of ions is discussed later in this section.
Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius ($r_{met}$) is defined as half the distance between the nuclei of two adjacent metal atoms in the solid (Figure $\PageIndex{2b}$). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius ($r_{vdW}$), which is half the internuclear distance between two nonbonded atoms in the solid (Figure $\PageIndex{2c}$). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a $\ce{Cl2}$ molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, $\ce{Cl2(s)}$ at low temperatures). These radii are generally not the same (Figure $\PageIndex{2d}$).
Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure $3$).
In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure $4$).
Trends in atomic size result from differences in the effective nuclear charges ($Z_{eff}$) experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always less than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius.
Atomic radii decrease from left to right across a row and increase from top to bottom down a column.
The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1s2 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2s and 2p orbitals, electrons in the same principal shell are not very effective at shielding one another from the nuclear charge. Thus the single 2s electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1s2 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of Zeff = +1.26 for Li.) In contrast, the two 2s electrons in beryllium do not shield each other very well, although the filled 1s2 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2s electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2s and 2p orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size (Figure $5$).
The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number n is not constant. As we saw in Chapter 2, the size of the orbitals increases as n increases, provided the nuclear charge remains the same. In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to ns orbitals with increasing values of n. However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number n increases from 2 to 6, but the nuclear charge increases from +3 to +55!
As a consequence the radii of the lower electron orbitals in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6s1 valence electron configuration, is much larger than lithium, with a 2s1 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1s22s22p63s23p64s23d104p65s24d105p6 shells, abbreviated as [Xe]5s24d105p6, which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge.
Not all Electrons shield equally
Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge.
Example $1$
On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon.
Given: three elements
Asked for: arrange in order of increasing atomic radius
Strategy:
1. Identify the location of the elements in the periodic table. Determine the relative sizes of elements located in the same column from their principal quantum number n. Then determine the order of elements in the same row from their effective nuclear charges. If the elements are not in the same column or row, use pairwise comparisons.
2. List the elements in order of increasing atomic radius.
Solution:
A These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater.
B Combining the two inequalities gives the overall order: C < Si < Al.
Exercise $1$
On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur.
Answer
O < S < P < K
Atomic Radius: Atomic Radius, YouTube(opens in new window) [youtu.be]
Ionic Radii and Isoelectronic Series
An ion is formed when either one or more electrons are removed from a neutral atom to form a positive ion (cation) or when additional electrons attach themselves to neutral atoms to form a negative one (anion). The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode.
Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it is possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure $6$, the internuclear distance corresponds to the sum of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na+ ion is essentially the same in NaCl and Na2S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed.
A comparison of ionic radii with atomic radii (Figure $7$) shows that a cation, having lost an electron, is always smaller than its parent neutral atom, and an anion, having gained an electron, is always larger than the parent neutral atom. When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases and the ion shrinks (compare Li at 167 pm with Li+ at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe2+ at 78 pm with Fe3+ at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases and the ion expands (compare F at 42 pm with F at 133 pm).
Cations are always smaller than the neutral atom and anions are always larger.
Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na ion, allowing comparison of its size with that of the far more familiar Na+ ion, which is found in many compounds. The radius of sodium in each of its three known oxidation states is given in Table $1$. All three species have a nuclear charge of +11, but they contain 10 (Na+), 11 (Na0), and 12 (Na) electrons. The Na+ ion is significantly smaller than the neutral Na atom because the 3s1 electron has been removed to give a closed shell with n = 2. The Na ion is larger than the parent Na atom because the additional electron produces a 3s2 valence electron configuration, while the nuclear charge remains the same.
Table $1$: Experimentally Measured Values for the Radius of Sodium in Its Three Known Oxidation States
Na+ Na0 Na
Electron Configuration 1s22s22p6 1s22s22p63s1 1s22s22p63s2
Radius (pm) 102 154* 202
*The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253.
Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of n lie at successively greater distances from the nucleus.
Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic series. For example, the isoelectronic series of species with the neon closed-shell configuration (1s22s22p6) is shown in Table $3$.
The sizes of the ions in this series decrease smoothly from N3− to Al3+. All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al3+) is the smallest, and the ion with the smallest nuclear charge (N3−) is the largest. The neon atom in this isoelectronic series is not listed in Table $3$, because neon forms no covalent or ionic compounds and hence its radius is difficult to measure.
Ion Radius (pm) Atomic Number
Table $3$: Radius of Ions with the Neon Closed-Shell Electron Configuration. Source: R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.
N3− 146 7
O2− 140 8
F 133 9
Na+ 98 11
Mg2+ 79 12
Al3+ 57 13
Example $2$
Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl, K+, S2−, and Se2.
Given: four ions
Asked for: order by increasing radius
Strategy:
1. Determine which ions form an isoelectronic series. Of those ions, predict their relative sizes based on their nuclear charges. For ions that do not form an isoelectronic series, locate their positions in the periodic table.
2. Determine the relative sizes of the ions based on their principal quantum numbers n and their locations within a row.
Solution:
A We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K+, Cl, and S2− form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K+ has the greatest nuclear charge (Z = 19), its radius is smallest, and S2− with Z = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se2 ion to be even larger than S2−.
B The order must therefore be K+ < Cl < S2− < Se2.
Exercise $2$
Based on their positions in the periodic table, arrange these ions in order of increasing size: Br, Ca2+, Rb+, and Sr2+.
Answer
Ca2+ < Sr2+ < Rb+ < Br
Summary
Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius (rcov) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius (rmet) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The van der Waals radius (rvdW) of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The ionic radii of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size.
Contributors and Attributions
Learning Objectives
• To understand the basics of electron shielding and penetration
For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of both electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies. These effects are the underlying basis for the periodic trends in elemental properties that we will explore in this chapter.
Electron Shielding and Effective Nuclear Charge
If an electron is far from the nucleus (i.e., if the distance $r$ between the nucleus and the electron is large), then at any given moment, many of the other electrons will be between that electron and the nucleus (Figure $1$). Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge ($Z_{eff}$) that is less than the actual nuclear charge $Z$. This effect is called electron shielding.
As the distance between an electron and the nucleus approaches infinity, $Z_{eff}$ approaches a value of 1 because all the other ($Z − 1$) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At $r ≈ 0$, the positive charge experienced by an electron is approximately the full nuclear charge, or $Z_{eff} ≈ Z$. At intermediate values of $r$, the effective nuclear charge is somewhere between 1 and $Z$:
$1 ≤ Z_{eff} ≤ Z. \nonumber$
Notice that $Z_{eff} = Z$ only for hydrogen (Figure $2$).
Definition: Shielding
Shielding refers to the core electrons repelling the outer electrons, which lowers the effective charge of the nucleus on the outer electrons. Hence, the nucleus has "less grip" on the outer electrons insofar as it is shielded from them.
$Z_{eff}$ can be calculated by subtracting the magnitude of shielding from the total nuclear charge and the effective nuclear charge of an atom is given by the equation:
$Z_{eff}=Z-S \label{4}$
where $Z$ is the atomic number (number of protons in nucleus) and $S$ is the shielding constant and is approximated by number of electrons between the nucleus and the electron in question (the number of nonvalence electrons). The value of $Z_{eff}$ will provide information on how much of a charge an electron actually experiences.
We can see from Equation \ref{4} that the effective nuclear charge of an atom increases as the number of protons in an atom increases (Figure $2$). As we will discuss later on in the chapter, this phenomenon can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge.
The shielding constant can be estimated by totaling the screening by all nonvalence electrons ($n$) except the one in question.
$S = \sum_{i}^{n-1} S_i \label{2.6.0}$
where $S_i$ is the shielding of the ith electron.
Electrons that are shielded from the full charge of the nucleus experience an effective nuclear charge ($Z_{eff}$) of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ion.
From Equations \ref{4} and \ref{2.6.0}, $Z_{eff}$ for a specific electron can be estimated is the shielding constants for that electron of all other electrons in species is known. A simple approximation is that all other non-valence electrons shield equally and fully:
$S_i=1 \label{simple}$
This crude approximation is demonstrated in Example $1$.
Example $1$: Fluorine, Neon, and Sodium
What is the effective attraction $Z_{eff}$ experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation?
Solution
Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence), but the effective nuclear charge varies because each has a different atomic number $A$. This is an application of Equations \ref{4} and \ref{2.6.0}. We use the simple assumption that all electrons shield equally and fully the valence electrons (Equation \ref{simple}).
The charge $Z$ of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals.
• $Z_\mathrm{eff}(\mathrm{F}^-) = 9 - 2 = 7+$
• $Z_\mathrm{eff}(\mathrm{Ne}) = 10 - 2 = 8+$
• $Z_\mathrm{eff}(\mathrm{Na}^+) = 11 - 2 = 9+$
So the sodium cation has the greatest effective nuclear charge. This also suggests that $\mathrm{Na}^+$ has the smallest radius of these species and that is correct.
Exercise $1$: Magnesium Species
What is the effective attraction $Z_{eff}$ experienced by the valence electrons in the magnesium anion, the neutral magnesium atom, and magnesium cation? Use the simple approximation for shielding constants. Compare your result for the magnesium atom to the more accurate value in Figure $2$ and proposed an origin for the difference.
Answer
• $Z_\mathrm{eff}(\ce{Mg}^{-}) = 12 - 10 = 2+$
• $Z_\mathrm{eff}(\ce{Mg}) = 12 - 10 = 2+$
• $Z_\mathrm{eff}(\ce{Mg}^{+}) = 12 - 10 = 2+$
Remember that the simple approximations in Equations \ref{2.6.0} and \ref{simple} suggest that valence electrons do not shield other valence electrons. Therefore, each of these species has the same number of non-valence electrons and Equation \ref{4} suggests the effective charge on each valence electron is identical for each of the three species.
This is not correct and a more complex model is needed to predict the experimental observed $Z_{eff}$ value. The ability of valence electrons to shield other valence electrons or in partial amounts (e.g., $S_i \neq 1$) is in violation of Equations \ref{2.6.0} and \ref{simple}. That fact that these approximations are poor is suggested by the experimental $Z_{eff}$ value shown in Figure $2$ for $\ce{Mg}$ of 3.2+. This is appreciably larger than the +2 estimated above, which means these simple approximations overestimate the total shielding constant $S$. A more sophisticated model is needed.
Electron Penetration
The approximation in Equation \ref{simple} is a good first order description of electron shielding, but the actual $Z_{eff}$ experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in $Z_{eff}$ for different elements, as shown in Figure $2$ for the elements of the first three rows of the periodic table.
Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom (Figure $3$). Electrons in different orbitals have different electron densities around the nucleus. In other words, penetration depends on the shell ($n$) and subshell ($l$).
For example, a 1s electron (Figure $3$; purple curve) has greater electron density near the nucleus than a 2p electron (Figure $3$; red curve) and has a greater penetration. This related to the shielding constants since the 1s electrons are closer to the nucleus than a 2p electron, hence the 1s screens a 2p electron almost perfectly ($S=1$. However, the 2s electron has a lower shielding constant ($S<1$ because it can penetrate close to the nucleus in the small area of electron density within the first spherical node (Figure $3$; green curve). In this way the 2s electron can "avoid" some of the shielding effect of the inner 1s electron.
For the same shell value ($n$) the penetrating power of an electron follows this trend in subshells (Figure $3$):
$s > p > d \approx f. \label{better1}$
for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:
$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...} \label{better2}$
Definition: Penetration
Penetration describes the proximity of electrons in an orbital to the nucleus. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity.
Table $1$: Effective Nuclear Charges for Selected Atoms
Atom Sublevel Z Zeff
H 1s 1 1
He 1s 2 1.69
Li 1s, 2s 3 2.69, 1.28
Be 1s, 2s 4 3.68, 1.91
B 1s, 2s, 2p 5 4.68, 2.58, 2.42
F 1s, 2s, 2p 9 8.65, 5.13, 5.10
Na 1s, 2s, 2p, 3s 11 10.63, 6.57, 6.80, 2.51
Data from E. Clementi and D. L. Raimondi; The Journal of Chemical Physics 38, 2686 (1963).
Because of the effects of shielding and the different radial distributions of orbitals with the same value of n but different values of l, the different subshells are not degenerate in a multielectron atom. For a given value of n, the ns orbital is always lower in energy than the np orbitals, which are lower in energy than the nd orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of n; for example, the 4s orbital is lower in energy than the 3d orbitals for most atoms.
A Better Estimation of Shielding: Slater Rules
The concepts of electron shielding, orbital penetration and effective nuclear charge were introduced above, but we did so in a qualitative manner (e.g., Equations \ref{better1} and \ref{better2}). A more accurate model for estimating electron shielding and corresponding effective nuclear charge experienced is Slater's Rules. However, the application of these rules is outside the scope of this text.
Zeff and Electron Shielding: Zeff and Electron Shielding(opens in new window) [youtu.be]
Summary
The calculation of orbital energies in atoms or ions with more than one electron (multielectron atoms or ions) is complicated by repulsive interactions between the electrons. The concept of electron shielding, in which intervening electrons act to reduce the positive nuclear charge experienced by an electron, allows the use of hydrogen-like orbitals and an effective nuclear charge ($Z_{eff}$) to describe electron distributions in more complex atoms or ions. The degree to which orbitals with different values of l and the same value of n overlap or penetrate filled inner shells results in slightly different energies for different subshells in the same principal shell in most atoms. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.06%3A_Periodic_Trends_in_the_Size_of_Atoms_and_Effective_Nuclear_Charge.txt |
Learning Objectives
• To correlate ionization energies with the chemistry of the elements
We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. Why is this so? In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation.
Ionization Energies
Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy ($I$) of an element as the amount of energy needed to remove an electron from the gaseous atom $E$ in its ground state. $I$ is therefore the energy required for the reaction
$E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ \text{energy required=I } \label{7.4.1}$
Because an input of energy is required, the ionization energy is always positive ($I > 0$) for the reaction as written in Equation $1$. Larger values of I mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV):
$1\; eV/atom = 96.49\; kJ/mol \nonumber$
If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy ($I_1$), a second ionization energy ($I_2$), and in general an nth ionization energy ($I_n$) according to the following reactions:
$\ce{E(g) \rightarrow E^+(g) +e^-} \;\;\ I_1=\text{1st ionization energy} \label{7.4.2}$
$\ce{E^{+}(g) \rightarrow E^{2+}(g) +e^-} \;\;\ I_2=\text{2nd ionization energy} \label{7.4.3}$
$\ce{E^{2+}(g) \rightarrow E^{3+}(g) +e^-} \;\;\ I_3=\text{3rd ionization energy} \label{7.4.4}$
Values for the ionization energies of $Li$ and $Be$ listed in Table $1$ show that successive ionization energies for an element increase as they go; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger.
Successive ionization energies for an element increase.
Table $1$: Ionization Energies (in kJ/mol) for Removing Successive Electrons from Li and Be. Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction Electronic Transition $I$ Reaction Electronic Transition $I$
$\ce{Li (g)\rightarrow Li^+ (g) + e^-}$ $1s^22s^1 \rightarrow 1s^2$ I1 = 520.2 $\ce{Be (g) \rightarrow Be^+(g) + e^-}$ $1s^22s^2 \rightarrow 1s^22s^1$ I1 = 899.5
$\ce{Li^+(g) \rightarrow Li^{2+}(g) +e^-}$ $1s^2 \rightarrow 1s^1$ I2 = 7298.2 $\ce{Be^+(g) \rightarrow Be^{2+}(g) + e^-}$ $1s^22s^1 \rightarrow 1s^2$ I2 = 1757.1
$\ce{Li^{2+} (g) \rightarrow Li^{3+}(g) + e^-}$ $1s^1 \rightarrow 1s^0$ I3 = 11,815.0 $\ce{Be^{2+}(g) \rightarrow Be^{3+}(g) + e^-}$ $1s^2 \rightarrow 1s^1$ I3 = 14,848.8
$\ce{Be^{3+}(g) \rightarrow Be^{4+}(g) + e^-}$ $1s^1 \rightarrow 1s^0$ I4 = 21,006.6
The increase in successive ionization energies, however, is not linear, but increases drastically when removing electrons in lower $n$ orbitals closer to the nucleus. The most important consequence of the values listed in Table $1$ is that the chemistry of $\ce{Li}$ is dominated by the $\ce{Li^+}$ ion, while the chemistry of $\ce{Be}$ is dominated by the +2 oxidation state. The energy required to remove the second electron from $\ce{Li}$:
$\ce{Li^+(g) \rightarrow Li^{2+}(g) + e^-} \label{7.4.5}$
is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the third electron from $\ce{Be}$:
$\ce{Be^{2+}(g) \rightarrow Be^{3+}(g) + e^-} \label{7.4.6}$
is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both $\ce{Li^+}$ and $\ce{Be^{2+}}$ have 1s2 closed-shell configurations, and much more energy is required to remove an electron from the 1s2 core than from the 2s valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large and simply cannot be achieved in normal chemical reactions.
The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions.
Ionization Energy: Ionization Energy, YouTube(opens in new window) [youtu.be] (opens in new window)
Ionization Energies of s- and p-Block Elements
Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of $Li$ and $Be$ (Table $2$): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3s or 3p, in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in Table $2$. Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from $Al$, the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons.
Table $2$: Successive Ionization Energies (in kJ/mol) for the Elements in the Third Row of the Periodic Table.Source: Data from CRC Handbook of Chemistry and Physics (2004).
Element $I_1$ $I_2$ $I_3$ $I_4$ $I_5$ $I_6$ $I_7$
*Inner-shell electron
Na 495.8 4562.4*
Mg 737.7 1450.7 7732.7
Al 577.4.4 1816.7 2744.8 11,577.4.4
Si 786.5 1577.1 3231.6 4355.5 16,090.6
P 1011.8 1907.4.4 2914.1 4963.6 6274.0 21,267.4.3
S 999.6 2251.8 3357 4556.2 7004.3 8495.8 27,107.4.3
Cl 1251.2 2297.7 3822 5158.6 6540 9362 11,018.2
Ar 1520.6 2665.9 3931 5771 7238 8781.0 11,995.3
Example $1$: Highest Fourth Ionization Energy
From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N.
Given: three elements
Asked for: element with highest fourth ionization energy
Strategy:
1. List the electron configuration of each element.
2. Determine whether electrons are being removed from a filled or partially filled valence shell. Predict which element has the highest fourth ionization energy, recognizing that the highest energy corresponds to the removal of electrons from a filled electron core.
Solution:
A These elements all lie in the second row of the periodic table and have the following electron configurations:
• B: [He]2s22p1
• C: [He]2s22p2
• N: [He]2s22p3
B The fourth ionization energy of an element ($I_4$) is defined as the energy required to remove the fourth electron:
$E^{3+}_{(g)} \rightarrow E^{4+}_{(g)} + e^- \nonumber$
Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1s2 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol.
Exercise $1$: Lowest Second Ionization Energy
From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar.
Answer
$\ce{Sr}$
The first column of data in Table $2$ shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons.
However, the first ionization energy decreases at Al ([Ne]3s23p1) and at S ([Ne]3s23p4). The electron configurations of these "exceptions" provide the answer why. The electrons in aluminum’s filled 3s2 subshell are better at screening the 3p1 electron than they are at screening each other from the nuclear charge, so the s electrons penetrate closer to the nucleus than the p electron does and the p electron is more easily removed. The decrease at S occurs because the two electrons in the same p orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements.
The first ionization energies of the elements in the first six rows of the periodic table are plotted in Figure $1$ and are presented numerically and graphically in Figure $2$. These figures illustrate three important trends:
1. The changes seen in the second (Li to Ne), fourth (K to Kr), fifth (Rb to Xe), and sixth (Cs to Rn) rows of the s and p blocks follow a pattern similar to the pattern described for the third row of the periodic table. The transition metals are included in the fourth, fifth, and sixth rows, however, and the lanthanides are included in the sixth row. The first ionization energies of the transition metals are somewhat similar to one another, as are those of the lanthanides. Ionization energies increase from left to right across each row, with discrepancies occurring at ns2np1 (group 13), ns2np4 (group 16), and ns2(n − 1)d10 (group 12).
2. First ionization energies generally decrease down a column. Although the principal quantum number n increases down a column, filled inner shells are effective at screening the valence electrons, so there is a relatively small increase in the effective nuclear charge. Consequently, the atoms become larger as they acquire electrons. Valence electrons that are farther from the nucleus are less tightly bound, making them easier to remove, which causes ionization energies to decrease. A larger radius typically corresponds to a lower ionization energy.
3. Because of the first two trends, the elements that form positive ions most easily (have the lowest ionization energies) lie in the lower left corner of the periodic table, whereas those that are hardest to ionize lie in the upper right corner of the periodic table. Consequently, ionization energies generally increase diagonally from lower left (Cs) to upper right (He).
Generally, $I_1$ increases diagonally from the lower left of the periodic table to the upper right.
Gallium (Ga), which is the first element following the first row of transition metals, has the following electron configuration: [Ar]4s23d104p1. Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3d10 subshell of gallium lies inside the 4p subshell, shielding the single 4p electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4s2 orbital, not the 3d10 subshell. The chemistry of gallium is dominated by the resulting Ga3+ ion, with its [Ar]3d10 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier p-block elements. They are sometimes referred to as pseudo noble gas configurations. In fact, for elements that exhibit these configurations, no chemical compounds are known in which electrons are removed from the (n − 1)d10 filled subshell.
Ionization Energies of Transition Metals & Lanthanides
As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the s- and p-block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the ns electrons before the (n − 1)d or (n − 2)f electrons, respectively. This means that transition metal cations have (n − 1)dn valence electron configurations, and lanthanide cations have (n − 2)fn valence electron configurations. Because the (n − 1)d and (n − 2)f shells are closer to the nucleus than the ns shell, the (n − 1)d and (n − 2)f electrons screen the ns electrons quite effectively, reducing the effective nuclear charge felt by the ns electrons. As Z increases, the increasing positive charge is largely canceled by the electrons added to the (n − 1)d or (n − 2)f orbitals.
That the ns electrons are removed before the (n − 1)d or (n − 2)f electrons may surprise you because the orbitals were filled in the reverse order. In fact, the ns, the (n − 1)d, and the (n − 2)f orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the d orbitals are filled, the effective nuclear charge causes the 3d orbitals to be slightly lower in energy than the 4s orbitals. The [Ar]3d2 electron configuration of Ti2+ tells us that the 4s electrons of titanium are lost before the 3d electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an (n − 2)fn valence electron configuration.
Because their first, second, and third ionization energies change so little across a row, these elements have important horizontal similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M2+ ions, whereas the lanthanides primarily form compounds in which they exist as M3+ ions.
Example $2$: Lowest First Ionization Energy
Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr.
Given: six elements
Asked for: element with lowest first ionization energy
Strategy:
Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy.
Solution:
These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb.
Exercise $2$: Highest First Ionization Energy
Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn.
Answer
$\ce{As}$
Summary
The tendency of an element to lose electrons is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for ionization energy (I), the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called pseudo noble gas configurations, in either the parent atom or the resulting ion. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.07%3A_Ions-_Configurations_Magnetic_Properties_Radii_and_Ionization_Energy.txt |
Learning Objectives
• To master the concept of electron affinity as a measure of the energy required to add an electron to an atom or ion.
• To recognize the inverse relationship of ionization energies and electron affinities
The electron affinity ($EA$) of an element $E$ is defined as the energy change that occurs when an electron is added to a gaseous atom or ion:
$E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released (Figure $1$).
The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
$\ce{ Cl(g) + e^- \rightarrow Cl^- (g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$\ce{ Be(g) + e^- \rightarrow Be^- (g)} \;\;\; EA \ge 0 \label{7.5.3}$
Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy:
$\ce{ N(g) + e^- \rightarrow N^- (g)} \;\;\; EA \approx 0 \label{7.5.4}$
Generally, electron affinities become more negative across a row of the periodic table.
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as n increases, the extra electrons enter orbitals that are increasingly far from the nucleus.
Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend:
1. The electron affinities of elements B through F in the second row of the periodic table are less negative than those of the elements immediately below them in the third row. Apparently, the increased electron–electron repulsions experienced by electrons confined to the relatively small 2p orbitals overcome the increased electron–nucleus attraction at short nuclear distances. Fluorine, therefore, has a lower affinity for an added electron than does chlorine. Consequently, the elements of the third row (n = 3) have the most negative electron affinities. Farther down a column, the attraction for an added electron decreases because the electron is entering an orbital more distant from the nucleus. Electron–electron repulsions also decrease because the valence electrons occupy a greater volume of space. These effects tend to cancel one another, so the changes in electron affinity within a family are much smaller than the changes in ionization energy.
2. The electron affinities of the alkaline earth metals become more negative from Be to Ba. The energy separation between the filled ns2 and the empty np subshells decreases with increasing n, so that formation of an anion from the heavier elements becomes energetically more favorable.
The equations for second and higher electron affinities are analogous to those for second and higher ionization energies:
$E_{(g)} + e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA_1 \label{7.5.5}$
$E^-_{(g)} + e^- \rightarrow E^{2-}_{(g)} \;\;\; \text{energy change=}EA_2 \label{7.5.6}$
As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol:
$O_{(g)} + e^- \rightarrow O^-_{(g)} \;\;\; EA_1=-141 \;kJ/mol \label{7.5.7}$
$O^-_{(g)} + e^- \rightarrow O^{2-}_{(g)} \;\;\; EA_2=+744 \;kJ/mol \label{7.5.8}$
Thus the formation of a gaseous oxide ($O^{2−}$) ion is energetically quite unfavorable (estimated by adding both steps):
$O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)} \;\;\; EA=+603 \;kJ/mol \label{7.5.9}$
Similarly, the formation of all common dianions (such as $S^{2−}$) or trianions (such as $P^{3−}$) is energetically unfavorable in the gas phase.
