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Learning Objectives
By the end of this section, you will be able to:
• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The Second Law of Thermodynamics
In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the systemS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:
$\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \nonumber$
To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following:
$\Delta S_{ sys }=\frac{-q_{ rev }}{T_{ sys }} \quad \text { and } \quad \Delta S_{ surr }=\frac{q_{ rev }}{T_{ surr }} \nonumber$
The magnitudes of −qrev and qrev are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:
\begin{aligned} & \left|\Delta S_{\text {sys }}\right|<\left|\Delta S_{\text {surr }}\right| \[4pt] & \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0 \end{aligned} \nonumber
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following:
$\Delta S_{ sys }=\frac{q_{ rev }}{T_{ sys }} \quad \text { and } \quad \Delta S_{ surr }=\frac{-q_{ rev }}{T_{ surr }} \nonumber$
The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes (that is, the direction of the heat flow) will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The objects are at essentially the same temperature, TsysTsurr, and so the magnitudes of the entropy changes are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium.
\begin{aligned} & \left|\Delta S_{\text {sys }}\right| \approx\left|\Delta S_{\text {surr }}\right| \[4pt] & \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=0 \end{aligned} \nonumber
These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $1$.
Table $1$: The Second Law of Thermodynamics
ΔSuniv > 0 spontaneous
ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction)
ΔSuniv = 0 at equilibrium
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr is a good approximation of qrev, and the second law may be stated as the following:
$\Delta S_{ univ }=\Delta S_{ sys }+\Delta S_{ surr }=\Delta S_{ sys }+\frac{q_{ surr }}{T} \nonumber$
We may use this equation to predict the spontaneity of a process as illustrated in Example $1$.
Example $1$: Will Ice Spontaneously Melt?
The entropy change for the process
$H_2O(s) \longrightarrow H_2O(l) \nonumber$
is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?
Solution
We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.
At −10.00 °C (263.15 K), the following is true:
\begin{aligned} \Delta S_{\text {univ }} & =\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \[4pt] & =22.1 J / K +\frac{-6.00 \times 10^3 J }{263.15 K }=-0.7 J / K \end{aligned} \nonumber
$S_{univ} < 0$, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
\begin{align*} \Delta S_{\text {univ }} &=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \[4pt] &=22.1 J / K +\frac{-6.00 \times 10^3 J }{283.15 K }=+0.9 J / K \end{align*} \nonumber
Suniv > 0, so melting is spontaneous at 10.00 °C.
Exercise $1$
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?
Answer
Entropy is a state function, so ΔSfreezing = −ΔSmelting = −22.1 J/K and qsurr = +6.00 kJ. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
The Third Law of Thermodynamics
The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero.
$S=k \ln W=k \ln (1)=0 \nonumber$
This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.
Careful calorimetric measurements can be made to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies (S°) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy changeS°) for a reaction may be computed using standard entropies as shown below:
$\Delta S^{\circ}=\sum \nu S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants }) \nonumber$
where ν represents stoichiometric coefficients in the balanced equation representing the process. For example, ΔS° for the following reaction at room temperature
$m A +n B \longrightarrow x C +y D \nonumber$
is computed as:
$=\left[x S^{\circ}( C )+y S^{\circ}( D )\right]-\left[m S^{\circ}( A )+n S^{\circ}( B )\right] \nonumber$
A partial listing of standard entropies is provided in Table $2$, and additional values are provided in Appendix G. The example exercises that follow demonstrate the use of S° values in calculating standard entropy changes for physical and chemical processes.
Table $2$: Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.)
Substance (J mol−1 K−1)
carbon
C(s, graphite) 5.740
C(s, diamond) 2.38
CO(g) 197.7
CO2(g) 213.8
CH4(g) 186.3
C2H4(g) 219.5
C2H6(g) 229.5
CH3OH(l) 126.8
C2H5OH(l) 160.7
hydrogen
H2(g) 130.57
H(g) 114.6
H2O(g) 188.71
H2O(l) 69.91
HCI(g) 186.8
H2S(g) 205.7
oxygen
O2(g) 205.03
Example $2$: Determination of ΔS°
Calculate the standard entropy change for the following process:
$H_2O(g) \longrightarrow H_2O(l) \nonumber$
Solution
Calculate the entropy change using standard entropies as shown above:
$\Delta S^{\circ}=(1 mol )\left(70.0 J mol^{-1} K^{-1}\right)-(1 mol )\left(188.8 J mol^{-1} K^{-1}\right)=-118.8 J / K \nonumber$
The value for ΔS° is negative, as expected for this phase transition (condensation), which the previous section discussed.
Exercise $2$
Calculate the standard entropy change for the following process:
$H_2(g)+ C_2 H_4(g) \longrightarrow C_2 H_6(g) \nonumber$
Answer
−120.6 J K–1 mol–1
Example $3$: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH:
$2 CH_3 OH (l)+3 O_2(g) \longrightarrow 2 CO_2(g)+4 H_2O(l) \nonumber$
Solution
Calculate the entropy change using standard entropies as shown above:
$\begin{gathered} \Delta S^{\circ}=\sum \nu S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants }) \[4pt] {\left[2 mol \times S^{\circ}\left( CO_2(g)\right)+4 mol \times S^{\circ}\left( H_2O(l)\right)\right]-\left[2 mol \times S^{\circ}\left( CH_3 OH (l)\right)+3 mol \times S^{\circ}\left( O_2(g)]\right)\right.} \[4pt] =\{[2(213.8)+4 \times 70.0]-[2(126.8)+3(205.2)]\}=-161.6 J / K \end{gathered} \nonumber$
Exercise $3$
Calculate the standard entropy change for the following reaction:
$Ca ( OH )_2( s ) \longrightarrow CaO (s)+ H_2O(l) \nonumber$
Answer
24.7 J/K | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/12%3A_Thermodynamics/12.04%3A_The_Second_and_Third_Laws_of_Thermodynamics.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate free energy change for a process using free energies of formation for its reactants and products
• Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
• Explain how temperature affects the spontaneity of some processes
• Relate standard free energy changes to equilibrium constants
One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy ($G$) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
$G=H-T S \nonumber$
Free energy is a state function, and at constant temperature and pressure, the free energy change ($ΔG$) may be expressed as the following:
$\Delta G=\Delta H-T \Delta S \nonumber$
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)
The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:
$\Delta S_{\text {univ }}=\Delta S+\frac{q_{\text {surr }}}{T} \nonumber$
The first law requires that $q_{surr} = −q_{sys}$, and at constant pressure $q_{sys} = ΔH$, so this expression may be rewritten as:
$\Delta S_{\text {univ }}=\Delta S-\frac{\Delta H}{T} \nonumber$
Multiplying both sides of this equation by $−T$, and rearranging yields the following:
$-T \Delta S_{\text {univ }}=\Delta H-T \Delta S \nonumber$
Comparing this equation to the previous one for free energy change shows the following relation:
$\Delta G=-T \Delta S_{\text {univ }} \nonumber$
The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $ΔS_{univ}$. Table $1$ summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
Table $1$: Relation between Process Spontaneity and Signs of Thermodynamic Properties
$ΔS_{univ} > 0$ ΔG < 0 moves spontaneously in the forward direction, as written, to reach equilibrium
$ΔS_{univ} < 0$ ΔG > 0
nonspontaneous in the forward direction, as written, but moves spontaneously in the reverse direction, as written, to reach equilibrium
$ΔS_{univ} = 0$ ΔG = 0 reversible (at equilibrium)
What’s “Free” about ΔG?
In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work ($w$) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.
For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by
$\Delta G=\Delta H-T \Delta S \nonumber$
may be interpreted as representing the difference between the energy produced by the process, ΔH, and the energy lost to the surroundings, TΔS. The difference between the energy produced and the energy lost is the energy available (or “free”) to do useful work by the process, ΔG. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:
$\Delta G=w_{max } \nonumber$
where $w_{max}$ refers to all types of work except expansion (pressure-volume) work.
However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., batteries) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.
Calculating Free Energy Change
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, $ΔG^o$, according to the following relation.
$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \nonumber$
It is important to understand that for phase changes, $\Delta G^º$ tells you if the phase change is spontaneous or not; will it happen, or not happen. For chemical reactions, $\Delta G^º$ tells you the extent of a reaction. In other words, $\Delta G^º$ for a reaction tells you how much product will be present at equilibrium. A reaction with $\Delta G^º$ < 0 is considered product-favored at equilibrium; there will be more products than reactants when the reaction reaches equilibrium. A reaction with $\Delta G^º$ > 0 is considered reactant-favored at equilibrium; there will be more reactants than products when the reaction reaches equilibrium.
Example $1$: Using Standard Enthalpy and Entropy Changes to Calculate ΔG°
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for $ΔG^o$ say about the spontaneity of this process?
Solution
The process of interest is the following:
$\ce{H_2O(l) \longrightarrow H_2O(g)} \nonumber$
The standard change in free energy may be calculated using the following equation:
$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \nonumber$
From Appendix G:
Substance $\Delta H_{ f }^{\circ} (\text{kJ/mol})$ $S^{\circ} (\text{kJ/K mol})$
H2O(l) −285.83 70.0
H2O(g) −241.82 188.8
Using the appendix data to calculate the standard enthalpy and entropy changes yields:
\begin{align*} \Delta H^{\circ} &=\Delta H_{ f }^{\circ}\left( \ce{H2O(g)} \right)-\Delta H_{ f }^{\circ}\left( \ce{H2O(l)} \right) \[4pt] &=[-241.82 ~\text{kJ/mol} -(-285.83)] ~\text{kJ/mol} =44.01 ~\text{kJ/mol} \[4pt] \Delta S^{\circ} &=1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(g)} \right)-1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(l)} \right) \[4pt] &=(1 ~\text{mol}) ~188.8 J / mol \cdot K -(1 ~\text{mol}) ~ 70.0 ~ \text{J} / \text{mol K} =118.8 ~ \text{J / K} \[4pt] \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ} \end{align*} \nonumber
Substitution into the standard free energy equation yields:
\begin{align*} \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ} \[4pt] &=44.01~\text{kJ} -(298 K \times 118.8~\text{J/K}) \times \frac{1 ~\text{kJ}}{1000~\text{J} } \[4pt] &=44.01~\text{kJ} -35.4~\text{kJ} \[4pt] &=8.6~\text{kJ} \end{align*} \nonumber
At 298 K (25 °C) so boiling is nonspontaneous (not spontaneous).
Exercise $1$
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?
$\ce{C2H6(g) \longrightarrow H2(g) + C2H4(g)} \nonumber$
Answer
the reaction is nonspontaneous (not spontaneous) at 25 °C.
The standard free energy change for a reaction may also be calculated from standard free energy of formation $ΔG_f^o$ values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, $ΔG_f^o$ is by definition zero for elemental substances in their standard states. The approach used to calculate $ΔG^o$ for a reaction from $ΔG_f^o$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
$m A +n B \longrightarrow x C +y D \nonumber$
the standard free energy change at room temperature may be calculated as
\begin{align*} \Delta G^{\circ} &= \sum \nu \Delta G^{\circ}(\text { products })-\sum \nu \Delta G^{\circ}(\text { reactants }) \[4pt] &= \left[x \Delta G_{ f }^{\circ}( C )+y \Delta G_{ f }^{\circ}( D )\right]-\left[m \Delta G_{ f }^{\circ}( A )+n \Delta G_{ f }^{\circ}( B )\right] . \end{align*} \nonumber
Example $2$: Using Standard Free Energies of Formation to Calculate ΔG°
Consider the decomposition of yellow mercury(II) oxide.
$\ce{HgO (s, \text { yellow }) -> Hg (l) + 1/2 O2(g)} \nonumber$
Calculate the standard free energy change at room temperature, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution
The required data are available in Appendix G and are shown here.
Compound $\Delta G_{ f }^{\circ}( \text{kJ / mol} )$ $\Delta H_{ f }^{\circ}( \text{kJ / mol} )$ $S^{\circ}( \text{kJ / K mol} )$
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2
(a) Using free energies of formation:
\begin{align*} \Delta G^{\circ} & =\sum \nu G_{ f }^{\circ}(\text { products })-\sum \nu \Delta G_{ f }^{\circ}(\text { reactants }) \[4pt] &=\left[1 \Delta G_{ f }^{\circ}~ \ce{Hg(l)} + \dfrac{1}{2} \Delta G_{f}^{\circ} ~\ce{O2(g)} \right]-1 \Delta G_{ f }^{\circ} ~\ce{HgO(s, yellow)} \[4pt] &=\left[1 ~\text{mol} (0 ~\text{kJ / mol} )+\frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right] - 1 ~\text{mol} (-58.43 ~\text{kJ / mol} )=58.43 ~\text{kJ / mol} \end{align*} \nonumber
(b) Using enthalpies and entropies of formation:
\begin{align*} \Delta H^{\circ}&=\sum \nu \Delta H_{ f }^{\circ}(\text { products })-\sum \nu \Delta H_{ f }^{\circ}(\text { reactants }) \[4pt] &=\left[1 \Delta H_{ f }^{\circ}~\ce{Hg(l)} +\frac{1}{2} \Delta H_{ f }^{\circ}~\ce{O2(g)} \right]-1 \Delta H_{ f }^{\circ} ~\ce{HgO (s, yellow )} \[4pt] &=\left[1 ~\text{mol} (0~ \text{kJ / mol} ) + \frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right]-1 ~\text{mol} (-90.46~\text{kJ / mol} ) = 90.46 ~\text{kJ / mol} \[8pt] \Delta S^{\circ} &=\sum \nu \Delta S^{\circ}(\text { products })-\sum \nu \Delta S^{\circ}(\text { reactants }) \[4pt] &=\left[1 ~\Delta S^{\circ} ~\ce{Hg (l)} + \frac{1}{2} \Delta S^{\circ} ~\ce{O2(g)} \right]-1 \Delta S^{\circ} ~\ce{HgO (s, yellow )} \[4pt] & =\left[1 ~\text{mol} (75.9~\text{J / mol K} ) + \frac{1}{2} ~\text{mol} (205.2~\text{J / mol K} )\right] -1 \text{mol} (71.13 ~\text{J / mol K} )=107.4 ~\text{ J / mol K} \[8pt] \Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ}=90.46 ~\text{kJ} - 298.15 ~\text{K} \times 107.4 ~\text{J / K} \cdot \text{mol} \times \frac{1 ~\text{ kJ} }{1000 ~\text{J} } \[4pt] & =(90.46 - 32.01) ~\text{kJ / mol} =58.45 ~\text{kJ / mol} \end{align*} \nonumber
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.
Exercise $1$
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
$\ce{C2H4(g) \longrightarrow H2(g) + C2H2(g)} \nonumber$
Answer
1. 140.8 kJ/mol, nonspontaneous
2. 141.5 kJ/mol, nonspontaneous
Free Energy Changes for Coupled Reactions
The use of free energies of formation to compute free energy changes for reactions as described above is possible because $ΔG$ is a state function, and the approach is analogous to the use of Hess’ Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:
$\ce{H_2O(l) -> H_2O(g)} \nonumber$
An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:
\begin{align*} \ce{H2(g) + 1/2 O2(g) &-> H2O(g)} && \Delta G_{ f }^{\circ} \text { gas } \[4pt] \ce{H2O(l) &-> H2(g) + 1/2 O2(g)} && - \Delta G_{ f }^{ o } \text { liquid } \[4pt] \hline \ce{H2O(l) &-> H2O(g)} && \Delta G^{\circ}=\Delta G_{ f }^{\circ} ~\text{gas} - \Delta G_{ f }^{\circ} ~\text{liquid} \end{align*}
This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for $ΔG^o$:
$\ce{ZnS (s) \rightarrow Zn(s) + S(s)} \quad \Delta G_1^{\circ}=201.3 kJ \nonumber$
The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:
$\ce{S(s) + O2(g) \rightarrow SO2(g)} \quad \Delta G_2^{\circ}=-300.1 kJ \nonumber$
The coupled reaction exhibits a negative free energy change and is spontaneous:
$\ce{ZnS(s) + O2(g) \rightarrow Zn(s) + SO2(g)} \nonumber$
$\Delta G^{\circ}=201.3 ~\text{kJ} + -300.1 ~\text{kJ} =-98.8 ~\text{kJ} \nonumber$
This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.
Example $3$: Calculating Free Energy Change for a Coupled Reaction
Is a reaction coupling the decomposition of $\ce{ZnS}$ to the formation of $\ce{H2S}$ expected to be spontaneous under standard conditions?
Solution
Following the approach outlined above and using free energy values from Appendix G:
\begin{align*} &\text{Decomposition of zinc sulfide:} & \ce{ZnS (s) \rightarrow Zn(s) + S(s)} && \Delta G_1^{\circ}=201.3 ~\text{kJ}\[4pt] &\text{Formation of hydrogen sulfide:} & \ce{S(s) + H2(g) \rightarrow H2S(g)} && \Delta G_1^{\circ}=-33.4 ~\text{kJ} \[4pt] \hline &\text{Coupled reaction:} & \ce{ZnS(s) + H2(g) \rightarrow Zn(s) + H2S(g)} && \Delta G^{\circ}=201.3 ~\text{kJ} +-33.4 ~\text{kJ} =167.9 ~\text{kJ} \end{align*}
The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.
Exercise $3$
What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?
$\ce{FeS(s) \, + O2(g) \rightarrow Fe(s) \, + SO2(g)} \nonumber$
Answer
−199.7 kJ; spontaneous
Temperature Dependence of Spontaneity and Extent of Reaction
As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. In a similar, but not identical fashion, some chemical reactions can switch from being product-favored at equilibrium, to being reactant-favored at equilibrium, depending on the temperature.
Note
The numerical value of $\Delta G^º$ is always dependent on the temperature. In this section we are determining whether or not the sign of $\Delta G^º$ is dependent on the temperature.
