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D30.1 ICE Table
An ICE table (for Initial concentration, Change in concentration, and Equilibrium concentration) is a good methodology for calculating an equilibrium constant from experimental data. ICE tables also help to solve many other types of equilibrium problems.
An ICE table begins with the balanced reaction equation, using the reactants and products as column headings. The second row lists the initial concentrations of the reactants and products; these can usually be obtained from experimental data based on the assumption that no reaction has yet taken place. The third row is the change in concentration that occurs as the system proceeds toward equilibrium; this row is derived from the stoichiometry of the reaction. The last row is the sum of the first two rows, which yields the equilibrium concentrations.
For example, consider determining the equilibrium constant for the reaction:
I2(aq) + I‾(aq) ⇌ I3‾(aq)
If a solution initially has [I2]0 = [I]0 = 1.000 × 10−3 M and no triiodide ions ([I3−] = 0), and then reacts to give an equilibrium concentration of [I2]e = 6.61 × 10−4 M, we can use an ICE table to determine the equilibrium constant for the reaction. First, write the balanced reaction at the top of the table. The next row gives the initial concentrations.
I2(aq) + I‾(aq) I3‾(aq)
Initial concentration (M) 1.000 × 10-3 1.000 × 10-3 0
Change in concentration (M) x x +x
Equilibrium concentration (M) (1.000 × 10-3) – x (1.000 × 10-3) – x x
In the “Change” row, the mathematical sign indicates the direction of change: the sign is positive when the concentration increases as the reaction proceeds toward equilibrium, and negative when the concentration decreases. For example, because the reaction starts with no I3‾, [I2]t must decrease as the reaction proceeds, so its (unknown) change in concentration is “-x“. In general, x is also multiplied by the stoichiometric coefficient, but in this case, all coefficients are 1, so that is not obvious.
Using the information that [I2]e = 6.61 × 10−4 M, we can solve for x:
(1.000 × 10−3) – x = 6.61 × 10−4 M
x = 3.39 × 10−4 M
and then calculate the equilibrium constant Kc:
$\begin{array}{rcl} K_c & = & \dfrac{[\text{I}_3^{-}]_e}{[\text{I}_2]_e[\text{I}^{-}]_e} \[0.5em] & = & \dfrac{x}{\{(1.00 \times 10^{-3})-x\} \{ (1.00 \times {10}^{-3})-x\}} \[0.5em] & = & \dfrac{3.39\;\times\;10^{-4}\;M}{(6.61\;\times\;10^{-4}\;M)(6.61\;\times\;10^{-4}\;M)} \[0.5em] & = & 776\;\text{M}^{-1}\;\text{reported without units as 776} \end{array} \nonumber$
Exercise 1: Concentration Changes During Reactions
Activity 1: Calculating Equilibrium Concentrations I
Query $2$
Activity 2: Calculating Equilibrium Concentrations II
D30.2 “All-Reactant” or “All-Product” Starting Point
The same equilibrium concentrations are achieved whether a reaction begins with only reactants present or only products present. Let’s use an example to consider this in further detail. HF is a deadly, but weak, acid. It ionizes partially in water :
HF(aq) ⇌ H+(aq) + F‾(aq) Kc = 6.80 × 10-4
What are the equilibrium concentrations of the various aqueous species in a solution of 0.150-M HF?
One way to determine these equilibrium concentrations is to start with only reactants. This is called the “all-reactant” starting point.
Activity 3: All-reactant Starting Point
Query $4$
Alternatively, we could solve the problem assuming that all the HF ionizes first, and then the system comes to equilibrium. This is called the “all-product” starting point.
Activity 4: All-product Starting Point
Query $5$
The two approaches give the same results, and show that starting with all products leads to the same equilibrium conditions as starting with all reactants. Note that this is true only if the temperature is the same and the same total number of atoms of each kind is present in both cases. Here we either started with all reactants, 0.150-M HF (which contains 0.150 mol/L H and 0.150 mol/L F), or with all products, 0.150M H+ and 0.150-M F (which also contains 0.150 mol/L H and 0.150 mol/L F). Had we started with 0.140-M H+ and 0.160-M F, the equilibrium concentrations would not be the same.
For the ionization of HF(aq) the equilibrium constant is small: Kc = 6.80 × 10-4. That is, the ionization of HF(aq) is reactant-favored. In Activity 3 the change in concentration, x3 = 0.00977, while in Activity 4 the change in concentration, x4 = 0.140. Because the process is reactant-favored, the all-reactant initial concentrations are much closer to the equilibrium concentrations than the all-product initial concentrations. Therefore the all-reactants situation involves only small changes in concentrations to reach equilibrium. Recognizing this allows making approximations that can significantly simplify the calculations in equilibrium problems.
We know that when Kc << 1, the equilibrium is significantly reactant-favored, and when Kc >> 1, the equilibrium is significantly product-favored. If the ICE table can be set up so that the “initial” concentrations are close to equilibrium (either all-reactant or all-product, depending on the size of Kc), then any change in concentration that is small compared to the initial concentrations can be neglected. “Small” is defined as resulting in an error that does not change the answer within the the number of significant figures involved.
Activity 5: Solving an Aqueous Equilibrium Problem
D30.3 Reaction Quotient
Starting with only reactants present means that the reaction must initially be spontaneous left to right, even if only very small concentrations of products are present when equilibrium is reached. Starting with only products present requires that the reaction go from right to left. Which direction does a reaction go when both reactants and products are present but the reaction is not at equilibrium? The reaction quotient can answer this question.
For a generic reaction,
mA + nB ⇌ xC + yD
the reaction quotient (Q) is defined as:
$Q_c = \dfrac{[\text{C}]_t^{\;x}[\text{D}]_t^{\;y}}{[\text{A}]_t^{\;m}[\text{B}]_t^{\;n}} \nonumber$
In Qc the subscript “c” indicates Q is in terms of concentrations; for a gas-phase reaction we could write Qp similarly in terms of partial pressures.
The concentrations are represented by “[…]t” to emphasize that Qc for a reaction depends on the concentrations present at the time when Qc is determined; this is usually not at equilibrium. When only reactants are present, Qc = 0. As the reaction proceeds, Qc increases because product concentrations increase and reactant concentrations decrease (Figure 1). When the reaction reaches equilibrium, Qc no longer changes over time because the concentrations no longer change, and, at equilibrium, Qc = Kc:
$Q_c \text{(at equilibrium)} = \dfrac{[\text{C}]_e^{\;x}[\text{D}]_e^{\;y}}{[\text{A}]_e^{\;m}[\text{B}]_e^{\;n}} = K_c \nonumber$
A system that is not at equilibrium proceeds spontaneously in the direction that establishes equilibrium (Qc changes until it equals Kc). Hence, we can predict directional shifts of a reaction by comparing Qc to Kc: when Qc < Kc, the reaction proceeds spontaneously from left to right (to the product side); when Qc > Kc, the reaction proceeds spontaneously to the left (reactant side).
For example, for the water-gas shift reaction,
CO(g) + H2O(g) ⇌ CO2(g) + H2(g) Kc = 0.64 T = 800 °C
different starting mixtures of CO, H2O, CO2, and H2 react (and the concentrations of reactants and products change) until the compositions reach the same value of Qc; that is, until Qc = Kc (Figure 2).
Query $7$
Figure 2. Concentrations of four different mixtures are shown before and after reaching equilibrium at 800 °C for the water gas shift reaction: CO(g) + H2O(g) ⇌ CO2(g) + H2(g).
It is important to recognize that Qc reaches the same equilibrium value (Kc) whether the reaction starts from all reactants, from all products, or from a mixture of both. In fact, one technique to determine whether a reaction is truly at equilibrium is to start with only reactants in one experiment and start with only products in another. If the same value of the reaction quotient is observed when the concentrations have stopped changing in both experiments, then it is highly likely that the system has reached equilibrium.
D30.4 Le Chatelier’s principle: Change in Concentration
Le Chatelier’s principle states that when a chemical system is at equilibrium and conditions are changed so that the reaction quotient, Q, changes, the chemical system will react to achieve new equilibrium concentrations or partial pressures; reaction occurs in a way that partially counteracts the change in conditions. To establish the new equilibrium, the reaction proceeds in the forward direction if Q < K or in the reverse direction if Q > K, until Q is again equal to K.
For a chemical system at equilibrium at constant temperature, if the concentration of a reactant or a product is changed, therefore changing Q, the system is no longer at equilibrium because QK. The concentrations of all reaction species will then undergo additional changes until the system reaches a new equilibrium with a different set of equilibrium concentrations. We say that the equilibrium shifts in a direction (forward or reverse) that partially counteracts the change.
For example, consider the chemical reaction:
H2(g) + I2(g) ⇌ 2HI(g) Kc = 50.0 at 400 °C
A mixture of gases at 400 °C with [H2]e1 = 0.221 M, [I2]e1 = 0.290 M, and [HI]e1 = 1.790 M is at equilibrium in a closed container. For this mixture, Qc = Kc = 50.0. If additional H2 is introduced into the container quickly such that [H2] doubles before it begins to react (that is, the new [H2]t = 0.442 M), Qc is now ½ of Kc:
$Q_c = \dfrac{[\text{HI}]_t^{\;2}}{[\text{H}_2]_t[\text{I}_2]_t} = \dfrac{(1.790 M)^2}{(0.442 M)(0.290 M)} = 25.0 = \frac{1}{2}K_c \nonumber$
The reaction then proceeds in the forward direction to reach a new equilibrium. Experimental measurements show that at the new equilibrium [H2]e2 = 0.362 M, [I2]e2 = 0.210 M, and [HI]e2 = 1.950 M. Notice that [H2]e2 (0.362 M) is less than the doubled concentration (0.442 M) but more than [H2]e1 (0.221 M). The equilibrium has shifted to partially counteract the change in H2 concentration. Because of the shift, the concentration of the other reactant decreases and the concentration of the product increases. To verify that these new concentrations are equilibrium concentrations, calculate Q:
$Q_c = \dfrac{(1.950 M)^2}{(0.362 M)(0.210 M)} = 50.0 = K_c \nonumber$
Another way to think about the shift in equilibrium is to recall that at equilibrium, rateforward = ratereverse. The forward reaction is first-order in H2, so doubling the concentration of H2 doubles rateforward, making rateforward > ratereverse. As the equilibrium shifts, the concentrations of both reactants decrease and the concentration of product increases until new equilibrium concentrations are reached, where forward and reverse reactions have reached new, but equal, rates.
Figure 3 illustrates graphically the effect of adding H2 to the reaction at equilibrium.
D30.5 Le Chatelier’s Principle: Change in Pressure or Volume
Changes in pressure have a measurable effect on equilibrium in systems involving gases if the chemical reaction produces a change in the total number of gas molecules (that is, if the number of gas molecules on the reactant side differs from the product side). The overall change in pressure must affect partial pressures of reactants and/or products: adding an inert gas that is not a reactant or product changes the total pressure but not the partial pressures of the gases in the equilibrium constant expression, and therefore does not perturb the equilibrium.
Consider what happens when the volume decreases for this equilibrium:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
Decreasing the volume increases the total pressure and increases the partial pressure of each gas. The equilibrium shifts to partially counteract the increase in pressure. Formation of additional NO2 decreases the total number of gaseous molecules in the system because each time two molecules of NO2 form, a total of three molecules of NO and O2 react away. This reduces the total pressure exerted by the system and partially counteracts increased pressure. LeChatelier’s principle predicts that the equilibrium shifts to the right, toward products.
We can also look at this by considering Qc. Reducing the system volume increases the partial pressure (or concentration) of all gaseous species. If the volume is reduced by half, then each partial pressure becomes twice what it was for the previous equilibrium:
$Q_c = \dfrac{[\text{NO}_2]_t^{\;2}}{[\text{NO}]_t^{\;2}[\text{O}_2]_t} = \dfrac{(2[\text{NO}_2]_{e1})^2}{(2[\text{NO}]_{e1})^2)(2[\text{O}_2]_{e1})} = \frac{1}{2}K_c \nonumber$
Because Qc < Kc, the reaction proceeds toward the product side (forming additional NO2) to re-establish equilibrium.
Now consider this reaction:
N2(g) + O2(g) ⇌ 2 NO(g)
Because there is no change in the total number of gaseous molecules in the system during reaction, a change in pressure does not shift the equilibrium towards either reactant or product side.
For reactions in solution, changing the solution volume changes the concentrations of all reaction species. Therefore, when solvent is added (the solution is diluted), the equilibrium shifts toward the side with more solute particles, partially compensating for the dilution of total concentration. If solvent is removed (by evaporation, for example) so that all solute concentrations increase, the equilibrium shifts toward the side with fewer solute particles, decreasing the total concentration of solute particles.
For example, when enough water is added to this equilibrium to double the solution volume:
C2H2(aq) + 2 Br2(aq) ⇌ C2H2Br4(aq)
the concentration of each solute is halved compared to the initial equilibrium concentration. Hence:
$Q_c = \dfrac{\frac{1}{2}[\text{C}_2\text{H}_2\text{Br}_4]_{e1}}{\left(\frac{1}{2}[\text{C}_2\text{H}_2]_{e1}\right)\left(\frac{1}{2}[\text{Br}_2]_{e1}\right)^2} = 4K_c \nonumber$
Because Qc > Kc the reaction proceeds toward the reactant side (the side with more solute particles) as equilibrium is re-established.
Exercise 3: Using Le Chatelier’s Principle at Constant Temperature
Podia Question
This question explores Le Chatelier’s principle applied to the equilibrium between hydrogen, nitrogen, and ammonia:
N2(g) + 3 H2(g) ⇌ 2 NH3(g) Kp = 5.8 × 106 atm−2 at 25 °C
a) Write the expression for Kp for this reaction.
b) Predict the effect on the equilibrium (at 25 °C) of an increase in each partial pressure:
i) PN2
ii) PH2
c) Suppose that 0.245 mol N2, 0.00145 mol H2, and 0.162 mol NH3 occupy a 10.0-L volume; the partial pressures are PN2= 0.600 atm, PH2= 0.00356 atm, and PNH3= 0.396 atm and the total pressure is 1.00 atm. Is the system at equilibrium? If not, in which direction would the reaction shift to reach equilibrium? Show a calculation to support your answer.
d) Suppose the total pressure is kept constant and that 0.010 mol N2 is added to the system described in part c so that the total amount of N2 is 0.255 mol and the total amount of all gases is 0.418 mol. If the total pressure stays constant, what must happen to the volume? Why?
e) Based on the ideal gas law it is possible to calculate the new volume of the mixture of gases and to calculate the partial pressures of the constituent gases. The partial pressures are PN2 = 0.610 atm, PH2 = 0.00347 atm, and PNH3 = 0.388 atm. Is the system at equilibrium? If not, in which direction does the reaction shift to reach equilibrium? Show a calculation to support your answer.
f) Does your answer in part (e) agree with your first answer in part b? Explain why or why not.
g) Have you discovered an exception to Le Chatelier’s principle? Explain why or why not.
Query $10$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.04%3A_Day_30-_ICE_Table_Reaction_Quotient_Le_Chateliers_Principle.txt |
D31.1 Le Chatelier’s Principle: Change in Temperature
When a chemical reaction is at equilibrium and the temperature changes, the reaction’s equilibrium constant is different at the new temperature. Le Chatelier’s principle can be used to predict which direction an equilibrium shifts and hence whether increasing temperature increases or decreases K. Remember that, according to Le Chatelier’s principle, an equilibrium shifts in a direction that partially counteracts the change in conditions.
Consider the reaction
N2(g) + O2(g) ⇌ 2 NO(g) ΔrH° = 180.5 kJ/mol at 25 °C
As shown by the enthalpy change, this reaction is endothermic: when the reaction takes place in the forward direction, the temperature is lowered. Because enthalpy change does not vary significantly with temperature, the forward reaction is endothermic at all temperatures where all reactants and products are in the gas phase. The reverse reaction is always exothermic.
Suppose that the reaction is at equilibrium at a particular temperature and the temperature is suddenly increased. To partially compensate for the temperature increase, the reaction shifts toward products (the endothermic direction), which lowers the temperature a bit. Thus, when equilibrium is reached at the higher temperature, the concentration of NO is larger and the concentrations of N2 and O2 are lower. This results in a larger value for Kc at the higher temperature.
If the temperature of the reaction is suddenly lowered, the reaction shifts to partially raise the temperature—in the exothermic direction. In this case, the shift is from products to reactants. Thus, at a lower temperature the concentrations of reactants are larger, the concentrations of products are smaller, and the equilibrium constant is smaller.
Summarizing,
• An increase in temperature shifts an equilibrium in the endothermic direction (the direction with positive ΔrH°) because the endothermic reaction partially counteracts the increase in temperature
• A decrease in temperature shifts an equilibrium in the exothermic direction (the direction with negative ΔrH°) because the exothermic reaction partially counteracts the decrease in temperature.
• The different concentrations in the new equilibrium system (after the shift resulting from the temperature change) correspond to a different value for the equilibrium constant.
• The larger the magnitude of ΔrH° is the larger the shift in the equilibrium is and the greater the change in the equilibrium constant is.
D31.2 Catalysts and Equilibrium
A catalyst speeds up the rate of a reaction, allowing the equilibrium to be reached more quickly (by speeding up both forward and reverse reactions). Hence, catalysts influence the kinetics of a reaction. However, a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.
D31.3 Gibbs Free Energy and Equilibrium
The standard Gibbs free energy change for a reaction indicates whether a reaction is product-favored at equilibrium (ΔrG° < 0) or reactant-favored at equilibrium (ΔrG° > 0). A strongly product-favored reaction (large negative ΔrG°) has a large equilibrium constant (K>> 1) and a strongly reactant-favored reaction (large positive ΔrG°) has a very small equilibrium constant (K<<1, a very small fraction because K cannot be negative). These qualitative statements suggest that there may be a quantitative relationship between the equilibrium constant and ΔrG° for a given reaction.
The reaction quotient, Q, was introduced as a convenient measure of the status of a reaction. When Q < K, the reaction proceeds spontaneously in the forward direction until equilibrium is reached (Q = K). Conversely, if Q > K, the reaction proceeds spontaneously in the reverse direction until equilibrium is achieved.
The relationship between ΔrG°, Q, and K is illustrated graphically in Figure 1 in graphs of G vs. reaction progress. In each graph, on the far left, the system is all reactants and Q = 0. On the far right, the system is all products and Q = ∞. The slope at any point on each graph is ΔrG/Δ(reaction progress). Because Δ(reaction progress) is always positive, the sign of the slope is the sign of ΔrG.
In the light cyan region where Q < K, the slope of the plot is negative, corresponding to negative ΔrG, which predicts spontaneous forward reaction. In the light pink region where Q > K, the slope is positive, corresponding to positive ΔrG which predicts spontaneous reverse reaction. Where the slope is zero (bottom of the curve), ΔrG = 0, and the system is at equilibrium with Q = K. Hence, we can think of reaction progress as rolling down the sides of a Gibbs free energy valley, with equilibrium at the bottom (minimum G).
Where equilibrium lies along the reaction progress depends on the sign of ΔrG°. When ΔrG° < 0, the equilibrium (minimum in the curve) is further to the right, indicating that there are more products than reactants when equilibrium is reached. When ΔrG° > 0, the equilibrium is further to the left, indicating that reactants predominate.
The Gibbs free energy change at any point along the reaction progress involves adjusting ΔrG° by the factor RT(lnQ):
ΔrG = ΔrG° + RT(lnQ)
At equilibrium, Q = K and ΔrG = 0, therefore:
0 = ΔrG° + RT(lnK°)
ΔrG° = −RT(lnK°) or K° = erG°/RT
Note that in these last equations the equilibrium constant is represented by K°. The standard equilibrium constant, Kº, is either the concentration equilibrium constant (Kc) with each concentration divided by the standard-state concentration of 1 M or the pressure equilibrium constant (Kp) with each pressure divided by the standard-state pressure of 1 bar. Hence, Kº is truly unitless. Dividing by the standard-state concentration or pressure means that if concentrations in Kc are expressed in M (mol/L) the numerical values of Kº and Kc are the same. Similarly, if partial pressures in Kp are expressed in bar, the numerical values of Kº and Kp are the same.
ΔrG°
> 1 < 0 Product-favored at equilibrium.
< 1 > 0 Reactant-favored at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.
D31.4 Effect of Temperature
Recall that:
ΔrG° = ΔrH° – TΔrS°
Therefore:
RT(lnK°) = ΔrH° – TΔrS°
This equation can be used to calculate K° at different temperatures, if we assume that ΔrH° and ΔrS° for a reaction have the same values at all temperatures. This is a good, but not perfect, assumption and we will use it in this course unless specified otherwise. It is not a good assumption if there is a phase change for a reactant or a product within the temperature range of interest.
Dividing both sides of the equation by -RT gives:
lnK° = -ΔᵣH°/RT + ΔᵣS°/R
y = mx + b
A plot of lnK° vs. 1/T is called a van’t Hoff plot. The graph has
$\text{slope} = -\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T}\right) \nonumber$
and
$\text{intercept} = \dfrac{\Delta_{r}S^{\circ}}{R} \nonumber$
If the concentrations of reactants and products are measured at various temperatures so that K° can be calculated at each temperature, both the reaction entropy change and enthalpy change can be obtained from a van’t Hoff plot.
Based on the equation for the van’t Hoff plot, an exothermic reaction (ΔrH° < 0) has K° decreasing with increasing temperature, and an endothermic reaction (ΔrH° > 0) has K° increasing with increasing temperature. This quantitative result agrees with the qualitative predictions made by applying Le Chatelier’s principle. It also shows that the magnitude of ΔrH° dictates how rapidly changes as a function of temperature. In contrast, ΔrS° affects the magnitude of but not its temperature dependence.
For example, suppose that K°1 and K°2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively:
$\text{ln}\;K_1^{\circ} = -\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_1}\right)\;+\;\dfrac{\Delta_{r}S^{\circ}}{R} \nonumber$
$\text{ln}\;K_2^{\circ} = -\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_2}\right)\;+\;\dfrac{\Delta_{r}S^{\circ}}{R} \nonumber$
Subtracting the two equations yields:
$\begin{array}{rcl} \text{ln}\;\text{K}_2^{\circ}\;-\;\text{ln}\;\text{K}_1^{\circ} &=& \left(-\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_2}\right)\;+\;\dfrac{\Delta_{r}S^{\circ}}{R}\right) - \left(-\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_1}\right)\;+\;\dfrac{\Delta_{r}S^{\circ}}{R}\right)\[0.5em] \text{ln}\;\dfrac{K_2^{\circ}}{K_1^{\circ}} &=& -\dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_2}\;-\;\dfrac{1}{T_1}\right)\[0.5em] &=& \dfrac{\Delta_{r}H^{\circ}}{R}\left(\dfrac{1}{T_1}\;-\;\dfrac{1}{T_2}\right)\end{array} \nonumber$
Thus calculating ΔrH° from tabulated ΔfH° and measuring the equilibrium constant at one temperature allows us to calculate the equilibrium constant at any other temperature (assuming that ΔrH° and ΔrS° are independent of temperature).
Exercise 4: Temperature Dependence of the Equilibrium Constant
Podia Question
Ammonium nitrate is an important fertilizer that supplies nitrogen to crops. It also can be used as an explosive to remove tree stumps from farm fields or in terrorist bombings. There are two reactions that can occur to produce the explosive effect. At lower temperatures ammonium nitrate decomposes to form dinitrogen monoxide and water vapor, but at higher temperatures it decomposes explosively to form nitrogen, water vapor, and oxygen. The reaction equations and thermodynamic parameters at 298 K are
(1) NH4NO3(s) → N2O(g) + 2 H2O(g) ΔrH° = −36.026 kJ/mol ΔrS° = 446.42 J K−1 mol−1
(2) NH4NO3(s) → N2(g) + ½ O2(g) + 2 H2O(g) ΔrH° = −118.076 kJ/mol ΔrS° = 520.749 J K−1 mol−1
a) Which of the reactions is(are) product-favored at room temperature? Which of the reactions is(are) product-favored at 1000 K? Explain briefly.
b) Is there a temperature below which one of the reactions is more product-favored and above which the other reaction is more product favored? If there is such a temperature for these two reactions, determine what that temperature is.
c) Given your results, write an appropriate scientific explanation of the fact that nitrogen and oxygen are produced explosively at higher temperatures but dinitrogen monoxide is produced at lower temperatures.
Query $5$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.05%3A_Day_31-_Le_Chateliers_Principle_Equilibrium_and_Gibbs_Free_Energy.txt |
D32.1 Gibbs Free Energy and Work
Recall that when we we talk about kinetics of a reaction, we are concerned with the rate of the reaction: how fast it goes from reactants to products. When we talk about the thermodynamics of a reaction, we are concerned primarily with the difference in energy between reactants and products, but not with the mechanism by which reactants change into products.
When the Gibbs free energy of the products is lower than that of the reactants, a reaction is said to be exergonic. Conversely, an endergonic reaction is one in which the products are higher in Gibbs free energy than the reactants.
When there is a decrease in Gibbs free energy as a reaction occurs, ΔrG equals the maximum useful work that can be done by the reaction system. ΔrG = wmax. (The negative sign reflects the fact that w is defined as work done on the system.) Conversely, if a reaction has positive ΔrG, work must be done on the system to force the reaction to occur. The minimum work that must be done is given by ΔrG.
Activity 1: Exergonic and Endergonic Reactions
Query \(1\)
When considering a reaction under standard-state conditions the relevant thermodynamic quantity is ΔrG°. If a reaction is exergonic under standard-state conditions, ΔrG° < 0.
Coupled Reactions
One way to allow a reactant-favored process to occur is to couple it with a reaction that is product-favored. For example, consider the recovery of aluminum from alumina (Al2O3) ore:
Al2O3(s) → 2 Al(s) + 3/2 O2(g) ΔrG° (298 K) = 1576.4 kJ/mol
At least 1576.4 kJ of work must be done to change 1 mol Al2O3(s) into 2 mol Al(s) and 1.5 mol O2(g) (at 1 bar). In a modern aluminum manufacturing plant, this work is supplied electrically and the electricity is often provided by burning coal. Assuming coal to be mainly carbon, the combustion reaction is:
C(s) + O2(g) → CO2(g) ΔrG° (298 K) = −394.4 kJ/mol
Thus the ΔrG° values indicate that, under standard-state conditions and ideal 100% efficiency, at least four moles of carbon/coal must burn to process each mole of Al2O3 ore. (In practice the aluminum smelting process is only 17% efficient, so it is necessary to burn nearly 6 times the theoretical amount of coal.) Coupled reactions occur simultaneously and there is a means of exchanging energy between them. The energy exchange occurs via the electric power grid in this specific case.
In other words, a reaction that is endergonic under standard-state conditions can be coupled to a separate exergonic reaction that drives the endergonic reaction (the thermodynamically unfavorable one) to occur. The ΔrG° values for the two coupled reactions are summed to yield the overall ΔrG°. For this example, multiply the second reaction by 4, add the reaction equations, and apply Hess’s Law, gives:
Al2O3(s) + 4 C(s) + 4 O2(g) → 2 Al(s) + 3/2 O2(g) + 4 CO2(g) ΔrG°(298 K) = -1.2 kJ/mol
The overall reaction now has a negative ΔrG° and is product-favored.
Under non-standard-state conditions a reaction with ΔrG < 0 can drive a reaction with ΔrG > 0, provided energy can be transferred from one to the other.
D32.2 Gibbs Free Energy in Biological Systems
Biological organisms often couple the product-favored hydrolysis of ATP (adenosine triphosphate) to a reactant-favored reaction. Thus ATP hydrolysis reaction can be used to drive a necessary, but thermodynamically unfavorable, reaction.
ATP + H2O ⇌ ADP + H2PO4 ΔrG° = −30.5 kJ/mol
The triphosphate part of ATP is an inorganic ester. It can be formed by condensation reaction of ADP and H2PO4 with water formed as a byproduct (the reverse of the reaction shown above). ATP can be made available in an organism where an endergonic reaction needs to occur. Its hydrolysis can then be coupled with the endergonic reaction to yield a thermodynamically favorable overall reaction.
For example, ATP hydrolysis can be used to drive condensation reactions of amino acids to generate proteins as graphically illustrated by Figure 1.
Figure 1 shows ATP formation being initially coupled to the glucose oxidization reaction:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ΔrG°=−2880 kJ/mol
which has close to 100x greater capability to do work than the hydrolysis of a single ATP. Hence, the equilibrium for this reaction so strongly favors the products that a single arrow is typically used in the chemical equation as it is essential irreversible. It may not be surprising that glucose and all sugars are very energetic molecules since they are the primary energy source for life.
D32.3 Kinetic Metastability
At a given temperature, the rate law and rate constant can be used to determine how rapidly reactants are converted to products. The equilibrium constant expression and the value of K°, on the other hand, can be used to determine the equilibrium concentrations of products relative to reactants. In other words, kinetics describes how fast equilibrium is reached, and thermodynamics describes where the equilibrium lies. However, there is not necessarily a correlation between a fast reaction and one that is product-favored at equilibrium. Both kinetics and thermodynamics are needed to characterize a chemical reaction because a useful reaction usually is one where significant quantities of products can be produced in a short time.
On the other hand, it is often true that a substance is valuable for some purpose when it is stable and does not change into some other substance. For example, iron and steel are useful for making automobiles and constructing buildings precisely because they are stable. When discussing the concept of stability, it is useful to distinguish between thermodynamic stability and kinetic metastability.
Consider a generic reaction:
A ⇌ B ΔrG° < 0
Here product B has lower ΔfG° than reactant A so that ΔrG° of the forward reaction is negative. When a reaction favors products at equilibrium, we say that products are thermodynamically stable relative to reactants. In the example above, product B is thermodynamically stable compared to reactant A. However, if the activation-energy barrier (Ea) is high, at a given temperature the reaction could proceed very slowly, and reactant A would be described as being inert (unreactive). We say that compound A is kinetically metastable (or kinetically stable) relative to compound B.
Notice that stability (and metastability) are defined by comparing one substance with another. It is possible that some other substance, say C, is even more stable than B and therefore B is thermodynamically unstable relative to C.
For example, diamond and graphite are two compounds composed solely of carbon atoms. The title of an old James Bond film, and an even older advertising slogan, says, “Diamonds Are Forever.” This implies some stability: we don’t expect the diamond in a ring to change anytime soon.
Exercise 3: Diamond and Graphite
C(s, diamond) ⟶ C(s, graphite) ΔrG° = −2.9 kJ/mol
Diamond is kinetically metastable. Thermodynamics says it should change to graphite, but the change is so slow as to be essentially undetectable at day-to-day temperatures in a human lifetime. The fact that diamond exists is due to a very large activation-energy barrier for conversion of diamond to graphite; diamond would convert to graphite at temperatures of >4500 K.
This very large Ea, as well as the facts that diamond is the hardest known solids and graphite is one of the softest, can be explained by differences in the way the atoms are bonded. In diamond, every carbon atom has sp3 hybridization and each sp3 carbon is bonded to 4 other sp3 carbon atoms at the corners of a tetrahedron. In graphite, every carbon atom has sp2 hybridization and each sp2 carbon is bonded to 3 other sp2 carbon atoms in planar sheets of connected benzene rings. Because the sheets can slide past one another relatively easily, graphite is soft and slippery.
Conversion of diamond to graphite requires breaking numerous C−C single bonds with bond energy of 356 kJ/mol. There is no easy mechanism for this conversion and so transforming diamond into graphite, or vice versa, requires almost as much energy as destroying the entire crystal lattice and rebuilding it. Therefore, once diamond is formed, it cannot convert back to graphite under normal conditions because Ea is too high: diamond is said to be metastable because its stability depends on kinetics, not thermodynamics.
D32.4 Haber-Bosch Process
The interplay of thermodynamics and kinetics is illustrated in the industrial synthesis of ammonia. It became possible to manufacture ammonia in useful quantities in the early 20th century after the factors that influence this reaction equilibrium were understood:
N2(g) + 3 H2(g) ⇌ 2 NH3(g) ΔrH° = -92.2 kJ/mol
Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. It plays a vital role in the global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for enhancing growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.
To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system, which shifts the reaction equilibrium towards the product side, increasing the concentration and partial pressure of ammonia.
At low temperatures, the rate of formation of ammonia is slow so equilibrium would be achieved more quickly at higher temperatures. However, the reaction is exothermic. Increasing the temperature to increase the rate shifts the equilibrium in the endothermic direction and lowers the product yield.
Part of the slower rate caused by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly.
Ammonia, the reaction product, has a higher boiling point than the reactants, nitrogen and hydrogen; thus, ammonia can be condensed to a liquid at temperatures where N2 and H2 remain gases. Condensing ammonia by refrigeration of the gaseous mixture removes product, shifting the equilibrium to the right.
Exercise 5: Intermolecular Forces and Boiling Point
Ammonia has a higher boiling point than nitrogen and hydrogen because of intermolecular forces.
Query \(7\)
In the commercial production of ammonia, conditions are typically 400 – 500 °C and 150–250 bar. The catalyst consists of Fe3O4 mixed with KOH, Al2O3, and SiO2. This gives the best compromise among rate, product yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 3).
Podia Question
Consider the addition of HBr to 1,3-butadiene. Two products are possible: 1,2-product and 1,4-product (see left side of figure).
On the right side of the figure is a diagram showing Gibbs free energy as a function of reaction progress. On the horizontal axis reactants are in the middle. The 1,2-product is reached by moving left. The 1,4-product is reached by moving right.
1. Write an explanation in scientifically appropriate language for the fact that at lower temperatures the reaction produces mainly 1,2-product while at higher temperatures the reaction produces mainly 1,4-product. If any assumptions need to be made, say what they are.
2. A student says that at lower temperatures the 1,2-product is kinetically metastable compared with the 1,4-product. Is this a correct statement? Why or why not?
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.06%3A_Day_32-_Gibbs_Free_Energy_and_Work_Kinetic_Metastability.txt |
Day 33: Acids and Bases
We will apply the knowledge we learned so far, e.g. thermodynamics, kinetics, molecular structure, etc., to explore two prevalent types of chemical reactions: acid-base reactions and redox/electrochemical reactions.
D33.1 Definition of Acids and Bases
According to the Brønsted-Lowry acid-base definition, a chemical species that donates a proton (hydrogen ion, H+) to another chemical species is called an acid and a chemical species that accepts a proton is a base. (Recall that when a H atom loses an electron, only the proton in the nucleus remains, so a proton is a H+ ion.) An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base).
Query $1$
The chemical species that remains after an acid has donated a proton is called the conjugate base of that acid. Consider these examples of the generic reaction acid + H2O ⇌ conjugate base + H3O+:
HF + H2O ⇌ F + H3O+
H2SO4 + H2O ⇌ HSO4 + H3O+
HSO4 + H2O ⇌ SO42- + H3O+
NH4+ + H2O ⇌ NH3 + H3O+
Similarly, the chemical species that forms after a base accepts a proton is called the conjugate acid of that base. Consider the following examples of the generic reaction base + H2O ⇌ conjugate acid + OH:
NH3 + H2O ⇌ NH4+ + OH
S2- + H2O ⇌ HS + OH
CO32- + H2O ⇌ HCO3 + OH
F + H2O ⇌ HF + OH
From these examples, we can see that the conjugate acid and conjugate base are paired: the conjugate base of an acid has that acid as its conjugate acid. For instance, NH3 is the conjugate base of NH4+, while NH4+ is the conjugate acid of NH3. Similarly, F‾ is the conjugate base of HF, while HF is the conjugate acid of F‾.
You may have noticed in earlier sections that we wrote H+(aq) to represent hydrogen ions in aqueous solution, while in the preceding reactions we have written H3O+. A H+ ion, a proton, is much smaller than any other cation and therefore is a highly concentrated positive charge that strongly attracts water-molecule dipoles and forms strong hydrogen bonds. Thus, when a Brønsted-Lowry acid transfers a proton to water, more than a single water molecule accepts the proton. Experiments show that as many as six water molecules may be involved, which would make the formula of H+(aq) H13O6+, but the structure changes continually as water molecules move about in the liquid. For Brønsted-Lowry acid-base reactions, we will use H3O+(aq) to emphasize that a proton has been transferred to water and to represent the more complicated actual structure. In general, H+(aq) is appropriate to represent a proton surrounded by many water molecules.
Finally, Brønsted-Lowry acid-base reactions are very fast. Their equilibrium is established quickly and this equilibrium is an important aspect when we consider acid and base strengths later.
D33.2 Autoionization of Water
In the example reactions above, there are also two other conjugate acid-base pairs: H2O is the conjugate base of its conjugate acid H3O+, and H2O is the conjugate acid of its conjugate base OH. (However, note that H3O+ is not the conjugate acid of OH; these two species are not a conjugate acid-base pair because their structures do not differ by a single H+.)
Hence, H2O can react as an acid or a base depending on the other species involved in the reaction. In pure water, H2O acts as both acid and base—a very small fraction of water molecules donate protons to other water molecules:
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 109 molecules are ionized at 25 °C. The [H3O+]e and [OH]e give an autoionization constant for water, Kw = 1.0 × 10−14 at 25 °C. Because it is the mathematical product of concentrations of two ions, it is also called the ion-product constant for water.
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH(aq) Kw = [H3O+]e[OH]e = 1.0 × 10−14 at 25 °C
Activity 1: Autoionization of Water
Exercise 2: Autoionization of Water
Water is an example of an amphiprotic chemical species a molecule that could either gain a proton or lose a proton in an acid-base reaction. Amphiprotic species are also amphoteric, 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 definition). For example, the bicarbonate ion is also amphoteric:
HCO3(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)
HCO3(aq) + H2O(l) ⇌ H2CO3(aq) + OH(aq)
Activity 2: Amphiprotic Species
D33.3 pH and pOH
The concentrations of H3O+ and OH in a solution are important for the solution’s acid-base properties and often affect the chemical behaviors of other solutes. A solution is neutral if its [H3O+]e = [OH]e; acidic if its [H3O+]e > [OH]e; and basic if its [H3O+]e < [OH]e.
A common means of expressing values that span many orders of magnitude is to use a logarithmic scale. One such scale is based on the p-function:
pX = -logX
where “X” is the quantity of interest and “log” is the base-10 logarithm. The pH of a solution is therefore defined as:
$\text{pH}\; =\; -\text{log}\left(\dfrac{[\text{H}_3\text{O}^+]_e}{mol/L}\right) \nonumber$
The reason for dividing by the units “mol/L” (M) is that [H3O+] has units of mol/L and taking the logarithm of a unit makes no sense. From here on we will assume that you are aware that only the numeric value of a concentration (or other quantity) needs to be used as the argument of a logarithm and we will not explicitly divide by the units.
If a pH value is known, the concentration of hydronium ions can be calculated:
[H3O+]e = 10-pH
Here we assume that you know that units are required for the concentration obtained from this equation and the units are mol/L. Similarly, the hydroxide ion concentration may be expressed as pOH:
pOH = -log[OH]e and [OH]e = 10−pOH
Finally, the relation between pH and pOH can be derived from the Kw expression:
Kw = [H3O+]e[OH]e
-log(Kw) = -log([H3O+]e[OH]e)
pKw = -log([H3O+]e) + (-log([OH]e))
pKw = pH + pOH
At 25 °C:
pKw = 14 = pH + pOH
Therefore, at this temperature:
Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] < 7
neutral [H3O+] = [OH] 7
basic [H3O+] < [OH] > 7
Because Kw is temperature dependent, the correlations between pH values and the acidic/neutral/basic adjectives are different at different temperatures. For example, [H3O+] in pure water at 80 °C is 4.9 × 10−7 M, which corresponds to pH and pOH values of:
pH = -log[H3O+]e = -log(4.9 × 10−7) = 6.31
pOH = -log[OH]e = -log(4.9 × 10−7) = 6.31
At this temperature, neutral solutions have pH = pOH = 6.31, acidic solutions have pH < 6.31 and pOH > 6.31, and basic solutions have pH > 6.31 and pOH < 6.31. This distinction can be important when studying certain processes that occur at temperatures other than 25 °C, such as acid-base reactions in the human body where temperatures are typically 37 °C.
Unless otherwise noted, references to pH values are presumed to be those at 25 °C. Figure 1 shows the relationships among [H3O+], [OH], pH, and pOH, and gives values for these properties for some common substances.
Activity 3: pH and Relative Strengths of Acids
Exercise 3: pH of Aqueous Solutions
The acidity of a solution is typically determined by measuring its pH. The pOH of a solution is not usually measured, but it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 2), or visually estimated using colored indicators (Figure 3).
D33.4 Acid Constant Ka and Base Constant Kb
The relative strengths of acids and bases may be determined by comparing the equilibrium constants for their ionization reactions. For the reaction of a generic acid, HA, in water:
HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)
we write the acid ionization constant (Ka) expression as:
$K_{\text{a}} = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^{-}]_e}{[\text{HA}]_e} \nonumber$
(Although water is a reactant in the reaction, it is also the solvent with its phase indicated as “l, so we do not include [H2O] in the expression.)
An acid with a larger Ka would have a larger concentration of H3O+ and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid, which ionizes to a greater extent, has a larger ionization constant than a weaker acid.
For example, these data on acid ionization constants:
CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq) Ka = 1.8 × 10-5
HNO2(aq) + H2O(l) NO2(aq) + H3O+(aq) Ka = 7.4 × 10-4
HSO4(aq) + H2O(l) SO42-(aq) + H3O+(aq) Ka = 1.1 × 10-2
indicate that the order of acid strength is: acetic acid (CH3COOH) is a weaker acid than nitrous acid (HNO2) which is itself weaker than hydrogen sulfate ion (HSO4).
We can consider the strength of a base (B) similarly by considering the extend that it will form hydroxide ions in an aqueous solution:
B(aq) + H2O(l) ⇌ HB+(aq) + OH(aq)
where the base ionization constant (Kb) expression is:
$K_{\text{b}} = \dfrac{[\text{HB}^{+}]_e[\text{OH}^{-}]_e}{[\text{B}]_e} \nonumber$
A stronger base ionizes to a greater extent than does a weaker base. Therefore, a stronger base has a larger Kb than a weaker base.
Notice that Ka and Kb provide a quantitative measure of acid and base strengths—significantly more accurate than qualitative descriptions of “strong acid” or “weak acid”.
Query $8$
Consider the ionization reactions for a conjugate acid base pair, HA and A:
HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)
$K_a = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^-]_e}{[\text{HA}]_e} \nonumber$
A(aq) + H2O(l) ⇌ HA(aq) + OH(aq)
$K_b = \dfrac{[\text{HA}]_e[\text{OH}^-]_e}{[\text{A}^-]_e} \nonumber$
Adding these two chemical equations yields the equation for the autoionization of water:
HA(aq) + H2O(l) + A(aq) + H2O(l) ⇌ A(aq) + H3O+(aq) + HA(aq) + OH(aq)
When two equilibria sum to a third equilibrium, the product of the first two equilibrium constants equals the third equilibrium constant:
$K_{\text{a}}\;\times\;K_{\text{b}} = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^{-}]_e}{[\text{HA}]_e}\;\times\;\dfrac{[\text{HA}]_e[\text{OH}^{-}]_e}{[\text{A}^{-}]_e} = [\text{H}_3\text{O}^{+}]_e[\text{OH}^{-}]_e = K_{\text{w}} \nonumber$
For example, at 25 °C, Ka of acetic acid (CH3COOH) is 1.8 × 10−5 M, and Kb of its conjugate base, acetate anion (CH3COO), is 5.6 × 10−10 M. The product of these two constants is indeed equal to Kw:
Ka × Kb = (1.8 × 10−5 M) × (5.6 × 10−10 M) = 1.0 × 10−14 = Kw
This relationship tells us that stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases.
Although “strong” and “weak” are relative terms, generally, we refer to acids stronger than H3O+ as strong acids, and bases stronger than OH as strong bases. Because strong acids and strong bases are completely ionized in aqueous solutions, the concentration of nonionized acid or base is essentially zero. For example, in a 0.10-M solution of HCl, [HCl]e = 0, [H30+]e = 0.10 M, and [Cl]e = 0.10 M.
A consequence of this complete ionization is that in aqueous solution there is no way to tell whether one strong acid is stronger than another: HCl, HBr, and HI all are completely ionized. This is known as the leveling effect of water. However, when dissolved in some other solvents, these acids do not ionize completely. The extent of ionization increases in the order HCl < HBr < HI, and so HI is the strongest of these acids. Water exerts a similar leveling effect on strong bases.
Many acids and bases are considered “weak”. A solution of a weak acid in water is a equilibrium mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid.
Activity 4: Determining Ka
Query $9$
Activity 5: Using Ka to Calculate Concentrations
Query $10$
The percent ionization of a weak acid is another measure of the strength of an acid, HA:
$\text{percent ionization} = \dfrac{[\text{H}_3\text{O}^{+}]_e}{[\text{HA}]_0}\;\times\;100\% \nonumber$
A stronger acid, with a higher Ka, has higher percent ionization than a weaker acid (for the same concentration).
The percent ionization for a solution of a weak acid increases with decreasing acid concentration; this can be seen by applying Le Chatelier’s principle to the ionization equilibrium:
HA(aq) + H2O(l) ⇌ A(aq) + H3O+(aq)
Increasing the solution volume for a given quantity of HA added causes the equilibrium to shift to the product side to partially counteract the decrease in total solute concentration.
D33.5 Acid Strength and Molecular Structure
Acid-base reactions, like many other chemical reactions, involve breaking and forming bonds. Hence, we can use our chemical understanding of molecular structure and stability to understand what makes some acids stronger than others.
We know that an equilibrium favors the thermodynamically more stable side of the reaction, and that the magnitude of the equilibrium constant reflects the energy difference (ΔrG°) between the reactants and products. Therefore, in an acid-base equilibrium, the equilibrium always favors the side with weaker acid and base (these are the more stable species). Consequently, anything that stabilizes the conjugate base of an acid will necessarily make that acid stronger, and anything that stabilizes the acid will make it a weaker acid. This idea is illustrated in Figure 6.
Bond Strength
Let’s consider a generic acid:
HA ⇌ A + H+
In general, the stronger the H–A bond is, the more stable the HA molecule is, and the less acidic the substance is. This effect is illustrated by the hydrogen halides:
Relative Acid Strength HF HCl HBr HI
H–X Bond Enthalpy (kJ/mol) 566 431 366 299
pKa 3.2 −6.1 −8.9 −9.3
Note that the “bond enthalpy” values are associated with a different bond breaking reaction, which produces a hydrogen atom instead of a hydrogen ion:
HA ⇌ A· + H·
Nonetheless, bond strengths correlate with acid strength. As you go down the halide group, the overlap between the hydrogen 1s orbital and the valence orbital of the halogen atom decreases, and the H-X bond enthalpy decreases, indicating weaker bonds. As a result, HX acid strength increases as you go down the group.
A similar trend is observed for other groups. For example, for group 16:
H2A ⇌ HA + H+
H2O H2S H2Se H2Te
pKa 14.0 7.0 3.9 2.6
Stabilizing the Excess Charge on Conjugate Base
Electronegativity
When the conjugate base is negatively charged, factors that stabilize the excess negative charge on the conjugate base favor the dissociation of the acid and make the acid a stronger acid. For example, the relative acidity of the acids with second row elements is:
HnA ⇌ Hn-1A + H+
CH4 NH3 H2O HF
pKa 50 36 14.0 3.2
Consider the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of CH4 is CH3, and the conjugate base of HF is F. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the extra negative charge in the F anion than carbon can stabilize the extra negative charge in the CH3 anion. Consequently, HF can dissociate and form H+ and F to a much greater extent than CH4 can form H+ and CH3, making HF a much stronger acid than CH4.
The same trend is predicted by analyzing the acids as well: as the electronegativity of A in HnA increases, the A–H bond becomes more polar, favoring dissociation to form Hn-1A and H+. Due to both the increasing stability of the conjugate base and the increasing polarization of the A–H bond in the acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Electron Delocalization
The pKa values of some carboxylic acids are shown in Figure 7.
The bond that breaks in a carboxylic acid is the O–H bond:
R-COOH ⇌ R-COO‾ + H+
An O–H bond also breaks when an alcohol acts as an acid in an acid-base reaction:
R-OH ⇌ R-O‾ + H+
However, when we compare the carboxylic acid pKa‘s with those of comparable alcohols, such as:
it is clear that carboxylic acids are stronger acids than alcohols by over ten orders of magnitude. Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton?
Both the carboxylic acid and its conjugate base, carboxylate anion, are stabilized by having additional resonance structures that delocalize electron densities. However, the stabilization in carboxylate anion is much greater because the two major resonance structures have equal contributions to the resonance hybrid, and the extra electron density (negative charge) is shared equally between the two oxygen atoms, which have high electronegativity. This stabilization leads to a markedly increased acidity of carboxylic acids.
Query $12$
Figure 8. Greater resonance stabilization of carboxylate ion compared to carboxylic acid results in increased acid strength. Click on each “i” for more information.
Inductive Effect
Atoms (or groups of atoms) in a molecule that are not directly bonded to the acidic H can also influence the molecule’s acidity. They can do so via an inductive effect, that is, they induce a polarization in the distribution of electrons within the molecule. This can be seen by studying the structures in Figure 7 above. Electronegative substituents (F, Cl, Br, I) near the carboxyl group act to increase the acidity of the carboxylic acid. For example, fluoroacetic acid, CH2FCOOH, is significantly more acidic than acetic acid, CH3COOH.
The magnitude of the inductive effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
CH3COOH CH2ClCOOH CHCl2COOH CCl3COOH CF3COOH
pKa 4.74 2.85 1.35 0.77 0.52
Fluorine, which is more electronegative than chlorine, causes a larger inductive effect as it draws electron density away from the acidic proton in the carboxyl group. Moreover, having three halogens causes a larger inductive effect than having two or one. Note that inductive effects can be quite significant. For instance, replacing the –CH3 group of acetic acid by a –CF3 group results in a almost 10,000-fold increase in acidity.
In another example, the acidity of hypohalous acids (HOX ⇌ OX‾ + H+) varies by about three orders of magnitude due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 2.73 7.2
HOBr 2.64 8.5
HOI 2.11 10.5
As the electronegativity of X increases, 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, enhancing ionization of hydrogen as H+ and making the acid stronger.
Oxoacids
The acidity of oxoacids, with the general formula HOXOn (with n = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom X (Figure 8) due to both an inductive effect and increased stabilization of the conjugate base.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms (oxygen atoms only bonded to the central atom) causes electron densities to be drawn away from the O–H bond, thereby increasing the strength of the acid. Figure 9 shows how the H atom becomes steadily more blue from HClO to HClO4, making it easier for the acid to lose the hydrogen as an H+ ion.
Also important is the effect of electron delocalization that stabilizes the extra negative charge in the conjugate base. For example, in the chlorite anion (ClO2‾), the excess electron density is delocalized equally over both oxygen atoms, whereas in the hypochlorite ion (ClO), the negative charge is largely localized on a single oxygen atom:
As a result of this stabilization of the conjugate base, as well as additional inductive effect, chlorous acid is more than 200,000 times stronger than hypochlorous acid.
The inductive effect is additionally 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. For example, H3PO4 is a weak acid, H2SO4 is a strong acid, and HClO4 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 more electron densities to be drawn from OH group to the central atom, weakening the O–H bond and increasing the strength of the oxoacid.
Podia Question
Arrange these acids in order of decreasing acid strength (smallest pKa first, largest pKa last). Analyze the factors that affect the strength of each acid and based on that analysis explain the order of acidity.
CH3CH2OH CF3COOH HOClO3 CH2BrCOOH HOClO2
Query $13$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.07%3A_Day_33-_Acids_and_Bases.txt |
D34.1 Polyprotic Acids
We can classify acids by the number of protons per molecule that they can donate in an acid-base reaction. Acids that contain one ionizable hydrogen atom per molecule are called monoprotic acids. Examples are HCl, HNO3, CH3COOH, and HCN.
Even though it contains four hydrogen atoms, acetic acid is also monoprotic because only the hydrogen atom from the carboxyl group (-COOH) reacts with bases:
The three hydrogen atoms in the methyl group are not reactive (the C–H bonds are similar those in alkanes, which are unreactive).
In the same vein, monoprotic bases are bases that accept a single proton.
Diprotic acids contain two ionizable hydrogen atoms per molecule. The dissociation of the first H+ always takes place to a greater extent than the dissociation of the second H+. For example, sulfuric acid ionizes in two steps:
H2SO4(aq) + H2O(l) HSO4(aq) + H3O+(aq) Ka,1 > 102
HSO4(aq) + H2O(l) SO42-(aq) + H3O+(aq) Ka,2 = 1.1 × 10-2
This stepwise ionization occurs for all polyprotic acids.
A solution of a weak diprotic acid contains a mixture of acids. For example, when carbonic acid loses one H+, it yields hydronium ions and bicarbonate ions in small quantities:
H2CO3(aq) + H2O(l) ⇌ HCO3(aq) + H3O+(aq)
$K_{\text{a, H}_2\text{CO}_3} = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{HCO}_3^{\;-}]_e}{[\text{H}_2\text{CO}_3]_e} = 4.3\;\times\;10^{-7} \nonumber$
The bicarbonate ion can lose an H+ to form hydronium ions and carbonate ions in even smaller quantities:
HCO3(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)
$K_{\text{a, HCO}_3^-} = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{CO}_3^{\;2-}]_e}{[\text{HCO}_3^{\;-}]_e} = 4.7\;\times\;10^{-11}\;\text{M} \nonumber$
Ka(H2CO3) is larger than Ka(HCO3‾) by about four orders of magnitude (104 times larger), so H2CO3 is the dominant producer of H3O+ in the solution. This means that the concentrations of H3O+ and HCO3 are practically equal in a pure aqueous solution of H2CO3.
If Ka,1 of a weak diprotic acid at least 20 times larger than Ka,2, 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.
Activity 1: Ionization of a Diprotic Acid
Query $1$
A triprotic acid is an acid that has three protons that undergo stepwise ionization: Phosphoric acid is an example:
H3PO4(aq) + H2O(l) ⇌ H2PO4(aq) + H3O+(aq) Ka,1 = 7.2 × 10-3 M
H2PO4(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq) Ka,2 = 6.3 × 10-8 M
HPO42-(aq) + H2O(l) ⇌ PO43-(aq) + H3O+(aq) Ka,3 = 4.6 × 10-13 M
Again, the differences in the ionization constants of these reactions tell us that the degree of ionization is significantly weaker in each successive step. This is a general characteristic of polyprotic acids. Here, because the successive ionization constants differ by a factor of 105-106, the calculations of equilibrium concentrations in a solution of H3PO4 can be broken down into a series of parts, similar to activity 1.
Polyprotic bases can accept more than one H+. The carbonate ion is an example of a diprotic base, because it can accept up to two protons. Solutions of alkali metal carbonates (e.g. K2CO3) are quite alkaline, due to the reactions:
CO32-(aq) + H2O(l) ⇌ HCO3(aq) + OH(aq)
HCO3(aq) + H2O(l) ⇌ H2CO3(aq) + OH(aq)
D34.2 Acid-Base Reactions
Mixing a solution of an acid with a solution of a base results in an acid-base neutralization reaction that produces a salt and water. The thermodynamics of an acid-base reaction dictates that the side with the weaker acid and weaker base is favored. In other words, if the weaker acid and weaker base are on the left side of an equilibrium reaction, the reaction is reactant-favored at equilibrium; if the weaker acid and weaker base are on the right side, the reaction is product-favored at equilibrium. Strengths of acids and bases are quantitatively comparable by their Ka and Kb values, which can be obtained from a reference table.
A strong acid reacts with a strong base to form a neutral solution (containing equal concentrations of H3O+ and OH‾) provided that stoichiometrically equivalent quantities of acid and base are mixed. For example:×
HCl(aq) + NaOH(aq) ⇌ NaCl(aq) + H2O(l)
The salt formed, NaCl(aq), consists of Na+(aq) and Cl(aq) , each of which has negligible acid or base strength. Hence, this equilibrium heavily favors the product side and goes essentially to completion. (Note that any soluble salt consists of aqueous ions, so the formula NaCl(aq) represents an aqueous solution consisting of the same number of Na+(aq) and Cl(aq) ions.) If the mixture has an excess of one of the reactants, then the concentration of leftover acid (HCl) or base (NaOH) determines the pH of the solution.
A weak acid reacts with a strong base to form a salt that contains the conjugate base of the weak acid, which is usually a weak base. For example, the reaction of acetic acid with sodium hydroxide forms sodium acetate:
CH3COOH(aq) + NaOH(aq) ⇌ CH3COONa(aq) + H2O(l)
The equilibrium of this reaction favors the product side, and the reaction can be approximated as going to completion. Mixing stoichiometrically equivalent amounts of reactants gives a solution containing Na+(aq), which has no effect on the pH of the solution, and CH3COO(aq), the conjugate base of acetic acid. Because the acetate anion is a weak base, the solution pH is >7 after acetic acid reacts stoichiometrically with a strong base. The weak-base reaction is:
CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq)
The equilibrium constant for this reaction is the ionization constant, Kb, for the acetate anion. (Some reference tables only report ionization constants for acids; Kb can be calculated from Kw and Ka of the conjugate acid—acetic acid in this case.) Generalizing this example, when a strong base reacts stoichiometrically with a weak acid, the solution that results is basic.
Exercise 2: Using a Base Ionization Constant
A strong acid reacting with a weak base forms a salt containing the conjugate acid of the weak base, which is usually a weak acid. For example, the reaction of HCl with ammonia forms ammonium chloride:
NH3(aq) + HCl(aq) ⇌ NH4Cl(aq)
The equilibrium of this reaction favors the product side, and the reaction can be approximated as going to completion. Mixing stoichiometrically equivalent amounts of reactants gives a solution that contains Cl(aq), which is the conjugate base of a strong acid and has no effect on the pH of the solution, and NH4+(aq), the conjugate acid of ammonia. Because the ammonium ion is a weak acid, the solution pH would be <7 after ammonia reacts stoichiometrically with a strong acid. The reaction is:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
The equilibrium constant for this reaction is the ionization constant, Ka, for the acid NH4+. Generalizing this example, when a weak base reacts stoichiometrically with a strong acid, the solution that results is acidic.
Activity 2: pH of an Ammonium Salt
Query $4$
To predict the pH of a solution resulting from the reaction between a weak acid and a weak base, we must know both the Ka of the weak acid and the Kb of the weak base. If Ka > Kb, the solution is acidic; if Kb > Ka, the solution is basic.
D34.3 Reaction Between Amphiprotic Species
Acid-base reactions can also occur between two amphiprotic species. For example, mixing a solution containing hydrogen sulfate ions (HSO4) and a solution containing hydrogen carbonate ions (HCO3) results in an acid-base reaction. However, if both reactants can act as either an acid or a base, which reactant is the acid and which is the base? In the example mixture, there are two possibilities:
possibility I: HSO4(aq) + HCO3(aq) ⇌ SO42-(aq) + H2CO3(aq)
possibility II: HSO4(aq) + HCO3(aq) ⇌ H2SO4(aq) + CO32-(aq)
Qualitatively, a product-favored acid-base reaction involves a stronger acid reacting with a stronger base to form a weaker acid and a weaker base. Acid and base strengths are comparable by Ka and Kb values.
In possibility I the acids are HSO4 (Ka = 1.1 × 10-2) and H2CO3 (Ka = 4.3 × 10-7) and the bases are HCO3 (Kb = 2.3 × 10-8) and SO42 (Kb = 9.1 × 10-13). The stronger acid and the stronger base are on the left side of the equation so this reaction is product-favored.
On the other hand, possibility II is reactant-favored because it produces H2SO4, a strong acid, and CO32-, a weak base with a relatively large Kb = 2.1 × 10−4 (significantly larger than the Kb for HSO4).
Exercise 4: Acid-Base Reactions
Quantitatively, we can make use of the ionization constants to determine which reaction occurs. In possibility I,
HSO4(aq) + H2O(l) ⇌ SO42-(aq) + H3O+(aq)
$K_1 = K_{\text{a, HSO}_4^-} = 1.1\;\times\;10^{-2} \nonumber$
HCO3(aq) + H3O+(aq) ⇌ H2CO3(aq) + H2O(l)
$K_2 = \dfrac{1}{K_{\text{a, H}_2\text{CO}_3}} = \dfrac{1}{4.3\;\times\;10^{-7}} \nonumber$
The sum of these two equilibria gives the overall reaction for possibility I: HSO4(aq) + HCO3(aq) ⇌ SO42-(aq) + H2CO3(aq), and the total equilibrium constant is:
$K_{\text{total, possibility I}} = K_1 \times K_2 = \dfrac{1.1\;\times\;10^{-2}}{4.3\;\times\;10^{-7}} = 2.6\;\times\;10^{4} \nonumber$
Clearly possibility I is product-favored at equilibrium because the equilibrium constant is much greater than 1.
In possibility II,
HSO4(aq) + H3O+(aq) ⇌ H2CO3(aq) + H2O(l)
$K_1 = \dfrac{1}{K_{\text{a, H}_2\text{SO}_4}} ≤ \dfrac{1}{20} \nonumber$
HCO3(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq)
$K_2 = K_{\text{a, HCO}_3^-} = 4.7\;\times\;10^{-11} \nonumber$
(Ka for H2SO4 is too large to measure in aqueous solution but is greater than Ka for HNO3, which is ≅20, so the value 20 is a minimum for Ka for H2SO4.) The sum of these two equilibria gives the overall reaction for possibility II: HSO4(aq) + HCO3(aq) ⇌ H2SO4(aq) + CO32-(aq), and the total equilibrium constant is:
$K_{\text{total, possibility II}} = K_1 \times K_2 = \dfrac{4.7\;\times\;10^{-11}}{20} = 2\;\times\;10^{-12} \nonumber$
Possibility II is heavily reactant-favored at equilibrium. Therefore, of the two possibilities, the reaction that proceeds (and produces products) is possibility I, where HSO4 acts as an acid and HCO3 acts as a base.
D34.4 Amino Acids
Amino acids are amphiprotic because each amino acid molecule contains a carboxylic acid group that can donate a proton and an amine group that can accept a proton. Carboxylic acids are moderately acidic, many with Ka of ~10-5. Organic amines are somewhat basic, many with Kb of ~10-4. This combination creates an interesting situation, where an acid-base reaction is possible within a single amino acid molecule:
The carboxylic acid group, with Ka = ~10-5, is a stronger acid than the protonated amine group, with Ka = Kw/Kb(amine) = 10-14/10-4 = ~10-10. The amine group (Kb = ~10-4) is a stronger base than the carboxylate anion (Kb = ~10-9). The stronger acid and stronger base are on the left side so this reaction is product-favored at pH = ~7. Hence, at the pH of a typical living organism, the amino acid is a zwitterion (German for “double ion”). A zwitterion is a species with no overall electrical charge but with separate parts that are positively and negatively charged.
The formation of a zwitterion is analogous to the acid-base reaction between methylamine (Kb = 4.4 × 10-4) and acetic acid (Ka = 1.8 × 10-5):
CH3NH2(aq) + CH3COOH(aq) ⇌ CH3NH3+(aq) + CH3COO(aq)
where the equilibrium favors products because:
$K_{total} = \dfrac{1.8\;\times\;10^{-5}}{2.3\;\times\;10^{-11}} = 7.8\;\times\;10^{5} \nonumber$
Increasing the pH of an amino acid solution by adding hydroxide ions can remove the hydrogen ion from the -NH3+ group:
The product molecule is no longer a zwitterion. Instead, it is an anion with an overall charge of -1.
Similarly, decreasing the pH by adding strong acid to an amino acid solution protonates the -COO part of the zwitterion:
Again, the product molecule is not a zwitterion, but a cation with an overall charge of +1.
Podia Question
Write a clear, concise explanation in scientifically appropriate language for each of these correct statements.
1. When a polyprotic acid donates a hydrogen ion, the species that remains is usually a much weaker acid than was the original polyprotic acid.
2. When trichloroacetic acid reacts with hydrogen carbonate ion the equilibrium is significantly more product-favored than when acetic acid reacts with hydrogen carbonate ion.
3. When a strong base reacts with a weak acid in stoichiometrically equivalent quantity, the pH of the solution is above 7.
Query $7$
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D36.1 Buffer Solutions
A mixture of a weak acid and its conjugate base, such as acetic acid and sodium acetate (CH3COOH + CH3COONa), or a mixture of a weak base and its conjugate acid, such as ammonia and ammonium chloride (NH3 + NH4Cl), is a buffer solution. A buffer solution resists changes in pH when small amounts of a strong acid or a strong base are added (Figure 1).
A solution of equal concentrations of CH3COOH and CH3COONa is slightly acidic because the Ka,acetic acid > Kb, acetate anion. When a strong base, such as NaOH, is added to this solution, the OH anions react with the few H3O+ cations, decreasing concentrations of H3O+. This shifts the following equilibrium to the right:
CH3COOH(aq) + H2O(l) ⇌ CH3COO(aq) + H3O+(aq)
restoring H3O+ concentration to almost the value it had before the NaOH was added. The net effect of most of the added NaOH is to convert some of the weak acid, CH3COOH, to a weak base, CH3COO:
CH3COOH(aq) + OH(aq) → CH3COO(aq) + H2O(l)
Hence, there is only a minimal decrease in H3O+ concentration.
When a strong acid, such as HCl, is added, the net effect of most of the added H3O+ is to convert acetate anions to acetic acid molecules:
CH3COO(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
And again, there is only a minimal increase in H3O+ concentration.
As illustrated in Figure 2, a buffer solution can moderate changes to pH because it consists of a weak acid that can react with added strong base as well as a weak base that can react with added strong acid.
The weak base and weak acid in a buffer solution are typically a conjugate acid-base pair, which maintains one dynamic equilibrium that responds to additions of other acids and bases. If they are not a conjugate acid-base pair, then there would be two dynamic equilibria at play, which significantly complicates the buffering actions.