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
Electron Affinity: Electron Affinity, YouTube(opens in new window) [youtu.be] (opens in new window)
If energy is required to form both positively charged cations and monatomic polyanions, why do ionic compounds such as $MgO$, $Na_2S$, and $Na_3P$ form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions in the crystalline salt.
Example $1$: Contrasting Electron Affinities of Sb, Se, and Te
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
1. Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
2. Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
A We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise $1$: Contrasting Electron Affinities of Rb, Sr, and Xe
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
Answer
Rb
Summary
The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/04%3A_Periodic_Properties_of_the_Elements/4.08%3A_Electron_Affinities_and_Metallic_Character.txt |
• 5.1: Hydrogen, Oxygen, and Water
• 5.2: Chemical Bonds
Ionic vs. Covalent vs. Metallic bonding.
• 5.3: Representing Compounds - Chemical Formulas and Molecular Models
• 5.4: Ionic Compounds- Formulas and Names
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K2O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge.
• 5.5: Covalent Bonding- Simple Lewis Structures
The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest.
• 5.6: The Lewis Model - Representing Valance Electrons with Dots
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of 8 valence electrons in their compounds.
• 5.7: Molecular Compounds- Formulas and Names
Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include SF6, sulfur hexafluoride, and N2O4, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the –ide suffix to –ic, and adding “acid;” HCl is hydrochloric acid.
• 5.8: Composition of Compounds
Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the molecular formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the molecular formula (other techniques can though).
• 5.9: Determining a Chemical Formula from Experimental Data
In this section, we will explore how to derive the chemical formulas of unknown substances from experimental mass measurements.
• 5.10: Formula Mass and the Mole Concept for Compounds
• 5.11: Organic Compounds
Organic chemistry is the study of carbon compounds, nearly all of which also contain hydrogen atoms.
05: Molecules and Compounds
Learning Objectives
• To quantitatively describe the energetic factors involved in the formation of an ionic bond.
Chemical bonds form when electrons can be simultaneously close to two or more nuclei, but beyond this, there is no simple, easily understood theory that would not only explain why atoms bind together to form molecules, but would also predict the three-dimensional structures of the resulting compounds as well as the energies and other properties of the bonds themselves. Unfortunately, no one theory exists that accomplishes these goals in a satisfactory way for all of the many categories of compounds that are known. Moreover, it seems likely that if such a theory does ever come into being, it will be far from simple.
When we are faced with a scientific problem of this complexity, experience has shown that it is often more useful to concentrate instead on developing models. A scientific model is something like a theory in that it should be able to explain observed phenomena and to make useful predictions. But whereas a theory can be discredited by a single contradictory case, a model can be useful even if it does not encompass all instances of the phenomena it attempts to explain. We do not even require that a model be a credible representation of reality; all we ask is that be able to explain the behavior of those cases to which it is applicable in terms that are consistent with the model itself. An example of a model that you may already know about is the kinetic molecular theory of gases. Despite its name, this is really a model (at least at the level that beginning students use it) because it does not even try to explain the observed behavior of real gases. Nevertheless, it serves as a tool for developing our understanding of gases, and as a starting point for more elaborate treatments.Given the extraordinary variety of ways in which atoms combine into aggregates, it should come as no surprise that a number of useful bonding models have been developed. Most of them apply only to certain classes of compounds, or attempt to explain only a restricted range of phenomena. In this section we will provide brief descriptions of some of the bonding models; the more important of these will be treated in much more detail in later parts of this chapter.
Ionic Bonding
Ions are atoms or molecules which are electrically charged. Cations are positively charged and anions carry a negative charge. Ions form when atoms gain or lose electrons. Since electrons are negatively charged, an atom that loses one or more electrons will become positively charged; an atom that gains one or more electrons becomes negatively charged. Ionic bonding is the attraction between positively- and negatively-charged ions. These oppositely charged ions attract each other to form ionic networks (or lattices). Electrostatics explains why this happens: opposite charges attract and like charges repel. When many ions attract each other, they form large, ordered, crystal lattices in which each ion is surrounded by ions of the opposite charge. Generally, when metals react with non-metals, electrons are transferred from the metals to the non-metals. The metals form positively-charged ions and the non-metals form negatively-charged ions.
Ionic bonds form when metals and non-metals chemically react. By definition, a metal is relatively stable if it loses electrons to form a complete valence shell and becomes positively charged. Likewise, a non-metal becomes stable by gaining electrons to complete its valence shell and become negatively charged. When metals and non-metals react, the metals lose electrons by transferring them to the non-metals, which gain them. Consequently, ions are formed, which instantly attract each other—ionic bonding.
Example $1$: Sodium Chloride
For example, in the reaction of Na (sodium) and Cl (chlorine), each Cl atom takes one electron from a Na atom. Therefore each Na becomes a Na+ cation and each Cl atom becomes a Cl- anion. Due to their opposite charges, they attract each other to form an ionic lattice. The formula (ratio of positive to negative ions) in the lattice is NaCl.
$2Na_{(s)} + Cl_{2(g)} \rightarrow 2NaCl_{(s)}$
These ions are arranged in solid NaCl in a regular three-dimensional arrangement (or lattice):
NaCl lattice. (left) 3-D structure and (right) simple 2D slice through lattes. Images used with permission from Wikipedia and Mike Blaber.
The chlorine has a high affinity for electrons, and the sodium has a low ionization potential. Thus the chlorine gains an electron from the sodium atom. This can be represented using electron-dot symbols (here we will consider one chlorine atom, rather than Cl2):
The arrow indicates the transfer of the electron from sodium to chlorine to form the Na+ metal ion and the Cl- chloride ion. Each ion now has an octet of electrons in its valence shell:
• Na+: 2s22p6
• Cl-: 3s23p6
Covalent Bonding
Formation of an ionic bond by complete transfer of an electron from one atom to another is possible only for a fairly restricted set of elements. Covalent bonding, in which neither atom loses complete control over its valence electrons, is much more common. In a covalent bond the electrons occupy a region of space between the two nuclei and are said to be shared by them. This model originated with the theory developed by G.N. Lewis in 1916, and it remains the most widely-used model of chemical bonding. The essential element s of this model can best be understood by examining the simplest possible molecule. This is the hydrogen molecule ion H2+, which consists of two nuclei and one electron. First, however, think what would happen if we tried to make the even simpler molecule H22+. Since this would consist only of two protons whose electrostatic charges would repel each other at all distances, it is clear that such a molecule cannot exist; something more than two nuclei are required for bonding to occur.
In the hydrogen molecule ion H2+ we have a third particle, an electron. The effect of this electron will depend on its location with respect to the two nuclei. If the electron is in the space between the two nuclei, it will attract both protons toward itself, and thus toward each other. If the total attraction energy exceeds the internuclear repulsion, there will be a net bonding effect and the molecule will be stable. If, on the other hand, the electron is off to one side, it will attract both nuclei, but it will attract the closer one much more strongly, owing to the inverse-square nature of Coulomb's law. As a consequence, the electron will now help the electrostatic repulsion to push the two nuclei apart.
We see, then, that the electron is an essential component of a chemical bond, but that it must be in the right place: between the two nuclei. Coulomb's law can be used to calculate the forces experienced by the two nuclei for various positions of the electron. This allows us to define two regions of space about the nuclei, as shown in the figure. One region, the binding region, depicts locations at which the electron exerts a net binding effect on the new nuclei. Outside of this, in the antibinding region, the electron will actually work against binding.Summary
The amount of energy needed to separate a gaseous ion pair is its bond energy. The formation of ionic compounds are usually extremely exothermic. The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance. The total energy of the system is a balance between the repulsive interactions between electrons on adjacent ions and the attractive interactions between ions with opposite charges.
Metallic Bonding
Metals have several qualities that are unique, such as the ability to conduct electricity, a low ionization energy, and a low electronegativity (so they will give up electrons easily, i.e., they are cations). Metallic bonding is sort of like covalent bonding, because it involves sharing electrons. The simplest model of metallic bonding is the "sea of electrons" model, which imagines that the atoms sit in a sea of valence electrons that are delocalized over all the atoms. Because there are not specific bonds between individual atoms, metals are more flexible. The atoms can move around and the electron sea will keep holding them together. Some metals are very hard and have very high melting points, while others are soft and have low melting points. This depends roughly on the number of valence electrons that form the sea.
A False Dichotomy: The Ionic vs. Colvalent
The covalent-ionic continuum described above is certainly an improvement over the old covalent -versus - ionic dichotomy that existed only in the textbook and classroom, but it is still only a one-dimensional view of a multidimensional world, and thus a view that hides more than it reveals. The main thing missing is any allowance for the type of bonding that occurs between more pairs of elements than any other: metallic bonding. Intermetallic compounds are rarely even mentioned in introductory courses, but since most of the elements are metals, there are a lot of them, and many play an important role in metallurgy. In metallic bonding, the valence electrons lose their association with individual atoms; they form what amounts to a mobile "electron fluid" that fills the space between the crystal lattice positions occupied by the atoms, (now essentially positive ions.) The more readily this electron delocalization occurs, the more "metallic" the element.
Thus instead of the one-dimension chart shown above, we can construct a triangular diagram whose corners represent the three extremes of "pure" covalent, ionic, and metallic bonding.
Contributors | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.02%3A_Chemical_Bonds.txt |
6.9: Binary Ionic Compounds and Their Properties
6.18: Ionic Compounds Containing Polyatomic Ions
Learning Objectives
• Derive names for common types of inorganic compounds using a systematic approach
Nomenclature, a collection of rules for naming things, is important in science and in many other situations. This module describes an approach that is used to name simple ionic and molecular compounds, such as NaCl, CaCO3, and N2O4. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids (subsequent chapters in this text will focus on these compounds in great detail). We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry.
Ionic Compounds
To name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly.
Compounds Containing Only Monatomic Ions
The name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix –ide). Some examples are given in Table \(2\).
Table \(1\): Names of Some Ionic Compounds
NaCl, sodium chloride Na2O, sodium oxide
KBr, potassium bromide CdS, cadmium sulfide
CaI2, calcium iodide Mg3N2, magnesium nitride
CsF, cesium fluoride Ca3P2, calcium phosphide
LiCl, lithium chloride Al4C3, aluminum carbide
Compounds Containing Polyatomic Ions
Compounds containing polyatomic ions are named similarly to those containing only monatomic ions, except there is no need to change to an –ide ending, since the suffix is already present in the name of the anion. Examples are shown in Table \(2\).
CL, ammonium chloride, C a S O subscript 4 calcium sulfate, and M g subscript 3 ( P O subscript 4 ) subscript 2 magnesium phosphate." data-quail-id="56" data-mt-width="1071">
Table \(2\): Names of Some Polyatomic Ionic Compounds
KC2H3O2, potassium acetate (NH4)Cl, ammonium chloride
NaHCO3, sodium bicarbonate CaSO4, calcium sulfate
Al2(CO3)3, aluminum carbonate Mg3(PO4)2, magnesium phosphate
Ionic Compounds in Your Cabinets
Every day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table. Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.
Everyday Ionic Compounds
Ionic Compound Use
NaCl, sodium chloride ordinary table salt
KI , potassium iodide added to “iodized” salt for thyroid health
NaF, sodium fluoride ingredient in toothpaste
NaHCO3, sodium bicarbonate baking soda; used in cooking (and as antacid)
Na2CO3, sodium carbonate washing soda; used in cleaning agents
NaOCl, sodium hypochlorite active ingredient in household bleach
CaCO3 calcium carbonate ingredient in antacids
Mg(OH)2, magnesium hydroxide ingredient in antacids
Al(OH)3, aluminum hydroxide ingredient in antacids
NaOH, sodium hydroxide lye; used as drain cleaner
K3PO4, potassium phosphate food additive (many purposes)
MgSO4, magnesium sulfate added to purified water
Na2HPO4, sodium hydrogen phosphate anti-caking agent; used in powdered products
Na2SO3, sodium sulfite preservative
Compounds Containing a Metal Ion with a Variable Charge
Most of the transition metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or 3+ (see [link]), and the two corresponding compound formulas are FeCl2 and FeCl3. The simplest name, “iron chloride,” will, in this case, be ambiguous, as it does not distinguish between these two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table \(3\).
Table \(3\): Names of Some Transition Metal Ionic Compounds
Transition Metal Ionic Compound Name
FeCl3 iron(III) chloride
Hg2O mercury(I) oxide
HgO mercury(II) oxide
Cu3(PO4)2 copper(II) phosphate
Out-of-date nomenclature used the suffixes –ic and –ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, FeCl3, was previously called ferric chloride, and iron(II) chloride, FeCl2, was known as ferrous chloride. Though this naming convention has been largely abandoned by the scientific community, it remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula SnF2, which is more properly named tin(II) fluoride. The other fluoride of tin is SnF4, which was previously called stannic fluoride but is now named tin(IV) fluoride.
Naming Ionic Compounds
Name the following ionic compounds, which contain a metal that can have more than one ionic charge:
1. Fe2S3
2. CuSe
3. GaN
4. CrCl3
5. Ti2(SO4)3
Solution
The anions in these compounds have a fixed negative charge (S2−, Se2, N3−, Cl, and \(\ce{SO4^2-}\)), and the compounds must be neutral. Because the total number of positive charges in each compound must equal the total number of negative charges, the positive ions must be Fe3+, Cu2+, Ga3+, Cr4+, and Ti3+. These charges are used in the names of the metal ions:
1. iron(III) sulfide
2. copper(II) selenide
3. gallium(III) nitride
4. chromium(III) chloride
5. titanium(III) sulfate
Exercise \(1\)
Write the formulas of the following ionic compounds:
1. (a) chromium(III) phosphide
2. (b) mercury(II) sulfide
3. (c) manganese(II) phosphate
4. (d) copper(I) oxide
5. (e) chromium(VI) fluoride
Answer
(a) CrP; (b) HgS; (c) Mn3(PO4)2; (d) Cu2O; (e) CrF6
Summary
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K2O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, FeCl2 is iron(II) chloride and FeCl3 is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include SF6, sulfur hexafluoride, and N2O4, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the –ide suffix to –ic, and adding “acid;” HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion to –ic, and adding “acid;” H2CO3 is carbonic acid. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.04%3A_Ionic_Compounds-_Formulas_and_Names.txt |
Learning Objectives
• To use Lewis dot symbols to explain the stoichiometry of a compound
We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H2 molecule, which contains a purely covalent bond. Each hydrogen atom in H2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure $1$):
• The electrons in the two atoms repel each other because they have the same charge (
• The electrons in the two atoms repel each other because they have the same charge (E > 0).
• Similarly, the protons in adjacent atoms repel each other (E > 0).
• The electron in one atom is attracted to the oppositely charged proton in the other atom and vice versa (E < 0). Recall that it is impossible to specify precisely the position of the electron in either hydrogen atom. Hence the quantum mechanical probability distributions must be used.
A plot of the potential energy of the system as a function of the internuclear distance (Figure $2$) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r0 (the observed internuclear distance in H2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figures $1$ and $2$, which described a system containing two oppositely charged ions. The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities.
At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms.
Using Lewis Dot Symbols to Describe Covalent Bonding
The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
2. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32, for example, we add two electrons to the total because of the −2 charge.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
The $H_2O$ Molecule
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
The $OCl^−$ Ion
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
The $CH_2O$ Molecule
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. There are no electrons left to place on the central atom.
6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.
Example $1$
Write the Lewis electron structure for each species.
1. NCl3
2. S22
3. NOCl
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
1. Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise $1$
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Using Lewis Electron Structures to Explain Stoichiometry
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of Group 17 (the halogens), this number is one; for the elements of Group 16 (the chalcogens), it is two; for Group 15 elements, three; and for Group 14 elements four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group:
Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule.
Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule.
Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively.
Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure $3$). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom.
Formal Charges
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.
To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.
For each atom, we then compute a formal charge:
$\begin{matrix} formal\; charge= & valence\; e^{-}- & \left ( non-bonding\; e^{-}+\frac{bonding\;e^{-}}{2} \right )\ & ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )} \end{matrix} \label{8.5.1}$ (atom in Lewis structure)
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e/2)]. Substituting into Equation $\ref{8.5.2}$, we obtain
$formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\dfrac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2}$
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e/2)]. Using Equation $\ref{8.5.2}$ to calculate the formal charge on hydrogen, we obtain
$formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\dfrac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3}$
The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.
An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species.
Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.
Example $2$: The Ammonium Ion
Calculate the formal charges on each atom in the NH4+ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation $\ref{8.5.2}$ to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the NH4+ ion is as follows:
The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation $\ref{8.5.1}$, the formal charge on the nitrogen atom is therefore
$formal\; charge\left ( N \right )=5-\left ( 0+\dfrac{8}{2} \right )=0 \nonumber$
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
$formal\; charge\left ( H \right )=1-\left ( 0+\dfrac{2}{2} \right )=0 \nonumber$
The formal charges on the atoms in the NH4+ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.
Exercise $2$
Write the formal charges on all atoms in BH4.
Answer
If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.05%3A_Covalent_Bonding-_Simple_Lewis_Structures.txt |
Learning Objectives
• To use Lewis electron dot symbols to predict the number of bonds an element will form.
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond wil be discussed and the general properties found in typical substances in which the bond type occurs
1. Ionic bonds results from electrostatic forces that exist between ions of opposite charge. These bonds typically involves a metal with a nonmetal
2. Covalent bonds result from the sharing of electrons between two atoms. The bonds typically involves one nonmetallic element with another
3. Metallic bonds These bonds are found in solid metals (copper, iron, aluminum) with each metal bonded to several neighboring groups and bonding electrons free to move throughout the 3-dimensional structure.
Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds.
Lewis Symbols
At the beginning of the 20th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols (often shortened to Lewis dot symbols) that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons.
Lewis Dot symbols:
• convenient representation of valence electrons
• allows you to keep track of valence electrons during bond formation
• consists of the chemical symbol for the element plus a dot for each valence electron
To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the electron configuration for atomic sulfur is [Ne]3s23p4, thus there are six valence electrons. Its Lewis symbol would therefore be:
Fluorine, for example, with the electron configuration [He]2s22p5, has seven valence electrons, so its Lewis dot symbol is constructed as follows:
Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure 8.1.2. The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds.
The Octet Rule
In 1904, Richard Abegg formulated what is now known as Abegg's rule, which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder.
When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s2p6.
Definition: Octet Rule
A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made. This is known as octet rule by Lewis.
1. Normally two electrons pairs up and forms a bond, e.g., \(\ce{H_2}\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(\ce{CH_4}\)
The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell.
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Example \(1\): Salt
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why?
Solution
Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:
No dots are shown on Cs+ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs+ ion, which has the valence electron configuration of Xe, and the F ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.
Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe elsewhere. These include molecules in which one or more atoms contain fewer or more than eight electrons.
Summary
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule. Hydrogen, with only two valence electrons, does not obey the octet rule.
Contributors and Attributions
• Wikipedia
• National Programme on Technology Enhanced Learning (India) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.06%3A_The_Lewis_Model__-_Representing_Valance_Electrons_with_Dots.txt |
Learning Objectives
• Derive names for common types of inorganic compounds using a systematic approach
The bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios.
Compounds Composed of Two Elements
When two nonmetallic elements form a molecular compound, several combination ratios are often possible. For example, carbon and oxygen can form the compounds CO and CO2. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix –ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table \(3\).
Table \(3\): Nomenclature Prefixes
Number Prefix Number Prefix
1 (sometimes omitted) mono- 6 hexa-
2 di- 7 hepta-
3 tri- 8 octa-
4 tetra- 9 nona-
5 penta- 10 deca-
When only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, CO is named carbon monoxide, and CO2 is called carbon dioxide. When two vowels are adjacent, the a in the Greek prefix is usually dropped. Some other examples are shown in Table \(4\).
Table \(4\): Names of Some Molecular Compounds Composed of Two Elements
Compound Name Compound Name
SO2 sulfur dioxide BCl3 boron trichloride
SO3 sulfur trioxide SF6 sulfur hexafluoride
NO2 nitrogen dioxide PF5 phosphorus pentafluoride
N2O4 dinitrogen tetroxide P4O10 tetraphosphorus decaoxide
N2O5 dinitrogen pentoxide IF7 iodine heptafluoride
There are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, N2O is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And H2O is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them.
Naming Covalent Compounds
Name the following covalent compounds:
1. SF6
2. N2O3
3. Cl2O7
4. P4O6
Solution
Because these compounds consist solely of nonmetals, we use prefixes to designate the number of atoms of each element:
1. sulfur hexafluoride
2. dinitrogen trioxide
3. dichlorine heptoxide
4. tetraphosphorus hexoxide
Exercise \(2\)
Write the formulas for the following compounds:
1. phosphorus pentachloride
2. dinitrogen monoxide
3. iodine heptafluoride
4. carbon tetrachloride
Answer:
(a) PCl5; (b) N2O; (c) IF7; (d) CCl4
Binary Acids
Some compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, H+, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound’s name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):
1. The word “hydrogen” is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word “acid” is added as a second word
For example, when the gas HCl (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table \(5\).
Table \(5\): Names of Some Simple Acids
Name of Gas Name of Acid
HF(g), hydrogen fluoride HF(aq), hydrofluoric acid
HCl(g), hydrogen chloride HCl(aq), hydrochloric acid
HBr(g), hydrogen bromide HBr(aq), hydrobromic acid
HI(g), hydrogen iodide HI(aq), hydroiodic acid
H2S(g), hydrogen sulfide H2S(aq), hydrosulfuric acid
Oxyacids
Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:
1. Omit “hydrogen”
2. Start with the root name of the anion
3. Replace –ate with –ic, or –ite with –ous
4. Add “acid”
For example, consider H2CO3 (which you might be tempted to call “hydrogen carbonate”). To name this correctly, “hydrogen” is omitted; the –ate of carbonate is replace with –ic; and acid is added—so its name is carbonic acid. Other examples are given in Table \(6\). There are some exceptions to the general naming method (e.g., H2SO4 is called sulfuric acid, not sulfic acid, and H2SO3 is sulfurous, not sulfous, acid).
Table \(6\): Names of Common Oxyacids
Formula Anion Name Acid Name
HC2H3O2 acetate acetic acid
HNO3 nitrate nitric acid
HNO2 nitrite nitrous acid
HClO4 perchlorate perchloric acid
H2CO3 carbonate carbonic acid
H2SO4 sulfate sulfuric acid
H2SO3 sulfite sulfurous acid
H3PO4 phosphate phosphoric acid
Summary
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K2O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, FeCl2 is iron(II) chloride and FeCl3 is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include SF6, sulfur hexafluoride, and N2O4, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the –ide suffix to –ic, and adding “acid;” HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion to –ic, and adding “acid;” H2CO3 is carbonic acid. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.07%3A_Molecular_Compounds-_Formulas_and_Names.txt |
Learning Objectives
• To understand the definition and difference between empirical formulas and chemical formulas
• To understand how combustion analysis can be used to identify chemical formulas
Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's chemical formula cannot be reduced any more, then the empirical formula is the same as the chemical formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the chemical formula (other techniques can though). Once known, the chemical formula can be calculated from the empirical formula.
Empirical Formulas
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.
Example $1$: Mercury Chloride
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Solution
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
$(73.9 \;g) \times \left(\dfrac{1\; mol}{200.59\; g}\right) = 0.368 \;moles \nonumber$
For Chlorine:
$(26.1\; g) \times \left(\dfrac{1\; mol}{35.45\; g}\right) = 0.736\; mol \nonumber$
What is the molar ratio between the two elements?
$\dfrac{0.736 \;mol \;Cl}{0.368\; mol\; Hg} = 2.0 \nonumber$
Thus, we have twice as many moles (i.e. atoms) of $\ce{Cl}$ as $\ce{Hg}$. The empirical formula would thus be (remember to list cation first, anion last):
$\ce{HgCl2} \nonumber$
Chemical Formula from Empirical Formula
The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). The general flow for this approach is shown in Figure $1$ and demonstrated in Example $2$.
Example $2$: Ascorbic Acid
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?
Solution
Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have:
• 40.92 grams C
• 4.58 grams H
• 54.50 grams O
This would give us how many moles of each element?
• Carbon
$(40.92\; \cancel{g\; C}) \times \left( \dfrac{1\; mol\; C}{12.011\; \cancel{g\; C}} \right) = 3.407\; mol \; C \nonumber$
• Hydrogen
$(4.58\; \cancel{g\; H}) \times \left( \dfrac{1\; mol\; H}{1.008\; \cancel{g\; H}} \right) = 4.544\; mol \;H \nonumber$
• Oxygen
$(54.50\; \cancel{g\; O}) \times \left( \dfrac{1\; mol\; O}{15.999\; \cancel{g\; O}} \right) = 3.406\; mol \; O \nonumber$
Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see oxygen):
• Carbon
$C= \dfrac{3.407\; mol}{3.406\; mol} \approx 1.0 \nonumber$
• Hydrogen
$C= \dfrac{4.5.44\; mol}{3.406\; mol} = 1.0 \nonumber$
• Oxygen
$C= \dfrac{3.406\; mol}{3.406\; mol} = 1.0 \nonumber$
The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.
C = (1.0)*3 = 3
H = (1.333)*3 = 4
O = (1.0)*3 = 3
or
$\ce{C3H4O3} \nonumber$
This is our empirical formula for ascorbic acid.
What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?
(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu
The molecular mass from our empirical formula is significantly lower than the experimentally determined value. What is the ratio between the two values?
(176 amu/88.062 amu) = 2.0
Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual chemical formula is:
2* C3H4O3 = C6H8O6
Empirical Formulas: Empirical Formulas, YouTube(opens in new window) [youtu.be]
Combustion Analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $2$). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure $3$ and a typical combustion analysis is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Isopropyl Alcohol
What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?