To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
The extent of a process, as reflected in the arithmetic sign of its standard free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (Kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
1. Both ΔHº and ΔSº are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is greater than ΔHº. If the TΔSº term is less than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at high temperatures and reactant-favored at equilibrium at low temperatures.
2. Both ΔHº and ΔSº are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is less than ΔHº. If the TΔSº term’s magnitude is greater than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at low temperatures and reactant-favored at equilibrium at high temperatures.
3. ΔHº is positive and ΔSº is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔGº will be positive regardless of the temperature. Such a process is reactant-favored at equilibrium at all temperatures.
4. ΔHº is negative and ΔSº is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔGº will be negative regardless of the temperature. Such a process is product-favored at equilibrium at all temperatures.
These four scenarios are summarized in Table $1$
Table $1$
Sign of $\Delta H^o$
Sign of $\Delta S^o$
Sign of $\Delta G^o$
Temperature Dependence of $\Delta G^o$
- + - The sign of $\Delta G^o$ does not depend on the temperature.The reaction is product-favored at equilibrium at all temperatures.
+ - + The sign of $\Delta G^o$ does not depend on the temperature.The reaction is reactant-favored at equilibrium at all temperatures.
- - - or + The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
+ + - or +
The sign of $\Delta G^o$ does depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
Example $3$: Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
$\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber$
Does the sign of $\Delta G^º$ of this process depend upon temperature?
Solution
Combustion processes are exothermic ($ΔH^º < 0$). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, $ΔS^º > 0$). The reaction is therefore product-favored at equilibrium ($ΔG^º < 0$) at all temperatures.
Exercise $3$
Popular chemical hand warmers generate heat by the air-oxidation of iron:
$\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber$
Does the sign of $\Delta G^o$ of this process depend upon temperature?
Answer
ΔHº and ΔSº are both negative; the reaction is product-favored at equilibrium at low temperatures.
When considering the conclusions drawn regarding the temperature dependence of the sign of ΔGº, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is reactant-favored at equilibrium at one temperature but product-favored at equilibrium at another temperature will necessarily undergo a change in “extent” (as reflected by its ΔGº) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔGº is plotted on the y axis versus T on the x axis:
$ΔG^º=ΔH^º−TΔS^º \nonumber$
$y=b+mx \nonumber$
Such a plot is shown in Figure $2$. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependence for the sign of ΔGº as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔGº) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔGº is zero:
$ΔG^º=0=ΔH^º−TΔS^º \nonumber$
$T=\dfrac{ΔH^º}{ΔS^º} \nonumber$
Thus, saying a process is product-favored at equilibrium at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔGº for the process is zero.
Note
In this discussion, we have used two different descriptions for the meaning of the sign of ΔGº. You should be aware of the meaning of each description.
a) Extent of Reaction: This description is used to predict the ratio of the product and reactant concentrations at equilibrium. In this description, we use the thermodynamic term ΔGº to tell us the same information as the equilibrium constant, K. When ΔGº < 0, K > 1, and the reaction will be product-favored at equilibrium. When ΔGº > 0, K< 1, and the reaction is reactant-favored at equilibrium. When ΔGº = 0, K =1, and the reaction will have roughly equal amounts of products and reactants at equilibrium. In all cases, the reaction will form a mixture of products and reactants at equilibrium. We use the sign and magnitude of ΔGº to tell us how much product will be made if the reaction is allowed to reach equilibrium.
b) Spontaneity: This description is much more complicated because it involves two different interpretations of how a reaction at standard state occurs. One interpretation involves the hypothetical process in which the reaction proceeds from a starting point of pure reactants to a finishing point of pure products, with all substances isolated in their own containers under standard state conditions. In the second, more realistic interpretation, the reaction starts with all reactants and all products in their standard state in one container. We then allow this specific mixture to react an infinitesimally small amount so that we can obtain a rate of change in free energy with respect to the extent of reaction when all reactants and products are mixed and (essentially) in their standard states. Although each interpretation describes a different reaction scenario, the value of the difference in free energy and the value of the rate of change in free energy are the same number. If ΔGº < 0, we say that the reaction is spontaneous, meaning that the reaction would proceed in the forward direction, as written, to form pure products in their standard state. If ΔGº > 0, we say that the reaction is nonspontaneous, meaning that the reaction would proceed in the reverse direction, as written, to form pure reactants in their standard state. If ΔGº = 0, we say that the neither the reactants nor the products are favored to be formed.
A detailed treatment of the meaning of ΔGº can be found in the paper, "Free Energy versus Extent of Reaction" by Richard S. Treptow, Journal of Chemical Education, 1996, Volume 73 (1), 51-54.
Example $4$: Equilibrium Temperature for a Phase Transition
As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Tables T1 or T2 to estimate the boiling point of water.
Solution
The process of interest is the following phase change:
$\ce{H2O}(l)⟶\ce{H2O}(g) \nonumber$
When this process is at equilibrium, ΔG = 0, so the following is true:
$0=ΔH°−TΔS°\hspace{40px}\ce{or}\hspace{40px}T=\dfrac{ΔH°}{ΔS°} \nonumber$
Using the standard thermodynamic data from Tables T1 or T2,
\begin{align*} ΔH°&=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\ &=\mathrm{−241.82\: kJ/mol−(−285.83\: kJ/mol)=44.01\: kJ/mol} \nonumber \end{align*} \nonumber
\begin{align*} ΔS°&=ΔS^\circ_{298}(\ce{H2O}(g))−ΔS^\circ_{298}(\ce{H2O}(l)) \nonumber\ &=\mathrm{188.8\: J/K⋅mol−70.0\: J/K⋅mol=118.8\: J/K⋅mol} \nonumber \end{align*} \nonumber
$T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber$
The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Tables T1 or T2.). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.
Exercise $4$
Use the information in Tables T1 or T2 to estimate the boiling point of CS2.
Answer
313 K (accepted value 319 K).
Free Energy and Equilibrium
The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).
In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.
The free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:
$\Delta G=\Delta G^{\circ}+R T \ln Q \nonumber$
R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. For gas phase equilibria, the pressure-based reaction quotient, QP, is used. The concentration-based reaction quotient, QC, is used for condensed phase equilibria. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example $6$.
Example $6$: Calculating ΔG under Nonstandard Conditions
What is the free energy change for the process shown here
$2 NH_3(g) \longrightarrow 3 H_2(g)+ N_2(g) \nonumber$
under the specified conditions?
• $T=25^{\circ} C ,$
• $P_{ N_2}=0.870 ~\text{atm}$
• $P_{ H_2}=0.250 ~\text{atm}$
• $P_{ NH_3}=12.9 ~\text{atm}$
$\Delta G^{\circ}=33.0 kJ / mol \nonumber$
Solution
The equation relating free energy change to standard free energy change and reaction quotient may be used directly:
\begin{align*} \Delta G &=\Delta G^{\circ}+R T \ln Q \[4pt] &=33.0 \frac{ \text{kJ} }{ \text{mol} }+\left(8.314 \frac{ \text{J} }{ \text{mol K} } \times 298 K \times \ln \frac{\left(0.250^3\right) \times 0.870}{12.9^2}\right) \[4pt] &=9680 \frac{ \text{J} }{ \text{mol} } \text { or } 9.68 ~\text{kJ/mol} \end{align*} \nonumber
Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.
Exercise $6$
Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?
Answer
ΔG = –123.5 kJ/mol; yes
For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as
$0=\Delta G^{\circ}+R T \ln K \quad \text { (at equilibrium) } \nonumber$
$\Delta G^{\circ}=-R T \ln K \nonumber$
or
$K=e^{-\frac{\Delta G}{R T}} \nonumber$
This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table $2$.
Table $2$: Relations between Standard Free Energy Changes and Equilibrium Constants
K ΔG° Composition of an Equilibrium Mixture
> 1 < 0 Products are more abundant
< 1 > 0 Reactants are more abundant
= 1 = 0 Reactants and products are comparably abundant
Example $7$: Calculating an Equilibrium Constant using Standard Free Energy Change
Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.
Solution
The reaction of interest is the following:
$AgCl (s) \rightleftharpoons Ag^{+}(aq)+ Cl^{-}(aq) \quad K_{ sp }=\left[ Ag^{+}\right]\left[ Cl^{-}\right] \nonumber$
The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:
\begin{align*} \Delta G^{\circ} &=\left[\Delta G_{ f }^{\circ}\left( \ce{Ag^{+}(aq)} \right)+\Delta G_{ f }^{\circ}\left( \ce{Cl^{-}(aq)} \right)\right]-\left[\Delta G_{ f }^{\circ}( \ce{AgCl(s)} )\right] \[4pt] & =[77.1 ~\text{kJ/mol} -131.2 ~\text{kJ/mol} ]-[-109.8 ~\text{kJ/mol} ]=55.7 ~\text{kJ/mol} \end{align*} \nonumber
The equilibrium constant for the reaction may then be derived from its standard free energy change:
$K_{ sp }=e^{-\frac{\Delta G^{\circ}}{R T}}=\exp \left(-\frac{\Delta G^{\circ}}{R T}\right)=\exp \left(-\frac{55.7 \times 10^3 J / mol }{8.314 J / mol \cdot K \times 298.15 K }\right) =\exp (-22.470)=e^{-22.470}=1.74 \times 10^{-10} \nonumber$
This result is in reasonable agreement with the value provided in Appendix J.
Exercise $7$
Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
$2 NO_2(g) \rightleftharpoons N_2 O_4(g) \nonumber$
Answer
K = 3.1
To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure $3$). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/12%3A_Thermodynamics/12.05%3A_Free_Energy.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
entropy (S)state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates; often described as a measure of the disorder of the system
Gibbs free energy change (G)thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
microstatepossible configuration or arrangement of matter and energy within a system
nonspontaneous processprocess that requires continual input of energy from an external source
reversible processprocess that takes place so slowly as to be capable of reversing direction in response to an infinitesimally small change in conditions; hypothetical construct that can only be approximated by real processes
second law of thermodynamicsall spontaneous processes involve an increase in the entropy of the universe
spontaneous changeprocess that takes place without a continuous input of energy from an external source
standard entropy (S°)entropy for one mole of a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K
standard entropy change (ΔS°)change in entropy for a reaction calculated using the standard entropies
standard free energy change (ΔG°)change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)
standard free energy of formation change in free energy accompanying the formation of one mole of substance from its elements in their standard states
third law of thermodynamicsentropy of a perfect crystal at absolute zero (0 K) is zero
12.07: Key Equations
S = k ln W
ΔSuniv = ΔSsys + ΔSsurr
ΔG = ΔHTΔS
12.08: Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system.
Entropy (S) is a state function that can be related to the number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system.
For a given substance, entropy depends on phase with Ssolid < Sliquid < Sgas. For different substances in the same physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions and physical changes may be reliably predicted.
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/12%3A_Thermodynamics/12.06%3A_Key_Terms.txt |
1.
What is a spontaneous reaction?
2.
What is a nonspontaneous reaction?
3.
Indicate whether the following processes are spontaneous or nonspontaneous.
(a) Liquid water freezing at a temperature below its freezing point
(b) Liquid water freezing at a temperature above its freezing point
(c) The combustion of gasoline
(d) A ball thrown into the air
(e) A raindrop falling to the ground
(f) Iron rusting in a moist atmosphere
4.
A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
5.
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
6.
In Figure 12.8 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
7.
In Figure 12.8 all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, ΔS, for the system when it is converted from distribution (b) to distribution (d).
8.
How does the process described in the previous item relate to the system shown in Figure 12.4?
9.
Consider a system similar to the one in Figure 12.8, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to $18.18.$ What does this comparison tell us about even larger systems?
10.
Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?
11.
Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
12.
Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.
(a) H2(g), HBrO4(g), HBr(g)
(b) H2O(l), H2O(g), H2O(s)
(c) He(g), Cl2(g), P4(g)
13.
At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.
14.
Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature but somewhat higher pressure).
$I 2 ( s ) ⟶ I 2 ( g ) I 2 ( s ) ⟶ I 2 ( g )$
$I 2 ( s ) ⟶ I 2 ( l ) I 2 ( s ) ⟶ I 2 ( l )$
Is ΔS positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
15.
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
(a) C2H5OH(l) or C3H7OH(l)
(b) C2H5OH(l) or C2H5OH(g)
(c) 2H(g) or H(g)
16.
Predict the sign of the entropy change for the following processes.
(a) An ice cube is warmed to near its melting point.
(b) Exhaled breath forms fog on a cold morning.
(c) Snow melts.
17.
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
(a) $Na+(aq)+Cl−(aq)⟶NaCl(s)Na+(aq)+Cl−(aq)⟶NaCl(s)$
(b) $2Fe(s)+32O2(g)⟶Fe2O2(s)2Fe(s)+32O2(g)⟶Fe2O2(s)$
(c) $2C6H14(l)+19O2(g)⟶14H2O(g)+12CO2(g)2C6H14(l)+19O2(g)⟶14H2O(g)+12CO2(g)$
18.
Write the balanced chemical equation for the combustion of methane, CH4(g), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether ΔS is positive or negative for this chemical reaction.
19.
Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?
20.
What is the difference between ΔS and ΔS° for a chemical change?
21.
Calculate $ΔS°ΔS°$ for the following changes.
(a) $SnCl4(l)⟶SnCl4(g)SnCl4(l)⟶SnCl4(g)$
(b) $CS2(g)⟶CS2(l)CS2(g)⟶CS2(l)$
(c) $Cu(s)⟶Cu(g)Cu(s)⟶Cu(g)$
(d) $H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$
(e) $2H2(g)+O2(g)⟶2H2O(l)2H2(g)+O2(g)⟶2H2O(l)$
(f) $2HCl(g)+Pb(s)⟶PbCl2(s)+H2(g)2HCl(g)+Pb(s)⟶PbCl2(s)+H2(g)$
(g) $Zn(s)+CuSO4(s)⟶Cu(s)+ZnSO4(s)Zn(s)+CuSO4(s)⟶Cu(s)+ZnSO4(s)$
22.
Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under the standard conditions to give gaseous carbon dioxide and liquid water.
23.
Determine the entropy change for the combustion of gaseous propane, C3H8, under the standard conditions to give gaseous carbon dioxide and water.
24.
“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is $Fe2O3(s)+2Al(s)⟶Al2O3(s)+2Fe(s).Fe2O3(s)+2Al(s)⟶Al2O3(s)+2Fe(s).$ Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
25.
Using the relevant $S°S°$ values listed in Appendix G, calculate $ΔS°298ΔS°298$ for the following changes:
(a) $N2(g)+3H2(g)⟶2NH3(g)N2(g)+3H2(g)⟶2NH3(g)$
(b) $N2(g)+52O2(g)⟶N2O5(g)N2(g)+52O2(g)⟶N2O5(g)$
26.
From the following information, determine $ΔS°ΔS°$ for the following:
$N ( g ) + O ( g ) ⟶ NO ( g ) Δ S ° = ? N ( g ) + O ( g ) ⟶ NO ( g ) Δ S ° = ?$
$N 2 ( g ) + O 2 ( g ) ⟶ 2 NO ( g ) Δ S ° = 24.8 J/K N 2 ( g ) + O 2 ( g ) ⟶ 2 NO ( g ) Δ S ° = 24.8 J/K$
$N 2 ( g ) ⟶ 2 N ( g ) Δ S ° = 115.0 J/K N 2 ( g ) ⟶ 2 N ( g ) Δ S ° = 115.0 J/K$
$O 2 ( g ) ⟶ 2 O ( g ) Δ S ° = 117.0 J/K O 2 ( g ) ⟶ 2 O ( g ) Δ S ° = 117.0 J/K$
27.
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
$SNaCl(s)°=72.11Jmol·KSNaCl(l)°=95.06Jmol·KΔHfusion°=27.95 kJ/molSNaCl(s)°=72.11Jmol·KSNaCl(l)°=95.06Jmol·KΔHfusion°=27.95 kJ/mol$
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
28.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
(a) $MnO2(s)⟶Mn(s)+O2(g)MnO2(s)⟶Mn(s)+O2(g)$
(b) $H2(g)+Br2(l)⟶2HBr(g)H2(g)+Br2(l)⟶2HBr(g)$
(c) $Cu(s)+S(g)⟶CuS(s)Cu(s)+S(g)⟶CuS(s)$
(d) $2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)$
(e) $CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)$
(f) $CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)$
29.
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
(a) $C(s, graphite)+O2(g)⟶CO2(g)C(s, graphite)+O2(g)⟶CO2(g)$
(b) $O2(g)+N2(g)⟶2NO(g)O2(g)+N2(g)⟶2NO(g)$
(c) $2Cu(s)+S(g)⟶Cu2S(s)2Cu(s)+S(g)⟶Cu2S(s)$
(d) $CaO(s)+H2O(l)⟶Ca(OH)2(s)CaO(s)+H2O(l)⟶Ca(OH)2(s)$
(e) $Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)$
(f) $CaSO4·2H2O(s)⟶CaSO4(s)+2H2O(g)CaSO4·2H2O(s)⟶CaSO4(s)+2H2O(g)$
30.
What is the difference between ΔG and ΔG° for a chemical change?
31.
A reaction has $ΔH°ΔH°$ = 100 kJ/mol and $ΔS°=250 J/mol·K.ΔS°=250 J/mol·K.$ Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
32.
Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
33.
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
(a) $MnO2(s)⟶Mn(s)+O2(g)MnO2(s)⟶Mn(s)+O2(g)$
(b) $H2(g)+Br2(l)⟶2HBr(g)H2(g)+Br2(l)⟶2HBr(g)$
(c) $Cu(s)+S(g)⟶CuS(s)Cu(s)+S(g)⟶CuS(s)$
(d) $2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)$
(e) $CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)$
(f) $CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)$
34.
Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
(a) $C(s, graphite)+O2(g)⟶CO2(g)C(s, graphite)+O2(g)⟶CO2(g)$
(b) $O2(g)+N2(g)⟶2NO(g)O2(g)+N2(g)⟶2NO(g)$
(c) $2Cu(s)+S(g)⟶Cu2S(s)2Cu(s)+S(g)⟶Cu2S(s)$
(d) $CaO(s)+H2O(l)⟶Ca(OH)2(s)CaO(s)+H2O(l)⟶Ca(OH)2(s)$
(e) $Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)$
(f) $CaSO4·2H2O(s)⟶CaSO4(s)+2H2O(g)CaSO4·2H2O(s)⟶CaSO4(s)+2H2O(g)$
35.
Given:
$P4(s)+5O2(g)⟶P4O10(s)ΔG°=−2697.0 kJ/mol 2H2(g)+O2(g)⟶2H2O(g)ΔG°=−457.18 kJ/mol 6H2O(g)+P4O10(s)⟶4H3PO4(l)ΔG°=−428.66 kJ/molP4(s)+5O2(g)⟶P4O10(s)ΔG°=−2697.0 kJ/mol 2H2(g)+O2(g)⟶2H2O(g)ΔG°=−457.18 kJ/mol 6H2O(g)+P4O10(s)⟶4H3PO4(l)ΔG°=−428.66 kJ/mol$
(a) Determine the standard free energy of formation, $ΔGf°,ΔGf°,$ for phosphoric acid.
(b) How does your calculated result compare to the value in Appendix G? Explain.
36.
Is the formation of ozone (O3(g)) from oxygen (O2(g)) spontaneous at room temperature under standard state conditions?
37.
Consider the decomposition of red mercury(II) oxide under standard state conditions.
$2HgO(s,red)⟶2Hg(l)+O2(g)2HgO(s,red)⟶2Hg(l)+O2(g)$
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
38.
Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.
(a) Ammonia: $2NH3(g)⟶N2(g)+3H2(g)2NH3(g)⟶N2(g)+3H2(g)$
(b) Diborane: $B2H6(g)⟶2B(g)+3H2(g)B2H6(g)⟶2B(g)+3H2(g)$
(c) Hydrazine: $N2H4(g)⟶N2(g)+2H2(g)N2H4(g)⟶N2(g)+2H2(g)$
(d) Hydrogen peroxide: $H2O2(l)⟶H2O(g)+12O2(g)H2O2(l)⟶H2O(g)+12O2(g)$
39.
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
$(a)N2(g)+O2(g)⟶2NO(g)T=2000°CKp=4.1×10−4 (b)H2(g)+I2(g)⟶2HI(g)T=400°CKp=50.0 (c)CO2(g)+H2(g)⟶CO(g)+H2O(g)T=980°CKp=1.67 (d)CaCO3(s)⟶CaO(s)+CO2(g)T=900°CKp=1.04 (e)HF(aq)+H2O(l)⟶H3O+(aq)+F−(aq)T=25°CKp=7.2×10−4 (f)AgBr(s)⟶Ag+(aq)+Br−(aq)T=25°CKp=3.3×10−13 (a)N2(g)+O2(g)⟶2NO(g)T=2000°CKp=4.1×10−4 (b)H2(g)+I2(g)⟶2HI(g)T=400°CKp=50.0 (c)CO2(g)+H2(g)⟶CO(g)+H2O(g)T=980°CKp=1.67 (d)CaCO3(s)⟶CaO(s)+CO2(g)T=900°CKp=1.04 (e)HF(aq)+H2O(l)⟶H3O+(aq)+F−(aq)T=25°CKp=7.2×10−4 (f)AgBr(s)⟶Ag+(aq)+Br−(aq)T=25°CKp=3.3×10−13$
40.
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
$(a)Cl2(g)+Br2(g)⟶2BrCl(g)T=25°CKp=4.7×10−2 (b)2SO2(g)+O2(g)⇌2SO3(g)T=500°CKp=48.2 (c)H2O(l)⇌H2O(g)T=60°CKp=0.196 (d)CoO(s)+CO(g)⇌Co(s)+CO2(g)T=550°CKp=4.90×102 (e)CH3NH2(aq)+H2O(l)⟶CH3NH3+(aq)+OH−(aq)T=25°CKp=4.4×10−4 (f)PbI2(s)⟶Pb2+(aq)+2I−(aq)T=25°CKp=8.7×10−9 (a)Cl2(g)+Br2(g)⟶2BrCl(g)T=25°CKp=4.7×10−2 (b)2SO2(g)+O2(g)⇌2SO3(g)T=500°CKp=48.2 (c)H2O(l)⇌H2O(g)T=60°CKp=0.196 (d)CoO(s)+CO(g)⇌Co(s)+CO2(g)T=550°CKp=4.90×102 (e)CH3NH2(aq)+H2O(l)⟶CH3NH3+(aq)+OH−(aq)T=25°CKp=4.4×10−4 (f)PbI2(s)⟶Pb2+(aq)+2I−(aq)T=25°CKp=8.7×10−9$
41.
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
$(a)O2(g)+2F2(g)⟶2OF2(g)ΔG°=−9.2 kJ (b)I2(s)+Br2(l)⟶2IBr(g)ΔG°=7.3 kJ (c)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)ΔG°=−79 kJ (d)N2O3(g)⟶NO(g)+NO2(g)ΔG°=−1.6 kJ (e)SnCl4(l)⟶SnCl4(l)ΔG°=8.0 kJ (a)O2(g)+2F2(g)⟶2OF2(g)ΔG°=−9.2 kJ (b)I2(s)+Br2(l)⟶2IBr(g)ΔG°=7.3 kJ (c)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)ΔG°=−79 kJ (d)N2O3(g)⟶NO(g)+NO2(g)ΔG°=−1.6 kJ (e)SnCl4(l)⟶SnCl4(l)ΔG°=8.0 kJ$
42.
Determine ΔGº for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
$SbCl5(g)⟶SbCl3(g)+Cl2(g)SbCl5(g)⟶SbCl3(g)+Cl2(g)$
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
(b) Chlorine molecules dissociate according to this reaction:
$Cl2(g)⟶2Cl(g)Cl2(g)⟶2Cl(g)$
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
43.
Given that the $ΔGf°ΔGf°$ for Pb2+(aq) and Cl(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).
44.
Determine the standard free energy change, $ΔGf°,ΔGf°,$ for the formation of S2−(aq) given that the $ΔGf°ΔGf°$ for Ag+(aq) and Ag2S(s) are 77.1 kJ/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 $××$ 10−51.
45.
Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
46.
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
$H2O(l)⇌H2O(g)ΔG°=8.58 kJH2O(l)⇌H2O(g)ΔG°=8.58 kJ$
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, KP, for this physical process.
(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, $PH2O,PH2O,$ is 0.011 atm.
(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of $PH2OPH2O$ in the air?
47.
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
$Glu+ATP⟶G6P+ADPΔG°=−17 kJGlu+ATP⟶G6P+ADPΔG°=−17 kJ$
In this process, ATP becomes ADP summarized by the following equation:
$ATP⟶ADPΔG°=−30 kJATP⟶ADPΔG°=−30 kJ$
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
$Glu⟶G6PΔG°=?Glu⟶G6PΔG°=?$
48.
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
$G6P⇌F6PΔG°=1.7 kJG6P⇌F6PΔG°=1.7 kJ$
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
49.
Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.
(a) $N2(g)+3H2(g)⟶2NH3(g)N2(g)+3H2(g)⟶2NH3(g)$
(b) $HCl(g)+NH3(g)⟶NH4Cl(s)HCl(g)+NH3(g)⟶NH4Cl(s)$
(c) $(NH4)2Cr2O7(s)⟶Cr2O3(s)+4H2O(g)+N2(g)(NH4)2Cr2O7(s)⟶Cr2O3(s)+4H2O(g)+N2(g)$
(d) $2Fe(s)+3O2(g)⟶Fe2O3(s)2Fe(s)+3O2(g)⟶Fe2O3(s)$
50.
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
51.
An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation:
$Cu2S(s)⟶Cu(s)+S(s)Cu2S(s)⟶Cu(s)+S(s)$
(a) Determine $ΔG°ΔG°$ for the decomposition of Cu2S(s).
(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine $ΔG°ΔG°$ for the process.
(c) The production of copper from chalcocite is performed by roasting the Cu2S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
52.
What happens to $ΔG°ΔG°$ (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
(a) $S(s)+O2(g)⟶SO2(g)S(s)+O2(g)⟶SO2(g)$
(b) $2SO2(g)+O2(g)⟶SO3(g)2SO2(g)+O2(g)⟶SO3(g)$
(c) $HgO(s)⟶Hg(l)+O2(g)HgO(s)⟶Hg(l)+O2(g)$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/12%3A_Thermodynamics/12.09%3A_Exercises.txt |
In this chapter, you will learn how to predict the position of the balance and the yield of a product of a reaction under specific conditions, how to change a reaction's conditions to increase or reduce yield, and how to evaluate an equilibrium system's reaction to disturbances.
• 13.1: Introduction
Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance. Reactions can occur in both directions simultaneously (reactants to products and products to reactants) and eventually reach a state of balance.
• 13.2: Chemical Equilibria
A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process, meaning the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
• 13.3: Equilibrium Constants
For any reaction that is at equilibrium, the reaction quotient Q is equal to the equilibrium constant K for the reaction. If a reactant or product is a pure solid, a pure liquid, or the solvent in a dilute solution, the concentration of this component does not appear in the expression for the equilibrium constant. At equilibrium, the values of the concentrations of the reactants and products are constant and the reaction quotient will always equal K.
• 13.4: Shifting Equilibria - Le Chatelier’s Principle
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Châtelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium.
• 13.5: Equilibrium Calculations
The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant.
• 13.6: Key Terms
• 13.7: Key Equations
• 13.8: Summary
• 13.9: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. 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.
13: Fundamental Equilibrium Concepts
Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure \(1\)). This chapter provides a thorough introduction to the essential aspects of chemical equilibria. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the nature of equilibrium systems
• Explain the dynamic nature of a chemical equilibrium
• Relate the magnitude of an equilibrium constant to properties of the chemical system
The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of “reactant” and “product,” a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.
Figure $1$ illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:
$N_2 O_4(g) \rightleftharpoons 2 NO_2(g) \nonumber$
Note that a special double arrow is used to emphasize the reversible nature of the reaction.
For this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:
\begin{aligned} &\text {rate}_f=k_f\left[ N_2 O_4\right]\[4pt] &\text { rate }_r=k_r\left[ NO_2\right]^2 \end{aligned} \nonumber
As the reaction begins (t = 0), the concentration of the N2O4 reactant is finite and that of the NO2 product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N2O4 is consumed and its concentration falls, while NO2 is produced and its concentration increases (Figure $\PageIndex{1b}$). The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure $\PageIndex{1c}$). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure $\PageIndex{1b}$ and Figure $\PageIndex{1c}$). It’s important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not “stopped,” but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.
Physical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:
$\ce{Br2(l) <=> Br2(g)} \nonumber$
When liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in Figure $3$. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.02%3A_Chemical_Equilibria.txt |
Learning Objectives
By the end of this section, you will be able to:
• Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
• Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
• Relate the magnitude of an equilibrium constant to properties of the chemical system
The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by
$m A +n B +\rightleftharpoons x C +y D \nonumber$
the reaction quotient is derived directly from the stoichiometry of the balanced equation as
$Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber$
where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:
$Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber$
Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients.
Example $1$: Writing Reaction Quotient Expressions
Write the concentration-based reaction quotient expression for each of the following reactions:
1. $\ce{3 O2(g) \rightleftharpoons 2 O3(g)}$
2. $\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}$
3. $\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}$
Solution
1. $Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber$
2. $Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber$
3. $Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber$
Exercise $1$
Write the concentration-based reaction quotient expression for each of the following reactions:
1. $\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}$
2. $\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}$
3. $\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}$
Answer
1. $Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber$
2. $Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber$
3. $Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber$
The numerical value of $Q$ varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:
$\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber$
Two different experimental scenarios are depicted in Figure $1$, one in which this reaction is initiated with a mixture of reactants only, SO2 and O2, and another that begins with only product, SO3. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:
$Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber$
As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Qc), product concentration increases (as does the numerator of Qc), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Qc.
If the reaction begins with only product present, the value of Qc is initially undefined (immeasurably large, or infinite):
$Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber$
In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Qc decrease with time, the reactant concentrations and the denominator of Qc increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.
The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, $K$:
$K \equiv Q \text { at equilibrium } \nonumber$
Comparison of the data plots in Figure $1$ shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action.
Definition: Law of Mass Action
At a given temperature, the reaction quotient for a system at equilibrium is constant.
Example $2$: Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
$\ce{2 NO2(g) <=> N2O4(g)} \nonumber$
When 0.10 mol NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 0.042 M.
1. What is the value of the reaction quotient before any reaction occurs?
2. What is the value of the equilibrium constant for the reaction?
Solution
As for all equilibrium calculations in this text, use the simplified equations for $Q$ and $K$ and disregard any concentration or pressure units, as noted previously in this section.
(a) Before any product is formed
$\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber$
$[\ce{N2O4}] = 0~\text{M} \nonumber$
Thus
$Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber$
(b) At equilibrium,
$K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber$
The equilibrium constant is $1.6 \times 10^{2}$.
Exercise $2$
For the reaction
$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber$
the equilibrium concentrations are [SO2] = 0.90 M, [O2] = 0.35 M, and [SO3] = 1.1 M. What is the value of the equilibrium constant, Kc?
Answer
Kc = 4.3
By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.
The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.
To further illustrate this important point, consider the reversible reaction shown below:
$\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber$
The bar charts in Figure $2$ represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.
Example $3$: Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber$
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
Reactants/Products Experiment 1 Experiment 2 Experiment 3
[CO]i 0.020 M 0.011 M 0.0094 M
[H2O]i 0.020 M 0.0011 M 0.0025 M
[CO2]i 0.0040 M 0.037 M 0.0015 M
[H2]i 0.0040 M 0.046 M 0.0076 M
Solution
Experiment 1:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber$
Qc < Kc (0.040 < 0.64)
The reaction will proceed in the forward direction.
Experiment 2:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber$
Qc > Kc (140 > 0.64)
The reaction will proceed in the reverse direction.
Experiment 3:
$Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber$
Qc < Kc (0.48 < 0.64)
The reaction will proceed in the forward direction.
Exercise $3$
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
1. A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl: $\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber$
2. A 5.0-L flask containing 17 g of NH3, 14 g of N2, and 12 g of H2: $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber$
3. A 2.00-L flask containing 230 g of SO3(g): $\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber$
Answer
(a) Qc = 6.45 \times 10^{3} DELMAR, forward. (b) Qc = 0.23, reverse. (c) Qc = 0, forward.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:
\begin{aligned} \ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \[4pt] \ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \[4pt] \ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \[4pt] \ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]} \end{aligned} \nonumber
These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
The equilibria below all involve gas-phase solutions:
\begin{aligned} \ce{C2H6(g) & \rightleftharpoons C2H4(g) + H2(g)} & K_c & =\frac{\left[ C_2 H_4\right]\left[ H_2\right]}{\left[ C_2 H_6\right]} \[4pt] \ce{3 O2(g) & \rightleftharpoons 2 O3(g)} & K_c & =\frac{\left[ O_3\right]^2}{\left[ O_2\right]^3} \[4pt] \ce{N2(g) + 3 H2(g) & \rightleftharpoons 2 NH3(g)} & K_c & =\frac{\left[ NH_3\right]^2}{\left[ N_2\right]\left[ H_2\right]^3} \[4pt] \ce{C3H8(g) + 5 O2(g) & \rightleftharpoons 3 CO2(g) + 4 H2O(g)} & K_c & =\frac{\left[ CO_2\right]^3\left[ H_2 O \right]^4}{\left[ C_3 H_8\right]\left[ O_2\right]^5} \end{aligned} \nonumber
For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:
\begin{align*} P V &=n R T \[4pt] P & =\left(\frac{n}{V}\right) R T \[4pt] & =M R T \end{align*} \nonumber
where $P$ is partial pressure, $V$ is volume, $n$ is molar amount, $R$ is the gas constant, $T$ is temperature, and $M$ is molar concentration.
For the gas-phase reaction
$a A +b B \rightleftharpoons c C +d D \nonumber$
\begin{align*} K_P &=\frac{\left(P_C\right)^c\left(P_D\right)^d}{\left(P_A\right)^a\left(P_B\right)^b} \[4pt] &= \dfrac{([ C ] \times R T)^c([ D ] \times R T)^d}{([ A ] \times R T)^a([ B ] \times R T)^b} \[4pt] &= \dfrac{[ C ]^c[ D ]^d}{[ A ]^a[ B ]^b} \times \frac{(R T)^{c+d}}{(R T)^{a+b}} \[4pt] &= K_c(R T)^{(c+d)-(a+b)} \[4pt] &= K_c(R T)^{\Delta n} \end{align*} \nonumber
And so, the relationship between Kc and KP is
$K_P=K_c(R T)^{\Delta n} \nonumber$
where $Δn$ is the difference in the molar amounts of product and reactant gases, in this case:
$\Delta n=(c+d)-(a+b) \nonumber$
Example $4$: Calculation of KP
Write the equations relating Kc to KP for each of the following reactions:
1. $\ce{C2H6(g) <=> C2H4(g) + H2(g)}$
2. $\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$
3. $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$
4. Kc is equal to 0.28 for the following reaction at 900 °C: $\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g)} \nonumber$ What is KP at this temperature?
Solution
1. $Δn = (2) − (1) = 1$ $K_P = K_c (RT)^{Δn} = K_c (RT)^1 = K_c (RT) \nonumber$
2. $Δn = (2) − (2) = 0$ $K_P = K_c (RT)^{Δn} = K_c (RT)^0 = K_c \nonumber$
3. $Δn = (2) − (1 + 3) = −2$ $K_P = K_c (RT)^{Δn} = K_c (RT)^{−2} = \dfrac{K_c}{(R T)^2} \nonumber$
4. $K_P = K_c (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 \times 10^{−5} \nonumber$
Exercise $4$
Write the equations relating Kc to KP for each of the following reactions:
1. $\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$
2. $\ce{N2O4(g) <=> 2 NO2(g)}$
3. $\ce{C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O (g)}$
4. At 227 °C, the following reaction has Kc = 0.0952: $\ce{CH3OH(g) <=> CO(g) + 2 H2(g)} \nonumber$ What would be the value of KP at this temperature?