Activity 1: pH of a Buffer Solution
D36.2 Henderson-Hasselbalch Equation
The ionization constant expression for a weak acid HA is:
$K_a = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{A}^{-}]_e}{[\text{HA}]_e} \nonumber$
Rearranging gives:
$[\text{H}_3\text{O}^{+}]_e = K_a\;\times\;\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \nonumber$
Taking the negative logarithm of both sides, we have:
$\begin{array}{rcl} -\text{log}[\text{H}_3\text{O}^{+}]_e &=& -\text{log}\;K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \[1 em] \text{pH} &=& \text{p}K_a\;-\;\text{log}\dfrac{[\text{HA}]_e}{[\text{A}^{-}]_e} \[1 em] \text{pH} &=& \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_e}{[\text{HA}]_e} \end{array} \nonumber$
It is much more convenient to deal with the initial concentrations of the weak acid and weak base when preparing a buffer solution. (The initial concentration is the amount of weak acid and weak base added to the solution mixture divide by the volume.) Therefore:
$\text{pH} = \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_0\;+\;x}{[\text{HA}]_0\;-\;x} \nonumber$
x” is the increase in concentration of H3O+ as the solution reaches equilibrium (see activity 1 above) and [HA]0 and [A]0 are the concentration of HA and A before any reaction occurs. When the approximation that x is at least 100 times smaller than the concentrations of HA and A is valid, we have the Henderson-Hasselbalch equation:
$\text{pH} = \text{p}K_a\;+\;\text{log}\dfrac{[\text{A}^{-}]_0}{[\text{HA}]_0} \nonumber$
Note that when [A−]0 = [HA]0, pH = pKa + log(1) = pKa.
The Henderson-Hasselbalch equation can be used to calculate the buffer solution pH, given the Ka and the initial concentrations, or it can be used to determine the ratio of initial concentrations of weak acid and base required to achieve a desired pH.
The Henderson-Hasselbalch equation applies only to buffer solutions in which the ratio
$\dfrac{[\text{A}^{-}]_0}{[\text{HA}_0} \nonumber$
is between 0.1 and 10. If enough strong acid or strong base is added to the buffer solution to exceed this range, the pH begins to change significantly (in other words, the solution is no longer a buffer solution).
D36.3 Selection of a Suitable Buffer
A buffer solution moderates changes in pH because it contains both a weak acid that can react with added strong base and a weak base that can react with added strong acid. This leads to several criteria for selecting a suitable buffer solution for a given purpose.
1. The pKa of the weak acid in the buffer should be close to the desired pH of the buffer solution. According to the Henderson-Hasselbalch equation, if the concentrations of weak acid and weak base are equal, the pH of the buffer solution equals the pKa of the weak acid involved.
2. A buffer solution should have approximately equal concentrations of the weak acid and weak base. A
$\dfrac{[\text{A}^{-}]_0}{[\text{HA}_0} \nonumber$
ratio of >10 or <0.1 makes for a poor buffer solution. Figure 3 shows how the pH of an acetic acid-acetate ion buffer increases as strong base is added. The initial pH is pKa = 4.74. When pH reaches 5.74, a change of 1 pH unit, the ratio
$\dfrac{[\text{acetic acid}]}{[\text{acetate anion}]} = 0.11 = 11\text{%} \nonumber$
After that the pH increases more rapidly and the solution no longer provides significant buffering.
1. The larger the amounts (mol) of weak acid and weak base are the greater is the amount (mol) of strong base or strong acid that can be added before there is a significant change in pH
When designing a buffer system, look for weak conjugate acid-base pairs that have pKa of the weak acid near the desired pH. Then adjust the ratio of the weak base to weak acid concentrations to achieve the exact pH desired. Make certain that the concentrations of weak base and weak acid are large enough to react with the quantities of acid or base that might be added to the buffer solution.
Activity 2: Preparing a Buffer Solution with a Desired pH
D36.4 Buffer Capacity
We can see how a buffer solution works by comparing quantitatively the pH of a buffered solution with the pH of a unbuffered solution upon addition of a strong acid or base.
Activity 3: Calculating pH Change for a Buffer Solution
Query $5$
Activity 4: pH Change in an Unbuffered Solution
Query $6$
We can see from the above activities that the change in pH is much more significant in the unbuffered solution compared to the buffered solution.
However, buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 4). For example, if we add sufficient strong base to a buffer that all the weak acid has reacted, no more buffering action toward the base is possible. Similarly, if we add an excess of strong acid, the weak base would all be reacted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to react away all the weak acid or base in a buffer to make significant change in pH: buffering action diminishes rapidly as a given component nears depletion. This was seen in Figure 3 in the above section, where reducing the concentration of weak acid to 11% of the concentration of weak base caused a change of 1 pH unit. The curve in Figure 3 goes up rapidly after that, indicating that the buffer has been “broken” and no longer resists changes in pH.
The buffer capacity is the amount (mol) of acid or base that can be added to a given volume of a buffer solution before the pH changes by ±1 from the pKa of the weak acid. (Recall that if equal concentrations of weak acid and conjugate base are in a buffer solution, pH = pKa.)
Buffer capacity depends on the amount (mol) of weak acid and its conjugate base that are in a buffer mixture. For example, a 1 L solution of 1.0 M CH3COOH and 1.0 M CH3COONa has a greater buffer capacity than a 1 L solution of 0.10 M CH3COOH and 0.10 M CH3COONa, even though both solutions have the same pH. The first solution has more buffer capacity because it contains more moles of acetic acid and acetate ion.
It takes 0.82 mol HCl to change the buffer pH from 4.74 to 3.74 in the first solution:
$\begin{array}{rcl} 3.74 &=& 4.74\;+\;\text{log}\dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0}\[0.5em] 10^{3.74-4.74} &=& \dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0} \[0.5em] 0.10 &=& \dfrac{(1.0\;\text{mol}\;-\;x)/(1\;\text{L})}{(1.0\;\text{mol}\;+\;x)/(1\;\text{L})} \[0.5em] 0.10(1.0\;\text{mol}\;+\;x) &=& 1.0\;\text{mol}\;-\;x \[0.5em] 1.1x &=& 0.90\[0.5em] x &=& 0.82\;\text{mol} \end{array} \nonumber$
On the other hand, for the solution where the concentrations of weak acid and conjugate base are 0.10 M, it takes only one-tenth as much HCl, 0.082 mol HCl, to change the buffer pH from 4.74 to 3.74:
$\dfrac{[\text{CH}_3\text{COO}^-]_0}{[\text{CH}_3\text{COOH}]_0}\;=\;10^{-1}\;=\;0.1\;=\;\dfrac{0.1\;\text{mol}\;-\;x}{0.1\;\text{mol}\;+\;x}\;\;\;\;\;x\;=\;0.082\;\text{mol} \nonumber$
If a buffer solution does not have equal concentrations of weak acid and weak base, the buffer capacity when strong acid is added is different from the buffer capacity when strong base is added.
Exercise 3: Calculating Buffer Capacity
Podia Question
A solution is prepared by adding 50.0 mL 0.50-M acetic acid and 50.0 mL 0.30-M NaOH to a beaker and stirring. Is this solution a buffer solution? If so, calculate the pH of the buffer. If not, explain why the solution does not resist change in pH when 1.0 mL 0.5-M HCl is added.
Query $8$
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D37.1 Titration
At this point, you have already carried out several titration experiments in lab. In a titration, one solution is added to a second solution in a way that allows the quantity of each solution to be measured accurately. Typically, one solution is added from a buret, a calibrated glass tube with a stopcock at the bottom. The buret allows additions of very small increments of volume of solution and determination of the total volume added. The solution added from the buret is the titrant. Titrant is added to a carefully measured volume of a second solution in a flask or beaker.
A chemical reaction occurs between solutes in the two solutions, and when just enough titrant has been added for complete reaction with the second solution, the equivalence point of the titration has been reached. From the balanced chemical equation for the titration reaction, the volume of each solution at the equivalence point, and the concentration of one of the two solutions, the concentration of the other solution can be determined.
The equivalence point of a titration may be detected visually if there is a color change that accompanies the completion of the reaction. Typically, a special dye called an indicator is added to the solution being titrated to provide a color change at or very near the equivalence point. Equivalence points may also be detected by measuring a solution property that changes in a predictable way during the course of the titration, such as pH.
Regardless of the approach taken to detect a titration’s equivalence point, the measured total volume of titrant added is called the end point. A properly designed titration typically ensure that the difference between the equivalence point and the end point is negligible.
Although any type of chemical reaction may serve as the basis for a titration analysis, precipitation, acid-base, and oxidation-reduction titrations are most common.
D37.2 Titration Curves
Titration is effective in quantitatively analyzing a solution’s acid (or base) concentration because pH changes rapidly near the equivalence point. In other words, there is a large observable change accompanying a small addition of titrant, which minimizes experimental uncertainty. For example, an acid-base indicator generally changes color over a range of about 2 pH units, so if pH increases (or decreases) by 2 or more pH units when 0.01 mL of titrant is added at the equivalence point, the color change would effectively signal the end point of the titration.
Figure 1 shows a titration curve, a graph of pH as a function of volume of titrant, for the titration of a 50.00-mL sample of 0.100-M hydrochloric acid with 0.100-M sodium hydroxide.
Query $2$
Figure 1. Titration curve for the titration of 50.00 mL of 0.100-M HCl (strong acid) with 0.100-M NaOH (strong base) has the equivalence point at pH = 7.00. Click on “i” for more information.
Exercise 2: Calculating pH for a Titration Curve
Now consider the titration of 50.00 mL of 0.100-M acetic acid (a weak acid) with 0.100-M sodium hydroxide (Figure 2). Comparing Figure 2 to Figure 1, we see that although the initial volumes and concentrations of the acids (acetic acid vs. HCl) are the same, the pH for acetic acid begins at a higher value and the titration curve maintains higher pH values up to the equivalence point. This is because, unlike HCl, acetic acid is only partially ionized.
Query $4$
Figure 2. Titration curve for the titration of 50.00 mL of 0.100-M CH3COOH (weak acid) with 0.100-M NaOH (strong base) has an equivalence point at pH = 8.72. Move the slider to the right for overlay of Figure 1 and Figure 2 titration curves.
The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the presence of acetate anion, a weak base that raises the pH:
CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq)
After the equivalence point, the two curves (Figure 1 and Figure 2) are identical because the pH depends on the excess hydroxide ion from NaOH added in both cases.
Activity 1: Titration Equivalence Point
Exercise 3: Calculating pH for a Weak-acid, Strong-base Titration
The midpoint of a titration is when we have added half the volume of titrant needed to reach the equivalence point. As part (b) in the above exercise shows, when titrating a weak acid with a strong base, the pH of the solution equals the pKa of the weak acid at the midpoint because we have added half the amount of strong base needed to react with all the weak acid. Therefore, according to the Henderson-Hasselbalch equation:
$\text{pH} = \text{p}K_a + \text{log}\dfrac{[\text{weak base}]_0}{[\text{weak acid}]_0} = \text{p}K_a + \text{log}(1) = \text{p}K_a + 0 = \text{p}K_a \nonumber$
Activity 2: Titration of a Weak Base with a Strong Acid
D37.3 Acid-Base Indicators
Acid-base indicators are substances with intense colors that change color when [H3O+] reaches a particular value. Acid-base indicators are either weak organic acids or weak organic bases and can be used to determine the pH of a solution. For example, phenolphthalein is colorless in an aqueous solution with pH < 8.3 ([H3O+] > 5.0 × 10−9M), and turns red or pink when pH > 8.3.
The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use “HIn” as a simple representation for the complex methyl orange molecule:
$\begin{array}{ccccccc} \text{HIn}(aq) & + & \text{H}_2\text{O}(l) & {\leftrightharpoons} & \text{H}_3\text{O}^{+}(aq) & + & \text{In}^{-}(aq) \[0.5em] \text{red} & & & & & & \text{yellow} \end{array} \nonumber$
$K_{\text{a}} = \dfrac{[\text{H}_3\text{O}^{+}][\text{In}^{-}]}{[\text{HIn}]} = 4.0\;\times\;10^{-4} \nonumber$
When we add acid to a solution of methyl orange, the increased [H3O+] shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, the equilibrium shifts towards the anionic yellow form. The overall color of the solution is the visible result of the ratio of the concentrations of the two species, In and HIn. If most of the indicator is present as In, then we see the color yellow. If most is present as HIn, then we see the color red. We can rearrange the equation for Ka and write:
$\dfrac{[\text{In}^{-}]}{[\text{HIn}]} = \dfrac{[\text{substance with yellow color}]}{[\text{substance with red color}]} = \dfrac{K_{\text{a}}}{[\text{H}_3\text{O}^{+}]} \nonumber$
When [H3O+] = Ka,HIn, 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In), and the solution appears orange in color. When [H3O+] increases to 8 × 10−4 M (pH = 3.1), the solution turns red. No change in color is visible for further increase in [H3O+]. When [H3O+] decreases to 4 × 10−5 M (pH = 4.4), most of the indicator is in the yellow ionic form, and further decrease in [H3O+] does not produce a visible color change. The pH range of 3.1 – 4.4 is the color-change interval of methyl orange, the range of pH values over which the color change takes place.
There are many different acid-base indicators. Their color changes have a wide range of pH values (Figure 3). Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pH values.
Titration curves can help us select an indicator that will provide a sharp color change at the equivalence point. The best selection is an indicator with a color change interval that brackets the pH at the equivalence point of the titration. (We can also base our choice of indicator on the calculated pH at the equivalence point.)
The color change intervals of three indicators are shown in Figure 4. The steep section of the titration curves of both the titration of HCl and of CH3COOH are located in the color-change interval of phenolphthalein. We can use it for titrations of either acid.
Litmus is a suitable indicator for the HCl titration. However, we should not use it for the CH3COOH titration because the pH is within the color-change interval of litmus when only 8 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 17 mL of NaOH, making litmus useless as an indicator of the equivalence point.
We can use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color during the addition of 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein. Methyl orange would be completely useless as an indicator for the CH3COOH titration. Its color change is completed long before the equivalence point is reached and hence provides no indication of the equivalence point.
D37.4 Titration of Polyprotic Acids and Bases
When a polyprotic acid is titrated, there are usually multiple equivalence points. For example, when H2SO3 is titrated with NaOH, there are two equivalence points corresponding to the two acidic protons from the H2SO3 molecule. There are also as many midpoints as there are equivalence points.
When a strong monoprotic base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic proton (Ka,1) is titrated first, followed by the next most acidic (Ka,2), and so forth. If the Ka values differ by at least three orders of magnitude, then the overall titration curve will show well-resolved “steps” corresponding to the titration of each acidic proton.
Consider the titration of a generic weak polyprotic acid H3A with NaOH as shown in Figure 5. The first equivalence point correspond to the point where 1 mole of NaOH has been added per mole of H3A in the solution being titrated. As more titrant is added, the titration curve crosses another midpoint and reaches the second equivalence point, corresponding to a total of 2 moles of NaOH being added per mole of H3A in the solution. The titration is finally complete when all three equivalence points have been reached.
Query $8$
Figure 5: Titration of a weak polyprotic acid by a strong base. Move the slider to see the different stages of titration. .
It is not always possible to detect all of the equivalence points in the titration of a polyprotic acid. An actual titration of the triprotic acid H3PO4 with NaOH is illustrated in Figure 6. It shows two well-defined steps: the first midpoint corresponds to pKa,1, and the second midpoint corresponds to pKa,2. Because HPO42 is such a weak acid, pKa,3 has such a high value that the third step cannot be resolved using 0.100 M NaOH as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is inverted on the pH scale. The initial pH is high; as acid is added, the pH decreases in steps if the successive pKb values are well separated.
Podia Question
Draw the structure of the amino acid alanine at pH = 2.
Sketch a titration curve for titration of a solution of alanine starting at pH = 2 and ending at pH = 13. The concentration of alanine is 0.100 M and there is 25.00 mL initially, at pH = 2. This solution is titrated with 0.100-M NaOH. pKa values for alanine are 2.85 and 9.87. No calculations are required, but the curve should be as accurate as possible.
There are two equivalence points and two midpoints in this titration. Label each midpoint with the volume and pH. Explain how you determined the pH at each midpoint.
Query $9$
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D38.1 Oxidation-reduction Reactions and Electrochemistry
Electrochemistry deals with chemical reactions that involve transfer of electron density; in other words, the reactions either produce electricity or are caused by passage of electrical current through matter. These reactions are called oxidation-reduction (abbreviated redox) reactions.
Here’s a brief list of units and definitions used when discussing redox reactions and their applications.
• The SI unit of electric charge is the coulomb, C.
• The elementary unit of charge is the charge of a single electron, which is equal to 1.602 × 10−19 C.
• Movement of electrons (or ions) carries electric charges from one place to another, and the quantity of such charge transferred per unit time is the electric current.
• Current has the SI unit ampere, A, which is the transfer of one coulomb per second (1 A = 1 C/s).
• Typically, electric current flows in a closed path, called an electric circuit.
• It is necessary to maintain a closed circuit for current to flow. If the circuit is open, current will not flow.
• Electrical potential, SI unit volt, V, is the ability of an electric field to do work on a charge.
• A flow of charge is caused by an electrical potential difference between two points in the circuit.
• When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule of energy (1 J = 1 C × 1 V).
• Electric power is the quantity of energy transferred per unit time and is measured in watts, W (1 W = 1 J / 1 s).
D38.2 Redox Reactions and Oxidation Number
Some redox reactions involve transfer of electrons between reactant atoms to yield ionic products. For example, formation of a crystal lattice of sodium chloride (a lattice of Na+ ions and Cl ions) requires transfer of electrons from sodium atoms to chlorine atoms:
2Na(s) + Cl2(g) ⟶ 2NaCl(s)
It is useful to consider the electron transfer as two concurrent processes. The processes are called half-reactions, one in which electrons are lost and one in which electrons are gained. The half-reactions for reaction of sodium with chlorine are:
2Na(s) 2Na+(s) + 2e
Cl2(g) + 2e 2Cl(s)
The half-reactions allow us to see that the number of electrons lost equals the number of electrons gained—we cannot create or destroy electrons in a chemical reaction. The half-reactions also show which species gains electrons (Cl atoms in Cl2) and which species loses electrons (Na atoms). The species that loses electrons is said to be oxidized and the loss of electrons is called oxidation; the species that gains electrons is said to be reduced and the gain of electrons is called reduction.
The species that causes reduction to occur is called the reducing agent (or reductant). In this reaction sodium is the reducing agent because it causes Cl2 to gain electrons. The species that causes oxidation to occur is called the oxidizing agent (or oxidant). In this reaction chlorine functions as an oxidant, causing sodium to lose electrons.
Some redox processes do not involve obvious transfer of electrons because no ions are involved in the reaction. For example:
H2(g) + Cl2(g) ⟶ 2HCl(g)
To systematically classify redox reactions of all types, we define oxidation number (or oxidation state) of an element in a compound as the charge its atoms would possess if the compound was ionic, that is, if all electrons in each polar covalent bond were assigned to the more electronegative atom.
The steps below can be used to assign oxidation numbers to each element in a compound.
1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monoatomic ion is equal to the ion’s charge.
3. The sum of oxidation numbers over all atoms in a neutral compound is zero.
4. The sum of the oxidation numbers over all atoms in a polyatomic ion equals the ion’s charge. If a compound includes more than one polyatomic ion, the oxidation number of an atom in one ion can differ from the oxidation number in the other polyatomic ion.
5. Atoms of some elements have the same oxidation number in almost all compounds.
1. Fluorine always has oxidation number −1 when present in a compound
2. Atoms of alkali metals (Li, Na, K, Rb, Cs) have oxidation number +1 in nearly all compounds
3. Atoms of alkaline earth metals (Be, Mg, Ca, Sr, Ba) have oxidation number +2 in nearly all compounds
4. Hydrogen has oxidation number +1 when combined with nonmetals, −1 when combined with metals
6. Apply these two rules only if rules 1-5 have not determined all oxidation numbers.
1. Oxygen has oxidation number −2 unless rules 1-5 have already given O a different oxidation number
2. In binary compounds of nonmetals, the more electronegative element is given a negative oxidation number equal to the charge on its monoatomic ion; for example in PCl3, the more electronegative Cl is assigned oxidation number −1 and P is assigned +3 (by rule 3).
Activity 1
Exercise 2: Assigning Oxidation Numbers
Using oxidation numbers we can identify redox reactions by looking for one or more elements whose oxidation numbers change during the course of the reaction. When its oxidation number increases, an element has been oxidized; when its oxidation number decreases, the element has been reduced.
In this reaction
2 Na(s) + Cl2(g) ⟶ 2 NaCl(s)
sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in NaCl).
In this reaction
H2(g) + Cl2(g) ⟶ 2 HCl(g)
hydrogen is oxidized (its oxidation number increases from 0 in H2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in HCl).
Several subclasses of redox reactions can be identified. One example is combustion reactions in which the reductant (fuel) and oxidant (often, O2) react vigorously and produce significant temperature increase, often in the form of a flame. Another class of redox reaction is a rocket propellant reaction such as this one in which solid aluminum is oxidized by ammonium perchlorate:
10 Al(s) + 6 NH4ClO4(s) ⟶ 4 Al2O3(s) + 2 AlCl3(s) + 12 H2O(g) + 3 N2(g)
Exercise 3: Recognizing Oxidation-Reduction Reactions
Exercise 4: Recognizing Oxidizing and Reducing Agents
For the reaction below, identify which species is oxidized and which is reduced, and identify the oxidizing agent and the reducing agent.
8 H+(aq) + MnO4(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
[Enter your answers without subscripts, superscripts, or states, and put charges in parentheses. For example, SO42-(aq) would be entered as SO4(2-).]
D38.3 Balancing Redox Reactions
Redox reactions frequently occur in aqueous solutions, which can be acidic, basic, or neutral. Moreover, H2O molecules may actively participate in the reaction, and depending on the conditions, H3O+ (present under acidic conditions) and OH (present under basic conditions) may also be a reactant or a product.
Half-reactions make it easier to balance redox reactions because you can balance the oxidation half-reaction separately from the reduction half-reaction, and make certain that the number of electrons lost in the oxidation process equals the number of electrons gained in the reduction process.
We will balance an example redox reaction in acidic solution and one in basic solution to illustrate the process of balancing redox reactions, and highlight how the nature of the solution can play a role. (A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.)
Acidic Solution
Consider the unbalanced reaction:
MnO4‾(aq) + Fe2+(aq) ⟶ Mn2+(aq) + Fe3+(aq)
where iron underwent oxidation because Fe2+ has lost an electron to become Fe3+, and manganese underwent reduction because it gained five electrons to change from an oxidation state of +7 to an oxidation state of +2.
Oxidation (unbalanced): Fe2+(aq) Fe3+(aq)
Reduction (unbalanced): MnO4‾(aq) Mn2+(aq)
For redox reactions, it is a useful simplification to represent H3O+(aq) as H+(aq)—where there is a need for balancing O atoms, we can involve H2O(l), and then use H+(aq) to balance the H atoms.
For instance, in the above reduction half-reaction, there are four O atoms on the left side and none on the right side. To balance the elements, we can add 4 H2O(l) to the product side, and then to balance H atoms, add 8 H+(aq) to the reactant side:
Reduction (charge not balanced): MnO4(aq) + 8H+(aq) ⟶ Mn2+(aq) + 4H2O(l)
Once the atoms have been balanced, we need to balance the electric charge for each half-reaction. For the oxidation half-reaction, the total charge on the left side is +2 and the total charge on the right side is +3, so charge is unbalanced. We use electrons to balance the charge. Adding a single electron on the right side gives a balanced oxidation half-reaction:
Oxidation (balanced): Fe2+(aq) ⟶ Fe3+(aq) + e‾
In oxidation half-reactions, electrons appear on the right side of the equation. Because iron is oxidized, iron is the reducing agent in this redox reaction.
You should always check that the half-reaction is balanced for the number of atoms of each element and the total charge:
Fe: (1 atom in Fe2+)·(1 Fe2+) = 1 (1 atom in Fe3+)·(1 Fe3+) = 1 1 = 1
Charge: 1·(+2) = +2 1·(+3) + 1·(-1) = +2 +2 = +2
If the atoms and charges are balanced, then the half-reaction itself is balanced.
For the reduction half-reaction, we have balanced the atoms but not the charge. The total charge on the reactant-side is +7, the total charge on the product-side is +2. Therefore, it is necessary to add five electrons to the left side to achieve charge balance:
Reduction (balanced): MnO4(aq) + 8H+(aq) + 5e ⟶ Mn2+(aq) + 4H2O(l)
In all reduction half-reactions, electrons appear on the left side. The species that was reduced, MnO4, is the oxidizing agent in this redox reaction.
Again, check that the half-reaction is balanced for the number of atoms of each element and the total charge:
Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in Mn2+)·(1 Mn2+) = 1 1 = 1
O: (4 atoms in MnO4)·(1 MnO4) = 4 (1 atom in H2O)·(4 H2O) = 4 4 = 4
H: (1 atom in H+)·(8 H+) = 8 (2 atoms in H2O)·(4 H2O) = 8 8 = 8
Charge: 1·(-1) + 8·(+1) + 5·(-1) = +2 1·(+2) + 4·(0) = +2 +2 = +2
We now have two balanced half-reactions:
Oxidation: Fe2+(aq) Fe3+(aq) + e‾
Reduction: MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l)
The key to combining the half-reactions is the electrons: the number of electrons generated by the oxidation half-reaction must equal the number of electrons consumed by the reduction half-reaction. Here, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five. Therefore, it is necessary to first multiply the oxidation half-reaction by five and the reduction half-reaction by one, then sum the resulting half-reactions:
Oxidation: 5 × (Fe2+(aq) Fe3+(aq) + e‾)
Reduction: MnO4(aq) + 8 H+(aq) + 5e Mn2+(aq) + 4 H2O(l)
overall: 5 Fe2+(aq) + MnO4(aq) + 8 H+(aq) 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
This is the overall balanced equation in acidic solution. Electrons do not appear in the overall reaction equation because all electrons lost in the oxidation half-reaction are gained in the reduction half-reaction. On each side of the overall equation there should be the same number of atoms of each element and the same total electric charge. Be sure to carefully check each side to verify everything has been balanced correctly.
Basic Solution
The simplest way to balance redox reaction in basic solution is to start with the balanced equation in acidic solution, then “convert” H+(aq) to OH(aq) (there is an excess of OH ions instead of H3O+ ions in basic solutions). For example, when balancing the following reaction in basic solution:
MnO4(aq) + Cr(OH)3(s) ⟶ MnO2(s) + CrO42-(aq)
start by collecting the species given into unbalanced oxidation and reduction half-reactions:
Oxidation (unbalanced): Cr(OH)3(s) CrO42-(aq)
Reduction (unbalanced): MnO4‾(aq) MnO2(s)
For the oxidation half-reaction, we can add one H2O molecule to the left side to balance oxygen atoms, and then balance hydrogen atoms with five H+(aq) on the right side (again, we do the initial balancing by assuming acidic solution):
Oxidation (charge not balanced): Cr(OH)3(s) + H2O(l) ⟶ CrO42-(aq) + 5H+(aq)
The left side of the equation has a total charge of 0, and the right side a total charge of +3. Adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
Oxidation (balanced): Cr(OH)3(s) + H2O(l) ⟶ CrO42-(aq) + 5H+(aq) + 3e‾
Checking the half-reaction:
Cr: (1 atom in Cr(OH)3)·(1 Cr(OH)3) = 1 (1 atom in CrO42-)·(1 CrO42-) = 1 1 = 1
O: (3 atoms in Cr(OH)3)·(1 Cr(OH)3) + (1 atom in H2O)·(1 H2O) = 4 (4 atoms in CrO42-)·(1 CrO42-) = 4 4 = 4
H: (3 atoms in Cr(OH)3)·(1 Cr(OH)3) + (2 atoms in H2O)·(1 H2O) = 5 (1 atom in H+)·(5 H+) = 5 5 = 5
Charge: 1·(0) + 1·(0) = 0 1·(-2) + 5·(+1) + 3·(-1) = 0 0 = 0
For the reduction half-reaction, we need two H2O molecules on the right to balance oxygen atoms. Then to balance hydrogen atoms, we need to add four H+(aq) on the left:
Reduction (charge not balanced): MnO4(aq) + 4H+(aq) ⟶ MnO2(s) + 2H2O(l)
Then add three electrons to the left side to balance the charge:
Reduction (balanced): MnO4(aq) + 4H+(aq) + 3e‾ ⟶ MnO2(s) + 2H2O(l)
Make sure to check the half-reaction:
Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in MnO2)·(1 MnO2) = 1 1 = 1
O: (4 atoms in MnO4)·(1 MnO4) = 4 (2 atoms in MnO2)·(1 MnO2) + (1 atom in H2O)·(2 H2O) = 4 4 = 4
H: (1 atom in H+)·(4 H+) = 4 (2 atoms in H2O)·(2 H2O) = 4 4 = 4
Charge: 1·(-1) + 4·(+1) + 3·(-1) = 0 1·(0) + 2·(0) = 0 0 = 0
Collecting what we have so far:
Oxidation: Cr(OH)3(s) + H2O(l) CrO42-(aq) + 5H+(aq) + 3e‾
Reduction: MnO4(aq) + 4H+(aq) + 3e‾ MnO2(s) + 2H2O(l)
In this case, both half-reactions involve the same number of electrons, and therefore we can simply add the two half-reactions together and simplify:
Cr(OH)3(s) + H2O(l) + MnO4(aq) + 4 H+(aq) + 3 e‾ CrO42-(aq) + 5 H+(aq) + 3 e‾ + MnO2(s) + 2 H2O(l)
Cr(OH)3(s) + MnO4(aq) CrO42-(aq) + H+(aq) + MnO2(s) + H2O(l)
This is the balanced redox equation in an acidic solution. To “convert” to basic solution, add OH(aq) to both sides of the equation to “react” with all the H+(aq). This converts H+(aq) to H2O(l) on one side and adds OH(aq) on the other side:
Cr(OH)3(s) + MnO4(aq) + OH(aq) CrO42-(aq) + H+(aq) + OH(aq) + MnO2(s) + H2O(l)
Cr(OH)3(s) + MnO4(aq) + OH(aq) CrO42-(aq) + MnO2(s) + 2H2O(l)
This is the balanced equation in basic solution. Checking each side of the equation:
Cr: (1 atom in Cr(OH)3)·(1 Cr(OH)3) = 1 (1 atom in CrO42-)·(1 CrO42-) = 1 1 = 1
Mn: (1 atom in MnO4)·(1 MnO4) = 1 (1 atom in MnO2)·(1 MnO2) = 1 1 = 1
O: (3 in Cr(OH)3)·(1) + (4 in MnO4)·(1) + (1 in OH)·(1) = 8 (4in CrO42-)·(1) + (2 in MnO2)·(1) + (1 in H2O)·(2) = 8 8 = 8
H: (3 in Cr(OH)3)·(1) + (1 in OH)·(1) = 4 (2 atoms in H2O)·(2 H2O) = 4 4 = 4
Charge: 1·(0) + 1·(-1) + 1·(-1) = -2 1·(-2) + 1·(0) + 2·(0) = -2 -2 = -2
D38.4 Introduction to Voltaic Cells
A voltaic cell (or galvanic cell) is an electrochemical cell in which a spontaneous redox reaction produces an electric current. Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate.
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When the copper metal contacts the silver nitrate solution, silver metal begins to form on the copper surface and Cu2+ ions pass into the solution (indicated by the blue-green color of the solution). The net ionic equation for the reaction is:
Cu(s) + 2 Ag+(aq) ⟶ Cu2+(aq) + 2 Ag(s)
which may be split into its two half-reactions that sum to the overall reaction:
Oxidation: Cu(s) Cu2+(aq) + 2 e‾
Reduction: 2 Ag+(aq) + 2 e‾ 2 Ag(s)
The half-reactions make clear that two electrons are transferred from a copper atom, which forms a Cu2+ ion. Each of the two electrons is then transferred to one Ag+ ion, which forms a silver atom. Because a flow of electrons constitutes an electric current, this electron transfer can generate an electric current if we devise a way to carry out the two half-reactions in separate vessels and connect them with a metal (electrically conductive) wire. This is in essence how a voltaic cell is designed. Figure 1 shows one way to do this.
The beaker on the left contains a 1-M solution of copper(II) nitrate [Cu(NO3)2] with a strip of copper metal partially submerged in the solution. The copper strip is an electrode, a means for conducting electrons into or out of the solution. At the surface of the copper strip, the oxidation half-reaction occurs:
Cu(s) ⟶ Cu2+(aq) + 2 e‾
The flow of electrons (electric current) passes out of the solution via the copper strip, flows through the light bulb, and moves into the silver strip in the beaker on the right. In the right-hand beaker, the reduction half-reaction occurs near the surface of the silver strip:
2Ag+(aq) + 2 e‾ ⟶ 2Ag(s)
Thus, with the two half-reactions occurring in separate beakers, an electric current can be generated. The container in which each half-reactions occurs is called a half-cell.
If this were all that happened, the electric current would not flow for long. In the left-hand beaker, one Cu2+ ion is added to the solution for every two electrons conducted into the external wire. This means that the solution is continuously accumulating excess positive ions as the reaction occurs and an electric charge is building up. Such a charge build up would prevent further oxidation reaction from occurring. This can be mitigated if some positive ions move out of the beaker or some negative ions move into the beaker. A similar issue is occurring in the right-hand beaker, where Ag+ ions are being removed from the solution. Balancing total ionic charges requires either negative ions move out of the right-hand solution or positive ions move in.
This balancing of ion charges in the two separated half-reactions is maintained by the salt bridge, a solution of a salt that does not mix with either half-cell solution but allows ions to pass into or out of the half-cells. By allowing ions to conduct charge into or out of the half-cells, the salt bridge completes the electrical circuit involving the two half-cells. Without it, current cannot flow for more than an instant. Notice that negative ions in the salt bridge move in the same direction as electrons around the circuit and positive ions move in the opposite direction.
The half-cell in which oxidation occurs is called the anode. The half-cell in which reduction occurs is called the cathode. It is easy to remember that the anode involves oxidation because both words begin with vowels. It is easy to remember that the cathode involves reduction because both words begin with consonants. These definitions, anode/oxidation and cathode/reduction, are true for any electrochemical cell, not just a voltaic cell.
Exercise 6: Reactions for a Voltaic Cell
Podia Question
Make a diagram of a voltaic cell that generates electric current from this product-favored chemical reaction:
Fe(s) + Cu(NO3)2(aq) → Fe(NO3)2(aq) + Cu(s)
Choose an appropriate salt solution for the salt bridge. Make sure that the oxidation half-reaction is in the left-hand beaker. Label each electrode with an element symbol. Label the anode half-cell and the cathode half-cell. Label the direction of movement of electrons, positive ions, and negative ions in the electrical circuit.
As this voltaic cell operates, one electrode gains mass and one electrode loses mass. Identify which electrode gains mass. Explain why that electrode gains mass and why the other electrode loses mass.
Query \(8\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/05%3A_Unit_Five/5.03%3A_Day_38-_Oxidation-Reduction_Reactions_Voltaic_Cells.txt |
D39.1 Voltaic Cell Potential
When a voltaic cell is connected to a load, such as a light bulb, an electric current flows because there is a difference in electrical potential between the two electrodes. That electrical potential difference can be measured with a potentiometer (that is, a voltmeter). For an accurate measurement, the voltmeter must have a very high resistance so that no current flows; otherwise the voltage would drop from its highest possible value.
The voltaic cell shown in Figure 1, involves the spontaneous reaction
Cu(s) + 2 Ag+(aq) ⟶ Cu2+(aq) + 2 Ag(s)
According to the reaction equation, copper loses electrons and is oxidized to copper(II) ions, so the half-cell with the copper electrode in Figure 1 is the anode. According to the reaction equation, silver ions gain electrons and are reduced to silver, so the half-cell with the silver electrode is the cathode. The copper electrode is more negative than the silver electrode.
When the more negative copper electrode is connected to the negative terminal of the voltmeter and the more positive silver electrode is connected to the positive terminal of the voltmeter, the meter reads +0.46 V. This reading is called the cell potential, Ecell. It is a measure of the energy per unit charge available from a redox reaction (V = J/C). A positive cell potential indicates how much electrical work a spontaneous reaction in a voltaic cell can do per unit electric charge moving though the circuit.
Under standard-state conditions (1 bar or 1 M, such as the 1-M concentrations in Figure 1), the cell potential is the standard cell potential, $E^{\circ}_{\text{cell}}$ (pronounced “E-standard-cell”). Thus, based on the voltmeter reading, $E^{\circ}_{\text{cell}}=0.46\ \text{V}$
A meter like the one in Figure 1 measures the difference in electrical potential between its positive terminal and its negative terminal. Because the positive meter terminal is on the right, the cell potential is the difference in electrical potential between the right-hand half-cell and the left-hand half-cell, and we can write,
$E_{\text{cell}} = E_{\text{right half-cell}} - E_{\text{left half-cell}} \nonumber$
In Figure 1, all concentrations are 1 M (standard-state conditions), so we can also write,
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{right half-cell}} - E^{\circ}_{\text{left half-cell}} \nonumber$
If the wire connections are reversed, a typical voltmeter would read −0.46 V. This provides an experimental way to determine which half-cell is the cathode and which is the anode: a positive voltmeter reading indicates that the meter’s negative terminal is connected to the anode and the positive terminal is connected to the cathode; a negative voltmeter reading indicates the negative terminal is connected to the cathode and the positive terminal is connected to the anode.
The cell potential of a voltaic cell depends on the substances in each half-cell and the concentrations of solutions and partial pressures of gases involved in the half-cell. The half-cell potential does not depend on whether a half-reaction is occurring or on the direction in which the half-reaction goes. That is, the potential of the half cell involving silver ions and silver is the same whether silver is oxidized to silver ions or silver ions are reduced to silver.
A salt bridge must be present to complete an electric circuit. In the salt bridge when the cell generates electric current, anions move toward the anode and cations move toward the cathode.
D39.2 Cell Notation
Drawing a diagram, like Figure 1 above, to define a voltaic cell takes a lot of time. Cell notation is an abbreviation that summarizes the important information about a voltaic cell. In cell notation, a vertical line, |, denotes a phase boundary and a double line, ||, a salt bridge. The anode electrode is written to the left, followed by the anode solution, then the salt bridge, then the cathode solution, and, finally, the cathode electrode to the right. Figure 2 shows how the cell notation for a voltaic cell relates to various components of the cell.
Note that spectator ions, such as NO3, are not included in the cell notation, and if there are coefficients in a half-reaction the coefficients are not included (that is, the coefficients of 2 in the silver half-reaction are do not appear in the cell notation). When known, the initial concentrations of ions are usually included in the cell notation, so a more complete cell notation for the cell in Figure 2 is Cu(s) | Cu2+(aq, 1 M) || Ag+(aq, 1 M) | Ag(s).
Some redox reactions involve species that are poor conductors of electricity, such as gases or ionic solids. For such substances, an inert electrode, that does not participate in the reactions, is used. An example of such a voltaic cell is shown in Figure 3.
The redox reaction involved is:
Oxidation (anode): Mg(s) Mg2+(aq) + 2e
Reduction (cathode): 2H+(aq) + 2e H2(g)
overall: Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
This voltaic cell uses an inert platinum wire for the cathode, so the cell notation is:
Mg(s) | Mg2+(aq) || H+(aq) | H2(g) | Pt(s)
The magnesium electrode is an active electrode because it participates in the redox reaction. Inert electrodes, like the platinum electrode in Figure 3, do not participate in the redox reaction but must be present so that there is a complete electrical circuit. Platinum and gold are among the least reactive metals so they are good choices for inert electrodes. Graphite, also inert to many chemical reactions, is another common option.
D39.3 Standard Half-Cell Potentials
The cell potential in Figure 1 (+0.46 V) results from the difference in the electrical potential between the half-cells. It is not possible to measure directly the potential of a single half-cell; one half-cell has to be connected to another half-cell to measure a voltage.
However, it is useful to tabulate potentials for individual half-cells, such that the potential for a voltaic cell constructed from any two half-cells can be calculated from the values in the table. To create such a table, all half-cell potentials need to be measured relative to the same reference half-cell. That half-cell is the standard hydrogen electrode (SHE) (Figure 4), which consists of hydrogen gas at 1 bar pressure bubbling through a 1 M H+(aq) solution (platinum is used as the inert electrode):
2 H+(aq, 1 M) + 2 e‾ ⇌ H2(g, 1 bar) E° = 0 V
If a cell is set up with the SHE on the left and the half-cell whose potential we want to measure on the right, with all concentrations 1 M and all gas partial pressures 1 bar, then the reading on the voltmeter is E°, the standard half-cell potential, for the half-cell on the right. (Unless specified, the temperature is typically assumed to be 25 ºC.) For highly accurate measurements the voltmeter must not allow any current to flow; that is, no chemical reaction takes place.
For example, a voltaic cell consisted of a SHE and a Cu2+ | Cu(s) half-cell (Figure 5) can be used to determine the standard half-cell potential for Cu2+ | Cu(s).
The cell notation for this voltaic cell is:
Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || Cu2+(aq, 1 M) | Cu(s)
As we noted in Section D39.1, the cell potential, E°cell, measured by the voltmeter, is the difference between the potential of the right-hand half-cell and the left-hand half-cell:
E°cell = E°right half-cellE°left half-cell
From the measured E°cell = +0.337 V and the defined potential of zero for the Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) half-cell, we can calculate E° of the Cu2+(aq, 1 M) | Cu(s) half-cell:
+0.337 V = Cu2+|CuH+|H2 = Cu2+|Cu – 0 = Cu2+|Cu
Sometimes, when a cell is set up with the SHE on the left, the reading on the voltmeter is negative. That is, for some half-cells the standard half-cell potential is lower than the potential for the Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) half-cell. Consider the cell shown in Figure 6, with cell notation
Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || Zn2+(aq, 1 M) | Zn(s)
Following the same reasoning as for the Cu2+(aq, 1 M) | Cu(s) half-cell, E° of the Zn2+(aq, 1 M) | Zn(s) half-cell can be calculated.
E°cell = E°right half-cellE°left half-cell
$-0.76\;\text{V} = E_{\text{Zn}^{2+}|\text{Zn}}^{\circ}\;-\;E_{\text{H}^{+}|\text{H}_2}^{\circ} = E_{\text{Zn}^{2+}|\text{Zn}}^{\circ}\;-\;0 = E_{\text{Zn}^{2+}|\text{Zn}}^{\circ} \nonumber$
It may seem strange that the standard half-cell potential is negative. This just reflects the fact that the electrical potential of the Zn2+(aq, 1 M) | Zn(s) half-cell is smaller than the electrical potential of the Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) half-cell. The potential of a half-cell depends only on the composition of the half-cell and does not depend on the direction of either a half-reaction or an overall reaction.
The standard hydrogen electrode is rather dangerous because H2(g) is very flammable. Hence, it is rarely used in the laboratory. Its main significance is that it establishes the “zero” for standard half-cell potentials. Most standard half-cell potentials are measured by setting up a voltaic cell with one half-cell of known standard potential and one half-cell of unknown (to be measured) standard potential.
D39.4 Using Standard Half-Cell Potentials
Based on the methods described in the preceding section, standard half-cell potentials have been determined for many half-cells. The table below gives half-cell potentials for selected half-cells. Click on the table to open a version that can be enlarged or printed.
There are several important aspects to note:
• The oxidizing agent is on the left (reactant) side of a reduction half-reaction equation.
• The strongest oxidizing agents (the substances most easily reduced) have the largest positive E° values and are at the top of the table. (For example, F2(g) is a very strong oxidizing agent.)
• The reducing agent is on the right (product) side of a reduction half-reaction equation.
• The strongest reducing agents (the substances most easily oxidized) have the most negative E° values and are at the bottom of the table. (For example, Li(s) and K(s) are very strong reducing agents.)
• A redox reaction is product-favored when a stronger oxidizing agent reacts with a stronger reducing agent. This results in a positive value for E°cell when the oxidation half-reaction is combined with the reduction half-reaction.
• Half-cell reactions are reversible and the direction a half-cell reaction goes depends on the potential of the other half-cell to which it is connected in a cell.
The table of standard half-cell potentials can be used to determine the E°cell for any voltaic cell and predict whether a specific redox reaction is product-favored. For example, for the cell:
Cu(s) | Cu2+(aq, 1 M) || Ag+(aq, 1 M) | Ag(s)
E°cell = E°right half-cellE°left half-cell = E°Ag+|AgE°Cu2+|Cu = 0.7991 V – 0.340 V = 0.459 V
Because the value of E°cell is positive, the redox reaction corresponding to this cell notation is product-favored and the voltaic cell can produce electrical energy. The overall reaction can be obtained from the cell notation by writing the oxidation and the reduction half-reactions, multiplying each half-reaction by an appropriate number to balance electrons, and summing the two half-reactions:
Oxidation (left): Cu(s) Cu2+(aq) + 2e E°anode = +0.340 V
Reduction (right): 2 × (Ag+(aq) + e Ag(s)) E°cathode = +0.7991 V
Overall: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) E°cell = +0.459 V
Note that:
• Even though the reduction half-reaction is multiplied by 2, the E°cathode is not multiplied by 2 when E°cell is calculated. This is because electrical potential is a ratio of energy per coulomb of charge transferred (V = J/C). If the half-reaction is doubled, both energy and charge transferred are doubled, leaving the ratio constant.
• The oxidation half-reaction is the reverse of the reaction as shown in the half-cell potential table, but E°right is not multiplied by -1 . The cell potential does not depend on the direction of a half-reaction; it depends only on the composition of the half-cell.
E° for many half-cells are included in the appendix. Tables like these make it possible to determine the E°cell for many redox reactions. Moreover, by comparing the standard potentials, we can discern which species is easier to reduce/harder to oxidize/is a stronger oxidizing agent (higher/more positive E°) and which species is easier to oxidize/harder to reduce/is a stronger reducing agent (lower/more negative E°).
Exercise 5: Predicting whether a Redox Reaction is Product-favored
Podia Question
Use data from the table of standard half-cell potentials to determine whether this redox reaction is product-favored.
2 NO3(aq) + 8 H+(aq) + 3 Cd(s) ⇌ 2 NO(g) + 3 Cd2+(aq) + 4 H2O(l)
Query $6$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/05%3A_Unit_Five/5.04%3A_Day_39-_Voltaic_Cells_Half-Cell_Potentials.txt |
D40.1 Relationships among ΔrG°, K°, and E°cell
A redox reaction can be described in terms of ΔrG° and , and these thermodynamic variables can be related to E°cell because they all describe whether the redox reaction is reactant-favored or product-favored at equilibrium.
In a voltaic cell, the difference in Gibbs free energy between products and reactants allows the cell to do electrical work. We represent electrical work done by the cell as welec.
welec (J) = charge transferred (C) × potential difference (V)
The charge of 1 mol electrons is known as the Faraday’s constant (F):
$F = \dfrac{6.0221\;\times\;10^{23}\;\text{e}^{-}}{\text{mol}}\;\times\;\dfrac{1.6022\;\times\;10^{-19}\;\text{C}}{\text{e}^{-}} = 96485\;\frac{\text{C}}{\text{mol}} = 96485\frac{\text{J}}{\text{V}{\cdot}\text{mol}} \nonumber$
Hence, the total quantity of charge transferred per mole of a redox reaction is:
$\dfrac{\text{total charge transferred}}{\text{mole of reaction}} = (\text{number of e}^-\;\text{transferred}) \;\times\; F = n\;\times\;F \nonumber$
In this equation, n is the number of electrons transferred in the balanced redox reaction, which can be obtained from the balanced half-reactions that are added to produce the overall redox reaction. For example, in the reaction:
2 Au3+(aq) + 3 Zn(s) → 3 Zn2+(aq) + 2 Au(s)
6 e are transferred because each half-reaction, after multiplying by a factor to balance electrons, involves 6 e:
Oxidation: 3 × (Zn(s) Zn2+(aq) + 2e)
Reduction: 2 × (Au3+(aq) + 3e Au(s))
Overall: 2 Au3+(aq) + 3 Zn(s) 3 Zn2+(aq) + 2 Au(s)
When this redox reaction happens once, 6 e are transferred. When a mole of this reaction takes place, 6 moles of e are transferred; that is, 6 mol × 96485 C/mol = 578910 C are transferred. Hence, we have:
welec = charge transferred × potential difference
welec = (nF) × (Ecell)
If we operate a voltaic cell such that the maximum possible electrical work, welec, is done and the only work done is electrical work, then:
ΔrG = −wmax = −welec = −nFEcell
The negative sign makes sense because a positive Ecell indicates a spontaneous redox reaction while for ΔrG, a negative value indicates a spontaneous reaction.
If all the reactants and products are in their standard states, then the equation becomes:
ΔrG° = –nFE°cell
This equation also links standard cell potentials to equilibrium constants, since:
nFE°cell = ΔrG° = –RTlnK°
Therefore:
$E_{\text{cell}}^{\circ} = \dfrac{RT}{nF}\;\text{ln}\;K^{\circ} \nonumber$
Thus, if any one of ΔrG°, , or E°cell is known or can be calculated for a redox reaction, the other two quantities can be determined using the relationships shown in Figure 1. Moreover, any of the three quantities can be used to determine whether a reaction is product-favored at equilibrium.
D40.2 Nernst Equation
Now that the connection has been made between standard Gibbs free energy change and standard cell potentials, we can consider the situation under nonstandard conditions. Recall that ΔrG is related to ΔrG° by the reaction quotient, Q, and a similar relationship can be applied to cell potentials:
ΔrG = ΔrG° + RT lnQ
nFEcell = nFE°cell + RTlnQ
Rearranging the variables gives the Nernst equation:
$E_{\text{cell}} = E_{\text{cell}}^{\circ}\;-\;\dfrac{RT}{nF}\;\text{ln}\;Q \nonumber$
The Nernst equation can be used to calculate the cell potential at concentrations or partial pressures that differ from standard conditions, which is often the case for voltaic cells.
Activity 1: Is a Redox Reaction Spontaneous?
D40.3 Concentration Cells
A concentration cell is a special type of voltaic cell where the electrodes are the same material but the half-cells have different concentrations of soluble species. Because one or both half-cells are not under standard-state conditions, the half-cell potentials are unequal, and there is a potential difference between the half-cells. That potential difference can be calculated using the Nernst equation.
For example, consider this concentration cell at 25°C:
Zn(s) | Zn2+(aq, 0.10 M) || Zn2+(aq, 0.50 M) | Zn(s)
The standard cell potential is 0 V because the anode and cathode involve the same reaction; however, the process is spontaneous because if equal volumes of the two half-cell solutions were mixed, the concentration of Zn2+ would change to the average of the initial concentrations, namely, to 0.30 M. The cell can do work because the concentrations of Zn2+ change.
Oxidation: Zn(s) Zn2+(aq, 0.10 M) + 2e‾ E°left half-cell = -0.763 V
Reduction : Zn2+(aq, 0.50 M) + 2e‾ Zn(s) E°right half-cell = -0.763 V
Overall: Zn2+(aq, 0.50 M) Zn2+(aq, 0.10 M) E°cell = 0 V
The Nernst equation verifies that the process is spontaneous at the given conditions, because it shows that Ecell > 0 V:
$E_{\text{cell}} = 0\;\text{V}\;-\;\dfrac{\left(8.314\frac{\text{J}}{\text{K}{\cdot}\text{mol}}\right)(298.15\;\text{K})}{(2)\left(96485\frac{\text{J}}{\text{V}{\cdot}\text{mol}}\right)}\;\text{ln}\left(\dfrac{0.10\;\text{M}}{0.50\;\text{M}}\right) = +0.021\;\text{V} \nonumber$
In a concentration cell, the standard cell potential (E°cell) is always zero. In order to have a spontaneous forward reaction, and hence have a positive cell potential (Ecell), the reaction quotient Q must be less than 1 (when Q < 1, ln(Q) < 0). As the reaction proceeds, the concentrations change, Q approaches 1 and Ecell approaches 0 V.
Podia Question
The gravity cell, a variant of the Daniell cell, was described in Exercise 3 in Day 39. The cell involves oxidation of Zn(s) by Cu2+(aq). Suppose a gravity cell is constructed with 1.00 L saturated copper(II) sulfate in the bottom and 1.00 L 0.100-M zinc sulfate in the top. (There is no solid copper(II) sulfate in the cell. Assume a temperature of 25 °C.)
Write half-reactions for the overall reaction that powers the gravity cell and label them as oxidation and reduction.
Write the balanced overall cell reaction.
Calculate the standard cell potential.
Calculate the cell potential under the initial (nonstandard) conditions. The solubility of CuSO4•5H2O in water is 32 g/100 mL.
Calculate the cell potential after 99% of the copper(II) ions in the bottom cell solution has reacted. (Assume there is an excess of Zn(s).)
Query $4$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/05%3A_Unit_Five/5.05%3A_Day_40-_Thermodynamic_Properties_of_Voltaic_and_Electrolytic_Cells.txt |
D41.1 Electrolysis
In an electrolytic cell, supplied electrical energy causes a nonspontaneous redox reaction to occur in a process known as electrolysis. An electrolytic cell is the opposite of a voltaic cell, where a spontaneous redox reaction produces electrical energy.
For example, consider electrolysis of molten sodium chloride. A simplified diagram of the electrolytic cell used for commercial manufacture of sodium metal and chlorine gas is shown in Figure 1. In the molten salt, sodium ions move toward the cathode and chloride ions move toward the anode (note that negative ions move through the circuit in the same overall direction as electrons). A porous screen allows movement of ions but not mixing of the product sodium metal and chlorine gas, which would react spontaneously upon contact.
An external source of electric current forces electrons into the electrode in the cathode compartment, forcing the reduction half-reaction to occur:
Reduction (cathode): 2 × (Na+(aq) + e‾ ⟶ Na(s) ) E°Na+|Na = -2.714 V
(At the temperature of molten NaCl, sodium is a liquid, so E°Na+|Na(s) = -2.714 V is an approximation. E°Na+|Na(l) for liquid sodium is not available in the appendix.)
In the anode compartment, the oxidation half-reaction occurs:
Oxidation (anode): 2 Cl‾(aq) ⟶ Cl2(g) + 2e‾ E°Cl2|Cl‾ = +1.358 V
The electrons formed here are conducted through a wire to the the positive side of the voltage source, completing the electric circuit.
We can calculate E°cell for this electrolysis cell using the same method we used for voltaic cells; that is,
E°cell = E°right half-cellE°left half-cell = E°cathodeE°anode = -2.714 V − (+1.358 V) = -4.072 V
The overall reaction is:
Overall: 2 Na+(aq) + 2 Cl‾(aq) ⟶ 2 Na(s) + Cl2(g) E°cell = -4.072 V
The negative E°cell indicates that this reaction is strongly reactant-favored, and under standard-state conditions, the power supply must provide at least 4.1 V to cause the electrolysis reaction to occur. In practice, the applied voltages are higher due to inefficiencies in the process itself and also to help increase the rate of reaction.
With the transition from fossil fuels to renewable energy supplies, an important application of electrolysis is the “splitting” of water into hydrogen gas and oxygen gas (Figure 2). Electric energy from solar panels or wind turbines can be used to synthesize hydrogen for use as a fuel. For current to pass through the solution efficiently, there must be ions present. Hence, acid is typically added to the reaction solution to increase the concentration of ions in solution.
In 1-M acidic solution:
Oxidation (anode): 2H2O(l) O2(g) + 4H+(aq) + 4e‾ E°anode = +1.229 V
Reduction (cathode): 2 × (2H+(aq) + 2e‾ H2(g)) E°cathode = 0 V
Overall: 2H2O(l) O2(g) + 2H2(g) E°cell = -1.229 V
At least 1.229 V is required to make this reactant-favored process occur in 1-M acidic solution.
Finally, consider what occurs during the electrolysis of 1-M aqueous potassium iodide solution at 25 °C. Present in the solution are H2O(l), K+(aq), and I‾(aq). This example differs from the previous examples because more than one species can be oxidized and more than one species can be reduced.
Considering the anode first, the possible oxidation reactions are
(i) 2I‾(aq) I2(s) + 2e‾ E°anode = +0.535 V
(ii) 2H2O(l) O2(g) + 4H+(aq) + 4e‾ E°anode = +1.229 V
(Oxidation of K+ to K2+ is not considered because K+ has a noble-gas electron configuration and a very high ionization energy, making it very difficult to oxidize. Although I2 is generated in aqueous solution, we approximate the E° by using the value for I2(s).)
Because E°cell = E°cathodeE°anode, the more positive the anode half-cell potential is, the more negative the cell potential would be. Therefore, iodide should be oxidized at the anode because it has a less positive half-cell potential. However, the pH of a 1-M KI solution is 7, so [H+] is far from standard-state conditions. Assuming that O2 is produced at 1 bar, applying the Nernst equation to half-reaction (ii) gives:
$\begin{array}{rcl} E &=& {E^ \circ } - \dfrac{RT}{nF}\ln \left( \dfrac{1}{[\text{O}_2][\text{H}^+]^4}\right)\[0.5em] &=& 1.229\text{ V} - \dfrac{\left(8.314 \frac{\text{J}}{\text{K}\cdot\text{mol}}\right) (298.15\;\text{K})}{4\left(96485\frac{\text{J}}{\text{V}\cdot\text{mol}}\right)} \ln \left(\dfrac{1}{(1)(1 \times 10^{-7})^4}\right)\[0.5em] &=& 1.229\;{\text{ V}} - 0.414\;\text{V}\; =\; 0.815\;\text{V} \end{array} \nonumber$
(Note that the reaction given in the standard half-cell potential table is the reduction reaction “O2(g) + 4H+(aq) + 4e‾ ⟶ 2H2O(l) E° = +1.229 V”, hence reaction quotient Q is expressed in accordance to the given reaction.)
The Nernst equation shows that Eanode = +0.815 V for reaction (ii) at pH = 7, which is still higher than E°anode for reaction (i). Therefore, reaction (i) is the process that occurs at the anode and I2 forms as a product.
Now consider the possible reactions at the cathode (reduction of I is not considered because I has a noble-gas electron configuration and it is not energetically favorable to add more electrons):
(iii) 2H2O(l) + 2e‾ H2(g) + 2OH‾(aq) E°cathode = -0.8277 V
(ii) K+(aq) + e‾ K(s) E°cathode = -2.925 V
For half-reaction (iii), we again need to apply the Nernst equation to calculate E at pH = 7, assuming H2 is produced at 1 bar:
$\begin{array}{rcl} E &=& {E^ \circ } - \dfrac{RT}{nF}\ln \left( \dfrac{[\text{H}_2][\text{OH}^-]^2}{1}\right)\[0.5em] &=& -0.8277\text{ V} - \dfrac{\left(8.314 \frac{\text{J}}{\text{K}\cdot\text{mol}}\right) (298.15\;\text{K})}{4\left(96485\frac{\text{J}}{\text{V}\cdot\text{mol}}\right)} \ln \left(\dfrac{(1)(1 \times 10^{-7})^4}{1}\right)\[0.5em] &=& -0.8277\;{\text{ V}} - (-0.4141\;\text{V})\; =\; -0.4136\;\text{V} \end{array} \nonumber$
Hence, reduction of water, with Ecathode = -0.4136 V at pH = 7, is much more likely to occur than reduction of K+(aq) with cathode = -2.925 V. (This conclusion is supported by the fact that potassium metal reacts vigorously with water to generate K+(aq), hydrogen gas, and hydroxide ions, so if K(s) formed it would immediately react with water.)
The overall reaction is then:
Oxidation (anode): 2I‾(aq) I2(s) + 2e‾ E°anode = +0.535 V
Reduction (cathode): 2H2O(l) + 2e‾ H2(g) + 2OH‾(aq) Ecathode = -0.4136 V
Overall: 2H2O(l) + 2I‾(aq) H2(g) + I2(s) + 2OH‾(aq) Ecell = -0.949 V
Electroplating
An important use for electrolytic cells is electroplating, which results in a thin coating of metal on top of a conducting surface. Metals typically used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. As an example of electroplating, consider how silver-plated tableware is produced (Figure 3).
The anode consists of a silver electrode. The cathode is a spoon made from a less expensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential from the voltage source is increased, current flows. Silver metal is lost at the anode as it goes into solution:
anode: Ag(s) ⟶ Ag+(aq) + e
The mass of the cathode increases as silver ions from the solution are deposited onto the spoon:
cathode: Ag+(aq) + e ⟶ Ag(s)
The net result is the transfer of silver metal from the anode to the cathode. The quality of the electroplated object depends on the thickness of the deposited silver and the rate of deposition.
Quantitative Aspects of Electrolysis
The quantity of current that flows in an electrolytic cell is dictated by the amount (mol) of electrons transferred in a redox reaction, which is in turn related to quantities of reactants and products via reaction stoichiometry. Recall that current, I, is related to the total charge, Q:
$I = \dfrac{Q}{t} \;\;\;\;\; \left(\text{SI units:}\;\ A = \dfrac{C}{s}\right) \nonumber$
Hence:
Q = (amount e transferred) × F = I × t
where F is the Faraday constant.
D41.2 Commercial Batteries
Many of the devices we use every day, such as laptops and smartphones, are powered by batteries. A battery is an electrochemical cell or series of cells that produces an electric current. In principle, any voltaic cell can be used as a battery. An ideal battery would never run down/drain, produce a constant voltage, and be capable of withstanding environmental extremes of temperature and humidity. Real batteries strike a balance between ideal characteristics and practical limitations.
For example, the mass of a car-starter battery is about 18 kg or ~1% of the mass of an average car. This type of battery would supply nearly unlimited energy if used in a smartphone, but would be completely impractical because of its mass and size. Thus, no single battery is “best” and different batteries are selected for particular applications, keeping things like its mass, cost, reliability, and current capacity in mind.
There are two basic types of batteries: primary and secondary. A few batteries of each type are described next.
D41.3 Primary Batteries
Primary batteries are single-use batteries that cannot be recharged.
Zinc-Carbon Battery
A common primary battery is the dry cell (Figure 4), which is a zinc-carbon battery. The zinc serves as both a container and the negative electrode. The positive electrode is a rod made of carbon that is surrounded by a paste of manganese(IV) oxide, zinc chloride, ammonium chloride, carbon powder, and a small quantity of water.
The reaction at the anode can be represented as the oxidation of zinc:
Zn(s) ⟶ Zn2+(aq) + 2 e‾ E°anode = -0.763 V
The reaction at the cathode is more complicated, in part because more than one reduction reaction is occurring. The series of reactions that occurs at the cathode is approximately:
2MnO2(s) + 2NH4Cl(aq) + 2e‾ ⟶ Mn2O3(s) + 2NH3(aq) + H2O(l) + 2Cl‾(aq)
The overall reaction for the zinc–carbon battery can be represented as:
2MnO2(s) + 2 NH4Cl(aq) + Zn(s) ⟶ Mn2O3(s) + 2NH3(aq) + H2O(l) + 2 Cl‾(aq) + Zn2+(aq)
The cell potential is about 1.5 V initially, and decreases as the battery is used. As the zinc container oxidizes, its contents eventually leak out, so this type of battery should not be left in any electrical device for extended periods.
The voltage delivered by a battery is the same regardless of the size of a battery. For this reason, D, C, A, AA, and AAA batteries all have the same voltage. However, larger batteries can deliver more moles of electrons and will therefore last longer if powering the same device.
Alkaline Battery
Alkaline batteries (Figure 5) were developed in the 1950s partly to address some of the performance issues with zinc–carbon dry cells, and are manufactured to be their exact replacements. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide.
The reactions are:
Oxidation (anode): Zn(s) + 2OH‾(aq) ZnO(s) + H2O(l) + 2e‾ anode = -1.28 V
Reduction (cathode): 2MnO2(s) + H2O(l) + 2e‾ Mn2O3(s) + 2OH‾(aq) cathode = +0.15 V
overall: Zn(s) + 2MnO2(s) ZnO(s) + Mn2O3(s) cell = +1.43 V
An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries sometimes leak potassium hydroxide, so these should also be removed from devices for long-term storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.
D41.4 Secondary Batteries
Secondary batteries are rechargeable; that is, the reaction that powers the battery can be reversed so that the original reactants can be regenerated. Secondary batteries are found in smartphones, electronic tablets, automobiles, and many other devices.
Lead-Acid Battery
The lead-acid battery (Figure 6) is the type of secondary battery used to start gasoline-powered automobiles. It is inexpensive and capable of producing the high current required by the starter motors when starting a car. They are heavy because of lead’s high density, they contain highly corrosive concentrated sulfuric acid, and must be disposed of properly to avoid lead-poisoning hazards. But they can produce a lot of current in a short time so for certain applications they are the best choice.
The reactions for a lead acid battery are:
Oxidation (anode): Pb(s) + HSO4‾(aq) PbSO4(s) + H+(aq) + 2e‾
Reduction (cathode): PbO2(s) + HSO4‾(aq) + 3H+(aq) + 2e‾ PbSO4(s) + 2H2O(l)
overall: Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l)
Each cell produces 2.05 V, so six cells can be connected in series to produce a 12-V car battery.
In each cell, the lead electrodes are immersed in sulfuric acid. The anodes are spongy lead metal and the cathodes are lead impregnated with lead oxide. As the battery is discharged, a powder of PbSO4 forms on the electrodes. When a lead-acid battery is recharged by a car’s alternator, electrons are forced to flow in the opposite direction which reverses the reactions at anode and cathode, in other words, the cell undergoes electrolysis reactions to replenish the substances that have reacted away.
Practically, the concentrated sulfuric acid becomes quite viscous when the temperature is low, inhibiting the flow of ions between the plates and reducing the current that can be delivered. This effect is well-known to anyone who has had difficulty starting a car in cold weather. These batteries also tend to slowly self-discharge, so a car left idle for several weeks might be unable to start. And after thousands of discharge-charge cycles, PbSO4 that does not get converted to PbO2 gradually changes to an inert form which limits the battery capacity. Also, “fast” charging causes rapid evolution of potentially explosive H2 gas from the water in the electrolyte (electrolysis of water); the gas bubbles form on the lead surface and can tear PbO2 off the electrodes. Eventually enough solid material accumulates at the bottom of the electrolyte to short-circuit the battery, leading to its permanent demise.