Solution
From this information quantitate the amount of C and H in the sample.
$(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2 \nonumber$
Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?
$(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C \nonumber$
How about the hydrogen?
$(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O \nonumber$
Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?
$(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O \nonumber$
Overall therefore, we have:
• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
• C = 3.05 atoms
• H = 8.1 atoms
• O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be $C_3H_8O$
Example $4$: Combustion of Naphalene
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
1. Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
2. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of $\ce{CO2}$ is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C} \nonumber$
$= 1.883 \times 10^{-2} \, g \, C \nonumber$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H} \nonumber$
$= 1.264 \times 10^{-3} \, g \, H \nonumber$
B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C \nonumber$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H \nonumber$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250 \nonumber$
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the chemical formula of naphthalene is C10H8, which is consistent with our results.
Exercise $4$
1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its chemical formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer a
The empirical formula is C4H5. (The chemical formula of xylene is actually C8H10.)
Answer b
33.81 mg of CO2; 6.92 mg of H2O
Combustion Analysis: Combustion Analysis, YouTube(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.08%3A_Composition_of_Compounds.txt |
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Derivation of Molecular Formulas
Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.
Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n:
$\mathrm{\dfrac{molecular\: or\: molar\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}{empirical\: formula\: mass\left(amu\: or\:\dfrac{g}{mol}\right)}= \mathit n\: formula\: units/molecule}$
The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown by the generic empirical formula AxBy:
$\mathrm{(A_xB_y)_n=A_{nx}B_{nx}}$
For example, consider a covalent compound whose empirical formula is determined to be CH2O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:
$\mathrm{\dfrac{180\:amu/molecule}{30\:\dfrac{amu}{formula\: unit}}=6\:formula\: units/molecule}$
Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:
$\ce{(CH2O)6}=\ce{C6H12O6}$
Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.
Determination of the Molecular Formula for Nicotine
Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?
Solution
Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:
\begin{alignat}{2} &\mathrm{(74.02\:g\: C)\left(\dfrac{1\:mol\: C}{12.01\:g\: C}\right)}&&= \:\mathrm{6.163\:mol\: C}\ &\mathrm{(8.710\:g\: H)\left(\dfrac{1\:mol\: H}{1.01\:g\: H}\right)}&&= \:\mathrm{8.624\:mol\: H}\ &\mathrm{(17.27\:g\: N)\left(\dfrac{1\:mol\: N}{14.01\:g\: N}\right)}&&= \:\mathrm{1.233\:mol\: N} \end{alignat}
Next, we calculate the molar ratios of these elements.
The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C5H7N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.
We calculate the molar mass for nicotine from the given mass and molar amount of compound:
$\mathrm{\dfrac{40.57\:g\: nicotine}{0.2500\:mol\: nicotine}=\dfrac{162.3\:g}{mol}} \nonumber$
Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:
$\mathrm{\dfrac{162.3\:g/mol}{81.13\:\dfrac{g}{formula\: unit}}=2\:formula\: units/molecule} \nonumber$
Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:
$\ce{(C5H7N)6}=\ce{C10H14N2} \nonumber$
Exercise $5$
What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?
Answer
C8H10N4O2
Summary
The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.
Combustion Analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $2$). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
Figure $2$: Combustion analysis apparatus
One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure $3$ and a typical combustion analysis is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Isopropyl Alcohol
What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?
Solution
From this information quantitate the amount of C and H in the sample.
$(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2$
Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?
$(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C$
How about the hydrogen?
$(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O$
Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?
$(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O$
Overall therefore, we have:
• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
• C = 3.05 atoms
• H = 8.1 atoms
• O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be $C_3H_8O$
Example $4$: Combustion of Naphalene
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
1. Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
2. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of CO2 is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C}$
$= 1.883 \times 10^{-2} \, g \, C$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H}$
$= 1.264 \times 10^{-3} \, g \, H$
B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250$
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
Exercise $4$
1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its molecular formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer a
The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.)
Answer b
33.81 mg of CO2; 6.92 mg of H2O | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.09%3A_Determining_a_Chemical_Formula_from_Experimental_Data.txt |
Learning Objectives
• To recognize the composition and properties typical of organic and inorganic compounds.
Organic substances have been used throughout this text to illustrate the differences between ionic and covalent bonding and to demonstrate the intimate connection between the structures of compounds and their chemical reactivity. You learned, for example, that even though NaOH and alcohols (ROH) both have OH in their formula, NaOH is an ionic compound that dissociates completely in water to produce a basic solution containing Na+ and OH ions, whereas alcohols are covalent compounds that do not dissociate in water and instead form neutral aqueous solutions. You also learned that an amine (RNH2), with its lone pairs of electrons, is a base, whereas a carboxylic acid (RCO2H), with its dissociable proton, is an acid.
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH4Cl), expecting to get ammonium cyanate (NH4OCN). What he expected is described by the following equation.
$AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{25.1.1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$.
Key Takeaway
• Organic chemistry is the study of carbon compounds, nearly all of which also contain hydrogen atoms.
Hydrocarbons
Learning Objectives
• Identify alkanes, alkenes, alkynes, and aromatic compounds.
• List some properties of hydrocarbons.
The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons. Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons.
Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanes are aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and Alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes (or cycloalkenes or cycloalkynes).
Aromatic hydrocarbons have a special six-carbon ring called a benzene ring. Electrons in the benzene ring have special energetic properties that give benzene physical and chemical properties that are markedly different from alkanes. Originally, the term aromatic was used to describe this class of compounds because they were particularly fragrant. However, in modern chemistry the term aromatic denotes the presence of a six-membered ring that imparts different and unique properties to a molecule.
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
The next-largest alkane has two C atoms that are covalently bonded to each other. For each C atom to make four covalent bonds, each C atom must be bonded to three H atoms. The resulting molecule, whose formula is C2H6, is ethane:
Propane has a backbone of three C atoms surrounded by H atoms. You should be able to verify that the molecular formula for propane is C3H8:
The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula, which lists the formula of each C atom in the backbone of the molecule. For example, the condensed structural formula for ethane is CH3CH3, while for propane it is CH3CH2CH3. Table $1$ - The First 10 Alkanes, gives the molecular formulas, the condensed structural formulas, and the names of the first 10 alkanes.
Table $1$ The First 10 Alkanes
Molecular Formula Condensed Structural Formula Name
CH4 CH4 methane
C2H6 CH3CH3 ethane
C3H8 CH3CH2CH3 propane
C4H10 CH3CH2CH2CH3 butane
C5H12 CH3CH2CH2CH2CH3 pentane
C6H14 CH3(CH2)4CH3 hexane
C7H16 CH3(CH2)5CH3 heptane
C8H18 CH3(CH2)6CH3 octane
C9H20 CH3(CH2)7CH3 nonane
C10H22 CH3(CH2)8CH3 decane
Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons.
Alkenes have a C–C double bond. Because they have less than the maximum number of H atoms possible, they are unsaturated hydrocarbons. The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The next largest alkene—propene—has three C atoms with a C–C double bond between two of the C atoms. It is also known as propylene:
What do you notice about the names of alkanes and alkenes? The names of alkenes are the same as their corresponding alkanes except that the ending is -ene, rather than -ane. Using a stem to indicate the number of C atoms in a molecule and an ending to represent the type of organic compound is common in organic chemistry, as we shall see.
With the introduction of the next alkene, butene, we begin to see a major issue with organic molecules: choices. With four C atoms, the C–C double bond can go between the first and second C atoms or between the second and third C atoms:
2 structural formulas for butene, with the first butene having the double bond on the first and second carbon from the left and the latter having its double bond on the second and third carbon from the left.
(A double bond between the third and fourth C atoms is the same as having it between the first and second C atoms, only flipped over.) The rules of naming in organic chemistry require that these two substances have different names. The first molecule is named 1-butene, while the second molecule is named 2-butene. The number at the beginning of the name indicates where the double bond originates. The lowest possible number is used to number a feature in a molecule; hence, calling the second molecule 3-butene would be incorrect. Numbers are common parts of organic chemical names because they indicate which C atom in a chain contains a distinguishing feature.
The compounds 1-butene and 2-butene have different physical and chemical properties, even though they have the same molecular formula—C4H8. Different molecules with the same molecular formula are called isomers. Isomers are common in organic chemistry and contribute to its complexity.
Example $1$
Based on the names for the butene molecules, propose a name for this molecule.
Solution
With five C atoms, we will use the pent- stem, and with a C–C double bond, this is an alkene, so this molecule is a pentene. In numbering the C atoms, we use the number 2 because it is the lower possible label. So this molecule is named 2-pentene.
Exercise $1$
Based on the names for the butene molecules, propose a name for this molecule.
A structural formula of a six carbon molecule with a double bond on the third and fourth carbon from the left. There are twelve hydrogen atoms in total.
Answer
3-hexene
Alkynes, with a C–C triple bond, are named similarly to alkenes except their names end in -yne. The smallest alkyne is ethyne, which is also known as acetylene:
Propyne has the structure
Structural formula showing three carbon molecules with a triple bond present between the first and second carbon atom. The appropriate number of hydrogen atoms is attached to each carbon atom.
With butyne, we need to start numbering the position of the triple bond, just as we did with alkenes:
Two structural formula of butyne. One butyne has a triple bond between the first and second carbon atom, while two butyne has the triple bond between the second and third carbon atom.
Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
The six carbons are arranged in a hexagon pattern with one hydrogen atom emerging outwards from each carbon atom. The presence of a double bond is alternated between every other carbon atom.
The alternating single and double C–C bonds give the benzene ring a special stability, and it does not react like an alkene as might be suspected. Benzene has the molecular formula C6H6; in larger aromatic compounds, a different atom replaces one or more of the H atoms.
As fundamental as hydrocarbons are to organic chemistry, their properties and chemical reactions are rather mundane. Most hydrocarbons are nonpolar because of the close electronegativities of the C and H atoms. As such, they dissolve only sparingly in H2O and other polar solvents. Small hydrocarbons, such as methane and ethane, are gases at room temperature, while larger hydrocarbons, such as hexane and octane, are liquids. Even larger hydrocarbons are solids at room temperature and have a soft, waxy consistency.
Hydrocarbons are rather unreactive, but they do participate in some classic chemical reactions. One common reaction is substitution with a halogen atom by combining a hydrocarbon with an elemental halogen. Light is sometimes used to promote the reaction, such as this one between methane and chlorine:
$CH_{4}+Cl_{2}\overset{light}{\rightarrow} CH_{3}Cl+HCl\nonumber$
Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction. One example is
The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne.
Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction. In this case, however, the reaction conditions may not be mild; high pressures of H2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace:
$CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3}\nonumber$
By far the most common reaction of hydrocarbons is combustion, which is the combination of a hydrocarbon with O2 to make CO2 and H2O. The combustion of hydrocarbons is accompanied by a release of energy and is a primary source of energy production in our society (Figure $2$ - Combustion). The combustion reaction for gasoline, for example, which can be represented by C8H18, is as follows:
$2C^{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O+\sim 5060kJ\nonumber$
Key Takeaways
• The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen.
• Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes.
• The combustion of hydrocarbons is a primary source of energy for our society.
Exercise $2$
1. Define hydrocarbon. What are the two general types of hydrocarbons?
2. What are the three different types of aliphatic hydrocarbons? How are they defined?
3. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
4. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
5. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
6. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
7. Name and draw the structural formulas for the four smallest alkanes.
8. Name and draw the structural formulas for the four smallest alkenes.
9. What does the term aromatic imply about an organic molecule?
10. What does the term normal imply when used for alkanes?
11. Explain why the name 1-propene is incorrect. What is the proper name for this molecule?
12. Explain why the name 3-butene is incorrect. What is the proper name for this molecule?
13. Name and draw the structural formula of each isomer of pentene.
14. Name and draw the structural formula of each isomer of hexyne.
15. Write a chemical equation for the reaction between methane and bromine.
16. Write a chemical equation for the reaction between ethane and chlorine.
17. Draw the structure of the product of the reaction of bromine with propene.
18. Draw the structure of the product of the reaction of chlorine with 2-butene.
19. Draw the structure of the product of the reaction of hydrogen with 1-butene.
20. Draw the structure of the product of the reaction of hydrogen with 1-butene.
21. Write the balanced chemical equation for the combustion of heptane.
22. Write the balanced chemical equation for the combustion of nonane.
Answers
1. an organic compound composed of only carbon and hydrogen; aliphatic hydrocarbons and aromatic hydrocarbons
2.
1. aliphatic; alkane
2. aromatic
3. aliphatic; alkene
3.
1. aliphatic; alkane
2. aliphatic; alkene
3. aromatic
4.
5.
6. Aromatic means that the molecule has a benzene ring.
7.
8. The 1 is not necessary. The name of the compound is simply propene.
9.
10.
11. CH4 + Br2 → CH3Br + HBr
12.
13.
14.
15. C7H16 + 11O2 → 7CO2 + 8H2O | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/05%3A_Molecules_and_Compounds/5.11%3A_Organic_Compounds.txt |
• 6.1: Morphine - A Molecular Imposter
• 6.2: Electronegativity and Bond Polarity
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average
• 6.3: Writing Lewis Structures for Molecular Compounds and Polyatomic Ions
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families.
• 6.4: Resonance and Formal Charge
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.
• 6.5: Exceptions to the Octet Rule- Odd-Electron Species, Incomplete Octets, and Expanded Octets
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons
• 6.6: Bond Energies and Bond Lengths
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. The bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms.
• 6.7: VSEPR Theory - The Five Basic Shapes
The Lewis electron-pair approach described previously can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however.
• 6.8: VSPER Theory- The Effect of Lone Pairs
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The VSEPR model is not a theory; it does not attempt to explain observations. Instead, it is a counting procedure that accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
• 6.9: VSPER Theory - Predicting Molecular Geometries
• 6.10: Molecular Shape and Polarity
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them.
06: Chemical Bonding I- Drawing Lewis Structures and Determining Molecular Shapes
Learning Objectives
• To define electronegativity and bond polarity
• To calculate the percent ionic character of a covalent polar bond
The electron pairs shared between two atoms are not necessarily shared equally. For example, while the bonding electron pair is shared equally in the covalent bond in $Cl_2$, in $NaCl$ the 3s electron is stripped from the Na atom and is incorporated into the electronic structure of the Cl atom - and the compound is most accurately described as consisting of individual $Na^+$ and $Cl^-$ ions (ionic bonding). For most covalent substances, their bond character falls between these two extremes. As demonstrated below, the bond polarity is a useful concept for describing the sharing of electrons between atoms within a covalent bond:
• A nonpolar covalent bond is one in which the electrons are shared equally between two atoms.
• A polar covalent bond is one in which one atom has a greater attraction for the electrons than the other atom. If this relative attraction is great enough, then the bond is an ionic bond.
Electronegativity
The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table. Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table.
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
Electronegativity is a function of:
1. the atom's ionization energy (how strongly the atom holds on to its own electrons) and
2. the atom's electron affinity (how strongly the atom attracts other electrons).
Both of these are properties of the isolated atom. An element will be highly electronegative if it has a large (negative) electron affinity and a high ionization energy (always endothermic, or positive for neutral atoms). Thus, it will attract electrons from other atoms and resist having its own electrons attracted away.
The Pauling Electronegativity Scale
The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0.
Periodic variations in Pauling’s electronegativity values are illustrated in Figures $1$ and $2$. If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine ($\chi = 3.98$) is the most electronegative element and cesium is the least electronegative nonradioactive element ($\chi = 0.79$). Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl and N, S, and Br).
Figure $2$: Pauling Electronegativity Values of the s-, p-, d-, and f-Block Elements. Values for most of the actinides are approximate. Elements for which no data are available are shown in gray. Source: Data from L. Pauling, The Nature of the Chemical Bond, 3rd ed. (1960).
Linus Pauling (1901-1994)
When he was nine, Pauling’s father died, and his mother tried to convince him to quit school to support the family. He did not quit school, but was later denied a high school degree, and had to work several jobs to put himself through college. Pauling would go on to become one of the most influential chemists of the century if not all time. He won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962.
Pauling’s method is limited by the fact that many elements do not form stable covalent compounds with other elements; hence their electronegativities cannot be measured by his method. Other definitions have since been developed that address this problem, e.g., the Mulliken, Allred-Rochow, and Allen electronegativity scales. The Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity, showing the relationship between electronegativity and these other periodic properties.
Electronegativity Differences between Metals and Nonmetals
An element’s electronegativity provides us with a single value that we can use to characterize the chemistry of an element. Elements with a high electronegativity (χ ≥ 2.2 in Figure $2$) have very negative affinities and large ionization potentials, so they are generally nonmetals and electrical insulators that tend to gain electrons in chemical reactions (i.e., they are oxidants). In contrast, elements with a low electronegativity ($\chi \le 1.8$) have electron affinities that have either positive or small negative values and small ionization potentials, so they are generally metals and good electrical conductors that tend to lose their valence electrons in chemical reactions (i.e., they are reductants). In between the metals and nonmetals, along the heavy diagonal line running from B to At is a group of elements with intermediate electronegativities (χ ~ 2.0). These are the metalloids (or semimetals), elements that have some of the chemical properties of both nonmetals and metals. The distinction between metals and nonmetals is one of the most fundamental we can make in categorizing the elements and predicting their chemical behavior. Figure $3$ shows the strong correlation between electronegativity values, metallic versus nonmetallic character, and location in the periodic table.
Electronegativity values increase from lower left to upper right in the periodic table.
The rules for assigning oxidation states(opens in new window) are based on the relative electronegativities of the elements; the more electronegative element in a binary compound is assigned a negative oxidation state. As we shall see, electronegativity values are also used to predict bond energies, bond polarities, and the kinds of reactions that compounds undergo.
Example $1$: Increasing Electronegativity
On the basis of their positions in the periodic table, arrange Cl, Se, Si, and Sr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a metalloid.
Given: four elements
Asked for: order by increasing electronegativity and classification
Strategy:
1. Locate the elements in the periodic table. From their diagonal positions from lower left to upper right, predict their relative electronegativities.
2. Arrange the elements in order of increasing electronegativity.
3. Classify each element as a metal, a nonmetal, or a metalloid according to its location about the diagonal belt of metalloids running from B to At.
Solution:
A Electronegativity increases from lower left to upper right in the periodic table (Figure 8.4.2). Because Sr lies far to the left of the other elements given, we can predict that it will have the lowest electronegativity. Because Cl lies above and to the right of Se, we can predict that χCl > χSe. Because Si is located farther from the upper right corner than Se or Cl, its electronegativity should be lower than those of Se and Cl but greater than that of Sr. B The overall order is therefore χSr < χSi < χSe < χCl.
C To classify the elements, we note that Sr lies well to the left of the diagonal belt of metalloids running from B to At; while Se and Cl lie to the right and Si lies in the middle. We can predict that Sr is a metal, Si is a metalloid, and Se and Cl are nonmetals.
Exercise $1$
On the basis of their positions in the periodic table, arrange Ge, N, O, Rb, and Zr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a metalloid.
Answer
Rb < Zr < Ge < N < O; metals (Rb, Zr); metalloid (Ge); nonmetal (N, O)
Percent Ionic Character of a Covalent polar bond
The two idealized extremes of chemical bonding: (1) ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding, in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure $4$ compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta ($\delta$) is used to indicate that a bonded atom possesses a partial positive charge, indicated by $\delta^+$, or a partial negative charge, indicated by $\delta^-$, and a bond between two atoms that possess partial charges is a polar bond.
Bond Polarity
The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. Electronegativity (χ) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B (χB) is greater than the electronegativity of A (χA), for example, is indicated with the partial negative charge on the more electronegative atom:
$\begin{matrix} _{less\; electronegative}& & _{more\; electronegative}\ A\; \; &-& B\; \; \; \; \ ^{\delta ^{+}} & & ^{\delta ^{-}} \end{matrix} \label{8.4.1}$
One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: Δχ = χB − χA.
To predict the polarity of the bonds in Cl2, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms: χCl = 3.16, χH = 2.20, and χNa = 0.93. Cl2 must be nonpolar because the electronegativity difference (Δχ) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, Δχ is 2.23. This high value is typical of an ionic compound (Δχ ≥ ≈1.5) and means that the valence electron of sodium has been completely transferred to chlorine to form Na+ and Cl ions. In HCl, however, Δχ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.
Bond polarity and ionic character increase with an increasing difference in electronegativity.
As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in NaCl, Cl2, ClF5, and HClO4 would be exactly the same.
Dipole Moments
The asymmetrical charge distribution in a polar substance such as HCl produces a dipole moment where $Qr$ in meters (m). is abbreviated by the Greek letter mu (µ). The dipole moment is defined as the product of the partial charge Q on the bonded atoms and the distance r between the partial charges:
$\mu=Qr \label{8.4.2}$
where Q is measured in coulombs (C) and r in meters. The unit for dipole moments is the debye (D):
$1\; D = 3.3356\times 10^{-30}\; C\cdot ·m \label{8.4.3}$
When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure $4$).
We can measure the partial charges on the atoms in a molecule such as HCl using Equation $\ref{8.4.2}$. If the bonding in HCl were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of HCl is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is
$Q=\dfrac{\mu }{r} =1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=2.901\times 10^{-20}\;C \label{8.4.4}$
By dividing this calculated value by the charge on a single electron (1.6022 × 10−19 C), we find that the electron distribution in HCl is asymmetric and that effectively it appears that there is a net negative charge on the Cl of about −0.18, effectively corresponding to about 0.18 e. This certainly does not mean that there is a fraction of an electron on the Cl atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount.
$\dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{8.4.5}$
To form a neutral compound, the charge on the H atom must be equal but opposite. Thus the measured dipole moment of HCl indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing HCl as
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
we can therefore indicate the charge separation quantitatively as
$\begin{matrix} _{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine χH = 2.20; χCl = 3.16, χCl − χH = 0.96), a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule. Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:
The arrow shows the direction of electron flow by pointing toward the more electronegative atom.
The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure $6$ shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as NaCl(g) and CsF(g) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure $6$ show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example $2$.
Example $2$
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
A Compute the charge on each atom using the information given and Equation $\ref{8.4.2}$.
B Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
$Q=\dfrac{\mu }{r} =9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=1.272\times 10^{-19}\;C \nonumber$
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
$\% \; ionic\; character=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right )=79.39\%\simeq 79\% \nonumber$
Exercise $2$
In the gas phase, silver chloride (AgCl) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer
55.5%
Summary
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity. Elements with a high electronegativity are generally nonmetals and electrical insulators and tend to behave as oxidants in chemical reactions. Conversely, elements with a low electronegativity are generally metals and good electrical conductors and tend to behave as reductants in chemical reactions.
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.02%3A_Electronegativity_and_Bond_Polarity.txt |
Learning Objectives
• To use Lewis dot symbols to explain the stoichiometry of a compound
We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H2 molecule, which contains a purely covalent bond. Each hydrogen atom in H2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure $1$):
• The electrons in the two atoms repel each other because they have the same charge (
• The electrons in the two atoms repel each other because they have the same charge (E > 0).
• Similarly, the protons in adjacent atoms repel each other (E > 0).
• The electron in one atom is attracted to the oppositely charged proton in the other atom and vice versa (E < 0). Recall that it is impossible to specify precisely the position of the electron in either hydrogen atom. Hence the quantum mechanical probability distributions must be used.
A plot of the potential energy of the system as a function of the internuclear distance (Figure $2$) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r0 (the observed internuclear distance in H2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figures $1$ and $2$, which described a system containing two oppositely charged ions. The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities.
At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms.
Using Lewis Dot Symbols to Describe Covalent Bonding
The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl2, they can each complete their valence shell:
Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons.
We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:
The structure on the right is the Lewis electron structure, or Lewis structure, for H2O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
2. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO32, for example, we add two electrons to the total because of the −2 charge.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
The $H_2O$ Molecule
1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.
4. Each H atom has a full valence shell of 2 electrons.
5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.
The $OCl^−$ Ion
1. With only two atoms in the molecule, there is no central atom.
2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.
3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.
4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:
Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.
The $CH_2O$ Molecule
1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:
2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
Six electrons are used, and 6 are left over.
4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. There are no electrons left to place on the central atom.
6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.
Example $1$
Write the Lewis electron structure for each species.
1. NCl3
2. S22
3. NOCl
Given: chemical species
Asked for: Lewis electron structures
Strategy:
Use the six-step procedure to write the Lewis electron structure for each species.
Solution:
1. Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:
Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.
2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:
3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:
Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:
Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:
All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.
Exercise $1$
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Using Lewis Electron Structures to Explain Stoichiometry
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of Group 17 (the halogens), this number is one; for the elements of Group 16 (the chalcogens), it is two; for Group 15 elements, three; and for Group 14 elements four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group:
Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule.
Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule.
Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively.
Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure $3$). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom.
Formal Charges
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.
To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.
For each atom, we then compute a formal charge:
$\begin{matrix} formal\; charge= & valence\; e^{-}- & \left ( non-bonding\; e^{-}+\frac{bonding\;e^{-}}{2} \right )\ & ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )} \end{matrix} \label{8.5.1}$ (atom in Lewis structure)
To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH3) whose Lewis electron structure is as follows:
A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e/2)]. Substituting into Equation $\ref{8.5.2}$, we obtain
$formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\dfrac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2}$
A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e/2)]. Using Equation $\ref{8.5.2}$ to calculate the formal charge on hydrogen, we obtain
$formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\dfrac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3}$
The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH3 molecule.
An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species.
Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.
Example $2$: The Ammonium Ion
Calculate the formal charges on each atom in the NH4+ ion.