Answer
(a) KP = Kc (RT)−1; (b) KP = Kc (RT); (c) KP = Kc (RT); (d) 160 or 1.6 \times 10^{2} DELMAR
Heterogeneous Equilibria
A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:
\begin{align*} \ce{PbCl2(s) & \rightleftharpoons Pb^{2+}(aq) + 2 Cl^{-}(aq)} & K_c & =\left[ \ce{Pb^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \[4pt] \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_c & =\frac{1}{\left[ \ce{CO_2}\right]} \[4pt] \ce{C(s) + 2S(g) & \rightleftharpoons CS2(g)} & K_c & =\frac{\left[ CS_2\right]}{\left[ \ce{S^2} \right.} \[4pt] \ce{Br2(l) & \rightleftharpoons Br2(g)} & K_c & =\left[ \ce{Br_2(g)} \right] \end{align*} \nonumber
Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.
Two of the above examples include terms for gaseous species only in their equilibrium constants, and so Kp expressions may also be written:
\begin{align*} \ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_P & =\frac{1}{P_{ \ce{CO2}}} \[4pt] \ce{C(s) + 2 S(g) & \rightleftharpoons CS2(g)} & K_P & =\frac{P_{ \ce{CS2}}}{\left(P_{ \ce{S} }\right)^2} \end{align*} \nonumber
Coupled Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.
$\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \[4pt] B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]} \end{array} \nonumber$
$K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber$
2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:
$\begin{array}{ll} A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \[4pt] xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }} \end{array} \nonumber$
$K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber$
3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:
$\begin{array}{ll} A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \[4pt] B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]} \end{array} \nonumber$
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:
\begin{aligned} & A + B \rightleftharpoons B + C \[4pt] & A + B \rightleftharpoons B + C \[4pt] & A \rightleftharpoons C \end{aligned} \nonumber
$K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber$
Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:
$K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber$
$K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber$
Example $5$ demonstrates the use of this strategy in describing coupled equilibrium processes.
Example $5$: Equilibrium Constants for Coupled Reactions
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
$\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber$
Use the information below to calculate Kc for this reaction.
\begin{align*} \ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \[4pt] \ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C \end{align*} \nonumber
Solution
The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.
Reverse the first coupled reaction equation:
$2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber$
Multiply the second coupled reaction by 3:
$3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber$
Finally, add the two revised equations:
\begin{align*} \ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \[4pt] \ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)} \end{align*} \nonumber
$K_{ c }= K_{ c 1}, K_{ c 2},=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber$
Exercise $5$
Use the provided information to calculate Kc for the following reaction at 550 °C:
\begin{align*} \ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \[4pt] \ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \[4pt] \ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67 \end{align*} \nonumber
Answer
Kc = 0.14 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.03%3A_Equilibrium_Constants.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Chatelier’s principle
Definition: Le Chatelier’s Principle
if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.
Effect of a Change in Concentration
If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction
$\ce{H2(g) + I2(g) <=> 2 HI (g)} \quad K_c=50.0 \text { at } 400{ }^{\circ} C \nonumber$
The rate laws for the forward and reverse reactions are
\begin{align*} &\text{forward} \quad \ce{H2(g) + I2(g) \rightarrow 2HI(g)} \quad &&\text{rate}_f=k_f[ \ce{H2}]^m [ \ce{I2} ]^n \[4pt] &\text{reverse} \quad \ce{2 HI(g) \rightarrow H2(g) + I2(g)} \quad &&\text{rate}_r=k_r[\ce{HI}]^x \end{align*} \nonumber
When this system is at equilibrium, the forward and reverse reaction rates are equal.
$\text { rate }_f=\text { rate }_r \nonumber$
If the system is stressed by adding reactant, either $\ce{H_2}$ or $\ce{I_2}$, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:
$\operatorname{rate}_f>\text { rate }_r \nonumber$
The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.
The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction
$\text { rate }_f<\text { rate }_r \nonumber$
and a temporary net reaction in the reverse direction to re-establish equilibrium.
As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,
$Q_c=\dfrac{[ \ce{HI} ]^2}{\left[ \ce{H2} \right]\left[ \ce{I2} \right]}=K_c \nonumber$
If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the concentrations of all three species will be lower when equilibrium is reestablished. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left.For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M,
$M=\dfrac{n}{V}=\dfrac{P}{R T} \nonumber$
and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since
all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.
At equilibrium, the reaction
$\ce{H2(g) + I2(g) <=> 2HI(g)} \nonumber$
is described by the reaction quotient
$Q_P=\dfrac{(P_{\ce{HI}})^2}{P_{\ce{H_2}} P_{\ce{I2}}}=K_p \nonumber$
If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3:
\begin{align*} Q_p{ }^{\prime} &=\dfrac{\left(3 P_{ HI }\right)^2}{3 P_{ H_2} 3 P_{ I_2}}=\dfrac{9 P_{ H^2}}{9 P_{ H_2} P_{ I_2}}=\dfrac{P_{ HI^2}}{P_{ H_2} P_{ I_2}} \[4pt] &=Q_P=K_P \end{align*} \nonumber
And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.
A similar treatment of a different system,
$\ce{2NO2(g) ⇌ 2 NO(g) + O2(g)} \nonumber$
however, yields a different result:
\begin{align*} Q_P &=\dfrac{(P_{NO})^2} {(P_{O_2})}{(P_{NO_2})^2} \[4pt] Q_P^{\prime} &= \dfrac{(3 P_{ \ce{NO}})^2 (3 P_{ \ce{O2}}) }{3 (P_{ \ce{NO2} })^2} = \dfrac{9 (P_{ \ce{NO}})^2 (3 P_{\ce{O_2}}) }{9 (P_{\ce{NO2}})^2} = \dfrac{27 (P_{\ce{NO}})^2 P_{ \ce{O_2}}}{9 (P_{ \ce{NO_2} })^2} \[4pt] &=3 Q_P>K_P \end{align*} \nonumber
In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.
These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.
Check out this link to see a dramatic visual demonstration of how equilibrium changes with pressure changes.
Chemistry in Everyday Life: Equilibrium and Soft Drinks
The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestley’s approach involved production of carbon dioxide by reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).
The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:
\begin{align*} &\text{dissolution} \quad &\ce{CO2(g) <=> CO2(aq)} \[4pt] &\text{hydrolysis} \quad &\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \[4pt] &\text{ionization} \quad &\ce{H2CO3(aq) <=> HCO3^{-}(aq) + H^{+}(aq)} \end{align*} \nonumber
These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to
the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.
When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”
Effect of a Change in Temperature
Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re-establish equilibrium without any change in the value of the
equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.
To understand this phenomenon, consider the elementary reaction
$A \rightleftharpoons B \nonumber$
Since this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation’s stoichiometry:
\begin{aligned} \operatorname{rate}_f & =k_f[ A ] \[4pt] \operatorname{rate}_r & =k_r[ B ] \end{aligned} \nonumber
When the system is at equilibrium,
$\text { rate }_r=\text { rate }_f \nonumber$
Substituting the rate laws into this equality and rearranging gives
\begin{aligned} & k_f[ A ]=k_r[ B ] \[4pt] & \dfrac{[B]}{[A]}=\dfrac{k_f}{k_r}=K_c \end{aligned} \nonumber
The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as
described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.
Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:
$\ce{N2O4(g) <=> 2 NO2(g)} \quad \Delta H=+57.20 ~\text{kJ} \nonumber$
For purposes of applying Le Chatelier’s principle, heat (q) may be viewed as a reactant:
$\ce{ heat + N2O4(g) <=> 2 NO2(g)} \nonumber$
Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.
Effect of a Catalyst
The kinetics chapter of this text identifies a catalyst as a substance that enables a reaction to proceed via a different mechanism with an accelerated rate. The catalyzed reaction mechanism involves a lower energy transition state than the uncatalyzed reaction, resulting in a lower activation energy, Ea, and a correspondingly greater rate constant.
To discern the effect of catalysis on an equilibrium system, consider the reaction diagram for a simple one-step (elementary) reaction shown in Figure $2$. The lowered transition state energy of the catalyzed reaction results in lowered activation energies for both the forward and the reverse reactions. Consequently, both forward and reverse reactions are accelerated, and equilibrium is achieved more quickly but without a change in the equilibrium constant.
An interesting case study highlighting these equilibrium concepts is the industrial production of ammonia, NH3. This substance is among the “top 10” industrial chemicals with regard to production, with roughly two billion pounds produced annually in the US. Ammonia is used as a chemical feedstock to synthesize a wide range of commercially useful compounds, including fertilizers, plastics, dyes, and explosives.
Most industrial production of ammonia uses the Haber-Bosch process based on the following equilibrium reaction:
$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad \Delta H=-92.2 ~\text{kJ} \nonumber$
The traits of this reaction present challenges to its use in an efficient industrial process. The equilibrium constant is relatively small (Kp on the order of $10^{−5}$ at 25 °C), meaning very little ammonia is present in an equilibrium mixture. Also, the rate of this reaction is relatively slow at low temperatures. To raise the yield of ammonia, the industrial process is designed to operate under conditions favoring product formation:
• High pressures (concentrations) of reactants are used, ~150−250 atm, to shift the equilibrium right, favoring product formation.
• Ammonia is continually removed (collected) from the equilibrium mixture during the process, lowering its concentration and also shifting the equilibrium right.
• Although low temperatures favor product formation for this exothermic process, the reaction rate at low temperatures is inefficiently slow. A catalyst is used to accelerate the reaction to reasonable rates at relatively moderate temperatures (400−500 °C).
A diagram illustrating a typical industrial setup for production of ammonia via the Haber-Bosch process is shown in Figure $3$. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.04%3A_Shifting_Equilibria-_Le_Chateliers_Principle.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
• Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches
Having covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology—for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.
Many of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:
$\ce{2 NH3(g) <=> N2(g) + 3 H2(g)} \nonumber$
As shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x:
$\Delta\left[ N_2\right]=+x \nonumber$
the corresponding changes in the other species concentrations are
\begin{align*} \Delta\left[ \ce{H2} \right] &=\Delta\left[ \ce{N2} \right]\left(\frac{3 ~\text{mol} ~\ce{H2} }{1~\text{mol}~\ce{N2} }\right)=+3 x \[4pt] \Delta\left[ \ce{NH3} \right] &=-\Delta\left[ \ce{N2} \right]\left(\frac{2 ~\text{mol}~\ce{NH3}}{1~\text{mol} ~\ce{N2} }\right)=-2 x \end{align*}
where the negative sign indicates a decrease in concentration.
Example $1$: Determining Relative Changes in Concentration
Derive the missing terms representing concentration changes for each of the following reactions.
1. $\underset{x}{\ce{C2H2(g)}} + \underset{?}{\ce{2 Br2(g)}} \ce{<=>} \underset{?}{\ce{C2H2Br4(g)}} \nonumber$
2. $\underset{?}{\ce{I2(aq)}} + \underset{?}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber$
3. $\underset{x}{\ce{C3H8(g)}} + \underset{?}{\ce{5O2(g)}} \ce{<=>} \underset{?}{\ce{3CO2(g)}} + \underset{?}{\ce{4H2O(g)}} \nonumber$
Solution
1. $\underset{x}{\ce{C2H2(g)}} + \underset{2x}{\ce{2 Br2(g)}} \ce{<=>} \underset{-x}{\ce{C2H2Br4(g)}} \nonumber$
2. $\underset{-x}{\ce{I2(aq)}} + \underset{-x}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber$
3. $\underset{x}{\ce{C3H8(g)}} + \underset{5x}{\ce{5O2(g)}} \ce{<=>} \underset{-3x}{\ce{3CO2(g)}} + \underset{-4x}{\ce{4H2O(g)}} \nonumber$
Exercise $1$
Complete the changes in concentrations for each of the following reactions:
1. $\underset{?}{\ce{2SO2(g)}} + \underset{x}{\ce{O2(g)}} \ce{<=>} \underset{?}{\ce{2SO3(g)}} \nonumber$
2. $\underset{?}{\ce{C4H8(g)}} \ce{<=>} \underset{-2x}{\ce{2C2H4(g)}} \nonumber$
3. $\underset{?}{\ce{4NH3(g)}} + \underset{x}{\ce{7O2(g)}} \ce{<=>} \underset{?}{\ce{4NO2(g)}} + \underset{?}{\ce{6H2O(g)}} \nonumber$
Answer
(a) 2x, x, −2x; (b) x, −2x; (c) 4x, 7x, −4x, −6x or −4x, −7x, 4x, 6x
Calculation of an Equilibrium Constant
The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example $1$. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.
Example $2$: Calculation of an Equilibrium Constant
Iodine molecules react reversibly with iodide ions to produce triiodide ions.
$\ce{I2(aq) + I^{-}(aq) \rightleftharpoons I_3^{-}(aq)} \nonumber$
If a solution with the concentrations of $\ce{I2}$ and $\ce{I^{−}}$ both equal to $1.000 \times 10^{−3}~\text{M}$ before reaction gives an equilibrium concentration of $\ce{I2}$ of $6.61 \times 10^{−4} ~\text{M}$, what is the equilibrium constant for the reaction?
Solution
To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:
$K_C=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \nonumber$
Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.
At equilibrium the concentration of I2 is $6.61 \times 10^{−4}~\text{M}$ so that
\begin{align*} 1.000 \times 10^{-3}-x & =6.61 \times 10^{-4}\[4pt] x &=1.000 \times 10^{-3}-6.61 \times 10^{-4}\[4pt] &=3.39 \times 10^{-4} M \end{align*} \nonumber
The ICE table may now be updated with numerical values for all its concentrations:
Finally, substitute the equilibrium concentrations into the K expression and solve:
\begin{align*} K_c &=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \[4pt] &=\frac{3.39 \times 10^{-4} M}{\left(6.61 \times 10^{-4} M\right)\left(6.61 \times 10^{-4} M\right)}=776 \end{align*} \nonumber
Exercise $2$
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.
$\ce{C2H5OH + CH3CO2H \rightleftharpoons CH3CO2C2H5 + H2O} \nonumber$
When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\frac{1}{3}$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)
Answer
Kc = 4
Calculation of a Missing Equilibrium Concentration
When the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise.
Example $3$: Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the Kc for the reaction,
$\ce{N2(g) + O2(g) \rightleftharpoons 2 NO(g)} \nonumber$
is $4.1 \times 10^{−4}$. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N2 and O2 at this pressure and temperature are 0.036 M and 0.0089 M, respectively.
Solution
Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:
\begin{align*} K_c &=\frac{[ NO ]^2}{\left[ N_2\right]\left[ O_2\right]} \[4pt] [ NO ]^2&=K_c\left[ N_2\right]\left[ O_2\right] \[4pt] [ NO ]&=\sqrt{K_c\left[ N_2\right]\left[ O_2\right]} \[4pt] &=\sqrt{\left(4.1 \times 10^{-4}\right)(0.036)(0.0089)} \[4pt] &=\sqrt{1.31 \times 10^{-7}} \[4pt] &=3.6 \times 10^{-4} \end{align*} \nonumber
Thus [NO] is $3.6 \times 10^{−4} ~\text{mol/L}$ at equilibrium under these conditions.
To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:
\begin{align*} K_c &= \frac{[ NO ]^2}{\left[ N_2\right]\left[ O_2\right]} \[4pt] &= \frac{\left(3.6 \times 10^{-4}\right)^2}{(0.036)(0.0089)} \[4pt] &= 4.0 \times 10^{-4} \end{align*} \nonumber
This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.
Exercise $3$
The equilibrium constant Kc for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is $6.00 \times 10^{−2}$. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.
Answer
1.53 mol/L
Calculation of Equilibrium Concentrations from Initial Concentrations
Perhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:
1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Develop an ICE table.
3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.
4. Confirm the calculated equilibrium concentrations.
The last two example exercises of this chapter demonstrate the application of this strategy.
Example $4$: Calculation of Equilibrium Concentrations
Under certain conditions, the equilibrium constant Kc for the decomposition of $\ce{PCl5(g)}$ into $\ce{PCl3(g)}$ and $\ce{Cl2(g)}$ is 0.0211. What are the equilibrium concentrations of $\ce{PCl5}$, $\ce{PCl3}$, and $\ce{Cl2}$ in a mixture that initially contained only $\ce{PCl5}$ at a concentration of 1.00 M?
Solution
Use the stepwise process described earlier.
Step 1.
Determine the direction the reaction proceeds.
The balanced equation for the decomposition of $\ce{PCl5}$ is
$\ce{PCl5(g) \rightleftharpoons PCl3(g) + Cl2(g)} \nonumber$
Because only the reactant is present initially Qc = 0 and the reaction will proceed to the right.
Step 2:
Develop an ICE table.
Step 3.
Solve for the change and the equilibrium concentrations.
Substituting the equilibrium concentrations into the equilibrium constant equation gives
\begin{align*} K_c &=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2} \right]}{\left[ \ce{PCl5} \right]}=0.0211 \[4pt] &=\frac{(x)(x)}{(1.00-x)} \[4pt] 0.0211&=\frac{(x)(x)}{(1.00-x)} \[4pt] 0.0211(1.00-x)7 &=x^2 \[4pt] x^2+0.0211 x - 0.0211 &=0 \end{align*}
Appendix B shows an equation of the form $ax^2 + bx + c = 0$ can be rearranged to solve for $x$:
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \nonumber$
In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:
\begin{align*} x&=\frac{-0.0211 \pm \sqrt{(0.0211)^2-4(1)(-0.0211)}}{2(1)} \[4pt] &=\frac{-0.0211 \pm \sqrt{\left(4.45 \times 10^{-4}\right)+\left(8.44 \times 10^{-2}\right)}}{2} \[4pt] &=\frac{-0.0211 \pm 0.291}{2} \end{align*}
The two roots of the quadratic are, therefore,
$x=\frac{-0.0211+0.291}{2}=0.135 \nonumber$
and
$x=\frac{-0.0211-0.291}{2}=-0.156 \nonumber$
For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M.