Lithium Ion Battery
Lithium ion batteries (Figure 7) are among the most popular rechargeable batteries and are used in many portable electronic devices because their advantages outweigh the disadvantage of higher cost. In a typical Li-ion battery the reactions are:
Oxidation (anode): LiCoO2 Li1-xCoO2 + xLi+ + xe‾
Reduction (cathode): xLi+ + xC6 + xe‾ xLiC6
overall: LiCoO2 + xC6 Li1-xCoO2 + xLiC6
(x is no more than about 0.5.) The battery voltage is about 3.7 V.
Lithium batteries are popular because they can provide a large amount of current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.
D41.5 Fuel Cells
Suppose a voltaic cell is constructed such that the substance that is oxidized at the anode and the substance that is reduced at the cathode (the reactants in the overall redox reaction) are both supplied continuously. Such a battery would never run down because reactant concentrations or partial pressures would never decrease. Such a device is a fuel cell, which produces electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines (Figure 8).
In a hydrogen-oxygen proton-exchange fuel cell, the cell potential is about 1 V, and the reactions involved are:
Oxidation (anode): 2 × (H2(g) 2H+(aq) + 2e‾)
Reduction (cathode): O2(g) + 4H+(aq) + 4e‾ 2H2O(l)
overall: O2(g) + 2H2(g) 2H2O(l)
The efficiency of fuel cells is typically about 40-60%, which is higher than the typical internal combustion engine (25-35%). Moreover, in the case of the hydrogen fuel cell, nearly pure water is produced as exhaust. Currently, fuel cells are comparably more expensive and contain features that may cause a higher failure rate.
Podia Question
Consider this graph, which shows cell potential on the vertical axis and fraction of reactants remaining (battery life remaining) on the horizontal axis for a commercial battery.
• Write an explanation in appropriate scientific language for why the graph has the shape it has.
• Describe why the shape of this graph is a positive feature when batteries power devices such as smartphones and laptop computers.
Query $4$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/05%3A_Unit_Five/5.06%3A_Day_41-_Electrolysis_Commercial_Batteries.txt |
John Moore, Jia Zhou, and Etienne Garand
[2016. Scaled to an atomic weight of 12 for carbon-12 (12C), where 12C is a neutral atom in its nuclear and electronic ground state, having the result that atomic weight values are dimensionless.]
The atomic weights of many elements are not invariant, but depend on the origin and treatment of the material. The standard atomic weights apply to elements of natural terrestrial origin. Although the atomic weights of some elements in some naturally occurring materials may differ from given values because of a variation in the mole fractions of an element’s stable isotopes, the last significant figure of each tabulated value is considered reliable to ±1 except for zinc, which is ±2. For twelve of these elements, both a conventional atomic weight and an atomic weight interval is given with the symbol [a, b] to denote the set of atomic weight values in normal materials; thus, a ≤ atomic weight ≤ b. For lithium, the conventional atomic weight is only three digits because of the large variation found in lithium-bearing reagents.
Atomic Number Element Name Symbol Standard Atomic Weight Conventional Atomic Weight
1 hydrogen H [1.007, 1.009] 1.008
2 helium He 4.003
3 lithium Li [6.938, 6.997] 6.94
4 beryllium Be 9.012
5 boron B [10.80, 10.83] 10.81
6 carbon C [12.00, 12.02] 12.01
7 nitrogen N [14.00, 14.01] 14.01
8 oxygen O [15.99, 16.00] 16.00
9 fluorine F 19.00
10 neon Ne 20.18
11 sodium Na 22.99
12 magnesium Mg [24.30, 24.31] 24.31
13 aluminium (aluminum) Al 26.98
14 silicon Si [28.08, 28.09] 28.09
15 phosphorus P 30.97
16 sulfur S [32.05, 32.08] 32.06
17 chlorine Cl [35.44, 35.46] 35.45
18 argon Ar 39.95
19 potassium K 39.10
20 calcium Ca 40.08
21 scandium Sc 44.96
22 titanium Ti 47.87
23 vanadium V 50.94
24 chromium Cr 52.00
25 manganese Mn 54.94
26 iron Fe 55.85
27 cobalt Co 58.93
28 nickel Ni 58.69
29 copper Cu 63.55
30 zinc Zn 65.38
31 gallium Ga 69.72
32 germanium Ge 72.63
33 arsenic As 74.92
34 selenium Se 78.97
35 bromine Br [79.90, 79.91] 79.90
36 krypton Kr 83.80
37 rubidium Rb 85.47
38 strontium Sr 87.62
39 yttrium Y 88.91
40 zirconium Zr 91.22
41 niobium Nb 92.91
42 molybdenum Mo 95.95
43 technetium* Tc
44 ruthenium Ru 101.1
45 rhodium Rh 102.9
46 palladium Pd 106.4
47 silver Ag 107.9
48 cadmium Cd 112.4
49 indium In 114.8
50 tin Sn 118.7
51 antimony Sb 121.8
52 tellurium Te 127.6
53 iodine I 126.9
54 xenon Xe 131.3
55 caesium (cesium) Cs 132.9
56 barium Ba 137.3
57 lanthanum La 138.9
58 cerium Ce 140.1
59 praseodymium Pr 140.9
60 neodymium Nd 144.2
61 promethium* Pm
62 samarium Sm 150.4
63 europium Eu 152.0
64 gadolinium Gd 157.3
65 terbium Tb 158.9
66 dysprosium Dy 162.5
67 holmium Ho 164.9
68 erbium Er 167.3
69 thulium Tm 168.9
70 ytterbium Yb 173.1
71 lutetium Lu 175.0
72 hafnium Hf 178.5
73 tantalum Ta 180.9
74 tungsten W 183.8
75 rhenium Re 186.2
76 osmium Os 190.2
77 iridium Ir 192.2
78 platinum Pt 195.1
79 gold Au 197.0
80 mercury Hg 200.6
81 thallium Tl [204.3, 204.4] 204.4
82 lead Pb 207.2
83 bismuth* Bi 209.0
84 polonium* Po
85 astatine* At
86 radon* Rn
87 francium* Fr
88 radium* Ra
89 actinium* Ac
90 thorium* Th 232.0
91 protactinium* Pa 231.0
92 uranium* U 238.0
93 neptunium* Np
94 plutonium* Pu
95 americium* Am
96 curium* Cm
97 berkelium* Bk
98 californium* Cf
99 einsteinium* Es
100 fermium* Fm
101 mendelevium* Md
102 nobelium* No
103 lawrencium* Lr
104 rutherfordium* Rf
105 dubnium* Db
106 seaborgium* Sg
107 bohrium* Bh
108 hassium* Hs
109 meitnerium* Mt
110 darmstadtium* Ds
111 roentgenium* Rg
112 copernicium* Cn
113 nihonium* Nh
114 flerovium* Fl
115 moscovium* Mc
116 livermorium* Lv
117 tennessine* Ts
118 oganesson* Og
*Element has no stable isotopes, only radioactive isotopes, and an atomic weight cannot be determined. However, four such elements (Bi, Th, Pa, and U) do have a characteristic terrestrial isotopic composition, and for these a standard atomic weight is tabulated. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.01%3A_Atomic_Weights.txt |
John Moore, Jia Zhou, and Etienne Garand
Elemental Abundance in Solar System
Atom Atomic No. Abundance* (atoms/106 atoms Si) Log(abund)
H 1 27900000000 10.45
He 2 2720000000 9.43
Li 3 57.1 1.76
Be 4 0.73 -0.14
B 5 21.2 1.33
C 6 10100000 7
N 7 313000 5.5
O 8 23800000 7.38
F 9 843 2.93
Ne 10 3440000 6.54
Na 11 57400 4.76
Mg 12 1074000 6.03
Al 13 84900 4.93
Si 14 1000000 6
P 15 10400 4.02
S 16 515000 5.71
Cl 17 5240 3.72
Ar 18 101000 5
K 19 3770 3.58
Ca 20 61100 4.79
Sc 21 34.2 1.53
Ti 22 2400 3.38
V 23 293 2.47
Cr 24 13500 4.13
Mn 25 9550 3.98
Fe 26 900000 5.95
Co 27 2250 3.35
Ni 28 49300 4.69
Cu 29 522 2.72
Zn 30 1260 3.1
Ga 31 37.8 1.58
Ge 32 119 2.08
As 33 6.56 0.82
Se 34 62.1 1.79
Br 35 11.8 1.07
Kr 36 45 1.65
Rb 37 7.09 0.85
Sr 38 23.5 1.37
Y 39 4.64 0.67
Zr 40 11.4 1.06
Nb 41 0.698 -0.16
Mo 42 2.55 0.41
Tc 43
Ru 44 1.86 0.27
Rh 45 0.344 -0.46
Pd 46 1.39 0.14
Ag 47 0.486 -0.31
Cd 48 1.61 0.21
In 49 0.184 -0.74
Sn 50 3.82 0.58
Sb 51 0.309 -0.51
Te 52 4.81 0.68
I 53 0.9 -0.05
Xe 54 4.7 0.67
Cs 55 0.372 -0.43
Ba 56 4.49 0.65
La 57 0.446 -0.35
Ce 58 1.136 0.06
Pr 59 0.1669 -0.78
Nd 60 0.8279 -0.08
Pm 61
Sm 62 0.2282 -0.64
Eu 63 0.0973 -1.01
Gd 64 0.33 -0.48
Tb 65 0.0603 -1.22
Dy 66 0.3942 -0.4
Ho 67 0.0889 -1.05
Er 68 0.2508 -0.6
Tm 69 0.0378 -1.42
Yb 70 0.2479 -0.61
Lu 71 0.0367 -1.44
Hf 72 0.154 -0.81
Ta 73 0.0207 -1.68
W 74 0.133 -0.88
Re 75 0.0517 -1.29
Os 76 0.675 -0.17
Ir 77 0.661 -0.18
Pt 78 1.34 0.13
Au 79 0.187 -0.73
Hg 80 0.34 -0.47
Tl 81 0.184 -0.74
Pb 82 3.15 0.5
Bi 83 0.144 -0.84
Po 84
At 85
Rn 86
Fr 87
Ra 88
Ac 89
Th 90 0.0335 -1.47
Pa 91
U 92 0.009 -2.05
Np 93
Pu 94
Am 95
Cm 96
Bk 97
Cf 98
Es 99
Fm 100
Md 101
No 102
Lr 103
*Anders, Edward; Grevesse; Nicolas, Geochimica et Cosmochimica Acta 1989, 53, 197-214.
Elemental Abundance in Earth Crust
Atom Atomic No. Abundance*
H 1 6.2
He 2 0
Li 3 4.1
Be 4 3.2
B 5 4
C 6 6.7
N 7 4.3
O 8 8.67
F 9 5.8
Ne 10 0
Na 11 7.36
Mg 12 7.51
Al 13 7.92
Si 14 8.43
P 15 6
S 16 5.8
Cl 17 5.3
Ar 18 1
K 19 6.96
Ca 20 7.72
Sc 21 4.5
Ti 22 6.7
V 23 5.4
Cr 24 5.3
Mn 25 6.1
Fe 26 7.84
Co 27 4.5
Ni 28 5
Cu 29 4.9
Zn 30 4.9
Ga 31 4.3
Ge 32 3.2
As 33 3.4
Se 34 1.7
Br 35 3.4
Kr 36 -1
Rb 37 4.5
Sr 38 5.4
Y 39 4.3
Zr 40 5
Nb 41 4
Mo 42 3
Tc 43
Ru 44 0
Rh 45 -1
Pd 46 0
Ag 47 1.9
Cd 48 2
In 49 1.7
Sn 50 3.4
Sb 51 2.3
Te 52 -1
I 53 2.7
Xe 54 -2
Cs 55 3
Ba 56 5.4
La 57 4.2
Ce 58 4.5
Pr 59 3.6
Nd 60 4.2
Pm 61
Sm 62 3.5
Eu 63 3
Gd 64 3.5
Tb 65 2.8
Dy 66 3.6
Ho 67 2.9
Er 68 3.3
Tm 69 2.5
Yb 70 3.3
Lu 71 2.5
Hf 72 3.5
Ta 73 3
W 74 3
Re 75 -0.3
Os 76 -1
Ir 77 -1
Pt 78 0
Au 79 0.5
Hg 80 2
Tl 81 2.6
Pb 82 3.9
Bi 83 -1
Po 84
At 85
Rn 86
Fr 87
Ra 88 -4
Ac 89
Th 90 3.6
Pa 91 -5
U 92 3
Np 93
Pu 94
Am 95
Cm 96
Bk 97
Cf 98
Es 99
Fm 100
Md 101
No 102
Lr 103
*log(mass fraction in ppb, that is μg/kg) | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.02%3A_Elemental_Abundances.txt |
John Moore, Jia Zhou, and Etienne Garand
Ground State Electron Configuration of Atoms
Atomic No. Element Configuration
1 H 1s¹
2 He 1s²
3 Li [He] 2s¹
4 Be [He] 2s²
5 B [He] 2s²2p¹
6 C [He] 2s²2p²
7 N [He] 2s²2
8 O [He] 2s²2p⁴
9 F [He] 2s²2p⁵
10 Ne [He] 2s²2p⁶
11 Na [Ne] 3s¹
12 Mg [Ne] 3s²
13 Al [Ne] 3s²3p¹
14 Si [Ne] 3s²3
15 P [Ne] 3s²3
16 S [Ne] 3s²3p⁴
17 Cl [Ne] 3s²3p⁵
18 Ar [Ne] 3s²3p⁶
19 K [Ar] 4s¹
20 Ca [Ar] 4s²
21 Sc [Ar] 3d¹4s²
22 Ti [Ar] 3d²4s²
23 V [Ar] 3d³4s²
24 Cr [Ar] 3d⁵4s¹
25 Mn [Ar] 3d⁵4s²
26 Fe [Ar] 3d⁶4s²
27 Co [Ar] 3d⁷4s²
28 Ni [Ar] 3d⁸4s²
29 Cu [Ar] 3d¹⁰4s¹
30 Zn [Ar] 3d¹⁰4s²
31 Ga [Ar] 3d¹⁰4s²4p¹
32 Ge [Ar] 3d¹⁰4s²4
33 As [Ar] 3d¹⁰4s²4
34 Se [Ar] 3d¹⁰4s²4p⁴
35 Br [Ar] 3d¹⁰4s²4p⁵
36 Kr [Ar] 3d¹⁰4s²4p⁶
37 Rb [Kr] 5s¹
38 Sr [Kr] 5s²
39 Y [Kr]4d¹5s²
40 Zr [Kr] 4d²5s²
41 Nb [Kr] 4d⁴5s¹
42 Mo [Kr] 4d⁵5s¹
43 Tc [Kr] 4d⁵5s²
44 Ru [Kr] 4d⁷5s¹
45 Rh [Kr] 4d⁸5s¹
46 Pd [Kr] 4d¹⁰
47 Ag [Kr] 4d¹⁰5s¹
48 Cd [Kr] 4d¹⁰5s²
49 In [Kr] 4d¹⁰5s²5p¹
50 Sn [Kr] 4d¹⁰5s²5
51 Sb [Kr] 4d¹⁰5s²5
52 Te [Kr] 4d¹⁰5s²5p⁴
53 I [Kr] 4d¹⁰5s²5p⁵
54 Xe [Kr] 4d¹⁰5s²5p⁶
55 Cs [Xe] 6s¹
56 Ba [Xe] 6s²
57 La [Xe] 5d¹6s²
58 Ce [Xe] 4f¹5d¹6s²
59 Pr [Xe] 4f³6s²
60 Nd [Xe] 4f⁴6s²
61 Pm [Xe] 4f⁵6s²
62 Sm [Xe] 4f⁶6s²
63 Eu [Xe] 4f⁷6s²
64 Gd [Xe] 4f⁷5d¹6s²
65 Tb [Xe] 4f⁹6s²
66 Dy [Xe] 4f¹⁰6s²
67 Ho [Xe] 4f¹¹6s²
68 Er [Xe] 4f¹²6s²
69 Tm [Xe] 4f¹³6s²
70 Yb [Xe] 4f¹⁴6s²
71 Lu [Xe] 4f¹⁴5d¹6s²
72 Hf [Xe] 4f¹⁴5d²6s²
73 Ta [Xe] 4f¹⁴5d³6s²
74 W [Xe] 4f¹⁴5d⁴6s²
75 Re [Xe] 4f¹⁴5d⁵6s²
76 Os [Xe] 4f¹⁴5d⁶6s²
77 Ir [Xe] 4f¹⁴5d⁷6s²
78 Pt [Xe] 4f¹⁴5d⁹6s¹
79 Au [Xe] 4f¹⁴5d¹⁰6s¹
80 Hg [Xe] 4f¹⁴5d¹⁰6s²
81 Tl [Xe] 4f¹⁴5d¹⁰6s²6p¹
82 Pb [Xe] 4f¹⁴5d¹⁰6s²6
83 Bi [Xe] 4f¹⁴5d¹⁰6s²6
84 Po [Xe] 4f¹⁴5d¹⁰6s²6p⁴
85 At [Xe] 4f¹⁴5d¹⁰6s²6p⁵
86 Rn [Xe] 4f¹⁴5d¹⁰6s²6p⁶
87 Fr [Rn] 7s¹
88 Ra [Rn] 7
89 Ac [Rn] 6d¹7
90 Th [Rn] 67
91 Pa [Rn] 56d¹7
92 U [Rn] 56d¹7
93 Np [Rn] 5f⁴6d¹7
94 Pu [Rn] 5f⁶7
95 Am [Rn] 5f⁷7
96 Cm [Rn] 5f⁷6d¹7
97 Bk [Rn] 5f⁹7
98 Cf [Rn] 5f¹⁰7
99 Es [Rn] 5f¹¹7
100 Fm [Rn] 5f¹²7
101 Md [Rn] 5f¹³7
102 No [Rn] 5f¹⁴7
103 Lr [Rn] 5f¹⁴77p¹
104 Rf [Rn] 5f¹⁴67
105 Db [Rn] 5f¹⁴67
106 Sg [Rn] 5f¹⁴6d⁴7
107 Bh [Rn] 5f¹⁴6d⁵7
108 Hs [Rn] 5f¹⁴6d⁶7
109 Mt [Rn] 5f¹⁴6d⁷7
110 Ds [Rn] 5f¹⁴6d⁹7s¹
111 Rg [Rn] 5f¹⁴6d¹⁰7s¹
112 Cn [Rn] 5f¹⁴6d¹⁰7
113 [Rn] 5f¹⁴677p¹
114 Fl [Rn] 5f¹⁴677
115 [Rn] 5f¹⁴677
116 Lv [Rn] 5f¹⁴677p⁴
117 [Rn] 5f¹⁴6d¹⁰77p⁵
118 [Rn] 5f¹⁴6d¹⁰7s²7p⁶
All electron configurations in this table are from experimental determinations except for elements 103-118, which are predicted theoretically. For elements 103-118, only a few atoms can be produced at a time and it is very difficult to determine electron configurations. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.03%3A_Electron_Configuration.txt |
John Moore, Jia Zhou, and Etienne Garand
Atomic Radius
Atom Atomic No. Atomic Radius (pm)
H 1 37
He 2 31
Li 3 157
Be 4 112
B 5 86
C 6 77
N 7 74
O 8 74
F 9 72
Ne 10 68
Na 11 191
Mg 12 160
Al 13 143
Si 14 117
P 15 110
S 16 104
Cl 17 99
Ar 18 91
K 19 235
Ca 20 197
Sc 21 164
Ti 22 147
V 23 135
Cr 24 129
Mn 25 137
Fe 26 126
Co 27 125
Ni 28 125
Cu 29 126
Zn 30 137
Ga 31 153
Ge 32 130.5
As 33 121
Se 34 117
Br 35 114
Kr 36 108
Rb 37 250
Sr 38 215
Y 39 182
Zr 40 160
Nb 41 147
Mo 42 140
Tc 43 135
Ru 44 134
Rh 45 134
Pd 46 137
Ag 47 144
Cd 48 152
In 49 167
Sn 50 149
Sb 51 151
Te 52 137
I 53 133
Xe 54 126
Cs 55 272
Ba 56 224
La 57 188
Ce 58 182
Pr 59 181
Nd 60 180
Pm 61 179
Sm 62 179
Eu 63 206
Gd 64 178
Tb 65 177
Dy 66 176
Ho 67 175
Er 68 174
Tm 69 173
Yb 70 194
Lu 71 172
Hf 72 159
Ta 73 147
W 74 141
Re 75 137
Os 76 135
Ir 77 136
Pt 78 139
Au 79 144
Hg 80 155
Tl 81 171
Pb 82 175
Bi 83 182
Po 84 174
At 85 166
Rn 86 159
Fr 87
Ra 88
Ac 89
Th 90 180
Pa 91 163
U 92 156
Np 93 156
Pu 94 164
Am 95 173
Cm 96 155
Bk 97 170
Cf 98
Es 99
Fm 100
Md 101
No 102
Lr 103
Wells, A. F. Structural Inorganic Chemistry, 5th ed., Clarendon: Oxford, 1984, p 289 and 1288.
Slater, J. C. J. Chem. Phys. 1964, 41, 3199.
Clementi, E; Raimondi, D. L.; Reinhardt, W. P. J. Chem. Phys. 1967, 47(4), 1300-1307.
Ionic Radius
Ion Species Atomic No. Ionic Radius (pm)
H +1 1
He 2
Li +1 3 90
Be +2 4 59
B +3 5 41
C +4 6 30
N -3 7
N +3 7 30
N +5 7 27
O -2 8 126
F -1 9 119
F +7 9 22
Ne 10
Na +1 11 116
Mg +2 12 86
Al +3 13 67.5
Si +4 14 54
P +3 15 58
P +5 15 52
S -2 16 170
S +4 16 51
S +6 16 43
Cl -1 17 167
Cl +5 17
Cl +7 17 41
Ar 18
K +1 19 152
Ca +2 20 114
Sc +3 21 88.5
Ti +2 22 100
Ti +3 22 81
Ti +4 22 74.5
V +2 23 93
V +3 23 78
V +4 23 72
V +5 23 68
Cr +2 24 94
Cr +3 24 75.5
Cr +4 24 69
Cr +5 24 63
Cr +6 24 58
Mn +2 25 97
Mn +3 25 78.5
Mn +4 25 67
Mn +5 25
Mn +6 25
Mn +7 25 60
Fe +2 26 92
Fe +3 26 78.5
Fe +4 26 72.5
Fe +6 26
Co +2 27 88.5
Co +3 27 75
Co +4 27 67
Ni +2 28 83
Ni +3 28 74
Ni +4 28 62
Cu +1 29 91
Cu +2 29 87
Cu +3 29 68
Zn +2 30 88
Ga +3 31 76
Ge +2 32 87
Ge +4 32 67
As +3 33 72
As +5 33 60
Se -2 34 184
Se +4 34 64
Se +6 34 56
Br -1 35 182
Br +3 35
Br +5 35
Br +7 35 53
Kr 36
Rb +1 37 166
Sr +2 38 132
Y +3 39 104
Zr +4 40 86
Nb +3 41 86
Nb +4 41 82
Nb +5 41 78
Mo +3 42 83
Mo +4 42 79
Mo +5 42 75
Mo +6 42 73
Tc +4 43 78.5
Tc +5 43 74
Tc +7 43 70
Ru +3 44 82
Ru +4 44 76
Ru +5 44 70.5
Ru +7 44
Ru +8 44
Rh +3 45 80.5
Rh +4 45 74
Rh +5 45 69
Pd +1 46
Pd +2 46 100
Pd +3 46 90
Pd +4 46 75.5
Ag +1 47 129
Ag +2 47 108
Ag +3 47 89
Cd +2 48 109
In +3 49 94
Sn +4 50 83
Sb +3 51 90
Sb +5 51 74
Te -2 52 207
Te +4 52 111
Te +6 52 70
I -1 53 206
I +5 53
I +7 53 67
Xe +8 54 62
Cs +1 55 181
Ba +2 56 149
La +3 57 117.2
Ce +3 58 115
Ce +4 58 101
Pr +3 59 113
Pr +4 59 99
Nd +2 60
Nd +3 60 112.3
Pm +3 61 111
Sm +2 62
Sm +3 62 109.8
Eu +2 63 131
Eu +3 63 108.7
Gd +3 64 107.8
Tb +3 65 106.3
Tb +4 65 90
Dy +2 66 121
Dy +3 66 105.2
Ho +3 67 104.1
Er +3 68 103
Tm +2 69 117
Tm +3 69 102
Yb +2 70 116
Yb +3 70 100.8
Lu +3 71 100.1
Hf +4 72 85
Ta +3 73 86
Ta +4 73 82
Ta +5 73 78
W +4 74 80
W +5 74 76
W +6 74 74
Re +4 75 77
Re +5 75 72
Re +6 75 69
Re +7 75 67
Os +4 76 77
Os +5 76 71.5
Os +6 76 68.5
Os +7 76 66.5
Os +8 76
Ir +3 77 82
Ir +4 77 76.5
Ir +5 77 71
Pt +2 78 94
Pt +4 78 76.5
Pt +5 78 71
Au +1 79 151
Au +3 79 99
Au +5 79 71
Hg +1 80 133
Hg +2 80 116
Tl +1 81 164
Tl +3 81 102.5
Pb +2 82 133
Pb +4 82 91.5
Bi +3 83 117
Bi +5 83 90
Po +4 84 108
Po +6 84 81
At +7 85 76
Rn 86
Fr +1 87 194
Ra +2 88
Ac+3 89 126
Th +4 90 108
Pa +3 91 118
Pa +4 91 104
Pa +5 91 92
U +3 92 116.5
U +4 92 103
U +5 92 90
U +6 92 87
Np +2 93 124
Np +3 93 115
Np +4 93 101
Np +5 93 89
Np +6 93 86
Np +7 93 85
Pu +3 94 114
Pu +4 94 100
Pu +5 94 88
Pu +6 94 85
Am +2 95
Am +3 95 111.5
Am +4 95 99
Cm +3 96 111
Cm +4 96 99
Bk +3 97 110
Bk +4 97 97
Cf +3 98 109
Cf +4 98 96.1
Es 99
Fm 100
Md 101
No +2 102 124
Lr 103
Shannon, R. D. Acta Cryst. 1976, A32, 751-767. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.04%3A_Radius.txt |
John Moore, Jia Zhou, and Etienne Garand
Ionization Energy
Atom Atomic No. IE1 (kJ/mol) IE2 (kJ/mol) IE3 (kJ/mol) IE4 (kJ/mol) IE5 (kJ/mol) IE6 (kJ/mol) IE7 (kJ/mol) IE8 (kJ/mol) IE9 (kJ/mol) IE10 (kJ/mol) IE11 (kJ/mol) IE12 (kJ/mol) IE13 (kJ/mol) IE14 (kJ/mol) IE15 (kJ/mol) IE16 (kJ/mol) IE17 (kJ/mol) IE18 (kJ/mol) IE19 (kJ/mol)
Atom
Atomic No.
IE1 (kJ/mol)
IE2 (kJ/mol)
IE3 (kJ/mol)
IE4 (kJ/mol)
IE5 (kJ/mol)
IE6 (kJ/mol)
IE7 (kJ/mol)
IE8 (kJ/mol)
IE9 (kJ/mol)
IE10 (kJ/mol)
IE11 (kJ/mol)
IE12 (kJ/mol)
IE13 (kJ/mol)
IE14 (kJ/mol)
IE15 (kJ/mol)
IE16 (kJ/mol)
IE17 (kJ/mol)
IE18 (kJ/mol)
IE19 (kJ/mol)
C. E. Moore, Ionization Potentials and Limits Derived from the Analyses of Optical Spectra, NSRDS-NBS 34, National Bureau of Standards, Washington, DC, 1970.
For elements 57-72, values were obtained from Martin, W. C.; Hagan, Lucy; Reader, Joseph; Sugar, Jack J. Phys. Chem. Ref. Data, 1974, 3, 771-779.
For elements 59-73, fifth IEs came from Sugar, Jack J. Opt. Soc. Am. 1975, 65, 1366-1367.
For Lr see C&EN, Volume 93 Issue 15 | p. 8 | News of The Week, Issue Date: April 13, 2015 | Web Date: April 9, 2015 Lawrencium Ionization Energy Measured.
*Values for La, Hf, and Lr have been overwritten with more recent data
H 1 1311.7
He 2 2371.6 5248.8
Li 3 520.1 7296.1 11811.7
Be 4 899.2 1756.6 14844.3 21000.7
B 5 800.4 2426.5 3658.7 25018.5 32817.3
C 6 1086.8 2351.9 4618.2 6221.0 37820.9 47262.4
N 7 1402.9 2856.4 4575.9 7472.8 9442.3 53252.7 64339.3
O 8 1313.6 3391.1 5300.3 7467.4 10987.1 13322.7 71313.7 84050.3
F 9 1681.0 3375.1 6050.4 8416.4 11020.0 15159.5 17863.2 92008.4
Ne 10 2080.1 3962.7 6175.1 9374.5 12195.7 15236.0
Na 11 495.7 4562.8 6913.2 9540.5 13372.9 16630.2 20111.8 25487.1 28924.4
Mg 12 737.5 1450.2 7730.4 10544.9 13626.6 18029.3 21739.1 25661.0 31637.5 35444.9 169932.9
Al 13 577.4 1816.1 2744.0 11574.4 14836.6 18372.7 23342.7 27510.9 31849.8 38449.4 42636.9 201216.3
Si 14 786.3 1576.6 3228.4 4354.4 16087.0 19790.1 23775.0 29319.0 33946.4 38719.6 45927.0 50481.1
P 15 1011.8 1895.9 2909.6 4954.9 6272.2 21266.7 25405.6 29839.1 35952.4 41050.7 46255.1 54060.7 58996.0
S 16 999.3 2257.8 3377.0 4562.8 6995.2 8493.5 27111.4 31724.4 36563.1 43244.7
Cl 17 1255.3 2296.4 3849.8 5162.0 6541.7 9330.1 11025.4 33605.8 38661.7 43929.8
Ar 18 1520.1 2664.9 3946.3 5768.9 7236.4 8809.1 11964.2 13841.8 40620.3
K 19 418.6 3069.2 4438.3 5876.0 7975 9619.6 11385.3 14955.2 16975.6 48609.3
Ca 20 589.6 1145.3 4941.0 6464.5 8142.4 10496 12350.1 14183.3 18139.2 20386.4 57100.0 63197.9
Sc 21 632.9 1243.7 2388.0 7130.3 8876.7 10719.5 13314 15341.2 17367.4 21757.4 24098.2 66246.8
Ti 22 659.0 1315.1 2715.1 4172.0 9629.2 11578.2 13585.1 18621.7 25665.1 28122.6 76069.0
V 23 650.3 1370.1 2865.6 4631.3 6290.8 12437.0 14569.3 16759.5 29814.0 32447.1 86557.0
Cr 24 652.5 1591.0 2986.2 4785.7 7043.4 8741.6 15543.8 17849.8 20223.3 34252.3 37069.7 97739.6
Mn 25 717.1 1509.0 3250.6 7332.9 11504.9 18911.1 21419.7 23928.4 38980.1 41971.1 109597.7
Fe 26 762.2 1561.1 2956.3 14569.3 22674.1 25279.2 27980.7 34252.3 37629.3 44093.8 47181.3 122150.4
Co 27 758.4 1645.1 3231.3 29428.0 49400.5 52777.5 135368.9
Ni 28 736.5 1751.2 3488.9 33769.9 43900.8 58566.6
Cu 29 745.3 1957.7 3553.6 64741.7
Zn 30 906.1 1732.9 3830.5
Ga 31 578.9 1981.8 2962.1 6194.4 8299 10874 13585
Ge 32 760.3 1537.0 3300.8 4409.4 9011.7 11183 13981
As 33 946.5 1949.0 2730.5 4833.9 6040.0 12301.9 14183
Se 34 940.7 2074.4 3087.5 4139.2 7053.1 7882.9 14955.2
Br 35 1142.4 2084.1 3463.8 18621.7
Kr 36 1350.4 2369.7 3560.3
Rb 37 402.9 2653.3 3859.4 26726.4
Sr 38 549.2 1063.9 0.0 5499.7 31261.3
Y 39 627.2 1196.4 3762.9 7429.4 36085.5
Zr 40 670.6 1353.7 2392.8 3277.6 9552.0
Nb 41 653.2 1350.8 2711.2 3695.4 4824.3 9938.0 12060.7
Mo 42 685.0 1558.2 2617.6 4476.9 5904.9 6561.0 12157.2 14762.3
Tc 43 702.4 1472.4 0.0
Ru 44 710.5 1617.1 2746.0
Rh 45 719.8 1743.5 2995.9
Pd 46 803.7 1873.7 3176.3
Ag 47 730.8 2072.5 3359.6
Cd 48 867.5 1631.0 3615.3
In 49 558.2 1819.7 2704.5 5248.8
Sn 50 708.4 1411.4 2941.8 3928.9 6975.9
Sb 51 833.5 1592.0 2441.1 3193.7 5403.2 10420.4
Te 52 869.3 1794.6 2991.0 3666.4 5789.1 6946.9 13218.5
I 53 1008.7 1841.9 0.0 16402.5
Xe 54 1170.1 2045.5 3097.2
Cs 55 375.6 2421.8 0.0
Ba 56 502.7 964.9 0.0
La 57 538.1 1067.1 1850.3 4819.4
Ce 58 527.4 1046.9 1948.8 3546.6
Pr 59 523.1 1017.9 2086.4 3761.0 5550.8
Nd 60 529.6 1035.3 2132.3 3899.0 5789.1
Pm 61 535.9 1051.7 2151.6 3965.5 5952.2
Sm 62 543.3 1068.1 2257.8 3994.5 6045.8
Eu 63 546.7 1084.6 2404.4 4110.3 6100.8
Gd 64 592.5 1166.5 1990.5 4245.4 6248.4
Tb 65 564.6 1111.5 2114.0 3839.2 6412.4
Dy 66 571.9 1126.0 2199.9 4001.2 5989.8
Ho 67 580.6 1138.5 2203.7 4100.6 6168.3
Er 68 588.7 1151.1 2194.1 4115.1 6281.2
Tm 69 596.7 1162.6 2284.8 4119.0 6312.1
Yb 70 603.4 1175.6 2415.0 4220.3 6327.5
Lu 71 523.5 1341.1 2022.3 4360.2 6444.3
Hf 72 641.6 1437.6 2248.1 3215.9 6595.4
Ta 73 760.3 4657.6
W 74 770.0
Re 75 759.3
Os 76 839.4
Ir 77 868.4
Pt 78 868.4 1790.8
Au 79 889.6 1977.9
Hg 80 1006.3 1809.2 3299.8
Tl 81 589.1 1970.2 2875.3 4891.8
Pb 82 715.4 1450.0 3080.8 4082.3 6638.2
Bi 83 703.1 1609.4 2466.2 4370.8 5403.2 8519.7
Po 84 813.4
Rn 86 1036.8
Fr 87 393.0
Ra 88 509.2 978.7
Ac 89 498.8 1167.5
Th 90 586.6
Pa 91 568.3
U 92 583.7
Np 93 597.2
Pu 94 584.7
Am 95 578.2
Cm 96 580.8
Bk 97 601.1
Cf 98 607.9
Es 99 619.4
Fm 100 627.2
Md 101 634.9
No 102 641.6
Lr 103 478.6
C. E. Moore, Ionization Potentials and Limits Derived from the Analyses of Optical Spectra, NSRDS-NBS 34, National Bureau of Standards, Washington, DC, 1970.
For elements 57-72, values were obtained from Martin, W. C.; Hagan, Lucy; Reader, Joseph; Sugar, Jack J. Phys. Chem. Ref. Data, 1974, 3, 771-779.
For elements 59-73, fifth IEs came from Sugar, Jack J. Opt. Soc. Am. 1975, 65, 1366-1367.
For Lr see C&EN, Volume 93 Issue 15 | p. 8 | News of The Week, Issue Date: April 13, 2015 | Web Date: April 9, 2015 Lawrencium Ionization Energy Measured.
*Values for La, Hf, and Lr have been overwritten with more recent data | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.05%3A_Ionization_Energies.txt |
John Moore, Jia Zhou, and Etienne Garand
Electron Affinity
Atom Atomic No. 1st Electron Affinity (kJ/mol)
H 1 -72.77
He 2 48.24
Li 3 -59.63
Be 4 48.24
B 5 -26.73
C 6 -121.85
N 7 6.75
O 8 -140.98
F 9 -327.95
Ne 10 115.78
Na 11 -52.87
Mg 12 38.59
Al 13 -42.55
Si 14 -133.63
P 15 -72.03
S 16 -200.41
Cl 17 -348.99
Ar 18 96.49
K 19 -48.38
Ca 20 -2.37
Sc 21 -18.14
Ti 22 -7.62
V 23 -50.65
Cr 24 -64.26
Mn 25 115.78
Fe 26 -15.73
Co 27 -63.78
Ni 28 -111.54
Cu 29 -118.48
Zn 30 57.89
Ga 31 -28.95
Ge 32 -115.78
As 33 -78.15
Se 34 -194.97
Br 35 -324.67
Kr 36 96.49
Rb 37 -46.88
Sr 38 48.24
Y 39 -29.62
Zr 40 -41.10
Nb 41 -86.16
Mo 42 -71.98
Tc 43 -53.07
Ru 44 -101.31
Rh 45 -109.70
Pd 46 -53.74
Ag 47 -125.62
Cd 48 67.54
In 49 -28.95
Sn 50 -115.78
Sb 51 -103.24
Te 52 -190.15
I 53 -295.16
Xe 54 77.19
Cs 55 -45.51
Ba 56 28.95
La 57 -48.24
Ce 58
Pr 59
Nd 60
Pm 61
Sm 62
Eu 63
Gd 64
Tb 65
Dy 66
Ho 67
Er 68
Tm 69
Yb 70
Lu 71 0.00
Hf 72 -9.65
Ta 73 -31.07
W 74 -78.64
Re 75 -14.47
Os 76 -106.13
Ir 77 -151.00
Pt 78 -205.32
Au 79 -222.75
Hg 80 48.24
Tl 81 -19.30
Pb 82 -35.12
Bi 83 -91.28
Po 84 -183.32
At 85 -270.16
Rn 86 67.54
Fr 87
Ra 88
Ac 89
Th 90
Pa 91
U 92
Np 93
Pu 94
Am 95
Cm 96
Bk 97
Cf 98
Es 99
Fm 100
Md 101
No 102
Lr 103
Anderson, T.; Haugen, H. K.; and Hotop, W.. J. Phys. Chem. Ref. Data 1999, 28(6), 1526-1527.
Bratsch, Steven G.; Lagowski, J. J. Polyhedron 1986, 5, 1763-1770. Note that in this table the convention is that EA = ΔE when an electron is added to an atom to form a negative ion.
6.07: Common Polyatomic Ions
John Moore, Jia Zhou, and Etienne Garand
Polyatomic Ions (common)
Cations
NH4+ Ammonium
Hg22+ Mercury(I)
Anions (1-)
OH Hydroxide
HSO4 Hydrogen sulfate
CH3COO Acetate
ClO Hypochlorite
ClO2 Chlorite
ClO3 Chlorate
ClO4 Perchlorate
NO2 Nitrite
NO3 Nitrate
MnO4 Permanganate
H2PO4 Dihydrogen phosphate
CN Cyanide
HCO3 Hydrogen carbonate
SCN Thiocyanate
Anions (2-)
O22 Peroxide
SO32 Sulfite
SO42 Sulfate
S2O32 Thiosulfate
Cr2O72 Dichromate
CrO42 Chromate
CO32 Carbonate
HPO42 Monohydrogen phosphate
C2O42 Oxalate
Anions (3-)
PO43 Phosphate
AsO43 Arsenate
6.08: Bond Enthalpy and Length
John Moore, Jia Zhou, and Etienne Garand
Average Bond Enthalpy (Scroll down for average bond length.)
In kJ/mol
I Br Cl S P Si F O N C H
H 299 366 431 347 322 323 566 467 391 416 436
C 213 285 327 272 264 301 485 358 285 346
N - - 193 - ~200 335 272 201 160
O 201 - 205 - ~340 368 190 146
F - 238 255 326 490 582 158
Si 234 310 391 226 - 226
P 184 264 319 - 209
S - 213 255 226 Multiple Bonds Multiple Bonds
Cl 209 217 242 N=N 418 C=C 598
Br 180 193 N≡N 946 C≡C 813
I 151 C=N 616 C=O in CO2 803
C≡N 866 C=O carbonyl 695
N=O 607 C≡O 1073
O=O in O2 498 N≡O 632
Data from Cotton, F.A., Wilkinson, G. and Gaus, P.L., Basic Inorganic Chemistry, 3rd ed., New York: Wiley, 1995.
Corrected values for C-C, and C-O from Cottrell, T.L., The Strengths of Chemical Bonds, 2ed., London:Butterworths, 1958.
Average Bond Length
In picometers, (pm)
I Br Cl S P Si F O N C H
H 161 142 127 132 138 145 92 94 98 110 74
C 210 191 176 181 187 194 141 143 147 154
N 203 184 169 174 180 187 134 136 140
O 199 180 165 170 176 183 130 148
F 197 178 163 168 174 181 128
Si 250 231 216 221 227 234
P 243 224 209 214 220
S 237 218 203 208 Multiple Bonds Multiple Bonds
Cl 232 213 200 N=N 120 C=C 134
Br 247 228 N≡N 110 C≡C 121
I 266 C=N 127 C=O in CO2 116
C≡N 115 C=O carbonyl 121
N=O 115 C≡O 113
O=O in O2 121 N≡O 115
6.09: Electronegativity
John Moore, Jia Zhou, and Etienne Garand
Electronegativity
Atom Atomic No. Electronegativity
H 1 2.2
He 2
Li 3 1.0
Be 4 1.6
B 5 2.1
C 6 2.6
N 7 3.1
O 8 3.5
F 9 4.0
Ne 10
Na 11 1.0
Mg 12 1.3
Al 13 1.6
Si 14 1.9
P 15 2.2
S 16 2.6
Cl 17 3.2
Ar 18
K 19 0.8
Ca 20 1.0
Sc 21 1.4
Ti 22 1.6
V 23 1.7
Cr 24 1.7
Mn 25 1.6
Fe 26 1.9
Co 27 1.9
Ni 28 1.9
Cu 29 1.9
Zn 30 1.7
Ga 31 1.8
Ge 32 2.0
As 33 2.2
Se 34 2.6
Br 35 3.0
Kr 36 3.4
Rb 37 0.8
Sr 38 1.0
Y 39 1.2
Zr 40 1.4
Nb 41
Mo 42 2.2
Tc 43
Ru 44
Rh 45 2.3
Pd 46 2.2
Ag 47 2.0
Cd 48 1.7
In 49 1.8
Sn 50 2.0
Sb 51 2.1
Te 52 2.1
I 53 2.7
Xe 54 3.0
Cs 55 0.8
Ba 56 0.9
La 57
Ce 58
Pr 59
Nd 60
Pm 61
Sm 62
Eu 63
Gd 64
Tb 65
Dy 66
Ho 67
Er 68
Tm 69
Yb 70
Lu 71
Hf 72
Ta 73
W 74 2.4
Re 75
Os 76
Ir 77 2.2
Pt 78 2.3
Au 79 2.6
Hg 80 2.0
Tl 81 2.1
Pb 82 2.4
Bi 83 2.0
Po 84 2.0
At 85 2.2
Rn 86
Allred, A. L. J. Inorg. Nucl. Chem. 1961, 17, 215. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.06%3A_Electron_Affinity.txt |
John Moore, Jia Zhou, and Etienne Garand
Amino Acids
Name Abbreviation (3 letter) Abbreviation (1 letter) Structure
Alanine Ala A
Arginine Arg R
Asparagine Asn N
Aspartic Acid Asp D
Cysteine Cys C
Glutamic Acid Glu E
Glutamine Gln Q
Glycine Gly G
Histidine His H
Isoleucine Ile I
Leucine Leu L
Lysine Lys K
Methionine Met M
Phenylalanine Phe F
Proline Pro P
Serine Ser S
Threonine Thr T
Tryptophan Trp W
Tyrosine Tyr Y
Valine Val V
6.11: Solubility Product Constant
John Moore, Jia Zhou, and Etienne Garand
Solubility Product Constants for Select Compounds
Ksp determined at 25 °C.
Substance Kₛₚ
Aluminum Compounds
AlAsO₄ 1.6 × 10⁻¹⁶
Al(OH)₃ amorphous 1.3 × 10⁻³³
AlPO₄ 6.3 × 10⁻¹⁹
Barium Compounds
Ba₃(AsO₄)₂ 8.0 × 10⁻¹⁵
BaCO₃ 5.1 × 10⁻⁹
BaC₂O₄ 1.6 × 10⁻⁷
BaCrO₄ 1.2 × 10⁻¹⁰
BaF₂ 1.0 × 10⁻⁶
Ba(OH)₂ 5 × 10⁻³
Ba₃(PO₄)₂ 3.4 × 10⁻²³
BaSeO₄ 3.5 × 10⁻⁸
BaSO₄ 1.1 × 10⁻¹⁰
BaSO₃ 8 × 10⁻⁷
BaS₂O₃ 1.6 × 10⁻⁵
Bismuth Compounds
BiAsO₄ 4.4 × 10⁻¹⁰
BiOCl 1.8 × 10⁻³¹
BiO(OH) 4 × 10⁻¹⁰
Bi(OH)₃ 4 × 10⁻³¹
BiI₃ 8.1 × 10⁻¹⁹
BiPO₄ 1.3 × 10⁻²³
Cadmium Compounds
Cd₃(AsO₄)₂ 2.2 × 10⁻³³
CdCO₃ 5.2 × 10⁻¹²
Cd(CN)₂ 1.0 × 10⁻⁸
Cd₂[Fe(CN)₆] 3.2 × 10⁻¹⁷
Cd(OH)₂ fresh 2.5 × 10⁻¹⁴
Calcium Compounds
Ca₃(AsO₄)₂ 6.8 × 10⁻¹⁹
CaCO₃ 2.8 × 10⁻⁹
CaCrO₄ 7.1 × 10⁻⁴
CaC₂O₄·H₂O§ 4 × 10⁻⁹
CaF₂ 5.3 × 10⁻⁹
Ca(OH)₂ 5.5 × 10⁻⁶
CaHPO₄ 1 × 10⁻⁷
Ca₃(PO₄)₂ 2.0 × 10⁻²⁹
CaSeO₄ 8.1 × 10⁻⁴
CaSO₄ 9.1 × 10⁻⁶
CaSO₃ 6.8 × 10⁻⁸
Chromium Compounds
CrAsO₄ 7.7 × 10⁻²¹
Cr(OH)₂ 2 × 10⁻¹⁶
Cr(OH)₃ 6.3 × 10⁻³¹
CrPO₄·4H₂O green§ 2.4 × 10⁻²³
CrPO₄·4H₂O violet§ 1.0 × 10⁻¹⁷
Cobalt Compounds
Co₃(AsO₄)₂ 7.6 × 10⁻²⁹
CoCO₃ 1.4 × 10⁻¹³
Co(OH)₂ fresh 1.6 × 10⁻¹⁵
Co(OH)₃ 1.6 × 10⁻⁴⁴
CoHPO₄ 2 × 10⁻⁷
Co₃(PO₄)₂ 2 × 10⁻³⁵
Copper Compounds
CuBr 5.3 × 10⁻⁹
CuCl 1.2 × 10⁻⁶
CuCN 3.2 × 10⁻²⁰
CuI 1.1 × 10⁻¹²
CuOH 1 × 10⁻¹⁴
CuSCN 4.8 × 10⁻¹⁵
Cu₃(AsO₄)₂ 7.6 × 10⁻³⁶
CuCO₃ 1.4 × 10⁻¹⁰
Cu₂[Fe(CN)₆] 1.3 × 10⁻¹⁶
Cu(OH)₂ 2.2 × 10⁻²⁰
Cu₃(PO₄)₂ 1.3 × 10⁻³⁷
Gold Compounds
AuCl 2.0 × 10⁻¹³
AuI 1.6 × 10⁻²³
AuCl₃ 3.2 × 10⁻²⁵
Au(OH)₃ 5.5 × 10⁻⁴⁶
AuI₃ 1 × 10⁻⁴⁶
Iron Compounds
FeCO₃ 3.2 × 10⁻¹¹
Fe(OH)₂ 8.0 × 10⁻¹⁶
FeC₂O₄·2H₂O§ 3.2 × 10⁻⁷
FeAsO₄ 5.7 × 10⁻²¹
Fe₄[Fe(CN)₆]₃ 3.3 × 10⁻⁴¹
Fe(OH)₃ 4 × 10⁻³⁸
FePO₄ 1.3 × 10⁻²²
Lead Compounds
Pb₃(AsO₄)₂ 4.0 × 10⁻³⁶
PbBr₂ 4.0 × 10⁻⁵
PbCO₃ 7.4 × 10⁻¹⁴
PbCl₂ 1.6 × 10⁻⁵
PbCrO₄ 2.8 × 10⁻¹³
PbF₂ 2.7 × 10⁻⁸
Pb(OH)₂ 1.2 × 10⁻¹⁵
PbI₂ 7.1 × 10⁻⁹
PbC₂O₄ 4.8 × 10⁻¹⁰
PbHPO₄ 1.3 × 10⁻¹⁰
Pb₃(PO₄)₂ 8.0 × 10⁻⁴³
PbSeO₄ 1.4 × 10⁻⁷
PbSO₄ 1.6 × 10⁻⁸
Pb(SCN)₂ 2.0 × 10⁻⁵
Magnesium Compounds
Mg₃(AsO₄)₂ 2.1 × 10⁻²⁰
MgCO₃ 3.5 × 10⁻⁸
MgCO₃·3H₂O§ 2.1 × 10⁻⁵
MgC₂O₄·2H₂O§ 1 × 10⁻⁸
From Patnaik, P., Dean's Analytical Chemistry Handbook, 2nd ed. New York: McGraw-Hill, 2004, Table 4.2.
No metal sulfides are listed in this table because sulfide ion is such a strong base that the usual solubility product equilibrium equation does not apply. See Myers, R. J. Journal of Chemical Education, Vol. 63, 1986, pp. 687-690.
From Meites, L. Ed. Handbook of Analytical Chemistry, 1st ed. New York: McGraw-Hill, 1963.
§Because [H2O] does not appear in equilibrium constants for equilibria in aqueous solution in general, it does not appear in the Ksp expressions for hydrated solids. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.10%3A_Amino_Acids.txt |
John Moore, Jia Zhou, and Etienne Garand
Ionization Constants for Select Acids (a table for bases is below)
Ka determined at 25 °C.
Acid Formula and Ionization Equation Ka pKa
Acetic CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO 1.8 × 10⁻⁵ 4.74
Arsenic H₃AsO₄ + H₂O ⇌ H₃O⁺ + H₂AsO₄ K₁ = 6.17 × 10⁻³ 2.210
H₂AsO₄ + H₂O ⇌ H₃O⁺ + HAsO₄²⁻ K₂ = 1.17 × 10⁻⁷ 6.932
HAsO₄²⁻ + H₂O ⇌ H₃O⁺ + AsO₄³ K₃ = 3.09 × 10⁻¹² 11.523
Benzoic C₆H₅COOH + H₂O ⇌ H₃O⁺ + C₆H₅COO 1.2 × 10⁻⁴ 3.92
Boric B(OH)₃(H₂O) + H₂O ⇌ H₃O⁺ + B(OH)₄ 5.8 × 10⁻¹⁰ 9.24
Butyric CH₃CH₂CH₂COOH + H₂O ⇌ H₃O⁺ + CH₃CH₂CH₂COO- 1.5 × 10⁻⁵ 4.82
Carbonic H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃ K₁ = 4.3 × 10⁻⁷ 6.37
HCO₃+ H₂O ⇌ H₃O⁺ + CO₃²⁻ K₂ = 4.7 × 10⁻¹¹ 10.33
Citric H₃C₆H₅O₇ + H₂O ⇌ H₃O⁺ + H₂C₆H₅O₇ K₁ = 1.4 × 10⁻³ 2.85
H₂C₆H₅O₇ + H₂O ⇌ H₃O⁺ + HC₆H₅O₇²⁻ K₂ = 4.5 × 10⁻⁵ 4.35
HC₆H₅O₇²⁻ + H₂O ⇌ H₃O⁺ + C₆H₅O₇³ K₃ = 1.5 × 10⁻⁶ 5.82
Chloroacetic CH₂ClCOOH + H₂O ⇌ H₃O⁺ + CH₂ClCOO- 1.4 × 10-3 2.85
4-chlorobutyric CH₂ClCH₂CH₂COOH + H₂O ⇌ H₃O⁺ + CH₂ClCH₂CH₂COO- 3.0 × 10-5 4.53
3-chlorobutyric acid CH₃CHClCH₂COOH + H₂O ⇌ H₃O⁺ + CH₃CHClCH₂COO- 8.9 × 10-5 4.05
2-chlorobutyric acid CH₃CH₂CHClCOOH + H₂O ⇌ H₃O⁺ + CH₃CH₂CHClCOO- 1.3 × 10-3 2.89
Chlorous HClO₂ + H₂O ⇌ H₃O⁺ + ClO₂ 1.1 × 10⁻² 1.96
Dichloroacetic CHCl2COOH + H₂O ⇌ H₃O⁺ + CHCl2COO- 4.5 × 10-2 1.35
Formic HCOOH + H₂O ⇌ H₃O⁺ + HCOO 1.8 × 10⁻⁴ 3.75
Hydrazoic HN₃ + H₂O ⇌ H₃O⁺ + N₃ 1.0 × 10⁻⁵ 5.00
Hydrochloric HCl + H₂O ⇌ H₃O⁺ + Cl- 1.2 × 106 -6.1
Hydrocyanic HCN + H₂O ⇌ H₃O⁺ + CN 3.3 × 10⁻¹⁰ 9.48
Hydrofluoric HF + H₂O ⇌ H₃O⁺ + F 6.8 × 10⁻⁴ 3.17
Hydrogen peroxide H₂O₂ + H₂O ⇌ H₃O⁺ + HO₂ 2.1 × 10⁻¹² 11.68
Hydrosulfuric H₂S + H₂O ⇌ H₃O⁺ + HS K₁ = 1 × 10⁻⁷ 7.0
HS + H₂O ⇌ H₃O⁺ + S²⁻ K₂ = 1 × 10⁻¹⁹ 19.0
Hypobromous HOBr + H₂O ⇌ H₃O⁺ + OBr- 3 × 10-9 8.5
Hypochlorous HOCl + H₂O ⇌ H₃O⁺ + OCl 6.8 × 10⁻⁸ 7.17
Hypoiodous HOI + H₂O ⇌ H₃O⁺ + OI- 3 × 10-11 10.5
Nitric HNO₃ + H₂O ⇌ H₃O⁺ + NO₃- 27 -1.43
Nitrous HNO₂ + H₂O ⇌ H₃O⁺ + NO₂ 7.41 × 10⁻⁴ 3.130
Oxalic H₂C₂O₄ + H₂O ⇌ H₃O⁺ + HC₂O₄ K₁ = 5.5 × 10⁻² 1.26
HC₂O₄ + H₂O ⇌ H₃O⁺ + C₂O₄²⁻ K₂ = 1.4 × 10⁻⁴ 3.85
Phenol HC₆H₅O + H₂O ⇌ H₃O⁺ + C₆H₅O 1.7 × 10⁻¹⁰ 9.77
Phosphoric H₃PO₄ + H₂O ⇌ H₃O⁺ + H₂PO₄ K₁ = 7.2 × 10⁻³ 2.14
H₂PO₄ + H₂O ⇌ H₃O⁺ + HPO₄²⁻ K₂ = 6.3 × 10⁻⁸ 7.20
HPO₄²⁻ + H₂O ⇌ H₃O⁺ + PO₄³⁻ K₃ = 4.6 × 10⁻¹³ 12.34
Phosphorous H₃PO₃ + H₂O ⇌ H₃O⁺ + H₂PO₃ K₁ = 2.4 × 10⁻² 1.62
H₂PO₃ + H₂O ⇌ H₃O⁺ + HPO₃²⁻ K₂ = 2.9 × 10⁻⁷ 6.54
Propanoic CH₃CH₂COOH + H₂O ⇌ H₃O⁺ + CH₃CH₂COO 1.33 × 10⁻⁵ 4.85
Selenic H₂SeO₄ + H₂O ⇌ H₃O⁺ + HSeO₄ K₁ = very large ---
HSeO₄ + H₂O ⇌ H₃O⁺ + SeO₄²⁻ K₂ = 2.2 × 10⁻² 1.66
Selenous H₂SeO₃ + H₂O ⇌ H₃O⁺ + HSeO₃ K₁ = 2.5 × 10⁻³ 2.60
HSeO₃ + H₂O ⇌ H₃O⁺ + SeO₃²⁻ K₂ = 1.6 × 10⁻⁹ 8.80
Sulfuric H₂SO₄ + H₂O ⇌ H₃O⁺ + HSO₄ K₁ = 4.0 × 103 -3.6
HSO₄ + H₂O ⇌ H₃O⁺ + SO₄²⁻ K₂ = 1.1 × 10⁻² 1.96
Sulfurous H₂SO₃ + H₂O ⇌ H₃O⁺ + HSO₃ K₁ = 1.7 × 10⁻² 1.77
HSO₃ + H₂O ⇌ H₃O⁺ + SO₃²⁻ K₂ = 6.3 × 10⁻⁸ 7.2
Tellurous H₂TeO₃ + H₂O ⇌ H₃O⁺ + HTeO₃ K₁ = 7.1 × 10⁻⁷ 6.15
HTeO₃ + H₂O ⇌ H₃O⁺ + TeO₃²⁻ K₂ = 4.0 × 10⁻⁹ 8.40
Trichloroacetic CCl3COOH + H₂O ⇌ H₃O⁺ + CCl3COO- 0.17 0.77
Trifluoroacetic CF3COOH + H₂O ⇌ H₃O⁺ + CF3COO- 0.30 0.52
Taken from Hogfeldt, E. and Perrin, D. D., Stability Constants of Metal-Ion complexes, 1st ed. Oxford: New York: Pergamon, 1979-1982. International Union of Pure and Applied Chemistry Commission on Equilibrium Data.
Also from Serjeant, E. P. and Dempsey, B. (eds.), “Ionization Constants of Organic Acids in Aqueous Solution,” IUPAC Chemical Data Series No. 23, Pergamon Press, Oxford, 1979.
From Myers, R., Journal of Chemical Education, Vol. 63, 1986, pp. 687-690.
Ionization Constants for Select Bases
Kb determined at 25 °C.
Base Formula and Ionization Equation Kb pKb
Acetylide C₂H- + H₂O ⇌ OH- + C₂H₂ 1 × 1011 -11
Amide NH₂- + H₂O ⇌ OH- + NH₃ 1 × 1020 -20
Ammonia NH₃ + H₂O ⇌ NH₄⁺ + OH- 1.77 × 10⁻⁵ 4.752
Aniline§ C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH- 3.9 × 10⁻¹⁰ 9.41
Dimethylamine§ (CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH- 5.8 × 10⁻⁴ 3.24
Ethoxide CH₃CH₂O- + H₂O ⇌ OH- + CH₃CH₂OH 1 × 102 -2
Ethylenediamine (CH₂)₂(NH₂)₂ + H₂O ⇌ (CH₂)₂(NH₂)₂H⁺ + OH- K₁ = 7.8 × 10⁻⁵ 4.11
(CH₂)₂(NH₂)₂H + H₂O ⇌ (CH₂)₂(NH₂)₂H₂²⁺ + OH- K₂ = 2.1 × 10⁻⁸ 7.68
Hydrazine N₂H₄ + H₂O ⇌ N₂H₅ + OH- K₁ = 1.2 × 10⁻⁶ 5.92
N₂H₅ + H₂O ⇌ N₂H₆²⁺ + OH- K₂ = 1.3 × 10-15 14.89
Hydride H- + H₂O ⇌ OH- + H₂ 1 × 1021 -21
Hydroxylamine NH₂OH + H₂O ⇌ NH₃OH + OH- 9.3 × 10⁻⁹ 8.03
Methylamine CH₃NH₂ + H₂O ⇌ CH₃NH₃ + OH- 5.0 × 10⁻⁴ 3.30
Pyridine C₅H₅N + H₂O ⇌ C₅H₅NH + OH- 1.6 × 10⁻⁹ 8.80
Trimethylamine§ (CH₃)₃N + H₂O ⇌ (CH₃)₃NH + OH- 6.2 × 10⁻⁵ 4.21
Taken from Hogfeldt, E. and Perrin, D. D., Stability Constants of Metal-Ion complexes, 1st ed. Oxford: New York: Pergamon, 1979-1982. International Union of Pure and Applied Chemistry Commission on Equilibrium Data.
Also from Serjeant, E. P. and Dempsey, B. (eds.), “Ionization Constants of Organic Acids in Aqueous Solution,” IUPAC Chemical Data Series No. 23, Pergamon Press, Oxford, 1979.
From Read, A. J., Journal of Solution Chemistry, Vol. 11, No. 9, 1982, pp. 649-664.
§From Meites, L., Ed. Handbook of Analytical Chemistry, 1st ed. New York: McGraw-Hill, 1963. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.12%3A_Acid-Base_Ionization_Constant.txt |
John Moore, Jia Zhou, and Etienne Garand
Thermodynamic Values for Select Compounds
at 298.15 K.