Given: chemical species
Asked for: formal charges
Strategy:
Identify the number of valence electrons in each atom in the NH4+ ion. Use the Lewis electron structure of NH4+ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation $\ref{8.5.2}$ to calculate the formal charge on each atom.
Solution:
The Lewis electron structure for the NH4+ ion is as follows:
The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation $\ref{8.5.1}$, the formal charge on the nitrogen atom is therefore
$formal\; charge\left ( N \right )=5-\left ( 0+\dfrac{8}{2} \right )=0 \nonumber$
Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore
$formal\; charge\left ( H \right )=1-\left ( 0+\dfrac{2}{2} \right )=0 \nonumber$
The formal charges on the atoms in the NH4+ ion are thus
Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.
Exercise $2$
Write the formal charges on all atoms in BH4.
Answer
If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.03%3A_Writing_Lewis_Structures_for_Molecular_Compounds_and_Polyatomic_Ions.txt |
Learning Objectives
• To understand the concept of resonance.
Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding of a single polyatomic species including fractional bonds and fractional charges. Resonance structures are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integral number of covalent bonds.
Sometimes one Lewis Structure is not Enough
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (\(\ce{O3}\)), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
Ozone (\(O_3\))
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The Carbonate (\(CO_3^{2−} \)) Ion
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32 is an average of three resonance structures.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
The Nitrate (\(NO_3^-\)) ion
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the \(\ce{NO_3^{-}}\) ion to be somewhat shorter than a single bond.
Example \(1\): Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (\(\ce{C6H6}\)) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Exercise \(1\): Nitrite Ion
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2).
Answer
Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.
Warning
If several reasonable resonance forms for a molecule exists, the "actual electronic structure" of the molecule will probably be intermediate between all the forms that you can draw. The classic example is benzene in Example \(1\). One would expect the double bonds to be shorter than the single bonds, but if one overlays the two structures, you see that one structure has a single bond where the other structure has a double bond. The best measurements that we can make of benzene do not show two bond lengths - instead, they show that the bond length is intermediate between the two resonance structures.
Resonance structures is a mechanism that allows us to use all of the possible resonance structures to try to predict what the actual form of the molecule would be. Single bonds, double bonds, triple bonds, +1 charges, -1 charges, these are our limitations in explaining the structures, and the true forms can be in between - a carbon-carbon bond could be mostly single bond with a little bit of double bond character and a partial negative charge, for example.
Summary
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.04%3A_Resonance_and_Formal_Charge.txt |
Learning Objectives
• To assign a Lewis dot symbol to elements not having an octet of electrons in their compounds.
Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions:
1. When there are an odd number of valence electrons
2. When there are too few valence electrons
3. When there are too many valence electrons
Exception 1: Species with Odd Numbers of Electrons
The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide also called nitric oxide ($\ce{NO}$. Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in.
No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitric oxide. nitric oxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitric oxide had ten valence electrons we would come up with the Lewis Structure (Figure $1$):
Let's look at the formal charges of Figure $2$ based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure $1$, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure $1$, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitric oxide (Figure $2$):
Free Radicals
There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with $\cdot OH$, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted $\cdot Cl$. Interestingly, an odd Number of Valence Electrons will result in the molecule being paramagnetic.
Exception 2: Incomplete Octets
The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH3 (Borane).
If one were to make a Lewis structure for BH3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure $2$):
The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.
Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure $4$).
If you look Figure $4$, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure $5$):
Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.
This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure $5$, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table ($\chi=4.0$). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.
However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure $6$:
None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure $4$), the one with the double bond (Figure $5$), and the one with the ionic bond (Figure $6$). The most contributing structure is probably the incomplete octet structure (due to Figure $5$ being basically impossible and Figure $6$ not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.
As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .
Example $1$: $NF_3$
Draw the Lewis structure for boron trifluoride (BF3).
Solution
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:
3. Add octets to outer atoms:
4. Add extra electrons (24-24=0) to central atom:
5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:
6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:
Exception 3: Expanded Valence Shells
More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, $PCl_5$ is a legitimate compound (whereas $NCl_5$) is not:
Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:
Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.
One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.
Example $2$: The $SO_4^{-2}$ ion
Such is the case for the sulfate ion, SO4-2. A strict adherence to the octet rule forms the following Lewis structure:
If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2.
If instead we made a structure for the sulfate ion with an expanded octet, it would look like this:
Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure $12$, as opposed to +2 and -1 (difference of 3) in Figure $12$) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case.
Example $3$: The $ICl_4^-$ Ion
Draw the Lewis structure for $ICl_4^-$ ion.
Solution
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:
3. Add octet of electrons to outer atoms:
4. Add extra electrons (36-32=4) to central atom:
5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)
Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies.
Summary
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.05%3A_Exceptions_to_the_Octet_Rule-_Odd-Electron_Species_Incomplete_Octets_and_E.txt |
Learning Objectives
• The define Bond-dissociation energy (bond energy)
• To correlate bond strength with bond length
• To define and used average bond energies
In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds. The stability of a molecule is a function of the strength of the covalent bonds holding the atoms together.
The Relationship between Bond Order and Bond Energy
Triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table $1$ lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
Table $1$: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds Multiple Bonds
H–H 432 C–C 346 N–N ≈167 O–O ≈142 F–F 155 C=C 602
H–C 411 C–Si 318 N–O 201 O–F 190 F–Cl 249 C≡C 835
H–Si 318 C–N 305 N–F 283 O–Cl 218 F–Br 249 C=N 615
H–N 386 C–O 358 N–Cl 313 O–Br 201 F–I 278 C≡N 887
H–P ≈322 C–S 272 N–Br 243 O–I 201 Cl–Cl 240 C=O 749
H–O 459 C–F 485 P–P 201 S–S 226 Cl–Br 216 C≡O 1072
H–S 363 C–Cl 327 S–F 284 Cl–I 208 N=N 418
H–F 565 C–Br 285 S–Cl 255 Br–Br 190 N≡N 942
H–Cl 428 C–I 213 S–Br 218 Br–I 175 N=O 607
H–Br 362 Si–Si 222 I–I 149 O=O 494
H–I 295 Si–O 452 S=O 532
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11).
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore $\ce{Cl–Cl > Br–Br > F–F > I–I} \nonumber$ Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
1. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
1. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO42 ion in Section 8.6. In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Bond strengths increase as bond order increases, while bond distances decrease.
The Relationship between Molecular Structure and Bond Energy
Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is not the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table $2$); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table $1$ are average values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%.
Table $2$: Energies for the Dissociation of Successive C–H Bonds in Methane. Source: Data from CRC Handbook of Chemistry and Physics (2004).
Reaction D (kJ/mol)
CH4(g) → CH3(g) + H(g) 439
CH3(g) → CH2(g) + H(g) 462
CH2(g) → CH(g) + H(g) 424
CH(g) → C(g) + H(g) 338
We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction ($ΔH_{rxn} < 0$):
$ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{$1$}$
The ≈ sign is used because we are adding together average bond energies; hence this approach does not give exact values for ΔHrxn.
Let’s consider the reaction of 1 mol of n-heptane (C7H16) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline:
$\ce{CH3(CH2)5CH3(l) + 11 O2(g) \rightarrow 7 CO2(g) + 8 H2O(g)} \label{$2$}$
In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of n-heptane, while 14 C=O bonds (two for each CO2) and 16 O–H bonds (two for each H2O) are formed. The energy changes can be tabulated as follows:
Binds Broken (kJ/mol) and Bonds Formed (kJ/mol)
Bonds Broken (kJ/mol) Bonds Formed (kJ/mol)
6 C–C 346 × 6 = 2076 14 C=O 799 × 14 = 11,186
16 C–H 411 × 16 = 6576 16 O–H 459 × 16 = 7344
11 O=O 494 × 11 = 5434 Total = 18,530
Total = 14,086
The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that $ΔH_{rxn}$ is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction).
If we compare this approximation with the value obtained from measured $ΔH_f^o$ values ($ΔH_{rxn} = −481\;7 kJ/mol$), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual $ΔH^ο_f$ values are unknown. These approximations can be important for predicting whether a reaction is exothermic or endothermic—and to what degree.
Example $1$: Explosives
The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table $1$ to estimate the $ΔH_{rxn}$ per mole of RDX.
Given: chemical reaction, structure of reactant, and Table $1$.
Asked for: $ΔH_{rxn}$ per mole
Strategy:
1. List the types of bonds broken in RDX, along with the bond energy required to break each type. Multiply the number of each type by the energy required to break one bond of that type and then add together the energies. Repeat this procedure for the bonds formed in the reaction.
2. Use Equation $1$ to calculate the amount of energy consumed or released in the reaction (ΔHrxn).
Solution:
We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO2) can be viewed as having one N–O single bond and one N=O double bond, as follows:
In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures.
A We can organize our data by constructing a table:
Bonds Broken (kJ/mol)
Bonds Broken (kJ/mol) Bonds Broken (kJ/mol)
6 C–H 411 × 6 = 2466 6 C=O 799 × 6 = 4794
3 N–N 167 × 3 = 501 6 O–H 459 × 6 = 2754
3 N–O 201 × 3 = 603 Total = 10,374
3 N=O 607 × 3 = 1821
1.5 O=O 494 × 1.5 = 741
Total = 7962
B From Equation $1$, we have
\begin{align*} ΔH_{rxn} &\approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \[4pt] &= 7962 \; kJ/mol − 10,374 \; kJ/mol \[4pt] &=−2412 \;kJ/mol \end{align*} \nonumber
Thus this reaction is also highly exothermic
Exercise $1$: Freon
The molecule HCFC-142b is a hydrochlorofluorocarbon that is used in place of chlorofluorocarbons (CFCs) such as the Freons and can be prepared by adding HCl to 1,1-difluoroethylene:
HCL to produce CH3Cf2Cl and HCFC142b." data-cke-saved-src="/@api/deki/files/129585/imageedit_31_2777381278.png" src="/@api/deki/files/129585/imageedit_31_2777381278.png" data-quail-id="371">
Use tabulated bond energies to calculate $ΔH_{rxn}$.
Answer
−54 kJ/mol
Bond Dissociation Energy
Bond Dissociation Energy (also referred to as Bond energy) is the enthalpy change ($\Delta H$, heat input) required to break a bond (in 1 mole of a gaseous substance)
What about when we have a compound which is not a diatomic molecule? Consider the dissociation of methane:
There are four equivalent C-H bonds, thus we can that the dissociation energy for a single C-H bond would be:
\begin{align*} D(C-H) &= (1660/4)\, kJ/mol \[4pt] &= 415 \,kJ/mol \end{align*} \nonumber
The bond energy for a given bond is influenced by the rest of the molecule. However, this is a relatively small effect (suggesting that bonding electrons are localized between the bonding atoms). Thus, the bond energy for most bonds varies little from the average bonding energy for that type of bond
Bond energy is always a positive value - it takes energy to break a covalent bond (conversely energy is released during bond formation)
Bond (kJ/mol)
Table $4$: Average bond energies:
C-F 485
C-Cl 328
C-Br 276
C-I 240
C-C 348
C-N 293
C-O 358
C-F 485
C-C 348
C=C 614
C=C 839
The more stable a molecule (i.e. the stronger the bonds) the less likely the molecule is to undergo a chemical reaction.
Bond Energies and the Enthalpy of Reactions
If we know which bonds are broken and which bonds are made during a chemical reaction, we can estimate the enthalpy change of the reaction ($\Delta H_{rxn}$) even if we do not know the enthalpies of formation (($\Delta H_{f}^o$)for the reactants and products:
$\Delta H = \sum \text{bond energies of broken bonds} - \sum \text{bond energies of formed bonds} \label{8.8.3}$
Example $2$: Chlorination of Methane
What is the enthalpy of reaction between 1 mol of chlorine and 1 mol methane?
Solution
We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed.
• Bonds broken: 1 mol of Cl-Cl bonds, 1 mol of C-H bonds
• Bonds formed: 1 mol of H-Cl bonds, 1 mol of C-Cl bonds
\begin{align*} \Delta H &= [D(Cl-Cl) + D(C-H)] - [D(H-Cl)+D(C-Cl)] \[4pt] &= [242 kJ + 413 kJ] - [431 kJ + 328 kJ] \[4pt] &= -104 \,kJ \end{align*} \nonumber
Thus, the reaction is exothermic (because the bonds in the products are stronger than the bonds in the reactants)
Example $3$: Combustion of Ethane
What is the enthalpy of reaction for the combustion of 1 mol of ethane?
Solution
We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed.
• bonds broken: 6 moles C-H bonds, 1 mol C-C bonds, 7/2 moles of O=O bonds
• bonds formed: 4 moles C=O bonds, 6 moles O-H bonds
\begin{align*} \Delta {H} &= [(6 \times 413) + (348) + (\frac{7}{2} \times 495)] - [(4 \times 799) + (6 \times 463)] \[4pt] &= 4558 - 5974 \[4pt] &= -1416\; kJ \end{align*} \nonumber
Therefor the reaction is exothermic.
Table $5$: Bond strength and bond length
Bond Bond Energy (kJ/mol) Bond Length (Å)
C-C 348 1.54
C=C 614 1.34
C=C 839 1.
As the number of bonds between two atoms increases, the bond grows shorter and stronger
Summary
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa.
The breakage and formation of bonds is similar to a relationship: you can either get married or divorced and it is more favorable to be married.
• Energy is always released to make bonds, which is why the enthalpy change for breaking bonds is always positive.
• Energy is always required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate octet electronic configurations). The enthalpy change is always negative because the system is releasing energy when forming bond. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.06%3A_Bond_Energies_and_Bond_Lengths.txt |
Learning Objectives
• To use the VSEPR model to predict molecular geometries.
• To predict whether a molecule has a dipole moment.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(1\) and \(2\).
In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.
VESPR Produce to predict Molecular geometry
This VESPR procedure is summarized as follows:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles.
4. Describe the molecular geometry.
We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(2\) and Figure \(3\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\).
AX2 Molecules: BeH2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
4. From Figure \(3\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2 Molecules: CO2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure \(3\)). The structure of \(\ce{CO2}\) is shown in Figure \(1\).
Three Electron Groups
AX3 Molecules: BCl3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.
4. From Figure \(3\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure \(2\).
AX3 Molecules: CO32−
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
4. We see from Figure \(3\) that the molecular geometry of CO32 is trigonal planar with bond angles of 120°.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX2E Molecules: SO2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(4\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(2\) and \(3\)). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4 Molecules: CH4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure \(2\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX4.
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure \(3\)).
AX3E Molecules: NH3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure \(3\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(3\) and Figure \(4\)).
AX2E2 Molecules: H2O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.
Five Electron Groups
In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.
AX5 Molecules: PCl5
1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl5 is
3. All electron groups are bonding pairs, so the structure is designated as AX5. There are no lone pair interactions.
4. The molecular geometry of PCl5 is trigonal bipyramidal, as shown in Figure \(3\). The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.
AX4E Molecules: SF4
1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(2\).
3. We designate SF4 as AX4E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.
At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.
4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw. The Faxial–S–Faxial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.
AX3E2 Molecules: BrF3
1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is
Once again, we have a compound that is an exception to the octet rule.
2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.
3. With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:
Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.
4. The three nuclei in BrF3 determine its molecular structure, which is described as T shaped. This is essentially a trigonal bipyramid that is missing two equatorial vertices. The Faxial–Br–Faxial angle is 172°, less than 180° because of LP–BP repulsions (Figure \(2\).1).
Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.
AX2E3 Molecules: I3−
1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is
2. There are five electron groups about the central atom in I3, two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.
3. With two bonding pairs and three lone pairs, I3 has a total of five electron pairs and is designated as AX2E3. We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.
The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.
4. With three nuclei and three lone pairs of electrons, the molecular geometry of I3 is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.
Six Electron Groups
Six electron groups form an octahedron, a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(2\).)
AX6 Molecules: SF6
1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(2\) that the geometry that minimizes repulsions is octahedral.
3. With only bonding pairs, SF6 is designated as AX6. All positions are chemically equivalent, so all electronic interactions are equivalent.
4. There are six nuclei, so the molecular geometry of SF6 is octahedral.
AX5E Molecules: BrF5
1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is
With its expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:
3. With five bonding pairs and one lone pair, BrF5 is designated as AX5E; it has a total of six electron pairs. The BrF5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all Faxial–Br–Fequatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.
4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal. The Faxial–B–Fequatorial angles are 85.1°, less than 90° because of LP–BP repulsions.
AX4E2 Molecules: ICl4−
1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is
2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is
3. ICl4 is designated as AX4E2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.
4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar, an octahedron with two opposite vertices missing.
The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(6\).
Figure \(6\): Overview of Molecular Geometries
Example \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
2. H3O+ (hydronium ion)
Given: two chemical species
Asked for: molecular geometry
Strategy:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
4. Describe the molecular geometry.
Solution:
1. A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is
C All electron groups are bonding pairs, so PF5 is designated as AX5. Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.
D The PF5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.
2. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is
B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH3, repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
C With three bonding pairs and one lone pair, the structure is designated as AX3E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal, in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:
Exercise \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. XeO3
2. PF6
3. NO2+
Answer a
trigonal pyramidal
Answer b
octahedral
Answer c
linear
Example \(2\)
Predict the molecular geometry of each molecule.
1. XeF2
2. SnCl2
Given: two chemical compounds
Asked for: molecular geometry
Strategy:
Use the strategy given in Example\(1\).
Solution:
1. A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is
B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.
C From B, XeF2 is designated as AX2E3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:
The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I3. All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.
D With two nuclei about the central atom, the molecular geometry of XeF2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.
2. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is
B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.
C From B we designate SnCl2 as AX2E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.
D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl2 is bent, like SO2, but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.
Exercise \(2\)
Predict the molecular geometry of each molecule.
1. SO3
2. XeF4
Answer a
trigonal planar
Answer b
square planar
Molecules with No Single Central Atom
The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AXmEn fragments. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.
We can treat methyl isocyanate as linked AXmEn fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX4. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example \(3\)
Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(3\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise \(3\)
Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.
Answer
The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.
Molecular Dipole Moments
You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure \(\PageIndex{8a}\)). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure \(\PageIndex{8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.
Other examples of molecules with polar bonds are shown in Figure \(9\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment.
Example \(4\)
Which molecule(s) has a net dipole moment?
1. \(\ce{H2S}\)
2. \(\ce{NHF2}\)
3. \(\ce{BF3}\)
Given: three chemical compounds
Asked for: net dipole moment
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
1. The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of \(\ce{H2S}\) is bent (Figure \(6\)). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
2. Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
3. The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:
Exercise \(4\)
Which molecule(s) has a net dipole moment?
• \(\ce{CH3Cl}\)
• \(\ce{SO3}\)
• \(\ce{XeO3}\)
Answer
\(\ce{CH3Cl}\) and \(\ce{XeO3}\)
Summary
Lewis electron structures give no information about molecular geometry, the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. From this we can describe the molecular geometry. The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds.
Molecules with polar covalent bonds can have a dipole moment, an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.07%3A_VSEPR_Theory_-_The_Five_Basic_Shapes.txt |
Learning Objectives
• To use the VSEPR model to predict molecular geometries.
• To predict whether a molecule has a dipole moment.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom, ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups, which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(1\) and \(2\).
In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.
VESPR Produce to predict Molecular geometry
This VESPR procedure is summarized as follows:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles.
4. Describe the molecular geometry.
We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(2\) and Figure \(3\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\).
AX2 Molecules: BeH2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
4. From Figure \(3\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2 Molecules: CO2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure \(3\)). The structure of \(\ce{CO2}\) is shown in Figure \(1\).
Three Electron Groups
AX3 Molecules: BCl3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.
4. From Figure \(3\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure \(2\).
AX3 Molecules: CO32−
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
4. We see from Figure \(3\) that the molecular geometry of CO32 is trigonal planar with bond angles of 120°.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX2E Molecules: SO2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(4\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(2\) and \(3\)). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4 Molecules: CH4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure \(2\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX4.
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure \(3\)).
AX3E Molecules: NH3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure \(3\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(3\) and Figure \(4\)).
AX2E2 Molecules: H2O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.
Five Electron Groups
In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.
AX5 Molecules: PCl5
1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl5 is
3. All electron groups are bonding pairs, so the structure is designated as AX5. There are no lone pair interactions.
4. The molecular geometry of PCl5 is trigonal bipyramidal, as shown in Figure \(3\). The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.
AX4E Molecules: SF4
1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(2\).
3. We designate SF4 as AX4E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position. We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.
At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.
4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw. The Faxial–S–Faxial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.
AX3E2 Molecules: BrF3
1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is
Once again, we have a compound that is an exception to the octet rule.
2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.
3. With three bonding pairs and two lone pairs, the structural designation is AX3E2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:
Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.
4. The three nuclei in BrF3 determine its molecular structure, which is described as T shaped. This is essentially a trigonal bipyramid that is missing two equatorial vertices. The Faxial–Br–Faxial angle is 172°, less than 180° because of LP–BP repulsions (Figure \(2\).1).
Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.
AX2E3 Molecules: I3−
1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is
2. There are five electron groups about the central atom in I3, two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.
3. With two bonding pairs and three lone pairs, I3 has a total of five electron pairs and is designated as AX2E3. We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.
The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.
4. With three nuclei and three lone pairs of electrons, the molecular geometry of I3 is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.
Six Electron Groups
Six electron groups form an octahedron, a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(2\).)
AX6 Molecules: SF6
1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(2\) that the geometry that minimizes repulsions is octahedral.
3. With only bonding pairs, SF6 is designated as AX6. All positions are chemically equivalent, so all electronic interactions are equivalent.
4. There are six nuclei, so the molecular geometry of SF6 is octahedral.
AX5E Molecules: BrF5
1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is
With its expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:
3. With five bonding pairs and one lone pair, BrF5 is designated as AX5E; it has a total of six electron pairs. The BrF5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all Faxial–Br–Fequatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.
4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal. The Faxial–B–Fequatorial angles are 85.1°, less than 90° because of LP–BP repulsions.
AX4E2 Molecules: ICl4−
1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is
2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is
3. ICl4 is designated as AX4E2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.
4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar, an octahedron with two opposite vertices missing.
The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(6\).
Figure \(6\): Overview of Molecular Geometries
Example \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. PF5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
2. H3O+ (hydronium ion)
Given: two chemical species
Asked for: molecular geometry
Strategy:
1. Draw the Lewis electron structure of the molecule or polyatomic ion.
2. Determine the electron group arrangement around the central atom that minimizes repulsions.
3. Assign an AXmEn designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
4. Describe the molecular geometry.
Solution:
1. A The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF5 is
C All electron groups are bonding pairs, so PF5 is designated as AX5. Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.
D The PF5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.
2. A The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is
B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH3, repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
C With three bonding pairs and one lone pair, the structure is designated as AX3E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal, in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:
Exercise \(1\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
1. XeO3
2. PF6
3. NO2+
Answer a
trigonal pyramidal
Answer b
octahedral
Answer c
linear
Example \(2\)
Predict the molecular geometry of each molecule.
1. XeF2
2. SnCl2
Given: two chemical compounds
Asked for: molecular geometry
Strategy:
Use the strategy given in Example\(1\).
Solution:
1. A Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is
B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.
C From B, XeF2 is designated as AX2E3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:
The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I3. All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.
D With two nuclei about the central atom, the molecular geometry of XeF2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.
2. A The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is
B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.
C From B we designate SnCl2 as AX2E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.
D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl2 is bent, like SO2, but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.
Exercise \(2\)
Predict the molecular geometry of each molecule.
1. SO3
2. XeF4
Answer a
trigonal planar
Answer b
square planar
Molecules with No Single Central Atom
The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AXmEn fragments. We will demonstrate with methyl isocyanate (CH3–N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.
We can treat methyl isocyanate as linked AXmEn fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX4. We can therefore predict the CH3–N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO2, so its geometry, like that of CO2, is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example \(3\)
Use the VSEPR model to predict the molecular geometry of propyne (H3C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(3\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise \(3\)
Predict the geometry of allene (H2C=C=CH2), a compound with narcotic properties that is used to make more complex organic molecules.
Answer
The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.
Molecular Dipole Moments
You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure \(\PageIndex{8a}\)). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure \(\PageIndex{8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.
Other examples of molecules with polar bonds are shown in Figure \(9\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment.
Example \(4\)
Which molecule(s) has a net dipole moment?
1. \(\ce{H2S}\)
2. \(\ce{NHF2}\)
3. \(\ce{BF3}\)
Given: three chemical compounds
Asked for: net dipole moment
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
1. The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of \(\ce{H2S}\) is bent (Figure \(6\)). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
2. Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
3. The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:
Exercise \(4\)
Which molecule(s) has a net dipole moment?
• \(\ce{CH3Cl}\)
• \(\ce{SO3}\)
• \(\ce{XeO3}\)
Answer
\(\ce{CH3Cl}\) and \(\ce{XeO3}\)
Summary
Lewis electron structures give no information about molecular geometry, the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AXmEn designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. From this we can describe the molecular geometry. The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds.
Molecules with polar covalent bonds can have a dipole moment, an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.08%3A_VSPER_Theory-_The_Effect_of_Lone_Pairs.txt |
Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.
VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.
As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF2 molecule. The Lewis structure of BeF2 (Figure \(2\)) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure \(2\)).
Figure \(2\): The BeF2 molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.
Figure \(3\) illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.
6.10: Molecular Shape and Polarity
Learning Objectives
• To calculate the percent ionic character of a covalent polar bond
Previously, we described the two idealized extremes of chemical bonding:
• ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and
• covalent bonding, in which electrons are shared equally between two atoms.
Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure $1$ compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta ($\delta$) is used to indicate that a bonded atom possesses a partial positive charge, indicated by $\delta^+$, or a partial negative charge, indicated by $\delta^-$, and a bond between two atoms that possess partial charges is a polar bond.
Bond Polarity
The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. Electronegativity ($\chi$) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B ($\chi_B$) is greater than the electronegativity of A ($\chi_A$), for example, is indicated with the partial negative charge on the more electronegative atom:
$\begin{matrix} _{less\; electronegative}& & _{more\; electronegative}\ A\; &-& B\; \; \ ^{\delta ^{+}} & & ^{\delta ^{-}} \end{matrix} \label{9.3.1}$
One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms:
$Δ\chi = \chi_B − \chi_A. \nonumber$
To predict the polarity of the bonds in Cl2, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms (Table A2): $\chi_{Cl} = 3.16$, $\chi_H = 2.20$, and $\chi_{Na} = 0.93$. $\ce{Cl2}$ must be nonpolar because the electronegativity difference ($Δ\chi$) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, $Δ\chi$ is 2.23. This high value is typical of an ionic compound ($Δ\chi ≥ ≈1.5$) and means that the valence electron of sodium has been completely transferred to chlorine to form Na+ and Cl ions. In HCl, however, $Δ\chi$ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix} \nonumber$
Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.
Bond polarity and ionic character increase with an increasing difference in electronegativity.
As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in $\ce{NaCl}$, $\ce{Cl2}$, $\ce{ClF5}$, and $\ce{HClO4}$ would be exactly the same.
Dipole Moments
The asymmetrical charge distribution in a polar substance such as $\ce{HCl}$ produces a dipole moment where $Qr$ in meters ($m$). is abbreviated by the Greek letter mu ($\mu$). The dipole moment is defined as the product of the partial charge $Q$ on the bonded atoms and the distance $r$ between the partial charges:
$\mu=Qr \label{9.3.2}$
where $Q$ is measured in coulombs ($C$) and $r$ in meters. The unit for dipole moments is the debye (D):
$1\; D = 3.3356\times 10^{-30}\; C\cdot m \label{9.7.3}$
When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure $2$).
We can measure the partial charges on the atoms in a molecule such as $\ce{HCl}$ using Equation \ref{9.3.2}. If the bonding in $\ce{HCl}$ were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of $\ce{HCl}$ is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is
\begin{align*} Q &=\dfrac{\mu }{r} \[4pt] &=1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \[4pt] &=2.901\times 10^{-20}\;C \label{9.7.4} \end{align*}
By dividing this calculated value by the charge on a single electron (1.6022 × 10−19 C), we find that the electron distribution in$\ce{HCl}$ is asymmetric and that effectively it appears that there is a net negative charge on the $\ce{Cl}$ of about −0.18, effectively corresponding to about 0.18 e. This certainly does not mean that there is a fraction of an electron on the $\ce{Cl}$ atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount.
$\dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{9.7.5}$
To form a neutral compound, the charge on the $\ce{H}$ atom must be equal but opposite. Thus the measured dipole moment of $\ce{HCl}$ indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing$\ce{HCl}$ as
$\begin{matrix} _{\delta ^{+}}& & _{\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
we can therefore indicate the charge separation quantitatively as
$\begin{matrix} _{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\ H\; \; &-& Cl \end{matrix}$
Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine ($\chi_H = 2.20$ and $\chi_{Cl} = 3.16$) so
$\chi_{Cl} − \chi_H = 0.96 \nonumber$
This is a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule.Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:
The arrow shows the direction of electron flow by pointing toward the more electronegative atom.
A warning about Dipole Moment arrows
As the figure above shows, we represent dipole moments by an arrow with a length proportional to $μ$ and pointing from the positive charge to the negative charge. However, the opposite convention is still widely used especially among physicists.
The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure $2$ shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as $NaCl_{(g)}$ and $CsF_{(g)}$ is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure $2$ show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example $1$.
Example $1$: percent ionic character
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
1. Compute the charge on each atom using the information given and Equation $\ref{9.3.2}$
2. Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
\begin{align*} Q &=\dfrac{\mu }{r} \[4pt] &=9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right ) \[4pt] &=1.272\times 10^{-19}\;C \end{align*} \nonumber
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
\begin{align*} \text{ ionic character} &=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right ) \[4pt] &=79.39\%\simeq 79\% \end{align*} \nonumber
Exercise $1$
In the gas phase, silver chloride ($\ce{AgCl}$) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer
55.5%
Summary
Bond polarity and ionic character increase with an increasing difference in electronegativity.
$\mu = Qr \nonumber$
Compounds with polar covalent bonds have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a dipole moment, which is the product of the partial charges on the bonded atoms and the distance between them. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/06%3A_Chemical_Bonding_I-_Drawing_Lewis_Structures_and_Determining_Molecular_Shapes/6.09%3A_VSPER_Theory_-_Predicting_Molecular_Geometries.txt |
• 7.1: Oxygen- A Magnetic Liquid
• 7.2: Valence Bond Theory- Orbital Overlap as a Chemical Bond
A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.
• 7.3: Valence Bond Theory- Hybridization of Atomic Orbitals
The localized valence bonding theory uses a process called hybridization, in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.
• 7.4: Molecular Orbital Theory- Electron Delocalization
A molecular orbital is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital, which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which describe overlapping atomic orbitals.
• 7.5: Molecular Orbital Theory- Polyatomic Molecules
• 7.6: Bonding in Metals and Semiconductors
07: Chemical Bonding II- Valance Bond Theory and Molecular Orbital Theory
Learning Objectives
• To describe the bonding in simple compounds using valence bond theory.
Although the VSEPR model is a simple and useful method for qualitatively predicting the structures of a wide range of compounds, it is not infallible. It predicts, for example, that H2S and PH3 should have structures similar to those of \(\ce{H2O}\) and \(\ce{NH3}\), respectively. In fact, structural studies have shown that the H–S–H and H–P–H angles are more than 12° smaller than the corresponding bond angles in \(\ce{H2O}\) and \(\ce{NH3}\). More disturbing, the VSEPR model predicts that the simple group 2 halides (MX2), which have four valence electrons, should all have linear X–M–X geometries. Instead, many of these species, including \(\ce{SrF2}\) and \(\ce{BaF2}\), are significantly bent. A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.
Introduction
As we have talked about using Lewis structures to depict the bonding in covalent compounds, we have been very vague in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together?
The valence bond theory is introduced to describe bonding in covalent molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H2 or HF, our present understanding of s and p atomic orbitals will suffice. To explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.
Example: The H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals. In simple terms, we can say that both electrons now spend more time between the two nuclei and thus hold the atoms together. As we will see, the situation is not quite so simple as that, because the electron pair must still obey quantum mechanics - that is, the two electrons must now occupy a shared orbital space. This will be the essential principle of valence bond theory.
How far apart are the two nuclei? That is a very important issue to consider. If they are too far apart, their respective 1s orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second. However, something else is happening at the same time: as the atoms get closer, the repulsive positive-positive interaction between the two nuclei also begins to increase.
At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have a very unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the bond length. For the H2 molecule, this distance is 74 x 10-12 meters, or 0.74 Å (Å means angstrom, or 10-10 meters). Likewise, the difference in potential energy between the lowest state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond energy. For the hydrogen molecule, this energy is equal to about 104 kcal/mol.
Every covalent bond in a given molecule has a characteristic length and strength. In general, carbon-carbon single bonds are about 1.5 Å long (Å means angstrom, or 10-10 meters) while carbon-carbon double bonds are about 1.3 Å, carbon-oxygen double bonds are about 1.2 Å, and carbon-hydrogen bonds are in the range of 1.0 – 1.1 Å. Most covalent bonds in organic molecules range in strength from just under 100 kcal/mole (for a carbon-hydrogen bond in ethane, for example) up to nearly 200 kcal/mole. You can refer to tables in reference books such as the CRC Handbook of Chemistry and Physics for extensive lists of bond lengths, strengths, and many other data for specific organic compounds.
Balls and Springs
Although we tend to talk about "bond length" as a specific distance, it is not accurate to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as springs which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important, when we study the analytical technique known as infrared (IR) spectroscopy.
Valence Bond Theory: A Localized Bonding Approach
You learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm (Figure \(1\)). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1s atomic orbital of one hydrogen atom overlaps with the singly occupied 1s atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a localized electron-pair bond (Figure \(1\)).
Although Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theory, which is built on two assumptions:
1. The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals; that is, the greater the overlap, the more stable the bond.
2. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms.
Figure \(2\) shows an electron-pair bond formed by the overlap of two ns atomic orbitals, two np atomic orbitals, and an ns and an np orbital where n = 2. Notice that bonding overlap occurs when the interacting atomic orbitals have the correct orientation (are "pointing at" each other) and are in phase (represented by colors in Figure \(2\) ).
Maximum overlap occurs between orbitals with the same spatial orientation and similar energies.
Let’s examine the bonds in BeH2, for example. According to the VSEPR model, BeH2 is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1s22s2 electron configuration, and each H atom has a 1s1 electron configuration. Because the Be atom has a filled 2s subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1s orbitals on the H atoms. If a singly occupied 1s orbital on hydrogen were to overlap with a filled 2s orbital on beryllium, the resulting bonding orbital would contain three electrons, but the maximum allowed by quantum mechanics is two. How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2s electrons into an empty 2p orbital and reverse its spin, in a process called promotion:
In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2s and one of the 2p orbitals), each of which could overlap with a singly occupied 1s orbital of an H atom to form an electron-pair bond. Although this would produce \(\ce{BeH2}\), the two Be–H bonds would not be equivalent: the 1s orbital of one hydrogen atom would overlap with a Be 2s orbital, and the 1s orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2p orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as \(\ce{BeH2}\) form, scientists developed the concept of hybridization. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/07%3A_Chemical_Bonding_II-_Valance_Bond_Theory_and_Molecular_Orbital_Theory/7.02%3A_Valence_Bond_Theory-_Orbital_Overlap_as_a_Chemical_Bond.txt |
The localized valence bond theory uses a process called hybridization, in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.
Hybridization of s and p Orbitals
In BeH2, we can generate two equivalent orbitals by combining the 2s orbital of beryllium and any one of the three degenerate 2p orbitals. By taking the sum and the difference of Be 2s and 2pz atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the z-axes, as shown in Figure $1$.
Because the difference A − B can also be written as A + (−B), in Figure $2$ and subsequent figures we have reversed the phase(s) of the orbital being subtracted, which is the same as multiplying it by −1 and adding. This gives us Equation \ref{9.5.1b}, where the value $\frac{1}{\sqrt{2}}$ is needed mathematically to indicate that the 2s and 2p orbitals contribute equally to each hybrid orbital.
$sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{9.5.1a}$
and
$sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{9.5.1b}$
The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called sp hybrids because they are formed from one s and one p orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure s and p orbitals, as illustrated in this diagram:
Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable only if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure $4$). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds.
The concept of hybridization also explains why boron, with a 2s22p1 valence electron configuration, forms three bonds with fluorine to produce BF3, as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2s electrons to an unoccupied 2p orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2s and two 2p orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here:
Looking at the 2s22p2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2p electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2s orbital and the three 2p orbitals on carbon to give a set of four degenerate sp3 (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron:
In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH2 or CF2), but these species are highly reactive, unstable intermediates that only form in certain chemical reactions.
Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths.
The bonding in molecules such as NH3 or H2O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH3, for example, N, with a 2s22p3 valence electron configuration, can hybridize its 2s and 2p orbitals to produce four sp3 hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons:
The three singly occupied sp3 lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H2O has an sp3 hybridized oxygen atom that uses two singly occupied sp3 lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH3 and H2O. Unfortunately, however, recent experimental evidence indicates that in NH3 and H2O, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research.
Example $1$
Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.
1. H2S
2. CHCl3
Given: two chemical compounds
Asked for: number of electron pairs and molecular geometry, hybridization, and bonding
Strategy:
1. Using the VSEPR approach to determine the number of electron pairs and the molecular geometry of the molecule.
2. From the valence electron configuration of the central atom, predict the number and type of hybrid orbitals that can be produced. Fill these hybrid orbitals with the total number of valence electrons around the central atom and describe the hybridization.
Solution:
1. A H2S has four electron pairs around the sulfur atom with two bonded atoms, so the VSEPR model predicts a molecular geometry that is bent, or V shaped. B Sulfur has a 3s23p4 valence electron configuration with six electrons, but by hybridizing its 3s and 3p orbitals, it can produce four sp3 hybrids. If the six valence electrons are placed in these orbitals, two have electron pairs and two are singly occupied. The two sp3 hybrid orbitals that are singly occupied are used to form S–H bonds, whereas the other two have lone pairs of electrons. Together, the four sp3 hybrid orbitals produce an approximately tetrahedral arrangement of electron pairs, which agrees with the molecular geometry predicted by the VSEPR model.
2. A The CHCl3 molecule has four valence electrons around the central atom. In the VSEPR model, the carbon atom has four electron pairs, and the molecular geometry is tetrahedral. B Carbon has a 2s22p2 valence electron configuration. By hybridizing its 2s and 2p orbitals, it can form four sp3 hybridized orbitals that are equal in energy. Eight electrons around the central atom (four from C, one from H, and one from each of the three Cl atoms) fill three sp3 hybrid orbitals to form C–Cl bonds, and one forms a C–H bond. Similarly, the Cl atoms, with seven electrons each in their 3s and 3p valence subshells, can be viewed as sp3 hybridized. Each Cl atom uses a singly occupied sp3 hybrid orbital to form a C–Cl bond and three hybrid orbitals to accommodate lone pairs.
Exercise $1$
Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.
1. the BF4 ion
2. hydrazine (H2N–NH2)
Answer a
B is sp3 hybridized; F is also sp3 hybridized so it can accommodate one B–F bond and three lone pairs. The molecular geometry is tetrahedral.
Answer b
Each N atom is sp3 hybridized and uses one sp3 hybrid orbital to form the N–N bond, two to form N–H bonds, and one to accommodate a lone pair. The molecular geometry about each N is trigonal pyramidal.
The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom.
Hybridization Using d Orbitals
Hybridization is not restricted to the ns and np atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence (n − 1)d orbitals as well as its ns and np orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF5 and SF6). Using the ns orbital, all three np orbitals, and one (n − 1)d orbital gives a set of five sp3d hybrid orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure $7$). In this case, the five hybrid orbitals are not all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other.
Similarly, the combination of the ns orbital, all three np orbitals, and two nd orbitals gives a set of six equivalent sp3d2 hybrid orbitals oriented toward the vertices of an octahedron (part (b) in Figure 9.5.6). In the VSEPR model, PF5 and SF6 are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which sp3d or sp3d2 hybrid orbitals are used for bonding.
Example $2$
What is the hybridization of the central atom in each species? Describe the bonding in each species.
1. XeF4
2. SO42
3. SF4
Given: three chemical species
Asked for: hybridization of the central atom
Strategy:
1. Determine the geometry of the molecule using the strategy in Example $1$. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization.
2. Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding.
Solution:
1. A Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp3d2 hybridized. B With 12 electrons around Xe, four of the six sp3d2 hybrid orbitals form Xe–F bonds, and two are occupied by lone pairs of electrons.
2. A The S in the SO42 ion has four electron pairs and has four bonded atoms, so the structure is tetrahedral. The sulfur must be sp3 hybridized to generate four S–O bonds. B Filling the sp3 hybrid orbitals with eight electrons from four bonds produces four filled sp3 hybrid orbitals.
3. A The S atom in SF4 contains five electron pairs and four bonded atoms. The molecule has a seesaw structure with one lone pair:
To accommodate five electron pairs, the sulfur atom must be sp3d hybridized. B Filling these orbitals with 10 electrons gives four sp3d hybrid orbitals forming S–F bonds and one with a lone pair of electrons.
Exercise $2$
What is the hybridization of the central atom in each species? Describe the bonding.
1. PCl4+
2. BrF3
3. SiF62
Answer a
sp3 with four P–Cl bonds
Answer a
sp3d with three Br–F bonds and two lone pairs
Answer a
sp3d2 with six Si–F bonds
Hybridization using d orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S).
Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF4 and SiF4), only SiF4 reacts with F to give a stable hexafluoro dianion, SiF62. Because there are no 2d atomic orbitals, the formation of octahedral CF62 would require hybrid orbitals created from 2s, 2p, and 3d atomic orbitals. The 3d orbitals of carbon are so high in energy that the amount of energy needed to form a set of sp3d2 hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF62 have never been prepared.
Example $3$: $\ce{OF4}$
What is the hybridization of the oxygen atom in OF4? Is OF4 likely to exist?
Given: chemical compound
Asked for: hybridization and stability
Strategy:
1. Predict the geometry of OF4 using the VSEPR model.
2. From the number of electron pairs around O in OF4, predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist.
Solution:
A The VSEPR model predicts that OF4 will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. B To accommodate five electron pairs, the O atom would have to be sp3d hybridized. The only d orbital available for forming a set of sp3d hybrid orbitals is a 3d orbital, which is much higher in energy than the 2s and 2p valence orbitals of oxygen. As a result, the OF4 molecule is unlikely to exist. In fact, it has not been detected.
Exercise $3$
What is the hybridization of the boron atom in $BF_6^{3−}$? Is this ion likely to exist?
Answer a
sp3d2 hybridization; no
Summary
Hybridization increases the overlap of bonding orbitals and explains the molecular geometries of many species whose geometry cannot be explained using a VSEPR approach. The localized bonding model (called valence bond theory) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals (hybrids) to maximize the overlap with adjacent atoms. The formation of hybrid atomic orbitals can be viewed as occurring via promotion of an electron from a filled ns2 subshell to an empty np or (n − 1)d valence orbital, followed by hybridization, the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an ns and an np orbital gives rise to two equivalent sp hybrids oriented at 180°, whereas the combination of an ns and two or three np orbitals produces three equivalent sp2 hybrids or four equivalent sp3 hybrids, respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two (n − 1)d orbitals to give sets of five sp3d or six sp3d2 hybrid orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/07%3A_Chemical_Bonding_II-_Valance_Bond_Theory_and_Molecular_Orbital_Theory/7.03%3A_Valence_Bond_Theory-_Hybridization_of_Atomic_Orbitals.txt |
Learning Objectives
• To use molecular orbital theory to predict bond order
None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model.
Previously, we described the electrons in isolated atoms as having certain spatial distributions, called orbitals, each with a particular orbital energy. Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs) A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a delocalized approach to bonding.
Molecular Orbital Theory: A Delocalized Bonding Approach
Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins.
Molecular Orbitals Involving Only ns Atomic Orbitals
We begin our discussion of molecular orbitals with the simplest molecule, H2, formed from two isolated hydrogen atoms, each with a 1s1 electron configuration. As we explained in Chapter 9, electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions. The 1s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1s wave functions, and the other produced by taking their difference:
$\begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{9.7.1}$
The molecular orbitals created from Equation $\ref{9.7.1}$ are called linear combinations of atomic orbitals (LCAOs) Molecular orbitals created from the sum and the difference of two wave functions (atomic orbitals). A molecule must have as many molecular orbitals as there are atomic orbitals.
A molecule must have as many molecular orbitals as there are atomic orbitals.
Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure $1$). In a sigma (σ) orbital, (i.e., a bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry), the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived: The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument.
$\sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \label{9.7.2}$
Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure $1$). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $\sigma _{1s}^{*}$ (“sigma one ess star”). In a sigma star (σ*) orbital An antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis., there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
$\sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \label{9.7.3}$
The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions. In contrast, electrons in the $\sigma _{1s}^{\star }$ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $\sigma _{1s}^{\star }$ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wave functions).
Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.
Energy-Level Diagrams
Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the $\sigma _{1s}^{\star }$ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the $\sigma _{1s}^{\star }$ molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure $2$
A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).
To describe the bonding in a homonuclear diatomic molecule (a molecule that consists of two atoms of the same element) such as H2, we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure $2$). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ1s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds.
Bond Order in Molecular Orbital Theory
In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order. In the molecular orbital approach, bond order One-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons:
$bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{9.7.4}$
To calculate the bond order of H2, we see from Figure $2$ that the σ1s (bonding) molecular orbital contains two electrons, while the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is empty. The bond order of H2 is therefore
$\dfrac{2-0}{2}=1 \label{9.7.5}$
This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure $2$ to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule.
Figure $\PageIndex{3a}$ shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the σ1s bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is
$\dfrac{1-0}{2}=1/2 \nonumber$
Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 1/2 the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in Table $1$, these predictions agree with the experimental data.
Figure $\PageIndex{3b}$ is the molecular orbital energy-level diagram for He2+. This ion has a total of three valence electrons. Because the first two electrons completely fill the σ1s molecular orbital, the Pauli principle states that the third electron must be in the $\sigma _{1s}^{\star}$ antibonding orbital, giving a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration. This electron configuration gives a bond order of
$\dfrac{2-1}{2}=1/2 \nonumber$
As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions (Table $1$).
Table $1$: Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions
Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ 1s)1 1/2 106 269
H2 1s)2 1 74 436
He2+ $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ 1/2 108 251
He2 $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ 0 not observed not observed
Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Figure $\PageIndex{3c}$ is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the σ1s bonding and $\sigma _{1s}^{\star }$ antibonding orbitals must contain two electrons. This gives a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He2 molecule has no net bond and is not a stable species. Experiments show that the He2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions.
In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H2+ ion.
In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons.
Example $1$
Use a molecular orbital energy-level diagram, such as those in Figure $2$, to predict the bond order in the He22+ ion. Is this a stable species?
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and stability
Strategy:
1. Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbital
2. s. Draw the molecular orbital energy-level diagram for the system.
3. Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
4. Calculate the bond order and predict whether the species is stable.
Solution:
A Two He 1s atomic orbitals combine to give two molecular orbitals: a σ1s bonding orbital at lower energy than the atomic orbitals and a $\sigma _{1s}^{\star }$ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:
B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:
The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ1s) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is
$\dfrac{2-0}{2} =1 \nonumber$
He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.
Exercise $1$
Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H22 ion. Is this a stable species?
Answer
H22 has a valence electron configuration of $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ with a bond order of 0. It is therefore predicted to be unstable.
So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.
A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in Figure $4$, where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σ*ns antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule in Figure $4$. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.
Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule in part (c) in Figure $2$. As shown in part (b) in Figure $4$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is stable.
Example $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability
Strategy:
1. Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.
2. Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals. B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* and a $\left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1}$ electron configuration.
C The bond order is (2-1)÷2=1/2 With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.
Exercise $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.
Answer
Ca2+ has a $\left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1}$ electron configurations and a bond order of 1/2 and should exist.
Molecular Orbitals Formed from ns and np Atomic Orbitals
Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are npx, npy, and npz orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations.
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.6}$
Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in Figure $5$, it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $\sigma _{np_{z}}$ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{9.7.7A}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{9.7.7}$
In this combination, shown in part (b) in Figure $5$, the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right )$ antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).
Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals.
The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure $6$, we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals). The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..
$\pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{9.7.8}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{9.7.9}$
The two npy orbitals can also combine using side-to-side interactions to produce a bonding $\pi _{np_{y}}$ molecular orbital and an antibonding $\pi _{np_{y}}^{\star }$ molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the $\pi _{np_{x}}$ and $\pi _{np_{y}}$molecular orbitals are a degenerate pair, as are the $\pi _{np_{x}}^{\star }$ and $\pi _{np_{y}}^{\star }$ molecular orbitals.
Figure $7$ is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.
Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in Figure $8$, the sum of the two atomic wave functions (ns + npz) produces a σ bonding molecular orbital. Their difference (nsnpz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.
Summary
Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach. A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals.
A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely antibonding molecular orbital contains at least one node perpendicular to the internuclear axis. A sigma (σ) orbital (bonding) or a sigma star (σ*) orbital (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a pi (π) orbital (bonding) and a pi star (π*) orbital (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane.
The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an energy-level diagram. The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the bond order, defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist.
Contributors and Attributions
1. Orbitals or orbital lobes with the same sign interact to give increased electron probability along the plane of the internuclear axis because of constructive reinforcement of the wave functions. Consequently, electrons in such molecular orbitals help to hold the positively charged nuclei together. Such orbitals are bonding molecular orbitals, and they are always lower in energy than the parent atomic orbitals.
2. Orbitals or orbital lobes with opposite signs interact to give decreased electron probability density between the nuclei because of destructive interference of the wave functions. Consequently, electrons in such molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. These orbitals are called antibonding molecular orbitals, and they are always higher in energy than the parent atomic orbitals.
3. Some atomic orbitals interact only very weakly, and the resulting molecular orbitals give essentially no change in the electron probability density between the nuclei. Hence electrons in such orbitals have no effect on the bonding in a molecule or ion. These orbitals are nonbonding molecular orbitals, and they have approximately the same energy as the parent atomic orbitals.
Learning Objectives
• To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period.
If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2.
four key points to remember when drawing molecular orbital diagrams:
1. The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the "law of conservation of orbitals").
2. As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases.
3. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized.
4. The interaction between atomic orbitals is greatest when they have the same energy.
We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in part (a) in Figure $1$; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is $\sigma _{2p_{z}}$ and the next most stable are the two degenerate orbitals, $\pi _{2p_{x}}$ and $\pi _{2p_{y}}$. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $\sigma ^{\star }_{2p_{z}}$ orbital is higher in energy than either of the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.
Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the $\sigma _{2p_{z}}$ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F2 is predicted to have a stable F–F single bond, in agreement with experimental data.
We now turn to a molecular orbital description of the bonding in O2. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O2, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure $1$. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ2s and σ2s* orbitals, 2 more to fill the $\sigma _{2p_{z}}$ orbital, and 4 to fill the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π* orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K.