The equilibrium concentrations are
\begin{align*} [ \ce{PCl5}] &= 1.00-0.135 = 0.87 ~\text{M} \[4pt] [ \ce{PCl3} ] &= x=0.135 ~\text{M}= \[4pt] [ \ce{Cl2} ] &= x =0.135 ~\text{M} \end{align*} \nonumber
Step 4.
Confirm the calculated equilibrium concentrations.
Substitution into the expression for Kc (to check the calculation) gives
$K_c=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2}\right]}{\left[ \ce{PCl5} \right]}=\frac{(0.135)(0.135)}{0.87}=0.021 \nonumber$
The equilibrium constant calculated from the equilibrium concentrations is equal to the value of Kc given in the problem (when rounded to the proper number of significant figures).
Exercise $\PageIndex{4A}$
Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5.
$\ce{CH3CO2H + C2H5OH <=> CH3CO2C2H5 + H2O} \nonumber$
The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O?
Answer
[CH3CO2H] = 0.18 M, [C2H5OH] = 0.18 M, [CH3CO2C2H5] = 0.37 M, [H2O] = 0.37 M
Exercise $\PageIndex{4B}$
A 1.00-L flask is filled with 1.00 mole of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2, I2, and HI in moles/L?
$\ce{H2(g) + I2(g) <=> 2 HI(g)} \nonumber$
Answer
[H2] = 0.06 M, [I2] = 1.06 M, [HI] = 1.88 M
Example $5$: Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption
What are the concentrations at equilibrium of a 0.15 M solution of HCN?
$\ce{HCN(aq) <=> H^{+}(aq) + CN^{-}(aq)} \quad \quad K_c=4.9 \times 10^{-10} \nonumber$
Solution
Using “x” to represent the concentration of each product at equilibrium gives this ICE table.
Substitute the equilibrium concentration terms into the Kc expression
$K_c=\frac{(x)(x)}{0.15-x} \nonumber$
rearrange to the quadratic form and solve for x
\begin{align*} &x^2+4.9 \times 10^{-10}-7.35 \times 10^{-11}=0 \[4pt] &x=8.56 \times 10^{-6} M(3 \text { sig. figs. })=8.6 \times 10^{-6} M(2 \text { sig. figs. }) \end{align*}
Thus $[\ce{H^{+}}] = [\ce{CN^{-}}] = x = 8.6 \times 10^{-6} ~\text{M}$ and $[\ce{HCN}] = 0.15 – x = 0.15 ~\text{M}$.
Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:
$\text { if } x \ll 0.15 M \text {, then }(0.15-x) \approx 0.15 \nonumber$
This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:
\begin{align*} K_c &=\frac{(x)(x)}{0.15-x} \approx \frac{x^2}{0.15}\[4pt] 4.9 \times 10^{-10}&=\frac{x^2}{0.15}\[4pt] x^2 &=(0.15)\left(4.9 \times 10^{-10}\right) \[4pt] &=7.4 \times 10^{-11} \[4pt] x&=\sqrt{7.4 \times 10^{-11}} \[4pt] &=8.6 \times 10^{-6} ~\text{M} \end{align*} \nonumber
The value of $x$ calculated is, indeed, much less than the initial concentration
$8.6 \times 10^{-6} \ll 0.15 \nonumber$
and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.
Exercise $5$
What are the equilibrium concentrations in a 0.25 M NH3 solution?
$\ce{NH3(aq) + H2O(l) <=> NH4^{+}(aq) + OH^{-}(aq)} \quad K_{ c }=1.8 \times 10^{-5} \nonumber$
Answer
$\left[ \ce{OH^{-}} \right]=\left[ \ce{NH_4^{+}}\right]=0.0021 ~\text{M}$
$[\ce{NH3}] = 0.25 ~\text{M}$
13.06: Key Terms
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
equilibriumstate of a reversible reaction in which the forward and reverse processes occur at equal rates
equilibrium constant (K)value of the reaction quotient for a system at equilibrium; may be expressed using concentrations (Kc) or partial pressures (Kp)
heterogeneous equilibriaequilibria in which reactants and products occupy two or more different phases
homogeneous equilibriaequilibria in which all reactants and products occupy the same phase
law of mass actionwhen a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant
Le Chatelier’s principlean equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium
reaction quotient (Q)mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations (Qc) or pressures (Qp)
reversible reactionchemical reaction that can proceed in both the forward and reverse directions under given conditions
13.07: Key Equations
$Qc=[C]x[D]y[A]m[B]nfor the reactionmA+nB⇌xC+yDQc=[C]x[D]y[A]m[B]nfor the reactionmA+nB⇌xC+yD$
$QP=(PC)x(PD)y(PA)m(PB)nfor the reactionmA+nB⇌xC+yDQP=(PC)x(PD)y(PA)m(PB)nfor the reactionmA+nB⇌xC+yD$
P = MRT
Kc = Qc at equilibrium
Kp = Qp at equilibrium
KP = Kc (RT)Δn
13.08: Summary
A reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species.
The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.
A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system’s response to these disturbances is described by Le Chatelier’s principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture’s composition (K does not change).
Calculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.05%3A_Equilibrium_Calculations.txt |
1.
What does it mean to describe a reaction as “reversible”?
2.
When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?
3.
If a reaction is reversible, when can it be said to have reached equilibrium?
4.
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
5.
If the concentrations of products and reactants are equal, is the system at equilibrium?
6.
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
7.
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 13.4.
8.
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4?
$2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)$
9.
Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation $AgCl(s)⇌Ag+(aq)+Cl−(aq).AgCl(s)⇌Ag+(aq)+Cl−(aq).$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $Pb2+(aq)+2Cl−(aq)⇌PbCl2(s).Pb2+(aq)+2Cl−(aq)⇌PbCl2(s).$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
10.
Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation $CaCO3(s)⇌Ca2+(aq)+CO32−(aq).CaCO3(s)⇌Ca2+(aq)+CO32−(aq).$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $3Ba2+(aq)+2PO43−(aq)⇌Ba3(PO4)2(s).3Ba2+(aq)+2PO43−(aq)⇌Ba3(PO4)2(s).$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
11.
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $3C2H2(g)⇌C6H6(g).3C2H2(g)⇌C6H6(g).$ Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer.
12.
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $KI(aq)+I2(aq)⇌KI3(aq)KI(aq)+I2(aq)⇌KI3(aq)$ give the same expression for the reaction quotient. KI3 is composed of the ions K+ and $I3−.I3−.$
13.
For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction?
14.
For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction?
15.
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
(a) $CH4(g)+Cl2(g)⇌CH3Cl(g)+HCl(g)CH4(g)+Cl2(g)⇌CH3Cl(g)+HCl(g)$
(b) $N2(g)+O2(g)⇌2NO(g)N2(g)+O2(g)⇌2NO(g)$
(c) $2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)$
(d) $BaSO3(s)⇌BaO(s)+SO2(g)BaSO3(s)⇌BaO(s)+SO2(g)$
(e) $P4(g)+5O2(g)⇌P4O10(s)P4(g)+5O2(g)⇌P4O10(s)$
(f) $Br2(g)⇌2Br(g)Br2(g)⇌2Br(g)$
(g) $CH4(g)+2O2(g)⇌CO2(g)+2H2O(l)CH4(g)+2O2(g)⇌CO2(g)+2H2O(l)$
(h) $CuSO4·5H2O(s)⇌CuSO4(s)+5H2O(g)CuSO4·5H2O(s)⇌CuSO4(s)+5H2O(g)$
16.
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
(a) $N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
(b) $4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)$
(c) $N2O4(g)⇌2NO2(g)N2O4(g)⇌2NO2(g)$
(d) $CO2(g)+H2(g)⇌CO(g)+H2O(g)CO2(g)+H2(g)⇌CO(g)+H2O(g)$
(e) $NH4Cl(s)⇌NH3(g)+HCl(g)NH4Cl(s)⇌NH3(g)+HCl(g)$
(f) $2Pb(NO3)2(s)⇌2PbO(s)+4NO2(g)+O2(g)2Pb(NO3)2(s)⇌2PbO(s)+4NO2(g)+O2(g)$
(g) $2H2(g)+O2(g)⇌2H2O(l)2H2(g)+O2(g)⇌2H2O(l)$
(h) $S8(g)⇌8S(g)S8(g)⇌8S(g)$
17.
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
$(a)2NH3(g)⇌N2(g)+3H2(g)Kc=17;[NH3]=0.20M,[N2]=1.00M,[H2]=1.00M (b)2NH3(g)⇌N2(g)+3H2(g)KP=6.8×104;NH3=3.0atm,N2=2.0atm,H2=1.0atm (c)2SO3(g)⇌2SO2(g)+O2(g)Kc=0.230;[SO3]=0.00M,[SO2]=1.00M,[O2]=1.00M (d)2SO3(g)⇌2SO2(g)+O2(g)KP=16.5;SO3=1.00atm,SO2=1.00atm,O2=1.00atm (e)2NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×104;[NO]=1.00M,[Cl2]=1.00M,[NOCl]=0M (f)N2(g)+O2(g)⇌2NO(g)KP=0.050;NO=10.0atm,N2=O2=5 atm (a)2NH3(g)⇌N2(g)+3H2(g)Kc=17;[NH3]=0.20M,[N2]=1.00M,[H2]=1.00M (b)2NH3(g)⇌N2(g)+3H2(g)KP=6.8×104;NH3=3.0atm,N2=2.0atm,H2=1.0atm (c)2SO3(g)⇌2SO2(g)+O2(g)Kc=0.230;[SO3]=0.00M,[SO2]=1.00M,[O2]=1.00M (d)2SO3(g)⇌2SO2(g)+O2(g)KP=16.5;SO3=1.00atm,SO2=1.00atm,O2=1.00atm (e)2NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×104;[NO]=1.00M,[Cl2]=1.00M,[NOCl]=0M (f)N2(g)+O2(g)⇌2NO(g)KP=0.050;NO=10.0atm,N2=O2=5 atm$
18.
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
$(a)2NH3(g)⇌N2(g)+3H2(g)Kc=17;[NH3]=0.50M,[N2]=0.15M,[H2]=0.12M (b)2NH3(g)⇌N2(g)+3H2(g)KP=6.8×104;NH3=2.00atm,N2=10.00 atm,H2=10.00 atm (c)2SO3(g)⇌2SO2(g)+O2(g)Kc=0.230;[SO3]=2.00M,[SO2]=2.00M,[O2]=2.00M (d)2SO3(g)⇌2SO2(g)+O2(g)KP=6.5atm;SO2=1.00 atm,O2=1.130 atm,SO3=0 atm (e)2NO(g)+Cl2(g)⇌2NOCl(g)KP=2.5×103;NO=1.00 atm,Cl2=1.00 atm,NOCl=0 atm (f)N2(g)+O2(g)⇌2NO(g)Kc=0.050;[N2]=0.100M,[O2]=0.200M,[NO]=1.00M (a)2NH3(g)⇌N2(g)+3H2(g)Kc=17;[NH3]=0.50M,[N2]=0.15M,[H2]=0.12M (b)2NH3(g)⇌N2(g)+3H2(g)KP=6.8×104;NH3=2.00atm,N2=10.00 atm,H2=10.00 atm (c)2SO3(g)⇌2SO2(g)+O2(g)Kc=0.230;[SO3]=2.00M,[SO2]=2.00M,[O2]=2.00M (d)2SO3(g)⇌2SO2(g)+O2(g)KP=6.5atm;SO2=1.00 atm,O2=1.130 atm,SO3=0 atm (e)2NO(g)+Cl2(g)⇌2NOCl(g)KP=2.5×103;NO=1.00 atm,Cl2=1.00 atm,NOCl=0 atm (f)N2(g)+O2(g)⇌2NO(g)Kc=0.050;[N2]=0.100M,[O2]=0.200M,[NO]=1.00M$
19.
The following reaction has KP = 4.50 $××$ 10−5 at 720 K.
$N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH3) = 93 atm, P(N2) = 48 atm, and P(H2) = 52 atm
20.
Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?
$SO2Cl2(g)⇌SO2(g)+Cl2(g)SO2Cl2(g)⇌SO2(g)+Cl2(g)$
[SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078.
21.
Which of the systems described in Exercise 13.15 are homogeneous equilibria? Which are heterogeneous equilibria?
22.
Which of the systems described in Exercise 13.16 are homogeneous equilibria? Which are heterogeneous equilibria?
23.
For which of the reactions in Exercise 13.15 does Kc (calculated using concentrations) equal KP (calculated using pressures)?
24.
For which of the reactions in Exercise 13.16 does Kc (calculated using concentrations) equal KP (calculated using pressures)?
25.
Convert the values of Kc to values of KP or the values of KP to values of Kc.
$(a)N2(g)+3H2(g)⇌2NH3(g)Kc=0.50at400°C (b)H2(g)+I2(g)⇌2HI(g)Kc=50.2at448°C (c)Na2SO4·10H2O(s)⇌Na2SO4(s)+10H2O(g)KP=4.08×10−25at25°C (d)H2O(l)⇌H2O(g)KP=0.122at50°C (a)N2(g)+3H2(g)⇌2NH3(g)Kc=0.50at400°C (b)H2(g)+I2(g)⇌2HI(g)Kc=50.2at448°C (c)Na2SO4·10H2O(s)⇌Na2SO4(s)+10H2O(g)KP=4.08×10−25at25°C (d)H2O(l)⇌H2O(g)KP=0.122at50°C$
26.
Convert the values of Kc to values of KP or the values of KP to values of Kc.
$(a)Cl2(g)+Br2(g)⇌2BrCl(g)Kc=4.7×10−2at25°C (b)2SO2(g)+O2(g)⇌2SO3(g)KP=48.2at500°C (c)CaCl2·6H2O(s)⇌CaCl2(s)+6H2O(g)KP=5.09×10−44at25°C (d)H2O(l)⇌H2O(g)KP=0.196at60°C (a)Cl2(g)+Br2(g)⇌2BrCl(g)Kc=4.7×10−2at25°C (b)2SO2(g)+O2(g)⇌2SO3(g)KP=48.2at500°C (c)CaCl2·6H2O(s)⇌CaCl2(s)+6H2O(g)KP=5.09×10−44at25°C (d)H2O(l)⇌H2O(g)KP=0.196at60°C$
27.
What is the value of the equilibrium constant expression for the change $H2O(l)⇌H2O(g)Appendix E.)$
28.
Write the expression of the reaction quotient for the ionization of HOCN in water.
29.
Write the reaction quotient expression for the ionization of NH3 in water.
30.
What is the approximate value of the equilibrium constant KP for the change $C2H5OC2H5(l)⇌C2H5OC2H5(g)C2H5OC2H5(l)⇌C2H5OC2H5(g)$ at 25 °C. (The equilibrium vapor pressure for this substance is 570 torr at 25 °C.)
31.
The following equation represents a reversible decomposition:
$CaCO3(s)⇌CaO(s)+CO2(g)CaCO3(s)⇌CaO(s)+CO2(g)$
Under what conditions will decomposition in a closed container proceed to completion so that no CaCO3 remains?
32.
Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium.
33.
What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
34.
The following reaction occurs when a burner on a gas stove is lit:
$CH4(g)+2O2(g)⇌CO2(g)+2H2O(g)CH4(g)+2O2(g)⇌CO2(g)+2H2O(g)$
Is an equilibrium among CH4, O2, CO2, and H2O established under these conditions? Explain your answer.
35.
A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of SO3 is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures.
$2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)$
(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
(b) Is the reaction endothermic or exothermic?
36.
Suggest four ways in which the concentration of hydrazine, N2H4, could be increased in an equilibrium described by the following equation:
$N2(g)+2H2(g)⇌N2H4(g)ΔH=95kJN2(g)+2H2(g)⇌N2H4(g)ΔH=95kJ$
37.
Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
$P4(g)+6H2(g)⇌4PH3(g)ΔH=110.5kJP4(g)+6H2(g)⇌4PH3(g)ΔH=110.5kJ$
38.
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) $2NH3(g)⇌N2(g)+3H2(g)ΔH=92kJ2NH3(g)⇌N2(g)+3H2(g)ΔH=92kJ$
(b) $N2(g)+O2(g)⇌2NO(g)ΔH=181kJN2(g)+O2(g)⇌2NO(g)ΔH=181kJ$
(c) $2O3(g)⇌3O2(g)ΔH=−285kJ2O3(g)⇌3O2(g)ΔH=−285kJ$
(d) $CaO(s)+CO2(g)⇌CaCO3(s)ΔH=−176kJCaO(s)+CO2(g)⇌CaCO3(s)ΔH=−176kJ$
39.
How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) $2H2O(g)⇌2H2(g)+O2(g)ΔH=484kJ2H2O(g)⇌2H2(g)+O2(g)ΔH=484kJ$
(b) $N2(g)+3H2(g)⇌2NH3(g)ΔH=−92.2kJN2(g)+3H2(g)⇌2NH3(g)ΔH=−92.2kJ$
(c) $2Br(g)⇌Br2(g)ΔH=−224kJ2Br(g)⇌Br2(g)ΔH=−224kJ$
(d) $H2(g)+I2(s)⇌2HI(g)ΔH=53kJH2(g)+I2(s)⇌2HI(g)ΔH=53kJ$
40.
Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
$2H2(g)+CO(g)⇌CH3OH(g)ΔH=−90.2kJ2H2(g)+CO(g)⇌CH3OH(g)ΔH=−90.2kJ$
(b) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added?
(c) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed?
(d) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added?