Species fH° (kJ/mol) S° (J K−1 mol−1) fG° (kJ/mol)
Aluminum
Al(s) 0 28.275 0
Al³⁺(aq) 531 321.7 485
AlCl₃(s) 704.2 110.67 628.8
Al₂O₃(s, corundum) 1675.7 50.92 1582.3
Argon
Ar(g) 0 154.843 0
Ar(aq) 12.1 59.4 16.4
Barium
BaCl₂(s) 858.6 123.68 810.4
BaO(s) 553.5 70.42 525.1
BaSO₄(s) 1473.2 132.2 1362.2
BaCO₃(s) 1216.3 112.1 85.35
Beryllium
Be(s) 0 9.5 0
Be(OH)₂(s) 902.5 51.9 815
Bromine
Br(g) 111.884 175.022 82.396
Br₂() 0 152.231 0
Br₂(g) 30.907 245.463 3.110
Br₂(aq) 2.59 130.5 3.93
Br⁻(aq) 121.55 82.4 103.96
BrCl(g) 14.64 240.10 0.98
BrF₃(g) 255.6 292.53 229.43
HBr(g) 36.40 198.695 53.45
Calcium
Ca(s) 0 41.42 0
Ca(g) 178.2 158.884 144.3
Ca²⁺(g) 1925.9
Ca²⁺(aq) 542.83 53.1 553.58
CaC₂(s) 59.8 69.96 64.9
CaCO₃(s, calcite) 1206.92 92.9 1128.79
CaCl₂(s) 795.8 104.6 748.1
CaF₂(s) 1219.6 68.87 1167.3
CaH₂(s) 186.2 42 147.2
CaO(s) 635.09 39.75 604.03
CaS(s) 482.4 56.5 477.4
Ca(OH)₂(s) 986.09 83.39 898.49
Ca(OH)₂(aq) 1002.82 74.5 868.07
CaSO₄(s) 1434.11 106.7 1321.79
Carbon
C(s, graphite) 0 5.74 0
C(s, diamond) 1.895 2.377 2.9
C(g) 716.682 158.096 671.257
CCl₄() 135.44 216.4 65.21
CCl₄(g) 102.9 309.85 60.59
CHCl₃() 134.47 201.7 73.66
CHCl₃(g) 103.14 295.71 70.34
CH₄(g, methane) 74.81 186.264 50.72
C₂H₂(g, ethyne) 226.73 200.94 209.2
C₂H₄(g, ethene) 52.26 219.56 68.15
C₂H₆(g, ethane) 84.68 229.6 32.82
C₃H₈(g, propane) 103.8 269.9 23.49
C₄H₁₀(g, butane) 126.148 310.227 16.985
C₆H₆(, benzene) 49.03 172.8 124.5
C₆H₁₄(, hexane) 198.782 296.018 4.035
C₈H₁₈(g, octane) 208.447 466.835 16.718
C₈H₁₈(, octane) 249.952 361.205 6.707
CH₃OH(, methanol) 238.66 126.8 166.27
CH₃OH(g, methanol) 200.66 239.81 161.96
CH₃OH(aq, methanol) 245.931 133.1 175.31
C₂H₅OH(, ethanol) 277.69 160.7 174.78
C₂H₅OH(g, ethanol) 235.1 282.7 168.49
C₂H₅OH(aq, ethanol) 288.3 148.5 181.64
C₆H₁₂O₆(s, glucose) 1274.4 235.9 917.2
CH₃COO⁻(aq) 486.01 86.6 369.31
CH₃COOH(aq) 485.76 178.7 396.46
CH₃COOH() 484.5 159.8 389.9
CO(g) 110.525 197.674 137.168
CO₂(g) 393.509 213.74 394.359
H₂CO₃(aq) 699.65 187.4 623.08
HCO₃⁻(aq) 691.99 91.2 586.77
CO₃²⁻(aq) 677.14 56.9 527.81
HCOO⁻(aq) 425.55 92.0 351.0
HCOOH(aq) 425.43 163 372.3
HCOOH() 424.72 128.95 361.35
CS₂(g) 117.36 237.84 67.12
CS₂() 89.70 151.34 65.27
COCl₂(g) 218.8 283.53 204.6
Cesium
Cs(s) 0 85.23 0
Cs⁺(g) 457.964
CsCl(s) 443.04 101.17 414.53
Chlorine
Cl(g) 121.679 165.198 105.68
Cl⁻(g) 233.13
Cl⁻(aq) 167.159 56.5 131.228
Cl₂(g) 0 223.066 0
Cl₂(aq) 23.4 121 6.94
HCl(g) 92.307 186.908 95.299
HCl(aq) 167.159 56.5 131.228
ClO₂(g) 102.5 256.84 120.5
Cl₂O(g) 80.3 266.21 97.9
ClO⁻(aq) 107.1 42.0 36.8
HClO(aq) 120.9 142 79.9
ClF₃(g) 163.2 281.61 123.0
Chromium
Cr(s) 0 23.77 0
Cr₂O₃(s) 1139.7 81.2 1058.1
CrCl₃(s) 556.5 123 486.1
Copper
Cu(s) 0 33.15 0
CuO(s) 157.3 42.63 129.7
CuCl₂(s) 220.1 108.07 175.7
CuSO₄(s) 771.36 109 661.8
Fluorine
F₂(g) 0 202.78 0
F(g) 78.99 158.754 61.91
F⁻(g) 255.39
F⁻(aq) 332.63 13.8 278.79
HF(g) 271.1 173.779 273.2
HF(aq, un-ionized) 320.08 88.7 296.82
HF(aq, ionized) 332.63 13.8 278.79
Hydrogen
H₂(g) 0 130.684 0
H₂(aq) 4.2 57.7 17.6
HD(g) 0.318 143.801 1.464
D₂(g) 0 144.960 0
H(g) 217.965 114.713 203.247
H⁺(g) 1536.202
H⁺(aq) 0 0 0
OH⁻(aq) 229.994 10.75 157.244
H₂O() 285.83 69.91 237.129
H₂O(g) 241.818 188.825 228.572
H₂O₂() 187.78 109.6 120.35
H₂O₂(aq) 191.17 143.9 134.03
HO₂⁻(aq) 160.33 23.8 67.3
HDO() 289.888 79.29 241.857
D₂O() 294.600 75.94 243.439
Iodine
I₂(s) 0 116.135 0
I₂(g) 62.438 260.69 19.327
I₂(aq) 22.6 137.2 16.40
I(g) 106.838 180.791 70.25
I⁻(g) 197
I⁻(aq) 55.19 111.3 51.57
I₃⁻(aq) 51.5 239.3 51.4
HI(g) 26.48 206.594 1.70
HI(aq, ionized) 55.19 111.3 51.57
IF(g) 95.65 236.17 118.51
ICl(g) 17.78 247.551 5.46
ICl() 23.89 135.1 13.58
IBr(g) 40.84 258.773 3.69
ICl₃(s) 89.5 167.4 22.29
Iron
Fe(s) 0 27.78 0
FeO(s, wustite) 266.27 57.49 245.12
Fe₂O₃(s, hematite) 824.2 87.4 742.2
Fe₃O₄(s, magnetite) 1118.4 146.4 1015.4
FeCl₂(s) 341.79 117.95 302.3
FeCl₃(s) 399.49 142.3 344
FeS₂(s, pyrite) 178.2 52.93 166.9
Fe(CO)₅() 774 338.1 705.3
Lead
Pb(s) 0 64.81 0
PbCl₂(s) 359.41 136 314.1
PbO(s, yellow) 217.32 68.7 187.89
PbS(s) 100.4 91.2 98.7
PbSO₄(s) 919.94 148.57 813.14
PbCl₄() 329.3
PbO₂(s) 277.4 68.6 217.33
Lithium
Li(s) 0 29.12 0
Li⁺(g) 685.783
LiOH(s) 484.93 42.8 438.95
LiOH(aq) 508.48 2.8 450.58
LiCl(s) 408.701 59.33 384.37
Magnesium
Mg(s) 0 32.68 0
Mg²⁺(aq) 466.85 138.1 454.8
MgCl₂(g) 400.4
MgCl₂(s) 641.32 89.62 591.79
MgCl₂(aq) 801.15 25.1 717.1
MgO(s) 601.70 26.94 569.43
Mg(OH)₂(s) 924.54 63.18 833.51
MgS(s) 346 50.33 341.8
MgSO₄(s) 1284.9 91.6 1170.6
MgCO₃(s) 1095.8 65.7 1012.1
Mercury
Hg() 0 76.02 0
HgCl₂(s) 224.3 146 178.6
HgO(s, red) 90.83 70.9 58.539
HgS(s, red) 58.2 82.4 50.6
Nickel
Ni(s) 0 29.87 0
NiO(s) 239.7 37.99 211.7
NiCl₂(s) 305.332 97.65 259.032
Nitrogen
N₂(g) 0 191.61 0
N₂(aq) 10.8
N(g) 472.704 153.298 455.53
NH₃(g) 46.11 192.45 16.45
NH₃(aq) 80.29 111.3 26.50
NH₄⁺(aq) 132.51 113.4 79.31
N₂H₄() 50.63 121.21 149.34
NH₄Cl(s) 314.43 94.6 202.87
NH₄Cl(aq) 299.66 169.9 210.52
NH₄NO₃(s) 365.56 151.08 183.87
NH₄NO₃(aq) 339.87 259.8 190.56
NO(g) 90.25 210.761 86.55
NO₂(g) 33.18 240.06 51.31
N₂O(g) 82.05 219.85 104.20
N₂O₄(g) 9.16 304.29 97.89
N₂O₄(ℓ) 19.50 209.2 97.54
NOCl(g) 51.71 261.69 66.08
HNO₃() 174.10 155.60 80.71
HNO₃(g) 135.06 266.38 74.72
HNO₃(aq) 207.36 146.4 111.25
NO₃⁻(aq) 205.0 146.4 108.74
NF₃(g) 124.7 260.73 83.2
Oxygen
O₂(g) 0 205.138 0
O₂(aq) 11.7 110.9 16.4
O(g) 249.170 161.055 231.731
O₃(g) 142.7 238.93 163.2
OH⁻(aq) 229.994 10.75 157.244
Phosphorus
P₄(s, white) 0 164.36 0
P₄(s, red) 70.4 91.2 48.4
P(g) 314.64 163.193 278.25
PH₃(g) 5.4 310.23 13.4
PCl₃(g) 287.0 311.78 267.8
PCl₃() 319.7 217.1 272.3
PCl₅(s) 443.5
P₄O₁₀(s) 2984.0 228.86 2697.7
H₃PO₄(s) 1279 110.5 1119.1
Potassium
K(s) 0 64.18 0
KF(s) 567.27 66.57 537.75
KCl(s) 436.747 82.59 409.14
KCl(aq) 419.53 159.0 414.49
KBr(s) 393.798 95.90 380.66
KI(s) 327.900 106.32 324.892
KClO₃(s) 397.73 143.1 296.25
KOH(s) 424.764 78.9 379.08
KOH(aq) 482.37 91.6 440.5
Silicon
Si(s) 0 18.83 0
SiBr₄() 457.3 277.8 443.8
SiC(s) 65.3 16.61 62.8
SiCl₄(g) 657.01 330.73 616.98
SiH₄(g) 34.3 204.62 56.9
SiF₄(g) 1614.94 282.49 1572.65
SiO₂(s, quartz) 910.94 41.84 856.64
Silver
Ag(s) 0 42.55 0
Ag⁺(aq) 105.579 72.68 77.107
Ag₂O(s) 31.05 121.3 11.2
AgCl(s) 127.068 96.2 109.789
AgI(s) 61.84 115.5 66.19
AgN₃(s) 620.60 99.22 591.0
AgNO₃(s) 124.39 140.92 33.41
AgNO₃(aq) 101.8 219.2 34.16
Sodium
Na(s) 0 51.21 0
Na(g) 107.32 153.712 76.761
Na⁺(g) 609.358
Na⁺(aq) 240.12 59.0 261.905
NaF(s) 573.647 51.46 543.494
NaF(aq) 572.75 45.2 540.68
NaCl(s) 411.153 72.13 384.138
NaCl(g) 176.65 229.81 196.66
NaCl(aq) 407.27 115.5 393.133
NaBr(s) 361.062 86.82 348.983
NaBr(aq) 361.665 141.4 365.849
NaI(s) 287.78 98.53 286.06
NaI(aq) 295.31 170.3 313.47
NaOH(s) 425.609 64.455 379.484
NaOH(aq) 470.114 48.1 419.15
NaClO₃(s) 365.774 123.4 262.259
NaHCO₃(s) 950.81 101.7 851.0
Na₂CO₃(s) 1130.68 134.98 1044.44
Na₂SO₄(s) 1387.08 149.58 1270.16
Sulfur
S(s, monoclinic) 0.33
S(s, rhombic) 0 31.80 0
S(g) 278.805 167.821 238.250
S²⁻(aq) 33.1 14.6 85.8
S₂Cl₂(g) 18.4 331.5 31.8
SF₆(g) 1209 291.82 1105.3
SF₄(g) 774.9 292.03 731.3
H₂S(g) 20.63 205.79 33.56
H₂S(aq) 39.7 121 27.83
HS⁻(aq) 17.6 62.8 12.08
SO₂(g) 296.830 248.22 300.194
SO₃(g) 395.72 256.76 371.06
SOCl₂(g) 212.5 309.77 198.3
SO₄²⁻(aq) 909.27 20.1 744.53
H₂SO₄() 813.989 156.904 690.003
H₂SO₄(aq) 909.27 20.1 744.53
HSO₄⁻(aq) 887.34 131.8 755.91
Tin
Sn(s, white) 0 51.55 0
Sn(s, gray) 2.09 44.14 0.13
SnCl₂(s) 325.1
SnCl₄() 511.3 258.6 440.1
SnCl₄(g) 471.5 365.8 432.2
SnO₂(s) 580.7 52.3 519.6
Titanium
Ti(s) 0 30.63 0
TiCl₄() 804.2 252.34 737.2
TiCl₄(g) 763.2 354.9 726.7
TiO₂(s) 939.7 49.92 884.5
Uranium
U(s) 0 50.21 0
UO₂(s) 1084.9 77.03 1031.7
UO₃(s) 1223.8 96.11 1145.9
UF₄(s) 1914.2 151.67 1823.3
UF₆(g) 2147.4 377.9 2063.7
UF₆(s) 2197.0 227.6 2068.5
Zinc
Zn(s) 0 41.63 0
ZnCl₂(s) 415.05 111.46 369.398
ZnO(s) 348.28 43.64 318.3
ZnS(s, sphalerite) 205.98 57.7 201.29
From Wagman, D. D., Evans, W. H., Parker, V. B., Schumm, R. H., Halow, I., Bailey, S. M., Churney, K. L., and Nuttall, R., Journal of Physical and Chemical Reference Data, Vol. 11, Suppl. 2, 1982.
Many hydrogen-containing and oxygen-containing compounds are listed only under other elements. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.13%3A_Thermodynamic.txt |
John Moore, Jia Zhou, and Etienne Garand
Standard Electrode Potentials in Acidic Aqueous Solution (a table for basic solution is below)
at 25 °C.
Acidic Solution Standard Electrode Potential, E° (volts)
F₂(g) + 2e⁻ ⟶ 2 F(aq) 2.87
Co³⁺(aq) + e⁻ ⟶ Co²⁺(aq) 1.92
Au⁺(aq) + e⁻ ⟶ Au(s) 1.83
H₂O₂(aq) + 2 H⁺(aq) + 2 e⁻ ⟶ 2 H₂O() 1.763
Ce⁴⁺(aq) + e⁻ ⟶ Ce³⁺(aq) 1.72
Pb⁴⁺ (aq) + 2 e⁻ ⟶ Pb²⁺(aq) 1.69
PbO₂(s) + SO₄²⁻ (aq) + 4 H⁺(aq) + 2 e⁻ ⟶ PbSO₄(s) + 2 H₂O() 1.690
NiO₂(s) + 4 H⁺(aq) + 2 e⁻ ⟶ Ni²⁺(aq) + 2 H₂O() 1.68
2 HClO(aq) + 2 H⁺(aq) + 2 e⁻ ⟶ Cl₂(g) + 2 H₂O() 1.63
Au³⁺(aq) + 3 e⁻ ⟶ Au(s) 1.52
MnO₄⁻ (aq) + 8 H⁺(aq) + 5 e⁻ ⟶ Mn²⁺(aq) + 4 H₂O() 1.51
BrO₃⁻(aq) + 6 H⁺(aq) + 5 e⁻ ⟶ 12 Br₂(aq) + 3 H₂O() 1.478
2 ClO₃⁻(aq) + 12 H⁺(aq) + 10 e⁻ ⟶ Cl₂(g) + 6 H₂O() 1.47
Cr₂O₇²⁻(aq) + 14 H⁺(aq) + 6 e⁻ ⟶ 2 Cr³⁺(aq) + 7 H₂O() 1.36
Cl₂(g) + 2 e⁻ ⟶ 2 Cl⁻(aq) 1.358
N₂H₅⁺(aq) + 3 H⁺(aq) + 2 e⁻ ⟶ 2 NH⁴⁺(aq) 1.275
MnO₂(s) + 4 H⁺(aq) + 2 e⁻ ⟶ Mn²⁺(aq) + 2 H₂O() 1.23
O₂(g) + 4 H⁺(aq) + 4 e⁻ ⟶ 2 H₂O() 1.229
ClO₄⁻(aq) + 2 H⁺(aq) + 2 e⁻ ⟶ ClO₃⁻(aq) + H₂O() 1.201
IO₃⁻(aq) + 6 H⁺(aq) + 5 e⁻ ⟶ 12 I₂(aq) + 3 H₂O() 1.195
Pt²⁺(aq) + 2 e⁻ ⟶ Pt(s) 1.188
Br₂() + 2 e⁻ ⟶ 2 Br⁻(aq) 1.066
[AuCl₄]⁻(aq) + 3 e⁻ ⟶ Au(s) + 4 Cl⁻(aq) 1.00
NO₃⁻(aq) + 4 H⁺(aq) + 3 e⁻ ⟶ NO(g) + 2 H₂O() 0.96
NO₃⁻(aq) + 3 H⁺(aq) + 2 e⁻ ⟶ HNO₂(aq) + H₂O() 0.94
Pd²⁺(aq) + 2 e⁻ ⟶ Pd(s) 0.915
2 Hg²⁺(aq) + 2 e⁻ ⟶ Hg₂²⁺(aq) 0.9110
Hg²⁺(aq) + 2 e⁻ ⟶ Hg() 0.8535
SbCl₆⁻(aq) + 2 e⁻ ⟶ SbCl₄⁻(aq) + 2 Cl⁻(aq) 0.84
Ag⁺(aq) + e⁻ ⟶ Ag(s) 0.7991
Hg₂²⁺(aq) + 2 e⁻ ⟶ 2 Hg() 0.7960
Fe³⁺(aq) + e⁻ ⟶ Fe²⁺(aq) 0.771
[PtCl₄]²⁻(aq) + 2 e⁻ ⟶ Pt(s) + 4 Cl⁻(aq) 0.758
[PtCl₆]²⁻(aq) + 2 e⁻ ⟶ [PtCl₄]²⁻(aq) + 2 Cl⁻(aq) 0.726
O₂(g) + 2 H⁺(aq) + 2 e⁻ ⟶ H₂O₂(aq) 0.695
TeO₂(s) + 4 H⁺(aq) + 4 e⁻ ⟶ Te(s) + 2 H₂O() 0.604
H₃AsO₄(aq) + 2 H⁺(aq) + 2 e⁻ ⟶ HAsO₂(aq) + 2 H₂O() 0.560
I₂(s) + 2 e⁻ ⟶ 2 I⁻(aq) 0.535
Cu⁺(aq) + e⁻ ⟶ Cu(s) 0.521
[RhCl₆]³⁻(aq) + 3 e⁻ ⟶ Rh(s) + 6 Cl⁻(aq) 0.5
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s) 0.340
Hg₂Cl₂(s) + 2e⁻ ⟶ 2 Hg() + 2 Cl⁻(aq) 0.27
AgCl(s) + e⁻ ⟶ Ag(s) + Cl⁻(aq) 0.222
Cu²⁺(aq) + e⁻ ⟶ Cu⁺(aq) 0.159
SO₄²⁻(aq) + 4 H⁺(aq) + 2 e⁻ ⟶ H₂SO₃(aq) + H₂O() 0.158
Sn⁴⁺(aq) + 2e⁻ ⟶ Sn²⁺(aq) 0.15
S(s) + 2 H⁺(aq) + 2 e⁻ ⟶ H₂S(aq) 0.144
AgBr(s) + e⁻ ⟶ Ag(s) + Br⁻(aq) 0.0713
2 H⁺(aq) + 2 e⁻ ⟶ H₂(g) (reference electrode) 0
N₂O(g) + 6 H⁺(aq) + H₂O() + 4 e⁻ ⟶ 2 NH₃OH⁺(aq) 0.05
HgS(s, black) + 2 H⁺(aq) + 2 e⁻ ⟶ Hg() + H₂S(g) 0.085
Se(s) + 2 H⁺(aq) + 2 e⁻ ⟶ H₂Se(aq) 0.115
Pb²⁺(aq) + 2 e⁻ ⟶ Pb(s) 0.125
Sn²⁺(aq) + 2 e⁻ ⟶ Sn(s) 0.1375
AgI(s) + e⁻ ⟶ Ag(s) + I⁻(aq) 0.1522
[SnF₆]²⁻(aq) + 4 e⁻ ⟶ Sn(s) + 6 F⁻(aq) 0.200
Ni²⁺(aq) + 2 e⁻ ⟶ Ni(s) 0.25
Co²⁺(aq) + 2 e⁻ ⟶ Co(s) 0.277
Tl⁺(aq) + e⁻ ⟶ Tl(s) 0.3363
PbSO₄(s) + 2 e⁻ ⟶ Pb(s) + SO₄²⁻(aq) 0.3505
Cd²⁺(aq) + 2 e⁻ ⟶ Cd(s) 0.403
Cr³⁺(aq) + e⁻ ⟶ Cr²⁺(aq) 0.424
Fe²⁺(aq) + 2 e⁻ ⟶ Fe(s) 0.44
2 CO₂(g) + 2 H⁺(aq) + 2 e⁻ ⟶ (COOH)₂(aq) 0.481
TiO₂(s)+ 4 H⁺(aq) + 2 e⁻ ⟶ Ti²⁺(aq) + 2 H₂O() 0.502
Ga³⁺(aq) + 3 e⁻ ⟶ Ga(s) 0.53
Cr³⁺(aq) + 3 e⁻ ⟶ Cr(s) 0.74
Zn²⁺(aq) + 2 e⁻ ⟶ Zn(s) 0.763
Cr²⁺(aq) + 2 e⁻ ⟶ Cr(s) 0.90
V²⁺(aq) + 2 e⁻ ⟶ V(s) 1.13
Mn²⁺(aq) + 2 e⁻ ⟶ Mn(s) 1.18
Zr⁴⁺(aq) + 4 e⁻ ⟶ Zr(s) 1.55
Al³⁺(aq) + 3 e⁻ ⟶ Al(s) 1.676
H₂(g) + 2 e⁻ ⟶ 2 H⁻(aq) 2.25
Mg²⁺(aq) + 2 e⁻ ⟶ Mg(s) 2.356
Na⁺(aq) + e⁻ ⟶ Na(s) 2.714
Ca²⁺(aq) + 2 e⁻ ⟶ Ca(s) 2.84
Sr²⁺(aq) + 2 e⁻ ⟶ Sr(s) 2.89
Ba²⁺(aq) + 2 e⁻ ⟶ Ba(s) 2.92
Rb⁺(aq) + e⁻ ⟶ Rb(s) 2.925
K⁺(aq) + e⁻ ⟶ K(s) 2.925
Li⁺(aq) + e⁻ ⟶ Li(s) 3.045
From Bard, A. J., Parsons, R., and Jordan, J., Standard Potentials in Aqueous Solution, New York: Marcel Dekker, 1985. International Union of Pure and Applied Chemistry, Commission on Electrochemistry and Electroanalytical Chemistry.
From Brown, R. A., and Swift, E. H., Journal of the American Chemical Society, Vol. 71, 1949, pp. 2719-2723.
Standard Electrode Potentials in Basic Aqueous Solution
at 25 °C.
Basic Solution Standard Electrode Potential, E° (volts)
ClO⁻(aq) + H₂O() + 2 e⁻ ⟶ Cl⁻(aq) + 2 OH⁻(aq) 0.89
OOH⁻(aq) + H₂O() + 2 e⁻ ⟶ 3 OH⁻(aq) 0.867
2 NH₂OH(aq) + 2 e⁻ ⟶ N₂H₄(aq) + 2 OH⁻(aq) 0.73
ClO₃⁻(aq) + 3 H₂O() + 6 e⁻ ⟶ Cl⁻(aq) + 6 OH⁻(aq) 0.622
MnO₄⁻(aq) + 2 H₂O() + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq) 0.60
MnO₄⁻(aq) + e⁻ ⟶ MnO₄²⁻(aq) 0.56
NiO₂(s) + 2 H₂O() + 2 e⁻ ⟶ Ni(OH)₂(s) + 2 OH⁻(aq) 0.49
Ag₂CrO₄(s) + 2 e⁻ ⟶ 2 Ag(s) + CrO₄²⁻(aq) 0.4491
O₂(g) + 2 H₂O() + 4 e⁻ ⟶ 4 OH⁻(aq) 0.401
ClO₄⁻(aq) + H₂O() + 2 e⁻ ⟶ ClO₃⁻(aq) + 2 OH⁻(aq) 0.374
Ag₂O(s) + H₂O() + 2 e⁻ ⟶ 2 Ag(s) + 2 OH⁻(aq) 0.342
2 NO₂⁻(aq) + 3 H₂O() + 4 e⁻ ⟶ N₂O(g) + 6 OH⁻(aq) 0.15
N₂H₄(aq) + 2 H₂O() + 2 e⁻ ⟶ 2 NH₃(aq) + 2 OH⁻(aq) 0.10
HgO(s) + H₂O() + 2 e⁻ ⟶ Hg() + 2 OH⁻(aq) 0.0977
O₂(g) + H₂O() + 2 e⁻ ⟶ OOH⁻(aq) + OH⁻(aq) 0.0649
[Co(NH₃)₆]³⁺(aq) + e⁻ ⟶ [Co(NH₃)₆]²⁺(aq) 0.058
NO₃⁻(aq) + H₂O() + 2 e⁻ ⟶ NO₂⁻(aq) + 2 OH⁻(aq) 0.01
MnO₂(s) + 2 H₂O() + 2 e⁻ ⟶ Mn(OH)₂(s) + 2 OH⁻(aq) 0.05
CrO₄²⁻(aq) + 4 H₂O() + 3 e⁻ ⟶ Cr(OH)₃(s) + 5 OH⁻(aq) 0.11
Cu₂O(s) + H₂O() + 2 e⁻ ⟶ 2 Cu(s) + 2 OH⁻(aq) 0.365
FeO₂⁻(aq) + H₂O() + e⁻ ⟶ HFeO₂⁻(aq) + OH⁻(aq) 0.69
HFeO₂⁻(aq) + H₂O() + 2 e⁻ ⟶ Fe(s) + 3 OH⁻(aq) 0.8
2 H₂O() + 2 e⁻ ⟶ H₂(g) + 2 OH⁻(aq) 0.8277
2 NO₃⁻(aq) + 2 H₂O() + 2 e⁻ ⟶ N₂O₄(g) + 4 OH⁻(aq) 0.86
SO₄²⁻(aq) + H₂O() + 2 e⁻ ⟶ SO₃²⁻(aq) + 2 OH⁻(aq) 0.936
N₂(g) + 4 H₂O() + 4 e⁻ ⟶ N₂H₄(aq) + 4 OH⁻(aq) 1.16
Zn(OH)₂(s) + 2 e⁻ ⟶ Zn(s) + 2 OH⁻(aq) 1.246
[Zn(OH)₄]²⁻(aq) + 2 e⁻ ⟶ Zn(s) + 4 OH⁻(aq) 1.285
Cr(OH)₃(s) + 3 e⁻ ⟶ Cr(s) + 3 OH⁻(aq) 1.33
[Zn(CN)₄]²⁻(aq) + 2 e⁻ ⟶ Zn(s) + 4 CN(aq) 1.34
SiO₃²⁻(aq) + 3 H₂O() + 4 e⁻ ⟶ Si(s) + 6 OH⁻(aq) 1.69
From Bard, A. J., Parsons, R., and Jordan, J., Standard Potentials in Aqueous Solution, New York: Marcel Dekker, 1985. International Union of Pure and Applied Chemistry, Commission on Electrochemistry and Electroanalytical Chemistry. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.14%3A_Standard_Potential.txt |
Alkanes
The International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules:
1. To name an alkane, first identify the longest chain of carbon atoms in its structure. A two-carbon chain is called ethane; a three-carbon chain, propane; and a four-carbon chain, butane. Longer chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10).
2. Add prefixes to the name of the longest chain to indicate the positions and names of substituents. Substituents are branches or functional groups that replace hydrogen atoms on a chain. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.
When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ide at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with -ide like chloride; in organic compounds, such atoms are treated as substituents and the -o ending is used). The number of substituents of the same type is indicated by the prefixes di- (two), tri- (three), tetra- (four), and so on (for example, difluoro- indicates two fluoride substituents).
Example 1
Naming Halogen-substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2-bromo-; this will come at the beginning of the name, since bromo- comes before chloro- alphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane.
Check Your Learning
Name the following molecule:
Answer:
3,3-dibromo-2-iodopentane
We call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an alkyl group is obtained by dropping the suffix -ane of the alkane name and adding -yl:
The open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom.
Example 2
Naming Substituted Alkanes
Name the molecule whose structure is shown here:
Solution
The longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right—this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)—so we take our name for two carbons eth- and attach -yl at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane.
Check Your Learning
Name the following molecule:
Answer:
4-propyloctane
Some hydrocarbons can form more than one type of alkyl group when the hydrogen atoms that would be removed have different “environments” in the molecule. This diversity of possible alkyl groups can be identified in the following way: The four hydrogen atoms in a methane molecule are equivalent; they all have the same environment. They are equivalent because each is bonded to a carbon atom (the same carbon atom) that is bonded to three hydrogen atoms. Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent and removing any one of these hydrogen atoms produces an ethyl group. Each of the six hydrogen atoms is bonded to a carbon atom that is bonded to two other hydrogen atoms and a carbon atom. However, in both propane and 2–methylpropane, there are hydrogen atoms in two different environments, distinguished by the adjacent atoms or groups of atoms:
Each of the six equivalent hydrogen atoms of the first type in propane and each of the nine equivalent hydrogen atoms of that type in 2-methylpropane (all shown in black) are bonded to a carbon atom that is bonded to only one other carbon atom. The two purple hydrogen atoms in propane are of a second type. They differ from the six hydrogen atoms of the first type in that they are bonded to a carbon atom bonded to two other carbon atoms. The green hydrogen atom in 2-methylpropane differs from the other nine hydrogen atoms in that molecule and from the purple hydrogen atoms in propane. The green hydrogen atom in 2-methylpropane is bonded to a carbon atom bonded to three other carbon atoms. Two different alkyl groups can be formed from each of these molecules, depending on which hydrogen atom is removed. The names and structures of these and several other alkyl groups are listed in Figure 1.
Note that alkyl groups do not exist as stable independent entities. They are always a part of some larger molecule. The location of an alkyl group on a hydrocarbon chain is indicated in the same way as any other substituent:
Alcohols
The name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the –OH group is bonded is indicated by a number placed before the name.
Consider the following example:
The carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the –OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol.
Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol. Alcohols may also be classified as primary, 1º, secondary, 2º, and tertiary, 3º. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration).
Carboxylic Acids
The –e ending is removed from the name of the parent carbon chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. Many carboxylic acids are called by the common names that were chosen by chemists to usually describe the origin of the compound.
Formula Common Name Source IUPAC Name
HCO2H formic acid ants (L. formica) methanoic acid
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain.
Esters
Esters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain from the carboxylic acid part of the ester with an –e remove and replaced with the ending –oate.
For example:
1. First, identify the oxygen that is part of the continuous chain and bonded to carbon on both sides. (On one side of this oxygen there will be a carbonyl present but on the other side there won’t be.)
2. Second, begin numbering the carbon chains on either side of the oxygen identified in step 1.
3. Next, use this format: [alkyl on side further from the carbonyl] (space) [alkane on the side with the carbonyl]
4. Finally, change the ending of the alkane on the same side as the carbonyl from -e to -oate.
5. When an ester group is attached to a ring, the ester is named as a substituent on the ring.
In both common and International Union of Pure and Applied Chemistry (IUPAC) nomenclature, the –ic ending of the parent acid is replaced by the suffix –ate.
: Nomenclature of Esters
Condensed Structural Formula Common Name IUPAC Name
HCOOCH3 methyl formate methyl methanoate
CH3COOCH3 methyl acetate methyl ethanoate
CH3COOCH2CH3 ethyl acetate ethyl ethanoate
CH3CH2COOCH2CH3 ethyl propionate ethyl propanoate
CH3CH2CH2COOCH(CH3)2 isopropyl butyrate isopropyl butanoate
ethyl benzoate ethyl benzoate
Amines
The word “amine” is derived from ammonia, and the class of compounds known as amines therefore are commonly named as substituted ammonias. In this system, primary amines (RNH2), having only one substituent on nitrogen, are named with the substituent as a prefix. More systematic nomenclature appends -amine to the longest chain, as for alcohols:
Secondary (R2NH) and tertiary amines (R3N), which have two and three substituents on nitrogen, commonly are named as N-substituted amines. N is included to indicate that the substituent is on the nitrogen atom unless there is no ambiguity as to where the substituent is located. Each of the R groups is named as a separate word, except when the groups are identical, in which case the prefix di or bis may be used (di is used for simple groups, bis for substituted groups):
The nomenclature of amines is complicated by the fact that several different nomenclature systems exist. The four compounds shown in the top row of the following diagram are all C4H11N isomers.
• The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group. For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently since these names are not based on a rational system.
Amides
Primary amides are named by changing the name of the acid by dropping the -oic acid or -ic acid endings and adding -amide. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain. For example, the following three amides are called methanamide or formamide, ethanamide or acetamide, and benzamide.
Further examples are shown below:
Secondary amides are named by using an upper case N to designate that the alkyl group is on the nitrogen atom. Alkyl groups attached to the nitrogen are named as substituents. The letter N is used to indicate they are attached to the nitrogen. For example, the molecule below is called N-methylpropanamide:
Tertiary amides are named in the same way as secondary amides, but with two N’s | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/06%3A_Appendix/6.15%3A_Basics_of_Organic_Nomenclature.txt |
Review: Waves
Much of modern technology is based on electromagnetic radiation. Radio waves from a mobile phone, X-rays used by dentists, the energy used to cook food in your microwave, the radiant heat from a toaster wire, and the colors emanating from your television screen are forms of electromagnetic radiation. All exhibit behaviors that can be explained by a wave theory.
A wave is an oscillation or periodic movement that can transfer energy from one point to another. For example, the expansion and compression of air that accompanies a lightning strike generates sound waves (thunder) that can travel several miles and cause your ear drums (and sometimes windows) to vibrate. Kinetic energy is transferred through matter (the air) while the matter remains essentially in place. An insightful example of a wave occurs in sports stadiums when fans in a narrow region of seats rise simultaneously and stand with their arms raised for a few seconds before sitting down again and fans in neighboring sections likewise stand up and sit down in sequence. While this wave can quickly encircle a large stadium, none of the fans actually travel with the wave—they all stay in or above their seats.
All waves are characterized by these properties: wavelength (λ), the distance between two consecutive peaks, troughs, or other equivalent points in a wave; frequency (ν), the number of wave cycles (a cycle corresponds to one complete wavelength) that pass a specified point in space in a specified time; and amplitude (A), the magnitude of the wave’s displacement. See examples shown in Figure 1.
The SI unit for λ is meter, and for ν is hertz (Hz), which is the number of cycles per second.
$[Hz = \dfrac{1}{s} = s^{-1}] \nonumber$
Common multiples are megahertz, (1 MHz = 1 × 106 Hz) and gigahertz (1 GHz = 1 × 109 Hz). The maximum amplitude of a wave shown in Figure 1 corresponds to one-half the height between the peaks and troughs, and is related to the intensity of the wave. For example, for sound waves, greater amplitude means louder sound.
Waves are not restricted to traveling through matter. James Clerk Maxwell proposed the existence of electromagnetic waves, which consist of an electric field oscillating in the x dimension in step with a magnetic field oscillating in the y dimension. The electromagnetic oscillations travel in the z dimension, where x, y, and z are Cartesian coordinate axes. Electromagnetic radiation travels through a vacuum at a constant speed of 2.998 × 108 m/s, the speed of light (c).
The product of a wave’s wavelength and its frequency, λv, is the speed of the wave. (The wave travels distance λ per cycle and the number of cycles traveled in a given time is v.) Thus, for electromagnetic radiation in a vacuum:
$c = 2.998 \times 10^8\;\dfrac{\text{m}}{\text{s}} = \lambda\nu \nonumber$
Because the product λv is constant, wavelength and frequency are inversely proportional: as the wavelength increases, the frequency decreases.
Exercise 1: Photons and Laser Light
For all exercises, before doing any calculation or looking at the hint, write in your class notebook an explanation of how you plan to work out the problem. Do all the steps in the calculation in your notebook. Once you have arrived at an answer, submit your results below and click the “Check” button to see if it is correct. If one or more parts of your answer is incorrect, go over your work in your notebook carefully and check for errors. “Retry” with your new answer. Look at the hint (click on it to expand for view) only after you have made attempts at answering the question.
Query $1$
Additional Practice
Query $2$
Figure 3 shows the electromagnetic spectrum, the range of all types of electromagnetic radiation. It spans an enormous range—wavelengths of kilometers (103 m) to picometers (10−12 m) have been observed, and the visible-light range makes up only a small portion of it. Given the expansive range of wavelengths, different units are typically used for different parts of the spectrum. For example, radio waves are usually specified as frequencies (typically in units of MHz or GHz), while the visible region is usually specified in wavelengths (typically in units of nm or Å).
D1.2 Wave Properties
When a wave passes through a pinhole or a very narrow slit, waves fan out on other side of the pinhole or slit. This is shown on the right for water waves moving from left to right through a hole about the same width as the wavelength.
When two or more waves occupy the same region, the total wave amplitude is the sum of the amplitudes of the individual waves. Figure 4 shows that this can result in a larger total amplitude if the amplitudes of the waves have the same mathematical sign (are in phase) or in a smaller total amplitude if the two waves have amplitudes with opposite sign (are out of phase).
Summation of amplitudes of waves leads to interference of one wave with another. Figure 5 shows the interference patterns that arise when blue or red laser light passes through a diffraction grating with many narrow, vertical slits spaced about one wavelength apart.
When laser light passes through closely spaced slits, each slit effectively acts as a new source with waves fanning out from it. This results in closely spaced waves coming into contact at the detector (the camera used to make Figure 5). The dark regions in Figure 5 correspond to regions where the peaks for the wave from one slit happen to coincide with the troughs for the wave from the other slit (destructive interference, Fig. 4b), while the brightest regions correspond to regions where the peaks for the two waves (or their two troughs) happen to coincide (constructive interference, Fig. 4a).
Exercise 2: Interference Patterns
Additional Practice 1
Query $4$
Additional Practice 2
Query $5$
Interference patterns are an inherent property of wave motion but they cannot be explained by particles moving according to the laws of classical mechanics. Particles would move in straight lines through a slit or pinhole and produce a single bright line or spot. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/07%3A_Review_Section/7.01%3A_Review-_Waves.txt |
D1.3 Blackbody Radiation
Have you noticed that some LED light bulbs are a different color from others? Some are “warm” and others are “cool” (see Figure 6). Candlelight is “warmer” (yellower) than fluorescent lights: the latter have a blue cast that can make people’s faces look sickly. These observations are related to blackbody radiation—the continuous spectrum of light emitted by matter as the matter is heated to higher temperatures.
Figure 7 shows the spectra, graphs of intensity versus wavelength, for blackbody radiation emitted by matter at several different temperatures.
Here, each spectrum depends only on temperature. The wavelength of the maximum in each curve,
$\lambda_{max} \nonumber$
shifts to shorter wavelengths as the temperature increases. This corresponds to a heated metal first becoming red hot, then brighter and white hot as the temperature increases.
Exercise 3: Interpreting Blackbody Spectra
Additional Practice 1
Query $2$
Additional Practice 2
D1.4 Planck’s Quantum Theory
In the late 1800s, physicists derived mathematical expressions for the blackbody curves using well-accepted concepts of mechanics and electromagnetism. They assumed that as the temperature increased the energies of atoms in a metal object would increase as the atoms vibrated more vigorously; these vibrations were assumed to create electromagnetic waves—the blackbody radiation. But no theory based on these ideas was able to predict the shapes of the curves shown in Figure 7. Even worse, the theory predicted that the intensity would become infinitely large for very short wavelengths, an absurd result.
In 1900, Max Planck introduced a revolutionary idea, from which he was able to derive a theoretical expression for blackbody spectra that fit the experimental observations within experimental error. Instead of assuming that the vibrating atoms could have a continuous set of energy values, Planck restricted the vibrational energies to discrete values—that is, he assumed that there must be some minimum quantity of energy that could be transferred between vibrating atoms. That quantity of energy is proportional to the frequency of vibration and is called a quantum.
$\text{E}_{\text{quantum}} = h\nu;\;\;\;\;\;\text{because}\;c = \lambda\nu,\;\nu = \dfrac{c}{\lambda},\;\text{E}_{\text{quantum}} = \dfrac{hc}{\lambda} \nonumber$
The proportionality constant “h” is now known as Planck’s constant. Its value is very small, 6.626 × 10−34 J·s (joule-seconds). According to Planck’s theory, electromagnetic radiation occurs in small, indivisible quantities (quanta), just as matter consists of small, chemically indivisible quantities (atoms).
Exercise 4: Radiation and Quanta
Additional Practice
Query $5$
Although Planck had developed a theory of blackbody radiation that worked, he was not satisfied with the assumption of quantized energies for the vibrating atoms. But not for long—a few years later Albert Einstein used the idea of quantization of electromagnetic radiation to explain another puzzling phenomenon: the photoelectric effect, which is a topic for the next section.
D2.1 The Photoelectric Effect
Activity 1: Preparation—Photoelectric Effect
In your course notebook, make a heading for Photoelectric Effect. After the heading write down what you recall about the photoelectric effect from courses you have already taken. If you remember having a question about the photoelectric effect or if there is anything you remember being puzzled about, write that down as well. We will ask you to refer back to what you have written when you complete this section.
Planck’s quantum theory was able to predict accurately the distribution of wavelengths emitted by a blackbody at various temperatures. However, Planck found it difficult to justify his assumption that vibration energies had to be multiples of a minimum energy—a quantum. When Albert Einstein used Planck’s quantum hypothesis to explain a different phenomenon, the photoelectric effect, the validity of quantum theory became clearer.
Exercise 1: Quanta and Laser Light (Review)
Before doing any calculation or looking at the hint, write in your class notebook an explanation of how you plan to work out the problem. Do all the steps in the calculation in your notebook. Once you have arrived at an answer, then submit your results below and click the “Check” button to see if it is correct. If one or more parts of your answer is incorrect, go over your work in your notebook carefully and check for errors. “Retry” with your new answer. Look at the hint (click on it to expand for view) only after you have attempted to answer the question at least twice.
Query $6$
When electromagnetic radiation shines on a metal, such as sodium, electrons can be emitted and an electric current (a flow of electrons) can occur. This is called the photoelectric effect. The effect is complicated: for some wavelengths no electrons are emitted, but at other wavelengths electrons are emitted.
Exercise 2: Photoelectric Effect Simulation
Watch this photoelectric effect animation, where sodium is already selected, to answer the questions below. Write down your observations as you watch the animation, and then answer the questions in your course notebook.
1. For which colors of visible light are electrons emitted by sodium? Determine the maximum wavelength at which electrons are emitted.
2. At a wavelength where electrons are emitted, describe the shape of a graph of number of electrons emitted vs. light intensity.
3. Determine the shape of a graph of electron energy vs. light frequency. How does this graph differ from the electrons vs. intensity graph?
4. Based on your observations, draw a rough graph with labeled axes to show how the number of electrons emitted varies with wavelength of light. How would the graph change if the intensity of light increased?
If you would like to experiment with the simulation further, download and save the simulation program, go to the location where you saved it, and double-click on the file name (photoelectric_en.jar) to install and run it. You need to have Java installed for this to work.
Query $7$
Additional Practice
Query $8$
Activity 2: Summary of Photoelectric Effect Results
In your course notebook, write a brief summary of the results you obtained from experimenting with the photoelectric effect simulation. Also write a summary of how your experimental results can be interpreted based on the idea that light consists of quanta. Compare what you wrote with the summary below [also available at this link] and, if necessary, revise what you wrote.
Exercise 3: Photoelectric Effect
Additional Practice
Query $11$
Activity 3: Wrap-up—Photoelectric Effect
Refer to what you wrote (including things that puzzled you) when you made the heading Photoelectric Effect in your class notebook. Revise what you wrote based on what you have just learned. Write a summary so that it will be a good study aid when you review for an exam. If you still have questions, ask them on Piazza. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/07%3A_Review_Section/7.02%3A_Review-_Photons.txt |
D3.1 Atomic-Sized Particles Have Unusual Properties
Bohr’s model predicted experimental data for the hydrogen-atom emission spectra and was widely accepted, but it also raised questions. Why did electrons have energies defined by a single quantum number n = 1, 2, 3, and so on, but never in between? How could the model work so well for a hydrogen atom and one-electron ions, but not accurately predict the emission spectrum for helium or any atom or ion with two or more electrons? To answer such questions, we need to explore the unusual properties of matter at the atomic scale.
It is natural to try to interpret the behavior of electrons in atoms using analogies to the behavior of things we can see and experience. By the turn of the 20th century, scientists had identified two different kinds of behavior that might serve as analogies: waves and particles. Waves, such as water waves, can bend around objects and exhibit interference. A particle, such as a pea or tennis ball, moves in a straight line unless its path is changed by a force (such as friction, gravity, or hitting something). It turns out that properties of atomic-scale matter are best explained using a combination of these analogies.
Let’s think about an electron as if it were a particle and choose a very simple, but artificial, situation. Figure 1 shows an electron that is constrained to move in a single dimension, say the x direction. There are walls to the left and right of the electron that prevent it from moving beyond them; the distance between the walls is d (the electron is in a box of length d). Given an initial push, the electron will move back and forth forever (assuming no friction). The electron’s kinetic energy would be given by the equation
$E = \dfrac{1}{2}mv^2 \nonumber$
where m is the mass and v the velocity.
Exercise 1: Electron in One-dimensional Box
Consider these questions regarding the electron moving in a one dimensional box:
Query $1$
But to explain the hydrogen-atom line spectrum, Bohr had to assume that an electron in an atom can only have certain energies. This assumption seemed very unsatisfactory based on the particle analogy—as we just saw, an electron should be able to have any energy value.
Let’s explore a wave analogy instead. In 1924, Louis de Broglie proposed that a wave of wavelength λ is associated with every particle. The larger the mass of the particle and the faster it is moving, the smaller this wavelength becomes. The relationship is given by the formula:
$\lambda = \dfrac{h}{p} \nonumber$
where p is the momentum of the electron, the product of its mass and velocity
$p = m \times v \nonumber$
and h is Planck’s constant (h = 6.626 × 10–34 J s).
Optional Information: Wavelike behavior of electrons was confirmed in 1927; read about it here. View this video for another description of the relevant experiments.
Suppose a wave, such as a vibrating guitar string, occupies the box. A guitar string is constrained by being fixed at each end so we assume the ends of the string are tied to the walls at the ends of the box.
Activity 2: Graphical Representation of Vibrating String
In your course notebook, make a drawing of what a vibrating string of length d, with its ends in fixed positions, would look like. If you are studying with someone else, each make a drawing individually and then compare.
Query $2$
If an electron behaves like a guitar string, then only certain wavelengths will fit within the one-dimensional box with length d (see the figure in Activity 2). The length of the box can thus correspond to: a single half-wavelength, two half-wavelengths, three half-wavelengths, etc., but not to any wave where either end of the string is moving. In other words, the length d must correspond to an integer number of half wavelengths:
$d = n\dfrac{\lambda}{2} \nonumber$
where n = 1, 2, 3, 4,… Solving the equation for λ, we obtain:
$\lambda = \dfrac{2d}{n} \nonumber$
Combining this with de Broglie’s equation,
$\lambda = \dfrac{h}{p} \nonumber$
we have:
$\lambda = \dfrac{2d}{n} = \dfrac{h}{p} = \dfrac{h}{mv} \nonumber$
which rearranges to give:
$v = \dfrac{nh}{2md}\;\;\;n = 1, 2, 3, 4, \dots \nonumber$
We can now calculate the kinetic energy of our wave-particle. It is given by the formula:
$E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dfrac{nh}{2md}\right)^2 = n^2\left(\dfrac{h^2}{8md^2}\right) \nonumber$
Because n is a positive whole number, this equation says that the kinetic energy of the electron can have only certain values and not others.
Exercise 2: Energy Levels for Electron in a Box
Query $3$
This result shows that if we think of an electron as a wave in a 100 pm box, its energy is automatically restricted to certain specific values: the electron can have an energy of 6.0 attojoules (aJ) or 24.0 aJ, but not an intermediate energy such as 7.3 aJ or 11.6 aJ. We describe this situation by saying that the energy of the electron is quantized. Because the energy of 6.0 aJ is the lowest possible energy, the electron is most stable at this energy, where the box contains half a wavelength. This is the ground state. If the energy has a higher value, such as 24 aJ or 54 aJ, the electron is in an excited state. Notice that for the excited states the electron waves have one or more nodes. A node is a point where the wave has zero amplitude; that is, where the wave is not moving up or down at all. The greater the number of nodes, the higher the energy, a generalization that is true for all kinds of waves.
This new way of approaching the behavior of electrons (and other atomic-sized particles) became known as wave mechanics or quantum mechanics. Using both wave and particle analogies to describe atomic-scale matter is referred to as wave-particle duality. Wave-particle duality implies that we can no longer state that the electron is located at a specific position within the box or is moving in one direction or the other: the electron seems to be all over the box at once! | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/07%3A_Review_Section/7.03%3A_In_Depth-_Atomic-scale_Particles_and_Waves.txt |
D3.2 The Heisenberg Uncertainty Principle
This inability to locate an electron precisely seems strange, but it is true of all atomic-scale particles. In 1927, Werner Heisenberg introduced the Heisenberg uncertainty principle: it is fundamentally impossible to determine simultaneously and exactly both the momentum and the position of a particle. Mathematically,
$(\Delta x)(m \Delta v_x) = (\Delta x)(\Delta p_x) \ge \dfrac{h}{4\pi} \nonumber$
where Δx is the uncertainty in the position and Δpx is the uncertainty in the momentum along the x direction. Hence, the more accurately we measure the momentum of a particle, the less accurately we can determine its position at that time, and vice versa.
Exercise 3: Uncertainty Principle
The result of your calculation shows that if an electron has Δx = 1 pm, the uncertainty in its velocity is at least 5 × 107 m/s. In other words, we essentially have no idea how fast it’s moving.
Heisenberg’s principle imposes ultimate limits on what is measurable in science. It is possible to talk about the probability that the electron is at a specific location, or the probability that it is moving at a given speed, but there will also be some probability of finding it somewhere else in the box or moving at some other speed.
Optional Information: For more details about the uncertainty principle, click here.
The uncertainty principle may seem strange, but we can say that the probability of finding a particle-like electron at a given location depends on the shape of the wave associated with the electron. The various wave shapes you drew in Activity 2 can be described mathematically using functions such as sines or cosines; that is, there is a mathematical wave function that describes each wave. The wave function is usually represented by a Greek letter
$\psi \nonumber$
Shortly after the uncertainty principle was proposed, the German physicist Max Born (1882 to 1969) suggested that the square of the magnitude of the wave function,
$|\psi|^2 \nonumber$
at any position is proportional to the probability of finding the electron (as a particle) at that same position. If we can determine the wave function associated with an electron, we can also determine the relative probability of the electron’s being located at one point as opposed to another. Thus, each wave form you drew in Activity 2 can be represented by a different function,
$\psi_n \nonumber$
each has a different distribution of electric charge throughout the box, and each is associated with a different, specific energy value.
A graphic way of indicating the probability of finding the electron at a particular location is by the density of shading or stippling; that is, where the probability is high we draw lots of dots or darker shading and where the probability is low we draw fewer dots. We say that where the probability is high we have a large electron probability density (or just electron density). (In the guitar-string analogy we would draw lots of dots at places where the string was vibrating quite far from its rest position. For example, consider the wave shown below in red along with its square shown in blue. Wherever there is a maximum in the square of the wave function, we would plot a lot of dots.)
Activity 3: Representing Electron Density by Density of Dots
In your course notebook, annotate each of your drawings of the vibrating string with dots such that the density of dots indicates the electron probability density.
Query $2$
In Bohr’s theory an electron had a well-defined orbit around the nucleus, but according to the uncertainty principle, the best we can do is indicate electron density in various regions. Consequently we use a slightly different word to refer to the wave function and its electron-density distribution: orbital. The electron density of orbitals of an atom or molecule is very useful because it indicates where there is negative charge (electron density) relative to positive charge (atomic nuclei). This helps us understand atomic properties, chemical bonds, and forces between molecules.
7.05: In Depth- In-Phase and Out-of-Phase
D6.2 Constructive and Destructive Interference
Forming a chemical bond by overlap of two atomic orbitals, which we just described, is equivalent to constructive interference of the electron waves for the two hydrogen atoms. It occurs when the two waves are in-phase.
For example, in a one-dimensional vibrating string, in-phase means when two waves are both vibrating up (or down) at the same time. In the video below, both ends of the wave device are moving up (or down) at the same time. This results in a bigger wave in the middle, where the two waves overlap.
https://mediaspace.wisc.edu/id/1_0vj...yerId=25717641
A second possibility is that the two waves are out-of-phase—one is moving up when the other is moving down. This results in destructive interference between the waves. This is shown in the video below. Note that in the destructive-interference video there is a point in the middle of the wave device that never moves up or down. This point has zero amplitude all the time and is a node.
https://mediaspace.wisc.edu/id/1_ycc...yerId=25717641
An atom is three dimensional, so it is difficult or impossible to visualize the waves corresponding to an electron in an atom. The principle of in-phase or out-of-phase remains the same, though. In-phase results in constructive interference and larger values of the wave function between two nuclei. Larger wave function corresponds to larger electron density, which attracts the two nuclei. Out-of-phase atomic wave functions result in a node between the nuclei, which leaves two unscreened nuclei that repel. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/07%3A_Review_Section/7.04%3A_In_Depth-_The_Heisenberg_Uncertainty_Principle.txt |
Chemistry is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. As you begin your study of college chemistry, those of you who do not intend to become professional chemists may well wonder why you need to study chemistry. You will soon discover that a basic understanding of chemistry is useful in a wide range of disciplines and career paths. You will also discover that an understanding of chemistry helps you make informed decisions about many issues that affect you, your community, and your world. A major goal of this text is to demonstrate the importance of chemistry in your daily life and in our collective understanding of both the physical world we occupy and the biological realm of which we are a part.
• 1.1: Atoms and Molecules
Chemistry deals with the composition, structure, and properties of matter, and the ways by which various forms of matter may be interconverted. Thus, it occupies a central place in the study and practice of science and technology. Chemists use the scientific method to perform experiments, pose hypotheses, and formulate laws and develop theories, so that they can better understand the behavior of the natural world. To do so, they operate in the macroscopic, microscopic, and symbolic domains.
• 1.2: The Scientific Approach to Knowledge
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.
• 1.3: The Classification of Matter
Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. 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 & properties.
• 1.4: Physical and Chemical Changes and Properties
All substances have distinct physical and chemical properties, and may undergo physical or chemical changes. Physical properties, such as hardness and boiling point, and physical changes, such as melting or freezing, do not involve a change in the composition of matter. Chemical properties, such flammability and acidity, and chemical changes, such as rusting, involve production of matter that differs from that present beforehand.
• 1.5: Energy - A Fundamental Part of Physical and Chemical Change
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.
• 1.6: The Units of Measurement
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.
• 1.7: 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).
• 1.8: Solving Chemical Problems
Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.
• 1.E: Matter, Measurement, and Problem Solving (Exercises)
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro.
Thumbnail: Two small test tubes held in Spring Clamps. (CC BY-SA 3.0; Amitchell125).
01: Matter Measurement and Problem Solving
Learning Objectives
• Outline the historical development of chemistry
• Provide examples of the importance of chemistry in everyday life
• Provide examples illustrating macroscopic, microscopic, and symbolic domains
Throughout human history, people have tried to convert matter into more useful forms. Our Stone Age ancestors chipped pieces of flint into useful tools and carved wood into statues and toys. These endeavors involved changing the shape of a substance without changing the substance itself. But as our knowledge increased, humans began to change the composition of the substances as well—clay was converted into pottery, hides were cured to make garments, copper ores were transformed into copper tools and weapons, and grain was made into bread.
Humans began to practice chemistry when they learned to control fire and use it to cook, make pottery, and smelt metals. Subsequently, they began to separate and use specific components of matter. A variety of drugs such as aloe, myrrh, and opium were isolated from plants. Dyes, such as indigo and Tyrian purple, were extracted from plant and animal matter. Metals were combined to form alloys—for example, copper and tin were mixed together to make bronze—and more elaborate smelting techniques produced iron. Alkalis were extracted from ashes, and soaps were prepared by combining these alkalis with fats. Alcohol was produced by fermentation and purified by distillation.
Attempts to understand the behavior of matter extend back for more than 2500 years. As early as the sixth century BC, Greek philosophers discussed a system in which water was the basis of all things. You may have heard of the Greek postulate that matter consists of four elements: earth, air, fire, and water. Subsequently, an amalgamation of chemical technologies and philosophical speculations were spread from Egypt, China, and the eastern Mediterranean by alchemists, who endeavored to transform “base metals” such as lead into “noble metals” like gold, and to create elixirs to cure disease and extend life (Figure \(1\)).
Figure \(1\): This portrayal shows an alchemist’s workshop circa 1580. Although alchemy made some useful contributions to how to manipulate matter, it was not scientific by modern standards. (credit: Chemical Heritage Foundation).
From alchemy came the historical progressions that led to modern chemistry: the isolation of drugs from natural sources, metallurgy, and the dye industry. Today, chemistry continues to deepen our understanding and improve our ability to harness and control the behavior of matter. This effort has been so successful that many people do not realize either the central position of chemistry among the sciences or the importance and universality of chemistry in daily life.
Chemistry: The Central Science
Chemistry is sometimes referred to as “the central science” due to its interconnectedness with a vast array of other STEM disciplines (STEM stands for areas of study in the science, technology, engineering, and math fields). Chemistry and the language of chemists play vital roles in biology, medicine, materials science, forensics, environmental science, and many other fields (Figure \(2\)). The basic principles of physics are essential for understanding many aspects of chemistry, and there is extensive overlap between many subdisciplines within the two fields, such as chemical physics and nuclear chemistry. Mathematics, computer science, and information theory provide important tools that help us calculate, interpret, describe, and generally make sense of the chemical world. Biology and chemistry converge in biochemistry, which is crucial to understanding the many complex factors and processes that keep living organisms (such as us) alive. Chemical engineering, materials science, and nanotechnology combine chemical principles and empirical findings to produce useful substances, ranging from gasoline to fabrics to electronics. Agriculture, food science, veterinary science, and brewing and wine making help provide sustenance in the form of food and drink to the world’s population. Medicine, pharmacology, biotechnology, and botany identify and produce substances that help keep us healthy. Environmental science, geology, oceanography, and atmospheric science incorporate many chemical ideas to help us better understand and protect our physical world. Chemical ideas are used to help understand the universe in astronomy and cosmology.
Figure \(2\): Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields.
What are some changes in matter that are essential to daily life? Digesting and assimilating food, synthesizing polymers that are used to make clothing, containers, cookware, and credit cards, and refining crude oil into gasoline and other products are just a few examples. As you proceed through this course, you will discover many different examples of changes in the composition and structure of matter, how to classify these changes and how they occurred, their causes, the changes in energy that accompany them, and the principles and laws involved. As you learn about these things, you will be learning chemistry, the study of the composition, properties, and interactions of matter. The practice of chemistry is not limited to chemistry books or laboratories: It happens whenever someone is involved in changes in matter or in conditions that may lead to such changes.
The Domains of Chemistry
Chemists study and describe the behavior of matter and energy in three different domains: macroscopic, microscopic, and symbolic. These domains provide different ways of considering and describing chemical behavior.
Macro is a Greek word that means “large.” The macroscopic domain is familiar to us: It is the realm of everyday things that are large enough to be sensed directly by human sight or touch. In daily life, this includes the food you eat and the breeze you feel on your face. The macroscopic domain includes everyday and laboratory chemistry, where we observe and measure physical and chemical properties, or changes such as density, solubility, and flammability.
The microscopic domain of chemistry is almost always visited in the imagination. Micro also comes from Greek and means “small.” Some aspects of the microscopic domains are visible through a microscope, such as a magnified image of graphite or bacteria. Viruses, for instance, are too small to be seen with the naked eye, but when we’re suffering from a cold, we’re reminded of how real they are.
However, most of the subjects in the microscopic domain of chemistry—such as atoms and molecules—are too small to be seen even with standard microscopes and often must be pictured in the mind. Other components of the microscopic domain include ions and electrons, protons and neutrons, and chemical bonds, each of which is far too small to see. This domain includes the individual metal atoms in a wire, the ions that compose a salt crystal, the changes in individual molecules that result in a color change, the conversion of nutrient molecules into tissue and energy, and the evolution of heat as bonds that hold atoms together are created.
The symbolic domain contains the specialized language used to represent components of the macroscopic and microscopic domains. Chemical symbols (such as those used in the periodic table), chemical formulas, and chemical equations are part of the symbolic domain, as are graphs and drawings. We can also consider calculations as part of the symbolic domain. These symbols play an important role in chemistry because they help interpret the behavior of the macroscopic domain in terms of the components of the microscopic domain. One of the challenges for students learning chemistry is recognizing that the same symbols can represent different things in the macroscopic and microscopic domains, and one of the features that makes chemistry fascinating is the use of a domain that must be imagined to explain behavior in a domain that can be observed.
A helpful way to understand the three domains is via the essential and ubiquitous substance of water. That water is a liquid at moderate temperatures, will freeze to form a solid at lower temperatures, and boil to form a gas at higher temperatures (Figure \(3\)) are macroscopic observations. But some properties of water fall into the microscopic domain—what we cannot observe with the naked eye. The description of water as comprised of two hydrogen atoms and one oxygen atom, and the explanation of freezing and boiling in terms of attractions between these molecules, is within the microscopic arena. The formula H2O, which can describe water at either the macroscopic or microscopic levels, is an example of the symbolic domain. The abbreviations (g) for gas, (s) for solid, and (l) for liquid are also symbolic.
Figure \(3\): (a) Moisture in the air, icebergs, and the ocean represent water in the macroscopic domain. (b) At the molecular level (microscopic domain), gas molecules are far apart and disorganized, solid water molecules are close together and organized, and liquid molecules are close together and disorganized. (c) The formula H2O symbolizes water, and (g), (s), and (l) symbolize its phases. Note that clouds are actually comprised of either very small liquid water droplets or solid water crystals; gaseous water in our atmosphere is not visible to the naked eye, although it may be sensed as humidity. (credit a: modification of work by “Gorkaazk”/Wikimedia Commons).
Summary
Chemistry deals with the composition, structure, and properties of matter, and the ways by which various forms of matter may be interconverted. Thus, it occupies a central place in the study and practice of science and technology. Chemists use the scientific method to perform experiments, pose hypotheses, and formulate laws and develop theories, so that they can better understand the behavior of the natural world. To do so, they operate in the macroscopic, microscopic, and symbolic domains. Chemists measure, analyze, purify, and synthesize a wide variety of substances that are important to our lives. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.01%3A_Atoms_and_Molecules.txt |
Learning Objectives
• To identify the components of 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 \(1\)).
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 and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 °C, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 °C. An example of a quantitative observation was the initial observation leading to the modern theory of the dinosaurs’ extinction: iridium concentrations in sediments dating to 66 million years ago were found to be 20–160 times higher than normal. The development of this theory is a good exemplar of the scientific method in action (see Figure \(2\) below).
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. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists collected additional data that either support or refute it.
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 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.
One example of a law, the Law of Definite Proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Some solid compounds do not strictly obey the law of definite proportions. The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure \(2\) summarizes the application of the scientific method in this case.
Example \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice were added to 100 mL of water at 25 °C, the temperature of the water decreased to 15.5 °C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Given: components of the scientific method
Asked for: statement classification
Strategy: Refer to the definitions in this section to determine which category best describes each statement.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are quantitative observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
5. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas.
6. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive.
Answer a
experiment
Answer b
law
Answer c
theory
Answer d
hypothesis
Answer e
qualitative observation
Answer f
quantitative observation
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
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. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.
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.
Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.02%3A_The_Scientific_Approach_to_Knowledge.txt |
Learning Objectives
• To classify matter.
Chemists study the structures, physical properties, and chemical properties of material substances. These consist of matter, which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter.
The mass of an object is the quantity of matter it contains. Do not confuse an object’s mass with its weight, which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 (Section 1.9). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth.
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.
Pure Substances 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!)
Crystallization separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration.
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.
Different Definitions of Matter: YouTube, Different Definitions of Matter(opens in new window) [youtu.be]
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 \(5\)). 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 \(6\).
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 Changes: YouTube, Different Definitions of Changes(opens in new window) [youtu.be] (opens in new window)
Summary
Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. 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. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.03%3A_The_Classification_of_Matter.txt |
Learning Objectives
• Describe the difference between physical and chemical properties or changed.
• Identify a property or transformation as either physical or chemical using symbolic, particulate, or macroscopic representations.
• Identify the properties of matter as extensive or intensive.
• Recognize and describe the parts of the NFPA hazard diamond.
The characteristics that enable us to distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change. A physical change is a change in the state or properties of matter without any accompanying change in its chemical composition (the identities of the substances contained in the matter). We observe a physical change when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water (Figure \(1\)). Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common antitheft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color). In each of these examples, there is a change in the physical state, form, or properties of the substance, but no change in its chemical composition.
The change of one type of matter into another type (or the inability to change) is a chemical property. Examples of chemical properties include flammability, toxicity, acidity, reactivity (many types), and heat of combustion. Iron, for example, combines with oxygen in the presence of water to form rust; chromium does not oxidize (Figure \(2\)). Nitroglycerin is very dangerous because it explodes easily; neon poses almost no hazard because it is very unreactive.
To identify a chemical property, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure \(3\)).
Properties of matter fall into one of two categories. If the property depends on the amount of matter present, it is an extensive property. The mass and volume of a substance are examples of extensive properties; for instance, a gallon of milk has a larger mass and volume than a cup of milk. The value of an extensive property is directly proportional to the amount of matter in question. If the property of a sample of matter does not depend on the amount of matter present, it is an intensive property. Temperature is an example of an intensive property. If the gallon and cup of milk are each at 20 °C (room temperature), when they are combined, the temperature remains at 20 °C. As another example, consider the distinct but related properties of heat and temperature. A drop of hot cooking oil spattered on your arm causes brief, minor discomfort, whereas a pot of hot oil yields severe burns. Both the drop and the pot of oil are at the same temperature (an intensive property), but the pot clearly contains much more heat (extensive property).
Hazard Diamond
You may have seen the symbol shown in Figure \(4\) on containers of chemicals in a laboratory or workplace. Sometimes called a “fire diamond” or “hazard diamond,” this chemical hazard diamond provides valuable information that briefly summarizes the various dangers of which to be aware when working with a particular substance.
The National Fire Protection Agency (NFPA) 704 Hazard Identification System was developed by NFPA to provide safety information about certain substances. The system details flammability, reactivity, health, and other hazards. Within the overall diamond symbol, the top (red) diamond specifies the level of fire hazard (temperature range for flash point). The blue (left) diamond indicates the level of health hazard. The yellow (right) diamond describes reactivity hazards, such as how readily the substance will undergo detonation or a violent chemical change. The white (bottom) diamond points out special hazards, such as if it is an oxidizer (which allows the substance to burn in the absence of air/oxygen), undergoes an unusual or dangerous reaction with water, is corrosive, acidic, alkaline, a biological hazard, radioactive, and so on. Each hazard is rated on a scale from 0 to 4, with 0 being no hazard and 4 being extremely hazardous.
While many elements differ dramatically in their chemical and physical properties, some elements have similar properties. We can identify sets of elements that exhibit common behaviors. For example, many elements conduct heat and electricity well, whereas others are poor conductors. These properties can be used to sort the elements into three classes: metals (elements that conduct well), nonmetals (elements that conduct poorly), and metalloids (elements that have properties of both metals and nonmetals).
The periodic table is a table of elements that places elements with similar properties close together (Figure \(5\)). You will learn more about the periodic table as you continue your study of chemistry.
Summary
All substances have distinct physical and chemical properties, and may undergo physical or chemical changes. Physical properties, such as hardness and boiling point, and physical changes, such as melting or freezing, do not involve a change in the composition of matter. Chemical properties, such flammability and acidity, and chemical changes, such as rusting, involve production of matter that differs from that present beforehand.
Measurable properties fall into one of two categories. Extensive properties depend on the amount of matter present, for example, the mass of gold. Intensive properties do not depend on the amount of matter present, for example, the density of gold. Heat is an example of an extensive property, and temperature is an example of an intensive property.
Glossary
chemical change
change producing a different kind of matter from the original kind of matter
chemical property
behavior that is related to the change of one kind of matter into another kind of matter
extensive property
property of a substance that depends on the amount of the substance
intensive property
property of a substance that is independent of the amount of the substance
physical change
change in the state or properties of matter that does not involve a change in its chemical composition
physical property
characteristic of matter that is not associated with any change in its chemical composition | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.04%3A_Physical_and_Chemical_Changes_and_Properties.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_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.05%3A_Energy-_A_Fundamental_Part_of_Physical_and_Chemical_Change.txt |
Learning Objectives
• Explain the process of measurement and describe the three basic parts of a quantity.
• Describe the properties and units of length, mass, volume, density, temperature, and time.
• Recognize the common unit prefixes and use them to describe the magnitude of a measurement.
• Describe and calculate the density of a substance.
• Perform basic unit calculations and conversions in the metric and other unit systems.
Measurements provide the macroscopic information that is the basis of most of the hypotheses, theories, and laws that describe the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: the size or magnitude of the measurement (a number); a standard of comparison for the measurement (a unit); and an indication of the uncertainty of the measurement. While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.