None of the other bonding models can predict the presence of two unpaired electrons in O2. Chemists had long wondered why, unlike most other substances, liquid O2 is attracted into a magnetic field. As shown in Figure $2$, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed.
The magnetic properties of O2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H2O, CO2, and N2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H2O, CO2, and N2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O2 with organic compounds to give H2O, CO2, and N2 would require that at least one of the electrons on O2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier. Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke!
For period 2 diatomic molecules to the left of N2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $\sigma _{2p_{z}}$ molecular orbital is slightly higher in energy than the degenerate $\pi ^{\star }_{np_{x}}$ and $\pi ^{\star }_{np_{y}}$ orbitals. The difference in energy between the 2s and 2p atomic orbitals increases from Li2 to F2 due to increasing nuclear charge and poor screening of the 2s electrons by electrons in the 2p subshell. The bonding interaction between the 2s orbital on one atom and the 2pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ2s orbital and increases the energy of the $\sigma _{2p_{z}}$ orbital. Thus for Li2, Be2, B2, C2, and N2, the $\sigma _{2p_{z}}$ orbital is higher in energy than the $\sigma _{3p_{z}}$ orbitals, as shown in Figure $3$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $3$ ). Thus for example, the $\sigma _{2p_{z}}$ molecular orbital is at a lower energy than the $\pi _{2p_{x,y}}$ pair.
Completing the diagram for N2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N2 versus 141.2 pm in F2), and the bond energy is much greater for N2 than for F2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N2 bond is much shorter and stronger than the F2 bond, consistent with what we would expect when comparing a triple bond with a single bond.
Example $3$: Diatomic Sulfur
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons
Strategy:
1. Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another.
2. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2.
3. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule.
4. Calculate the bond order and describe the bonding.
Solution:
A Sulfur has a [Ne]3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $1$ and Figure $3$, we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the $\pi _{3p_{x,y}}$ or the $\sigma _{3p_{z}}$> molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases (Figure $3$), the $\sigma _{3p_{z}}$ molecular orbital will be lower in energy than the $\pi _{3p_{x,y}}$ pair.
B The molecular orbital energy-level diagram is as follows:
Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons.
C Ten valence electrons are used to fill the orbitals through $\pi _{3p_{x}}$ and $\pi _{3p_{y}}$, leaving 2 electrons to occupy the degenerate $\pi ^{\star }_{3p_{x}}$ and $\pi ^{\star }_{3p_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S2 is $\left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2}$ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond.
Exercise $3$: The Peroxide Ion
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22).
Answer
$\left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4}$ bond order of 1; no unpaired electrons
Molecular Orbitals for Heteronuclear Diatomic Molecules
Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $4$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.
A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.
An Odd Number of Valence Electrons: NO
Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.
Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $13$) shows that the general pattern is similar to that for the O2 molecule (Figure $11$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.
Note that electronic structure studies show the ground state configuration of $\ce{NO}$ to be $\left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{1}$ in order of increasing energy. Hence, the $\pi _{2p_{x,y}}$ orbitals are lower in energy than the $\sigma _{2p_{z}}$ orbital. This is because the $\ce{NO}$ molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond (Figure $11$).
Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $13$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond.
Nonbonding Molecular Orbitals
Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $6$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $H^{\delta +} -- Cl^{\delta -}$.
Electrons in nonbonding molecular orbitals have no effect on bond order.
Example $4$: The Cyanide Ion
Use a “skewed” molecular orbital energy-level diagram like the one in Figure $4$ to describe the bonding in the cyanide ion (CN). What is the bond order?
Given: chemical species
Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order
Strategy:
1. Calculate the total number of valence electrons in CN. Then place these electrons in a molecular orbital energy-level diagram like Figure $4$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
2. Calculate the bond order and describe the bonding in CN.
Solution:
A The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $4$ fills the five lowest-energy orbitals, as shown here:
Because $\chi_N > \chi_C$, the atomic orbitals of N (on the right) are lower in energy than those of C.
B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N2.
Exercise $4$: The Hypochlorite Ion
Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl). What is the bond order?
Answer
All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1.
Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods.
Summary
Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules, molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules, using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/07%3A_Chemical_Bonding_II-_Valance_Bond_Theory_and_Molecular_Orbital_Theory/7.04%3A_Molecular_Orbital_Theory-_Electron_Delocalization.txt |
• 8.1: Climate Change and the Combustion of Fossil Fuels
• 8.2: Chemical Change
• 8.3: Writing and Balancing Chemical Equations
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right.
• 8.4: Reaction Stoichiometry - How Much Carbon Dioxide?
A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.
• 8.5: Limiting Reactant, Theoretical Yield, and Percent Yield
The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained. The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first is the limiting reagent.
• 8.6: Three Examples of Chemical Reactions- Combustion, Alkali Metals, and Halogens
Thumbnail: Copper from a wire is displaced by silver in a silver nitrate solution it is dipped into, and solid silver precipitates out. (CC BY-SA 3.0; Toby Hudson)
08: Chemical Reactions and Chemical Quantities
Prince George's Community College
General Chemistry for Engineering
CHM 2000
Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium
Unit VI: Thermo & Electrochemistry
Unit VII: Nuclear Chemistry
Learning Objectives
• To use thermochemical concepts to discuss environmental issues.
Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 106 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States.
In contrast, the average energy consumption elsewhere in the world is about 105 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of fossil fuel energy and their impact on the environment.
Driven by environmental concerns about climate change and pollution, the world is undergoing a transformation from fossil fuels to renewable resources such as solar, and wind. The role that hydro and nuclear energy will play is uncertain and especially in the later case a policy rather than a scientific issue.
Fossil Fuels
According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another.
Fossil fuels, coal, oil and natural gas are the result of anaerobic decay of dead plants and animals laid down hundreds of millions of years ago, most of which took place well before the dinosaurs strode the earth. Fossil fuels slowly formed as further geological layers compressed and heated the dead organic matter. The energy content of fossil fuels results from the transformation of sunlight into vegetation and the chemical transformation brought about by anaerobic cooking at high pressures and temperatures over geological times.
Figure 15.7.1 represents a plant for generating electricity using oil or coal where the fuel is burned in a boiler, superheating steam which then powers a turbine for electrical generation. Oil derived fuels are seldom used in large power plants but diesel is used commonly in small electrical generators either in remote locations or as back up for when electrical distribution systems fail. Natural gas fueled power plants burn the fuel directly in the turbine which is similar to a jet engine. Coal power plants can convert ~40% of the energy released from combustion to electricity. In comparison, nuclear power plants can be more than 50% efficient and gas turbines can approach 60% mostly due to higher operating temperatures. Co-generation, using the plant to produce not only electricity but also heat for industrial or other purposes can raise overall efficiency by 10 - 15 % or so.
Figure 15.7.1 Electricity from Coal A coal-powered electric power plant uses the combustion of coal to produce steam, which drives a turbine to produce electricity.
The total expenditure of energy in the world each year is about 3 × 1017 kJ. Today, more than 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas (The sources of the energy consumed in the United States in 2009 are shown in Figure 15.7.2.) but as Table 15.7.1 from the Wikipedia shows, energy usage is a complex issue. Petroleum dominates as a source of energy for transportation because gasoline is easy to transport, but is very little used for electrical generation, whereas 91% of coal is used for electrical generation. The other major use of coal is as a reducing agent for metal refining from ores. The former is called thermal coal, the latter metallurgical coal.
Table 15.7.1: Energy useage in the United States for 2008
Supply Sources Percent of Source Demand Sectors Percent of Sector
Petroleum
37.1%
71% Transportation
23% Industrial
5% Residential and Commercial
1% Electric Power
Transportation
27.8%
95% Petroleum
2% Natural Gas
3% Renewable Energy
Natural Gas
23.8%
3% Transportation
34% Industrial
34% Residential and Commercial
29% Electric Power
Industrial
20.6%
42% Petroleum
40% Natural Gas
9% Coal
10% Renewable Energy
Coal
22.5%
8% Industrial
<1% Residential and Commercial
91% Electric Power
Residential and Commercial
10.8%
16% Petroleum
76% Natural Gas
1% Coal
1% Renewable Energy
Renewable Energy
7.3%
11% Transportation
28% Industrial
10% Residential and Commercial
51% Electric Power
Electric Power
40.1%
1% Petroleum
17% Natural Gas
51% Coal
9% Renewable Energy
21% Nuclear Electric Power
Nuclear Electric Power
8.5%
100% Electric Power
Figure 15.7.2 Energy Consumption in the United States by Source, 2009
More than 80% of the total energy expended is provided by the combustion of fossil fuels, such as oil, coal, and natural gas.
Coal
CoalA complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. It is used as a fuel. was primarily laid down from the large swamp forests of the Carboniferous Period. Coal deposits are found today where those forests were Coal is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of cellulose, derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure 15.7.3 ). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one.
Figure 15.7.3 The Structures of Cellulose and Coal (a) Cellulose consists of long chains of cyclic glucose molecules linked by hydrogen bonds. (b) When cellulose is subjected to high pressures and temperatures for long periods of time, water is eliminated, and bonds are formed between the rings, eventually producing coal. This drawing shows some of the common structural features of coal; note the presence of many different kinds of ring structures.
There are four distinct classes of coal (Table 15.7.2); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest ΔHcomb. Lignite is extensively mined in Germany and Poland. Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest ΔHcomb and is the highest grade of coal. Anthracite is the first choice for metallurgical refining. The most abundant form in the Western United States in anthracite while that in the Eastern United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS2). Combustion of coal releases the sulfur in FeS2 as SO2, which is a major contributor to acid rain. Table 15.7.3 compares the ΔHcomb per gram of oil, natural gas, and coal with those of selected organic compounds.
Table 15.7.2 Properties of Different Types of Coal
Type % Carbon Hydrogen:Carbon Mole Ratio % Oxygen % Sulfur Heat Content US Deposits
anthracite 92 0.5 3 1 high Pennsylvania, New York
bituminous 80 0.6 8 5 medium Appalachia, Midwest, Utah
subbituminous 77 0.9 16 1 medium Rocky Mountains
lignite 71 1.0 23 1 low Montana
Table 15.7.3 Enthalpies of Combustion of Common Fuels and Selected Organic Compounds
Fuel ΔHcomb (kJ/g)
dry wood −15
peat −20.8
bituminous coal −28.3
charcoal −35
kerosene −37
C6H6 (benzene) −41.8
crude oil −43
natural gas −50
C2H2 (acetylene) −50.0
CH4 (methane) −55.5
gasoline −84
hydrogen −143
Peat, a precursor to coal, is the partially decayed remains of plants that grow in the swampy areas of the Carboniferous Period. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poor-burning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure 15.7.4). If a peat bog were buried under many layers of sediment for a few million years, the peat would eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel.
Figure 15.7.4 A Peat Bog Peat is a smoky fuel that burns poorly and produces little heat, but it has been used as a fuel since ancient times.
Converting Coal to Gaseous and Liquid Fuels
As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, more than 75% of the coal produced each year is simply burned in power plants to produce electricity. Methods to convert coal to gaseous fuels (coal gasification) or liquid fuels (coal liquefaction) exist, but are not particularly economical unless the prices of oil and natural gas are high. With the development of fracking and the subsequent fall in oil and natural gas prices interest in these processes has fallen however they have played an important role in the past. In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H2 known as synthesis gas, or syngas:Because coal is 70%–90% carbon by mass, it is approximated as C in Equation 15.7.1
$C\left ( s \right ) + H_{2}O \left ( g \right ) \rightarrow CO \left ( g \right ) + H_{2} \left ( g \right ) \; \; \; \; \Delta H= 131 \; kJ \tag{15.8.1}$
Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases. Syngas or town gas was used for cooking until the 1960s when natural gas pipelines were built. Because syngas contains carbon monoxide (CO) it is poisonous, which accounts for scenes in old movies where people were killed by sticking their heads into an oven and allowing the gas to flow.
Syngas is can also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions:
$\begin{matrix} 2C\left ( s \right )+2H_{2}O\left ( g \right ) \rightarrow \cancel{2CO\left ( g \right )}+\cancel{2H_{2}\left ( g \right )} & \Delta H_{1}=262 \; kJ \ \cancel{CO\left ( g \right )}+\cancel{H_{2}O \left ( g \right )} \rightarrow CO_{2}\left ( g \right )+\cancel{H_{2}\left ( g \right )} & \Delta H_{2}=-41 \; kJ \ \cancel{CO\left ( g \right )}+\cancel{3H_{2}\left ( g \right )} \rightarrow CH_{4}\left ( g \right )+\cancel{H_{2}O\left ( g \right )} & \Delta H_{3}=-206 \; kJ \ -------------------&--------\ 2C\left ( s \right )+2H_{2}O\left ( g \right ) \rightarrow CH_{4}\left ( g \right ) + CO_{2}\left ( g \right ) & \Delta H_{3}=15 \; kJ \end{matrix} \tag{15.8.2}$
Techniques available for converting coal to liquid fuels are not economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Subsequent distillation, cracking, and reforming can be used to create products similar to those obtained from petroleum.
Petroleum
The petroleum that is pumped out of the ground is a complex mixture of several thousand organic compounds including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundences of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.
Gasoline
Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 15.7.5 shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 15.7.5 shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs.
Figure 15.7.5: The Distillation of Petroleum. (a) This is a diagram of a distillation column used for separating petroleum fractions. (b) Petroleum fractions condense at different temperatures, depending on the number of carbon atoms in the molecules, and are drawn off from the column. The most volatile components (those with the lowest boiling points) condense at the top of the column, and the least volatile (those with the highest boiling points) condense at the bottom.
The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking, in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Metal catalysts such as platinum are used to drive the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.
Octane Ratings
The quality of a fuel is indicated by its octane rating, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (Figure 15.8.6), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.
Figure 15.76: The Burning of Gasoline in an Internal Combustion Engine. (a) Normally, fuel is ignited by the spark plug, and combustion spreads uniformly outward. (b) Gasoline with an octane rating that is too low for the engine can ignite prematurely, resulting in uneven burning that causes knocking and pinging.
The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture:
$0.89(100)+0.11(0)=89 \tag{4.7.1}$
A gasoline that performs at the same level as a blend of 89% isooctane and 11% n-heptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher.
As shown in Figure 4.7.3, many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyllead [(C2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic.
Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place that combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing.
Figure 15.7.6: The Octane Ratings of Some Hydrocarbons and Common Additives
Natural Gas
Natural gas is a (mostly) combustible gas found underground. While primarily composed of methane (70-90%) the gas from each well has a different composition and the value of the other components affects the value of the gas. The gas from wells that are rich in methane is called dry and wells that have a considerable amount of higher hydrocarbons produce wet gas. The higher hydrocarbons have value above that of methane so stripping them out is important. Some wells are sour because their gas has hydrogen sulfide which must be removed before the gas can be used for heating or generating electricity.
Finally, a few wells in Texas and nearby Oklahoma have a relatively high amount of helium (0.3 - 2.7%). The Helium Act of 1925 established a national helium reserve at Cliffside near Amarillo TX. Political pressure and costs pushed laws to privatize the reserve, but other policy considerations including the need for helium for scientific research has slowed the process.
Table 15.8.4: Composition of Natural Gas
Gas Molecular Formula Composition
Methane CH4 70-95%
Ethane C2H6 0-20%
Carbon Dioxide CO2 0-8%
Nitrogen N2 0-5%
Hydrogen Sulfide H2S 0-5%
Propane C3H8 Traces
Butane C4H10 Traces
Rare Gases He (also Ne) 0-3% (only in Texas)
The purification of natural gas is a complex process with many steps as each of the impurities is stripped out
Figure 15.7.7: Purification of natural gas: Prospective Chemical Engineers can find details about each process by searching the net. The flow chart is from the Wikipedia
Gas turbine power plants to generate electricity are coming increasingly into use as fracking and other advanced drilling technologies have driven the cost of natural gas down and the supply up. While on a continental scale natural gas is transported by pipelines, natural gas can be cooled and compressed to be transported as liquified natural gas. Gas turbine power plants are small and quickly built. They can be rapidly spun up to meet peak demand. More detailed information can be found at the Department of Energy Fossil Fuel web site
The Carbon Cycle and the Greenhouse Effect
Since 1850 the burning of fossil fuels has increased the concentration of carbon dioxide in the atmosphere from 280 to just over 400 ppmV. A continued increase in the CO2 burden in the atmosphere will have serious negative effects and this requires shifting our entire energy producing economy from fossil fuels to non-carbon sources such as hydro, solar, wind and nuclear. These include sea level rise that will threaten low lying cities including but not limited to Miami Beach and Norfolk in the US, even a meter or more coupled with storm surge and high tides can cause massive damage as was seen during Hurricane Sandy. Increased carbon dioxide in the atmosphere has already measurably decreased the pH of the oceans. Sea life is adapted to a narrow range of pH.
Higher global temperatures of 2 or 3 C may not seem much, but one should keep in mind that the average global temperature during the ice ages was only ~6 C lower than it is today. During the Eemian interglacial the average temperature was only a few degrees higher than the present and the sea level was 6-9 m higher. Finally, humans are mammals who maintain a core temperature within a few degrees of 37.0 C. In hot weather we do so by evaporation of sweat however there are limits to this and by 2100 there is a significant probability even in the US that at least a few days a year will reach this limit by 2100. Given that most people on earth do not have access to air conditioning, parts of the planet may become uninhabitable. Indeed a worst case and a serious problem that merits attention.
Given the constraints of this text it is difficult to provide the level of detail needed to understand why this is so. A good source for those interested in learning more is David Archer's Global Warming, Understanding the Forecast and An Introduction to Modern Climate Change by Andrew Dessler
There are a few basic facts that anyone starting to learn about the issue need to know. First, that the Earth gains energy from the Sun, and that it must radiate the energy at the same rate. If more energy is absorbed than radiated the Earth will warm up, if less energy is absorbed than radiated it will cool. As discussed in Section 15.1 solar radiation follows a 5500 K blackbody distribution while radiation from the surface of the earth is also black body but at ~290K. This is shown schematically in Figure 15.8.8
Figure 15.7.8: Origin of the greenhouse effect. The top box show the intensity of solar radiation at the top of the Earth's atmosphere to the left and the infrared emission from the atmosphere to the left. The next shows the net absorption in the atmosphere across the spectrum. The bottom show the absorption by different atmospheric species at their current concentrations. Figure from the Wikipedia
The atmosphere cools with altitude up to about 15 km where it starts warming again because of absorption of UV radiation by ozone in the stratosphere. The transition between the troposphere and the stratosphere is called the tropopause and is the coldest part of the atmosphere.
Figure 15.7.9 The thermal structure of the atmosphere. The troposphere is the lowest level,extending from the the surface to the tropopause which is at 12-15 km altitude, more in the tropics and less less at the poles. The troposphere is the section of the atmosphere relevant to the greenhouse effect. The ozone layer is found in the stratosphere and one observes the aurora borealis or northern lights in the mesosphere Taken from Wikimedia
Figure 15.7.10 shows the IR emission spectrum observed looking down on the earth from a high altitude balloon
Figure 15.7.10: High resolution far infrared emission spectrum of the atmosphere looking down from a high altitude balloon at 35 km. The strong band between 600 and 700 cm-1 is associated with CO2, the sharp band at ~ 1100 cm-1 is associated with ozone and the feature near 1300 cm-1 is due to methane. The sharp lines (the hair) are due to water vapor. Spectrum is from the Federal Earth Observation Portal and was taken by an instrument operated by NASA/Langeley Research Center
The dotted lines in Figure 15.8.9 are blackbody curves. The IR window shown schematically in blue in Figure 15.8.8 is the region between the ozone and the carbon dioxide band, where the emission from the hot, 320 K ground follows the blackbody curve with a few sharp water vapor absorption lines. At those wavelengths, in the IR window, the emission comes directly from the surface. The CO2 band extends down to about a 220 K blackbody curve. What this means is that radiation from CO2 only escapes to space from the level in the troposphere where that is the temperature. The rate of emission is proportional to T4 so the rate of emission from higher, therefore colder levels, is slower. Radiation in this area of the spectrum from the surface is blocked and only the greatly reduced emission from the upper troposphere escapes to space. The same is true for the ozone and methane bands as well as the water lines.
The net effect is that the surface must warm in order to maintain the balance between incoming solar radiation and the outgoing emission. There is a simple calculation which models the atmosphere as a one dimensional problem and calculates what the temperature of the surface would be if there were no greenhouse gases. The result is 255 K, rather cold. In fact if one attempts a more complex calculation the effective temperature without greenhouse gases would be even colder.
What happens if we increase the carbon dioxide in the atmosphere? The altitude at which the atmosphere can emit radiation to space will rise because of increased absorption by the CO2. Since in the troposphere the temperature decreases with altitude, the rate of emission from a higher level must decrease. Again, in order to maintain the balance between incoming solar radiation and the outgoing emission the surface will have to warm even more, thus the term global warming. The change is not linear with increasing CO2 but logarithmic. But, of course it is not so simple, because increasing the surface temperature will increase the water vapor pressure in the atmosphere, which will increase the temperature further.
We also have to understand the flow of carbon between the atmosphere, the biosphere, the upper oceans and the deep. Observations to date show that natural emissions of CO2 from these reservoirs are in balance with absorption, while only about half of fossil fuel emissions remain in the atmosphere, the rest being absorbed by the upper ocean and the biosphere. The three upper reservoirs equilibrate in a decade or less, but flow into the deep ocean requires roughly a thousand years.
Figure 15.7.11: The carbon cycle. Figure from the Wikipedia
There is no doubt that atmospheric CO2 levels are increasing, and the major reason for this increase is the combustion of fossil fuels. An extremely conservative statement of the situation today can be found in the 2014 Synthesis Report from the IPCC, a consensus between scientists and policymakers. The report starkly states that
Cumulative emissions of CO2 largely determine global mean surface warming by the late 21st century and beyond.
and concludes that
Continued emission of greenhouse gases will cause further warming and long-lasting changes in all components of the climate system, increasing the likelihood of severe, pervasive and irreversible impacts for people and ecosystems. Limiting climate change would require substantial and sustained reductions in greenhouse gas emissions which, together with adaptation, can limit climate change risks.
The situation is serious, but we can work together to limit and even reverse damage while maintaining our standard of living in the developed world while helping the developing world to a better future. However, the issues are complex and we can only touch on some of the basics here.
Summary
More than 80% of the energy used by modern society (about 3 × 1017 kJ/yr) is from the combustion of fossil fuels. Because of their availability, ease of transport, and facile conversion to convenient fuels, natural gas and petroleum are currently the preferred fuels. Coal is primarily used for electricity generation. The combustion of fossil fuels releases large amounts of CO2 that upset the balance of the carbon cycle and result in a steady increase in atmospheric CO2 levels. Because CO2 is a greenhouse gas, which absorbs heat before it can be radiated from Earth into space, CO2 in the atmosphere results in increased surface temperatures (the greenhouse effect).
Key Takeaway
• Thermochemical concepts can be used to calculate the efficiency of various forms of fuel, which can then be applied to environmental issues.
Conceptual Problems
1. What is meant by the term greenhouse gases? List three greenhouse gases that have been implicated in global warming.
2. Name three factors that determine the rate of planetary CO2 uptake.
3. The structure of coal is quite different from the structure of gasoline. How do their structural differences affect their enthalpies of combustion? Explain your answer.
Numerical Problems
1. One of the side reactions that occurs during the burning of fossil fuels is
4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)
1. How many kilojoules of energy are released during the combustion of 10 lb of FeS2?
2. How many pounds of SO2 are released into the atmosphere?
3. Discuss the potential environmental impacts of this combustion reaction.
2. How many kilograms of CO2 are released during the combustion of 16 gal of gasoline? Assume that gasoline is pure isooctane with a density of 0.6919 g/mL. If this combustion was used to heat 4.5 × 103 L of water from an initial temperature of 11.0°C, what would be the final temperature of the water assuming 42% efficiency in the energy transfer?
3. A 60 W light bulb is burned for 6 hours. If we assume an efficiency of 38% in the conversion of energy from oil to electricity, how much oil must be consumed to supply the electrical energy needed to light the bulb? (1 W = 1 J/s)
4. How many liters of cyclohexane must be burned to release as much energy as burning 10.0 lb of pine logs? The density of cyclohexane is 0.7785 g/mL, and its ΔHcomb = −46.6 kJ/g.
Contributors
• Anonymous
Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/08%3A_Chemical_Reactions_and_Chemical_Quantities/8.01%3A_Climate_Change_and_the_Combustion_of_Fossil_Fuels.txt |
Learning Objectives
• To describe a chemical reaction.
• To calculate the quantities of compounds produced or consumed in a chemical reaction
What happens to matter when it undergoes chemical changes? The Law of conservation of mass says that "Atoms are neither created, nor destroyed, during any chemical reaction." Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements").
Chemical Equations
As shown in Figure $1$, applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. This reaction is described with a chemical equation, an expression that gives the identities and quantities of the substances in a chemical reaction.
Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H2O). The chemical equation for this reaction is written as:
$\ce{2H_2 + O_2 \rightarrow 2H_2O} \nonumber$
Chemical formulas and other symbols are used to indicate the starting materials, or reactants, which by convention are written on the left side of the equation, and the final compounds, or products, which are written on the right. An arrow points from the reactant to the products. The chemical reaction for the ammonium dichromate volcano in Figure $1$ is
$\underbrace{\ce{(NH_4)_2Cr_2O_7}}_{ reactant } \rightarrow \underbrace{\ce{Cr_2O_3 + N_2 + 4H_2O}}_{products }\label{3.1.1}$
The arrow is read as “yields” or “reacts to form.” Equation $\ref{3.1.1}$ indicates that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products). The equation for this reaction is even more informative when written as follows:
$\ce{ (NH4)2Cr2O7(s) \rightarrow Cr2O3(s) + N2(g) + 4H2O(g)} \label{3.1.2}$
Equation $\ref{3.1.2}$ is identical to Equation $\ref{3.1.1}$ except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water.
Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of Equations $\ref{3.1.1}$ and $\ref{3.1.2}$. Each side of the reaction has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms.
In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equations $\ref{3.1.1}$ and $\ref{3.1.2}$ are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms, but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.
A chemical reaction changes only the distribution of atoms, not the number of atoms.
Introduction to Chemical Reaction Equations: Introduction to Chemical Reaction Equations, YouTube(opens in new window) [youtu.be]
Balancing Simple Chemical Equations
When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane ($C_7H_{16}$), an important component of gasoline:
$\ce{C_7H_{16} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.3}$
The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water.
Equation $\ref{3.1.3}$ is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, the coefficients of the reactants and products must be adjusted to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $3$.
Balancing Combustion Reactions: Balancing Combustions Reactions, YouTube(opens in new window) [youtu.be]
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation
1. Identify the most complex substance.
2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
3. Balance polyatomic ions (if present) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Check your work by counting the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
Example $\PageIndex{1A}$: Combustion of Heptane
To demonstrate this approach, let’s use the combustion of n-heptane (Equation $\ref{3.1.3}$) as an example.
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $\ce{C_7H_{16}}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:
$\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O } \label{3.1.4}$
1. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
2. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: $\ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O} \label{3.1.5}$ The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:$\ce{C_7H_{16} (l) + 11O_2 (g) \rightarrow 7CO_2 (g) + 8H_2O (g)} \label{3.1.6}$
3. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start.
Example $\PageIndex{1B}$: Combustion of Isooctane
Consider, for example, a similar reaction, the combustion of isooctane ($\ce{C8H18}$). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows:
$\ce{C_8H_{18} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.7}$
1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane.
2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + H_2O} \label{3.1.8}$
3. Balance polyatomic ions as a unit. This step does not apply to this equation.
4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:$\ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + 9H_2O } \label{3.1.9}$The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: $\ce{C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O} \label{3.1.10}$Equation $\ref{3.1.10}$ is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: $\ce{2C_8H_{18} (l) + 25O_2 (g) \rightarrow 16CO_2 (g) + 18H_2O (g) }\label{3.11}$
5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.
Balancing Complex Chemical Equations: Balancing Complex Chemical Equations, YouTube(opens in new window) [youtu.be]
Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
Example $\PageIndex{1C}$: Hydroxyapatite
The reaction of the mineral hydroxyapatite ($\ce{Ca5(PO4)3(OH)}$) with phosphoric acid and water gives $\ce{Ca(H2PO4)2•H2O}$ (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.
Given: reactants and product
Asked for: balanced chemical equation
Strategy:
1. Identify the product and the reactants and then write the unbalanced chemical equation.
2. Follow the steps for balancing a chemical equation.
Solution:
A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem (recall that phosphoric acid is H3PO4). The initial (unbalanced) equation is as follows:
$\ce{ Ca5(PO4)3(OH)(s) + H_3PO4 (aq) + H_2O_{(l)} \rightarrow Ca(H_2PO_4)_2 \cdot H_2O_{(s)} } \nonumber$
1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, $\ce{Ca5(PO4)3(OH)}$, appears in the balanced chemical equation.
2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of $\ce{Ca5(PO4)3(OH)}$ contains 5 calcium atoms, so we need 5 Ca(H2PO4)2•H2O on the right side:
$\ce{Ca5(PO4)3(OH) + H3PO4 + H2O \rightarrow 5Ca(H2PO4)2 \cdot H2O} \nonumber$
3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43), shows up in three places. In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2 • H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4:
$\ce{Ca_5(PO_4)_3(OH) + 7H_3PO_4 + H_2O \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O } \nonumber$
Although OH is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately.
4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules:
$\ce{Ca_5(PO_4)_3(OH) (s) + 7H_3PO_4 (aq) + 4H_2O (l) \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O (s) } \nonumber$
The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 41 oxygen atoms on each side.
5. Check your work. Both sides of the equation contain 5 calcium atoms, 10 phosphorus atoms, 30 hydrogen atoms, and 41 oxygen atoms.
Exercise $1$: Fermentation
Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol ($CH_3CH_2OH$ and carbon dioxide $CO_2$. Write a balanced chemical reaction for the fermentation of glucose.
Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.
Answer
$C_6H_{12}O_6(s) \rightarrow 2C_2H_5OH(l) + 2CO_2(g) \nonumber$
Balancing Reactions Which Contain Polyatomics: Balancing Reactions Which Contain Polyatomics, YouTube(opens in new window) [youtu.be]
Interpreting Chemical Equations
In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it gives the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficient of that species (e.g., the 4 preceding H2O in Equation $\ref{3.1.1}$). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure $4$, the coefficients allow Equation $\ref{3.1.1}$ to be interpreted in any of the following ways:
• Two NH4+ ions and one Cr2O72 ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules.
• One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O.
• A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O.
• A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O.
These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of $H_2O$ to $N_2$ in Equation $\ref{3.1.1}$ is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:
$252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \nonumber$
yield
$152 + 28 + 72 = 252 \; g \; \text{of products.} \nonumber$
The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.
An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose ($C_6H_{12}O_6$), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.
Example $2$: Combustion of Glucose
The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:
$\ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)} \nonumber$
Construct a table showing how to interpret the information in this equation in terms of
1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.
Given: balanced chemical equation
Asked for: molecule, mole, and mass relationships
Strategy:
1. Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.
2. Use the molar masses of the reactants and products to convert from moles to grams.
3. Use Avogadro’s number to convert from moles to the number of molecules.
Solution:
This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.
1. One molecule of glucose reacts with 6 molecules of O2 to yield 6 molecules of CO2 and 6 molecules of H2O.
2. One mole of glucose reacts with 6 mol of O2 to yield 6 mol of CO2 and 6 mol of H2O.
3. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O2, 31.9988; CO2, 44.010; and H2O, 18.015.
\begin{align*} \text{mass of reactants} &= \text{mass of products} \[4pt] g \, glucose + g \, O_2 &= g \, CO_2 + g \, H_2O \end{align*} \nonumber
$1\,mol\,glucose \left ( {180.16 \, g \over 1 \, mol \, glucose } \right ) + 6 \, mol \, O_2 \left ( { 31.9988 \, g \over 1 \, mol \, O_2} \right ) \nonumber$
$= 6 \, mol \, CO_2 \left ( {44.010 \, g \over 1 \, mol \, CO_2} \right ) + 6 \, mol \, H_2O \left ( {18.015 \, g \over 1 \, mol \, H_2O} \right ) \nonumber$
$372.15 \, g = 372.15 \, g \nonumber$
C One mole of glucose contains Avogadro’s number (6.022 × 1023) of glucose molecules. Thus 6.022 × 1023 glucose molecules react with (6 × 6.022 × 1023) = 3.613 × 1024 oxygen molecules to yield (6 × 6.022 × 1023) = 3.613 × 1024 molecules each of CO2 and H2O.
In tabular form:
Solution to Example 3.1.2
$C_6H_{12}O_{6\;(s)}$ + $6O_{2\;(g)}$ $6CO_{2\;(g)}$ $6H_2O_{(l)}$
a. 1 molecule 6 molecules 6 molecules 6 molecules
b. 1 mol 6 mol 6 mol 6 mol
c. 180.16 g 191.9928 g 264.06 g 108.09 g
d. 6.022 × 1023 molecules 3.613 × 1024 molecules 3.613 × 1024 molecules 3.613 × 1024 molecule
Exercise $2$: Ammonium Nitrate Explosion
Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas.
The explosion resulted from the following reaction:
$2NH_4NO_{3\;(s)} \rightarrow 2N_{2\;(g)} + 4H_2O_{(g)} + O_{2\;(g)} \nonumber$
Construct a table showing how to interpret the information in the equation in terms of
1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
Answer:
Answer to Exercise 3.1.2
$2NH_4NO_{3\;(s)}$ $2N_{2\;(g)}$ + $4H_2O_{(g)}$ + $O_{2\;(g)}$
a. 2NH4+ ions and 2NO3 ions 2 molecules 4 molecules 1 molecule
b. 2 mol 2 mol 4 mol 1 mol
c. 160.0864 g 56.0268 g 72.0608 g 31.9988 g
d. 1.204 × 1024 formula units 1.204 × 1024 molecules 2.409 × 1024 molecules 6.022 × 1023 molecules
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be]
Summary
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/08%3A_Chemical_Reactions_and_Chemical_Quantities/8.03%3A_Writing_and_Balancing_Chemical_Equations.txt |
Learning Objectives
• Explain the concept of stoichiometry as it pertains to chemical reactions
• Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
• Perform stoichiometric calculations involving mass, moles, and solution molarity
A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.
The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Cooking, for example, offers an appropriate comparison. Suppose a recipe for making eight pancakes calls for 1 cup pancake mix, $\dfrac{3}{4}$ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is
$\mathrm{1\:cup\: mix+\dfrac{3}{4}\:cup\: milk+1\: egg \rightarrow 8\: pancakes} \label{4.4.1}$
If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is
$\mathrm{24\: \cancel{pancakes} \times \dfrac{1\: egg}{8\: \cancel{pancakes}}=3\: eggs} \label{4.4.2}$
Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:
$\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g) \label{4.4.3}$
This equation shows that ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:
$\ce{\dfrac{2NH3 \: molecules}{3H2 \: molecules}\: or \: \dfrac{2 \: doz \: NH3\: molecules}{3\: doz\:H2 \:molecules} \: or \: \dfrac{2\: mol\: NH3\: molecules}{3\: mol\: H2\: molecules}} \label{4.4.4}$
These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.
Example $1$: Moles of Reactant Required in a Reaction
How many moles of I2 are required to react with 0.429 mol of Al according to the following equation (see Figure $2$)?
$\ce{2Al + 3I2 \rightarrow 2AlI3} \label{4.4.5}$
Solution
Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $\ce{\dfrac{3\: mol\: I2}{2\: mol\: Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:
\begin{align*} \mathrm{mol\: I_2} &=\mathrm{0.429\: \cancel{mol\: Al}\times \dfrac{3\: mol\: I_2}{2\:\cancel{mol\: Al}}} \[4pt] &=\mathrm{0.644\: mol\: I_2} \end{align*} \nonumber
Exercise $1$
How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation $\ce{3Ca(OH)2 + 2H3PO4 \rightarrow Ca3(PO4)2 + 6H2O}$ ?
Answer
2.04 mol
Example $2$: Number of Product Molecules Generated by a Reaction
How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?
$\ce{C3H8 + 5O2 \rightarrow 3CO2 + 4H2O} \label{4.4.6}$
Solution
The approach here is the same as for Example $1$, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number.
The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:
$\ce{\dfrac{3\: mol\: CO2}{1\: mol\: C3H8}} \label{4.4.7}$
Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,
$\mathrm{0.75\: \cancel{mol\: C_3H_8}\times \dfrac{3\: \cancel{mol\: CO_2}}{1\:\cancel{mol\:C_3H_8}}\times \dfrac{6.022\times 10^{23}\:CO_2\:molecules}{\cancel{mol\:CO_2}}=1.4\times 10^{24}\:CO_2\:molecules} \label{4.4.8}$
Exercise $1$
How many NH3 molecules are produced by the reaction of 4.0 mol of Ca(OH)2 according to the following equation:
$\ce{(NH4)2SO4 + Ca(OH)2 \rightarrow 2NH3 + CaSO4 + 2H2O} \label{4.4.9}$
Answer
4.8 × 1024 NH3 molecules
These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.
Example $3$: Relating Masses of Reactants and Products
What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?
$\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)$
Solution
The approach used previously in Examples $1$ and $2$ is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:
$\mathrm{16\:\cancel{g\: Mg(OH)_2} \times \dfrac{1\:\cancel{mol\: Mg(OH)_2}}{58.3\:\cancel{g\: Mg(OH)_2}}\times \dfrac{2\:\cancel{mol\: NaOH}}{1\:\cancel{mol\: Mg(OH)_2}}\times \dfrac{40.0\: g\: NaOH}{\cancel{mol\: NaOH}}=22\: g\: NaOH} \nonumber$
Exercise $3$
What mass of gallium oxide, Ga2O3, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $\ce{4Ga + 3O2 \rightarrow 2Ga2O3}$.
Answer
39.0 g
Example $4$: Relating Masses of Reactants
What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
$\ce{2C8H18 + 25O2 \rightarrow 16CO2 + 18H2O} \nonumber$
Solution
The approach required here is the same as for the Example $3$, differing only in that the provided and requested masses are both for reactant species.
$\mathrm{702\:\cancel{g\:\ce{C8H18}}\times \dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114.23\:\cancel{g\:\ce{C8H18}}}\times \dfrac{25\:\cancel{mol\:\ce{O2}}}{2\:\cancel{mol\:\ce{C8H18}}}\times \dfrac{32.00\: g\:\ce{O2}}{\cancel{mol\:\ce{O2}}}=2.46\times 10^3\:g\:\ce{O2}}$
Exercise $4$
What mass of CO is required to react with 25.13 g of Fe2O3 according to the equation $\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}$?
Answer
13.22 g
These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure $2$ provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.
Airbags
Airbags (Figure $3$) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN3 to initiate its decomposition:
$\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber$
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN3 will generate approximately 50 L of N2.
Summary
A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.
Glossary
stoichiometric factor
ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products
stoichiometry
relationships between the amounts of reactants and products of a chemical reaction | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/08%3A_Chemical_Reactions_and_Chemical_Quantities/8.04%3A_Reaction_Stoichiometry_-_How_Much_Carbon_Dioxide.txt |
Learning Objectives
• To understand the concept of limiting reactants and quantify incomplete reactions
In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess.
Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus
$1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}$
If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies.
Introduction to Limiting Reactant Problems: Introduction to Limiting Reactant Problems, YouTube(opens in new window) [youtu.be]
Now consider a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride ($\ce{TiCl4}$) and carbon dioxide.
$\ce{TiO2 (s) + Cl2 (g) \rightarrow TiCl4 (g) + CO2 (g)} \nonumber$
Titanium tetrachloride is then converted to metallic titanium by reaction with molten magnesium metal at high temperature:
$\ce{ TiCl4 (g) + 2 \, Mg (l) \rightarrow Ti (s) + 2 \, MgCl2 (l)} \label{3.7.2}$
Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about \$100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.
With 1.00 kg of titanium tetrachloride and 200 g of magnesium metal, how much titanium metal can be produced according to Equation \ref{3.7.2}?
Solving this type of problem requires that you carry out the following steps
1. Determine the number of moles of each reactant.
2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
3. Calculate the number of moles of product that can be obtained from the limiting reactant.
4. Convert the number of moles of product to mass of product.
Step 1: To determine the number of moles of reactants present, calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows: \begin{align} \text{moles} \; \ce{TiCl4} &= \dfrac{\text{mass} \, \ce{TiCl4}}{\text{molar mass} \, \ce{TiCl4}}\nonumber \[4pt] &= 1000 \, \cancel{g} \; \ce{TiCl4} \times {1 \, mol \; TiCl_4 \over 189.679 \, \cancel{g} \; \ce{TiCl4}}\nonumber \[4pt] &= 5.272 \, mol \; \ce{TiCl4} \[4pt] \text{moles }\, \ce{Mg} &= {\text{mass }\, \ce{Mg} \over \text{molar mass }\, \ce{Mg}}\nonumber \[4pt] &= 200 \, \cancel{g} \; \ce{Mg} \times {1 \; mol \, \ce{Mg} \over 24.305 \, \cancel{g} \; \ce{Mg} }\nonumber \[4pt] &= 8.23 \, \text{mol} \; \ce{Mg} \end{align}\nonumber
Step 2: There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following: ${mol \, \ce{Mg} \over mol \, \ce{TiCl4}} = {8.23 \, mol \over 5.272 \, mol } = 1.56 \nonumber$ Because the ratio of the coefficients in the balanced chemical equation is, ${ 2 \, mol \, \ce{Mg} \over 1 \, mol \, \ce{TiCl4}} = 2 \nonumber$ there is not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, there are 8.23 mol of $\ce{Mg}$, so (8.23 ÷ 2) = 4.12 mol of $\ce{TiCl4}$ are required for complete reaction. Because there are 5.272 mol of $\ce{TiCl4}$, titanium tetrachloride is present in excess. Conversely, 5.272 mol of $\ce{TiCl4}$ requires 2 × 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Therefore, magnesium is the limiting reactant.
Step 3: Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed: $mol \; \ce{Ti} = 8.23 \, mol \; \ce{Mg} = {1 \, mol \; \ce{Ti} \over 2 \, mol \; \ce{Mg}} = 4.12 \, mol \; \ce{Ti} \nonumber$ Thus only 4.12 mol of Ti can be formed.
Step 4. To calculate the mass of titanium metal that can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol): \begin{align} \text{moles }\, \ce{Ti} &= \text{mass }\, \ce{Ti} \times \text{molar mass } \, \ce{Ti}\nonumber \[6pt] &= 4.12 \, mol \; \ce{Ti} \times {47.867 \, g \; \ce{Ti} \over 1 \, mol \; \ce{Ti}}\nonumber\[6pt] &= 197 \, g \; \ce{Ti}\nonumber \end{align} \nonumber
Here is a simple and reliable way to identify the limiting reactant in any problem of this sort:
1. Calculate the number of moles of each reactant present: 5.272 mol of $\ce{TiCl4}$ and 8.23 mol of Mg.
2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation: $TiCl_4 : { 5.272 \, mol \, (actual) \over 1 \, mol \, (stoich)} = 5.272\nonumber \[6pt] Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12\nonumber$
3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant.
Density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example $1$ demonstrates.
Example $1$: Fingernail Polish Remover
Ethyl acetate ($\ce{CH3CO2C2H5}$) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol ($\ce{C2H5OH}$) with acetic acid ($\ce{CH3CO2H}$); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.
Given: reactants, products, and volumes and densities of reactants
Asked for: mass of product
Strategy:
1. Balance the chemical equation for the reaction.
2. Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.
3. Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.
4. Convert from moles of product to mass of product.
Solution:
A Always begin by writing the balanced chemical equation for the reaction:
$\ce{ C2H5OH (l) + CH3CO2H (aq) \rightarrow CH3CO2C2H5 (aq) + H2O (l)}\nonumber$
B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall that the density of a substance is the mass divided by the volume:
$\text{density} = {\text{mass} \over \text{volume} }\nonumber$
Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm3):
\begin{align*} \text{moles} \; \ce{C2H5OH} & = { \text{mass} \; \ce{C2H5OH} \over \text{molar mass } \; \ce{C2H5OH} }\nonumber \[6pt] & = {( \text{volume} \; \ce{C2H5OH} ) \times (\text{density} \, \ce{C2H5OH}) \over \text{molar mass } \; \ce{C2H5OH}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{C2H5OH} \times {0.7893 \, \cancel{g} \; \ce{C2H5OH} \over 1 \, \cancel{ml} \, \ce{C2H5OH} } \times {1 \, mol \; \ce{C2H5OH} \over 46.07 \, \cancel{g}\; \ce{C2H5OH}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \[6pt] \text{moles} \; \ce{CH3CO2H} &= {\text{mass} \; \ce{CH3CO2H} \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= { (\text{volume} \; \ce{CH3CO2H} )\times (\text{density} \; \ce{CH3CO2H}) \over \text{molar mass} \, \ce{CH3CO2H}}\nonumber \[6pt] &= 10.0 \, \cancel{ml} \; \ce{CH3CO2H} \times {1.0492 \, \cancel{g} \; \ce{CH3CO2H} \over 1 \, \cancel{ml} \; \ce{CH3CO2H}} \times {1 \, mol \; \ce{CH3CO2H} \over 60.05 \, \cancel{g} \; \ce{CH3CO2H} } \[6pt] &= 0.175 \, mol \; \ce{CH3CO2H}\nonumber \end{align*} \nonumber
C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:
\begin{align*} moles \; \text{ethyl acetate} &= mol \, \text{ethanol} \times {1 \, mol \; \text{ethyl acetate} \over 1 \, mol \; \text{ethanol}}\nonumber \[6pt] &= 0.171 \, mol \; \ce{C2H5OH} \times {1 \, mol \, \ce{CH3CO2C2H5} \over 1 \, mol \; \ce{C2H5OH}} \[6pt] &= 0.171 \, mol \; \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:
\begin{align*} \text{ mass of ethyl acetate} &= mol \; \text{ethyl acetate} \times \text{molar mass}\; \text{ethyl acetate}\nonumber \[6pt] &= 0.171 \, mol \, \ce{CH3CO2C2H5} \times {88.11 \, g \, \ce{CH3CO2C2H5} \over 1 \, mol \, \ce{CH3CO2C2H5}}\nonumber \[6pt] &= 15.1 \, g \, \ce{CH3CO2C2H5}\nonumber \end{align*} \nonumber
Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced:
\begin{align*} \text{volume of ethyl acetate} & = 15.1 \, g \, \ce{CH3CO2C2H5} \times { 1 \, ml \; \ce{CH3CO2C2H5} \over 0.9003 \, g\; \ce{CH3CO2C2H5}} \[6pt] &= 16.8 \, ml \, \ce{CH3CO2C2H5} \end{align*} \nonumber
Exercise $1$
Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound $P_4S_{10}$. How much $P_4S_{10}$ can be prepared starting with 10.0 g of $\ce{P4}$ and 30.0 g of $S_8$?
Answer
35.9 g
Determining the Limiting Reactant and Theoretical Yield for a Reaction: Determining the Limiting Reactant and Theoretical Yield for a Reaction, YouTube(opens in new window) [youtu.be]
Limiting Reactants in Solutions
The concept of limiting reactants applies to reactions carried out in solution as well as to reactions involving pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example $2$.
Example $2$: Breathalyzer reaction
Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:
$\ce{3CH_3 CH_2 OH(aq)} + \underset{yellow-orange}{\ce{2Cr_2 O_7^{2 -}}}(aq) + \ce{16H^+ (aq)} \underset{\ce{H2SO4 (aq)}}{\xrightarrow{\hspace{10px} \ce{Ag^{+}}\hspace{10px}} } \ce{3CH3CO2H(aq)} + \underset{green}{\ce{4Cr^{3+}}}(aq) + \ce{11H2O(l)}\nonumber$
When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the $\ce{Cr2O7^{2−}}$ ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a minute. Because the $\ce{Cr2O7^{2−}}$ ion (the reactant) is yellow-orange and the Cr3+ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.
A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K2Cr2O7 in 50% H2SO4 as well as a fixed concentration of AgNO3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr6+ to Cr3+?
Given: volume and concentration of one reactant
Asked for: mass of other reactant needed for complete reaction
Strategy:
1. Calculate the number of moles of $\ce{Cr2O7^{2−}}$ ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass.
2. Find the total number of moles of $\ce{Cr2O7^{2−}}$ ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).
3. Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of $\ce{Cr2O7^{2−}}$ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass.
Solution:
A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K2Cr2O7 in 1 mL of solution, which can be used to calculate the number of moles of K2Cr2O7 contained in 1 mL:
$\dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles\nonumber$
B Because 1 mol of K2Cr2O7 produces 1 mol of $\ce{Cr2O7^{2−}}$ when it dissolves, each milliliter of solution contains 8.5 × 10−7 mol of Cr2O72. The total number of moles of Cr2O72 in a 3.0 mL Breathalyzer ampul is thus
$moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–}\nonumber$
C The balanced chemical equation tells us that 3 mol of C2H5OH is needed to consume 2 mol of $\ce{Cr2O7^{2−}}$ ion, so the total number of moles of C2H5OH required for complete reaction is
$moles\: of\: \ce{C2H5OH} = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: \ce{Cr2O7^{2-}}} ) \left( \dfrac{3\: mol\: \ce{C2H5OH}} {2\: \cancel{mol\: \ce{Cr2O7^{2 -}}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: \ce{C2H5OH}\nonumber$
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:
$mass\: \ce{C2H5OH} = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: \ce{C2H5OH}} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: \ce{C2H5OH}}} \right) = 1 .8 \times 10 ^{-4}\: g\: \ce{C2H5OH}\nonumber$
Thus 1.8 × 10−4 g or 0.18 mg of C2H5OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.
Exercise $2$
The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction
Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess $\ce{NaOH}$ is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10−4 g of para-nitrophenol to ensure that formation of the yellow anion is complete?
Answer
4.93 × 10−5 L or 49.3 μL
In Examples $1$ and $2$, the identities of the limiting reactants are apparent: [Au(CN)2], LaCl3, ethanol, and para-nitrophenol. When the limiting reactant is not apparent, it can be determined by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation. The only difference is that the volumes and concentrations of solutions of reactants, rather than the masses of reactants, are used to calculate the number of moles of reactants, as illustrated in Example $3$.
Example $3$
When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:
$\ce{2AgNO3(aq) + K2Cr2O7(aq) \rightarrow Ag2Cr2O7(s) + 2KNO3(aq) }\nonumber$
What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M $\ce{K2Cr2O7}$ are mixed with 250 mL of 0.57 M AgNO3?
Given: balanced chemical equation and volume and concentration of each reactant
Asked for: mass of product
Strategy:
1. Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.
2. Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.
3. Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.
Solution:
A The balanced chemical equation tells us that 2 mol of AgNO3(aq) reacts with 1 mol of K2Cr2O7(aq) to form 1 mol of Ag2Cr2O7(s) (Figure 8.3.2). The first step is to calculate the number of moles of each reactant in the specified volumes:
$moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7\nonumber$
$moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3\nonumber$
B Now determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:
\begin{align*} \ce{K2Cr2O7}: \: \dfrac{0 .085\: mol} {1\: mol} &= 0.085 \[4pt] \ce{AgNO3}: \: \dfrac{0 .14\: mol} {2\: mol} &= 0 .070 \end{align*} \nonumber
Because 0.070 < 0.085, we know that $\ce{AgNO3}$ is the limiting reactant.