(e) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased?
41.
Nitrogen and oxygen react at high temperatures.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
$N2(g)+O2(g)⇌2NO(g)ΔH=181kJN2(g)+O2(g)⇌2NO(g)ΔH=181kJ$
(b) What will happen to the concentrations of N2, O2, and NO at equilibrium if more O2 is added?
(c) What will happen to the concentrations of N2, O2, and NO at equilibrium if N2 is removed?
(d) What will happen to the concentrations of N2, O2, and NO at equilibrium if NO is added?
(e) What will happen to the concentrations of N2, O2, and NO at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentrations of N2, O2, and NO at equilibrium if the temperature of the system is increased?
42.
Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.
(a) Write the expression for the equilibrium constant for the reversible reaction
$C(s)+H2O(g)⇌CO(g)+H2(g)ΔH=131.30kJC(s)+H2O(g)⇌CO(g)+H2(g)ΔH=131.30kJ$
(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
43.
Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.
(a) Write the expression for the equilibrium constant (Kc) for the reversible reaction
$Fe2O3(s)+3H2(g)⇌2Fe(s)+3H2O(g)ΔH=98.7kJFe2O3(s)+3H2(g)⇌2Fe(s)+3H2O(g)ΔH=98.7kJ$
(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
(c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
(d) What will happen to the concentration of each reactant and product at equilibrium if H2 is added?
(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?
(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?
44.
Ammonia is a weak base that reacts with water according to this equation:
$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)$
Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water?
(a) Addition of NaOH
(b) Addition of HCl
(c) Addition of NH4Cl
45.
Acetic acid is a weak acid that reacts with water according to this equation:
$CH3CO2H(aq)+H2O(aq)⇌H3O+(aq)+CH3CO2−(aq)CH3CO2H(aq)+H2O(aq)⇌H3O+(aq)+CH3CO2−(aq)$
Will any of the following increase the percent of acetic acid that reacts and produces $CH3CO2−CH3CO2−$ ion?
(a) Addition of HCl
(b) Addition of NaOH
(c) Addition of NaCH3CO2
46.
Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na+, Cl, Ag+, and $NO3−,NO3−,$ in contact with solid AgCl.
$Na+(aq)+Cl−(aq)+Ag+(aq)+NO3−(aq)⇌AgCl(s)+Na+(aq)+NO3−(aq)Na+(aq)+Cl−(aq)+Ag+(aq)+NO3−(aq)⇌AgCl(s)+Na+(aq)+NO3−(aq)$
$ΔH=−65.9kJΔH=−65.9kJ$
47.
How can the pressure of water vapor be increased in the following equilibrium?
$H2O(l)⇌H2O(g)ΔH=41kJH2O(l)⇌H2O(g)ΔH=41kJ$
48.
A solution is saturated with silver sulfate and contains excess solid silver sulfate:
$Ag2SO4(s)⇌2Ag+(aq)+SO42−(aq)Ag2SO4(s)⇌2Ag+(aq)+SO42−(aq)$
A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive Ag+ ions. Explain this observation.
49.
When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?
(a) HCl
(b) H + + Cl
(c) HOCl
(d) H + + OCl
50.
A reaction is represented by this equation: $A(aq)+2B(aq)⇌2C(aq)Kc=1×103A(aq)+2B(aq)⇌2C(aq)Kc=1×103$
(a) Write the mathematical expression for the equilibrium constant.
(b) Using concentrations ≤1 M, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.
51.
A reaction is represented by this equation: $2W(aq)⇌X(aq)+2Y(aq)Kc=5×10−42W(aq)⇌X(aq)+2Y(aq)Kc=5×10−4$
(a) Write the mathematical expression for the equilibrium constant.
(b) Using concentrations of ≤1 M, identify two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.
52.
What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
$N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g )$
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 $××$ 10−1 M NH3.
53.
Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
$CH 4 ( g ) + H 2 O ( g ) ⇌ 3 H 2 ( g ) + CO ( g ) CH 4 ( g ) + H 2 O ( g ) ⇌ 3 H 2 ( g ) + CO ( g )$
What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2 1.15 M, at a temperature of 760 °C?
54.
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature.
55.
At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.
$2 NO 2 ( g ) ⇌ 2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g ) ⇌ 2 NO ( g ) + O 2 ( g )$
56.
Calculate the value of the equilibrium constant KP for the reaction $2NO(g)+Cl2(g)⇌2NOCl(g)2NO(g)+Cl2(g)⇌2NOCl(g)$ from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm.
57.
When heated, iodine vapor dissociates according to this equation:
$I 2 ( g ) ⇌ 2 I ( g ) I 2 ( g ) ⇌ 2 I ( g )$
At 1274 K, a sample exhibits a partial pressure of I2 of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, KP, for the decomposition at 1274 K.
58.
A sample of ammonium chloride was heated in a closed container.
$NH 4 Cl ( s ) ⇌ NH 3 ( g ) + HCl ( g ) NH 4 Cl ( s ) ⇌ NH 3 ( g ) + HCl ( g )$
At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant KP for the decomposition at this temperature?
59.
At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant KP for the vaporization equilibrium at 60 °C?
$H 2 O ( l ) ⇌ H 2 O ( g ) H 2 O ( l ) ⇌ H 2 O ( g )$
60.
Complete the following partial ICE tables.
(a)
$2SO3(g)⇌2SO2(g)+O2(g) change______+x 2SO3(g)⇌2SO2(g)+O2(g) change______+x$
(b)
$4NH3(g)+3O2(g)⇌2N2(g)+6H2O(g) change___+x______ 4NH3(g)+3O2(g)⇌2N2(g)+6H2O(g) change___+x______$
(c)
$2CH4(g)⇌C2H2(g)+3H2(g) change___+x___ 2CH4(g)⇌C2H2(g)+3H2(g) change___+x___$
(d)
$CH4(g)+H2O(g)⇌CO(g)+3H2(g) change___+x______ CH4(g)+H2O(g)⇌CO(g)+3H2(g) change___+x______$
(e)
$NH4Cl(s)⇌NH3(g)+HCl(g) change+x___ NH4Cl(s)⇌NH3(g)+HCl(g) change+x___$
(f)
$Ni(s)+4CO(g)⇌Ni(CO)4(g) change+x___ Ni(s)+4CO(g)⇌Ni(CO)4(g) change+x___$
61.
Complete the following partial ICE tables.
(a)
$2H2(g)+O2(g)⇌2H2O(g) change______+x 2H2(g)+O2(g)⇌2H2O(g) change______+x$
(b)
$CS2(g)+4H2(g)⇌CH4(g)+2H2S(g) change+x_________ CS2(g)+4H2(g)⇌CH4(g)+2H2S(g) change+x_________$
(c)
$H2(g)+Cl2(g)⇌2HCl(g) change+x______ H2(g)+Cl2(g)⇌2HCl(g) change+x______$
(d)
$2NH3(g)+2O2(g)⇌N2O(g)+3H2O(g) change_________+x 2NH3(g)+2O2(g)⇌N2O(g)+3H2O(g) change_________+x$
(e)
$NH4HS(s)⇌NH3(g)+H2S(g) change+x___ NH4HS(s)⇌NH3(g)+H2S(g) change+x___$
(f)
$Fe(s)+5CO(g)⇌Fe(CO)5(g) change___+x Fe(s)+5CO(g)⇌Fe(CO)5(g) change___+x$
62.
Why are there no changes specified for Ni in Exercise 13.60, part (f)? What property of Ni does change?
63.
Why are there no changes specified for NH4HS in Exercise 13.61, part (e)? What property of NH4HS does change?
64.
Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400 °C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.
$N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) K c = 0.50 at 400 ° C N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) K c = 0.50 at 400 ° C$
Calculate the equilibrium molar concentration of NH3.
65.
Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00−L flask at 448 °C.
$H 2 + I 2 ⇌ 2 HI K c = 50.2 at 448 ° C H 2 + I 2 ⇌ 2 HI K c = 50.2 at 448 ° C$
66.
What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?
$Cl 2 ( g ) + Br 2 ( g ) ⇌ 2 BrCl ( g ) K P = 4.7 × 10 −2 Cl 2 ( g ) + Br 2 ( g ) ⇌ 2 BrCl ( g ) K P = 4.7 × 10 −2$
67.
What is the pressure of CO2 in a mixture at equilibrium that contains 0.50 atm H2, 2.0 atm of H2O, and 1.0 atm of CO at 990 °C?
$H 2 ( g ) + CO 2 ( g ) ⇌ H 2 O ( g ) + CO ( g ) K P = 1.6 at 990 °C H 2 ( g ) + CO 2 ( g ) ⇌ H 2 O ( g ) + CO ( g ) K P = 1.6 at 990 °C$
68.
Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
$CoO ( s ) + CO ( g ) ⇌ Co ( s ) + CO 2 ( g ) K c = 4.90 × 10 2 at 550 °C CoO ( s ) + CO ( g ) ⇌ Co ( s ) + CO 2 ( g ) K c = 4.90 × 10 2 at 550 °C$
What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M?
69.
Carbon reacts with water vapor at elevated temperatures.
$C ( s ) + H 2 O ( g ) ⇌ CO ( g ) + H 2 ( g ) K c = 0.2 at 1000 °C C ( s ) + H 2 O ( g ) ⇌ CO ( g ) + H 2 ( g ) K c = 0.2 at 1000 °C$
Assuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with [H2O] = 0.500 M at 1000 °C?
70.
Sodium sulfate 10−hydrate, Na2SO4·10H2O, dehydrates according to the equation
$Na 2 SO 4 · 10 H 2 O ( s ) ⇌ Na 2 SO 4 ( s ) + 10 H 2 O ( g ) K P = 4.08 × 10 −25 at 25 °C Na 2 SO 4 · 10 H 2 O ( s ) ⇌ Na 2 SO 4 ( s ) + 10 H 2 O ( g ) K P = 4.08 × 10 −25 at 25 °C$
What is the pressure of water vapor at equilibrium with a mixture of Na2SO4·10H2O and NaSO4?
71.
Calcium chloride 6−hydrate, CaCl2·6H2O, dehydrates according to the equation
$CaCl 2 · 6 H 2 O ( s ) ⇌ CaCl 2 ( s ) + 6 H 2 O ( g ) K P = 5.09 × 10 −44 at 25 °C CaCl 2 · 6 H 2 O ( s ) ⇌ CaCl 2 ( s ) + 6 H 2 O ( g ) K P = 5.09 × 10 −44 at 25 °C$
What is the pressure of water vapor at equilibrium with a mixture of CaCl2·6H2O and CaCl2 at 25 °C?
72.
A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
$2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) K c = 4.32 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) K c = 4.32$
What are the equilibrium concentrations of all species in a mixture that was prepared with [SO3] = 0.500 M, [SO2] = 0 M, and [O2] = 0.350 M?
73.
A student solved the following problem and found [N2O4] = 0.16 M at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N2O4 in a mixture formed from a sample of NO2 with a concentration of 0.10 M?
$2 NO 2 ( g ) ⇌ N 2 O 4 ( g ) K c = 160 2 NO 2 ( g ) ⇌ N 2 O 4 ( g ) K c = 160$
74.
Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.
$N2O4(g)⇌2NO2(g)Kc=1.07×10−5N2O4(g)⇌2NO2(g)Kc=1.07×10−5$ in chloroform
(b) Confirm that the change is small enough to be neglected.
75.
Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M.
$COCl 2 ( g ) ⇌ CO ( g ) + Cl 2 ( g ) K c = 2.2 × 10 −10 COCl 2 ( g ) ⇌ CO ( g ) + Cl 2 ( g ) K c = 2.2 × 10 −10$
(b) Confirm that the change is small enough to be neglected.
76.
Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
$2 H 2 S ( g ) ⇌ 2 H 2 ( g ) + S 2 ( g ) K P = 2.2 × 10 −6 2 H 2 S ( g ) ⇌ 2 H 2 ( g ) + S 2 ( g ) K P = 2.2 × 10 −6$
(b) Confirm that the change is small enough to be neglected.
77.
What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl2O] = 1.00 M comes to equilibrium at 25 °C?
$H 2 O ( g ) + Cl 2 O ( g ) ⇌ 2 HOCl ( g ) K c = 0.0900 H 2 O ( g ) + Cl 2 O ( g ) ⇌ 2 HOCl ( g ) K c = 0.0900$
78.
Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.
$H 2 + I 2 ⇌ 2 HI K c = 50.2 at 448 °C H 2 + I 2 ⇌ 2 HI K c = 50.2 at 448 °C$
79.
Butane exists as two isomers, n−butane and isobutane.
KP = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
80.
What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.50 for the decomposition reaction of CaCO3 at that temperature?
$CaCO 3 ( s ) ⇌ CaO ( s ) + CO 2 ( g ) CaCO 3 ( s ) ⇌ CaO ( s ) + CO 2 ( g )$
81.
The equilibrium constant (Kc) for this reaction is 1.60 at 990 °C:
$H 2 ( g ) + CO 2 ( g ) ⇌ H 2 O ( g ) + CO ( g ) H 2 ( g ) + CO 2 ( g ) ⇌ H 2 O ( g ) + CO ( g )$
Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00-L container at 990 °C.
82.
In a 3.0-L vessel, the following equilibrium partial pressures are measured: N2, 190 torr; H2, 317 torr; NH3, 1.00 $××$ 103 torr.
$N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g )$
(a) How will the partial pressures of H2, N2, and NH3 change if H2 is removed from the system? Will they increase, decrease, or remain the same?
(b) Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.
83.
The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.
$CO ( g ) + H 2 O ( g ) ⇌ CO 2 ( g ) + H 2 ( g ) CO ( g ) + H 2 O ( g ) ⇌ CO 2 ( g ) + H 2 ( g )$
(a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture?
(b) Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.
84.
Antimony pentachloride decomposes according to this equation:
$SbCl 5 ( g ) ⇌ SbCl 3 ( g ) + Cl 2 ( g ) SbCl 5 ( g ) ⇌ SbCl 3 ( g ) + Cl 2 ( g )$
An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?
85.
Consider the equilibrium
$4 NO 2 ( g ) + 6 H 2 O ( g ) ⇌ 4 NH 3 ( g ) + 7 O 2 ( g ) 4 NO 2 ( g ) + 6 H 2 O ( g ) ⇌ 4 NH 3 ( g ) + 7 O 2 ( g )$
(a) What is the expression for the equilibrium constant (Kc) of the reaction?
(b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
(c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of NO2?
(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?
86.
The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as
$HbO 2 ( a q ) + H 3 O + ( a q ) + CO 2 ( g ) ⇌ CO 2 − Hb − H + + O 2 ( g ) + H 2 O ( l ) HbO 2 ( a q ) + H 3 O + ( a q ) + CO 2 ( g ) ⇌ CO 2 − Hb − H + + O 2 ( g ) + H 2 O ( l )$
(a) Write the equilibrium constant expression for this reaction.
(b) Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle.
87.
Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g).
88.
A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N2, 1.00 M; H2, 0.50 M; and NH3, 0.25 M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M? The equilibrium reaction is
$N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g )$
89.
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
(a) $I2(s)+Cl2(g)⟶2ICl(g)ΔG°=−10.88 kJI2(s)+Cl2(g)⟶2ICl(g)ΔG°=−10.88 kJ$
(b) $H2(g)+I2(s)⟶2HI(g)ΔG°=3.4 kJH2(g)+I2(s)⟶2HI(g)ΔG°=3.4 kJ$
(c) $CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)ΔG°=−39 kJCS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)ΔG°=−39 kJ$
(d) $2SO2(g)+O2(g)⟶2SO3(g)ΔG°=−141.82 kJ2SO2(g)+O2(g)⟶2SO3(g)ΔG°=−141.82 kJ$
(e) $CS2(g)⟶CS2(l)ΔG°=−1.88 kJCS2(g)⟶CS2(l)ΔG°=−1.88 kJ$
90.
Calculate the equilibrium constant at the temperature given.
(a) $O2(g)+2F2(g)⟶2F2O(g)(T=100°C)O2(g)+2F2(g)⟶2F2O(g)(T=100°C)$
(b) $I2(s)+Br2(l)⟶2IBr(g)(T=0.0°C)I2(s)+Br2(l)⟶2IBr(g)(T=0.0°C)$
(c) $2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)(T=575°C)2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)(T=575°C)$
(d) $N2O3(g)⟶NO(g)+NO2(g)(T=−10.0°C)N2O3(g)⟶NO(g)+NO2(g)(T=−10.0°C)$
(e) $SnCl4(l)⟶SnCl4(g)(T=200°C)SnCl4(l)⟶SnCl4(g)(T=200°C)$
91.
Calculate the equilibrium constant at the temperature given.
(a) $I2(s)+Cl2(g)⟶2ICl(g)(T=100°C)I2(s)+Cl2(g)⟶2ICl(g)(T=100°C)$
(b) $H2(g)+I2(s)⟶2HI(g)(T=0.0°C)H2(g)+I2(s)⟶2HI(g)(T=0.0°C)$
(c) $CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)(T=125°C)CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)(T=125°C)$
(d) $2SO2(g)+O2(g)⟶2SO3(g)(T=675°C)2SO2(g)+O2(g)⟶2SO3(g)(T=675°C)$
(e) $CS2(g)⟶CS2(l)(T=90°C)CS2(g)⟶CS2(l)(T=90°C)$
92.
Consider the following reaction at 298 K:
$N2O4(g)⇌2NO2(g)KP=0.142N2O4(g)⇌2NO2(g)KP=0.142$
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
93.
Determine the normal boiling point (in kelvin) of dichloroethane, CH2Cl2. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
94.
Under what conditions is $N2O3(g)⟶NO(g)+NO2(g)N2O3(g)⟶NO(g)+NO2(g)$ spontaneous?
95.
At room temperature, the equilibrium constant (Kw) for the self-ionization of water is 1.00 $××$ 10−14. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
96.