The number in the measurement can be represented in different ways, including decimal form and scientific notation. For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as 2.98 $\times$ 105 kg. The mass of the average mosquito is about 0.0000025 kilograms, which can be written as 2.5 $\times$ 10−6 kg.
Units, such as liters, pounds, and centimeters, are standards of comparison for measurements. When we buy a 2-liter bottle of a soft drink, we expect that the volume of the drink was measured, so it is two times larger than the volume that everyone agrees to be 1 liter. The meat used to prepare a 0.25-pound hamburger is measured so it weighs one-fourth as much as 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.
We usually report the results of scientific measurements in SI units, an updated version of the metric system, using the units listed in Table $1$. Other units can be derived from these base units. The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Système International d’Unités). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964.
Table $1$: Base Units of the SI System
Property Measured Name of Unit Symbol of Unit
length meter m
mass kilogram kg
time second s
temperature kelvin K
electric current ampere A
amount of substance mole mol
luminous intensity candela cd
Sometimes we use units that are fractions or multiples of a base unit. Ice cream is sold in quarts (a familiar, non-SI base unit), pints (0.5 quart), or gallons (4 quarts). We also use fractions or multiples of units in the SI system, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means “one thousand,” which in scientific notation is 103 (1 kilometer = 1000 m = 103 m). The prefixes used and the powers to which 10 are raised are listed in Table $2$.
NG, equals 4 times ten to the negative 9, or 0.000000004 g. The prefix micro has the greek letter mu as its symbol and a factor of 10 to the negative sixth power. Therefore, 1 microliter, or mu L, is equal to one times ten to the negative 6 or 0.000001 L. The prefix milli has a lowercase M as its symbol and a factor of 10 to the negative third power. Therefore, 2 millimoles, or M mol, are equal to two times ten to the negative 3 or 0.002 mol. The prefix centi has a lowercase C as its symbol and a factor of 10 to the negative second power. Therefore, 7 centimeters, or C M, are equal to seven times ten to the negative 2 meters or 0.07 M O L. The prefix deci has a lowercase D as its symbol and a factor of 10 to the negative first power. Therefore, 1 deciliter, or lowercase D uppercase L, are equal to one times ten to the negative 1 meters or 0.1 L. The prefix kilo has a lowercase K as its symbol and a factor of 10 to the third power. Therefore, 1 kilometer, or K M, is equal to one times ten to the third meters or 1000 M. The prefix mega has an uppercase M as its symbol and a factor of 10 to the sixth power. Therefore, 3 megahertz, or M H Z, are equal to three times 10 to the sixth hertz, or 3000000 H Z. The prefix giga has an uppercase G as its symbol and a factor of 10 to the ninth power. Therefore, 8 gigayears, or G Y R, are equal to eight times 10 to the ninth years, or 800000000 G Y R. The prefix tera has an uppercase T as its symbol and a factor of 10 to the twelfth power. Therefore, 5 terawatts, or T W, are equal to five times 10 to the twelfth watts, or 5000000000000 W." data-quail-id="64" data-mt-width="1076">
Table $2$: Common Unit Prefixes
Prefix Symbol Factor Example
femto f 10−15 1 femtosecond (fs) = 1 $\times$ 10−15 s (0.000000000000001 s)
pico p 10−12 1 picometer (pm) = 1 $\times$ 10−12 m (0.000000000001 m)
nano n 10−9 4 nanograms (ng) = 4 $\times$ 10−9 g (0.000000004 g)
micro µ 10−6 1 microliter (μL) = 1 $\times$ 10−6 L (0.000001 L)
milli m 10−3 2 millimoles (mmol) = 2 $\times$ 10−3 mol (0.002 mol)
centi c 10−2 7 centimeters (cm) = 7 $\times$ 10−2 m (0.07 m)
deci d 10−1 1 deciliter (dL) = 1 $\times$ 10−1 L (0.1 L )
kilo k 103 1 kilometer (km) = 1 $\times$ 103 m (1000 m)
mega M 106 3 megahertz (MHz) = 3 $\times$ 106 Hz (3,000,000 Hz)
giga G 109 8 gigayears (Gyr) = 8 $\times$ 109 yr (8,000,000,000 Gyr)
tera T 1012 5 terawatts (TW) = 5 $\times$ 1012 W (5,000,000,000,000 W)
SI Base Units
The initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters.
Length
The standard unit of length in both the SI and original metric systems is the meter (m). A meter was originally specified as 1/10,000,000 of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in 1/299,792,458 of a second. A meter is about 3 inches longer than a yard (Figure $1$); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers (1 km = 1000 m = 103 m), whereas shorter distances can be reported in centimeters (1 cm = 0.01 m = 10−2 m) or millimeters (1 mm = 0.001 m = 10−3 m).
Mass
The standard unit of mass in the SI system is the kilogram (kg). A kilogram was originally defined as the mass of a liter of water (a cube of water with an edge length of exactly 0.1 meter). It is now defined by a certain cylinder of platinum-iridium alloy, which is kept in France (Figure 1.4.2). Any object with the same mass as this cylinder is said to have a mass of 1 kilogram. One kilogram is about 2.2 pounds. The gram (g) is exactly equal to 1/1000 of the mass of the kilogram (10−3 kg).
Temperature
Temperature is an intensive property. The SI unit of temperature is the kelvin (K). The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word “degree” nor the degree symbol (°). The degree Celsius (°C) is also allowed in the SI system, with both the word “degree” and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at 273.15 K (0 °C) and boils at 373.15 K (100 °C) by definition, and normal human body temperature is approximately 310 K (37 °C). The conversion between these two units and the Fahrenheit scale will be discussed later in this chapter.
Time
The SI base unit of time is the second (s). Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds = 0.000003 s = 3 $\times$ 10−6 and 5 megaseconds = 5,000,000 s = 5 $\times$ 106 s. Alternatively, hours, days, and years can be used.
Derived SI Units
We can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.
Volume
Volume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure $3$). The standard volume is a cubic meter (m3), a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.
A more commonly used unit of volume is derived from the decimeter (0.1 m, or 10 cm). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter (dm3). A liter (L) is the more common name for the cubic decimeter. One liter is about 1.06 quarts. A cubic centimeter (cm3) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for cubic centimeter) is often used by health professionals. A cubic centimeter is also called a milliliter (mL) and is 1/1000 of a liter.
Density
We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.
The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m3). For many situations, however, this as an inconvenient unit, and we often use grams per cubic centimeter (g/cm3) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm3 (the density of gasoline) to 19 g/cm3 (the density of gold). The density of air is about 1.2 g/L. Table $3$ shows the densities of some common substances.
Table $3$: Densities of Common Substances
Solids Liquids Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3 water 1.0 g/cm3 dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3 ethanol 0.79 g/cm3 oxygen 1.31 g/L
iron 7.9 g/cm3 acetone 0.79 g/cm3 nitrogen 1.14 g/L
copper 9.0 g/cm3 glycerin 1.26 g/cm3 carbon dioxide 1.80 g/L
lead 11.3 g/cm3 olive oil 0.92 g/cm3 helium 0.16 g/L
silver 10.5 g/cm3 gasoline 0.70–0.77 g/cm3 neon 0.83 g/L
gold 19.3 g/cm3 mercury 13.6 g/cm3 radon 9.1 g/L
While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.
$\mathrm{density=\dfrac{mass}{volume}} \nonumber$
Example $1$
Calculation of Density Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm3. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?
Solution
The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.
$\mathrm{volume\: of\: lead\: cube=2.00\: cm\times2.00\: cm\times2.00\: cm=8.00\: cm^3} \nonumber$
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{90.7\: g}{8.00\: cm^3}=\dfrac{11.3\: g}{1.00\: cm^3}=11.3\: g/cm^3} \nonumber$
(We will discuss the reason for rounding to the first decimal place in the next section.)
Exercise $1$
1. To three decimal places, what is the volume of a cube (cm3) with an edge length of 0.843 cm?
2. If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?
Answer a
0.599 cm3;
Answer b
8.91 g/cm3
Example $2$: Using Displacement of Water to Determine Density
This PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.
Solution
When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density = 1.00 kg/L), and the water level rises to 101.25 L.
The red block therefore displaces 1.25 L water, an amount equal to the volume of the block. The density of the red block is:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{5.00\: kg}{1.25\: L}=4.00\: kg/L} \nonumber$
Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{5.00\: kg}{10.00\: L}=0.500\: kg/L} \nonumber$
Exercise $1$
Remove all of the blocks from the water and add the green block to the tank of water, placing it approximately in the middle of the tank. Determine the density of the green block.
Answer
2.00 kg/L
Summary
Measurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison, and an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use the SI (International System) or metric systems. We use base SI units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and g/cm3 (for density). In many cases, we find it convenient to use unit prefixes that yield fractional and multiple units, such as microseconds (10−6 seconds) and megahertz (106 hertz), respectively.
Key Equations
• $\mathrm{density=\dfrac{mass}{volume}}$
Glossary
Celsius (°C)
unit of temperature; water freezes at 0 °C and boils at 100 °C on this scale
cubic centimeter (cm3 or cc)
volume of a cube with an edge length of exactly 1 cm
cubic meter (m3)
SI unit of volume
density
ratio of mass to volume for a substance or object
kelvin (K)
SI unit of temperature; 273.15 K = 0 ºC
kilogram (kg)
standard SI unit of mass; 1 kg = approximately 2.2 pounds
length
measure of one dimension of an object
liter (L)
(also, cubic decimeter) unit of volume; 1 L = 1,000 cm3
meter (m)
standard metric and SI unit of length; 1 m = approximately 1.094 yards
milliliter (mL)
1/1,000 of a liter; equal to 1 cm3
second (s)
SI unit of time
SI units (International System of Units)
standards fixed by international agreement in the International System of Units (Le Système International d’Unités)
unit
standard of comparison for measurements
volume
amount of space occupied by an object | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.06%3A_The_Units_of_Measurement.txt |
Learning Objectives
• Compare and contrast exact and uncertain numbers.
• Correctly represent uncertainty in quantities using significant figures.
• Identify the number of significant figures in value.
• Solve problems that involve various calculations and report the results with the appropriate number of significant figures.
• Apply proper rounding rules to computed quantities
• Define accuracy and precision, and use accuracy and precision to describe data sets.
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.
Significant Figures in Measurement
The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. 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.
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.
Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.
Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.
Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.
The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 $\times$ 10−3; then the number 8.32407 contains all of the significant figures, and 10−3 locates the decimal point.
The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 $\times$ 103 (two significant figures), 1.30 $\times$ 103 (three significant figures, if the tens place was measured), or 1.300 $\times$ 103 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.
When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely—that is, whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 317 million, or $3.17 \times 10^8$ people.
Significant Figures in Calculations
A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:
1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)
The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:
• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)
Let’s work through these rules with a few examples.
Example $1$: Rounding Numbers
Round the following to the indicated number of significant figures:
1. 31.57 (to two significant figures)
2. 8.1649 (to three significant figures)
3. 0.051065 (to four significant figures)
4. 0.90275 (to four significant figures)
Solution
1. 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
2. 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
3. 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
4. 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Exercise $1$
Round the following to the indicated number of significant figures:
1. 0.424 (to two significant figures)
2. 0.0038661 (to three significant figures)
3. 421.25 (to four significant figures)
4. 28,683.5 (to five significant figures)
Answer a
0.42
Answer b
0.00387
Answer c
421.2
Answer d
28,684
Example $2$: Addition and Subtraction with Significant Figures Rule:
When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
1. Add 1.0023 g and 4.383 g.
2. Subtract 421.23 g from 486 g.
Solution
(a)
\begin{align*} &\mathrm{1.0023\: g}\ +\: &\underline{\mathrm{4.383\: g}\:\:}\ &\mathrm{5.3853\: g} \end{align*} \nonumber
Answer is 5.385 g (round to the thousandths place; three decimal places)
(b)
\begin{align*} &\mathrm{486\: g}\ -\: &\underline{\mathrm{421.23\: g}}\ &\mathrm{\:\:64.77\: g} \end{align*} \nonumber
Answer is 65 g (round to the ones place; no decimal places)
Exercise $2$
1. Add 2.334 mL and 0.31 mL.
2. Subtract 55.8752 m from 56.533 m.
Answer a
2.64 mL
Answer b
0.658 m
Example $3$: Multiplication and Division with Significant Figures
Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
1. Multiply 0.6238 cm by 6.6 cm.
2. Divide 421.23 g by 486 mL.
Solution
(a)
$\mathrm{0.6238\: cm\times6.6\:cm=4.11708\:cm^2\rightarrow result\: is\:4.1\:cm^2}\:\textrm{(round to two significant figures)} \nonumber$
$\textrm{four significant figures}\times \textrm{two significant figures}\rightarrow \textrm{two significant figures answer} \nonumber$
(b)
$\mathrm{\dfrac{421.23\: g}{486\: mL}=0.86728...\: g/mL\rightarrow result\: is\: 0.867\: g/mL} \: \textrm{(round to three significant figures)} \nonumber$
$\mathrm{\dfrac{five\: significant\: figures}{three\: significant\: figures}\rightarrow three\: significant\: figures\: answer} \nonumber$
Exercise $3$
1. Multiply 2.334 cm and 0.320 cm.
2. Divide 55.8752 m by 56.53 s.
Answer a
0.747 cm2
Answer b
0.9884 m/s
In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.
Example $4$: Calculation with Significant Figures
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
Solution
\begin{align*} V&=l\times w\times d\ &=\mathrm{13.44\: dm\times 5.920\: dm\times 2.54\: dm}\ &=\mathrm{202.09459...dm^3}\:\textrm{(value from calculator)}\ &=\mathrm{202\: dm^3,}\textrm{ or 202 L (answer rounded to three significant figures)} \end{align*} \nonumber
Exercise $4$: Determination of Density Using Water Displacement
What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm3?
Answer
1.034 g/mL
Example $4$
A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
1. Use these values to determine the density of this piece of rebar.
2. Rebar is mostly iron. Does your result in (a) support this statement? How?
Solution
The volume of the piece of rebar is equal to the volume of the water displaced:
$\mathrm{volume=22.4\: mL-13.5\: mL=8.9\: mL=8.9\: cm^3}\nonumber$
(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
The density is the mass-to-volume ratio:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{69.658\: g}{8.9\: cm^3}=7.8\: g/cm^3}\nonumber$
(rounded to two significant figures, per the rule for multiplication and division)
The density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.
Exercise $4$
An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.
1. Use these values to determine the density of this material.
2. Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.
Answer a
19 g/cm3
Answer b
It is likely gold; it has the right appearance for gold and very close to the density given for gold.
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 $2$).
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_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.07%3A_The_Reliability_of_a_Measurement.txt |
Learning Objectives
• Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities.
• Describe how to use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties.
• Convert between the three main temperature units: Fahrenheit, Celsius, and Kelvin.
It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:
$\mathrm{speed=\dfrac{distance}{time}} \nonumber$
An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of
$\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s} \nonumber$
Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:
$\mathrm{time=\dfrac{distance}{speed}} \nonumber$
The time can then be computed as:
$\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s} \nonumber$
Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”
These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.
Conversion Factors and Dimensional Analysis
A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,
$\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}} \nonumber$
Several other commonly used conversion factors are given in Table $1$.
Table $1$: Common Conversion Factors
Length Volume Mass
1 m = 1.0936 yd 1 L = 1.0567 qt 1 kg = 2.2046 lb
1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g
1 km = 0.62137 mi 1 ft3 = 28.317 L 1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m 1 tbsp = 14.787 mL 1 (troy) oz = 31.103 g
When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:
$\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm} \nonumber$
Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield
$\mathrm{\dfrac{in.\times cm}{in.}}. \nonumber$
Just as for numbers, a ratio of identical units is also numerically equal to one,
$\mathrm{\dfrac{in.}{in.}=1} \nonumber$
and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.
Example $1$: Using a Unit Conversion Factor
The mass of a competition Frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table $1$).
Solution
If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.
$x\:\mathrm{oz=125\: g\times unit\: conversion\: factor}\nonumber$
We write the unit conversion factor in its two forms:
$\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}}\nonumber$
The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.
\begin{align*} x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\ &=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\ &=\mathrm{4.41\: oz\: (three\: significant\: figures)} \end{align*} \nonumber
Exercise $1$
Convert a volume of 9.345 qt to liters.
Answer
8.844 L
Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.
Example $2$: Computing Quantities from Measurement Results
What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.
Solution
Since $\mathrm{density=\dfrac{mass}{volume}}$, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A $\times$ unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:
$\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g}\nonumber$
We need to use two steps to convert volume from quarts to milliliters.
1. Convert quarts to liters.
$\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L}\nonumber$
1. Convert liters to milliliters.
$\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL}\nonumber$
Then,
$\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL}\nonumber$
Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:
$\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL}\nonumber$
Exercise $2$
What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?
Answer
$\mathrm{2.956\times10^{-2}\:L}$
Example $3$: Computing Quantities from Measurement Results
While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.
1. What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
Answer a
51 mpg
Answer b
\$62
Conversion of Temperature Units
We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler.
To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).
Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:
$\mathrm{length\: in\: feet=\left(\dfrac{1\: ft}{12\: in.}\right)\times length\: in\: inches} \nonumber$
where
• y = length in feet,
• x = length in inches, and
• the proportionality constant, m, is the conversion factor.
The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one ($y = mx + b$). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points ($b$).
The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as $x$ and the Fahrenheit temperature as $y$, the slope, $m$, is computed to be:
\begin{align*} m &=\dfrac{\Delta y}{\Delta x} \[4pt] &= \mathrm{\dfrac{212\: ^\circ F - 32\: ^\circ F}{100\: ^\circ C-0\: ^\circ C}} \[4pt] &= \mathrm{\dfrac{180\: ^\circ F}{100\: ^\circ C}} \[4pt] &= \mathrm{\dfrac{9\: ^\circ F}{5\: ^\circ C} }\end{align*} \nonumber
The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:
\begin{align*} b&=y-mx \[4pt] &= \mathrm{32\:^\circ F-\dfrac{9\:^\circ F}{5\:^\circ C}\times0\:^\circ C} \[4pt] &= \mathrm{32\:^\circ F} \end{align*} \nonumber
The equation relating the temperature scales is then:
$\mathrm{\mathit{T}_{^\circ F}=\left(\dfrac{9\:^\circ F}{5\:^\circ C}\times \mathit{T}_{^\circ C}\right)+32\:^\circ C} \nonumber$
An abbreviated form of this equation that omits the measurement units is:
$\mathrm{\mathit{T}_{^\circ F}=\dfrac{9}{5}\times \mathit{T}_{^\circ C}+32} \nonumber$
Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:
$\mathrm{\mathit{T}_{^\circ C}=\dfrac{5}{9}(\mathit{T}_{^\circ F}+32)} \nonumber$
As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).
The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of $\mathrm{1\:\dfrac{K}{^\circ\:C}}$. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:
$T_{\ce K}=T_{\mathrm{^\circ C}}+273.15 \nonumber$
$T_\mathrm{^\circ C}=T_{\ce K}-273.15 \nonumber$
The 273.15 in these equations has been determined experimentally, so it is not exact. Figure $1$ shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.
Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.
Example $4$: Conversion from Celsius
Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?
Solution
$\mathrm{K= {^\circ C}+273.15=37.0+273.2=310.2\: K}\nonumber$
$\mathrm{^\circ F=\dfrac{9}{5}\:{^\circ C}+32.0=\left(\dfrac{9}{5}\times 37.0\right)+32.0=66.6+32.0=98.6\: ^\circ F}\nonumber$
Exercise $4$
Convert 80.92 °C to K and °F.
Answer
354.07 K, 177.7 °F
Example $5$: Conversion from Fahrenheit
Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?
Solution
$\mathrm{^\circ C=\dfrac{5}{9}(^\circ F-32)=\dfrac{5}{9}(450-32)=\dfrac{5}{9}\times 418=232 ^\circ C\rightarrow set\: oven\: to\: 230 ^\circ C}\hspace{20px}\textrm{(two significant figures)}\nonumber$
$\mathrm{K={^\circ C}+273.15=230+273=503\: K\rightarrow 5.0\times 10^2\,K\hspace{20px}(two\: significant\: figures)}\nonumber$
Exercise $5$
Convert 50 °F to °C and K.
Answer
10 °C, 280 K
Summary
Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.
Key Equations
• $T_\mathrm{^\circ C}=\dfrac{5}{9}\times T_\mathrm{^\circ F}-32$
• $T_\mathrm{^\circ F}=\dfrac{9}{5}\times T_\mathrm{^\circ C}+32$
• $T_\ce{K}={^\circ \ce C}+273.15$
• $T_\mathrm{^\circ C}=\ce K-273.15$
Glossary
dimensional analysis
(also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities
Fahrenheit
unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale
unit conversion factor
ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.08%3A_Solving_Chemical_Problems.txt |
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. 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.
Q1.35
A household receives a $125 electricity bill. The cost of electricity is$0.150 /kWh. How much energy, in joules (J), did the household use?
Strategy
Step 1: You must first multiply the total cost of the bill ($125) by 1 kWh and then divide that answer by the cost of each kW ($0.150 kWh). You should end up with an equation that looks like this:
$(125)\times \dfrac{1\;kWh}{0.150\;kWh}\nonumber$
Step 2: Once you do this math, then you should receive the answer of 833.33 kWh. This is the number of kWh used during the billing cycle.
Step 3: Next, you will then take your solution found during step 1 and multiply it by $\dfrac{1}{2.78\times 10^{-7}}$ in order to convert kWh to kWh/J. You should have the equation:
$(833.33)\times \dfrac{1}{2.78\cdot 10^{-7}}\nonumber$
Step 4: After completing step 3, you should receive an answer of $3.0 \times 10^{9} J\nonumber$
Step 5: YOU ARE FINISHED!
Q1.36
Determine whether the mixtures are Homogeneous or Heterogeneous. State whether it is a compound or pure substance?
1. Mud and Water
2. Salt and water
3. Chicken noodle soup
4. Gold
5. $H_{2}O$
Strategy
• Homogeneous: mixtures are uniformly distributed, easily dissolved, of a single phase, also they are easily dispersed through the membrane.
• Heterogeneous: mixtures are unequally distributed, do not dissolve in water, and could be of different phases. A pure substance: has a definite and constant composition, it can be a compound or and element.
• A element: is composed of a single atom.
• A compound is composed of two or more elements. For example, $H_{2}O$ is a ratio of two hydrogen atoms and one oxygen atom to form $H_{2}O$. So with this strategy you can solve the following question.
1. Mud and Water: Heterogeneous mixture
2. Salt and water: Homogeneous mixture
3. Chicken noodle soup: Heterogeneous mixture
4. Gold: (element)
5. $H_{2}O$: Compound
Q1.36
Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous.
1. Plain yogurt
2. Chicken noodle soup
3. Titanium
4. Table salt
What we know:
• A pure substance is made up of only one type of particle.
• A mixture is a substance composed of two or more particles.
• An element is a substance that cannot be chemically broken down into simpler substances.
• A compound is a substance composed of two or more elements.
• A heterogeneous mixture is one in which the composition varies from one region of the mixture to another.
• A homogeneous mixture is one with the same composition throughout.
Solution
1. mixture, homogeneous
2. mixture, heterogeneous
3. pure substance, element
4. pure substance, compound
Q1.41
Classify Each statement as an Observation, Theory, or Law.
1. All matter that exists is made up of Submicroscopic particles known as atoms.
2. If an iron rod is placed in a closed container and it rusts, the mass of the container and its contents do not change.
3. Matter is neither created nor destroyed
4. When a candle is burned, heat is released
Strategy
You need to understand what an Observation, Theory or law is before you classify it in the question.
• Observation: An outcome of events that is being viewed
• Theory: Preserved explanation for observations and laws
• Law: Summarizes past observations and predicts future ones
Now that we have defined them, we need to apply it to the question (AnswerS IN BOLD)
1. This statement is proposing all matter is made up of atoms so it would become a theory because it is an idea that studies propose is true.
2. This statement is saying, without matter being added to the container, As you See it rust, the mass doesn't change so this is assumed by Observation.
3. This statement is called the Law of conservation of mass, so because this is true, the statement is a Law.
4. If you are looking at a candle and see the flame, then you should understand that heat is being released. You tell this from Observation.
Q1.45
When 12 g of sodium reacts with 23.5 g of chlorine to form sodium chloride, how many grams of sodium chloride is formed? (Assume that sodium chloride is the only product).
Solution:
What we know: We have a chemical reaction between sodium and chlorine to form only sodium chloride. We are also given the masses of the reactants, 12 g of sodium and 23.5 g of chlorine. We also know that the Law of Conservation of Mass tells us that matter is neither created nor destroyed in a chemical reaction.
What we're asked for: The mass, in grams, of sodium chloride formed from the reaction.
Strategy:
A. Find the sum of the masses of the reactants from the chemical equation.
Solution
12 g + 23.5 g = 35.5 g
35.5 g of sodium chloride is formed from the reaction.
Q1.55
According to Dalton's Atomic Theory, what are compounds made up of?
Strategy
1. Find the summary of Dalton's Atomic Theory
2. Read through it
3. Locate where it talks about what compounds are made of
4. Write your answer
Solution
Compounds are made up of two or more different atoms.
Q1.64
Isotopes are atoms that have the same number of protons but differ in the number of neutrons. Write the chemical symbols for each of the following isotopes:
1. the chlorine isotope with 18 neutrons
2. the chlorine isotope with 20 neutrons
3. the calcium isotope with 20 neutrons
4. the neon isotope with 11 neutrons
Strategy:
1. Determine the given element's atomic number from the periodic table.
2. Add the number of neutrons given in the question to the atomic number to calculate the mass number of the isotope.
3. The mass number will be represented in the upper left corner of the chemical symbol, while the atomic number will be represented in the lower left corner.
Helpful Hints:
• A chemical symbol is written as $_{Z}^{A}\textrm{X}\nonumber$ . The A stands for the number of protons and neutrons, also known as the mass number. The Z stands for the number of protons, also known as the atomic number.
• Subtracting Z from A will give you the number of neutrons. Example: $_{19}^{39}\textrm{K}\nonumber$ --> 39-19= 20 neutrons
• Therefore, adding the given number of neutrons to the atomic number, Z, gives you the mass number.
Solution:
1. $_{17}^{35}\textrm{Cl}\nonumber$
2. $_{17}^{37}\textrm{Cl}\nonumber$
3. $_{20}^{40}\textrm{Ca}\nonumber$
4. $_{10}^{21}\textrm{Ne}\nonumber$
Q1.65
Give the number of protons and neutrons in each isotope listed
1. $_{19}^{41}\textrm{K}\nonumber$
2. $_{4}^{11}\textrm{Be}\nonumber$
3. $_{22}^{45}\textrm{Ti}\nonumber$
4. $_{8}^{16}\textrm{O}\nonumber$
Strategy
Take the larger number and subtract the smaller number from it. The smaller number is the atomic number (z), which indicates the number of protons the element has. The larger number indicates the number of protons and neutron present in the element.
1. 41K has an atomic mass of 41. The atomic mass is the number of protons and neutrons present in the element. To get the number of neutrons, subtract the atomic number (19) from the atomic mass (41). When done so you find that K41 has 19 protons and 22 neutrons. Keep in mind the atomic number is the number of protons present in the element, so the only thing you are trying to find through subtraction is the number of neutrons.
2. For 11Be, the same thing occurs. Take the atomic number and subtract it from the atomic mass. The atomic mass is 11 and the atomic number is 4, so, 11 minus for equals 7. The answer for part b. is 4 protons and 7 neutrons.
3. For 45 Ti, you subtract 22 from 45, which gives you 22 protons and 23 neutrons.
4. For 16O, the same method is applied. Subtract 8 from 16 and you get 8 protons and 8 electrons.
Q1.66
Question: Find the amount of protons and neutrons in the following atoms:
1. $_{6}^{14}\textrm{C}\nonumber$
2. $_{8}^{18}\textrm{O}\nonumber$
3. $_{25}^{55}\textrm{Mn}\nonumber$
4. $_{30}^{64}\textrm{Zn}\nonumber$
Strategy
1. First identify the subscript, the bottom number on the left hand side of the element's symbol. This is the atomic number and indicates how many protons the atom's nucleus contains. All isotopes of the same element contain the same number of protons.
2. Next, identify the superscript, the top number left hand side of the element symbol. This is the atomic mass number, the total combined number of protons and neutrons. Different isotopes of a given element differ in the number of neutrons contained in their nuclei.
3. The number of protons can be found simply by identifying the atomic number. To find the number of neutrons in the atom, the atomic number is subtracted from the atomic mass number.
Solution
A The atomic number of carbon is 6, the atomic number of oxygen is 8, the atomic number of manganese is 25, and the atomic number of zinc is 30. Therefore, we now know the number of protons contained within each atom since this number is represented by the atomic number.
B The atomic mass number of the carbon isotope is 14, the mass number of the oxygen isotope is 18, the mass number of the manganese isotope is 55, and the mass number of the zinc isotope is 64. This is the combined number of protons and neutrons in each respective isotope.
C Carbon isotope: $14-6=8\, neutrons\nonumber$
Oxygen isotope: $18-8=10\, neutrons\nonumber$
Manganese isotope: $55-25=30\, neutrons\nonumber$
Zinc isotope: $64-30=34\, neutrons\nonumber$
Q1.64
Write the isotopic symbols in the form $\ce{^A_ZX}$ for each isotope.
1. The oxygen isotope with 9 neutrons.
2. The neon isotope with 10 neutrons.
3. The chlorine isotope with 18 neutrons.
4. The carbon isotope with 8 neutrons.
Strategy
1. Identify each element’s atomic number (number of protons).
2. Find the atomic mass of each element by adding the number of protons and the number of neutrons.
3. Properly write the isotopic symbol of each element by placing the atomic mass as the superscript and the atomic number (number of protons) as the subscript.
Solution
1. Atomic Numbers (Number of Protons):
Oxygen: 8 protons
Neon: 10 protons
Chlorine: 17 protons
Carbon: 6 protons
1. Atomic Mass:
Oxygen: 8+9=17
Neon: 10+10=20
Chlorine: 17+18=35
Carbon: 6+8=14
1. Isotopic Symbols:
Oxygen: 178O
Neon: 2010Ne
Chlorine: 3517Cl
Carbon: 146C
Q1.69
Question: Determine the number of protons and electrons in each of the following ions.
1. $\small N^{3-}$
2. $\small Be^{2+}$
3. $\small K^{+}$
4. $\small Se^{2-}$
Strategy
1. First locate each element on the periodic table. Identify the atomic number associated with each element, or the number in between the element's symbol and scientific name. The atomic number is the number of protons as well as the number of electrons an element possesses in it's neutral state.
2. Next, identify the superscript of each element, or the number to the top right of the element's symbol. This number should include a "+" or "-" sign next to it, indicating the element's charge. A "+" indicates a positively charged ion and a "-" indicates a negatively charged ion. An element becomes positively charged when it loses electrons, and negative when it gains electrons.
3. To solve for each element's total number of electrons, take that element's atomic number, and inversely add or subtract the number in the superscript according to the element's charge. If an element has a "+" sign, it is losing the amount of electrons indicated by the superscript number, so subtract this number from the element's atomic number. The same goes if an element has a "-" sign, instead adding the number in the superscript to the element's atomic number.
Solution
(1) N3-
A. Nitrogen's atomic number is 7, so in a neutral state, Nitrogen has 7 protons and electrons.
B. Nitrogen's superscript is "3-" indicating that it is gaining 3 electrons
C. Because it gains 3 electrons, N3- now has a total of 10 electrons.
(2) Be2+
A. Beryllium's atomic number is 4, so Beryllium possesses 4 protons and electrons
B. The superscript "2+" indicates beryllium is losing 2 electrons
C. 2 electrons are subtracted from Beryllium's total 4, so it now has 2 electrons.
(3) K+
A. Potassium as a neutral element possesses 19 protons and electrons, indicated by it's atomic number of 19.
B. The superscript "+" indicates Potassium will lose 1 electron from it's total.
C. 19 total electrons minus 1 makes K+ number of electrons to 18.
(4) Se2-
A. Selenium's atomic number is 34, indicating 34 protons and electrons at ground state.
B. It' superscript "2-" indicates that it will be gaining 2 electrons to it's total.
C. 34 electrons plus 2 makes selenium's total number of electrons 34.
Q 1.73
The atomic mass for Helium is 4.003 amu and the mass spectrum for Helium shows a spike at this mass. Meanwhile, the atomic mass for Bromine is 79.904 amu; however, the mass spectrum for Bromine does not show a peak at that mass. What causes the difference?
Solution
1. The best way to start this problem is to recall what the atomic mass means in the context of the problem. This is the standard atomic mass, which is the weighted average of all the isotopes. "Weighted" in this situation means that the abundance of each isotope factors in to how it factors in to the overall the standard atomic mass for that particular element.
2. Next step is to look at Helium. Helium has two isotopes and is stable at either; however, in atmospheric values, Helium is found most abundantly as 4He (99.999863%) and much less commonly found as 3He (0.000137%). Consequently, 3He does not have a strong influence on the standard atomic mass of Helium or much expression on the mass spectrum which measures relative abundance.
3. Examining Bromine reveals two major isotopes for that element as well. Both 79Br as well as 81Br are rather prevalent (79Br with 50.69% and 81Br with 49.31%). This means that their relative abundance is high enough that they would make a significant impact on both the mass spectrum as well as the standard atomic weight of Bromine.
4. To connect the dots, the dichotomy between the two elements comes from the abundance of their isotopes. Bromine's two isotopes almost equally comprise the atomic mass so the atomic mass they produce does not match up with an actual relatively abundant isotope of Br so therefore the line for that particular atomic mass is absent on the mass spectrum for Br. Helium, meanwhile, is primarily influenced by the 3He because of its incredibly high abundance and therefore the mass spectrum and the standard atomic mass are very similar.
Q1.74
The atomic mass of chlorine is 35.453 amu. There are two stable isotopes for chlorine. Would either of these isotopes have a mass of 35.453 amu? Why or why not?
Important things to consider
• The atomic mass is the sum of the protons and neutrons.
• Isotopes vary in the amount of neutrons of the same element.
• Each isotope will have a different mass number depending upon how many neutrons are present.
• amu = atomic mass unit
• unit used for indicating mass on an atomic scale
Answer
No, because isotopes are variants of an element that differ in the number of neutrons present, but they have the same number of protons (atomic number the same). | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/01%3A_Matter_Measurement_and_Problem_Solving/1.E%3A_Matter_Measurement_and_Problem_Solving_%28Exercises%29.txt |
• 2.2: Early Ideas about the Building Blocks of Matter
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass.
• 2.3: Modern Atomic Theory and the Laws That Led to It
Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass.
• 2.4: The Discovery of the Electron
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. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays.
• 2.5: The Structure of The Atom
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $1/12$ of the mass of a carbon-12 atom and is equal to 1.6605 $\times$ 10−24 g.
• 2.6: Protons, Neutrons, and Electrons in Atoms
• 2.7: Finding Patterns - 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.
• 2.8: The Average Mass of an Element’s Atoms
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, and 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 different mass number.
• 2.9: Molar Mass - Counting Atoms by Weighing Them
The chemical changes we observe always involve discrete numbers of atoms that rearrange themselves into new configurations. These numbers are far too large in magnitude for us to count , but they are still numbers, and we need to have a way to deal with them. We also need a bridge between these numbers, which we are unable to measure directly, and the weights of substances, which we do measure and observe. The mole concept provides this bridge, and is key to all of quantitative chemistry.
• 2.E: Atoms and Elements (Exercises)
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro.
02: Atoms and Elements
Chemistry is the study of the material world. What are different materials made of? How is their composition and structure related to their properties? How does one material become transformed into another? These are the sorts of questions that have driven the development of chemistry. People have been using chemistry for a very long time. Medicines were obtained from plants in early societies all over the world. People made dyes for clothing and paints for houses. Metallurgy was practiced in India and the Sahel, in Africa, before 1000 BC.
The Greek philosopher, Democritus, is often cited as the earliest person to formulate an idea of atoms, although similar ideas were recorded in India around the same time. Democritus thought that all things were made of atoms. Atoms were very small, he thought. They were also indivisible. Although you could cut a piece of wood in half, and cut each of those pieces in half, at some point you would reach the stage at which the wood could not be cut any longer, because you had a slice that was one atom thick.
There were an infinite variety of atoms, Democritus thought, making up an infinite variety of materials in the world. The properties of those atoms were directly responsible for the properties of materials. Water was made of water atoms, and water atoms were slippery. Iron was made of iron atoms, and iron atoms were strong and hard.
All of the materials in the world around us are made from atoms.
A more practical aspect of chemistry has its roots in the Islamic Golden Age. Practitioners such as Jabir Ibn Hayyan developed laboratory apparatus and experimental methods to recrystallize and distill compounds from natural sources. Like Democritus, these early chemists wanted to know what the world is made of, but they were also trying to make improvements in practical applications such as tanning leather, making glass or rust-proofing iron.
The translation of Arabic texts into Latin helped spur the European Renaissance. Practical observations from the Islamic period, such as the fact that matter could be converted into different forms but did not disappear, gave rise to some of the most fundamental ideas of modern chemistry.
Conservation of mass: matter can be converted from one form to another, but it does not disappear.
As is usually true in science, new developments in chemistry built on earlier work as well as the work of contemporary colleagues, continually improving our understanding of nature in small steps. In the 1600's, Robert Boyle adopted the Islamic emphasis on experimental work. Among his experiments, he was able to isolate the hydrogen gas formed by reacting metals with acids, as other scientists were doing at that time. In the 1700's, Joseph Priestley isolated several different "airs" or gases, including oxygen. Henry Cavendish showed that hydrogen combined with oxygen to form water. Antoine Lavoisier argued that oxygen and nitrogen, the other major component of air, are elements. The free exchange of ideas allowed people to rapidly advance our understanding of the material world.
Lavoisier, in particular, was important in bringing a number of important ideas together. He clearly stated that elements were the basic unit of matter, that could not be obtained from other materials. Compounds were made by combining different elements. His careful use of a balance to weigh reactants and products of an experiment clearly confirmed the idea of conservation of mass: the total mass of products after a reaction equals the total mass of reactants. These conclusions were more sophisticated versions of earlier ideas, and Lavoisier was able to present them in a way that eventually gained wide acceptance.
• A compound is a mixture of different elements bonded together.
• Conservation of mass, revised: the total mass of products after a reaction equals the total mass of reactants.
In the 1800's one of the principle proponents of the developing atomic theory was John Dalton. He advanced the idea that all atoms of a particular element are identical (as far as he could tell at that time). An element is a fundamental atomic building block from which other materials are made. Dalton performed analyses to try to deduce the atomic weights of different elements. Taking these ideas together, he showed that a particular compound always contained the same elements in the same ratio.
• An element is a fundamental atomic building block from which other materials are made.
• A compound is a mixture of different elements bonded together in a specific ratio.
• A compound may have a specific number of atoms of one type combined with a specific number of atoms of another type.
• Because all atoms have weight, we can also think of a compound as a specific weight of one kind of element combined with a specific weight of another element.
For example, water is a compound made from hydrogen and oxygen. It is crucial for life, of course. Water is about 1/9th hydrogen by weight; the other 8/9ths are oxygen. However, a different compound, hydrogen peroxide, is a rocket fuel. Hydrogen peroxide is only about 1/19th hydrogen by weight. Those specific ratios of hydrogen to oxygen are inherent qualities of each compound.
Furthermore, Dalton found that he could make compounds through different methods. For example, he could make cupric oxide (CuO) by heating copper in air, or he could make it through various reactions involving copper and acids. It didn't matter how he made the cupric oxide; the ratio of copper to oxygen was always the same in the product.
There is one other compound containing copper and oxygen in a different ratio; it is called cuprous oxide, and it has the formula Cu2O. However, it is very different from cupric oxide. The most obvious difference is that cuprous oxide is red whereas cupric oxide is black. Once again, when elements are combined in different ratios, different materials are produced, and they have properties that differ from each other and from they elements of which they are comprised. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.02%3A_Early_Ideas_about_the_Building_Blocks_of_Matter.txt |
Learning Objectives
By the end of this section, you will be able to:
• State the postulates of Dalton’s atomic theory
• Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions
The language used in chemistry is seen and heard in many disciplines, ranging from medicine to engineering to forensics to art. The language of chemistry includes its own vocabulary as well as its own form of shorthand. Chemical symbols are used to represent atoms and elements. Chemical formulas depict molecules as well as the composition of compounds. Chemical equations provide information about the quality and quantity of the changes associated with chemical reactions.
This chapter will lay the foundation for our study of the language of chemistry. The concepts of this foundation include the atomic theory, the composition and mass of an atom, the variability of the composition of isotopes, ion formation, chemical bonds in ionic and covalent compounds, the types of chemical reactions, and the naming of compounds. We will also introduce one of the most powerful tools for organizing chemical knowledge: the periodic table.
Atomic Theory through the Nineteenth Century
The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four “elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and “elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas.
The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton’s atomic theory.
1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.
2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure $1$). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties.
1. Atoms of one element differ in properties from atoms of all other elements.
2. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure $2$).
1. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure $3$).
Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter).
Example $1$: Testing Dalton’s Atomic Theory
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
Solution
The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)
Exercise $1$
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one
Answer
The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.
Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant composition. 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.
Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers. For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.
$\mathrm{\dfrac{\dfrac{1.116\: g\: Cl}{1\: g\: Cu}}{\dfrac{0.558\: g\: Cl}{1\: g\: Cu}}=\dfrac{2}{1}} \nonumber$
This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.
This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure $4$).
Example $2$: Laws of Definite and Multiple Proportions
A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?
Solution
In compound A, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{1.33\: g\: O}{1\: g\: C}} \nonumber$
In compound B, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{2.67\: g\: O}{1\: g\: C}} \nonumber$
The ratio of these ratios is:
$\mathrm{\dfrac{\dfrac{1.33\: g\: O}{1\: g\: C}}{\dfrac{2.67\: g\: O}{1\: g\: C}}=\dfrac{1}{2}} \nonumber$
This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO2 and B = CO.
Exercise $1$
A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?
Answer
In compound X, the mass ratio of carbon to hydrogen is $\mathrm{\dfrac{14.13\: g\: C}{2.96\: g\: H}}$.
In compound Y, the mass ratio of carbon to oxygen is $\mathrm{\dfrac{19.91\: g\: C}{3.34\: g\: H}}$.
The ratio of these ratios is
$\mathrm{\dfrac{\dfrac{14.13\: g\: C}{2.96\: g\: H}}{\dfrac{19.91\: g\: C}{3.34\: g\: H}}=\dfrac{4.77\: g\: C/g\: H}{5.96\: g\: C/g\: H}=0.800=\dfrac{4}{5}}. \nonumber$
This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.
Summary
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.
Glossary
Dalton’s atomic theory
set of postulates that established the fundamental properties of atoms
law of constant composition
(also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass
law of multiple proportions
when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers
law of definite proportions
(also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.03%3A_Modern_Atomic_Theory_and_the_Laws_That_Led_to_It.txt |
Learning Objectives
• To become familiar with the components and structure of the atom.
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 (Figure $2$). 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.
Subsequently, 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. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:
$\dfrac {mass}{charge} \times {charge} ={mass} \nonumber$
It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy. The video below shows how JJ Thompson used such a tube to measure the ratio of charge over mass of an electron
Measuring e/m For an Electron. Video from Davidson College demonstrating Thompson's e/m experiment.
Radioactivity
The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivity (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning.
Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to the lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.
These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As Figure $3$ illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. Figure $5$ shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.
The Atomic 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 $6$).
In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (Figure $\PageIndex{7a}$) 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 $\PageIndex{7b}$). 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 $\PageIndex{7c}$). 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.”
The Nuclear Atom: The Nuclear Atom, YouTube(opens in new window) [youtu.be]
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.
Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that 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 protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.”
The historical development of the different models of the atom’s structure is summarized in Figure $8$. 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.
Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure $9$ shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.
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. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.04%3A_The_Discovery_of_the_Electron.txt |
Learning Objectives
• Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
• Define the atomic mass unit and average atomic mass
• Calculate average atomic mass and isotopic abundance
The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom’s mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom’s volume. The diameter of an atom is on the order of 10−10 m, whereas the diameter of the nucleus is roughly 10−15 m—about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure $1$).
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 $\times$ 10−23 g, and an electron has a charge of less than 2 $\times$ 10−19 C (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu. (This isotope is known as “carbon-12” as will be discussed later in this module.) Thus, one amu is exactly $1/12$ of the mass of one carbon-12 atom: 1 amu = 1.6605 $\times$ 10−24 g. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 $\times$ 10−19 C.
A proton has a mass of 1.0073 amu and a charge of 1+. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1− and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton. The properties of these fundamental particles are summarized in Table $1$. (An observant student might notice that the sum of an atom’s subatomic particles does not equal the atom’s actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu, slightly larger than the 12.00 amu of an actual carbon-12 atom. This “missing” mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)
Table $1$: Properties of Subatomic Particles
Name Location Charge (C) Unit Charge Mass (amu) Mass (g)
electron outside nucleus $−1.602 \times 10^{−19}$ 1− 0.00055 $0.00091 \times 10^{−24}$
proton nucleus $1.602 \times 10^{−19}$ 1+ 1.00727 $1.67262 \times 10^{−24}$
neutron nucleus 0 0 1.00866 $1.67493 \times10^{−24}$
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: A – Z = number of neutrons.
\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*} \nonumber
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:
Atomic charge = number of protons − 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+). A neutral oxygen atom (Z = 8) has eight electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).
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 (Figure $2$).
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+
Chemical Symbols
A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure $3$). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).
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 Cl for chlorine. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (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
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure $4$). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as 24Mg, 25Mg, and 26Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, 24Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” 25Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.
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 $2$: 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 9.25
Atomic Mass
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.
The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.
$\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance\times isotopic\: mass})_i \nonumber$
For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are 10B with a mass of 10.0129 amu, and the remaining 80.1% are 11B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:
\begin{align*} \textrm{boron average mass} &=\mathrm{(0.199\times10.0129\: amu)+(0.801\times11.0093\: amu)}\ &=\mathrm{1.99\: amu+8.82\: amu}\ &=\mathrm{10.81\: amu} \end{align*} \nonumber
It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.
Example $2$: Calculation of Average Atomic Mass
A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu), 0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?
Solution
\begin{align*} \mathrm{average\: mass} &=\mathrm{(0.9184\times19.9924\: amu)+(0.0047\times20.9940\: amu)+(0.0769\times21.9914\: amu)}\ &=\mathrm{(18.36+0.099+1.69)\:amu}\ &=\mathrm{20.15\: amu} \end{align*} \nonumber
The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)
Exercise $2$
A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25Mg atoms (mass 24.99 amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.
Answer
24.31 amu
We can also do variations of this type of calculation, as shown in the next example.
Example $3$: Calculation of Percent Abundance
Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
Solution
The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37Cl times the mass of 37Cl.
$\mathrm{average\: mass=(fraction\: of\: ^{35}Cl\times mass\: of\: ^{35}Cl)+(fraction\: of\: ^{37}Cl\times mass\: of\: ^{37}Cl)} \nonumber$
If we let x represent the fraction that is 35Cl, then the fraction that is 37Cl is represented by 1.00 − x.
(The fraction that is 35Cl + the fraction that is 37Cl must add up to 1, so the fraction of 37Cl must equal 1.00 − the fraction of 35Cl.)
Substituting this into the average mass equation, we have:
\begin{align*} \mathrm{35.453\: amu} &=(x\times 34.96885\: \ce{amu})+[(1.00-x)\times 36.96590\: \ce{amu}]\ 35.453 &=34.96885x+36.96590-36.96590x\ 1.99705x &=1.513\ x&=\dfrac{1.513}{1.99705}=0.7576 \end{align*} \nonumber
So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% 35Cl and 24.24% 37Cl.
Exercise $3$
Naturally occurring copper consists of 63Cu (mass 62.9296 amu) and 65Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes?
Answer
69.15% Cu-63 and 30.85% Cu-65
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure $5$), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.
Video $1$: Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.
Summary
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $1/12$ of the mass of a carbon-12 atom and is equal to 1.6605 $\times$ 10−24 g.
Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.
Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.
Key Equations
• $\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance \times isotopic\: mass})_i$
Glossary
anion
negatively charged atom or molecule (contains more electrons than protons)
atomic mass
average mass of atoms of an element, expressed in amu
atomic mass unit (amu)
(also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to $\dfrac{1}{12}$ of the mass of a 12C atom
atomic number (Z)
number of protons in the nucleus of an atom
cation
positively charged atom or molecule (contains fewer electrons than protons)
chemical symbol
one-, two-, or three-letter abbreviation used to represent an element or its atoms
Dalton (Da)
alternative unit equivalent to the atomic mass unit
fundamental unit of charge
(also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 $\times$ 10−19 C
ion
electrically charged atom or molecule (contains unequal numbers of protons and electrons)
mass number (A)
sum of the numbers of neutrons and protons in the nucleus of an atom
unified atomic mass unit (u)
alternative unit equivalent to the atomic mass unit | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.05%3A_The_Structure_of_The_Atom.txt |
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_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.07%3A_Finding_Patterns_-_The_Periodic_Law_and_the_Periodic_Table.txt |
Learning Objectives
• to know the meaning of isotopes and atomic masses.
Atomic and Molecular Weights
The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. $\ce{H_2O}$, for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation:
$\ce{C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H2O(l)} \label{Eq1}$
not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved.
Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen:
$\text{100 grams Water} \rightarrow \text{11.1 grams Hydrogen} + \text{88.9 grams Oxygen} \label{Eq2}$
Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom:
$\dfrac{\dfrac{88.9\;g\;Oxygen}{1\; atom}}{\dfrac{111\;g\;Hydrogen}{2\;atoms}} = 16 \label{Eq3}$
Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16. We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. The atomic mass unit (amu) was not standardized against hydrogen, but rather, against the 12C isotope of carbon (amu = 12).
Thus, the mass of the hydrogen atom (1H) is 1.0080 amu, and the mass of an oxygen atom (16O) is 15.995 amu. Once the masses of atoms were determined, the amu could be assigned an actual value:
1 amu = 1.66054 x 10-24 grams conversely: 1 gram = 6.02214 x 1023 amu
Mass Numbers and Atomic Mass of Elements: Mass Numbers and Atomic Mass of Elements, YouTube(opens in new window) [youtu.be]
Average Atomic Mass
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 2.3.1), 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 $1$). 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 listed in Table 2.3.2 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.
The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes in Table 2.3.2 reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. 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 $1$: Bromine
Naturally occurring bromine consists of the two isotopes listed in the following table:
Solutions to Example 2.4.1
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. In general, we can write
atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + …
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 $1$
Magnesium has the three isotopes listed in the following table:
Solutions to Example 2.4.1
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
Finding the Averaged Atomic Weight of an Element: Finding the Averaged Atomic Weight of an Element(opens in new window) [youtu.be]
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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.08%3A_The_Average_Mass_of_an_Elements_Atoms.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Define Avogadro's number and explain why it is important to know.
• Define the mole. Be able to calculate the number of moles in a given mass of a substance, or the mass corresponding to a given number of moles.
• Define molecular weight, formula weight, and molar mass; explain how the latter differs from the first two.
• Be able to find the number of atoms or molecules in a given weight of a substance.
• Find the molar volume of a solid or liquid, given its density and molar mass.
• Explain how the molar volume of a metallic solid can lead to an estimate of atomic diameter.
The chemical changes we observe always involve discrete numbers of atoms that rearrange themselves into new configurations. These numbers are HUGE— far too large in magnitude for us to count or even visualize, but they are still numbers, and we need to have a way to deal with them. We also need a bridge between these numbers, which we are unable to measure directly, and the weights of substances, which we do measure and observe. The mole concept provides this bridge, and is central to all of quantitative chemistry.
Counting Atoms: Avogadro's Number
Owing to their tiny size, atoms and molecules cannot be counted by direct observation. But much as we do when "counting" beans in a jar, we can estimate the number of particles in a sample of an element or compound if we have some idea of the volume occupied by each particle and the volume of the container. Once this has been done, we know the number of formula units (to use the most general term for any combination of atoms we wish to define) in any arbitrary weight of the substance. The number will of course depend both on the formula of the substance and on the weight of the sample. However, if we consider a weight of substance that is the same as its formula (molecular) weight expressed in grams, we have only one number to know: Avogadro's number.
Avogadro's number
Avogadro's number is known to ten significant digits:
$N_A = 6.022141527 \times 10^{23}.$
However, you only need to know it to three significant figures:
$N_A \approx 6.02 \times 10^{23}. \label{3.2.1}$
So $6.02 \times 10^{23}$ of what? Well, of anything you like: apples, stars in the sky, burritos. However, the only practical use for $N_A$ is to have a more convenient way of expressing the huge numbers of the tiny particles such as atoms or molecules that we deal with in chemistry. Avogadro's number is a collective number, just like a dozen. Students can think of $6.02 \times 10^{23}$ as the "chemist's dozen".
Before getting into the use of Avogadro's number in problems, take a moment to convince yourself of the reasoning embodied in the following examples.
Example $1$: Mass ratio from atomic weights
The atomic weights of oxygen and carbon are 16.0 and 12.0 atomic mass units ($u$), respectively. How much heavier is the oxygen atom in relation to carbon?
Solution
Atomic weights represent the relative masses of different kinds of atoms. This means that the atom of oxygen has a mass that is
$\dfrac{16\, \cancel{u}}{12\, \cancel{u}} = \dfrac{4}{3} ≈ 1.33 \nonumber$
as great as the mass of a carbon atom.
Example $2$: Mass of a single atom
The absolute mass of a carbon atom is 12.0 unified atomic mass units ($u$). How many grams will a single oxygen atom weigh?
Solution
The absolute mass of a carbon atom is 12.0 $u$ or
$12\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} \nonumber$
The mass of the oxygen atom will be 4/3 greater (from Example $1$):
$\left( \dfrac{4}{3} \right) 1.99 \times 10^{–23} \, g = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber$
Alternatively we can do the calculation directly like with carbon:
$16\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber$
Example $3$: Relative masses from atomic weights
Suppose that we have $N$ carbon atoms, where $N$ is a number large enough to give us a pile of carbon atoms whose mass is 12.0 grams. How much would the same number, $N$, of oxygen atoms weigh?
Solution
We use the results from Example $1$ again. The collection of $N$ oxygen atoms would have a mass of
$\dfrac{4}{3} \times 12\, g = 16.0\, g. \nonumber$
Exercise $1$
What is the numerical value of $N$ in Example $3$?
Answer
Using the results of Examples $2$ and $3$.
$N \times 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} = 12\, g \nonumber$
or
$N = \dfrac{12\, \cancel{g}}{1.99 \times 10^{–23} \, \cancel{g} \text{ (per carbon atom)}} = 6.03 \times 10^{23} \text{atoms} \nonumber$
There are a lot of atoms in 12 g of carbon.
Things to understand about Avogadro's number
• It is a number, just as is "dozen", and thus is dimensionless.
• It is a huge number, far greater in magnitude than we can visualize
• Its practical use is limited to counting tiny things like atoms, molecules, "formula units", electrons, or photons.
• The value of NA can be known only to the precision that the number of atoms in a measurable weight of a substance can be estimated. Because large numbers of atoms cannot be counted directly, a variety of ingenious indirect measurements have been made involving such things as Brownian motion and X-ray scattering.
• The current value was determined by measuring the distances between the atoms of silicon in an ultrapure crystal of this element that was shaped into a perfect sphere. (The measurement was made by X-ray scattering.) When combined with the measured mass of this sphere, it yields Avogadro's number. However, there are two problems with this:
• The silicon sphere is an artifact, rather than being something that occurs in nature, and thus may not be perfectly reproducible.
• The standard of mass, the kilogram, is not precisely known, and its value appears to be changing. For these reasons, there are proposals to revise the definitions of both NA and the kilogram.
Moles and their Uses
The mole (abbreviated mol) is the the SI measure of quantity of a "chemical entity", which can be an atom, molecule, formula unit, electron or photon. One mole of anything is just Avogadro's number of that something. Or, if you think like a lawyer, you might prefer the official SI definition:
Definition: The Mole
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12
Avogadro's number (Equation \ref{3.2.1}) like any pure number, is dimensionless. However, it also defines the mole, so we can also express NA as 6.02 × 1023 mol–1; in this form, it is properly known as Avogadro's constant. This construction emphasizes the role of Avogadro's number as a conversion factor between number of moles and number of "entities".
Example $4$: number of moles in N particles
How many moles of nickel atoms are there in 80 nickel atoms?
Solution
$\dfrac{80 \;atoms}{6.02 \times 10^{23} \; atoms\; mol^{-1}} = 1.33 \times 10^{-22} mol \nonumber$
Is this answer reasonable? Yes, because 80 is an extremely small fraction of $N_A$.
Molar Mass
The atomic weight, molecular weight, or formula weight of one mole of the fundamental units (atoms, molecules, or groups of atoms that correspond to the formula of a pure substance) is the ratio of its mass to 1/12 the mass of one mole of C12 atoms, and being a ratio, is dimensionless. But at the same time, this molar mass (as many now prefer to call it) is also the observable mass of one mole (NA) of the substance, so we frequently emphasize this by stating it explicitly as so many grams (or kilograms) per mole: g mol–1.
It is important always to bear in mind that the mole is a number and not a mass. But each individual particle has a mass of its own, so a mole of any specific substance will always correspond to a certain mass of that substance.
Example $5$: Boron content of borax
Borax is the common name of sodium tetraborate, $\ce{Na2B4O7}$.
1. how many moles of boron are present in 20.0 g of borax?
2. how many grams of boron are present in 20.0 g of borax?
Solution
The formula weight of $\ce{Na2B4O7}$ so the molecular weight is:
$(2 \times 23.0) + (4 \times 10.8) + (7 \times 16.0) = 201.2 \nonumber$
1. 20 g of borax contains (20.0 g) ÷ (201 g mol–1) = 0.10 mol of borax, and thus 0.40 mol of B.
2. 0.40 mol of boron has a mass of (0.40 mol) × (10.8 g mol–1) = 4.3 g.
Example $6$: Magnesium in chlorophyll
The plant photosynthetic pigment chlorophyll contains 2.68 percent magnesium by weight. How many atoms of Mg will there be in 1.00 g of chlorophyll?
Solution
Each gram of chlorophyll contains 0.0268 g of Mg, atomic weight 24.3.
• Number of moles in this weight of Mg: (.0268 g) / (24.2 g mol–1) = 0.00110 mol
• Number of atoms: (0.00110 mol) × (6.02E23 mol–1) = $6.64 \times 10^{20}$
Is this answer reasonable? (Always be suspicious of huge-number answers!) Yes, because we would expect to have huge numbers of atoms in any observable quantity of a substance.
Molar Volume
This is the volume occupied by one mole of a pure substance. Molar volume depends on the density of a substance and, like density, varies with temperature owing to thermal expansion, and also with the pressure. For solids and liquids, these variables ordinarily have little practical effect, so the values quoted for 1 atm pressure and 25°C are generally useful over a fairly wide range of conditions. This is definitely not the case with gases, whose molar volumes must be calculated for a specific temperature and pressure.
Example $7$: Molar Volume of a Liquid
Methanol, CH3OH, is a liquid having a density of 0.79 g per milliliter. Calculate the molar volume of methanol.
Solution
The molar volume will be the volume occupied by one molar mass (32 g) of the liquid. Expressing the density in liters instead of mL, we have
$V_M = \dfrac{32\; g\; mol^{–1}}{790\; g\; L^{–1}}= 0.0405 \;L \;mol^{–1} \nonumber$
The molar volume of a metallic element allows one to estimate the size of the atom. The idea is to mentally divide a piece of the metal into as many little cubic boxes as there are atoms, and then calculate the length of each box. Assuming that an atom sits in the center of each box and that each atom is in direct contact with its six neighbors (two along each dimension), this gives the diameter of the atom. The manner in which atoms pack together in actual metallic crystals is usually more complicated than this and it varies from metal to metal, so this calculation only provides an approximate value.
Example $8$: Radius of a Strontium Atom
The density of metallic strontium is 2.60 g cm–3. Use this value to estimate the radius of the atom of Sr, whose atomic weight is 87.6.
Solution
The molar volume of Sr is:
$\dfrac{87.6 \; g \; mol^{-1}}{2.60\; g\; cm^{-3}} = 33.7\; cm^3\; mol^{–1}$
The volume of each "box" is"
$\dfrac{33.7\; cm^3 mol^{–1}} {6.02 \times 10^{23}\; mol^{–1}} = 5.48 \times 10^{-23}\; cm^3$
The side length of each box will be the cube root of this value, $3.79 \times 10^{–8}\; cm$. The atomic radius will be half this value, or
$1.9 \times 10^{–8}\; cm = 1.9 \times 10^{–10}\; m = 190 pm$
Note: Your calculator probably has no cube root button, but you are expected to be able to find cube roots; you can usually use the xy button with y=0.333. You should also be able estimate the magnitude of this value for checking. The easiest way is to express the number so that the exponent is a multiple of 3. Take $54 \times 10^{-24}$, for example. Since 33=27 and 43 = 64, you know that the cube root of 55 will be between 3 and 4, so the cube root should be a bit less than 4 × 10–8.
So how good is our atomic radius? Standard tables give the atomic radius of strontium is in the range 192-220 pm. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.09%3A_Molar_Mass_-_Counting_Atoms_by_Weighing_Them.txt |
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. 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.
Q2.15
The length of a Nickel coin is measured using a ruler with markings at every 0.1 mm. What is the correct measurement report for the Nickel?
What We Know: According to the US government, the size of a Nickel is 0.835 inches or 21.21 mm
What It's Asking For: Using significant figures, determine the correct measurement of the Nickel.
Strategy: Every digit in a scientific measurement is certain except for the last digit. Unless stated otherwise, the last digit is estimated with an uncertainty of +/- 1.
Solution:
1. 21.21 mm
2. 21.2 mm
3. 21 mm
Q2.16
When measuring the mass of a cat that weighs around 4530 g that is placed on a scale that has markings at every 0.01 kg, which is the correct measurement reported on this scale?
1. 4.5 kg
2. 4.53 kg
3. 5 kg
4. 4.534 kg
Strategy:
First you have to convert grams to kg because the scale is measured in kg. Once it is concerted to kg, you have to remember that when reading a scale, you need to read on sig fig past where the markings read.
$4534\;\cancel{g} \left( \dfrac{1\;kg}{1000\;\cancel{g}} \right) = 4.53\;kg$
The grams will cross out to leave you with kg as the units. Since the the scale reads to the 0.01 kg, it would read as a clear marking at 4.53, and you would estimate the next number at the 0.001 kg. That being said the only answer that reads to that is D) 4.534 kg.
Q2.20
A silver picture frame displaces 0.454 L of water and has a mass of 2.78 kg. Calculate the density of the silver in g/cm3.
Strategy
1. Convert L to cm3 using dimensional analysis and conversion factors.$0.454\;L \left( \dfrac{1,000\; mL }{1\; L} \right)\left( \dfrac{1 cm^{3}}{1 \;mL} \right) = 454cm^{^{3}}$
2. Convert kg to g using dimensional analysis and conversion factors.$2.78\;kg \left( \dfrac{1,000 \;g }{1\; kg} \right)= 2,780\;g$
3. Calculate the density $\rho$.
$\rho=\dfrac{m}{V}$
$\rho=\dfrac{2,780g}{454cm^{^{3}}}$
$\rho=6.12\dfrac{g}{cm^{^{3}}}$
Q2.23
The density of liquid gold (Au) is 19.3g/cm3.
1. If given a volume of 450 mL of this liquid, what is the mass?
2. If given a mass of 24.4 kg of this liquid, what is its volume in L?
Solutions For Part a:
First we must start with how to approach the problem.
1. We are given the density. So we must recall the formula we learned this year for density.
Density= Mass(g)/ Volume(mL)
2. Part a of the question asks for us to solve for the mass. Therefore, we must rearrange the previous equation and solve for mass. We do this by multiplying both side by volume. This leaves you with the following equation:
Mass(g)= Density(g/mL) x Volume(mL)
3. Before plugging in all of our data into this equation we must make sure that all of our units match up in order to be able to perform demensional analysis. We see that the density is given to us in g/cm3 however, we are given a volume of liquid in mL. Therefore we must convert cm3 to mL. We must recall that 1mL=1cm3. This makes our conversion very simple:
1g/cm3= 1g/mL
19.3g/cm3= 19.3g/mL
4. Now that all of our units are in alignment we can go ahead and plug all of our data into the equation solved for mass and solve the problem. Our density is 19.3 g/mL and our volume is 450 mL. When we multiply these two numbers we get an answer of 8685. Our units also end up cancelling out because we have 1/mL x mL. This leaves us with grams on the top of the equation. Our answer is then 8685 g. We are not finished yet because we have to take into account significant figures. The rule for multiplication is that there are as many significant figures in the answer as there are in the value in the equation with the least number of significant figures. In this problem, this is 450, having two significant figures. Therefore, our final answer will also have two significant figures. Since our number is 8685, we round up to 8700 giving us only two significant figures and our final answer of 8700 g.
M=D x V
M= 19.3g/mL x 450 mL
M = 8685 g
M= 8700 g
Solutions For part b:
1. We are given the mass and density and are asked to solve for the volume. So we must again recall the formula of density and rearrange it, in order that it will solve for volume. We do this by multiplying both sides of the equation by volume. Once this is done, in order to isolate volume on one side of the equation we divide both sides of the equation by density to leave us with an equation solved for volume.
D(g/mL) =M(g) / V(mL)
D x V= M
V = M / D
2. Now that we have an equation that solves for what we want, we now need to look at our units. We are given the mass in kilograms and the density in g/cm3. First we must convert kilograms to grams in order that when we divide. The two will cancel each other out when we divide and this will leave us with volume in the units of milliliters. Then we must convert g/cm3 to g/mL.