C Each mole of $\ce{Ag2Cr2O7}$ formed requires 2 mol of the limiting reactant ($\ce{AgNO3}$), so we can obtain only 0.14/2 = 0.070 mol of $\ce{Ag2Cr2O7}$. Finally, convert the number of moles of $\ce{Ag2Cr2O7}$ to the corresponding mass:
$mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7\nonumber$
The Ag+ and Cr2O72 ions form a red precipitate of solid $\ce{Ag2Cr2O7}$, while the $\ce{K^{+}}$ and $\ce{NO3^{−}}$ ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.)
Exercise $3$
Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:
$\ce{2NaHCO3(aq) + H2SO4(aq) \rightarrow 2CO2(g) + Na2SO4(aq) + 2H2O(l)}\nonumber$
If 13.0 mL of 3.0 M H2SO4 are added to 732 mL of 0.112 M NaHCO3, what mass of CO2 is produced?
Answer
3.4 g
Limiting Reactant Problems Using Molarities: Limiting Reactant Problems Using Molarities, YouTube(opens in new window) [youtu.be]eOXTliL-gNw (opens in new window)
Theoretical Yields
When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield, the amount obtained if the reaction occurred perfectly and the purification method were 100% efficient.
In reality, less product is always obtained than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not run to completion, resulting in a mixture of products and reactants; this last possibility is a common occurrence. Therefore, the actual yield, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:
$\text{percent yield} = {\text{actual yield } \; (g) \over \text{theoretical yield} \; (g) } \times 100\% \label{3.7.3}$
The method used to calculate the percent yield of a reaction is illustrated in Example $4$.
Example $4$: Novocain
Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction
$\underset {\text{p-amino benzoic acid}}{\ce{C7H7NO2}} + \underset {\text{2-diethylaminoethanol}}{\ce{C6H15NO}} \ce{->[\ce{H2SO4}]} \underset {\text{procaine}}{\ce{C13H20N2O2}} + \ce{H2O}\nonumber$
If this reaction were carried out with 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and 15.7 g of procaine were isolated, what is the percent yield?
The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.
Given: masses of reactants and product
Asked for: percent yield
Strategy:
1. Write the balanced chemical equation.
2. Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.
3. Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.
Solution:
A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.
B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:
$mol \; \text{p-aminobenzoic acid} = 10.0 \, g \, \times \, {1 \, mol \over 137.14 \, g } = 0.0729 \, mol \; \text{p-aminbenzoic acid}\nonumber$
$mol \; \text{2-diethylaminoethanol} = 10.0 \, g \times {1 \, mol \over 117.19 \, g} = 0.0853 \, mol \; \text{2-diethylaminoethanol}\nonumber$
The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.
$\text{theoretical yield of procaine} = 0.0729 \, mol \times {236.31 \, g \over 1 \, mol } = 17.2 \, g\nonumber$
C The actual yield was only 15.7 g of procaine, so the percent yield (via Equation \ref{3.7.3}) is
$\text{percent yield} = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \%\nonumber$
(If the product were pure and dry, this yield would indicate very good lab technique!)
Exercise $4$: Extraction of Lead
Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena ($\ce{PbS}$), which is easily converted to lead oxide ($\ce{PbO}$) in 100% yield by roasting in air via the following reaction:
$\ce{ 2PbS (s) + 3O2 \rightarrow 2PbO (s) + 2SO2 (g)}\nonumber$
The resulting $\ce{PbO}$ is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:
$\ce{PbO (s) + C(s) \rightarrow Pb (l) + CO (g)}\nonumber$
If 93.3 kg of $\ce{PbO}$ is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?
Answer
89.2%
Percent yield can range from 0% to 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.
Summary
The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained. The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/08%3A_Chemical_Reactions_and_Chemical_Quantities/8.05%3A_Limiting_Reactant_Theoretical_Yield_and_Percent_Yield.txt |
• 9.1: Molecular Gastronomy
• 9.2: Solution Concentration
Solution concentrations are typically expressed as molarities and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution. The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
• 9.3: Solution Stoichiomentry
Solution concentrations are typically expressed as molarities and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution. The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
• 9.4: Types of Aqueous Solutions and Solubility
Electrolytic solutions are those that are capable of conducting an electric current. A substance that, when added to water, renders it conductive, is known as an electrolyte. A common example of an electrolyte is ordinary salt, sodium chloride. Solid NaCl and pure water are both non-conductive, but a solution of salt in water is readily conductive. A solution of sugar in water, by contrast, is incapable of conducting a current; sugar is therefore a non-electrolyte.
• 9.5: Precipitation Reactions
A complete ionic equation consists of the net ionic equation and spectator ions. Predicting the solubility of ionic compounds gives insight into feasibility of reactions occuring. The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances in their undissociated forms; the complete ionic equation shows substances in the form in which they actually exist in solution; and the net ionic equation omits all spectator ions.
• 9.6: Representing Aqueous Reactions- Molecular, Ionic, and Complete Ionic Equations
The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients.
• 9.7: Acid-Base
An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt. Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton and a base is a substance that can accept a proton. Acids also differ in their tendency to donate a proton, a measure of their acid strength. The acidity or basicity of an aqueous solution is described quantitatively using the pH scale.
• 9.8: Gas-Evolution Reactions
An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt. Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton and a base is a substance that can accept a proton. Acids also differ in their tendency to donate a proton, a measure of their acid strength. The acidity or basicity of an aqueous solution is described quantitatively using the pH scale.
• 9.9: Oxidation-Reduction Reactions
Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method, in which the overall reaction is separated into an oxidation equation and a reduction equation. The outcome of these reactions can be predicted using the activity series.
09: Introduction to Solutions and Aqueous Reactions
Learning Objectives
• To describe the concentrations of solutions quantitatively
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
$molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.5.1}$
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as $M$. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,
$[\rm{sucrose}] = 1.00\: M \nonumber$
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
$V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.5.2}$
or
$V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.5.3}$
Figure $1$ illustrates the use of Equations $\ref{4.5.2}$ and $\ref{4.5.3}$.
Example $1$: Calculating Moles from Concentration of NaOH
Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.
Given: identity of solute and volume and molarity of solution
Asked for: amount of solute in moles
Strategy:
Use either Equation \ref{4.5.2} or Equation \ref{4.5.3}, depending on the units given in the problem.
Solution:
Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation \ref{4.5.2} is more useful:
$moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH$
Exercise $1$: Calculating Moles from Concentration of Alanine
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
Answer
41.6 mmol
Calculations Involving Molarity (M): https://youtu.be/TVTCvKoSR-Q
Concentrations are also often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in Table $1$.
Table $1$: Common Units of Concentration
Concentration Units
m/m g of solute/g of solution
m/v g of solute/mL of solution
ppm g of solute/106 g of solution
μg/mL
ppb g of solute/109 g of solution
ng/mL
The Preparation of Solutions
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure $1$ illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure $2$, for some substances this effect can be significant, especially for concentrated solutions.
Example $2$
The solution contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2•2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of $\ce{CoCl2•2H2O}$?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of $\ce{CoCl2•2H2O}$, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.
Solution:
The molar mass of CoCl2•2H2O is 165.87 g/mol. Therefore,
$moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \nonumber$
The volume of the solution in liters is
$volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \nonumber$
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
$molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \nonumber$
Exercise $2$
The solution shown in Figure $2$ contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?
Answer
$(NH_4)_2Cr_2O_7 = 1.43\: M \nonumber$
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation $\ref{4.5.2}$. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example $3$.
Example $3$: D5W Solution
The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.
Given: molarity, volume, and molar mass of solute
Asked for: mass of solute
Strategy:
1. Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.
2. Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.
Solution:
A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
$V_L M_{mol/L} = moles$
$500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose$
B We then convert the number of moles of glucose to the required mass of glucose:
$mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose$
Exercise $3$
Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.
Answer
2.3 g NaCl
A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution is a commercially prepared solution of known concentration and is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.
The procedure for preparing a solution of known concentration from a stock solution is shown in Figure $3$. It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore
$(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\label{4.5.4}$
where the subscripts s and d indicate the stock and dilute solutions, respectively. Example $4$ demonstrates the calculations involved in diluting a concentrated stock solution.
Example $4$
What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example $3$?
Given: volume and molarity of dilute solution
Asked for: volume of stock solution
Strategy:
1. Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.
2. To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.
Solution:
A The D5W solution in Example 4.5.3 was 0.310 M glucose. We begin by using Equation 4.5.4 to calculate the number of moles of glucose contained in 2500 mL of the solution:
$moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \nonumber$
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
$volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \nonumber$
In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).
We could also have solved this problem in a single step by solving Equation 4.5.4 for Vs and substituting the appropriate values:
$V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \nonumber$
As we have noted, there is often more than one correct way to solve a problem.
Exercise $4$
What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?
Answer
16 mL
Example $5$
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.21 M NaOH
2. 3.7 M (CH3)2CHOH
3. 0.032 M In(NO3)3
Given: molarity
Asked for: concentrations
Strategy:
A Classify each compound as either a strong electrolyte or a nonelectrolyte.
B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.
Solution:
1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution: $NaOH(s) \xrightarrow {H_2 O(l)} Na^+ (aq) + OH^- (aq)$
B Because each formula unit of NaOH produces one Na+ ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH] = 0.21 M.
2. A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.
B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M.
3. A Indium nitrate is an ionic compound that contains In3+ ions and NO3 ions, so we expect it to behave like a strong electrolyte in aqueous solution:
$In(NO _3 ) _3 (s) \xrightarrow {H_ 2 O(l)} In ^{3+} (aq) + 3NO _3^- (aq)$
B One formula unit of In(NO3)3 produces one In3+ ion and three NO3 ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3—that is, [In3+] = 0.032 M and [NO3] = 0.096 M.
Exercise $5$
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.0012 M Ba(OH)2
2. 0.17 M Na2SO4
3. 0.50 M (CH3)2CO, commonly known as acetone
Answer a
$[Ba^{2+}] = 0.0012\: M; \: [OH^-] = 0.0024\: M$
Answer b
$[Na^+] = 0.34\: M; \: [SO_4^{2-}] = 0.17\: M$
Answer c
$[(CH_3)_2CO] = 0.50\: M$
Summary
Solution concentrations are typically expressed as molarities and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.
• definition of molarity: $molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \nonumber$
• relationship among volume, molarity, and moles: $V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \nonumber$
• relationship between volume and concentration of stock and dilute solutions: $(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d) \nonumber$
The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.02%3A_Solution_Concentration.txt |
Learning Objectives
• How to calculate the concentrations of ions when a salt is dissolved
In Example $2$, the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL were calculated to be 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72 ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72 ions:
$(NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\label{4.5.5}$
Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72 anions and 2 mol of NH4+ cations (see Figure $4$).
When carrying out a chemical reaction using a solution of a salt such as ammonium dichromate, it is important to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72 must also be 1.43 M because there is one Cr2O72 ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.
Concentration of Ions in Solution from a Soluble Salt: https://youtu.be/qsekSJBLemc
9.04: Types of Aqueous Solutions and Solubility
8.9A: Electrolytes and Electrolytic Solutions
8.9B: The nature of ions in aqueous solution
https://chem.libretexts.org/Textbook...A_Electrolytes
In Binary Ionic Compounds and Their Properties we point out that when an ionic compound dissolves in water, the positive and negative ions originally present in the crystal lattice persist in solution. Their ability to move nearly independently through the solution permits them to carry positive or negative electrical charges from one place to another. Hence the solution conducts an electrical current.
Electrolytes
Substances whose solutions conduct electricity are called electrolytes. All soluble ionic compounds are strong electrolytes. They conduct very well because they provide a plentiful supply of ions in solution. Some polar covalent compounds are also strong electrolytes. Common examples are HCl, HBr, HI and H2SO4, all of which react with H2O to form large concentrations of ions. A solution of HCl, for example, conducts even better than one of NaCl having the same concentration.
The effect of the concentration of ions on the electrical current flowing through a solution is illustrated in Figure $1$. Part a of the figure shows what happens when a battery is connected through an electrical meter to two inert metal strips (electrodes) dipping in ethanol. Each cubic decimeter of such a solution contains 0.10 mol NaCl (that is, 0.10 mol Na+ and 0.10 mol Cl). An electrical current is carried through the solution both by the Na+ ions moving toward the negative electrode and by the Cl- ions which are attracted toward the positive electrode. The dial on the meter indicates the quantity of current.
Figure 1b shows that if we replace the 0.10-M NaCl solution with a 0.05-M NaCl solution, the meter reading falls to about one-half its former value. Halving the concentration of NaCl halves the number of ions between the electrodes, and half as many ions can only carry half as much electrical charge. Therefore the current is half as great. Because it responds in such a direct way to the concentration of ions, conductivity of electrical current is a useful tool in the study of solutions.
Conductivity measurements reveal that most covalent compounds, if they dissolve in water at all, retain their original molecular structures. Neutral molecules cannot carry electrical charges through the solution, and so no current flows. A substance whose aqueous solution conducts no better than water itself is called a nonelectrolyte. Some examples are oxygen, O2, ethanol, C2H5OH, and sugar, C12H22O11.
Some covalent substances behave as weak electrolytes—their solutions allow only a small current flow, but it is greater than that of the pure solvent. An example is mercury(II) chloride (seen in the Figure above). For a 100-M HgCl2 solution the meter reading shows only about 0.2 percent as much current as for 0.10 M NaCl. A crystal of HgCl2 consists of discrete molecules, like those shown for HgBr2 in Figure $2$. When the solid dissolves, most of these molecules remain intact, but a few dissociate into ions according to the equation
$\underbrace{HgCl_2}_{99.8\%} \rightleftharpoons \underbrace{HgCl^+}_{0.2\%} + Cl^- \nonumber$
(The double arrows indicate that the ionization proceeds only to a limited extent and an equilibrium state is attained.) Since only 0.2 percent of the HgCl2 forms ions, the 0.10 M solution can conduct only about 0.2 percent as much current as 0.10 M NaCl.
Conductivity measurements can tell us more than whether a substance is a strong, a weak, or a nonelectrolyte. Consider, for instance, the data in Table $1$ which shows the electrical current conducted through various aqueous solutions under identical conditions. At the rather low concentration of 0.001 M, the strong electrolyte solutions conduct between 2500 and 10 000 times as much current as pure H2O and about 10 times as much as the weak electrolytes HC2H3O2 (acetic acid) and NH3 (ammonia).
Closer examination of the data for strong electrolytes reveals that some compounds which contain H or OH groups [such as HCl or Ba(OH)2] conduct unusually well. If these compounds are excluded, we find that 1:1 electrolytes (compounds which consist of equal numbers of +1 ions and –1 ions) usually conduct about half as much current as 2:2 electrolytes (+2 and -2 ions), 1:2 electrolytes (+1 and -2 ions), or 2:1 electrolytes (+2 and -1 ions).
TABLE $1$: Electrical Current Conducted Through Various 0.001 M Aqueous Solutions at 18°C.*
Substance Current /mA Substance Current /mA
Pure Water 1:2 Electrolytes
H2O 3.69 x 10-4 Na2SO4 2.134
Weak Electrolytes Na2CO3 2.24
HC2H3O2 0.41 K2CO3 2.660
NH3 0.28 2:1 Electrolytes
1:1 Electrolytes MgCl2 2.128
NaCl 1.065 CaCl2 2.239
NaI 1.069 SrCl2 2.290
KCl 1.273 BaCl2 2.312
KI 1,282 Ba(OH)2 4.14
AgNO3 1.131 2:2 Electrolytes
HCl 3.77 MgSO4 2.00
HNO3 3.75 CaSO4 2.086
NaOH 2.08 CuSO4 1.97
KOH 2.34 ZnSO4 1.97
* All measurements refer to a cell in which the distance between the electrodes is 1.0 mm and the area of each electrode is 1.0 cm². A potential difference of 1.0 V is applied to produce the tabulated currents.
There is a simple reason for this behavior. Under similar conditions, most ions move through water at comparable speeds. This means that ions like Mg2+ or SO42, which are doubly charged, will carry twice as much current through the solution as will singly charged ions like Na+ or Cl. Consequently, a 0.001 M solution of a 2:2 electrolyte like MgSO4 will conduct about twice as well as a 0.001 M solution of a 1:1 electrolyte like NaCl.
A similar argument applies to solutions of 1:2 and 2:1 electrolytes. A solution like 0.001 M Na2SO4 conducts about twice as well as 0.001 M NaCl partly because there are twice as many Na ions available to move when a battery is connected, but also because SO42 ions carry twice as much charge as Cl ions when moving at the same speed. These differences in conductivity between different types of strong electrolytes can sometimes be very useful in deciding what ions are actually present in a given electrolyte solution as the following example makes clear.
A second, slightly more subtle, conclusion can be drawn from the data in Table $1$. When an electrolyte dissolves, each type of ion makes an independent contribution to the current the solution conducts. This can be seen by comparing NaCl with KCl, and NaI with KI. In each case the compound containing K+ conducts about 0.2 mA more than the one containing Na+. If we apply this observation to Na2CO3 and K2CO3, each of which produces twice as many Na+ or K+ ions in solution, we find that the difference in current is also twice as great—about 0.4 mA.
Thus conductivity measurements confirm our statement that each ion exhibits its own characteristic properties in aqueous solutions, independent of the presence of other ions. One such characteristic property is the quantity of electrical current that a given concentration of a certain type of ion can carry.
Example $1$: Ions
At 18°C a 0.001-M aqueous solution of potassium hydrogen carbonate, KHCO3, conducts a current of 1.10 mA in a cell of the same design as that used to obtain the data in Table 11.1. What ions are present in solution?
Solution
Referring to Table 6.2 which lists possible polyatomic ions, we can arrive at three possibilities for the ions from which KHCO3 is made:
1. K+ and H+ and C4+ and three O2–
2. K+ and H+ and CO32
3. K+ and HCO3
Since the current conducted by the solution falls in the range of 1.0 to 1.3 mA characteristic of 1:1 electrolytes, possibility c is the only reasonable choice. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.03%3A_Solution_Stoichiomentry.txt |
Learning Objectives
• To identify a precipitation reaction and predict solubilities.
Exchange (Double-Displacement) Reactions
A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate:
$\ce{AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)} \label{4.2.1}$
This unbalanced equation has the general form of an exchange reaction:
$\overbrace{\ce{AC}}^{\text{soluble}} + \overbrace{\ce{BD}}^{\text{soluble}} \rightarrow \underbrace{\ce{AD}}_{\text{insoluble}} + \overbrace{\ce{BC}}^{\text{soluble}} \label{4.2.2}$
The solubility and insoluble annotations are specific to the reaction in Equation \ref{4.2.1} and not characteristic of all exchange reactions (e.g., both products can be soluble or insoluble). Precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.
Video $1$: Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation $\ref{4.2.1}$).
While full chemical equations show the identities of the reactants and the products and give the stoichiometries of the reactions, they are less effective at describing what is actually occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.
Let’s consider the reaction of silver nitrate with potassium dichromate above. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electrically neutral compounds:
$\ce{2AgNO_3(aq)} + \ce{K_2Cr_2O_7(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s) }+ \ce{2KNO_3(aq)} \label{4.2.1a}$
Although Equation $\ref{4.2.1a}$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as $\ce{AgNO3}$ and $\ce{K2Cr2O7}$ are strong electrolytes (i.e., they dissociate completely in aqueous solution to form ions). In contrast, because $\ce{Ag2Cr2O7}$ is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases:
$\ce{2Ag^{+}(aq) + 2NO_3^{-} (aq) + 2K^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^{+}(aq) + 2NO_3^{-}(aq)}\label{4.2.2a}$
Note that $\ce{K^+ (aq)}$ and $\ce{NO3^{−} (aq)}$ ions are present on both sides of Equation $\ref{4.2.2a}$ and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{4.2.3}$
Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $\ref{4.2.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral $\ce{Ag2Cr2O7}$ formula unit on the right side.
By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{4.2.4}$
The complete ionic equation for this reaction is as follows:
$\ce{2Ag^{+}(aq)} + \cancel{\ce{2F^{-}(aq)}} + \cancel{\ce{2NH_4^{+}(aq)}} + \ce{Cr_2O_7^{2-}(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s)} + \cancel{\ce{2NH_4^{+}(aq)}} + \cancel{\ce{2F^{-}(aq)}} \label{4.2.5}$
Because two $\ce{NH4^{+}(aq)}$ and two $\ce{F^{−} (aq)}$ ions appear on both sides of Equation $\ref{4.2.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{4.2.6}$), which is identical to Equation $\ref{4.2.3}$:
$\ce{2Ag^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)} \label{4.2.6}$
If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.
Determining the Products for Precipitation Reactions: Determining the Products for Precipitation Reactions, YouTube(opens in new window) [youtu.be]
Example $1$: Balancing Precipitation Equations
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.
Given: reactants and products
Asked for: overall, complete ionic, and net ionic equations
Strategy:
Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.
Solution:
From the information given, we can write the unbalanced chemical equation for the reaction:
$\ce{Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)} \nonumber$
Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43 ions per formula unit, we can balance the equation by inspection:
$\ce{3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)} \nonumber$
This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:
$\ce{3Ba^{2+}(aq)} + \cancel{\ce{6NO_3^{-}(aq)}} + \cancel{\ce{6Na^{+} (aq)}} + \ce{2PO_4^{3-} (aq)} \rightarrow \ce{Ba_3(PO_4)_2(s)} + \cancel{\ce{6Na^+(aq)}} + \cancel{\ce{6NO_3^{-}(aq)}} \nonumber$
The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:
$\ce{3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)} \nonumber$
Exercise $1$: Mixing Silver Fluoride with Sodium Phosphate
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.
Answer
overall chemical equation:
$\ce{3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq) } \nonumber$
complete ionic equation:
$\ce{3Ag^+(aq) + 3F^{-}(aq) + 3Na^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^{+}(aq) + 3F^{-}(aq) } \nonumber$
net ionic equation:
$\ce{3Ag^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)} \nonumber$
So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.
The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.
Predicting Solubilities
Table $1$ gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table $1$ to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.
Table $1$: Guidelines for Predicting the Solubility of Ionic Compounds in Water
Soluble Exceptions
Rule 1 most salts that contain an alkali metal (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+)
Rule 2 most salts that contain the nitrate (NO3) anion
Rule 3 most salts of anions derived from monocarboxylic acids (e.g., CH3CO2) but not silver acetate and salts of long-chain carboxylates
Rule 4 most chloride, bromide, and iodide salts but not salts of metal ions located on the lower right side of the periodic table (e.g., Cu+, Ag+, Pb2+, and Hg22+).
Insoluble Exceptions
Rule 5 most salts that contain the hydroxide (OH) and sulfide (S2−) anions but not salts of the alkali metals (group 1), the heavier alkaline earths (Ca2+, Sr2+, and Ba2+ in group 2), and the NH4+ ion.
Rule 6 most carbonate (CO32) and phosphate (PO43) salts but not salts of the alkali metals or the NH4+ ion.
Rule 7 most sulfate (SO42) salts that contain main group cations with a charge ≥ +2 but not salts of +1 cations, Mg2+, and dipositive transition metal cations (e.g., Ni2+)
Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na+(aq), 0.50 M Cl(aq), 0.50 M K+(aq), and 0.50 M Br(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure $1$).
Example $2$
Using the information in Table $1$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. Aqueous solutions of barium chloride and lithium sulfate are mixed.
2. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
3. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
4. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.
Given: reactants
Asked for: reaction and net ionic equation
Strategy:
1. Identify the ions present in solution and write the products of each possible exchange reaction.
2. Refer to Table $1$ to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.
Solution:
A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba2+, Cl, Li+, and SO42 ions. The only possible exchange reaction is to form LiCl and BaSO4:
B We now need to decide whether either of these products is insoluble. Table $1$ shows that LiCl is soluble in water (rules 1 and 4), but BaSO4 is not soluble in water (rule 5). Thus BaSO4 will precipitate according to the net ionic equation
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) \nonumber$
Although soluble barium salts are toxic, BaSO4 is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO4 particles in water.
1. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb+, OH, Co2+, and Cl ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):
B According to Table $1$, RbCl is soluble (rules 1 and 4), but Co(OH)2 is not soluble (rule 5). Hence Co(OH)2 will precipitate according to the following net ionic equation:
$Co^{2+}(aq) + 2OH^-(aq) \rightarrow Co(OH)_2(s)$
2. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr2+, Br, Al3+, and NO3 ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:
B According to Table $1$, both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur.
1. A According to Table $1$, lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb2+ and CH3CO2 ions. Because the solution also contains NH4+ and I ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:
B According to Table $1$, ammonium acetate is soluble (rules 1 and 3), but PbI2 is insoluble (rule 4). Thus Pb(C2H3O2)2 will dissolve, and PbI2 will precipitate. The net ionic equation is as follows:
$Pb^{2+} (aq) + 2I^-(aq) \rightarrow PbI_2(s)$
Exercise $2$
Using the information in Table $1$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.
Answer a
$Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$
Answer b
$2PO_4^{3-}(aq) + 3Hg^{2+}(aq) \rightarrow Hg_3(PO_4)_2(s)$
Answer c
$NaF(s)$ dissolves; no net reaction
Answer d
$Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$
Predicting the Solubility of Ionic Compounds: Predicting the Solubility of Ionic Compounds, YouTube(opens in new window) [youtu.be] (opens in new window) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.05%3A_Precipitation_Reactions.txt |
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