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction $2H2S(g)+SO2(g)⇌38S8(s,rhombic)+2H2O(l).2H2S(g)+SO2(g)⇌38S8(s,rhombic)+2H2O(l).$ What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
97.
Consider the decomposition of CaCO3(s) into CaO(s) and CO2(g). What is the equilibrium partial pressure of CO2 at room temperature?
98.
In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH3(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH4Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH3 in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
99.
Benzene can be prepared from acetylene. $3C2H2(g)⇌C6H6(g).3C2H2(g)⇌C6H6(g).$ Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
100.
Carbon dioxide decomposes into CO and O2 at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO2 was 1.15 atm?
101.
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
$CH4(g)+4Cl2(g)⟶CCl4(g)+4HCl(g)CH4(g)+4Cl2(g)⟶CCl4(g)+4HCl(g)$
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
102.
Acetic acid, CH3CO2H, can form a dimer, (CH3CO2H)2, in the gas phase.
$2CH3CO2H(g)⟶(CH3CO2H)2(g)2CH3CO2H(g)⟶(CH3CO2H)2(g)$
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
At 25 °C, the equilibrium constant for the dimerization is 1.3 $××$ 103 (pressure in atm). What is ΔS° for the reaction? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/13%3A_Fundamental_Equilibrium_Concepts/13.09%3A_Exercises.txt |
This chapter will illustrate the chemistry of acid-base reactions and equilibria, and provide you with tools for quantifying the concentrations of acids and bases in solutions.
• 14.1: Introduction
• 14.2: Brønsted-Lowry Acids and Bases
Compounds that donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species.
• 14.3: pH and pOH
The concentration of hydronium ion in a solution of an acid in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H3O+ in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH− can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7 and pOH = 7.
• 14.4: Relative Strengths of Acids and Bases
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are compete successfully with water for possession of protons.
• 14.5: Hydrolysis of Salt
The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.
• 14.6: Polyprotic Acids
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations sequentially.
• 14.7: Buffers
A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).
• 14.8: Acid-Base Titrations
A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.
• 14.9: Key Terms
• 14.10: Key Equations
• 14.11: Summary
• 14.12: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
14: Acid-Base Equilibria
Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, \(\ce{H^{+}}\) and \(\ce{OH^{–}}\), is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases). These \(\ce{H^{+}}\) transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation (Figure \(1\)) to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
• Write equations for acid and base ionization reactions
• Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
• Describe the acid-base behavior of amphiprotic substances
The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be $\ce{CO2}$), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Johannes Brønsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1H.) A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).
The concept of conjugate pairs is useful in describing Brønsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates $\ce{H^{+}}$, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts $\ce{H^{+}}$, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, $\ce{OH^{−}}$, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, $\ce{NH4^{+}}$, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
Base ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, $\ce{C5NH5}$, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:
The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Brønsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:
$\ce{HCO3^{–}(aq) + H2O(l) <=> CO3^{2–}(aq) + H3O^{+} (aq)} \nonumber$
$\ce{HCO3^{-}(aq) + H2O(l) <=> H2CO3(aq) + OH^{-} (aq)} \nonumber$
The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.
In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:
The process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at 25 °C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, $K_w$:
$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{autoionization}$
with
$K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \label{Kw}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, $K_w$ has a value of $1.0 \times 10^{−14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for $K_w$ is about $5.6 \times 10^{−13}$, roughly 50 times larger than the value at 25 °C.
Example $1$: Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3O+] = [OH] = x. At 25 °C:
$K_{ w }=\left[ H_3 O^{+}\right]\left[ OH^{-}\right]=(x)(x)=x^2=1.0 \times 10^{-14} \nonumber$
So:
$x=\left[ H_3 O^{+}\right]=\left[ OH^{-}\right]=\sqrt{1.0 \times 10^{-14}}=1.0 \times 10^{-7} M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, $1.0 \times 10^{−7}\, M$.
Exercise $1$
The ion product of water at 80 °C is 2.4 \times 10^{−13}. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer
$[\ce{H3O^{+}}] = [\ce{OH^{-}}] = 4.9 \times 10^{−7}\, M \nonumber$
Example $2$: The Inverse Relation between [H3O+] and [OH−]
A solution of an acid in water has a hydronium ion concentration of $2.0 \times 10^{−6} M$. What is the concentration of hydroxide ion at 25 °C?
Solution
Use the value of the ion-product constant for water at 25 °C (Equations \ref{autoionization} and \ref{Kw}) to calculate the missing equilibrium concentration.
$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \nonumber$
with $K_w=[\ce{H3O^{+}}][\ce{OH^{-}}] \nonumber$.
Rearrangement of the Kw expression shows that [OH] is inversely proportional to [H3O+]:
$K_{ w }=\left[\ce{H3O^{+}}\right]\left[\ce{OH^{-}}\right]=\left(2.0 \times 10^{-6}\right)\left(5.0 \times 10^{-9}\right)=1.0 \times 10^{-14} \nonumber$
Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Chatelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.
Substituting the ion concentrations into the Kw expression confirms this calculation, resulting in the expected value:
$K_w=[\ce{H3O^{+}}][\ce{OH^{-}}]=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber$
Exercise $2$
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
$\ce{[H3O+] = 1 \times 10^{−11}\, M} \nonumber$
Example $3$: Representing the Acid-Base Behavior of an Amphoteric Substance
Write separate equations representing the reaction of $\ce{HSO3^{-}}$
1. as an acid with $\ce{OH^{-}}$
2. as a base with $\ce{HI}$
Solution
1. $\ce{HSO3^{-}(aq) + OH^{-}(aq) <=> SO3^{2-}(aq)+H2O(l)}$
2. $\ce{HSO3^{-}(aq) + HI(aq) <=> H2SO3(aq) + I^{-}(aq)}$
Exercise $3$
Write separate equations representing the reaction of $\ce{H2PO4^{-}}$
1. as a base with $\ce{HBr}$
2. as an acid with $\ce{OH^{-}}$
Answer
1. $\ce{H2PO4^{-}(aq) + HBr(aq) <=> H3PO4(aq) + Br^{-}(aq)}$
2. $\ce{H2PO4^{-}(aq) + OH^{-}(aq) <=> HPO4^{2-}(aq) + H2O(l)}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.02%3A_Brnsted-Lowry_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the characterization of aqueous solutions as acidic, basic, or neutral
• Express hydronium and hydroxide ion concentrations on the pH and pOH scales
• Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ($K_w$). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “$X$” is the quantity of interest and “log” is the base-10 logarithm:
$pX =-\log X \nonumber$
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
$pH =-\log \left[ H_3 O^{+}\right] \nonumber$
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
$\left[ H_3 O^{+}\right]=10^{- pH } \nonumber$
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
$pOH =-\log \left[ OH^{-}\right] \nonumber$
or
$\left[ OH^{-}\right]=10^{- pOH } \nonumber$
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_w$ expression:
\begin{align} K_{ w } &=\left[ \ce{H3O^{+}} \right]\left[\ce{OH^{-}}\right] \nonumber \[4pt] -\log K_{ w } &=-\log \left(\left[ \ce{H_3O^{+}}\right]\left[\ce{OH^{-}}\right]\right) \nonumber \[4pt] &=-\log \left[ \ce{H3O^{+}}\right] + -\log \left[\ce{OH^{-}}\right] \nonumber \[4pt] pK_{ w } &= pH + pOH \label{pHpOH} \end{align}
At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$, and so:
$14.00= pH + pOH \nonumber$
As was demonstrated previously, the hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times 10^{−7}~\text{M}$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$pH =-\log \left[ H_3 O^{+}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber$
$pOH =-\log \left[ OH^{-}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{−7} ~\text{M}$ and hydroxide ion molarities less than $1.0 \times 10^{−7} ~\text{M}$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{−7} ~\text{M}$ and hydroxide ion molarities greater than $1.0 \times 10^{−7} ~\text{M}$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, Exercise $1$ showed the hydronium molarity of pure water at 80 °C is $4.9 \times 10^{−7} ~\text{M}$, which corresponds to pH and pOH values of:
$pH =-\log \left[ H_3 O^{+}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber$
$pOH =-\log \left[ OH^{-}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber$
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (Table $1$).
Table $1$: Summary of Relations for Acidic, Basic and Neutral Solutions
Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7
Figure $1$: shows the relationships between [H3O+], [OH], pH, and pOH for solutions classified as acidic, basic, and neutral.
Example $1$: Calculation of pH from [H3O+]
What is the pH of stomach acid, a solution of $\ce{HCl}$ with a hydronium ion concentration of $1.2 \times 10^{−3}\, M$?
Solution
\begin{align*} pH &=-\log \left[\ce{H3O^{+}}\right] \[4pt] &=-\log \left(1.2 \times 10^{-3}\right) \[4pt] &=-(-2.92)=2.92 \end{align*} \nonumber
(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)
Exercise $1$
Water exposed to air contains carbonic acid, $\ce{H2CO3}$, due to the reaction between carbon dioxide and water:
$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \nonumber$
Air-saturated water has a hydronium ion concentration caused by the dissolved $\ce{CO2}$ of $2.0 \times 10^{−6} ~\text{M}$, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
Answer
5.70
Example $2$: Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3.
Solution
\begin{align*} pH =-\log \left[ H_3 O^{+}\right] &=7.3 \[4pt] \log \left[ H_3O^{+}\right] &=-7.3 \[4pt] \left[ H_3 O^{+}\right] &=10^{-7.3} \end{align*} \nonumber
or
\begin{align*} [ \ce{H3O^{+}} ] &= \text {antilog of } -7.3 \[4pt] &=5 \times 10^{-8} M \end{align*} \nonumber
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate $10^{−7.3}$.)
Exercise $2$
Calculate the hydronium ion concentration of a solution with a pH of −1.07.
Answer
12 M
How Sciences Interconnect: Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved $\ce{CO2}$ which forms carbonic acid:
\begin{align*} \ce{H2O(l) + CO2(g) & -> H2CO3(aq)} \[4pt] \ce{H2CO3(aq) &<=> H^{+}(aq) + HCO3^{-}(aq)}\end{align*} \nonumber
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including $\ce{CO2}$, $\ce{SO2}$, $\ce{SO3}$, $\ce{NO}$, and $\ce{NO2}$ being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
\begin{align*} \ce{H2O(l) + SO3(g) &-> H2SO4(aq)} \[4pt] \ce{H2SO4(aq) &-> H^{+}(aq) + HSO_4^{-}(aq)} \end{align*} \nonumber
Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $2$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
For further information on acid rain, visit this website hosted by the US Environmental Protection Agency.
Example $3$: Calculation of pOH
What are the $\text{pOH}$ and the $\text{pH} of a 0.0125-M solution of potassium hydroxide, \(\ce{KOH}$?
Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:
\begin{align*} \text{pOH} & =-\log \left[ OH^{-}\right]=-\log 0.0125 \[4pt] & =-(-1.903)=1.903 \end{align*} \nonumber
The pH can be found from the pOH:
\begin{align*} \text{pH} + \text{pOH} &=14.00 \[4pt] pH &=14.00- pOH \[4pt] &=14.00-1.903=12.10 \end{align*} \nonumber
Exercise $3$
The hydronium ion concentration of vinegar is approximately $4 \times 10^{−3} ~\text{M}$. What are the corresponding values of $\text{pOH}$ and $\text{pH}$?
Answer
pOH = 11.6, pH = 2.4
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure $3$).
The pH of a solution may also be visually estimated using colored indicators (Figure $4$). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.03%3A_pH_and_pOH.txt |
Learning Objectives
By the end of this section, you will be able to:
• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
Acid and Base Ionization Constants
The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure $1$.
The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, $K_a$. For the reaction of an acid $\ce{HA}$:
$\ce{HA(aq) + H2O(l) <=> H3O^{+}(aq) + A^{-}(aq)} \nonumber$
the acid ionization constant is written
$K_{ a }=\frac{[\ce{H3O^{+}}] [\ce{A^{-}}]}{[\ce{HA}]} \nonumber$
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include $[\ce{H2O}]$ in the equation.
Note
Equilibrium constant expressions are actually ratios of activities, and the value of K is determined at the limit of infinite dilution of the solutes. In these very dilute solutions, the activity of the solvent has a value of unity (1) and the activity of each solute can be approximated by its molar concentration.
The larger the Ka of an acid, the larger the concentration of $\ce{H3O^{+}}$ and $\ce{A^{-}}$ relative to the concentration of the nonionized acid, $\ce{HA}$, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large ($K_a ≈ ∞$). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.
To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order:
$\ce{CH3CO2H < HNO2 < HSO4^{−}} \nonumber$
as demonstrated with the equations below:
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad \quad Ka=1.8 \times 10^{−5} \nonumber$
$\ce{HNO2(aq) + H2O(l) <=> H3O^{+}(aq) + NO2^{-}(aq)} \quad \quad K_a=4.6 \times 10^{-4} \nonumber$
$\ce{HSO4^{-}(aq) + H2O(aq) <=> H3O^{+}(aq) + SO4^{2-}(aq) } \quad\quad K_a=1.2 \times 10^{-2} \nonumber$
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:
$\% \text { ionization }=\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{[ \ce{HA} ]_0} \times 100 \nonumber$
where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A] = [H3O+]). Unlike the Ka value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.
Example $1$: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
$\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{\left[ \ce{HNO2} \right]_0} \times 100 \nonumber$
Converting the provided pH to hydronium ion molarity yields
$\left[ \ce{H3O^{+}} \right]=10^{-2.09}=0.0081 M \nonumber$
Substituting this value and the provided initial acid concentration into the percent ionization equation gives
$\frac{8.1 \times 10^{-3}}{0.125} \times 100=6.5 \% \nonumber$
(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)
Exercise $1$
Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.
Answer
1.3% ionized
View the simulation of strong and weak acids and bases at the molecular level.
Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, $\ce{B}$:
$\ce{B(aq) + H2O(l) <=> HB^{+}(aq) + OH^{-}(aq)}, \nonumber$
the ionization constant is written as
$K_{ b }=\frac{\left[ \ce{HB^{+}} \right]\left[ \ce{OH^{-}} \right]}{[ \ce{B} ]} \nonumber$
Inspection of the data for three weak bases presented below shows the base strength increases in the order: $\ce{NO2^{−} < CH2CO2^{−} < NH3}$
\begin{align*} \ce{NO_2^{-}(aq) + H2O(l) & <=> HNO_2(aq) + OH^{-}(aq)} & K_{ b }=2.17 \times 10^{-11} \[4pt] \ce{CH_3 CO_2^{-}(aq) + H2O(l) & <=> CH_3 CO_2 H(aq) + OH^{-}(aq)} & K_{ b }=5.6 \times 10^{-10} \[4pt] \ce{NH_3(aq) + H2O(l) & <=> NH_4^{+}(aq) + OH^{-}(aq)} & K_{ b }=1.8 \times 10^{-5} \end{align*} \nonumber
A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as
$\% \text { ionization }= \left( \dfrac{[ \ce{OH^{-}}]_{eq}}{[ \ce{B} ]_0} \right) \times 100 \% \nonumber$
but will vary depending on the base ionization constant and the initial concentration of the solution.
Relative Strengths of Conjugate Acid-Base Pairs
Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, $K_a$ or $K_b$, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair $\ce{HA / A^{-}}$, ionization equilibrium equations and ionization constant expressions are
$\ce{HA(aq) + H2O(l) <=> H3O^{+}(aq) + A^{-}(aq)} \nonumber$
with
$K_a=\ce{\dfrac{[H3O^{+}][A^{−}]}{[HA]}} \nonumber$
and
$\ce{A^{−}(aq) + H2O(l) <=> OH^{-}(aq) + HA(aq)} \nonumber$
with
$K_b=\ce{\dfrac{[HA][OH–]}{[A−]}} \nonumber$
Adding these two chemical equations yields the equation for the autoionization for water:
\begin{align*} \ce{\cancel{HA(aq)} + H2O(l) + \cancel{A^{−}(aq)} + H2O(l) &<=> H3O^{+}(aq) + \cancel{A^{-}(aq)} + OH^{-}(aq) + \cancel{HA(aq)}} \[4pt] \ce{2H2O(l) &<=> H3O+(aq) + OH^{−}(aq)} \end{align*} \nonumber
As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so
$K_a \times K_b= \dfrac{[\ce{H3O^{+}}][\ce{A^{−}}]}{[\ce{HA}]} × \dfrac{[\ce{HA}][\ce{OH^{-}}]}{[\ce{A^{-}}]}=[\ce{H3O^{+}}][\ce{OH^{-}}]=K_w \nonumber$
This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, $K_w$. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:
$K_{ a }= \dfrac{K_{ w }}{ K_{ b }} \nonumber$
or
$K_{ b }=\dfrac{K_{ w }}{K_{ a }} \nonumber$
The inverse proportional relation between $K_a$ and $K_b$ means the stronger the acid or base, the weaker its conjugate partner. Figure $2$: illustrates this relation for several conjugate acid-base pairs.
The listing of conjugate acid–base pairs shown in Figure $3$: is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.
If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H3O+(aq), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for “strong” acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be “weak,” and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl < HBr < HI).
The right column of Figure $3$ lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don’t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large Ka, and so its conjugate base will exhibit a Kb that is essentially zero:
• strong acid: $\quad K_{ a } \approx \infty$
• conjugate base: $K_{ b }=K_{ w } / K_{ a }=K_{ w } / \infty \approx 0$
A similar approach can be used to support the observation that conjugate acids of strong bases (Kb ≈ ∞) are of negligible strength (Ka ≈ 0).
Example $2$: Calculating Ionization Constants for Conjugate Acid-Base Pairs
Use the Kb for the nitrite ion, $\ce{NO2^{-}}$, to calculate the Ka for its conjugate acid.