Hints:
1 cm3= 1mL
1000 g = 1kg
Solution:
24.4kg x 1000g/ 1kg = 24400g
1cm3= 1mL
19.3 g/cm3= 19.3 g/mL
3. We now have all of our units in line and we are able to plug in the values into the equation and solve for volume.
V= M / D
V= 24400 g / 19.3 g/mL
V= 1264.2487mL
4. In order to complete this problem we have two last steps. The first thing we must take into account after doing division is significant figures. The rule for significant figures when dividing is that the answer will have as many significant figures as the number in the problem with the least amount of significant figures. So we must find the number with the least amount of sig figs and then apply this number to our final answer
Hints
• trailing zeros do not count as significant figures
• -5 and above rounds up, 4 and below rounds down
Solution
24400/19.3=1264.2487
=1260 mL
5. Our last step in the problem is to make sure our answer is what the question is asking for. when we go back and look at the problem, it asks for us to solve for the volume in Liters. By looking at our dimensional analysis, we end up with volume in milliliters. We must convert this to Liters an then our answer will be complete.
Hint: 1000mL= 1L
Solution:
1260 mL x 1L / 1000 mL= 1.26 Liters
Q2.24
1. Sodium Hypochlorite (bleach) has a density of 1.11 g/cm3 and a volume of 48.21 mL, what is the mass?
2. What is the volume of 8.31 g of bleach?
Use the correct units in your answers.
Answers
1. m= 53.51g
2. V= 7.49mL
Strategy for part A
1. Recognize the equation for density is density (d) equals mass (m) divided by volume (V)
$\rho =\dfrac{m}{V}$
2. Plug in the provided information from the question above
$1.11 \;g/cm^3 =\dfrac{m}{48.21 \;ml}$
3. Convert cm3 into mL using the conversion factor 1 cm3=1 mL, these units should then cancel each other out leaving only grams as the unit.
4.Now solve for $m$.
$m=53.51\;g$
Strategy for part B
1. Use the density formula from part a to plug in the information given in the question above
$\rho =\dfrac{m}{V}$
2. Plug in the values from the question above
$1.11\; g/cm^3 =\dfrac{8.31\;g}{V}$
3. To get volume in mL, convert cm3 to mL use the conversion factor 1 cm3= 1 mL.
4. Solve for $V$
$V= 7.49\; mL$
Q2.25
A ship has a fuel tank that can hold 900 L of gasoline. The fuel has a density of 0.0432 g/cm3. How much weight is the ship carrying in fuel?
Strategy
First, you want to look at what the problem has given you. Look at the units and determine if you have been given the density, the mass, or the volume. Also, check to see if any of the units need to be converted to match the units of the density formula. Then you are going to plug the numbers you have into the formula and solve for what is being asked for.
Solving
The problem has given us the volume and the density of the gasoline, so now we must solve for the mass of the fuel. First, we must make sure that none of the numbers need to be converted to different units, but they do! The 900 L needs to be converted to cm3. First we must convert L to mL.
$900 \; \cancel{L} = \dfrac{1000\; ml}{1\;\cancel{L}}$
which gives us 900,000 mL. Now mL=cm3 so you do not need to convert for that. So now we have 900000 cm3 and we can plug all our numbers into the density formula. The density formula is
$\rho=\dfrac{mass}{volume}$
However, since we are looking for the mass we will switch the formula around and use it in this manner
$Mass=Density*Volume$
So now we will plug our values into the second formula
$Mass=\left ( 0.432 g/cm^3 \right )*\left ( 900000 cm^{3} \right )$
After the cm3 cancel each other out, we are left with $3.8 \times 10^4$ grams of gasoline.
Q2.40
A children's cough syrup contains 25.0mg of Benzocaine (a numbing agent) per 5.00 milliliters of syrup. The bottle instructs a parent to give a max of 5.00mg per kilogram of body weight per day. How many mL of the cough syrup can a 32.0 lb toddler take in one day?
Strategy
What do we know?
1. The cough syrup is concentrated at 25.0mg Benzocaine per 5.0 mL► 25.0mg/5mL. We can divide by 5 to simplify this ratio so we know how many mL contain 5.0mg, the recommended per kg dose. ► 5.0mg/1mL
2. The recommended maximum dose is 5.0mg per kilogram of the child's weight. ► 5.0mg/kg
What are we trying to find?
• The amount of mL needed to supply 5mg per kilogram of body weight.
• Child's weight in kg
Since the ratio of Benzocaine to kilogram of weight is 5:1, we can find the weight in kilograms and use dimensional analysis to find the correct ratio.
Solving:
A To convert, we can use dimensional analysis. Remember that each "fraction" represents a ratio of two things to each other. In order for dimensional analysis to work, units for the given information have to match the units in the denominator of the ratio to the right of it. (Think of it this way, we cant divide apples by oranges!)
$32\;lb \cdot \dfrac{1lb}{2.204kg}= 14.519\;kg$
B When calculating sig figs, we round our answer at the end. We are going to use this number in our dimensional analysis question.
C Using dimensional analysis, we can multiply the weight by the ratio of Benzocaine per kg. The kg unit cancels because it is present both as a numerator and denominator. Cancelling them is equivalent to simplifying them to an implied 1. The units on the far right should be the answer we are looking for.
$14.519kg \cdot \dfrac{5mL}{1kg}= 72.595mL$
Answer: The child can take a maximum of 72.6mL of the cough syrup.
Q2.45
Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or compound. If it is a mixture, classify it as homogenous or heterogeneous.
1. tears
2. lithium hydroxide
3. gold
4. bowl of chili
Strategy
1. You must know the difference between a:
1. pure substance - A substance that is made up of only one type of atom or molecule.
2. mixture - A material system made up of two or more different substances that are mixed but not combined together chemically.
2. You must know the difference between:
1. element - Cannot be broken down into simpler substances.
2. compound - Substance formed when two or more chemical elements are bonded together chemically and can be broken down into simpler substances.
3. homogeneous - Mixture that has a uniform composition and properties throughout (cannot see ingredients)
4. heterogeneous - Mixture that remains physically separated throughout (can see ingredients)
3. Label either pure substance or mixture
4. Label each pure substance as either an element or compound and each mixture as either homogeneous or heterogeneous.
Solution
1. homogeneous mixture
2. pure compound
3. pure element
4. heterogeneous mixture
Q2.47
How many Magnesium atoms are there in 4.67 mol of Magnesium?
What we know: We know that we have 4.67 mol of Magnesium.
What we are asked for: The number of Magnesium atoms present.
Strategy:
1. Familiarize yourself with Avogadro's number which is 6.022 x 1023.
2. Set up a dimensional analysis equation to convert Magnesium moles to Magnesium atoms.
3. Carry out the equation.
Solution
1. 1 mol is equal to 6.022 x 1023 of anything, but in this case Magnesium atoms.
2. $4.67 mol Mg \times \dfrac{6.022\times 10^{23} atoms Mg}{1 mol Mg}$
3. The "mol Mg" units cancel so you're left with:
$4.67 \times (6.022\times 10^{23} atoms Mg) = 2.81\times 10^{24} atoms Mg$
2.81 x 1024 atoms of Magnesium are in 4.67 mol of Magnesium
Q2.55
Find the mass (in grams) of each of the following:
1. 2.1 x 1022 Fe molecules
2. 2.9 x 1023 Al molecules
3. 3.9 x 1022 Ca molecules
4. 4.3 x 1022 Ga molecules
Solution
1. Firstly, because the question requires the same steps for each part, the best way to begin would be to break down the steps necessary into a basic equation:
$\dfrac{number\; of\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{g}{1\; mol}$
This equation is formed by the fact there are 6.022•1023 molecules per mole. Next, the g/mol is the atomic weight of the respective elements in order to finalize the conversion from molecules to grams.
1. Next is to fill in the numbers for each part and each element.
a. $\dfrac{2.1\cdot 10^{22}Fe\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{55.845\; g}{1\; mol}$
$=1.9 g$
b.$\dfrac{2.9\cdot 10^{23}Al\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{26.982\; g}{1\; mol}$
$=13.0 g ( from 12.99 g)]$
c. $\dfrac{3.9\cdot 10^{22}Ca\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{40.078\; g}{1\; mol}$
$=2.6 g$
d. $\dfrac{4.3\cdot 10^{22}Ga\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{69.723\; g}{1\; mol}$
$=5.0 g ( from 4.98 g)$ | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/02%3A_Atoms_and_Elements/2.E%3A_Atoms_and_Elements_%28Exercises%29.txt |
• 3.1: Hydrogen, Oxygen, and Water
Under construction
• 3.2: Chemical Bonds
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.
• 3.3: Representing Compounds- Chemical Formulas and Molecular Models
A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic co
• 3.4: An Atomic-Level Perspective of Elements and Compounds
Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table.
• 3.5: 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.
• 3.6: 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.
• 3.7: Summary of Inorganic Nomenclature
Chemical nomenclature is the names we use for chemicals. For instance, H2O is called "water", and CH4 (the gas you burn in a stove) is called "methane." You should learn the chemical nomenclature here on this page now, so that you will be able to understand when it is used.
• 3.8: 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).
• 3.9: 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.
• 3.10: 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.
• 3.11: Organic Compounds
Organic chemistry is the study of carbon compounds, nearly all of which also contain hydrogen atoms.
• 3.E: Molecules, Compounds and Chemical Equations (Exercises)
03: Molecules Compounds and Chemical Equations
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
Ever since the discovery early in the 19th century that solutions of salts and other electrolytes conduct electric current, there has been general agreement that the forces that hold atoms together must be electrical in nature. Electrolytic solutions contain ions having opposite electrical charges, opposite charges attract, so perhaps the substances from which these ions come consist of positive and negatively charged atoms held together by electrostatic attraction.
It turns out that this is not true generally, but a model built on this assumption does a fairly good job of explaining a rather small but important class of compounds that are called ionic solids. The most well known example of such a compound is sodium chloride, which consists of two interpenetrating lattices of Na+ and Cl ions arranged in such as way that every ion of one type is surrounded (in three dimensional space) by six ions of opposite charge.
The main limitation of this model is that it applies really well only to the small class of solids composed of Group 1 and 2 elements with highly electronegative elements such as the halogens. Although compounds such as CuCl2 dissociate into ions when they dissolve in water, the fundamental units making up the solid are more like polymeric chains of covalently-bound CuCl2 molecules that have little ionic character.
According to the ionic electrostatic model, solids such as NaCl consist of positive and negative ions arranged in a crystal lattice. Each ion is attracted to neighboring ions of opposite charge, and is repelled by ions of like charge; this combination of attractions and repulsions, acting in all directions, causes the ion to be tightly fixed in its own location in the crystal lattice.
Since electrostatic forces are nondirectional, the structure of an ionic solid is determined purely by geometry: two kinds of ions, each with its own radius, will fall into whatever repeating pattern will achieve the lowest possible potential energy. Surprisingly, there are only a small number of possible structures
Covalent Bonding
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.
This simple picture illustrates the number one rule of chemical bonding: chemical bonds form when electrons can be simultaneously close to two or more nuclei. It should be pointed out that this principle applies also to the ionic model; as will be explained later in this chapter, the electron that is "lost" by a positive ion ends up being closer to more nuclei (including the one from whose electron cloud it came) in the compound.
• The polar covalent model: A purely covalent bond can only be guaranteed when the electronegativities (electron-attracting powers) of the two atoms are identical. When atoms having different electronegativities are joined, the electrons shared between them will be displaced toward the more electronegative atom, conferring a polarity on the bond which can be described in terms of percent ionic character. The polar covalent model is thus an generalization of covalent bonding to include a very wide range of behavior.
• The Coulombic model: This is an extension of the ionic model to compounds that are ordinarily considered to be non-ionic. Combined hydrogen is always considered to exist as the hydride ion H, so that methane can be treated as if it were C4+ H–4. This is not as bizarre as it might seem at first if you recall that the proton has almost no significant size, so that it is essentially embedded in an electron pair when it is joined to another atom in a covalent bond. This model, which is not as well known as it deserves to be, has considerable predictive power, both as to bond energies and structures.
• The VSEPR model: The "valence shell electron repulsion" model is not so much a model of chemical bonding as a scheme for explaining the shapes of molecules. It is based on the quantum mechanical view that bonds represent electron clouds- physical regions of negative electric charge that repel each other and thus try to stay as far apart as possible.
Summary
The covalent bond is formed when two atoms are able to share electrons:
whereas the ionic bond is formed when the "sharing" is so unequal that an electron from atom A is completely lost to atom B, resulting in a pair of ions:
The two extremes of electron sharing represented by the covalent and ionic models appear to be generally consistent with the observed properties of molecular and ionic solids and liquids. But does this mean that there are really two kinds of chemical bonds, ionic and covalent? | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.02%3A_Chemical_Bonds.txt |
Chemistry is the experimental and theoretical study of materials on their properties at both the macroscopic and microscopic levels. Understanding the relationship between properties and structures/bonding is also a hot pursuit. Chemistry is traditionally divided into organic and inorganic chemistry. The former is the study of compounds containing at least one carbon-hydrogen bonds. By default, the chemical study of all other substances is called inorganic chemistry, a less well defined subject.
However, the boundary between organic and inorganic compounds is not always well defined. For example, oxalic acid, H2C2O4, is a compound formed in plants, and it is generally considered an organic acid, but it does not contain any C-H bond. Inorganic chemistry is also closely related to other disciplines such as materials sciences, physical chemistry, thermodynamics, earth sciences, mineralogy, crystallography, spectroscopy etc.
A chemical formula is a format used to express the structure of atoms. The formula tells which elements and how many of each element are present in a compound. Formulas are written using the elemental symbol of each atom and a subscript to denote the number of elements. This notation can be accredited to Swedish chemist Jons Jakob Berzeliu. The most common elements present in organic compounds are carbon, hydrogen, oxygen, and nitrogen. With carbon and hydrogen present, other elements, such as phosphorous, sulfur, silicon, and the halogens, may exist in organic compounds. Compounds that do not pertain to this rule are called inorganic compounds.
Molecular Geometry and Structural Formula
Understanding how atoms in a molecules are arranged and how they are bonded together is very important in giving the molecule its identity. Isomers are compounds in which two molecules can have the same number of atoms, and thus the same molecular formula, but can have completely different physical and chemical properties because of differences in structural formula.
Methylpropane and butane have the same molecular formula of C4H10, but are structurally different (methylpropane on the left, butane on the right).
Polymers
A polymer is formed when small molecules of identical structure, monomers, combine into a large cluster. The monomers are joined together by covalent bonds. When monomers repeat and bind, they form a polymer. While they can be comprised of natural or synthetic molecules, polymers often include plastics and rubber. When a molecule has more than one of these polymers, square parenthesis are used to show that all the elements within the polymer are multiplied by the subscript outside of the parenthesis. The subscript (shown as n in the example below) denotes the number of monomers present in the macromolecule (or polymer).
Ethylene becomes the polymer polyethylene.
Molecular Formula
The molecular formula is based on the actual makeup of the compound. Although the molecular formula can sometimes be the same as the empirical formula, molecular compounds tend to be more helpful. However, they do not describe how the atoms are put together. Molecular compounds are also misleading when dealing with isomers, which have the same number and types of atoms (see above in molecular geometry and structural formula).
Ex. Molecular Formula for Ethanol: C2H6O.
Empirical Formula
An empirical formula shows the most basic form of a compound. Empirical formulas show the number of atoms of each element in a compound in the most simplified state using whole numbers. Empirical formulas tend to tell us very little about a compound because one cannot determine the structure, shape, or properties of the compound without knowing the molecular formula. Usefulness of the empirical formula is decreased because many chemical compounds can have the same empirical formula.
Ex. Find the empirical formula for C8H16O2.
Answer: C4H8O (divide all subscripts by 2 to get the smallest, whole number ratio).
Structural Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent.
Ex. Structural Formula for Ethanol:
Condensed Structural Formula
Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for Ethanol: CH3CH2OH (Molecular Formula for Ethanol C2H6O).
Line-Angle Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Ex. Line-Angle Formula for Ethanol:
Formulas of Inorganic Compounds
Inorganic compounds are typically not of biological origin. Inorganic compounds are made up of atoms connected using ionic bonds. These inorganic compounds can be binary compounds, binary acids, or polyatomic ions.
Binary compounds
Binary compounds are formed between two elements, either a metal paired with a nonmetal or two nonmetals paired together. When a metal is paired with a nonmetal, they form ionic compounds in which one is a negatively charged ion and the other is positvely charged. The net charge of the compound must then become neutral. Transition metals have different charges; therefore, it is important to specify what type of ion it is during the naming of the compound. When two nonmetals are paired together, the compound is a molecular compound. When writing out the formula, the element with a positive oxidation state is placed first.
Ex. Ionic Compound: BaBr2(Barium Bromide)
Ex. Molecular Compound: N2O4 (Dinitrogen Tetroxide)
Binary acids
Binary acids are binary compounds in which hydrogen bonds with a nonmetal forming an acid. However, there are exceptions such as NH3, which is a base. This is because it shows no tendency to produce a H+. Because hydrogen is positively charged, it is placed first when writing out these binary acids.
Ex. HBr (Hydrobromic Acid)
Polyatomic ions
Polyatomic ions is formed when two or more atoms are connected with covalent bonds. Cations are ions that have are postively charged, while anions are negatively charged ions. The most common polyatomic ions that exists are those of anions. The two main polyatomic cations are Ammonium and Mercury (I). Many polyatomic ions are typically paired with metals using ionic bonds to form chemical compounds.
Ex. MnO4- (Polyatomic ion); NaMnO4 (Chemical Compound)
Oxoacids
Many acids have three different elements to form ternary compounds. When one of those three elements is oxygen, the acid is known as a oxoacid. In other words, oxacids are compounds that contain hydrogen, oxgygen, and one other element.
Ex. HNO3 (Nitric Acid)
Complex Compounds
Certain compounds can appear in multiple forms yet mean the same thing. A common example is hydrates: water molecules bond to another compound or element. When this happens, a dot is shown between H2O and the other part of the compound. Because the H2O molecules are embedded within the compound, the compound is not necessarily "wet". When hydrates are heated, the water in the compound evaporates and the compound becomes anhydrous. These compounds can be used to attract water such as CoCl2. When CoCl2 is dry, CoCl2 is a blue color wherease the hexahydrate (written below) is pink in color.
Ex. CoCl2 ·6 H2O
Formulas of Organic Compounds
Organic compounds contain a combination carbon and hydrogen or carbon and hydrogen with nitrogen and a few other elements, such as phosphorous, sulfur, silicon, and the halogens. Most organic compounds are seen in biological origin, as they are found in nature.
Hydrocarbons
Hydrocarbons are compounds that consist of only carbon and hydrogen atoms. Hydrocarbons that are bonded together with only single bonds are alkanes. The simplest example is methane (shown below). When hydrocarbons have one or more double bonds, they are called alkenes. The simplest alkene is Ethene (C2H4) which contains a double bond between the two carbon atoms.
Ex. Methane on left, Ethene on right
Functional Groups
Functional groups are atoms connected to carbon chains or rings of organic molecules. Compounds that are within a functional group tend to have similar properties and characteristics. Two common functional groups are hydroxyl groups and carboxyl groups. Hydroxyl groups end in -OH and are alcohols. Carboxyl groups end in -COOH, making compounds containing -COOH carboxylic acids. Functional groups also help with nomenclature by using prefixes to help name the compounds that have similar chemical properties.
Ex. Hydroxyl Group on top; Carboxyl Group on bottom
Problems
1. Which of the following formulas are organic?
1. HClO
2. C5H10
3. CO2
2. What is the name of the following formula?
1. Classify the following formulas into their appropriate functional group
1. Acetic acid
2. Butanol
3. Oxalic acid
1. What are the empirical formulas for the following compounds?
1. C12H10O6
2. CH3CH2CH2CH2CH2CH2CH3
3. H3O
1. What is the name of the following figure and what is the molecular formula of the following figure?
Answer Key:
1. b and c. 2. Propane. 3. a. carboxyl group, b. hydroxyl group, c. carboxyl group. 4. a. C6H5O3, b. C7H16, c. H3O. 5. Methylbutane, C5H12
Contributors and Attributions
• Jean Kim (UCD), Kristina Bonnett (UCD) | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.03%3A_Representing_Compounds-_Chemical_Formulas_and_Molecular_Models.txt |
Learning Objectives
• Define ionic and molecular (covalent) compounds
• Predict the type of compound formed from elements based on their location within the periodic table
• Determine formulas for simple ionic compounds
In ordinary chemical reactions, the nucleus of each atom (and thus the identity of the element) remains unchanged. Electrons, however, can be added to atoms by transfer from other atoms, lost by transfer to other atoms, or shared with other atoms. The transfer and sharing of electrons among atoms govern the chemistry of the elements. During the formation of some compounds, atoms gain or lose electrons, and form electrically charged particles called ions (Figure $1$).
You can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a 2+ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a 2+ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized Ca2+. The name of a metal ion is the same as the name of the metal atom from which it forms, so Ca2+ is called a calcium ion.
When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1− charge; atoms of group 16 gain two electrons and form ions with a 2− charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1− charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized Br. (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)
Note the usefulness of the periodic table in predicting likely ion formation and charge (Figure $2$). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form 1+ ions; group 2 elements form 2+ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge equal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1− ions; group 16 elements (two groups left) form 2− ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a 1+ or 2+ charge, and iron can form ions with a 2+ or 3+ charge.
Example $1$: Composition of Ions
An ion found in some compounds used as antiperspirants contains 13 protons and 10 electrons. What is its symbol?
Solution
Because the number of protons remains unchanged when an atom forms an ion, the atomic number of the element must be 13. Knowing this lets us use the periodic table to identify the element as Al (aluminum). The Al atom has lost three electrons and thus has three more positive charges (13) than it has electrons (10). This is the aluminum cation, Al3+.
Exercise $1$
Give the symbol and name for the ion with 34 protons and 36 electrons.
Answer
Se2, the selenide ion
Example $2$: Formation of Ions
Magnesium and nitrogen react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
Solution
Magnesium’s position in the periodic table (group 2) tells us that it is a metal. Metals form positive ions (cations). A magnesium atom must lose two electrons to have the same number electrons as an atom of the previous noble gas, neon. Thus, a magnesium atom will form a cation with two fewer electrons than protons and a charge of 2+. The symbol for the ion is Mg2+, and it is called a magnesium ion.
Nitrogen’s position in the periodic table (group 15) reveals that it is a nonmetal. Nonmetals form negative ions (anions). A nitrogen atom must gain three electrons to have the same number of electrons as an atom of the following noble gas, neon. Thus, a nitrogen atom will form an anion with three more electrons than protons and a charge of 3−. The symbol for the ion is N3−, and it is called a nitride ion.
Exercise $2$
Aluminum and carbon react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
Answer
Al will form a cation with a charge of 3+: Al3+, an aluminum ion. Carbon will form an anion with a charge of 4−: C4−, a carbide ion.
The ions that we have discussed so far are called monatomic ions, that is, they are ions formed from only one atom. We also find many polyatomic ions. These ions, which act as discrete units, are electrically charged molecules (a group of bonded atoms with an overall charge). Some of the more important polyatomic ions are listed in Table $1$. Oxyanions are polyatomic ions that contain one or more oxygen atoms. At this point in your study of chemistry, you should memorize the names, formulas, and charges of the most common polyatomic ions. Because you will use them repeatedly, they will soon become familiar.
Table $1$: Common Polyatomic Ions
Name Formula Related Acid Formula
ammonium $\ce{NH4+}$
hydronium $\ce{H_3O^+}$
oxide $\ce{O^{2-}}$
peroxide $\ce{O_2^{2-}}$
hydroxide $\ce{OH^-}$
acetate $\ce{CH_3COO^-}$ acetic acid $\ce{CH_3COOH}$
cyanide $\ce{CN^-}$ hydrocyanic acid $\ce{HCN}$
azide $\ce{N_3^-}$ hydrazoic acid $\ce{HN_3}$
carbonate $\ce{CO_3^{2-}}$ carbonic acid $\ce{H_2CO_3}$
bicarbonate $\ce{HCO_3^-}$
nitrate $\ce{NO_3^-}$ nitric acid $\ce{HNO_3}$
nitrite $\ce{NO_2^-}$ nitrous acid $\ce{HNO_2}$
sulfate $\ce{SO_4^{2-}}$ sulfuric acid $\ce{H_2SO_4}$
hydrogen sulfate $\ce{HSO_4^-}$
sulfite $\ce{SO_3^{2-}}$ sulfurous acid $\ce{H_2SO_3}$
hydrogen sulfite $\ce{HSO_3^-}$
phosphate $\ce{PO_4^{3-}}$ phosphoric acid $\ce{H_3PO_4}$
hydrogen phosphate $\ce{HPO_4^{2-}}$
dihydrogen phosphate $\ce{H_2PO_4^-}$
perchlorate $\ce{ClO_4^-}$ perchloric acid $\ce{HClO_4}$
chlorate $\ce{ClO_3^-}$ chloric acid $\ce{HClO_3}$
chlorite $\ce{ClO_2^-}$ chlorous acid $\ce{HClO_2}$
hypochlorite $\ce{ClO^-}$ hypochlorous acid $\ce{HClO}$
chromate $\ce{CrO_4^{2-}}$ chromic acid $\ce{H_2CrO_4}$
dichromate $\ce{Cr_2O_7^{2-}}$ dichromic acid $\ce{H_2Cr_2O7}$
permanganate $\ce{MnO_4^-}$ permanganic acid $\ce{HMnO_4}$
Note that there is a system for naming some polyatomic ions; -ate and -ite are suffixes designating polyatomic ions containing more or fewer oxygen atoms. Per- (short for “hyper”) and hypo- (meaning “under”) are prefixes meaning more oxygen atoms than -ate and fewer oxygen atoms than -ite, respectively. For example, perchlorate is $\ce{ClO4-}$, chlorate is $\ce{ClO3-}$, chlorite is $\ce{ClO2-}$ and hypochlorite is ClO. Unfortunately, the number of oxygen atoms corresponding to a given suffix or prefix is not consistent; for example, nitrate is $\ce{NO3-}$ while sulfate is $\ce{SO4^{2-}}$. This will be covered in more detail in the next module on nomenclature.
The nature of the attractive forces that hold atoms or ions together within a compound is the basis for classifying chemical bonding. When electrons are transferred and ions form, ionic bonds result. Ionic bonds are electrostatic forces of attraction, that is, the attractive forces experienced between objects of opposite electrical charge (in this case, cations and anions). When electrons are “shared” and molecules form, covalent bonds result. Covalent bonds are the attractive forces between the positively charged nuclei of the bonded atoms and one or more pairs of electrons that are located between the atoms. Compounds are classified as ionic or molecular (covalent) on the basis of the bonds present in them.
Ionic Compounds
When an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal), a transfer of electrons usually occurs, producing ions. The compound formed by this transfer is stabilized by the electrostatic attractions (ionic bonds) between the ions of opposite charge present in the compound. For example, when each sodium atom in a sample of sodium metal (group 1) gives up one electron to form a sodium cation, Na+, and each chlorine atom in a sample of chlorine gas (group 17) accepts one electron to form a chloride anion, Cl, the resulting compound, NaCl, is composed of sodium ions and chloride ions in the ratio of one Na+ ion for each Cl ion. Similarly, each calcium atom (group 2) can give up two electrons and transfer one to each of two chlorine atoms to form CaCl2, which is composed of Ca2+ and Cl ions in the ratio of one Ca2+ ion to two Cl ions.
A compound that contains ions and is held together by ionic bonds is called an ionic compound. The periodic table can help us recognize many of the compounds that are ionic: When a metal is combined with one or more nonmetals, the compound is usually ionic. This guideline works well for predicting ionic compound formation for most of the compounds typically encountered in an introductory chemistry course. However, it is not always true (for example, aluminum chloride, AlCl3, is not ionic).
You can often recognize ionic compounds because of their properties. Ionic compounds are solids that typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at 801 °C and boils at 1413 °C. (As a comparison, the molecular compound water melts at 0 °C and boils at 100 °C.) In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (“electricity” is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure $3$).
In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.
Example $3$: Predicting the Formula of an Ionic Compound
The gemstone sapphire (Figure $4$) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al3+, and oxygen anions, O2−. What is the formula of this compound?
Solution Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2−, would give us six negative charges. The formula would be Al2O3.
Exercise $3$
Predict the formula of the ionic compound formed between the sodium cation, Na+, and the sulfide anion, S2−.
Answer
Na2S
Many ionic compounds contain polyatomic ions (Table $1$) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca3(PO4)2. This formula indicates that there are three calcium ions (Ca2+) for every two phosphate $\left(\ce{PO4^{3-}}\right)$ groups. The $\ce{PO4^{3-}}$ groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3−. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.
Example $4$: Predicting the Formula of a Compound with a Polyatomic Anion
Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca2+ and $\ce{H2PO4-}$. What is the formula of this compound?
Solution
The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca2+ ion to two $\ce{H2PO4-}$ ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H2PO4)2.
Exercise $4$
Predict the formula of the ionic compound formed between the lithium ion and the peroxide ion, $\ce{O2^2-}$ (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)
Answer
Li2O2
Because an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a molecular formula. Instead, ionic compounds must be symbolized by a formula indicating the relative numbers of its constituent ions. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO4), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na+ and $\ce{C2O4^2-}$ ions combined in a 2:1 ratio, and its formula is written as Na2C2O4. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO2. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound’s polyatomic anion, $\ce{C2O4^2-}$.
Molecular Compounds
Many compounds do not contain ions but instead consist solely of discrete, neutral molecules. These molecular compounds (covalent compounds) result when atoms share, rather than transfer (gain or lose), electrons. Covalent bonding is an important and extensive concept in chemistry, and it will be treated in considerable detail in a later chapter of this text. We can often identify molecular compounds on the basis of their physical properties. Under normal conditions, molecular compounds often exist as gases, low-boiling liquids, and low-melting solids, although many important exceptions exist.
Whereas ionic compounds are usually formed when a metal and a nonmetal combine, covalent compounds are usually formed by a combination of nonmetals. Thus, the periodic table can help us recognize many of the compounds that are covalent. While we can use the positions of a compound’s elements in the periodic table to predict whether it is ionic or covalent at this point in our study of chemistry, you should be aware that this is a very simplistic approach that does not account for a number of interesting exceptions. Shades of gray exist between ionic and molecular compounds, and you’ll learn more about those later.
Example $5$: Predicting the Type of Bonding in Compounds
Predict whether the following compounds are ionic or molecular:
1. KI, the compound used as a source of iodine in table salt
2. H2O2, the bleach and disinfectant hydrogen peroxide
3. CHCl3, the anesthetic chloroform
4. Li2CO3, a source of lithium in antidepressants
Solution
1. Potassium (group 1) is a metal, and iodine (group 17) is a nonmetal; KI is predicted to be ionic.
2. Hydrogen (group 1) is a nonmetal, and oxygen (group 16) is a nonmetal; H2O2 is predicted to be molecular.
3. Carbon (group 14) is a nonmetal, hydrogen (group 1) is a nonmetal, and chlorine (group 17) is a nonmetal; CHCl3 is predicted to be molecular.
4. Lithium (group 1) is a metal, and carbonate is a polyatomic ion; Li2CO3 is predicted to be ionic.
Exercise $5$
Using the periodic table, predict whether the following compounds are ionic or covalent:
1. SO2
2. CaF2
3. N2H4
4. Al2(SO4)3
Answer a
molecular
Answer b
ionic
Answer c
molecular
Answer d
ionic
Summary
Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table. Thus, nonmetals tend to form negative ions. Positively charged ions are called cations, and negatively charged ions are called anions. Ions can be either monatomic (containing only one atom) or polyatomic (containing more than one atom).
Compounds that contain ions are called ionic compounds. Ionic compounds generally form from metals and nonmetals. Compounds that do not contain ions, but instead consist of atoms bonded tightly together in molecules (uncharged groups of atoms that behave as a single unit), are called covalent compounds. Covalent compounds usually form from two or more nonmetals.
Glossary
covalent bond
attractive force between the nuclei of a molecule’s atoms and pairs of electrons between the atoms
covalent compound
(also, molecular compound) composed of molecules formed by atoms of two or more different elements
ionic bond
electrostatic forces of attraction between the oppositely charged ions of an ionic compound
ionic compound
compound composed of cations and anions combined in ratios, yielding an electrically neutral substance
molecular compound
(also, covalent compound) composed of molecules formed by atoms of two or more different elements
monatomic ion
ion composed of a single atom
polyatomic ion
ion composed of more than one atom
oxyanion
polyatomic anion composed of a central atom bonded to oxygen atoms | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.04%3A_An_Atomic-Level_Perspective_of_Elements_and_Compounds.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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.05%3A_Ionic_Compounds-_Formulas_and_Names.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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.06%3A_Molecular_Compounds-_Formulas_and_Names.txt |
Chemical nomenclature is the names we use for chemicals. For instance, H2O is called "water", and CH4 (the gas you burn in a stove) is called "methane." You should learn the chemical nomenclature here on this page now, so that you will be able to understand when it is used.
The Basics
Here is some important info about how we write chemicals.
1. Elements have symbols of one or two letters. The first is a capital letter (ABC). If there is a second letter, it is a lower-case letter (abc). For instance, "m" is one unit, and "M" is a different unit. "K" is the symbol for one type of constant, and "k" is the symbol for a different type. You need to remember that capital letter symbols are usually different from lower-case symbols. For instance, Co is cobalt, a metal element next to iron, and CO is carbon monoxide, a poisonous gas made of one carbon atom and one oxygen atom.
1. We write the charge of a chemical using a superscript, which looks like this: H+or H. If we write just H, that means an H atom, which is one proton and one electron. H+ means 1 H atom – 1 electron, so it means just one proton, also called hydrogen ion. If we write H this means one hydrogen atom + one electron, so a proton and 2 electrons, also called hydride ion. If there's an number in the superscript, that says how many electrons are added or removed. For instance, Ca2+ is a calcium atom – 2 electrons, or calcium ion. S2– is sulfide, or sulfur + 2 electrons.
1. We indicate the number of atoms of a particular type using a subscript, like this: CO2. This means one carbon atom and 2 oxygen atoms. If we write O2 that means the oxygen molecule, which is two atoms of oxygen connected together. Sometimes people might write O2 to mean the same thing. If the number comes first, though, it has a different meaning. 2 O means 2 atoms of oxygen that aren't connected to anything.
1. If we want to show how many protons and neutrons are present in an atom, we can use the mass number, as a superscript before the element symbol, such as13C. This means carbon with (protons + neutrons) = 13. You can tell that this is different from the charge, because the charge will always include + or – and come after the symbol.
1. Most molecules or ions that are stable have an even number of electrons. If they have an odd number of electrons, this is called a radical. For instance, H is a radical, because it has one electron. Because this is unusual, it might be indicated with a dot, like this: H. For instance, water is H2O, and if you remove hydrogen ion, you are left with hydroxide ion, OH. If you remove Hfrom water, you are left with OH molecule, which is neutral. This is also called hydroxyl radical, written OH.
1. The phase of a substance is often indicated by a letter in () after the symbol. For instance, He is almost always a gas, written He(g). If it's a liquid (4.2K or below, less than -269° C) that is written He(l). You'll probably never hear about He(s), since it would be very hard to make it a solid. You might also see something written with (aq), which means "dissolved in water." For instance, NaCl(aq) means salt dissolved in water so there is no solid left. Or you might just see K+(aq), meaning potassium ions dissolved in water.
Elements
There are lots of elements and you don't need to memorize them all. Here are a few that you should learn right now, though, because they are common or important, so that you won't be confused when they are mentioned later. They are organized by their type.
1. Non-metals
• Light elements: the elements with smallest mass
• Hydrogen (H): exists as H2 or in combination with other elements, such as in water
• Helium (He): named after the sun, because it was discovered in the sun before being discovered on Earth (we'll explain how later); it doesn't react with anything
• Major gases in air
• Oxygen (O): we get most of our energy from reactions with oxygen, when we breathe or when we burn fuel; O2 is 21% of air
• Nitrogen (N): often the limiting factor for agriculture or population growth, even though N2 is 78% of air, because it only reacts under special circumstances.
• Halogens: reactive elements that make salts; common negative ions
• Fluorine (F): the lightest halogen and most reactive element in the periodic table, people say that it killed the first two chemists who tried to isolate F2
• Chlorine (Cl): part of normal salt, NaCl, it is common in the ocean and in your body
• Bromine (Br): one of only two elements that are liquid at room temperature, bromine is also found in salts and minerals
• Iodine (I): a soft, shiny silver solid that easily evaporates to a purple gas, iodine can be used to disinfect cuts and is essential for human brains, suggesting that humans may have evolved to live near the ocean, which provides sources of iodine in fish and seaweed
• Main group solid non-metals: non-conductive and usually soft materials
• Carbon (C): the element on which biology is based, also found in diamond, graphite, coal, and charcoal
• Silicon (Si): the basis of the electronics industry; also a main component of sand, glass, and most rocks
• Sulfur (S): a smelly yellow solid, used to make strong acid in industry, also common in minerals and essential to life
• Phosphorus (P): first isolated from urine, although common in minerals; essential for life, often glows
1. Metals: soft or hard, light or heavy, usually solid electric conductors
• Alkali metals: soft, light, common soluble positive ions
• Lithium (Li): the lightest alkali, used in batteries and anti-depressants
• Sodium (Na, you may know it as natrium): common in the ocean and salt
• Potassium (K, you may know it as kalium): also common, high concentration inside cells
• Alkaline Earth metals: like alkalis, but less so, less reactive, less soluble, positive ions more common in rocks, but also abundant in ocean
• Magnesium (Mg): common in rocks, essential for life, especially photosynthesis
• Calcium (Ca): common in biomaterials such as bone, tooth, and shells, also essential for muscles
• Main group metals
• Aluminum (Al, also called aluminium): requires lots of energy to produce the metal from the mineral sources, but very common and useful metal
• Tin (Sn, from Latin stannum): used since ancient times, especially in alloys such as bronze; still used in solder and many other applications
• Lead (Pb, from Latin plumbum): very heavy, soft, sweet-tasting toxic metal, commonly used since ancient times, now used to shield radiation and in bullets, among many other uses
• Transition Metals: a widely varied group, often characterized by complex chemical properties
• Iron (Fe, from Latin ferrum): most abundant element on earth, essential in steel, with complex reaction properties essential to life
• Copper (Cu, from Latin cuprum): less reactive metal, with characteristic colors, commonly used in coins and electronics
• Silver (Ag, from Latin argentum): used in jewelry, coins and other ornaments and utensils since ancient times, it didn't tarnish until after the industrial revolution, and now also used in electronics
• Gold (Au, from Latin aurum): used since ancient times in coins and jewelry, to color stained glass, also in dentistry and other applications
• Mercury (Hg, from Latin hydrargyrum): also called quicksilver, because it is a silver liquid, it is toxic but very important in the history of science; it may be familiar from thermometers
Common Positive Ions (Cations)
"[element name](charge in Roman numerals if needed) ion"
Cation is another word for positive ion. The common positive ions are the ions of the alkali and alkaline earth metals and ammonium, NH4+. The alkali metals form +1 cations, such as Na+ and K+. The alkaline earth metals form +2 cations, such as Ca2+ and Mg2+. The hydrogen ion, H+ is a very common cation. For these cations, you can call them "[element name] ion", such as sodium ion or calcium ion.
You'll also see transition metal cations or main group metal cations, but it is harder to predict what charge they will have, especially because some of them can have different charges, like iron, which is commonly Fe2+ or Fe3+. The charge on a transition metal cation can also be indicated using Roman numerals in parentheses, which looks like Fe(II) or Fe(III). The Roman numerals you will need to know for chemistry are:
1 2 3 3 5 6 7 8 9 10
I II III IV V VI VII VIII IX X
For cations that have uncertain charge, you should call them "[element name](charge in Roman numerals) ion." For instance, iron(II) ion or sometimes just Fe(II).
Sometimes people use special names for these ions, in which the higher charge ion is called "[name]-ic ion" and the lower charge ion is called "[name]-ous ion," such as ferrous for Fe(II) and ferric for Fe(III), or cuprous ion for Cu(I) and cupric ion for Cu(II). I think this is most common for Fe, and I've never heard anyone call nickel(II) nickelous ion because that sounds ridiculous.
Here's a list of common cations with less predictable charges:
• Ag+
• Cu+
• Cu2+
• Fe2+
• Fe3+
• Hg22+
• Hg2+
• Pb2+
• Sn2+
• Al3+
Elements not on the list above, that you may see soon anyway: zinc(II): Zn2+, cadmium(II): Cd2+, cobalt(II): Co2+, manganese(II): Mn2+, nickel(II): Ni2+, chromium(III): Cr3+.
Common Negative Ions (Anions)
"[base name] + (-ide,-ate, or -ite)"
Anion is another word for negative ion. Common negative ions are the halide ions, formed from the halogen elements: fluoride, F; chloride, Cl; bromide, Br; and iodide, I. As you may have noticed, the names of anions have "-ide" at the end when they are formed from elements. Other examples include oxide, O2–, sulfide, S2–, and nitride, N3–.
There are also many important polyatomic anions, which means anions that include more than one atom. These include toxic cyanide ion, CN, common hydroxide ion, OH, and peroxide ion, O22. Other important anions include acetate ion (C2H3O2), which is in vinegar, the chlorate ion (ClO3), the perchlorate ion (ClO4) which is often explosive, the nitrate ion (NO3), the carbonate ion (CO32) found in shells, the sulfate ion (SO42), and the phosphate ion (PO43). All of these end in "-ate", which means that they have more oxygen. Also, notice that "per-___-ate" means more oxygen than just "-ate", as in perchlorate.
Less common but still important are some "-ite" anions, which have less oxygen, such as nitrite (NO2), sulfite (SO32), chlorite (ClO2) and hypochlorite (ClO). Notice that "hypo-___-ite" means less oxygen than just "-ite" as in hypochlorite. Sulfite and nitrite are used to preserve foods. Sulfite salts are used in wine, dried fruit and preserved radish (mu). Nitrite salts are used in preserved meats.
One more rule says that if you take an anion like carbonate or sulfate and add one hydrogen ion, then you call that "bicarbonate" (HCO3) or "bisulfate" (HSO4). Or you might see it called "hydrogen carbonate" or "hydrogen sulfate." Note that because we added a hydrogen ion, the charge on the bicarbonate ion is one less than the charge on the carbonate ion. Also, note that "disulfate" is S2O72, quite different from bisulfate.
Chemical Nomenclature for Ionic Compounds
"[cation name] + [anion name]"
Ionic compounds are compounds that include at least two components, a positive ion and a negative ion. Often the positive ion is a metal element ion and the negative ion is a non-metal ion. To name an ionic compound, you usually just give the cation followed by the anion, such as "sodium chloride" or "ammonium nitrate." If the cation is the type that could have different charges, than you should say what the charge is, such as "mercury(I) iodide" or "cupric sulfate."
Chemical Nomenclature for Acids
"(hydro if -ide)[anion base name] + (-ic if -ide, -ate; -ous if -ite) + acid"
Acid usually means an anion combined with the hydrogen ion as the cation. For instance, HCl is a common acid, which is the hydrogen ion and the chloride anion. If the anion ends in "-ide" then usually the acid is called "hydro-___-ic acid" such as "hydrochloric acid" for HCl. You'll see this for all the "hydrohalic acids" which are H + a halogen, such as "hydrofluoric acid" or "hydroiodic acid." You might also see "hydrocyanic acid" for HCN. If the anion ends in "-ate" than you call the acid "___-ic acid," such as "sulfuric acid," which is H2SO4, or "nitric acid." HNO3. If the anion ends in "-ite" than the acid name is "___-ous acid." such as "hypochlorous acid" for HClO. Notice that earlier "-ic" and "-ous" meant more and less charge for cations, such as ferric and ferrous ions of iron. Now it also means more and less oxygen in acids.
Chemical Nomenclature for Non-metal Compounds
"(prefix, not mono)[less anion-like atom name] + (prefix)[more anion-like atom name]-ide"
Non-metal compounds are often called covalent compounds. They are named following a different rule from ionic compounds. You will need these "prefixes" which indicate how many of each type of atom are present:
1 2 3 3 5 6 7 8 9 10
mono di tri tetra penta hexa hepta octa nona deca
The prefixes come from Greek. You will put the element that is more left on the periodic table first, unless it is oxygen, which is always last unless it is in a compound with fluorine. This follows the same pattern as ionic compounds. In ionic compounds, the cation is written first, and you will notice that it is usually more to the left in the periodic table than the anion, which is written last. When you name covalent compounds, the atom that's more like an anion is written last. Fluorine is always most "anionic," and oxygen is next most "anionic," so they will always be last. (Fluorine is actually most electronegative, but we will study this concept much later, which is why right now I'm calling it "anion-like.") If both atoms are in the same group (same column of the periodic table) then the lower one is named first. Notice that the two most "anion-like," F and O, are in the upper right of the periodic table. The atom written second, that's more "anion-like" is named like an anion, with the "-ide" ending. For example, CO: carbon is on the left, so we can write "monocarbon monoxide." Actually people usually just call it "carbon monoxide." You can skip "mono" for the first element. For instance, SO3 is called "sulfur trioxide" and N2O4 is called "dinitrogen tetroxide." XeO2 is xenon dioxide, even though xenon is more to the right than oxygen, because oxygen is more like an anion than anything except fluorine. If the compound involves hydrogen, then you can leave out the prefixes, such as "hydrogen chloride" for HCl or "hydrogen sulfide" for H2S, because the numbers of each atom can be predicted as if it were an ionic substance. But actually many compounds of hydrogen have special names, such as "ammonia" for NH3, "methane" for CH4, "borane" for BH3, "silane" for SiH4 and "phosphine" for PH3. You should learn the first two of these now.
Summary
Key Info for Common Elements
Element name Symbol Atomic number Commonly found as...
Hydrogen H 1 H2(g), water (H2O), acid (H+(aq))
Helium He 2 He(g), He(l) if you want to make things very cold
Lithium Li 3 Li+ always, (aq) or in solids with anions, lithium metal Li(s) only in chem class
Carbon C 6 Covalent compounds, making 4 bonds
Nitrogen N 7 N2(g) in air, in ammonia (NH3(g or l)), in basic covalent compounds, in proteins
Oxygen O 8 O2(g) in air, in water, in rock and glass, usually combined with Si
Fluorine F 9 F(aq) or with cations in rock, in covalent compounds with carbon (non-stick pans)
Sodium (Natrium) Na 11 Na+(aq) or with anions in salts, sodium metal (Na(s)) only in chem class
Magnesium Mg 12 Mg2+(aq) or with anions in salts and rocks
Aluminum (Aluminium) Al 13 Al3+ with anions in rocks and salts, industrially made Al(s) metal
Silicon Si 14 Industrially made Si(s) in computer chips, Si(IV) oxides in sand, glass, most rocks
Phosphorus P 15 Phosphates: PO43, P2O74, etc. in rock, DNA
Sulfur S 16 S8(s), S2– or sulfate (SO42) in salts or rocks
Chlorine Cl 17 Cl(aq) or with cations in salts, Cl2(g) or ClO(aq) in disinfectants
Potassium (kalium) K 19 K+(aq) or with anions in salts, potassium metal (K(s)) only in chem class
Calcium Ca 20 Ca2+(aq) or with anions in salts and rocks
Iron Fe 26 Fe(s) metal industrially made, Fe(II) or Fe(III) oxides or sulfides and other minerals
Copper Cu 29 Natural Cu(s) metal, Cu(I) or Cu(II) salts or minerals, usually blue or green
Bromine Br 35 Br(aq) or with cations in salts
Silver Ag 47 Natural Ag(s) metal, Ag(I) or Ag(II) in salts or sulfide minerals
Tin Sn 50 Industrially made Sn(s) in alloys, Sn(II) or Sn(IV) salts and oxide or sulfide minerals
Iodine I 53 I(aq) or with cations in salts
Gold Au 79 Natural Au(s) metal, rarely Au(I) or Au(III) salts
Mercury Hg 80 Natural (but rare) metal Hg(l), Hg(II) sulfides and halides, Hg(I) exists as Hg22+
Lead Pb 82 Pb(s) in alloys, Pb(II) sulfide, carbonate and sulfate minerals, sometimes Pb(IV) salts or minerals
Common Polyatomic Ions
Name Formula Where you find it
Ammonium NH4+ Soluble salts, fertilizer
Cyanide CN Toxic, in some plant products and dyes
Hydroxide OH In bases and some minerals
Peroxide O22 In bleaches and disinfectants
Acetate C2H3O2 In vinegar
Perchlorate ClO4 Soluble salts and explosives, a strong acid
Chlorate ClO3 Similar to perchlorates but less stable
Chlorite ClO2 Disinfectants and bleaches, some explosive salts
Hypochlorite ClO Disinfectants and bleaches, most salts unstable
Nitrate NO3 Soluble salts, fertilizer, explosives, a strong acid
Nitrite NO2 Preservatives for food
Carbonate CO32 In rock, seashells, cement; a base
Bicarbonate HCO3 A base (baking soda), in soda, in blood
Sulfate SO42 In salts, plaster, detergent, a strong acid
Bisulfate HSO4 Food additives
Sulfite SO32 Salts, food preservatives
Phosphate PO43 Salts, rock, fertilizers, a strong acid, in ATP and DNA | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.07%3A_Summary_of_Inorganic_Nomenclature.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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.08%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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.09%3A_Determining_a_Chemical_Formula_from_Experimental_Data.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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.10%3A_Writing_and_Balancing_Chemical_Equations.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_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.11%3A_Organic_Compounds.txt |
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. 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.
Q5.29
Classify each compound as ionic or molecular
1. $Fe_2O_3$
2. $SO_4$
3. $H_2O$
4. $NaCl$
Strategy
1. First you need to have good understanding of the difference between Ionic and Molecular.
2. Ionic compound is formed between a metal and a non-metal
3. Molecular compound is formed between two non-metals
4. Then you need to figure out which elements are metals and non-metals by looking at the periodic table
Answers
1. Ionic
2. Molecular
3. Molecular
4. Ionic
Q5.33
Question: How many types of each atom are in each formula?
1. $(NH_4)_2SO_4$
2. $CaCO_3$
3. $Mg(OH)_2$
4. $NaHCO_3$
What We Know: A chemical formula gives the number of atoms in a compound and the element ratio within a compound.
What It's Asking For: Determine the number of each atom in the given compounds.
Strategy: The subscripts after each atom give the number of atoms in the compound. If the subscript is outside of the parenthesis, the subscript is applied to each atom within the parenthesis.
Solution:
1. 2 N, 8 H, 1 S, 4 O
2. 1 Ca, 1 C, 3 O
3. 1 Mg, 2 O, 2 H
4. 1 Na, 1 H, 1 C, 3 O
Q5.48
Write four different ionic compounds that can be formed using the following elements: Sodium, Chlorine, Aluminum and Sulfur.
Strategy:
1. Identify which elements are the metals and which are the nonmetals that will be arranged to form the ionic compounds.
2. Ionic compounds always contain positive and negative ions; therefore identify the positive charge of each metal identified in the above step and the negative charge of each nonmetal identified in the above step using the Periodic Table for reference.
3. Combine each metal with each nonmetal to form the initial foundation for the ionic compound. The sum of the positive charges must equal the sum of the negative charges within the compound which means that the number of each ion may have to be adjusted to achieve the balance of ions. If so, then the formula of the ionic compound will always reflect the smallest whole number ratio of ions. For example: 2:1, 2:3, etc.
Solution
• $NaCl$
• $Na_2S$
• $AlCl_3$
• $Al_2S_3$
Q5.49
Write the formula for the compound that forms between Potassium and each of the polyatomic ions listed below:
1. Sulfate
2. Perchlorate
3. Nitrate
4. Phosphate
Strategy
1. Identify the compound listed
2. Write out the chemical equation for each reaction
3. Balance the equations
Hints
• The formula for Sulfate is SO42-
• The formula for Perchlorate is ClO4-
• The formula for Nitrate is NO3-
• The formula for Phosphate is PO43-
Solution
1. $2K^+ + SO_4^{2-} \rightarrow K_2SO_4$
2. $K^+ + ClO_4^- \rightarrow K_2CLO_4$
3. $K^+ + NO_3^- \rightarrow KNO_3$
4. $3K^+ + PO_4^{3-} \rightarrow K_3PO_4$
Q5.60
Write the name from the formula of the formula from the name for each hydrated ionic compound
1. Iron(III) phosphate tetrahydrate
2. $CsBr \cdot 4H_2O$
3. Cobalt(II) chloride hexahydrate
4. $LiI \cdot 3H_2O$
Strategy
1. Name the cation and anion combination in the first part of the equation, or if the name is given, name the element combination with the anion first, then the cation. The anion is capitalized, while the cation remains lowercase.
2. If the compound involves a metal that forms more than one cation, the number of cations it forms goes in between parentheses in roman numeral form. If the compound does not involve a metal, than list cation charges as normal.
3. For each hydrate at the end of the compound, the prefix at the beginning indicates the number of hydrates. Name the appropriate prefix or number needed for each formula or compound name. Make sure the hydrate is lowercase.
Solutions
(1) Iron(III) phosphate tetrahydrate
1. Iron phosphate as an element is written as FePO4, since both parts have a charge of 3, they are not needed in the equation.
2. The prefix tetra stands for "4", so therefore tetrahydrate as an element is written as 4H2O
3. The two parts combined make the final formula $FePO_4 \cdot 4H_2O$
(2) $CsBr \cdot 4H_2O$
1. Cs and Br with 1 charge stands for Cesium Bromide
2. 4 as a prefix is Tetra, so the end of the formula is tetrahydrate
3. The final name of the compound is Cesium bromide tetrahydrate
(3) Cobalt(II) chloride hexahydrate
1. Cobalt Chloride has a compound formula of $CoCl$, and with Cobalt's charge of 2 involved, a subscript of 2 is added to the end to make $CoCl_2$
2. The "hexa" prefix stands for 6, meaning there are 6 hydrates or 6H2O
3. The two parts put together makes the final compound, $CoCl_2 \cdot 6H_2O$
(4) $LiI \cdot 3H_2O$
1. Li and I both with single charges make up the compound Lithium iodide
2. 3 as a prefix i labled as "tri", so with 3 hydrates, the compound is named as trihydrate
3. The final compound put together makes Lithium iodide trihydrate
Help with naming compounds: http://chemwiki.ucdavis.edu/Wikitext...onic_Compounds
Q5.65
Write the name for each molecular compound.
1. $CO$
2. $H_2S$
3. $SF_6$
4. $N_2O_2$
Strategy
First, you have to know how to name a molecular compound. To name a compound, its the prefix, name of first element, prefix and name of the second element with the suffix -ide. The prefixes are as follows:
• mono=1
• di=2
• tri=3
• tetra=4
• penta=5
• hexa=6
• hepta=7
• octa=8
• nona=9
• deca=10
After finding the elements and suffixes, just put it together and you have a molecular compound.
Solution
1. CO: Since we only have one of each, the carbon does not have a suffix, so we would just use Carbon and there is only one oxygen so the suffix is mono. So the answer is Carbon Monoxide
2. H2S: The hydrogen has a coefficient of 2, so the suffix would be di and the sulfur has no coefficients. Therefore, the answer would be dihydrogen monosulfide.
3. SF6: The sulfur has no coefficients, so it does not change. The fluoride has a coefficient of 6. Therefore, the answer is Sulfur Hexafluoride
4. N2O2: The nitrogen has a coefficient of 2, while the oxygen has a coefficient of 3. The compounds name is Dinitrogen trioxide.
Q5.75a
Find the amount of moles contained in each specimen.
1. 7.87 kg H2O2
2. 2.34 kg NaCl
3. 12.5 g C2H6O
4. 85.72 g NH3
Strategy
A. First, determine the units that are given for each sample. In order to convert these measurements to moles, each sample should first be written in grams. Two of the given samples are measured in kilograms. Use the following conversion factor to convert kilograms to grams, with $x$ representing the given mass:
$x\, kg\cdot \dfrac{1000\, g}{1\, kg}$
B. Next, find the atomic masses of all the atoms in each compound by using the Periodic Table. Then, add these atomic masses together for each compound. The resulting value will be the number of grams of each sample that make up one mole.
C. Convert the mass in grams of each sample to moles by multiplying it by the following conversion factor, called the molar mass, with x representing the mass of the given specimen and $y$ representing the calculated atomic mass found in step B:
$x\, g\cdot \dfrac{1\, mole}{y\, g}$
A Numbers 1 and 2 first need to be converted into grams.
1. $7.87\, kg\, H_2O_2\cdot \dfrac{1000\, g}{1\, kg}=7,870\, g\, H_2O_2$
2. $2.34\, kg\, NaCl\cdot \dfrac{1000\, g}{1\, kg}=2,340\, g\, NaCl$
B The following are the calculated molar masses of each of the given compounds. The atomic mass of each element has been rounded to 4 significant figures in these calculations.
1. H2O2: $2\, (1.008\, g\, H)+2\, (16.00\, g\, O)=\dfrac{34.02\, g}{1\, mole}\, H_2O_2$
2. NaCl: $(22.99\, g\, Na)+(35.45\, g\, Cl)=\dfrac{58.44\, g}{1\, mole}\, NaCl$
3. C2H6O: $2\, (12.01\, g\, C)+6\, (1.008\, g\, H)+\left ( 16.00\, g\, O \right )=\dfrac{46.07\, g}{1\, mole}\, C_2H_6O$
4. NH3: $(14.01\, g\, N)+3\, (1.008\, g\, H)=\dfrac{17.03\, g}{1\, mole}\, NH_3$
C The number of moles in each specimen can now be calculated by multiplying the mass in grams of each sample by its molarity.
1. $7,870\, g\, H_2O_2\cdot \dfrac{1\, mole\, H_2O_2}{34.02\, g\, H_2O_2}=231\, moles\, H_2O_2$
2. $2,340\, g\, NaCl\cdot \dfrac{1\, mole\, NaCl}{58.44\, g\, NaCl}=40.0\, moles\, NaCl$
3. $12.5\, g\, C_2H_6O\cdot \dfrac{1\, mole\, C_2H_6O}{46.07\, g\, C_2H_6O}=0.271\, moles\, C_2H_6O$
4. $85.72\, g\, NH_3\cdot \dfrac{1\, mole\, NH_3}{17.03\, g\, NH_3}=5.033\, moles\, NH_3$
Q5.75b
Calculate the number of moles in each of the following examples.
1. 402.5 mg of NO2
2. 2.7 kg of H2O
3. 323 g of CBr4
4. 2.9 kg of CaO
Solution
a)
First, calculate the molar mass of NO2.
14.007g/mol N + 2 (15.999 g/mol O) = 46.0055 g/mol
Now, convert the sample from mg to g.
402.5 mg NO2 1 g 0.4025 g NO2
1000 mg
Finally, use the molar mass to determine how many moles your sample size contains.
0.4025 g NO2 1 mol 0.0087 moles of NO2
46.0055 g
This gives us the final answer, 0.0087 moles of NO2.
From this, we can deduce that:
$\text{Moles of substance} = \text{Mass of substance (g)}{Molar mass of substance}$
b) 150 moles of H2O
c) 0.974 moles of CBr4
d) 52 moles of CaO
Q5.80
For each substance, convert the given molecules to mass in grams:
1. 3.2 x 1024 Cl2 molecules
2. 8.25 x 1018 CH2O molecules
3. 1 carbon dioxide molecule
Strategy:
Step 1: Convert the given molecules to moles by dividing by Avogadro's number.
Step 2: Multiply the number of moles by the Molar Mass of the substance to determine the grams.
Step 3: Through correct Dimensional Analysis, the molecules and moles will cancel to leave grams.
$\dfrac{molecules}{Avogadro's}\rightarrow moles\rightarrow moles \cdot MM\rightarrow grams$
Helpful Hints:
• Avogadro's number: 6.022 x 1023 mol of a substance
• Molar mass (MM): the weight of a substance in the units g/mol, found by adding the atomic masses of each element.
• Example: MM of H2O --> H(1.008 x 2) + O(15.999)= 18 g/mol
Solution
a) 3.8 x 1051 grams
b) 4.11 x 1045 grams
c) 7.31 x 1025 grams
Q5.105
Write the empirical formula and molecular formula for the compound with the given percent composition and molecular weight.
• 40.0% Carbon
• 6.70% Hydrogen
• 53.3% Oxygen
Molecular Weight ≈ 240 g/mol
Solution
Strategy:
1. Calculate amount of grams of each element we have.
2. Convert amount of grams into number of moles of each element.
3. Obtain molecular formula.
4. Find the common factor among the number of atoms each elements has, then obtain the empirical formula.
Solution:
1. Calculate amount of grams in each element:
2. Convert amount of grams into number of moles of each elements:
Carbon-
Hydrogen-
Oxygen-
3. Obtain molecular formula:
$C_8H_{16}O_8$
4. Find common factor and obtain empirical formula:
Common factor is 8
Empirical formula is
$CH_2O$
Q5.106
For several compounds both the molar mass and and empirical formulas are listed. What is the molecular formula for each compound.
1. C2OH4, 88 g/mol
2. C2H8N, 46 g/mol
3. NH2, 32 g/mol
Strategy
1. A. Find the empirical formula mass of the compound
2. B. Determine the ratio between the empirical formula mass and the molar mass of the compound
3. C.Multiply the empirical formula by the factor found in part B.
Solution
C2OH4, 88 g/mol
A.
Carbon: 2(12.0g/mol)
Oxygen: 1(16.0g/mol)
Hydrogen: 4(1.0g/mol)
=44.0 g/mol
B.
$\dfrac{88.0g/mol}{44.0g/mol}=2$
C.
2(C2OH4)= C4O2H8
C2H8N, 46 g/mol
A.
Carbon: 2(12.0g/mol)
Hydrogen: 8(1.0g/mol)
Nitrogen: 1(14.0g/mol)
=46.0 g/mol
B.
$\dfrac{46.0g/mol}{46.0g/mol}=1.0$
C.
1(C2H8N)=C2H8N
NH2
A.
Nitrogen: 1(14.0g/mol)
Hydrogen: 2(1.0g/mol)
=16.0g/mol
B.
$\dfrac{32.0g/mol}{16.0g/mol}=2$
C.
2(NH2)=N2H4
Q5.107
After the combustion of a hydrocarbon, we find 66.02g of CO2 and 27.02g of H2O. What is the empirical formula of this hydrocarbon?
Solutions
66.02 x 1 mol/ 44.0 g= 1.500 mol CO2= 1.5 mol C
27.02 x 1 mol/ 18g = 1.501 mol H20= 3 mol H
C1.5 H3= C3 H6
You start by taking both, divide by the molar mass, use the mol ratio of the compound and multiple to get rid of decimals.
Additional Question
Additional Questions 1. For the following molecules; write the chemical formula, determine how many atoms are present in one molecule, determine the molecular weight, determine the number of moles in exactly 1 gram, and the number of moles in exactly 10-6 gram. a. carbon dioxide b. iron (II) chloride c. dinitrogen pentoxide d. iron (III) sulfate 2. Name the following compounds, determine the molecular weight, determine the percent composition, and determine how many moles in 8.35 grams of the compound. a. KI b. CaF2 c. Cu2SO4 d. N2O e. LiOH 3. Give the chemical formula (or atomic symbol), molecular (or atomic) weight, and charge for the following ions: a. sulfate b. sulfite c. nitrate d. chloride e. nitride f. acetate g. carbonate 4. Calculate the number of moles in: a. 20.1797 g of Ne b. 45.3594 g of Ne c. 0.198669 g of Ne 5. Calculate the mass of: a. 2.00 mole of Fe b. 4.362 x 10-5 mol of Fe c. 4.362 x 10-5 mol of Li | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/03%3A_Molecules_Compounds_and_Chemical_Equations/3.E%3A_Molecules_Compounds_and_Chemical_Equations_%28Exercises%29.txt |
• 4.1: Global Warming and the Combustion of Fossil Fuels
• 4.2: 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.
• 4.3: 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.
• 4.4: Solution Concentration and 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.
• 4.5: 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.
• 4.6: 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.
• 4.7: 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.
• 4.8: Acid-Base and 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.
• 4.9: 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.
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 au; Toby Hudson via Wikipedia).
04: Chemical Reactions and Aqueous Reactions
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_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.02%3A_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_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.03%3A_4.3_Limiting_Reactant_Theoretical_Yield_and_Percent_Yield.txt |
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): Calculations Involving Molarity (M), YouTube(opens in new window) [youtu.be]
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
Ion Concentrations in Solution
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: Concentration of Ions in Solution from a Soluble Salt, YouTube(opens in new window) [youtu.be]
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_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.04%3A_Solution_Concentration_and_Solution_Stoichiomentry.txt |
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_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.05%3A_Types_of_Aqueous_Solutions_and_Solubility.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_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.06%3A_Precipitation_Reactions.txt |
The chemical equations discussed in Chapter 7 showed the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was 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. As you learned in Example 9, when aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equationA chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. for the reaction shows each reactant and product as undissociated, electrically neutral compounds:
$2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq)\label{8.4.1}$
Although Equation $\ref{8.4.1}$ 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 AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because 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:
$2Ag^+(aq) + 2NO_3^-(aq) + 2K^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^+(aq) + 2NO_3^-(aq)\label{8.4.2}$
Note that K+(aq) and NO3(aq) ions are present on both sides of the equation, 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:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.3}$
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 $\ref{8.4.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag2Cr2O7 formula unit.
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:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{8.4.4}$
The complete ionic equation for this reaction is as follows:
$2Ag^+(aq) + 2F^-(aq) + 2NH_4^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4^+(aq) + 2F^-(aq)\label{8.4.5}$
Because two NH4+(aq) and two F(aq) ions appear on both sides of Equation $\ref{8.4.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{8.4.6}$), which is identical to Equation $\ref{8.4.3}$:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.6}$
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.
Example $1$
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:
$Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)$
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:
$3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)$
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:
$3Ba^{2+}(aq) + 6NO_3^-(aq) + 6Na^+(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s) + 6Na^+(aq) + 6NO_3^-(aq)$
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:
$3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)$
Exercise $1$
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:
$3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq)$
complete ionic equation:
$3Ag^+(aq) + 3F^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3F^-(aq)$
net ionic equation:
$3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)$
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 introduced in Chapter 3 (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.
Summary
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. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.
Key Takeaway
• A complete ionic equation consists of the net ionic equation and spectator ions.
Conceptual Problem
1. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation? | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.07%3A_Representing_Aqueous_Reactions-_Molecular_Ionic_and_Complete_Ionic_Equations.txt |
Learning Objectives
• To know the characteristic properties of acids and bases.
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.
Definitions of Acids and Bases
We can define acids as substances that dissolve in water to produce H+ ions, whereas bases are defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.
The Arrhenius Definition of Acids and Bases
The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; Equation $\ref{4.3.1}$), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions (Equation $\ref{4.3.2}$):
$\underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{4.3.1}$
$\underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{4.3.2}$
According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations:
1. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution.
2. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce $H^+$ and $OH^−$ ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation $\ref{4.3.3}$) is not an acid–base reaction because it does not involve $H^+$ and $OH^−$:
$NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{4.3.3}$
The Brønsted–Lowry Definition of Acids and Bases
Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H+ ion) from one substance to another.
According to Brønsted and Lowry, an acid (A substance with at least one hydrogen atom that can dissociate to form an anion and an $H^+$ ion (a proton) in aqueous solution, thereby forming an acidic solution) is any substance that can donate a proton, and a base (a substance that produces one or more hydroxide ions ($OH^-$ and a cation when dissolved in aqueous solution, thereby forming a basic solution) is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form $NH_4^+$, so in Equation $\ref{4.3.3}$, $NH_3$ is a Brønsted–Lowry base and $HCl$ is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified.
Polyprotic Acids
Acids differ in the number of protons they can donate. For example, monoprotic acids (a compound that is capable of donating one proton per molecule) are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2 and H+. A compound that can donate more than one proton per molecule is known as a polyprotic acid. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acid (a compound that can donate two protons per molecule in separate steps) and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acid (a compound that can donate three protons per molecule in separate steps), (Equation $\ref{4.3.4}$, Equation $\ref{4.3.5}$, and Equation $\ref{4.3.6}$):
$H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{4.3.4}$
$H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{4.3.5}$
$HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{4.3.6}$
In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium (the point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time).
Strengths of Acids and Bases
We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strong acids react essentially completely with water to give $H^+$ and the corresponding anion. Similarly, strong bases dissociate essentially completely in water to give $OH^−$ and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids and weak bases react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.
In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows:
$CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{4.3.7}$
Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates). Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation $\ref{4.3.8}$) but a weak acid when it donates its second proton (Equation $\ref{4.3.9}$) as indicated by the single and double arrows, respectively:
$\underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{4.3.8}$
$\underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{4.3.9}$
Consequently, an aqueous solution of sulfuric acid contains $H^+_{(aq)}$ ions and a mixture of $HSO^-_{4\;(aq)}$ and $SO^{2−}_{4\;(aq)}$ ions, but no $H_2SO_4$ molecules. All other polyprotic acids, such as H3PO4, are weak acids.
The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:
$NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{4.3.10}$
Most of the ammonia (>99%) is present in the form of NH3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−).
There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
Definition of Strong/Weak Acids & Bases: Definition of Strong/Weak Acids & Bases, YouTube (opens in new window) [Definition of Strong] [Definition of Strong] [youtu.be] (opens in new window)
Table $1$ lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.
Table $1$: Common Strong Acids and Bases
Strong Acids Strong Bases
Hydrogen Halides Oxoacids Group 1 Hydroxides Hydroxides of Heavy Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH
Example $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. CH3CH2CO2H
2. CH3OH
3. Sr(OH)2
4. CH3CH2NH2
5. HBrO4
Given: compound
Asked for: acid or base strength
Strategy:
A Determine whether the compound is organic or inorganic.
B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids.
Solution:
1. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid.
2. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH ion. Because it does not contain a carboxylic acid (−CO2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.
3. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2.
4. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (Equation $\ref{4.3.7}$), reacting with water to produce small amounts of the OH ion. Ethylamine is therefore a weak base.
5. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in Table $1$ as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is.
Exercise $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. Ba(OH)2
2. HIO4
3. CH3CH2CH2CO2H
4. (CH3)2NH
5. CH2O
Answer a
strong base
Answer b
strong acid
Answer c
weak acid
Answer d
weak base
Answer e
none of these; formaldehyde is a neutral molecule | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.08%3A_Acid-Base_and_Gas-Evolution_Reactions.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.
Any oxidation must ALWAYS be accompanied by a reduction and vice versa.
Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows:
$\ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1}$
Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or "oil rig". The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is
$\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2}$
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements):
$4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3}$
Equation $\ref{4.4.1}$ and Equation $\ref{4.4.2}$ are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation $\ref{4.4.3}$, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
\begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*}
\begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*}
The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in Figure $1$.
In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Assigning Oxidation States
Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure $1$), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.
A set of rules for assigning oxidation states to atoms in chemical compounds follows.
Rules for Assigning Oxidation States
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.
In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.
Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6. Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.
The reduction of copper(I) oxide shown in Equation $\ref{4.4.5}$ demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:
$\overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:
$\text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a}$
$\text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b}$
Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
Example $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. sulfur hexafluoride (SF6)
2. methanol (CH3OH)
3. ammonium sulfate [(NH4)2SO4]
4. magnetite (Fe3O4)
5. ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula
Asked for: oxidation states
Strategy:
Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.
Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:
[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:
[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42) (see Table 2.4). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:
[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion
For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:
[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
[(4 O atoms)(−2)]+[(3 Fe atoms)$\left (+{8 \over 3} \right )$]= 0
Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”
e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of
[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of
[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3
To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. barium fluoride (BaF2)
2. formaldehyde (CH2O)
3. potassium dichromate (K2Cr2O7)
4. cesium oxide (CsO2)
5. ethanol (CH3CH2OH)
Answer a
Ba, +2; F, −1
Answer b
C, 0; H, +1; O, −2
Answer c
K, +1; Cr, +6; O, −2
Answer d
Cs, +1; O, −½
Answer e
C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Types of Redox Reactions
Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include:
• Synthesis reactions: The formation of any compound directly from the elements is a redox reaction, for example, the formation of water from hydrogen and oxygen: $\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber$
• Decomposition reactions: Conversely, the decomposition of a compound to its elements is also a redox reaction, as in the electrolysis of water: $\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \nonumber$
• Combustion reactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water as the products: $\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \nonumber$
The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution.
Redox Reactions of Solid Metals in Aqueous Solution
• A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure $2$). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81}$
In subsequent steps, $\ce{FeCl2}$ undergoes oxidation to form a reddish-brown precipitate of $\ce{Fe(OH)3}$.
Many metals dissolve through reactions of this type, which have the general form
$\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$\ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
• Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation $\ref{4.4.84}$) and the reduction of silver salts by copper (Equation $\ref{4.4.85}$ and Figure $3$):
$\ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84}$
$\ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85}$
The reaction in Equation $\ref{4.4.84}$ is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
• The Activity Series
By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing $\ce{Zn2+}$. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:
$\ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10}$
$\ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11}$
Magnesium has a greater tendency to be oxidized than zinc does.
Pairwise reactions of this sort are the basis of the activity series (Figure $4$), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure $4$, magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example $2$ demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.
Example $2$: Activity
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.
Given: reactants
Asked for: overall reaction and net ionic equation
Strategy:
1. Locate the reactants in the activity series in Figure $4$ and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
2. Write the net ionic equation for the redox reaction.
Solution:
1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
$\ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber$
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
2. A Mercury lies below lead in the activity series, so no reaction will occur.
3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42 form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
$\ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber$
Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
Answer a
$no\: reaction$
Answer b
$3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
Answer c | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/04%3A_Chemical_Reactions_and_Aqueous_Reactions/4.09%3A_Oxidation-Reduction_Reactions.txt |
• 5.1: Water from Wells- Atmospheric Pressure at Work
t impossible to pump water from very deep in the ground with a surface pump. The key to understanding why is realizing that suction is not a force, but simply removing an opposing force to the force of air pressure which is already there.
• 5.2: Pressure- The Result of Particle Collisions
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
• 5.3: The Simple Gas Laws- Boyle’s Law, Charles’s Law and Avogadro’s Law
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature at constant pressure (Charles’s law), and Avogadro showed that the volume of a gas is directly proportional to the number of moles of gas (Avogadro’s law).
• 5.4: The Ideal Gas Law
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm.
• 5.5: Applications of the Ideal Gas Law- Molar Volume, Density and Molar Mass of a Gas
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can be calculated from the volume of water displaced.
• 5.6: Mixtures of Gases and Partial Pressures
The pressure exerted by each gas in a gas mixture is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas in a mixture may be described by its partial pressure or its mole fraction. In a mixture, the partial pressure of each gas is the product of the total pressure and the mole fraction.
• 5.7: Gases in Chemical Reactions- Stoichiometry Revisited
• 5.8: Kinetic Molecular Theory- A Model for Gases
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square speed. The actual values of speed and kinetic energy are not the same for all gas particles.
• 5.9: Mean Free Path, Diffusion, and Effusion of Gases
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.
• 5.10: Real Gases- The Effects of Size and Intermolecular Forces
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces.
• 5.E: Gases (Exercises)
05: Gases
A water well is an excavation or structure created in the ground by digging, driving, boring, or drilling to access groundwater in underground aquifers. The well water is often drawn by a pump (Figure $1$). Unfortunately, it impossible to pump water from very deep in the ground with just a surface pump. The key to understanding why is realizing that suction generated by the pump is not a force, but simply removing an opposing force to the force of air pressure which is already there. When you stick a pipe down a deep hole into a pool of water at the bottom of a well, air inside the pipe is pushing down on the water in the pipe, and air outside the pipe is pushing down on the water outside the pipe, which in turn pushes up on water inside the pipe - all is in a balance.
Figure $1$: The manual water pump draws water up from a well by creating a vacuum that water rushes in to fill. Hand pump to pump water from a well in a village near Chennai in India. (CC by 2.0; Sustainable Sanitation Alliance).
Let's say you suck out the air inside the pipe. The water is pushed up the the same as it was before, but there is no counter acting force pushing the water down, so it begins to rise inside the pipe (Figure $2$). So far so good, but the water stops rising at some height since the water is pulled down by gravity (i.e., the more water in the pipe, the more it weighs). Since the force of the air outside the pipe is not changing, eventually the weight of the water is equal to the air pressure outside the pipe. When this happens, the system is in balance again and water stops flowing.
Suction is not a force, the atmospheric pressure is.
Water is pumped from a well by creating a partial vacuum above the water by the pump. The amount of vacuum is equal to the weight of the column of water from the water table to the surface. Atmospheric pressure at sea level is 760 mm of mercury ($1.01 \times 10^5 \,Pascals$), which is equivalent to a 10.3-meter column of water. This is how deep water can be pumped from (with a surface pump; other pressurized pumps can go deeper).
Figure $2$: Cross section and details of a surface pump used in a well. (CC BY-SA 3.0; Manco Capac). | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.01%3A_Water_from_Wells-_Atmospheric_Pressure_at_Work.txt |
Learning Objectives
• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe the operation of common tools for measuring gas pressure
• Calculate pressure from manometer data
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure $1$). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.
Pressure is defined as the force exerted on a given area:
$P=\dfrac{F}{A} \label{9.2.1}$
Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area.
Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure $2$.—the elephant or the figure skater?
A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in2), so the pressure exerted by each foot is about 14 lb/in2:
$\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2}$
The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in2, so the pressure exerted by each blade is about 30 lb/in2:
$\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3}$
Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:
$\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4}$
The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m2, where N is the newton, a unit of force defined as 1 kg m/s2. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table $1$ provides some information on these and a few other common units for pressure measurements
Table $1$: Pressure Units
Unit Name and Abbreviation Definition or Relation to Other Unit Comment
pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit
kilopascal (kPa) 1 kPa = 1000 Pa
pounds per square inch (psi) air pressure at sea level is ~14.7 psi
atmosphere (atm) 1 atm = 101,325 Pa air pressure at sea level is ~1 atm
bar (bar, or b) 1 bar = 100,000 Pa (exactly) commonly used in meteorology
millibar (mbar, or mb) 1000 mbar = 1 bar
inches of mercury (in. Hg) 1 in. Hg = 3386 Pa used by aviation industry, also some weather reports
torr $\mathrm{1\: torr=\dfrac{1}{760}\:atm}$ named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg) 1 mm Hg ~1 torr
Example $1$: Conversion of Pressure Units
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
1. torr
2. atm
3. kPa
4. mbar
Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1.
1. $\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}$
2. $\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}$
3. $\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}$
4. $\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}$
Exercise $1$
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?
Answer
0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar
We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure $3$). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.
If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be $\dfrac{1}{13.6}$ as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, p:
$p=hρg \label{9.2.5}$
where
• $h$ is the height of the fluid,
• $ρ$ is the density of the fluid, and
• $g$ is acceleration due to gravity.
Example $2$: Calculation of Barometric Pressure
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = $13.6 \,g/cm^3$.
Solution
The hydrostatic pressure is given by Equation \ref{9.2.5}, with $h = 760 \,mm$, $ρ = 13.6\, g/cm^3$, and $g = 9.81 \,m/s^2$. Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:)
$\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber$
\begin {align*} p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\[4pt] &\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber
Exercise $2$
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm3.
Answer
10.3 m
A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure $3$) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.
Example $3$: Calculation of Pressure Using an Open-End Manometer
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Solution
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)
1. In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg
2. $\mathrm{897\cancel{mm Hg}×\dfrac{1\: atm}{760\cancel{mm Hg}}=1.18\: atm}$
3. $\mathrm{1.18\cancel{atm}×\dfrac{101.325\: kPa}{1\cancel{atm}}=1.20×10^2\:kPa}$
Exercise $3$
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
642 mm Hg
Answer b
0.845 atm
Answer c
85.6 kPa
Application: Measuring Blood Pressure
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure $5$). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure—the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure—the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).
Meteorology, Climatology, and Atmospheric Science
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure $5$) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.
In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.
The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure $7$: the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.
Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.
Summary
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
Key Equations
• $P=\dfrac{F}{A}$
• p = hρg
Glossary
atmosphere (atm)
unit of pressure; 1 atm = 101,325 Pa
bar
(bar or b) unit of pressure; 1 bar = 100,000 Pa
barometer
device used to measure atmospheric pressure
hydrostatic pressure
pressure exerted by a fluid due to gravity
manometer
device used to measure the pressure of a gas trapped in a container
pascal (Pa)
SI unit of pressure; 1 Pa = 1 N/m2
pounds per square inch (psi)
unit of pressure common in the US
pressure
force exerted per unit area
torr
unit of pressure; $\mathrm{1\: torr=\dfrac{1}{760}\,atm}$ | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.02%3A_Pressure-_The_Result_of_Particle_Collisions.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{10.3.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{10.3.2}$
or
$V \propto \dfrac{1}{P} \label{10.3.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equations $\ref{10.3.1}$ and $\ref{10.3.3}$. Dividing both sides of Equation $\ref{10.3.1}$ by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. This law in practice is shown in Figure $2$.
At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{10.3.4}$
or
$V \propto T \label{10.3.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{10.3.6}$
or
$V \propto.n \text{@ constant T and P} \label{10.3.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount, but decreases with increasing pressure.
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.03%3A_The_Simple_Gas_Laws-_Boyles_Law_Charless_Law_and_Avogadros_Law.txt |
Learning Objectives
• Derive the ideal gas law from the constituent gas laws
• To use the ideal gas law to describe the behavior of a gas.
In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as $V$) and several other variables ($P$, $T$, and $n$) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously:
• Boyle’s law
$V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber$
• Charles’s law
$V \propto T \;\; \text{@ constant n and P} \nonumber$
• Avogadro’s law
$V \propto n \;\; \text{@ constant T and P} \nonumber$
Combining these three expressions gives
$V \propto \dfrac{nT}{P} \label{10.4.1}$
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
$V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2}$
By convention, the proportionality constant in Equation $\ref{10.4.1}$ is called the gas constant, which is represented by the letter $R$. Inserting R into Equation $\ref{10.4.2}$ gives
$V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3}$
Clearing the fractions by multiplying both sides of Equation $\ref{10.4.4}$ by $P$ gives
$PV = nRT \label{10.4.4}$
This equation is known as the ideal gas law.
An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
$R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5}$
Because the product PV has the units of energy, R can also have units of J/(K•mol):
$R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6}$
Standard Conditions of Temperature and Pressure
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and $\rm1\; bar = 100 \;kPa = 10^5\;Pa$ pressure, referred to as standard temperature and pressure (STP).
$\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber$
Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm.
We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation $\ref{10.4.4}$:
$V=\dfrac{nRT}{P}\label{10.4.7}$
Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm, approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table $1$: Molar Volumes of Selected Gases at 0°C and 1 atm
Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example $1$
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation $\ref{10.4.4}$) for $n$, we obtain
$\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber$
B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
$T=273+30=303{\rm K}\nonumber$
Substituting these values into the expression we derived for n, we obtain
\begin{align*} n &=\dfrac{PV}{RT} \[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber
Exercise $1$
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer
1.5 atm
In Example $1$, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example $5$.
General Gas Equation
When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:
$\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber$
Both equations can be rearranged to give:
$R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber$
The two equations are equal to each other since each is equal to the same constant $R$. Therefore, we have:
$\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8}$
The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.
Example $2$
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example $1$?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
1. Use the results from Example $1$ for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case $P$ and $n$.
3. Solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
Solution to Example 10.4.2
Initial (August) Final (January)
$T_i=30\, °C = 303\, K$ $T_f=−10\,°C = 263\, K$
$P_i= 0.980 \, atm$ $P_f= 0.980\, atm$
$n_i=1.23 × 10^3\, mol$ $n_f= 1.23 × 10^3\, mol$
$V_i=31150\, L$ $V_f=?$
B Both $n$ and $P$ are the same in both cases ($n_i=n_f,P_i=P_f$). Therefore, Equation \ref{10.4.8} can be simplified to:
$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber$
This is the relationship first noted by Charles.
C Solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \[4pt] &=2.70\times10^4\;L \end{align*} \nonumber
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise $2$
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer
0.52 L
Example $1$ illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example $1$ can be applied in any such case, as we demonstrate in Example $2$ (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example $3$
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise $1$ (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example $2$.
Solution:
Prepare a table to determine which parameters change and which are held constant:
Solution to Example 10.4.3
Initial Final
$V_i=0.406\;\rm L$ $V_f=0.406\;\rm L$
$n_i=0.025\;\rm mol$ $n_f=0.025\;\rm mol$
$T_i=\rm25\;^\circ C=298\;K$ $T_i=\rm750\;^\circ C=1023\;K$
$P_i=1.5\;\rm atm$ $P_f=?$
Both $V$ and $n$ are the same in both cases ($V_i=V_f,n_i=n_f$). Therefore, Equation can be simplified to:
$\dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber$
By solving the equation for $P_f$, we get:
\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \[4pt] &=5.1\;atm \end{align*} \nonumber
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise $3$
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer
23.4 atm
In Examples $1$ and $2$, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.
Example $4$
We saw in Example $1$ that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example $3$.
Solution:
Begin by setting up a table of the two sets of conditions:
Solution to Example 10.4.4
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm750-30\;^\circ C=243\;K$
$n_i=\rm1.2\times10^3\;mol$ $n_i=\rm1.2\times10^3\;mol$
$V_i=\rm31150\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, Equation $\ref{10.4.8}$ is simplified to:
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber$
By solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \[4pt] &=5.96\times10^4\;L \end{align*} \nonumber
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise $4$
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer
4.07 × 103
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
$\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9}$
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ($m$, in grams) divided by its molar mass ($M$, in grams per mole):
$n=\dfrac{m}{M}\label{10.4.10}$
Substituting this expression for $n$ into Equation $\ref{10.4.9}$ gives
$\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11}$
Because $m/V$ is the density $d$ of a substance, we can replace $m/V$ by $d$ and rearrange to give
$\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12}$
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example $5$
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation $\ref{10.4.12}$ to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
$M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber$
Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
$P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber$
B Substituting these values into Equation $\ref{10.4.12}$ gives
$\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber$
Exercise $5$: Density of Radon
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer
radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation $\ref{10.4.12}$ is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example $6$.
Example $6$
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
1. Solve Equation $\ref{10.4.12}$ for the molar mass of the gas and then calculate the density of the gas from the information given.
2. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
3. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation $\ref{10.4.12}$ for the molar mass gives
$M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber$
Density is the mass of the gas divided by its volume:
$\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber$
B We must convert the other quantities to the appropriate units before inserting them into the equation:
$T=18+273=291 K\nonumber$
$P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber$
The molar mass of the unknown gas is thus
$M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber$
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
$M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber$
$M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber$
$M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber$
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise $6$
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer
44 g/mol; $CO_2$
Summary
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Ideal gas equation: $PV = nRT$,
where $R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}$
General gas equation: $\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}$
Density of a gas: $\rho=\dfrac{MP}{RT}$
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.04%3A_The_Ideal_Gas_Law.txt |
Learning Objectives
• To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.
• To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.
With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing.
Gas Densities and Molar Mass
The ideal-gas equation can be manipulated to solve a variety of different types of problems. For example, the density, $\rho$, of a gas, depends on the number of gas molecules in a constant volume. To determine this value, we rearrange the ideal gas equation to
$\dfrac{n}{V}=\dfrac{P}{RT}\label{10.5.1}$
Density of a gas is generally expressed in g/L (mass over volume). Multiplication of the left and right sides of Equation \ref{10.5.1} by the molar mass in g/mol ($M$) of the gas gives
$\rho= \dfrac{g}{L}=\dfrac{PM}{RT} \label{10.5.2}$
This allows us to determine the density of a gas when we know the molar mass, or vice versa.
The density of a gas INCREASES with increasing pressure and DECREASES with increasing temperature
Example $1$
What is the density of nitrogen gas ($\ce{N_2}$) at 248.0 Torr and 18º C?
Step 1: Write down your given information
• P = 248.0 Torr
• V = ?
• n = ?
• R = 0.0820574 L•atm•mol-1 K-1
• T = 18º C
Step 2: Convert as necessary.
$(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm} \nonumber$
$18\,^oC + 273 = 291 K\nonumber$
Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation.
Write down all known equations:
$PV = nRT \nonumber$
$\rho=\dfrac{m}{V} \nonumber$
where $\rho$ is density, $m$ is mass, and $V$ is volume.
$m=M \times n \nonumber$
where $M$ is molar mass and $n$ is the number of moles.
Now take the definition of density (Equation \ref{10.5.1})
$\rho=\dfrac{m}{V} \nonumber$
Keeping in mind $m=M \times n$...replace $(M \times n)$ for $mass$ within the density formula.
\begin{align*} \rho &=\dfrac{M \times n}{V} \[4pt] \dfrac{\rho}{M} &= \dfrac{n}{V} \end{align*} \nonumber
Now manipulate the Ideal Gas Equation
\begin{align*} PV &= nRT \[4pt] \dfrac{n}{V} &= \dfrac{P}{RT} \end{align*} \nonumber
$(n/V)$ is in both equations.
\begin{align*} \dfrac{n}{V} &= \dfrac{\rho}{M} \[4pt] &= \dfrac{P}{RT} \end{align*} \nonumber
Now combine them please.
$\dfrac{\rho}{M} = \dfrac{P}{RT}\nonumber$
Isolate density.
$\rho = \dfrac{PM}{RT} \nonumber$
Step 4: Now plug in the information you have.
\begin{align*} \rho &= \dfrac{PM}{RT} \[4pt] &= \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206\, L\, atm/K mol)(291 \; \rm{K})} \[4pt] &= 0.3828 \; g/L \end{align*} \nonumber
An example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation $\ref{10.5.2}$), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose.
Determining Gas Volumes in Chemical Reactions
The ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations.
Example $\PageIndex{2A}$
What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g $\ce{CaCO_3}$ via the equation:
$\ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)} \nonumber$
Solution
Begin by converting the mass of calcium carbonate to moles.
$\dfrac{0.150\;g}{100.1\;g/mol} = 0.00150\; mol \nonumber$
The stoichiometry of the reaction dictates that the number of moles $\ce{CaCO_3}$ decomposed equals the number of moles $\ce{CO2}$ produced. Use the ideal-gas equation to convert moles of $\ce{CO2}$ to a volume.
\begin{align*} V &= \dfrac{nRT}{PR} \[4pt] &= \dfrac{(0.00150\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} \[4pt] &= 0.0336\;L \; or \; 33.6\;mL \end{align*} \nonumber
Example $\PageIndex{2B}$
A 3.00 L container is filled with $\ce{Ne(g)}$ at 770 mmHg at 27oC. A $0.633\;\rm{g}$ sample of $\ce{CO2}$ vapor is then added.
• What is the partial pressure of $\ce{CO2}$ and $\ce{Ne}$ in atm?
• What is the total pressure in the container in atm?
Solution
Step 1: Write down all given information, and convert as necessary.
Before:
• $P = 770\,mmHg \rigtharrow 1.01 \,atm \nonumber \] • \(V = 3.00\,L$
• $n_{\ce{Ne}} = ?$
• $T = 27^o C \rightarrow 300\; K$
Other Unknowns: $n_{\ce{CO2}}$= ?
$n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2 \nonumber$
Step 2: After writing down all your given information, find the unknown moles of $\ce{Ne}$.
\begin{align*} n_{Ne} &= \dfrac{PV}{RT} \[4pt] &= \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})} \[4pt] &= 0.123 \; \rm{mol} \end{align*} \nonumber
Because the pressure of the container before the $\ce{CO2}$ was added contained only $\ce{Ne}$, that is your partial pressure of $Ne$. After converting it to atm, you have already answered part of the question!
$P_{Ne} = 1.01\; \rm{atm} \nonumber$
Step 3: Now that have pressure for $\ce{Ne}$, you must find the partial pressure for $CO_2$. Use the ideal gas equation.
$\dfrac{P_{Ne}\cancel{V}}{n_{Ne}\cancel{RT}} = \dfrac{P_{CO_2}\cancel{V}}{n_{CO_2}\cancel{RT}} \nonumber$
but because both gases share the same Volume ($V$) and Temperature ($T$) and since the Gas Constant ($R$) is constants, all three terms cancel.
\begin{align*} \dfrac{P}{n_{Ne}} &= \dfrac{P}{n_{CO_2}} \[4pt] \dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} &= \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2} \[4pt] P_{CO_2} &= 0.118 \; \rm{atm} \end{align*} \nonumber
This is the partial pressure $\ce{CO_2}$.
Step 4: Now find total pressure.
\begin{align*} P_{total} &= P_{Ne} + P_{CO_2} \[4pt] &= 1.01 \; \rm{atm} + 0.118\; \rm{atm} \[4pt] &= 1.128\; \rm{atm} \[4pt] &\approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)} \end{align*} \nonumber
Example $\PageIndex{2C}$: Sulfuric Acid
Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give $\ce{SO2}$, followed by the reaction of $\ce{SO2}$ with $\ce{O2}$ in the presence of a catalyst to give $\ce{SO3}$, which reacts with water to give $\ce{H2SO4}$. The overall chemical equation is as follows:
$\ce {2S(s) + 3O2(g) + 2H2O(l) \rightarrow 2H2SO4(aq)} \nonumber$
What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?
Given: reaction, temperature, pressure, and mass of one product
Asked for: volume of gaseous reactant
Strategy:
A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of $\ce{O2}$ required.
B Use the ideal gas law to determine the volume of $\ce{O2}$ required under the given conditions. Be sure that all quantities are expressed in the appropriate units.
Solution:
mass of $\ce{H2SO4}$ → moles $\ce{H2SO4}$ → moles $\ce{O2}$ → liters $\ce{O2}$
A We begin by calculating the number of moles of H2SO4 in 1.00 ton:
$\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4 \nonumber$
We next calculate the number of moles of $\ce{O2}$ required:
$\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2 \nonumber$
B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:
\begin{align*} V&=\dfrac{nRT}{P} \[4pt] &=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}} \[4pt] &=3.43\times10^5\;L \end{align*} \nonumber
The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.
Exercise $2$
Charles used a balloon containing approximately 31,150 L of $\ce{H2}$ for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:
$\ce{ Fe(s) + 2 HCl(aq) \rightarrow H2(g) + FeCl2(aq)} \nonumber$
How much iron (in kilograms) was needed to produce this volume of $\ce{H2}$ if the temperature were 30°C and the atmospheric pressure was 745 mmHg?
Answer
68.6 kg of Fe (approximately 150 lb)
Example $3$: Emergency Air bags
Sodium azide ($\ce{NaN_3}$) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:
$\ce{ 2NaN3 \rightarrow 2Na(s) + 3N2(g)} \nonumber$
This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the $\ce{N_2}$ gas that results from the decomposition of a 5.00 g sample of $\ce{NaN_3}$ could be collected by displacing water from an inverted flask, what volume of gas would be produced at 21°C and 762 mmHg?
Given: reaction, mass of compound, temperature, and pressure
Asked for: volume of nitrogen gas produced
Strategy:
A Calculate the number of moles of $\ce{N_2}$ gas produced. From the data in Table S3, determine the partial pressure of $\ce{N_2}$ gas in the flask.
B Use the ideal gas law to find the volume of $\ce{N_2}$ gas produced.
Solution:
A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of $\ce{N_2}$ gas produced:
$\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2 \nonumber$
The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. Table S3 tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the $\ce{N_2}$ gas in the flask is only
\begin{align*} \rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;mmHg} &= 743.4\; \cancel{mmHg} \times\dfrac{1\;atm}{760\;\cancel{mmHg}} \[4pt] &= 0.978\; atm. \end{align*} \nonumber
B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get
$V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L \nonumber$
Exercise$3$
A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce $\ce{H2}$ gas according to the equation
$\ce{ Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)}. \nonumber$
The resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?
Answer
0.397 L
Summary
The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.05%3A_Applications_of_the_Ideal_Gas_Law-_Molar_Volume_Density_and_Molar_Mass_of_a_Gas.txt |
Learning Objectives
• To determine the contribution of each component gas to the total pressure of a mixture of gases
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture.
Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
$P=n \left(\dfrac{RT}{V}\right) = n \times \rm const. \label{10.6.1}$
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures. We can write it mathematically as
\begin{align} P_{tot} &= P_1+P_2+P_3+P_4 \ldots \[4pt] &= \sum_{i=1}^n{P_i} \label{10.6.2} \end{align}
where $P_{tot}$ is the total pressure and the other terms are the partial pressures of the individual gases (up to $n$ component gases).
For a mixture of two ideal gases, $A$ and $B$, we can write an expression for the total pressure:
\begin{align} P_{tot} &=P_A+P_B \[4pt] &=n_A\left(\dfrac{RT}{V}\right) + n_B\left(\dfrac{RT}{V}\right) \[4pt] &=(n_A+n_B)\left(\dfrac{RT}{V}\right) \label{10.6.3} \end{align}
More generally, for a mixture of $n$ component gases, the total pressure is given by
\begin{align} P_{tot} &=(P_1+P_2+P_3+ \; \cdots +P_n)\left(\dfrac{RT}{V}\right)\label{10.6.2a} \[4pt] &=\sum_{i=1}^n{P_i}\left(\dfrac{RT}{V}\right)\label{10.6.2b} \end{align}
Equation $\ref{10.6.2b}$ restates Equation $\ref{10.6.3}$ in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation $\ref{10.6.2b}$ to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example $1$.
Example $1$: The Bends
Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of $O_2$ and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
1. Calculate the number of moles of $\ce{He}$ and $\ce{O_2}$ present.
2. Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.
Solution:
A The number of moles of $\ce{He}$ is
$n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \nonumber$
The number of moles of $\ce{O_2}$ is
$n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \nonumber$
B We can now use the ideal gas law to calculate the partial pressure of each:
$P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \nonumber$
$P_{\rm O_2}=\dfrac{n_{\rm O_2} RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \nonumber$
The total pressure is the sum of the two partial pressures:
$P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \nonumber$
Exercise $1$
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer
$P_{CH_4}=137 \; atm$; $P_{C_2H_6}=13.4\; atm$; $P_{tot}=151\; atm$
Mole Fractions of Gas Mixtures
The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ($\chi$) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ($n_{tot}$):
$\chi_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\label{10.6.5}$
The mole fraction is a dimensionless quantity between 0 and 1. If $\chi_A = 1.0$, then the sample is pure $A$, not a mixture. If $\chi_A = 0$, then no $A$ is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas $A$ to the total pressure of a gas mixture that contains $A$. We can use the ideal gas law to describe the pressures of both gas $A$ and the mixture: $P_A = n_ART/V$ and $P_{tot} = n_tRT/V$. The ratio of the two is thus
$\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=\chi_A \label{10.6.6}$
Rearranging this equation gives
$P_A = \chi_AP_{tot} \label{10.6.7}$
That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% $N_2$, 21% $O_2$, and 0.9% $Ar$, with trace amounts of gases such as $CO_2$, $H_2O$, and others. This means that 78% of the particles present in the atmosphere are $N_2$; hence the mole fraction of $N_2$ is 78%/100% = 0.78. Similarly, the mole fractions of $O_2$ and $Ar$ are 0.21 and 0.009, respectively. Using Equation \ref{10.6.7}, we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of $O_2$ and $Ar$ are 0.21 and 0.009 atm, respectively.
Example $2$: Exhaling Composition
We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.
the mole fractions of the gases
Inhaled Air / mmHg Exhaled Air / mmHg
$P_{\rm N_2}$ 597 568
$P_{\rm O_2}$ 158 116
$P_{\rm H_2O}$ 0.3 28
$P_{\rm CO_2}$ 5 48
$P_{\rm Ar}$ 8 8
$P_{tot}$ 767 767
Given: pressures of gases in inhaled and exhaled air
Asked for: mole fractions of gases in exhaled air
Strategy:
Calculate the mole fraction of each gas using Equation $\ref{10.6.7}$.
Solution:
The mole fraction of any gas $A$ is given by
$\chi_A=\dfrac{P_A}{P_{tot}} \nonumber$
where $P_A$ is the partial pressure of $A$ and $P_{tot}$ is the total pressure. For example, the mole fraction of $\ce{CO_2}$ is given as:
$\chi_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063 \nonumber$
The following table gives the values of $\chi_A$ for the gases in the exhaled air.
Solutions to Example 10.6.2
Gas Mole Fraction
${\rm N_2}$ 0.741
${\rm O_2}$ 0.151
${\rm H_2O}$ 0.037
${\rm CO_2}$ 0.063
${\rm Ar}$ 0.010
Exercise $2$
Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.
Answer
$P_{\rm CO_2}=\rm86\; atm \nonumber$
$P_{\rm N_2}=\rm2.7\;atm \nonumber$
Summary
The partial pressure of each gas in a mixture is proportional to its mole fraction. The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas.
5.07: Gases in Chemical Reactions- Stoichiometry Revisited
Earlier in the course we performed stoichiometric calculations with chemical reactions using quantities of moles and mass (typically in grams). These same principles can be applied to chemical reactions involving gases except that we first have to convert volumes of gases into moles.
Example 5.7
Example
Hydrogen gas reacts with oxygen gas to produce water vapor via the following balanced chemical equation:
2H2(g) + O2(g) -> 2H2O(g)
If the temperature is 320 K and pressure is 1.34 atm, what volume of oxygen is required to produce 65.0 g of water?
Strategy: Since we are given the temperature and pressure, to find the volume of oxygen using the ideal gas law we need to first calculate the moles of oxygen. To find the moles of oxygen required, we can first calculate the moles of water in 65.0g.
$n_{water}=\frac{m_{water}}{mm_{water}}=\frac{\text{65.0 g}}{\text{18.0g/mol}}=\text{3.61 mol of water}$
From the balanced chemical equation, we can see that 1 equivalent of oxygen produces 2 equivalents of water. We can therefore write the following ratio:
1 mol O2 : 2 mol H2O
We can now solve for the amount of oxygen:
$(\text{3.61 mol } H_{2}O)\times(\frac{ \text{1 mol } O_{2}}{\text{2 mol } H_{2}O })= \text{1.81 mol } O_{2}$
Finally, now that we know how many moles of oxygen are required, we can calculate the volume of the oxygen using the ideal gas law and the temperature&pressure provided in question:
$PV=nRT$
$V=\frac{nRT}{P}$
${V = \rm \frac{1.81 mol\ \cdot 0.08206 \frac{L atm}{mol K}\cdot 320K}{1.34 atm }}$
$V = 35.5 \rm L$ | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.06%3A_Mixtures_of_Gases_and_Partial_Pressures.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
five postulates of Kinetic Molecular Theory
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In the following sections, we explain how this theory must be modified to account for the behavior of real gases.
Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a momentum transfer (impulse) from molecules to the walls (Figure $2$).
The momentum transfer to the wall perpendicular to $x$ axis as a molecule with an initial velocity $u_x$ in $x$ direction hits is expressed as:
$\Delta p_x=2mu_x \label{10.7.1}$
The collision frequency, a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume.
$f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{10.7.2}$
The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency.
$P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{10.7.3}$
At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace $u_x^2$ in the expression above with the average value of $u_x^2$, which is denoted by $\overline{u_x^2}$. The overbar designates the average value over all molecules.
The exact expression for pressure is given as :
$P=\dfrac{N}{V}m\overline{u_x^2} \label{10.7.4}$
Finally, we must consider that there is nothing special about $x$ direction. We should expect that
$\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}. \nonumber$
Here the quantity $\overline{u^2}$ is called the mean-square speed defined as the average value of square-speed ($u^2$) over all molecules. Since
$u^2=u_x^2+u_y^2+u_z^2 \nonumber$
for each molecule, then
$\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}. \nonumber$
By substituting $\dfrac{1}{3}\overline{u^2}$ for $\overline{u_x^2}$ in the expression above, we can get the final expression for the pressure:
$P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{10.7.5}$
Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.
Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the average translational kinetic energy of the molecules of a gas $(\overline{e_K})$, which can be represented as and states that at a given Kelvin temperature $(T)$, all gases have the same value of
$\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{10.7.6}$
where $N_A$ is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by $N_A$:
$N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{10.7.7}$
where $M$ is the molar mass of the gas molecules and is related to the molecular mass by $M=N_Am$. By rearranging the equation, we can get the relationship between the root-mean square speed ($u_{\rm rms}$) and the temperature. The rms speed ($u_{\rm rms}$) is the square root of the sum of the squared speeds divided by the number of particles:
$u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{10.7.8}$
where $N$ is the number of particles and $u_i$ is the speed of particle $i$.
The relationship between $u_{\rm rms}$ and the temperature is given by:
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{10.7.9}$
In Equation $\ref{10.7.9}$, $u_{\rm rms}$ has units of meters per second; consequently, the units of molar mass $M$ are kilograms per mole, temperature $T$ is expressed in kelvins, and the ideal gas constant $R$ has the value 8.3145 J/(K•mol). Equation $\ref{10.7.9}$ shows that $u_{\rm rms}$ of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones.
The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation $\ref{10.7.8}$ tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.
At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.
Example $1$
The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed ($v_{\rm av}$) root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$).
Given: particle speeds
Asked for: average speed ($v_{\rm av}$), root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$)
Strategy:
Use Equation $\ref{10.7.6}$ to calculate the average speed and Equation $\ref{10.7.8}$ to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.
Solution:
The average speed is the sum of the speeds divided by the number of particles:
$v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \nonumber$
The rms speed is the square root of the sum of the squared speeds divided by the number of particles:
$v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \nonumber$
The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence $v_{\rm m}=6.0$ m/s. The $v_{\rm rms}$ of the particles, which is related to the average kinetic energy, is greater than their average speed.
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure $3$. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure $3$ were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• Pressure versus Volume: At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.
• Volume versus Temperature: Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
• Pressure of Gas Mixtures: Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Example $2$
The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the
1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?
Given: temperatures and volume
Asked for: effect of increase in temperature
Strategy:
Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.
Solution:
1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.
Exercise $2$
A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the
1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?
Answer a
no change
Answer b
no change
Answer c
no change
Answer d
no change
Answer e
decreases
Answer f
decreases
Answer g
decreases
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T n\nonumber$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \nonumber$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \nonumber$
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.08%3A_Kinetic_Molecular_Theory-_A_Model_for_Gases.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
Diffusion and Effusion
As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment.
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{10.8.1}$
Helium (M = 4.00 g/mol) effuses much more rapidly than ethylene oxide (M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $\sqrt{\dfrac{29}{4.00}}=2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days.
At a given temperature, heavier molecules move more slowly than lighter molecules.
Example $1$
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C).
1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?
Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
Strategy:
1. Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.
2. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.
Solution:
A The first step is to calculate the molar mass of UF6 containing 235U and 238U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235UF6 is
234.04 + (6)(18.998) = 349.03 g/mol
The molar mass of 238UF6 is
238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation $\ref{10.8.1}$: $\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043 \nonumber$
B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated:
final purity = (initial purity)(separation)n
In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give
$1.0043^n=\dfrac{0.990}{0.00720}=137.50 \nonumber$
Taking the logarithm of both sides gives
\begin{align} n\ln(1.0043)&=\ln(137.50) \[4pt] n &=\dfrac{\ln(137.50)}{\ln(1.0043)} \[4pt]&=1148 \end{align} \nonumber
Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Below is a small part of a system that is used to prepare enriched uranium on a large scale.
Exercise $1$
Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.
1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?
Answer a
ratio of effusion rates = 1.15200; one step gives 0.000154% 3He
Answer b
96 steps
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{10.8.2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{10.8.3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{10.8.4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation $\ref{10.8.4}$ shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure $3$ for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure $4$.
The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10−8 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 1010 m (about 6 million miles).
The denser the gas, the shorter the mean free path.
Example $2$
Calculate the rms speed of a sample -butene (C4H8) at 20°C.
Given: compound and temperature
Asked for: rms speed
Strategy:
Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.8.5 to calculate the rms speed of the gas.
Solution:
To use Equation 10.8.4, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus
\begin{align} u_{\rm rms} &= \sqrt{\dfrac{3RT}{M}} \[4pt] &=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}\[4pt] &=361\;m/s \end{align} \nonumber
or approximately 810 mi/h.
Exercise $1$
Calculate the rms speed of a sample of radon gas at 23°C.
Answer
$1.82 \times 10^2\; m/s$ (about 410 mi/h)
The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas.
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T, \nonumber$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}}, \nonumber$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}. \nonumber$
• Graham’s law for effusion: $\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}} \nonumber$
Diffusion is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, effusion is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The mean free path of a molecule is the average distance it travels between collisions. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.09%3A_Mean_Free_Path_Diffusion_and_Effusion_of_Gases.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure $\PageIndex{1a}$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure $\PageIndex{1b}$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected (Figure $4$). Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$:: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ((L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{10.9.1}$ corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the $an^2/V^2$ term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation ($\ref{10.9.1}$) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
\begin{align} n &=\dfrac{m}{M} \[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \[4pt] &=7.052\;mol\nonumber \end{align} \nonumber
Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure:
\begin{align} P &=\dfrac{nRT}{V} \[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \[4pt] &= 43.1\;atm \end{align} \nonumber
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for $P$ gives
\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{align} \nonumber
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer a
77 atm
Answer b
67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.10%3A_Real_Gases-_The_Effects_of_Size_and_Intermolecular_Forces.txt |
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. 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.
Q5.38
What volume does 41.2 g of sodium gas at a pressure of 6.9 atm and a temperature of 514 K occupy? Would the volume be different if the sample were 41.2 g of calcium (under identical conditions)?
Strategy
1. Know the equation of Ideal Gas Law.
2. Rewrite the equation to V=nRT/P
3. Write down all of the known data with units.
4. Find out the atomic mass.
5. Transfer grams to mols.
6. Put all numbers into the equation and calculate.
Hints
• PV=nRT, so V=nRT/P
• Mol= mass/ atomic mass
Solution
Based on $PV=nRT$, and we need to know V, so the equation is rearranged to $V= nRT/P$.
• R = 0.08206 L (atm)/K (mol).
• P = 6.9 atm.
• T = 514 K
Then, the atomic mass of Sodium =22.99 g/mol
$n = 41.2 \;g = \dfrac{mass}{atomic\; mass} = \dfrac{41.2\;g}{22.99\; g/mol} = 1.79\; mol$
Finally, put every number into the equation
$V=\dfrac{nRT}{P}=\dfrac{(1.79\;mol)(0.0820\;6L (atm)/(K mol))(514\;K}{6.9\; atm} = 10.94\; L$
Because Calcium has different atomic mass than Sodium, so the volume is different.
Q5.40
We have a 20.0 L cylinder that is filled with 28.6 g of oxygen gas at the temperature of 401 K. What is the pressure that the oxygen gas is exerting on the cylinder?
What we know:
• Molar mass of O2= 32 g/mol
• Temperature is 401 K
• There is 28.6 g of Oxygen
• The volume is 20.0 L
What its asking for: The pressure of the gas
Strategy
1. Find an equation that goes with what information you are given
2. plug in the information into the formula and get your answer
Solution
Molar mass of O2= 32 g/mol
Moles of O2 in 28.6 g= $\small \frac{28.6g}{32\frac{g}{mol}}$ = 0.894 mol O2
PV=nRT
$\small R=0.08206\frac{L\cdot atm}{mol\cdot K}$
$\small P(20.0 L)=(0.894 mol O_{2})\cdot (0.08406\frac{L\cdot atm}{mol\cdot K})\cdot (401K)$
$\small P(20.0L)=(30.14 L\cdot atm)$
$\small P=\frac{(30.14 L\cdot atm)}{20.0L}$
$\small P=1.51 atm$
Q5.43
A car tire has a maximum rating of 37.0 psi (gauge pressure). While at the temperature of $11.0^{\circ}C$, the tire is inflated to a volume of 10.4 L and a gauge pressure of 31.0 psi. When driving on a hot day, the tire warms to $62.0^{\circ}C$ and its volume expands to 10.9 L. Does the pressure in the tire exceed its maximum rating on the hot day? Note: Gauge pressure is defined as the absolute pressure minutes the atmospheric pressure (14.7 psi).
Solution
• Ideal Gas Law $PV=nRT$
• P=pressure, V=volume, n=number of moles, R=gas constant, T=temperature
What we know
• Cold tire: gauge pressure (31.0 psi), volume (10.4 L), temperature ($11^{\circ}C$)
• Hot tire: volume (10.9), temperature ($62^{\circ}C$)
• Both- gas constant ($0.082057\dfrac{L\cdot atm}{mol\cdot K}$)
What we need to find:
• Number of moles (same amount in cold and hot tire)
• Pressure in the hot tire
Strategy
1. Convert gauge pressure to total pressure (must be total pressure for Ideal Gas Law). Convert total pressure from psi to atm. Convert temperature from Celsius to Kelvin. Keep volume in L.
2. Find the number of moles in the cold tire by using the Ideal Gas Law equation. $n=\dfrac{PV}{RT}$
3. Find the pressure of the hot tire by plugging in the number of moles found in part B.
4. Convert the total pressure to gauge pressure.
5. Determine if the pressure exceeds its maximum rating.
Solution
A.
Cold tire:
Gauge pressure=total pressure-atmospheric pressure
31.0 psi= X -14.7 psi
X= 45.7 psi
$P=45.7 psi\cdot \dfrac{1 atm}{14.7 psi}=3.11 atm$
$T=11^{\circ}C+273= 284 K$
V=10.4 L
$R=0.082057\dfrac{L\cdot atm}{mol\cdot K}$
Hot tire:
$T=62^{\circ}C+273= 335 K$
V=10.9 L
$R=0.082057\dfrac{L\cdot atm}{mol\cdot K}$
B.
$n=\dfrac{3.11 atm\cdot 10.4 L}{0.082057\dfrac{L\cdot atm}{mol\cdot K}\cdot 284 K}$
n= 1.39 mol
C.
$P=\dfrac{1.39 mol\cdot 0.082057\dfrac{L\cdot atm}{mol\cdot K}\cdot 335 K}{10.9 L}$
P= 3.51 atm
D.
$3.51 atm\cdot \dfrac{14.7 psi}{1 atm}= 51.6 psi$
51.6 psi-14.7 psi= 36.9 psi
The gauge pressure is 36.9 psi and since the maximum rating is 37.0 psi, the pressure does not exceed the maximum rating (although close).
Q5.44
A balloon is floating on top of an ocean at a volume of 2.04 L at a pressure of 730 mmHg and a temperature at 20 °C. A sea creature then gets hungry and pulls the balloon down to the bottom of the ocean where the temperature decreases to 5°C while the pressure increases to 1510 mmHg. Suppose the balloon can freely increase, calculate the volume of the balloon.
Strategy
1. Figure out which formula corresponds with the Ideal Gas Law
2. Write down everything that is given to you in the problem
3. Convert If you need to
4. Solve for what you're looking for
Solution
Step 1. Equation for Ideal Gas Law:
$PV=nRT$
• $P$: Pressure
• $V$: Volume
• $n$: Moles
• $R$: Ideal Gas Constant (0.08206 atm. mol / (L K)
• $T$: Temperature
Step 2. What do we know ?
$P_{1}=730mmHg$ $P_{2}=1510mmHg$
$V_1{}=2.04L$ $V_2{}=?$
$N=?$ $R=0.08206\frac{atm*mol}{L*K}$
$T_{1}=20�C$ $T_{2}=5�C$
Step 3. Convert. The first thing we need to do is to convert the Pressure and the Temperature.
$730mmHg*\frac{1atm}{760mmHg}=0.961atm$ $1510mmHg*\frac{1atm}{760mmHg}=1.97atm$
$20�C+273K=293K$ $5�C+273K=278K$
Step 4. The next thing we need to do is find $n$ and cancel out units.
$PV=NRT\rightarrow N=\frac{PV}{RT}\rightarro$ ---------> $N=\frac{(0.961atm)(2.04L)}\left ( 0.08206\frac{atm*mol}\left L*K \right \fr \right )(293K)$ $=0.082mol$
Step 5. Now that we have found our moles, we can now plug it into our equation to find the volume. Don't forget to cancel out your units.
$PV - nRT \rightarrow V = \dfrac{nRT}{P} = 0.950\;L$
Answer
$V_2 =0.950\;L$
Q5.55
Calculate the density of Ne gas at 143 ºC and 4.3 atm.
What we know
• Pressure (4.3 atm),
• temperature (143 ºC ),
• the identity of the gas (Ne),
• the molar mass of Ne from the periodic table (20.2 g/mol).
Asked for
• Density ($\rho$) of Ne under specific pressure and temperature conditions
Strategy:
1. The temperature is given in degrees Celsius, which must be converted to Kelvin.
2. Solve for the density. $\rho = \dfrac{mass}{volume}$
Solution:
A Calculate Temperature in Kelvin:
$T = 143\; ºC + 273 = 416\; K$
B Calculate the density of Ne
$\rho = \dfrac{MM \cdot P}{RT}$
$\rho = \dfrac{\rm 20.2 g/mol \cdot 4.3 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 416 K}$
$\rho= \rm 2.54 g/L$
Q.11.56
Determine the volume in liters occupied by 64.5 grams of Argon at STP.
Hint
• Use the molar volume of a gas at STP to help determine the answer.
Strategy
1. Determine the molar volume using the condition of standard temperature and pressure, or STP. The standard temperature is 273 K and the standard pressure is 1 atm. We need to determine what formula would be best for the problem.
2. After we figure out the formula to use for this problem, we need to figure out what numbers to plug in. We already know the standard temperature of pressure of a gas which are the letters P and T. We are trying to find V, so that leaves us to figure out n and R.
3. Last, we plug in the numbers we have found and work out the equation to determine the volume.
Solution
A. The formula to use to determine the molar volume for this problem is $PV=nRT$.
B. Since we know the numbers to plug in for P and T, we need to find numbers for $n$ and $R$. The rate, R, is $0.08206 \dfrac{L\cdot atm}{mol\cdot K}$. This will always be the R for this equation. To determine n, the number of moles, we must use what is given to us, which is the grams or Ar.mol of
$Ar= 64.5\; g \;of \;Ar \cdot \dfrac{1 \;mol\; of\; Ar}{39.948\; g \;of \;Ar} =1.61 \;mol\; of Ar$
C. Now, we have all the numbers except the one we are looking for, which is V. Plug them in and work out the equation.
$1 atm\cdot V=1.61 mol of Ar \cdot 0.08206 \dfrac{L\cdot atm}{mol\cdot K} \cdot 273 K$
Q5.61
What is the total pressure in (atm) of a mixture of gases in a closed container with the partial pressures as indicated: H, 115 torr; Ar, 105 torr; and N2, 204 torr? What is the mass of H, Ar, and N2 at 28°C and 1.45 L?
Solution and Strategy
1. Find total pressure of the system by adding all of the partial pressures. $115\; torr +105\;torr+204\;torr=total pressure$424\;torr=total pressure$
2. Convert total pressure to atmospheres through dimensional analysis. $\left(\dfrac{424torr}{1}\right) \times \left(\dfrac{1atm}{760\;torr})\right)=0.558atm$
3. Now we must try and find the masses of each of the substances, to do this we must first find the fractional pressures, and convert them to atmospheres by multiplying by the total pressure.
$H=(\dfrac{115\;torr}{424\;torr})=0.2712$
$Ar=(\dfrac{105\;torr}{424\;torr})=0.2476$
$N_{2}=(\dfrac{204\;torr}{424\;torr})=0.4811$
1. Next we multiply by the total pressure to get our fractional pressures in atmospheres.
$H=(0.2712)(0.558\;atm)=0.151\;atm$
$Ar=(0.2476)(0.558\;atm)=0.138\;atm$
$N_{2}=(0.4811)(0.558\;atm)=0.268\;atm$
1. Before we plug in our numbers we must find the volume for each element, to do this we multiply our fractional pressures by the total volume.
$H=(0.2712)(1.45\;L)=0.393L$
$Ar=(0.2476)(1.45\;L)=0.359L$
$N_2=(0.4811)(1.45\;L)=0.698L$
1. The final step before we can use our equation is to convert our temperature from Celsius to Kelvin. $28 °C + 273 = 30\;1K$
2. Now we must utilize the equation $PV=nRT, where P=pressure(atm), V=volume(L), n=moles, R=.0821\dfrac{L*atm}{mol*K}, T=Temperature(K))$ to find mole of the substance. We modify this equation to give us. $n=\dfrac{PV}{RT}$
3. Now we use this equation to find our moles of each gas.
$n(H)=\dfrac{(0.151atm)(0.393L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0024mol$
$n(Ar)=\dfrac{(0.138atm)(0.359L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0020mol$
$n(N_{2})=\dfrac{(0.268atm)(0.698L)}{(0.0821\dfrac{L*atm}{mol*K})(301K)}=0.0076mol$
1. Now that we have the moles we must multiply this by the molar mass of each gas to find its value in grams.
$H=(0.0024mol)(1.008\dfrac{g}{mol})=0.0024g$
$Ar=(0.0020mol)(39.95\dfrac{g}{mol})=0.0799g$
$N_{2}=(0.0076mol)(28.02\dfrac{g}{mol})=0.2130g$
Answer:
$Total Pressure=0.558atm$ $H=0.0024g$ $Ar=0.0799g$ $N_{2}=0.2130g$
Q5.67a
Hydrogen gas produced by a chemical reaction is collected in the vapor space of a container holding water. The hydrogen partial pressure is 605 mmHg and the system temperature is 55 °C. What is the total pressure in the container? If 0.1 mole of H2 was produced in the reaction, what is the total volume of the gas above the water in the container in L? Use the gas constant R of 62.3637 $\dfrac{L\cdot mmHg}{mol\cdot K}$
Solution
Answer, Part 1
Knowing the system temperature is 55 °C, the vapor pressure of the water in the hydrogen/water vapor mixture can be determined. Using Table 11.3 on page 412, the water partial pressure is found to be 118.2 mmHg.
From Equation 11.8 we know that
$P_{Total} = P_{H_2O} + P_{H_2}$
PH2 is given as 605 mmHg and PH2O was found to be 118.2 mmHg
PTotal = 605 mmHg + 118.2 mmHg = 723.2 mmHg
Answer, Part 2
It is given there is 0.1 mole of H2 and hydrogen partial pressure 605 mmHg. In the first part of this problem, the total system pressure was found to be 723.2 mmHg.
Solving Equation 11.10 for the total number of moles, we find $n_{total}=\left ( \dfrac{n_{H_{2}}}{P_{H_{2}}} \right )\cdot P_{total}$
Therefore
$n_{total}=\left ( \dfrac{0.1 mol_{H_{2}}}{605mmHg_} \right )\cdot 723.2 mmHg_=0.12 mol$
From the Ideal Gas Law, we know $PV=n_{total}RT$
Solving for V, we find
$V=n_{total}\left ( \dfrac{RT}{P} \right )$
Convert the system temperature to Kelvin. K = 55 °C + 273.2 = 328.2 K
Therefore V is:
$V=0.12mol\left ( \dfrac{62.3637\dfrac{L\cdot mmHg}{mol\cdot K}\cdot 328.2K}{723.2mmHg} \right )=3.4L$
Q5.67b
Hydrogen gas is collected over water at 35 °C at a total pressure of 745 mmHg. What is the partial pressure of the hydrogen gas collected? Given that the total volume of the gas collected is 750 ml, calculate the mass of the hydrogen gas collected. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.E%3A_Gases_%28Exercises%29.txt |
All chemical changes are accompanied by the absorption or release of heat. The intimate connection between matter and energy has been a source of wonder and speculation from the most primitive times; it is no accident that fire was considered one of the four basic elements (along with earth, air, and water) as early as the fifth century BCE. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes. This will build the groundwork for a more ambitious goal: to predict the direction and extent of change itself.
• 6.1: Light the Furnace- The Nature of Energy and Its Transformations
• 6.2: The Nature and Types of Energy
All chemical changes are accompanied by the absorption or release of heat. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes.
• 6.3: Quantifying Heat and Work
Heat is the amount of energy that is transferred from one system to its surroundings because of a temperature difference. 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.
• 6.4: The First Law of Thermodynamics
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy.
• 6.5: Constant Volume Calorimetry- Measuring ΔU for Chemical Reactions
A bomb calorimeter operates at constant volume and is particularly useful for measuring energies of combustion.
• 6.6: Enthalpy- The Heat Evolved in a Chemical Reaction at Constant Pressure
Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on t
• 6.7: Constant Pressure Calorimetry- Measuring ΔH for Chemical Reactions
a constant-pressure calorimeter, which gives ΔH values directly
• 6.8: Relationships Involving Enthalpy of Reactions
Hess's law argues that for a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporiz
• 6.9: Enthalpies of Reaction from Standard Heats of Formation
The standard state for measuring and reporting enthalpies of formation or reaction is 25 oC and 1 atm. The elemental form of each atom is that with the lowest enthalpy in the standard state. The standard state heat of formation for the elemental form of each atom is zero. The enthalpy of formation (ΔHf) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation (ΔHof) are determined under standard conditions: a pressure of 1 atm for ga
• 6.10: Energy Use and The Environment
under construction
Thumbnail: Dancing Flames of burning charcoal in the dark (CC BY-SA 3.0; Oscar via Wikipedia).
06: Thermochemistry
Chemical changes and their accompanying changes in energy are important parts of our everyday world (Figure \(1\)). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges.
Over 90% of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world’s energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter.
This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes—an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science.
6.02: The Nature and Types of Energy
; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of and molecules. "" usually refers to the energy that is stored in the of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an chemical reaction, the electrons and nuclei within the reactants undergo rearrangement into products possessing lower energies, and the difference is released to the environment in the form of heat. chemical reaction heats surrounding , the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as . We observe the effects of this as a rise in the temperature of the . The temperature of a body is direct measure of the quantity of thermal energy is contains. , energy comes in different types. Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the : (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons. The is also one version of the first law of thermodynamics, as you will learn later. in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of to other forms. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal. ). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed ( or first law of thermodynamics). has thermal energy due to the KE of its molecules and temperature that corresponds to the average KE of its molecules. Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat () or release heat (). The SI unit of energy, heat, and work is the . | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.01%3A_Light_the_Furnace-_The_Nature_of_Energy_and_Its_Transformations.txt |
Learning Objectives
• To calculate changes in internal energy
• Distinguish the related properties of heat, thermal energy, and temperature
• Define and distinguish specific heat and heat capacity, and describe the physical implications of both
• Perform calculations involving heat, specific heat, and temperature change
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold” (Figure $1$). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.
Figure $1$: (a) The molecules in a sample of hot water move more rapidly than (b) those in a sample of cold water.
Heat (q) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance (L). The atoms and molecules in H have a higher average KE than those in L. If we place substance H in contact with substance L, the thermal energy will flow spontaneously from substance H to substance L. The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance L will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure $2$).
Figure $2$: (a) Substances H and L are initially at different temperatures, and their atoms have different average kinetic energies. (b) When they are put into contact with each other, collisions between the molecules result in the transfer of kinetic (thermal) energy from the hotter to the cooler matter. (c) The two objects reach “thermal equilibrium” when both substances are at the same temperature, and their molecules have the same average kinetic energy.
Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process—this process also releases energy in the form of light as evidenced by the torch’s flame (Figure $\PageIndex{3a}$). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.
Figure $3$: (a) An oxyacetylene torch produces heat by the combustion of acetylene in oxygen. The energy released by this exothermic reaction heats and then melts the metal being cut. The sparks are tiny bits of the molten metal flying away. (b) A cold pack uses an endothermic process to create the sensation of cold. (credit a: modification of work by “Skatebiker”/Wikimedia commons).
Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m2/s2, which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.
$1\,cal \equiv 4.184\, J$
Application: Heat can do more than increase temperature
Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure $4$. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.
Direction of Heat Flow: Endothermic vs. Exothermic Processes
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel. The balanced chemical equation for the reaction is as follows:
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) }\label{12.2.1}$
We can also write this chemical equation as
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{12.2.2}$
to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat ($q$) is transferred from a system to its surroundings is described as exothermic. By convention, $q < 0$ for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
$\text{heat} + \ce{H_2O_{(s)} \rightarrow H_2O_{(l)}} \label{12.2.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
By convention, $q < 0$ for an exothermic reaction and $q > 0$ for an endothermic reaction.
Exercise $1$
Decide whether the following are endothermic or exothermic processes
1. water evaporates off a shower door
2. an acid tablet being added to a pool and the surrounding water heats up
3. $\ce{NH_4Cl}$ is dissolved in water and the solution cools
4. the burning of a log in a campfire
Hint
During an endothermic process heat is absorbed from surroundings, causing them to cool, so in every case where there is cooling there is most likely an endothermic process taking place. For exothermic reactions energy is being released to the surroundings and so the surroundings feel like they have been heated by the process.
Answer a
endothermic
Answer b
exothermic
Answer c
endothermic
Answer d
exothermic
Heat is technically not a component in Chemical Reactions
Technically, it is poor form to have a $heat$ term in the chemical reaction like in Equations $\ref{12.2.2}$ and $\ref{12.2.3}$ since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists.
Contributors and Attributions
Learning Objectives
• To know the relationship between energy, work, and heat.
One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas.
Mechanical Work
The easiest form of work to visualize is mechanical work (Figure $5$), which is the energy required to move an object a distance $d$ when opposed by a force $F$, such as gravity:
$w=F\,d \label{12.3.1}$
with $w$ is work, $F$ is opposing force, and $d$ is distance.
Because the force ($F$) that opposes the action is equal to the mass ($m$) of the object times its acceleration ($a$), Equation $\ref{12.3.1}$ can be rewritten to:
$w = m\,a\,d \label{12.3.2}$
with $w$ is work, $m$ is mass, $a$ is acceleration, and $d$ is distance.
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Hence for works against gravity (on Earth), $a$ can be set to $g=9.8\; m/s^2)$. 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 opposing 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.
Pressure-Volume (PV) Work
To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions. Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure $P_{int}$ and initial volume $V_i$ (Figure $6$). If $P_{ext} = P_{int}$, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston ($P_{ext}$) is less than $P_{int}$, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume ($V_f$) will be greater than $V_i$. If $P_{ext} > P_{int}$, then the gas will be compressed, and the surroundings will perform work on the system.
Figure $6$: PV Work demonstrated with a frictionless piston. (a) if the external pressure is less than $P_{int}$, the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings. The final volume ($V_f$) will be greater than $V_i$. (b) Alternatively, if the external pressure is greater than $P_{int}$, the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area $A$, the external pressure exerted by the piston is, by definition, the force per unit area:
$P_{ext} = \dfrac{F}{A} \label{eq5}$
The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height ($V = Ah$). Rearranging Equation \ref{eq5} to give
$F = P_{ext}A$
and defining the distance the piston moves ($d$) as $Δh$, we can calculate the magnitude of the work performed by the piston by substituting into Equation $\ref{12.3.1}$:
$w = F d = P_{ext}AΔh \label{12.3.3}$
The change in the volume of the cylinder ($ΔV$) as the piston moves a distance d is $ΔV = AΔh$, as shown in Figure $7$.
Figure $7$: Work Performed with a change in volume. The change in the volume ($ΔV$) of the cylinder housing a piston is $ΔV = AΔh$ as the piston moves. The work performed by the surroundings on the system as the piston moves inward is given by Equation \ref{12.3.4}.
The PV work performed is thus
$w = P_{ext}ΔV \label{12.3.4}$
The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
$w=\left(\dfrac{F}{A}\right)_{\textrm{ext}}(\Delta V)=\dfrac{\textrm{newton}}{\textrm m^2}\times \textrm m^3=\mathrm{newton\cdot m}=\textrm{joule}$
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
$R=\dfrac{0.08206\;\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}=\dfrac{8.314\textrm{ J}}{\mathrm{mol\cdot K}}$
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J.
Exercise: Expansion (PV) work
1. How much work is done by a gas that expands from 2 liters to 5 liters against an external pressure of 750 mmHg?
2. How much work is done by 0.54 moles of a gas that has an initial volume of 8 liters and expands under the following conditions: 30 oC and 1.3 atm?
3. How much work is done by a gas (P=1.7 atm, V=1.56 L) that expands against an external pressure of 1.8 atm?
Solution a
$w = − PΔV$
$ΔV = V_{final} - V_{Initial} = 5 \,L - 2\, L = 3 L$
Convert 750 mmHg to atm:
$750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm.$
$W = − pΔV = -(.9868\, atm)(3\,L) = -2.96 \,L\, atm.$
Solution b
First we must find the final volume using the ideal gas law:
$pV = nRT$
or
$V = \dfrac{nRT}{P} = \dfrac{ [(0.54 \,moles)(0.082057 (L\, atm)/ (mol\, K))(303\,K)] }{1.3\, atm} = 10.33\, L$
$ΔV = V_{final} - V_{initial} = 10.3\, L - 8\, L = 2.3\, L$
$w = − pΔV = - (1.3\, atm)(2.3 \,L) = -3\, L\, atm.$
Solution c
$w = - p * ΔV\) = - (1.8 \,atm )\, ΔV.$
Given $p_1$, $V_1$, and $p_2$, find $V_2$: $p_1V_1=p_2V_2$ (at constant $T$ and $n$)
$V_2= (V_1* P_1) / P_2$ = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L
Now,
$ΔV = V_2 - V_1=1.47\, L - 1.56\, L = -0.09$
$w = - (1.8 atm) * (-0.09 L) = 0.162 L atm.$
Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, Equation $\ref{12.3.4}$ must be written with a negative sign to describe PV work done by the system as negative:
$w = −P_{ext}ΔV \label{12.3.5}$
The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings.
A Matter of Convention
• Heat flow is defined from the system to its surroundings as negative
• Work is defined as by the system on its surroundings as negative
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining Figure $8$, in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from ($V_1$, $P_1$) to ($V_2$, $P_2$), which means that work is not a state function.
Internal energy is a state function, whereas work is not.
Example $1$: Internal Combustion Engine
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Asked for: work done
Strategy:
Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio. Use Equation $\ref{12.3.5}$ to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution:
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
\begin{align} w &= −P_{ext}ΔV \nonumber \[4pt] &= −(40.0\, atm)(0.400\, L − 0.0400 \,L) \nonumber \[4pt] &= −14.4\, L·atm \nonumber \end{align} \nonumber
Converting from liter-atmospheres to joules,
\begin{align} w &=-(14.4\;\mathrm{L\cdot atm})[101.3\;\mathrm{J/(L\cdot atm)}] \nonumber \[4pt] &= -1.46\times10^3\textrm{ J} \nonumber \end{align} \nonumber
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
Exercise $1$: Work to Breath
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?
Answer
−0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
Work and Chemical Reactions
We have stated that the change in energy ($ΔU$) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just PV work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_3)_{2(aq)} + 2H_2O_{(l)} + 2NO_{2(g)} \label{12.3.5a}$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $9$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure)
$w = −PΔV \label{12.3.6}$
The negative sign associated with $PV$ work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
The symbol $U$ represents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $U$ plus the product of its pressure $P$ and volume $V$:
$H =U + PV \label{12.3.7}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
If a chemical change occurs at constant pressure (i.e., for a given P, ΔP = 0), the change in enthalpy (ΔH) is
$ΔH = Δ(U + PV) = ΔU + ΔPV = ΔU + PΔV \label{12.3.8}$
Substituting $q + w$ for $ΔU$ (Equation $\ref{12.3.8}$) and $−w$ for $PΔV$ (Equation $\ref{12.3.6}$), we obtain
$ΔH = ΔU + PΔV = q_p + w − w = q_p \label{12.3.9}$
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $\ref{12.3.9}$ we see that at constant pressure the change in enthalpy, $ΔH$ of the system, defined as $H_{final} − H_{initial}$, is equal to the heat gained or lost.
$ΔH = H_{final} − H_{initial} = q_p \label{12.3.10}$
Just as with $ΔU$, because enthalpy is a state function, the magnitude of $ΔH$ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find $ΔH$ for a reaction, measure $q_p$ under constant pressure.
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.
Outside Links
• Gasparro, Frances P. "Remembering the sign conventions for q and w in deltaU = q - w." J. Chem. Educ. 1976: 53, 389.
• Koubek, E. "PV work demonstration (TD)." J. Chem. Educ. 1980: 57, 374. ' | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.03%3A_Quantifying_Heat_and_Work.txt |
Learning Objectives
• To calculate changes in internal energy
To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure $1$). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa.
Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved.
The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure $2$). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose.
Direction of Heat Flow
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (Figure $3$). The balanced chemical equation for the reaction is as follows:
$\ce{ 2Al(s) + Fe_2O_3(s) -> 2Fe(s) + Al_2O_3(s)} \label{5.2.1}$
We can also write this chemical equation as
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{5.2.2}$
to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat (q) is transferred from a system to its surroundings is described as exothermic. By convention, $q < 0$ for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
$\ce{heat + H_2O(s) \rightarrow H_2O(l)} \label{5.2.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
Heat is technically not a component in Chemical Reactions
Technically, it is poor form to have a $heat$ term in the chemical reaction like in Equations $\ref{5.2.2}$ and $\ref{5.2.3}$ since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists.
The First Law
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows:
$U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{5.2.4a}$
$\Delta{U_{sys}}=−ΔU_{surr} \label{5.2.4b}$
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the $CO_2/H_2O$ product mixture is less than that of the isooctane/ $O_2$ reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):
$ΔU_{sys} = q + w \label{5.2.5}$
Although $q$ and $w$ are not state functions on their own, their sum ($ΔU_{sys}$) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed.
Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
Although $q$ and $w$ are not state functions, their sum ($ΔU_{sys}$) is independent of the path taken and therefore is a state function.
Thus, because of the first law, we can determine $ΔU$ for any process if we can measure both $q$ and $w$. Heat, $q$, may be calculated by measuring a change in temperature of the surroundings. Work, $w$, may come in different forms, but it too can be measured. One important form of work for chemistry is pressure-volume work done by an expanding gas. At a constant external pressure (for example, atmospheric pressure)
$w = −PΔV \label{5.2.6}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. That is, an expanding gas does work on its surroundings, while a gas that is compressed has work done on it by the surroundings.
Example $1$
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
1. Determine the sign of $q$ to use in Equation $\ref{5.2.5}$.
2. From Equation $\ref{5.2.6}$ calculate $w$ from the values given. Substitute this value into Equation $\ref{5.2.5}$ to calculate $ΔU$.
Solution
A From Equation $\ref{5.2.5}$, we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so $w$ must be positive. From Equation $\ref{5.2.5}$,
$w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J} \nonumber$
Thus
\begin{align*} ΔU &= q + w \[4pt] &= −140 \,J + 284\, J \[4pt] &= 144\, J\end{align*} \nonumber
In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise $1$
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules?
Answer
−216 J
By convention (to chemists), both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Summary
In chemistry, the small part of the universe that we are studying is the system, and the rest of the universe is the surroundings. Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state, not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic. A reaction or process in which heat is transferred to a system from its surroundings is endothermic. The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.04%3A_The_First_Law_of_Thermodynamics.txt |
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter(A device used to measure energy changes in chemical processes. shown schematically in Figure $1$). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in internal energyE) rather than the enthalpy change (ΔH); ΔE is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔE < ΔH, the relationship between the measured temperature change and ΔHcomb is given in Equation $\ref{5.5.9}$, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
$\Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{5.5.9}$
To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation $\ref{5.42}$ to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example $4$.
Example $4$: Combustion of Glucose
The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample
Asked for: ΔHcomb of glucose
Strategy:
1. Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use Equation $ref{5.5.9}$ to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT.
2. Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.
Solution:
The first step is to use Equation $\ref{5.5.9}$ and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate qcomb from the mass of benzoic acid:
$q{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ$
From Equation $\ref{5.5.9}$,
$-C{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C$
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
$q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ$
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
$\Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol$
This result is in good agreement (< 1% error) with the value of ΔHcomb = −2803 kJ/mol that calculated using enthalpies of formation.
Exercise $4$: Combustion of Benzoic Acid
When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer
−1.30 × 103 kJ/mol | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.05%3A_Constant_Volume_Calorimetry-_Measuring_U_for_Chemical_Reactions.txt |
Learning Objectives
• To understand how enthalpy pertains to chemical reactions
We have stated that the change in energy ($ΔU$) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just $PV$ work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$\ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(l) + 2NO2(g)} \nonumber$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $1$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of $PV$ work done by multiplying the external pressure $P$ by the change in volume caused by movement of the piston ($ΔV$). At a constant external pressure (here, atmospheric pressure),
$w = −PΔV \label{5.4.2}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure ($ΔV > 0$), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases ($ΔV < 0$), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
The internal energy $U$ of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ($H$) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $U$ plus the product of its pressure $P$ and volume $V$:
$H =U + PV \label{5.4.3}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. So we can define a change in enthalpy ($\Delta H$) accordingly
$ΔH = H_{final} − H_{initial} \nonumber$
If a chemical change occurs at constant pressure (i.e., for a given $P$, $ΔP = 0$), the change in enthalpy ($ΔH$) is
\begin{align} ΔH &= Δ(U + PV) \[4pt] &= ΔU + ΔPV \[4pt] &= ΔU + PΔV \label{5.4.4} \end{align}
Substituting $q + w$ for $ΔU$ (First Law of Thermodynamics) and $−w$ for $PΔV$ (Equation $\ref{5.4.2}$) into Equation $\ref{5.4.4}$, we obtain
\begin{align} ΔH &= ΔU + PΔV \[4pt] &= q_p + \cancel{w} −\cancel{w} \[4pt] &= q_p \label{5.4.5} \end{align}
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $\ref{5.4.5}$ we see that at constant pressure the change in enthalpy, $ΔH$ of the system, is equal to the heat gained or lost.
\begin{align} ΔH &= H_{final} − H_{initial} \[4pt] &= q_p \label{5.4.6} \end{align}
Just as with $ΔU$, because enthalpy is a state function, the magnitude of $ΔH$ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find $ΔH$ for a reaction, measure $q_p$.
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction ($ΔH_{rxn}$), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so $ΔH_{rxn}$ is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so $ΔH_{rxn}$ is positive. Thus:
• $ΔH_{rxn} < 0$ for an exothermic reaction, and
• $ΔH_{rxn} > 0$ for an endothermic reaction.
In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:
sign conventions for heat flow and enthalpy changes
Reaction Type q ΔHrxn
exothermic < 0 < 0 (heat flows from a system to its surroundings)
endothermic > 0 > 0 (heat flows from the surroundings to a system)
If ΔHrxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Figure $2a$). Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure $\PageIndex{2b}$). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.y.
• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):
$\begin{matrix} heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0 \end{matrix} \label{5.4.7}$
$\begin{matrix} H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0 \end{matrix} \label{5.4.8}$
In both cases, the magnitude of the enthalpy change is the same; only the sign is different.
• Enthalpy is an extensive property (like mass). The magnitude of $ΔH$ for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation $\ref{5.4.9}$, which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:
$\ce{2Al(s) + Fe2O3(s) -> 2Fe (s) + Al2O3 (s) } + 815.5 \; kJ \label{5.4.9}$
Thus ΔH = −851.5 kJ/mol of Fe2O3. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation $\ref{5.4.10}$, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
$\ce{ 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10}$
If 4 mol of Al and 2 mol of $\ce{Fe2O3}$ react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
$- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a}$
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example $1$.
Example $1$: Melting Icebergs
Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If $ΔH$ is 6.01 kJ/mol for the reaction at 0°C and constant pressure:
$\ce{H2O(s) → H_2O(l)} \nonumber$
How much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.)
Given: energy per mole of ice and mass of iceberg
Asked for: energy required to melt iceberg
Strategy:
1. Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
2. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice.
Solution:
A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):
\begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{\text{metric ton }} \ce{H2O} \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{\text{metric ton}}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right ) \[4pt] & = 5.55\times 10^{10} \; mol \,\ce{H2O} \end{align*} \nonumber
B The energy needed to melt the iceberg is thus
$\left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber$
Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown below.
Possible sources of the approximately $3.34 \times 10^{11}\, kJ$ needed to melt a $1.00 \times 10^6$ metric ton iceberg
• Combustion of 3.8 × 103 ft3 of natural gas
• Combustion of 68,000 barrels of oil
• Combustion of 15,000 tons of coal
• $1.1 \times 10^8$ kilowatt-hours of electricity
Alternatively, we can rely on ambient temperatures to slowly melt the iceberg. The main issue with this idea is the cost of dragging the iceberg to the desired place.
Exercise $1$: Thermite Reaction
If 17.3 g of powdered aluminum are allowed to react with excess $\ce{Fe2O3}$, how much heat is produced?
Answer
273 kJ
Enthalpies of Reaction
One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, since enthalpy is a state function, all we have to know is the initial and final states of the reaction. This allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:
• Enthalpy of combustion (ΔHcomb) The change in enthalpy that occurs during a combustion reaction. Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance.
• Enthalpy of fusion (ΔHfus) The enthalpy change that acompanies the melting (fusion) of 1 mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds.
• Enthalpy of vaporization (ΔHvap) The enthalpy change that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds.
• Enthalpy of solution (ΔHsoln) The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent. The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent.
Table $1$: Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points
Substance ΔHvap (kJ/mol) ΔHfus (kJ/mol)
argon (Ar) 6.3 1.3
methane (CH4) 9.2 0.84
ethanol (CH3CH2OH) 39.3 7.6
benzene (C6H6) 31.0 10.9
water (H2O) 40.7 6.0
mercury (Hg) 59.0 2.29
iron (Fe) 340 14
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Enthalpy of Reaction: Enthalpy of Reaction, YouTube(opens in new window) [youtu.be]
Summary
For a chemical reaction, the enthalpy of reaction ($ΔH_{rxn}$) is the difference in enthalpy between products and reactants; the units of $ΔH_{rxn}$ are kilojoules per mole. Reversing a chemical reaction reverses the sign of $ΔH_{rxn}$. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.06%3A_Enthalpy-_The_Heat_Evolved_in_a_Chemical_Reaction_at_Constant_Pressure.txt |
Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure $1$), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6 °C). Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and $ΔH_{rxn}$ is
$\Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{5.5.8}$
The use of a constant-pressure calorimeter is illustrated in Example $3$.
Example $1$
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:
1. Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
2. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.5.8.
3. Use the molar mass of KOH to calculate ΔHsoln.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
$\left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g$
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
$q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ$
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation $\ref{5.5.8}$, we see that
ΔHrxn = −qcalorimeter = −5.13 kJ
This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
$\Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol$
Exercise $1$
A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example $1$, find ΔHsoln for NH4Br (in kilojoules per mole).
16.6 kJ/mol | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.07%3A_Constant_Pressure_Calorimetry-_Measuring_H__for_Chemical_Reactions.txt |
Learning Objectives
• To use Hess’s law and thermochemical cycles to calculate enthalpy changes of chemical reactions.
Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s law, after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
Hess's Law argues that ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This is nothing more than arguing that ΔH is a state function.
We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation $\ref{5.6.1}$ can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.6.1, the first reaction produces 1 mol of solid aluminum oxide (Al2O3) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −732.5 kJ/mol of Fe2O3. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.6.1); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.6.1, the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation 5.6.1 gives the overall reaction, shown in part (d):
\small \newcommand{\Celsius}{^{\circ}\text{C}} \begin{align*} \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> 2 Fe (l, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius)} & \Delta H = - 732.5\,\text{kJ} && \text{(a)} \ \ce{2 Fe (l, 1758 \Celsius) &-> 2 Fe (s, 1758 \Celsius)} & \Delta H = -\phantom{0}27.6\,\text{kJ} && \text{(b)} \ \ce{2 Fe (s, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius) &-> 2 Fe (s, 25 \Celsius) + Al2O3 (s, 25 \Celsius) } & \Delta H = -\phantom{0}91.0\,\text{kJ} && \text{(c)} \[2ex] \hline \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> Al2O3 (s, 25 \Celsius) + 2 Fe (s, 25 \Celsius) } & \Delta H = -851.1\,\text{kJ} && \text{(d)} \ \end{align*} \label{5.6.1} \tag{5.6.1}
The net reaction in part (d) in Equation $\ref{5.6.1}$ is identical to the equation for the thermite reaction that we saw in a previous section. By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure $1$.
Comparing parts (a) and (d) in Equation $\ref{5.6.1}$ also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation $\ref{5.6.1}$), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation $\ref{5.6.1}$). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation.
When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:
1. Identify the equation whose ΔH value is unknown and write individual reactions with known ΔH values that, when added together, will give the desired equation. We illustrate how to use this procedure in Example $1$.
2. Arrange the chemical equations so that the reaction of interest is the sum of the individual reactions.
3. If a reaction must be reversed, change the sign of ΔH for that reaction. Additionally, if a reaction must be multiplied by a factor to obtain the correct number of moles of a substance, multiply its ΔH value by that same factor.
4. Add together the individual reactions and their corresponding ΔH values to obtain the reaction of interest and the unknown ΔH.
Example $1$
When carbon is burned with limited amounts of oxygen gas (O2), carbon monoxide (CO) is the main product:
$\left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ} \nonumber$
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
$\left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O2 to give CO2.
Given: two balanced chemical equations and their ΔH values
Asked for: enthalpy change for a third reaction
Strategy:
1. After balancing the chemical equation for the overall reaction, write two equations whose ΔH values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of ΔH.)
2. Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each.
Solution:
A We begin by writing the balanced chemical equation for the reaction of interest:
$\left ( 3 \right ) \; \ce{CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H_{rxn}=? \nonumber$
There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give
$\ce{2 CO (g) -> 2 C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ} \nonumber$
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO2, is the product we want in Equation 3:
$\ce{C (s) + O2 (g) -> CO2 (s)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
B Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:
$\ce{2 C (s) + 2 O2 (g) -> 2 CO2 (s)} \quad \Delta H=-787.0 \; \text{kJ} \nonumber$
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
\begin{align*} \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = -\Delta H_1 = +221.0 \; \text{kJ} \ \ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & = -2\Delta H_2 =-787.0 \; \text{kJ} \[2ex] \hline \ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ} \end{align*} \nonumber
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O2, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:
$\ce{ CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H = -283.0 \; \text{kJ} \nonumber$
An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:
\begin{alignat*}{3} \text{(A)} \quad && \ce{ 2 C (s) + O2 (g) &-> \cancel{2 CO (g)}} \qquad & \Delta H_A &= \Delta H_1 &&= + 221.0 \; \text{kJ} \ \text{(B)} \quad && \ce{ \cancel{2 CO (g)} + O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H_B && &= ? \ \text{(C)} \quad && \ce{2 C (s) + 2 O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H &= 2 \Delta H_2 &= 2 \times \left ( -393.5 \; \text{kJ} \right ) &= -787.0 \; \text{kJ} \end{alignat*}
The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO2. From Hess’s law, ΔHA + ΔHB = ΔHC, and we are given ΔH for reactions A and C. Substituting the appropriate values gives
$\begin{matrix} -221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \ \Delta H_{B} = -566.0 \end{matrix} \nonumber$
This is again the enthalpy change for the conversion of 2 mol of CO to CO2. The enthalpy change for the conversion of 1 mol of CO to CO2 is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem.
Exercise $1$
The reaction of acetylene (C2H2) with hydrogen (H2) can produce either ethylene (C2H4) or ethane (C2H6):
$\begin{matrix} C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right ) & \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \ C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2} \end{matrix} \nonumber$
What is ΔH for the reaction of C2H4 with H2 to form C2H6?
Answer
−136.3 kJ/mol of C2H4
Hess’s Law: Hess's Law, YouTube(opens in new window) [youtu.be]
Summary
Hess's law is arguing the overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions. For a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporization (ΔHvap), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion (ΔHcomb) is the enthalpy change that occurs when a substance is burned in excess oxygen. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.08%3A_Relationships_Involving_Enthalpy_of_Reactions.txt |
Learning Objectives
• To understand Enthalpies of Formation and be able to use them to calculate Enthalpies of Reaction
One way to report the heat absorbed or released by chemical reactions would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, starting from the elemental forms of each atom at 25 oC and 1 atm pressure.
Enthalpy of formation ($ΔH_f$) is the enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The formation of any chemical can be as a reaction from the corresponding elements:
$\text{elements} \rightarrow \text{compound} \nonumber$
which in terms of the the Enthalpy of formation becomes
$\Delta H_{rxn} = \Delta H_{f} \label{7.8.1}$
For example, consider the combustion of carbon:
$\ce{ C(s) + O2 (g) -> CO2 (g)} \nonumber$
then
$\Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \nonumber$
The sign convention for ΔHf is the same as for any enthalpy change: $ΔH_f < 0$ if heat is released when elements combine to form a compound and $ΔH_f > 0$ if heat is absorbed.
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Standard Enthalpies of Formation
The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.
The standard conditions for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state, which is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ($ΔH^o_f$) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.
The standard enthalpy of formation of any element in its standard state is zero by definition.
For example, although oxygen can exist as ozone (O3), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H2(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure $1$). Therefore, $\ce{O2(g)}$, $\ce{H2(g)}$, and graphite have $ΔH^o_f$ values of zero.
The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
$6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \label{7.8.2}$
It is not possible to measure the value of $ΔH^oo_f$ for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, $\ce{O2}$, and $\ce{H2}$ and measuring the heat evolved as glucose is formed since the reaction shown in Equation $\ref{7.8.2}$ does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of $ΔH^oo_f$ are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of $ΔH^o_f$ for an extensive list of compounds are given in Table T1. Note that $ΔH^o_f$ values are always reported in kilojoules per mole of the substance of interest. Also notice in Table T1 that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.
Example $1$: Enthalpy of Formation
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{HCl(g)}$
2. $\ce{MgCO3(s)}$
3. $\ce{CH3(CH2)14CO2H(s)}$ (palmitic acid)
Given:
compound formula and phase.
Asked for:
balanced chemical equation for its formation from elements in standard states
Strategy:
Use Table T1 to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made.
Solution:
To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in Table T1: by a $ΔH^o_f$ value of 0 kJ/mol.
Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are $\ce{H2(g)}$ and $\ce{Cl2(g)}$, respectively, the unbalanced chemical equation is
$\ce{H2(g) + Cl2(g) \rightarrow HCl(g)} \nonumber$
Fractional coefficients are required in this case because ΔHof values are reported for 1 mol of the product, $\ce{HCl}$. Multiplying both $\ce{H2(g)}$ and $\ce{Cl2(g)}$ by 1/2 balances the equation:
$\ce{1/2 H_{2} (g) + 1/2 Cl_{2} (g) \rightarrow HCl (g)} \nonumber$
The standard states of the elements in this compound are $\ce{Mg(s)}$, $\ce{C(s, graphite)}$, and $\ce{O2(g)}$. The unbalanced chemical equation is thus
$\ce{Mg(s) + C (s, graphite) + O2 (g) \rightarrow MgCO3 (s)} \nonumber$
This equation can be balanced by inspection to give
$\ce{Mg (s) + C (s, graphite ) + 3/2 O2 (g)\rightarrow MgCO3 (s)} \nonumber$
Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows:
$\ce{C(s, graphite) + H2(g) + O2(g) \rightarrow CH3(CH2)14CO2H(s)} \nonumber$
There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is
$\ce{16C (s, graphite) + 16 H2(g) + O2(g) -> CH3(CH2)14CO2H(s) } \nonumber$
Exercise $1$
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{NaCl(s)}$
2. $\ce{H2SO4(l)}$
3. $\ce{CH3CO2H(l)}$ (acetic acid)
Answer a
$\ce{ Na (s) + 1/2 Cl2 (g) \rightarrow NaCl (s)} \nonumber$
Answer b
$\ce{H_{2} (g) + 1/8 S8 (s) + 2O2 ( g) \rightarrow H2 SO4( l) } \nonumber$
Answer c
$\ce{2C(s) + O2(g) + 2H2(g) -> CH3CO2H(l)} \nonumber$
Definition of Heat of Formation Reactions: https://youtu.be/A20k0CK4doI
Standard Enthalpies of Reaction
Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose $\Delta{H_f^o}$ values are known. The standard enthalpy of reaction $\Delta{H_{rxn}^o}$ is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction
$aA + bB \rightarrow cC + dD \label{7.8.3}$
where $A$, $B$, $C$, and $D$ are chemical substances and $a$, $b$, $c$, and $d$ are their stoichiometric coefficients. The magnitude of $ΔH^ο$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
$\Delta H_{rxn}^{o} = \underbrace{ \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] }_{\text{products} } - \underbrace{ \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ]}_{\text{reactants }} \label{7.8.4}$
More generally, we can write
$\Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \label{7.8.5}$
where the symbol $\sum$ means “sum of” and $m$ and $n$ are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation $\ref{7.8.5}$ arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.
"Products minus reactants" summations are typical of state functions.
To demonstrate the use of tabulated ΔHο values, we will use them to calculate $ΔH_{rxn}$ for the combustion of glucose, the reaction that provides energy for your brain:
$\ce{ C6H12O6 (s) + 6O2 (g) \rightarrow 6CO2 (g) + 6H2O (l)} \label{7.8.6}$
Using Equation $\ref{7.8.5}$, we write
$\Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ \Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \label{7.8.7}$
From Table T1, the relevant ΔHοf values are ΔHοf [CO2(g)] = -393.5 kJ/mol, ΔHοf [H2O(l)] = -285.8 kJ/mol, and ΔHοf [C6H12O6(s)] = -1273.3 kJ/mol. Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol. Inserting these values into Equation $\ref{7.8.7}$ and changing the subscript to indicate that this is a combustion reaction, we obtain
\begin{align} \Delta H_{comb}^{o} &= \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] - \left [-1273.3 + 6\left ( 0 \; kJ\;mol \right ) \right ] \label{7.8.8} \[4pt] &= -2802.5 \; kJ/mol \end{align}
As illustrated in Figure $2$, we can use Equation $\ref{7.8.8}$ to calculate $ΔH^ο_f$ for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled $ΔH^ο_{comb}$. The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the $ΔH^ο_f$ values of the reactants. Consequently, the enthalpy changes are
\begin{align} \Delta H_{1}^{o} &= \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] \nonumber \[4pt] &= -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) \nonumber \[4pt] &= +1273.3 \; kJ \nonumber \[4pt] \Delta H_{2}^{o} &= 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] \nonumber \[4pt] & =6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) \nonumber \[4pt] &= 0 \; kJ \end{align} \label{7.8.9}
Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change (Equation \ref{7.8.4} since the products are now reactants and vice versa.
The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O2) to the elements is therefore +1273.3 kJ.
The reactions that convert the elements to final products (downward purple arrows in Figure $2$) are identical to those used to define the ΔHοf values of the products. Consequently, the enthalpy changes (from Table T1) are
$\begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix}$
The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):
$\Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \label{7.8.10}$
This is the same result we obtained using the “products minus reactants” rule (Equation $\ref{7.8.5}$) and ΔHοf values. The two results must be the same because Equation $\ref{7.8.10}$ is just a more compact way of describing the thermochemical cycle shown in Figure $1$.
Example $2$: Heat of Combustion
Long-chain fatty acids such as palmitic acid ($\ce{CH3(CH2)14CO2H}$) are one of the two major sources of energy in our diet ($ΔH^o_f$ =−891.5 kJ/mol). Use the data in Table T1 to calculate ΔHοcomb for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel — glucose or palmitic acid?
Given: compound and $ΔH^ο_{f}$ values
Asked for: $ΔH^ο_{comb}$ per mole and per gram
Strategy:
1. After writing the balanced chemical equation for the reaction, use Equation $\ref{7.8.5}$ and the values from Table T1 to calculate $ΔH^ο_{comb}$ the energy released by the combustion of 1 mol of palmitic acid.
2. Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation $\ref{7.8.8}$ for the combustion of glucose to determine which is the better fuel.
Solution:
A To determine the energy released by the combustion of palmitic acid, we need to calculate its $ΔH^ο_f$. As always, the first requirement is a balanced chemical equation:
$C_{16}H_{32}O_{2(s)} + 23O_{2(g)} \rightarrow 16CO_{2(g)} + 16H_2O_{(l)} \nonumber$
Using Equation $\ref{7.8.5}$ (“products minus reactants”) with ΔHοf values from Table T1 (and omitting the physical states of the reactants and products to save space) gives
\begin{align*} \Delta H_{comb}^{o} &= \sum m \Delta H^o_f\left( {products} \right) - \sum n \Delta H^o_f \left( {reactants} \right) \[4pt] &= \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \[4pt] & - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \[4pt] &= -9977.3 \; kJ/mol \nonumber \end{align*}
This is the energy released by the combustion of 1 mol of palmitic acid.
B The energy released by the combustion of 1 g of palmitic acid is
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \nonumber$
As calculated in Equation $\ref{7.8.8}$, $ΔH^o_f$ of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \nonumber$
The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight.
Exercise $2$: Water–gas shift reaction
Use Table T1 to calculate $ΔH^o_{rxn}$ for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H2(g):
$\ce{ CO ( g ) + H2O (g ) -> CO2 (g) + H2 ( g )} \nonumber$
Answer
−41.2 kJ/mol
We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s $ΔH^ο_f$ which we cannot obtain otherwise. This procedure is illustrated in Example $3$.
Example $3$: Tetraethyllead
Beginning in 1923, tetraethyllead [$\ce{(C2H5)4Pb}$] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are $\ce{CO2(g)}$, $\ce{H2O(l)}$, and red $\ce{PbO(s)}$. What is the standard enthalpy of formation of tetraethyllead, given that $ΔH^ο_f$ is −19.29 kJ/g for the combustion of tetraethyllead and $ΔH^ο_f$ of red PbO(s) is −219.0 kJ/mol?
Given: reactant, products, and $ΔH^ο_{comb}$ values
Asked for: $ΔH^ο_f$ of the reactants
Strategy:
1. Write the balanced chemical equation for the combustion of tetraethyl lead. Then insert the appropriate quantities into Equation $\ref{7.8.5}$ to get the equation for ΔHοf of tetraethyl lead.
2. Convert $ΔH^ο_{comb}$ per gram given in the problem to $ΔH^ο_{comb}$ per mole by multiplying $ΔH^ο_{comb}$ per gram by the molar mass of tetraethyllead.
3. Use Table T1 to obtain values of $ΔH^ο_f$ for the other reactants and products. Insert these values into the equation for $ΔH^ο_f$ of tetraethyl lead and solve the equation.
Solution:
A The balanced chemical equation for the combustion reaction is as follows:
$\ce{2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)} \nonumber$
Using Equation $\ref{7.8.5}$ gives
$\Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \nonumber$
Solving for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
$\Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \nonumber$
The values of all terms other than $ΔH^o_f [\ce{(C2H5)4Pb}]$ are given in Table T1.
B The magnitude of $ΔH^o_{comb}$ is given in the problem in kilojoules per gram of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get $ΔH^o_{comb}$ for 1 mol of tetraethyl lead:
\begin{align*} \Delta H_{comb}^{o} &= \left ( \dfrac{-19.29 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) \[4pt] &= -6329 \; kJ/mol \end{align*}
Because the balanced chemical equation contains 2 mol of tetraethyllead, $ΔH^o_{rxn}$ is
\begin{align*} \Delta H_{rxn}^{o} &= 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6329 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) \[4pt] &= -12,480 \; kJ \end{align*}
C Inserting the appropriate values into the equation for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
\begin{align*} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\[4pt] &= -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{align*}
Exercise $3$
Ammonium sulfate, $\ce{(NH4)2SO4}$, is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid:
$\ce{2NH3(g) + H2SO4(aq) \rightarrow (NH4)2SO4(s)} \nonumber$
The value of $ΔH^o_{rxn}$ is -179.4 kJ/mole $\ce{H2SO4}$. Use the data in Table T1 to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).
Answer
−1181 kJ/mol
Calculating DH° using DHf°: https://youtu.be/Y3aJJno9W2c
Summary
• The standard state for measuring and reporting enthalpies of formation or reaction is 25 oC and 1 atm.
• The elemental form of each atom is that with the lowest enthalpy in the standard state.
• The standard state heat of formation for the elemental form of each atom is zero.
The enthalpy of formation ($ΔH_{f}$) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation ($ΔH^o_{f}$) are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The standard enthalpy of reaction ($ΔH^o_{rxn}$) can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy of solution ($ΔH_{soln}$) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/06%3A_Thermochemistry/6.09%3A_Enthalpies_of_Reaction_from_Standard_Heats_of_Formation.txt |
• 7.1: Schrödinger's Cat
Schrödinger's cat is a thought experiment, sometimes described as a paradox, where a scenario is conceived such that a cat that may be simultaneously both alive and dead. This is a state known as a quantum superposition, as a result of being linked to a random subatomic event that may or may not occur. The thought experiment is also often featured in theoretical discussions of the interpretations of quantum mechanics.
• 7.2: The Nature of Light
Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of 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.
• 7.3: Atomic Spectroscopy and The Bohr Model
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e
• 7.4: The Wavelength Nature of Matter
An electron possesses both particle and wave properties. 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. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. 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.
• 7.5: Quantum Mechanics and The Atom
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 to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies.
• 7.6: 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.
• 7.E: The Quantum-Mechanical Model of the Atom (Exercises)
07: The Quantum-Mechanical Model of the Atom
In the early 1930's Erwin 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 describes the world in terms of a wave function. DeWitt wrote about the cat that "at the end of [one half-life] the total wave function for the system will have a form in which the living cat and dead cat are mixed in equal portions." (Reference: B.S. DeWitt and N. Graham, eds., The Many-Worlds Interpretation of Quantum Mechanics (Princeton, 1973), pg. 156.) When we open the box, we "collapse the wave function" or "collapse the state" and have either a live cat or a dead cat.
Of course, this is just a thought experiment. So far as I know nobody has actually every done this experiment. In a sense the cat is a "red herring" [sorry!]. The paradox is just an illuminating way of thinking about the consequences of radioactive decay being totally random. Imagine we have a friend waiting outside when we open the box. For us the wave function collapses and we have, say, a live cat. But our friend's wave function does not collapse until he comes into the room. This leads to a strong solipsism, since our friend can they say that we owe our objective existence to his kind intervention in coming into the room and collapsing our state.
As Heisenberg said, then, "The wave function represents partly a fact and partly our knowledge of a fact." Our friend needn't have come into the room to collapse his wave function: if we have a cell phone we can call him and tell him the result of the experiment. Of course, this assumes that we don't lie to him and tell him the cat is dead when it is alive. Unexplained but apparently true is the fact that when a state collapses, it collapses into the same state for everybody. If we see a live cat everybody sees a live cat (unless they or us are hallucinating). | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.01%3A_Schrodinger%27s_Cat.txt |
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 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. 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 passes a particular point on the surface of the water, anything floating there moves up and down. —between the midpoints of two peaks, for example, or two troughs—is the ($λ$, lowercase Greek lambda). Wavelengths are described by a unit of distance, typically meters. The ($u$, lowercase Greek nu) of a 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), which in the SI system is called the hertz (Hz). is defined as half the peak-to-trough height; as the amplitude of a 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 per unit time is its speed ($v$), which is typically measured in meters per second (m/s). The speed of a is equal to the product of its wavelength and frequency: }} \right )\left ( \dfrac{\cancel{\text{}}}{\text{second}} \right ) &=\dfrac{\text{meters}}{\text{second}} \label{6.1.1b} \end{align} \] (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 . (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 (), which turns out to be a physical constant with a value of 2.99792458 × 10 m/s (about 3.00 ×10 m/s or 1.86 × 10 mi/s). This is about a million times faster than the speed of sound. , with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the of the atmosphere absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. requires an understanding of the properties of waves and electromagnetic radiation. A is a periodic oscillation by which energy is transmitted through space. All waves are , repeating regularly in both space and time. Waves are characterized by several interrelated properties: ($λ$), the distance between successive waves; ($u$), the number of waves that pass a fixed point per unit time; ($v$), the rate at which the propagates through space; and , the magnitude of the oscillation about the mean position. The speed of a is equal to the product of its wavelength and frequency. consists of two perpendicular waves, one electric and one magnetic, propagating at the ($c$). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.02%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 $4$
The Bohr Atom:
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.
The Energy States of the Hydrogen Atom
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_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.03%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. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom.
The Wave Character of Matter
Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the famous equation:
$E=h u=h\dfrac{c}{\lambda }=mc^{2} \label{6.4.1}$
According to this theory, a photon of wavelength $λ$ and frequency $u$ has a nonzero mass, which is given as follows:
$m=\dfrac{E}{c^{2}}=\dfrac{h u }{c^{2}}=\dfrac{h}{\lambda c} \label{6.4.2}$
That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle.
In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie 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
• $h$ is Planck’s constant,
• $m$ is the mass of the particle, and
• $v$ is the velocity of the particle.
This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. 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, 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 too short (less than the diameter of an atomic nucleus) to be noticeable.
The de Broglie Equation:
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
$\lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \nonumber$
(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 $1$).
An Important Wave Property: Phase And Interference
A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the relative displacement of two waves, called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure $\PageIndex{2A}$ shows an arbitrary phase difference between two wave and Figure $\PageIndex{2B}$ shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure $\PageIndex{2C}$ shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities. A pattern of constructive and destructive interference is obtained when two (or more) diffracting waves interact with each other. This principle of diffraction and interference was used to prove the wave properties of electrons and is the basis for how electron microscopes work.
Photograph of an interference pattern produced by circular water waves in a ripple tank.
For a mathematical analysis of phase aspects in sinusoids, check the math Libretexts library.
Standing Waves
De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure $3$). If the length of the string is $L$, then the lowest-energy vibration (the fundamental) has wavelength
\begin{align} \dfrac{\lambda }{2} & =L \nonumber \ \lambda &= 2L \nonumber \end{align} \label{6.4.4}
Higher-energy vibrations are called overtones (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by
$\lambda=\dfrac{2L}{n} \label{6.4.5}$
where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure $3$ that all overtones have one or more nodes, points where the string does not move. The amplitude of the wave at a node is zero.
Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength
$2πr = nλ \label{6.4.6}$
where $r$ is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (Figure $4$). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference. Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. The nonresonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to $n = 1$ in Equation $\ref{6.4.6}$, one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
Like all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far, can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we now know to be incorrect: in the currently accepted model, the electron in a given orbit is not always at the same distance from the nucleus.
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. A characteristic of light is that is can be bent or spread out by passing through a narrow slit. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction.
This behavior of waves is captured in Maxwell's equations (1870 or so) for electromagnetic waves and was and is well understood. An "uncertainty principle" for light is, if you will, 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 wavelike characteristics will also undergo diffraction from slits whose size is on the order of the electron wavelength.
This situation 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.
Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify simultaneously both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the most probable radius of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model.
Example $1$: Quantum Nature of Baseballs
Calculate the minimum uncertainty in the position of the pitched baseball from Example $\ref{6.4.1}$ that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h.
Given: mass and speed of object
Asked for: minimum uncertainty in its position
Strategy:
1. Rearrange the inequality that describes the Heisenberg uncertainty principle (Equation $\ref{6.4.7}$) to solve for the minimum uncertainty in the position of an object (Δx).
2. Find Δv by converting the velocity of the baseball to the appropriate SI units: meters per second.
3. Substitute the appropriate values into the expression for the inequality and solve for Δx.
Solution:
A The Heisenberg uncertainty principle (Equation \ref{6.4.7}) tells us that $(Δx)(Δ(mv)) = h/4π \nonumber$. Rearranging the inequality gives
$\Delta x \ge \left( {\dfrac{h}{4\pi }} \right)\left( {\dfrac{1}{\Delta (mv)}} \right)$
B We know that h = 6.626 × 10−34 J•s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ(mv) = mΔv and Δv = ±1 mi/h. We have
$\Delta u =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \nonumber$
C Therefore,
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \nonumber$
Inserting the definition of a joule (1 J = 1 kg•m2/s2) gives
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \nonumber$
$\Delta x \ge 7.92 \pm \times 10^{-34}\; m \nonumber$
This is equal to $3.12 \times 10^{−32}$ inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.
Exercise $2$
Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest.
Answer
6 × 10−10 m, or 0.6 nm (about the diameter of a benzene molecule)
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_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.04%3A_The_Wavelength_Nature_of_Matter.txt |
Learning Objectives
• To apply the results of quantum mechanics to chemistry.
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.
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.
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.
Wavefunctions
A wavefunction (Ψ) 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 example, if you were the captain of a ship trying to intercept an enemy submarine, you would need to know its latitude, longitude, and depth, as well as the time at which it was going to be at this position (Figure $1$). 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.
Quantum Numbers
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.
Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell:
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.
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:
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
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): Magnetic Quantum Number (ml) & Spin Quantum Number (ms) YouTube(opens in new window) [youtu.be] (opens in new window)
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_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.05%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_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.06%3A_The_Shape_of_Atomic_Orbitals.txt |
These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. 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.
Q7.45
What wavelength of light in meters produces this amount of energy? Round to two significant figures.
Formula for Energy is E=hc/lambda
h is Planck's constant it is equivalent to 6.626 x 10-34
c is the speed of light constant it is equivalent to 2.998 x 108
lambda is the wavelength, which in this problem you are trying to find
E is the energy which is what you will be given in this problem
Examples:
a) 1.33 x 10-19J
1.33 x 10-19J =hc/lambda
1.33 x 10-19 = (6.626 x 10-34)(2.998 x 108)/lambda
1.33 x 10-19= 1.98 x 10-25/lambda
1.33 x 10-19(lambda)= 1.986 x 10-25
lambda =1.98 x 10-25/1.33 x 10-19
lambda= 1.4936 x 10-6
lambda = 1.5 x 10-6
lambda = 1500 x 10 -9 m
b) 3.98 x 10-19 J
3.98 x 10-19J = hc/lambda
3.98 x 10 -19 = (6.626 x 10-34)(2.998 x 108)/lambda
3.98 x 10-19 = 1.986 x 10-25/lambda
3.98 x 10-19(lambda) = 1.986 x 10-25
lambda = 1.986 x 10-25/3.98 x 10-19
lambda = 4.989 x 10-7
lambda = 5.0 x 10-7
lambda = 500 x 10-9 m
c) 1.32 x 10-18J
1.32 x 10-18J = hc/lambda
1.32 x 10-18 = (6.626 x 10-34)(2.998 x 108)/lambda
1.32 x 10-18 = 1.986 x 10-25/lambda
1.32 x 10-18(lambda) = 1.986 x 10-25
lambda = 1.5045 x 10-7
lambda = 1.5 x 10-7
lambda = 150 x 10-9 m
Q7.57a
How many electrons are in a s orbital verses a p orbital?
Notation to know:
• e- is the notation for an electron
• Orbital- the energy state in an atom where an electron will be found
1. Figure out where an s orbital and p orbital would be located
2. Determine which level each one is
3. Use which level they are to determine how many boxes electrons can occupy
4. Realize that two opposite spinning electrons can occupy one box
1. Using the number of boxes and the knowledge that two electrons can occupy a box, multiply the number of boxes by the two orbitals
2. Write your answer
Solution
The s orbital has one box that could have electrons in it since s is the lowest orbital. p is the next level up, which has three boxes electrons can occupy. Each box has the capacity for two electrons to occupy the space. Since s has one box, and p has three boxes, the number of electrons that could possibly be in an s orbital is two, while a p orbital has 6.
Q7.57b
If an electron is excited from a 3p orbital to a 2p orbital, does it give off energy or gain energy? Why?
Strategy
1. Identify the location of each orbital and how this relates to their energy levels
2. Explain why your answer is correct, referring to the movement of photons
Answer
The 3p orbital is further away from the nucleus than the 2p orbital.The closer an electron is to the nucleus the less amount of energy it has.Thus, the electron will give off energy due to its movement from a higher level orbital to a lower level orbital releasing a photon of energy in the process. When the electron gets excited it will jump back up to a higher level orbital.
Q7.59
When each value of n is given, what are the possible values for l?
1. 2
2. 4
3. 6
4. 8
Strategy:
• The relationship between the quantum number (n) and the angular momentum quantum number (l).
• The principle quantum number is an integer that determines the overall size and energy of an orbital. The possible values for n are 1, 2, 3, … and so on.
• The angular momentum quantum number is an integer that determines the shape of the orbital. Every value of n has a certain l value. In other words, for a given value of n, l can be any integer up to n-1.
Example:
If n=3, what are the possible values of l?
Since l = n-1, and n=3. Substitute 3 with n (l=3-1), which l equals to 2 maximum values or 0, 1, 2.
Solution:
1. n= 2
l= n-1
l= 0, 1
1. n= 4
l= n-1
l= 0, 1, 2, 3
1. n= 6
l= n-1
l= 0, 1, 2, 3, 4, 5
1. n= 8
l= n-1
l= 0, 1, 2, 3, 4, 5, 6, 7
Q7.78a
A photon of wavelength 0.787 nm strikes a surface. The surface has a binding energy of 1.35 *1010 10^-19 kJ/mol. What is the kinetic energy of the emitted electron in eV?
Strategy
Solution
1. Convert kJ/mol à J to keep units consistent.
2. Solve For energy of the photon
3. Subtract binding energy from the energy of the photon
4. Convert J --> eV
1. 1.35*10^-19 kJ/mol = 1.35*10-16 J/mol
2. Energy=(Planck's Constant)(Speed of light)/ wavelength
1. h= 6.626* 10^-34 (Planck’s Constant)
2. c=2.99*1010 10^-16 m/s (speed of light)
3. E=Energy
4. wavelength (convert to meters)
3. Energy of photon= binding energy + kinetic energy
1. 1.67*r10^-16 J=Kinetic energy
4. 1.167*10^-16 J* * 1 eV/1.603*10^-19 J
1. =728.5 eV
Q7.78b
A photon which has a wave length of 0.875 nm strikes a surface. The emitted electron has a kinetic energy of 900 eV. Calculate the binding energy of the electron in kJ/mol.
Solution
First we must change our value in units of eV to joules using the conversion 1 eV=1.602x10-19 J. This gives us a value of 1.4418x10-16J | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.E%3A_The_Quantum-Mechanical_Model_of_the_Atom_%28Exercises%29.txt |
Electric currents in the vastly complex system of billions of nerves in our body allow us to sense the world, control parts of our body, and think. These are representative of the three major functions of nerves. First, nerves carry messages from our sensory organs and others to the central nervous system, consisting of the brain and spinal cord. Second, nerves carry messages from the central nervous system to muscles and other organs. Third, nerves transmit and process signals within the central nervous system. The sheer number of nerve cells and the incredibly greater number of connections between them makes this system the subtle wonder that it is. Nerve conduction is a general term for electrical signals carried by nerve cells. It is one aspect of bioelectricity, or electrical effects in and created by biological systems.
Nerve cells, properly called neurons, look different from other cells—they have tendrils, some of them many centimeters long, connecting them with other cells. (See Figure 1.) Signals arrive at the cell body across synapses or through dendrites, stimulating the neuron to generate its own signal, sent along its long axon to other nerve or muscle cells. Signals may arrive from many other locations and be transmitted to yet others, conditioning the synapses by use, giving the system its complexity and its ability to learn.
The method by which these electric currents are generated and transmitted is more complex than the simple movement of free charges in a conductor, but it can be understood with principles already discussed in this text. The most important of these are the Coulomb force and diffusion.
Figure 2 illustrates how a voltage (potential difference) is created across the cell membrane of a neuron in its resting state. This thin membrane separates electrically neutral fluids having differing concentrations of ions, the most important varieties being \(Na^{+}\), \(K^{+}\), and \(Cl^{-}\). (these are sodium, potassium, and chlorine ions with single plus or minus charges as indicated). As discussed in Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes, free ions will diffuse from a region of high concentration to one of low concentration. But the cell membrane is semipermeable, meaning that some ions may cross it while others cannot. In its resting state, the cell membrane is permeable to \(K^{+}\) and \(Cl^{-}\), and impermeable to \(Na^{+}\). Diffusion of \(K^{+}\) and \(Cl^{-}\) thus creates the layers of positive and negative charge on the outside and inside of the membrane. The Coulomb force prevents the ions from diffusing across in their entirety. Once the charge layer has built up, the repulsion of like charges prevents more from moving across, and the attraction of unlike charges prevents more from leaving either side. The result is two layers of charge right on the membrane, with diffusion being balanced by the Coulomb force. A tiny fraction of the charges move across and the fluids remain neutral (other ions are present), while a separation of charge and a voltage have been created across the membrane.
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8.02: The Development of the Periodic Table
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_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.01%3A_Nerve_Signal_Transmission.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_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.03%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
8.05: The Explanatory Power of the Quantum-Mechanical Model
The chemical properties of elements are largely determined by the number of valence electrons they contain | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.04%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_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.06%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_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.07%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_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.08%3A_Electron_Affinities_and_Metallic_Character.txt |
The elements within the same group of the periodic table tend to exhibit similar physical and chemical properties. Four major factors affect reactivity of metals: nuclear charge, atomic radius, shielding effect and sublevel arrangement (of electrons). Metal reactivity relates to ability to lose electrons (oxidize), form basic hydroxides, form ionic compounds with non-metals. In general, the bigger the atom, the greater the ability to lose electrons. The greater the shielding, the greater the ability to lose electrons. Therefore, metallic character increases going down the table, and decreases going across -- so the most active metal is towards the left and down.
Group 1: The Alkali Metals
The word "alkali" is derived from an Arabic word meaning "ashes". Many sodium and potassium compounds were isolated from wood ashes ($\ce{Na2CO3}$ and $\ce{K2CO3}$ are still occasionally referred to as "soda ash" and "potash"). In the alkali group, as we go down the group we have elements Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr). Several physical properties of these elements are compared in Table $1$. These elements have all only one electron in their outermost shells. All the elements show metallic properties and have valence +1, hence they give up electron easily.
Table $1$: General Properties of Group I Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Lithium $[He]2s^1$ 181 0.53 1.52 520
Sodium $[Ne]3s^1$ 98 0.97 1.86 496
Potassium $[Ar]4s^1$ 63 0.86 2.27 419
Rubidium $[Kr]5s^1$ 39 1.53 2.47 403
Cesium $[Xe]6s^1$ 28 1.88 2.65 376
As we move down the group (from Li to Fr), the following trends are observed (Table $1$):
• All have a single electron in an 's' valence orbital
• The melting point decreases
• The density increases
• The atomic radius increases
• The ionization energy decreases (first ionization energy)
The alkali metals have the lowest $I_1$ values of the elements
This represents the relative ease with which the lone electron in the outer 's' orbital can be removed.
The alkali metals are very reactive, readily losing 1 electron to form an ion with a 1+ charge:
$M \rightarrow M^+ + e- \nonumber$
Due to this reactivity, the alkali metals are found in nature only as compounds. The alkali metals combine directly with most nonmetals:
• React with hydrogen to form solid hydrides
$2M_{(s)} + H_{2(g)} \rightarrow 2MH(s) \nonumber$
(Note: hydrogen is present in the metal hydride as the hydride H- ion)
• React with sulfur to form solid sulfides
$2M_{(s)} + S_{(s)} \rightarrow M_2S_{(s)} \nonumber$
React with chlorine to form solid chlorides
$2M_{(s)} + Cl_{2(g)} \rightarrow 2MCl_{(s)} \nonumber$
Alkali metals react with water to produce hydrogen gas and alkali metal hydroxides; this is a very exothermic reaction (Figure $1$).
$2M_{(s)} + 2H_2O_{(l)} \rightarrow 2MOH_{(aq)} + H_{2(g)} \nonumber$
The reaction between alkali metals and oxygen is more complex:
• A common reaction is to form metal oxides which contain the O2- ion
$4Li_{(s)} + O_{2 (g)} \rightarrow \underbrace{2Li_2O_{(s)}}_{\text{lithium oxide}} \nonumber$
Other alkali metals can form metal peroxides (contains O22- ion)
$2Na(s) + O_{2 (g)} \rightarrow \underbrace{Na_2O_{2(s)}}_{\text{sodium peroxide}} \nonumber$
K, Rb and Cs can also form superoxides (O2- ion)
$K(s) + O_{2 (g)} \rightarrow \underbrace{KO_{2(s)}}_{\text{potassium superoxide}} \nonumber$
Colors via Absorption
The color of a chemical is produced when a valence electron in an atom is excited from one energy level to another by visible radiation. In this case, the particular frequency of light that excites the electron is absorbed. Thus, the remaining light that you see is white light devoid of one or more wavelengths (thus appearing colored). Alkali metals, having lost their outermost electrons, have no electrons that can be excited by visible radiation. Alkali metal salts and their aqueous solution are colorless unless they contain a colored anion.
Colors via Emission
When alkali metals are placed in a flame the ions are reduced (gain an electron) in the lower part of the flame. The electron is excited (jumps to a higher orbital) by the high temperature of the flame. When the excited electron falls back down to a lower orbital a photon is released. The transition of the valence electron of sodium from the 3p down to the 3s subshell results in release of a photon with a wavelength of 589 nm (yellow)
Flame colors:
• Lithium: crimson red
• Sodium: yellow
• Potassium: lilac
Group 2: The Alkaline Earth Metals
Compared with the alkali metals, the alkaline earth metals are typically harder, more dense, melt at a higher temperature. The first ionization energies ($I_1$) of the alkaline earth metals are not as low as the alkali metals. The alkaline earth metals are therefore less reactive than the alkali metals (Be and Mg are the least reactive of the alkaline earth metals). Several physical properties of these elements are compared in Table $2$.
Table $2$: General Properties of Group 2 Metals
Element Electronic Configuration Melting Point (°C) Density (g/cm3) Atomic Radius Ionization Energy (kJ/mol)
Beryllium $[He]2s^2$ 1278 1.85 1.52 899
Magnesium $[Ne]3s^2$ 649 1.74 1.60 738
Calcium $[Ar]4s^2$ 839 1.54 1.97 590
Strontium $[Kr]5s^2$ 769 2.54 2.15 549
Barium $[Xe]6s^2$ 725 3.51 2.17 503
Calcium, and elements below it, react readily with water at room temperature:
$Ca_{(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(aq)} + H_{2(g)} \nonumber$
The tendency of the alkaline earths to lose their two valence electrons is demonstrated in the reactivity of Mg towards chlorine gas and oxygen:
$Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \nonumber$
$2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)} \nonumber$
The 2+ ions of the alkaline earth metals have a noble gas like electron configuration and are thus form colorless or white compounds (unless the anion is itself colored). Flame colors:
• Calcium: brick red
• Strontium: crimson red
• Barium: green
Contributors and Attributions
Learning Objectives
• To understand the trends in properties and reactivity of the group 18 elements: the noble gases.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
J. W. Strutt (Lord Rayleigh) (1842–1919)
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Sir William Ramsay (1852–1916)
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Preparation and General Properties of the Group 18 Elements
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $1$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table $1$: Selected Properties of the Group 18 Elements
Property Helium Neon Argon Krypton Xenon Radon
*The configuration shown does not include filled d and f subshells. This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given.
atomic symbol He Ne Ar Kr Xe Rn
atomic number 2 10 18 36 54 86
atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222
valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6
triple point/boiling point (°C) —/−269 −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62
density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07
atomic radius (pm) 31 38 71 88 108 120
first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037
normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2)
electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0
electronegativity 2.6
product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none
type of oxide acidic
product of reaction with N2 none none none none none none
product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2
product of reaction with H2 none none none none none none
Reactions and Compounds of the Noble Gases
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $1$).
Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved.
Methane Clathrates
In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (Figure $3$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion:
$\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}$
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
$\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}$
The ionization energies of helium, neon, and argon are so high (Table $1$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
$\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}$
$\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}$
The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF8 ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid:
$\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}$
Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3), and XeF82− is isoelectronic with the IF8 ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
Example $1$
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise $1$
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer
$\ce{KXeF7}$; the xenon atom in XeF7 has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
Summary
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/08%3A_Periodic_Properties_of_the_Elements/8.09%3A_Examples_of_Periodic_Chemical_Behavior.txt |
• 9.1: Bonding Models and AIDS Drugs
• 9.2: Types of Chemical Bonds
Ionic vs. Covalent vs. Metallic bonding.
• 9.3: 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.
• 9.4: Ionic Bonding
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.
• 9.5: Covalent Bonding- Lewis Structure
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.
• 9.6: 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
• 9.7: Lewis Structures
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.
• 9.8: 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.
• 9.9: 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. 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
• 9.10: 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.
• 9.11: Bonding in Metals
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). Their physical properties include a lustrous (shiny) appearance, and they are malleable and ductile. In the 1900's, Paul Drüde came up with the sea of electrons theory by modeling metals as a mixture of atomic cores (atomic cores = positive nuclei + inner shell of electrons) and valence ele
• 9.E: Chemical Bonding I (Exercises)
Thumbnail: Ball and Stick model for Methane (\(\ce{CH4}\)). (CC BY-SA-NC; anonymous by request).
09: Chemical Bonding I- Lewis Structures and Determining Molecular Shapes
under construction | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.01%3A_Bonding_Models_and_AIDS_Drugs.txt |
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_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.02%3A_Types_of_Chemical_Bonds.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_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.03%3A_Representing_Valance_Electrons_with_Dots.txt |
Learning Objectives
• To describe the characteristics of ionic bonding.
• To quantitatively describe the energetic factors involved in the formation of an ionic bond.
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.
Generating Ionic Bonds
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.
In the overall ionic compound, positive and negative charges must be balanced, because electrons cannot be created or destroyed, only transferred. Thus, the total number of electrons lost by the cationic species must equal the total number of electrons gained by the anionic species.
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 $\ce{NaCl}$.
$\ce{2Na (s) + Cl 2(g) \rightarrow 2NaCl (s)} \nonumber$
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 energy. Thus the chlorine gains an electron from the sodium atom. This can be represented using ewis 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
Energetics of Ionic Bond Formation
Ionic bonds are formed when positively and negatively charged ions are held together by electrostatic forces. Consider a single pair of ions, one cation and one anion. How strong will the force of their attraction be? According to Coulomb's Law, the energy of the electrostatic attraction ($E$) between two charged particles is proportional to the magnitude of the charges and inversely proportional to the internuclear distance between the particles ($r$):
$E \propto \dfrac{Q_{1}Q_{2}}{r} \label{Eq1a}$
$E = k\dfrac{Q_{1}Q_{2}}{r} \label{Eq1b}$
where each ion’s charge is represented by the symbol Q. The proportionality constant k is equal to 2.31 × 10−28 J·m. This value of k includes the charge of a single electron (1.6022 × 10−19 C) for each ion. The equation can also be written using the charge of each ion, expressed in coulombs (C), incorporated in the constant. In this case, the proportionality constant, k, equals 8.999 × 109 J·m/C2. In the example given, Q1 = +1(1.6022 × 10−19 C) and Q2 = −1(1.6022 × 10−19 C). If Q1 and Q2 have opposite signs (as in NaCl, for example, where Q1 is +1 for Na+ and Q2 is −1 for Cl), then E is negative, which means that energy is released when oppositely charged ions are brought together from an infinite distance to form an isolated ion pair.
Energy is always released when a bond is formed and correspondingly, it always requires energy to break a bond.
As shown by the green curve in the lower half of Figure $1$, the maximum energy would be released when the ions are infinitely close to each other, at r = 0. Because ions occupy space and have a structure with the positive nucleus being surrounded by electrons, however, they cannot be infinitely close together. At very short distances, repulsive electron–electron interactions between electrons on adjacent ions become stronger than the attractive interactions between ions with opposite charges, as shown by the red curve in the upper half of Figure $1$. The total energy of the system is a balance between the attractive and repulsive interactions. The purple curve in Figure $1$ shows that the total energy of the system reaches a minimum at r0, the point where the electrostatic repulsions and attractions are exactly balanced. This distance is the same as the experimentally measured bond distance.
Consider the energy released when a gaseous $Na^+$ ion and a gaseous $Cl^-$ ion are brought together from r = ∞ to r = r0. Given that the observed gas-phase internuclear distance is 236 pm, the energy change associated with the formation of an ion pair from an $Na^+_{(g)}$ ion and a $Cl^-_{(g)}$ ion is as follows:
\begin{align*} E &= k\dfrac{Q_{1}Q_{2}}{r_{0}} \[4pt] &= (2.31 \times {10^{ - 28}}\rm{J}\cdot \cancel{m} ) \left( \dfrac{( + 1)( - 1)}{236\; \cancel{pm} \times 10^{ - 12} \cancel{m/pm}} \right) \[4pt] &= - 9.79 \times 10^{ - 19}\; J/ion\; pair \label{Eq2} \end{align*}
The negative value indicates that energy is released. Our convention is that if a chemical process provides energy to the outside world, the energy change is negative. If it requires energy, the energy change is positive. To calculate the energy change in the formation of a mole of NaCl pairs, we need to multiply the energy per ion pair by Avogadro’s number:
$E=\left ( -9.79 \times 10^{ - 19}\; J/ \cancel{ion pair} \right )\left ( 6.022 \times 10^{ 23}\; \cancel{ion\; pair}/mol\right )=-589\; kJ/mol \label{Eq3}$
This is the energy released when 1 mol of gaseous ion pairs is formed, not when 1 mol of positive and negative ions condenses to form a crystalline lattice. Because of long-range interactions in the lattice structure, this energy does not correspond directly to the lattice energy of the crystalline solid. However, the large negative value indicates that bringing positive and negative ions together is energetically very favorable, whether an ion pair or a crystalline lattice is formed.
We summarize the important points about ionic bonding:
• At r0, the ions are more stable (have a lower potential energy) than they are at an infinite internuclear distance. When oppositely charged ions are brought together from r = ∞ to r = r0, the energy of the system is lowered (energy is released).
• Because of the low potential energy at r0, energy must be added to the system to separate the ions. The amount of energy needed is the bond energy.
• The energy of the system reaches a minimum at a particular internuclear distance (the bond distance).
Example $2$: LiF
Calculate the amount of energy released when 1 mol of gaseous Li+F ion pairs is formed from the separated ions. The observed internuclear distance in the gas phase is 156 pm.
Given: cation and anion, amount, and internuclear distance
Asked for: energy released from formation of gaseous ion pairs
Strategy:
Substitute the appropriate values into Equation $\ref{Eq1b}$ to obtain the energy released in the formation of a single ion pair and then multiply this value by Avogadro’s number to obtain the energy released per mole.
Solution:
Inserting the values for Li+F into Equation $\ref{Eq1b}$ (where Q1 = +1, Q2 = −1, and r = 156 pm), we find that the energy associated with the formation of a single pair of Li+F ions is
\begin{align*} E &=k \dfrac{Q_1Q_2}{r_0} \[4pt] &=\left(2.31 \times 10^{−28} J⋅\cancel{m} \right) \left(\dfrac{\text{(+1)(−1)}}{156\; pm \times 10^{−12} \cancel{m/pm}} \right)\[4pt] &=−1.48 \times 10^{−18} \end{align*} \nonumber
Then the energy released per mole of Li+F ion pairs is
\begin{align*} E&= \left(−1.48 \times 10^{−18} J/ \cancel{\text{ion pair}}\right) \left(6.022 \times 10^{23} \cancel{\text{ion pair}}/mol\right)\[4pt] &−891 \;kJ/mol \end{align*} \nonumber
Because Li+ and F are smaller than Na+ and Cl (see Section 7.3), the internuclear distance in LiF is shorter than in NaCl. Consequently, in accordance with Equation $\ref{Eq1b}$, much more energy is released when 1 mol of gaseous Li+F ion pairs is formed (−891 kJ/mol) than when 1 mol of gaseous Na+Cl ion pairs is formed (−589 kJ/mol).
Exercise $2$: Magnesium oxide
Calculate the amount of energy released when 1 mol of gaseous $\ce{MgO}$ ion pairs is formed from the separated ions. The internuclear distance in the gas phase is 175 pm.
Answer
−3180 kJ/mol = −3.18 × 103 kJ/mol
Electron Configuration of Ions
How does the energy released in lattice formation compare to the energy required to strip away a second electron from the Na+ ion? Since the Na+ ion has a noble gas electron configuration, stripping away the next electron from this stable arrangement would require more energy than what is released during lattice formation (Sodium I2 = 4,560 kJ/mol). Thus, sodium is present in ionic compounds as Na+ and not Na2+. Likewise, adding an electron to fill a valence shell (and achieve noble gas electron configuration) is exothermic or only slightly endothermic. To add an additional electron into a new subshell requires tremendous energy - more than the lattice energy. Thus, we find Cl- in ionic compounds, but not Cl2-.
Table $1$: Lattice energies range from around 700 kJ/mol to 4000 kJ/mol:
Compound Lattice Energy (kJ/mol)
LiF 1024
LiI 744
NaF 911
NaCl 788
NaI 693
KF 815
KBr 682
KI 641
MgF2 2910
SrCl2 2130
MgO 3938
This amount of energy can compensate for values as large as I3 for valence electrons (i.e. can strip away up to 3 valence electrons). Because most transition metals would require the removal of more than 3 electrons to attain a noble gas core, they are not found in ionic compounds with a noble gas core. A transition metal always loses electrons first from the higher 's' subshell, before losing from the underlying 'd' subshell. (The remaining electrons in the unfilled d subshell are the reason for the bright colors observed in many transition metal compounds!) For example, iron ions will not form a noble gas core:
• Fe: [Ar]4s23d6
• Fe2+: [Ar] 3d6
• Fe3+: [Ar] 3d5
Some metal ions can form a pseudo noble gas core (and be colorless), for example:
• Ag: [Kr]5s14d10 Ag+ [Kr]4d10 Compound: AgCl
• Cd: [Kr]5s24d10 Cd2+ [Kr]4d10 Compound: CdS
The valence electrons do not adhere to the "octet rule" in this case (a limitation of the usefulness of this rule). Note: The silver and cadmium atoms lost the 5s electrons in achieving the ionic state.
When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest principle quantum number
Polyatomic Ions
Not all ionic compounds are formed from only two elements. Many polyatomic ions exist, in which two or more atoms are bound together by covalent bonds. They form a stable grouping which carries a charge (positive or negative). The group of atoms as a whole acts as a charged species in forming an ionic compound with an oppositely charged ion. Polyatomic ions may be either positive or negative, for example:
• NH4+ (ammonium) = cation
• SO42- (sulfate) = anion
The principles of ionic bonding with polyatomic ions are the same as those with monatomic ions. Oppositely charged ions come together to form a crystalline lattice, releasing a lattice energy. Based on the shapes and charges of the polyatomic ions, these compounds may form crystalline lattices with interesting and complex structures.
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. | textbooks/chem/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.04%3A_Ionic_Bonding.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_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.05%3A_Covalent_Bonding-_Lewis_Structure.txt |
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_A_Molecular_Approach_(Tro)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.06%3A_Electronegativity_and_Bond_Polarity.txt |
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