Solution
$K_b$ for $\ce{NO2^{-}}$ is given in this section as $2.17 \times 10^{−11}$. The conjugate acid of $\ce{NO2^{-}}$ is $\ce{HNO2}$; $K_a$ for $\ce{HNO2}$ can be calculated using the relationship:
$K_{ a } \times K_{ b }=1.0 \times 10^{-14}=K_{ w } \nonumber$
Solving for Ka yields
$K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{2.17 \times 10^{-11}}=4.6 \times 10^{-4} \nonumber$
This answer can be verified by finding the Ka for $\ce{HNO2}$ in Appendix H.
Exercise $2$
Determine the relative acid strengths of $\ce{ NH 4^{ +}}$ and $\ce{HCN}$ by comparing their ionization constants. The ionization constant of $\ce{HCN}$ is given in Appendix H as $4.9 \times 10^{−10}$. The ionization constant of $\ce{ NH 4^{ +}}$ is not listed, but the ionization constant of its conjugate base, $\ce{NH3}$, is listed as $1.8 \times 10^{−5}$.
Answer
$\ce{NH4^{+}}$ is the slightly stronger acid ($K_a$ for $\ce{NH4^{+}}$ is $5.6 \times 10^{−10}$).
Acid-Base Equilibrium Calculations
The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.
Example $3$: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar (Figure $4$) that provides its sour taste. At equilibrium, a solution contains $\ce{[CH3CO2H]} = 0.0787\, M$: and $\ce{H3O^+}]=[\ce{CH3CO2^{-}}]=0.00118\,M$. What is the value of $K_a$ for acetic acid?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Ka for acetic acid.
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \nonumber$
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{CH3CO2^{-}} \right]}{\left[ \ce{CH3CO2H} \right]}=\frac{(0.00118)(0.00118)}{0.0787}=1.77 \times 10^{-5} \nonumber$
Exercise $3$
The $\ce{HSO4^{−}}$ ion, weak acid used in some household cleansers:
$\ce{HSO4^{-}(aq) + H2O(l) <=> H3O^{+}(aq) + SO4^{2-}(aq)} \nonumber$
What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $[\ce{H3O^{+}}] = 0.027\, M$; $[\ce{HSO4^{-}}]=0.29\,M$; $[\ce{HSO4^{-}}] =0.29\,M$; and $[\ce{SO4^{2-}}]=0.13\,M$?
Answer
Ka for $\ce{HSO4^{-}} = 1.2 \times 10^{−2}$
Example $4$: Determination of Kb from Equilibrium Concentrations
Caffeine, $\ce{C8H10N4O2}$ is a weak base. What is the value of $K_b$ for caffeine if a solution at equilibrium has $[\ce{C8H10N4O2}] = 0.050~\text{M}$, $[\ce{C8H10N4O2H^{+}}] = 5.0 \times 10^{−3} ~\text{M}$, and $[\ce{OH^{-}}] = 2.5 \times 10^{−3} ~\text{M}$?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_b$ for caffeine.
$\ce{C8H10N4O2(aq) + H2O(l)⇌C8H10N4O2H^{+}(aq) + OH^{−}(aq)} \nonumber$
\begin{align*} K_b &=\ce{[C8H10N4O2H^{+}][OH^{-}][C8H10N4O2]} \[4pt] &=(5.0 \times 10^{-3})(2.5 \times 10^{-3}) (0.050) \[4pt]&=2.5 \times 10^{-4} \end{align*} \nonumber
Exercise $4$
What is the equilibrium constant for the ionization of the $\ce{HPO4^{2−}}$ ion, a weak base
$\ce{HPO4^{2-}(aq) + H2O(l) <=> H2PO4^{−}(aq) + OH^{-}(aq)} \nonumber$
if the composition of an equilibrium mixture is as follows: $\ce{[OH^{-}}] = 1.3 \times 10^{−6} M}$; $\ce{H2PO4^{-}}]=0.042\,M$; and $[\ce{HPO4^{2-}}]=0.341\,M$?
Answer
$K_b$ for $\ce{HPO4^{2-}}$ is $1.6 \times 10^{−7}$
Example $5$: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, $\ce{HNO2}$, is 2.34. What is its $K_a$?
$\ce{HNO2(aq) + H2O(l) <=> H3O^{+}(aq) + NO2^{−}(aq)} \nonumber$
Solution
The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H3O+ is present ($1 × \times 10^{−7}~ \text{M}$) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.
The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:
$\left[ \ce{H3O^{+}} \right]=10^{-2.34}=0.0046 ~\text{M} \nonumber$
The ICE table for this system is then
Finally, calculate the value of the equilibrium constant using the data in the table:
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{NO2^{-}} \right]}{\left[ \ce{HNO2} \right]}=\frac{(0.0046)(0.0046)}{(0.0470)}=4.6 \times 10^{-4} \nonumber$
Exercise $5$
The pH of a solution of household ammonia, a 0.950-M solution of $\ce{NH3}$, is 11.612. What is $K_b$ for $\ce{NH3}$.
Answer
$K_b = 1.8 \times 10^{−5}$
Example $6$: Calculating Equilibrium Concentrations in a Weak Acid Solution
Formic acid, HCO2H, is one irritant that causes the body’s reaction to some ant bites and stings (Figure $5$).
What is the concentration of hydronium ion and the pH of a 0.534-M solution of formic acid?
$\ce{HCO2H(aq) + H2O(l) <=> H3O^{+}(aq) + HCO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-4} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Ka expression gives
\begin{align*} K_{ a }&=1.8 \times 10^{-4}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{HCO2^{-}} \right]}{\left[ \ce{HCO2H} \right]}\[4pt] &=\frac{(x)(x)}{0.534-x}=1.8 \times 10^{-4} \end{align*} \nonumber
The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes
$K_{ a }=1.8 \times 10^{-4}=\frac{x^2}{0.534} \nonumber$
Solving the equation for $x$ yields
\begin{align*} x^2 &=0.534 \times\left(1.8 \times 10^{-4}\right)=9.6 \times 10^{-5}\[4pt] x&=\sqrt{9.6 \times 10^{-5}}\[4pt] &=9.8 \times 10^{-3} M \end{align*} \nonumber
To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:
$\frac{x}{0.534}=\frac{9.8 \times 10^{-3}}{0.534}=1.8 \times 10^{-2}(1.8 \% \text { of } 0.534) \nonumber$
Because x is less than 5% of the initial concentration, the assumption is valid.
As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:
$x=\left[ \ce{H3O^{+}} \right]=0.0098 ~\text{M} \nonumber$
Finally, the pH is calculated to be
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0098)=2.01 \nonumber$
Exercise $6$
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic acid, CH3CO2H?
$\ce{CH3CO2H(aq) + H2O(l) <=> H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-5} \nonumber$
Answer
percent ionization = 1.3%
Example $7$: Calculating Equilibrium Concentrations in a Weak Base Solution
Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25-M solution of trimethylamine, a weak base:
$\ce{(CH3)3N(aq) + H2O(l) <=> (CH3)3NH^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=6.3 \times 10^{-5} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Kb expression gives
$K_{ b }=\frac{\left[ \ce{(CH3)3NH^{+}} \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{(CH3)3N} \right]}=\frac{(x)(x)}{0.25-x}=6.3 \times 10^{-5} \nonumber$
Assuming $x \ll 0.25$ and solving for $x$ yields
$x=4.0 \times 10^{-3}~\text{M}\nonumber$
This value is less than 5% of the initial concentration (0.25), so the assumption is justified. As defined in the ICE table, $x$ is equal to the equilibrium concentration of hydroxide ion:
\begin{align*} \left[ OH^{-}\right] &=\sim 0+x \[4pt] &=x=4.0 \times 10^{-3} M\[4pt] \end{align*} \nonumber
The pOH is calculated to be
$\text{pOH} =-\log \left(4.0 \times 10^{-3}\right)=2.40 \nonumber$
Using the relation introduced in the previous section of this chapter:
$\text{pH} + \text{pOH} = p K_{ w }=14.00 \nonumber$
permits the computation of pH:
$\text{pH} = 14.00 - \text{pOH} =14.00-2.40=11.60 \nonumber$
Exercise $7$
Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of $1.76 \times 10^{−5}$.
Answer
$7.56 \times 10^{−4}~ \text{M}$, 2.33%
In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible cannot be made. Calculations of this sort are demonstrated in Example $8$ below.
Example $8$: Calculating Equilibrium Concentrations without Simplifying Assumptions
Sodium bisulfate, NaHSO4, is used in some household cleansers as a source of the $\ce{HSO4^{−}}$ ion, a weak acid. What is the pH of a 0.50-M solution of $\ce{HSO4^{-}}$?
$\ce{HSO4^{-}(aq) + H2O(l) <=> H3O^{+}(aq) + SO4^{2-}(aq)} \quad K_{ a }=1.2 \times 10^{-2} \nonumber$
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the Ka expression gives
$K_{ a }=1.2 \times 10^{-2}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{SO4^{2-}} \right]}{\left[ \ce{HSO4^{-}} \right]}=\frac{(x)(x)}{0.50-x} \nonumber$
If the assumption that $x \ll 0.5$ is made, simplifying and solving the above equation yields
$x=0.077~\text{M} \nonumber$
This value of $x$ is clearly not significantly less than 0.50 M; rather, it is approximately 15% of the initial concentration:
When we check the assumption, we calculate:
$\frac{x}{\left[ \ce{HSO4^{-}} \right]_{ i }} \nonumber$
$\frac{x}{0.50}=\frac{7.7 \times 10^{-2}}{0.50}=0.15(15 \%) \nonumber$
Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:
$K_{ a }=1.2 \times 10^{-2}=\frac{(x)(x)}{0.50-x} \nonumber$
Rearranging this equation yields
$6.0 \times 10^{-3}-1.2 \times 10^{-2} x=x^2 \nonumber$
Writing the equation in quadratic form gives
$x^2+1.2 \times 10^{-2} x-6.0 \times 10^{-3}=0 \nonumber$
Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x. As defined in the ICE table, $x$ is equal to the hydronium concentration.
$x=\left[ H3O^{+}\right]=0.072~\text{M} \nonumber$
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.072)=1.14 \nonumber$
Exercise $8$
Calculate the pH in a 0.010-M solution of caffeine, a weak base:
$\ce{C8H{10}N4O2(aq) + H2O(l) <=> C8H{10}N4O2H^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=2.5 \times 10^{-4} \nonumber$
Answer
pH = 11.16
Effect of Molecular Structure on Acid-Base Strength
Binary Acids and Bases
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is
$\ce{HF < HCl < HBr < HI}. \nonumber$
Likewise, for group 16, the order of increasing acid strength is
$\ce{H2O < H2S < H2Se < H2Te.} \nonumber$
Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\ce{CH4 < NH3 < H2O < HF}$; across the third row, it is $\ce{SiH4 < PH3 < H2S < HCl}$ (see Figure $6$).
Ternary Acids and Bases
Ternary compounds composed of hydrogen, oxygen, and some third element (“E”) may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula OmE(OH)n. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure $7$).
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\ce{Al(H2O)3(OH)3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\ce{Al(H2O)3(OH)3}$, is converted into the soluble ion, $\ce{[Al(H2O)2(OH)4]^{-}}$, by reaction with hydroxide ion:
$\ce{Al(H_2O)_3(OH)3(aq) + OH^{-}(aq) <=> H2O(l) + [Al(H2O)2(OH)4]^{-}(aq)} \nonumber$
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The $\ce{Al(H2O)3(OH)3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\ce{[Al(H2O)6]^{3+}}$ by reaction with hydronium ion:
$\ce{3 H3O^{+}(aq) + Al(H2O)3(OH)3(aq) <=> Al(H2O)6^{3+}(aq) + 3 H2O(l)} \nonumber$
In this case, protons are transferred from hydronium ions in solution to $\ce{Al(H2O)3(OH)3}$, and the compound functions as a base. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.04%3A_Relative_Strengths_of_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Predict whether a salt solution will be acidic, basic, or neutral
• Calculate the concentrations of the various species in a salt solution
• Describe the acid ionization of hydrated metal ions
Salts with Acidic Ions
Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation
$\ce{NH4Cl(s) <=> NH4^{+}(aq) + Cl^{-}(aq)} \nonumber$
The ammonium ion is the conjugate acid of the base ammonia, NH3; its acid ionization (or acid hydrolysis) reaction is represented by
$\ce{NH4^{+}(aq) + H2O(l) <=> H3O^{+}(aq) + NH3(aq)} \quad K_{ a }=K_{ w } / K_{ b } \nonumber$
Since ammonia is a weak base, Kb is measurable and Ka > 0 (ammonium ion is a weak acid).
The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by
$\ce{Cl^{-}(aq) + H2O(l) <=> HCl(aq) + OH^{-}(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber$
Since $\ce{HCl}$ is a strong acid, Ka is immeasurably large and Kb ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).
Thus, dissolving ammonium chloride in water yields a solution of weak acid cations () and inert anions (Cl), resulting in an acidic solution.
Example $1$: Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride
$\ce{C6H5NH3^{+}(aq) + H2O(l) <=> H3O^{+}(aq) + C6H5NH2(aq)} \nonumber$
Solution
The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline (see Appendix H):
$K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-10}}=2.3 \times 10^{-5} \nonumber$
Using the provided information, an ICE table for this system is prepared:
Substituting these equilibrium concentration terms into the Ka expression gives
\begin{align*} K_a & =\dfrac{\left[ \ce{C6H5NH2} \right]\left[ \ce{H3O^{+}} \right] }{\left[ \ce{C6H5NH3^{+}}\right]} \[4pt] 2.3 \times 10^{-5} & = \dfrac{(x)(x)}{0.233-x} \end{align*} \nonumber
Assuming $x \ll 0.233$, the equation is simplified and solved for $x$:
\begin{align*} & 2.3 \times 10^{-5}= \dfrac{x^2}{0.233} \[4pt] & x=0.0023 ~\text{M} \end{align*} \nonumber
The ICE table defines $x$ as the hydronium ion molarity, and so the pH is computed as
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0023)=2.64 \nonumber$
Exercise $1$
What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, $\ce{NH4NO3}$, a salt composed of the ions $NH_4^{+}$ and $NO_3^{-}$ and.
Which is the stronger acid $\ce{C6H5NH3+}$ or $\ce{NH4+}$?
Answer
$[\ce{H3O^{+}}] = 7.5 \times 10^{−6} ~\text{M}$; is the stronger acid.
Salts with Basic Ions
As another example, consider dissolving sodium acetate in water:
$\ce{NaCH_3CO_2(s) <=> Na^{+}(aq) + CH_3CO_2^{-}(aq)} \nonumber$
The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.
The acetate ion, $\ce{CH3CO2^{-}(aq)}$, is the conjugate base of acetic acid, $\ce{CH3CO2H}$, and so its base ionization (or base hydrolysis) reaction is represented by
$\ce{CH3CO2^{-}(aq) + H2O(l) <=> CH_3 CO_2 H (aq) + OH -(aq)} \quad K_{ b }=K_{ w } / K_{ a } \nonumber$
Because acetic acid is a weak acid, its Ka is measurable and Kb > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions $( \ce{CH3CO2^{-}})$, resulting in a basic solution.
Example $2$: Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with and $[\ce{OH^{-}}] = 0.050~\text{M}$ and $[\ce{OH^{-}}] = 2.5 \times 10^{−6} M$ at equilibrium. The reaction is:
$\ce{CH_3CO_2^{-}(aq) + H2O(l) <=> CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber$
Solution
The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which
$\ce{CH_3CO_2^{-}(aq) + H2O(l) <=> CH_3CO_2H(aq) + OH^{-}(aq)} \nonumber$
Substituting the available values into the Kb expression gives
\begin{align*} K_{ b } &=\frac{\left[ CH_3 CO_2 H \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{CH3CO2^{-}} \right]}=5.6 \times 10^{-10} \[4pt] &=\frac{\left[ \ce{CH3CO2H} \right]\left(2.5 \times 10^{-6}\right)}{(0.050)}=5.6 \times 10^{-10} \end{align*} \nonumber
Solving the above equation for the acetic acid molarity yields $[\ce{CH3CO2H}] = 1.1 \times 10^{−5}~ \text{M}$.
Exercise $2$
What is the pH of a 0.083-M solution of $\ce{NaCN}$?
Answer
11.11
Salts with Acidic and Basic Ions
Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.
Example $3$: Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. KBr
2. NaHCO3
3. Na2HPO4
4. NH4F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
(a) The $\ce{K^{+}}$ cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.
(b) The $\ce{Na^{+}}$ cation is inert and will not affect the pH of the solution; while the $\ce{HCO3^{−}}$ anion is amphiprotic. The $K_a$ of $\ce{HCO3^{-}}$ is $4.7 \times 10^{−11}$, and its $K_b$ is:
$\frac{1.0 \times 10^{-14}}{4.3 \times 10^{-7}}=2.3 \times 10^{-8} \nonumber$
Since $K_b \gg K_a$, the solution is basic.
(c) The $\ce{Na^{+}}$ cation is inert and will not affect the pH of the solution, while the $\ce{HPO4^{2−}}$ anion is amphiprotic. The $K_a$ of $\ce{HPO4^{2−}}$ is $4.2 \times 10^{−13}$, and its $K_b$ is:
$\frac{1.0 \times 10^{-14}}{6.2 \times 10^{-8}}=1.6 \times 10^{-7} \nonumber$
Because $K_b \gg K_a$, the solution is basic.
(d) The $\ce{NH4^{+}}$ ion is acidic (see above discussion) and the $\ce{F^{−}}$ ion is basic (conjugate base of the weak acid $\ce{HF}$). Comparing the two ionization constants: $K_a$ of $\ce{NH4^{+}}$ is 5.6 × 10−10 and the $K_b$ of $\ce{F^{−}}$ is $1.6 \times 10^{−11}$, so the solution is acidic, since $K_a > K_b$.
Exercise $3$
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
1. K2CO3
2. CaCl2
3. KH2PO4
4. (NH4)2CO3
Answer
(a) basic; (b) neutral; (c) acidic; (d) basic
The Ionization of Hydrated Metal Ions
Unlike the group 1 and 2 metal ions of the preceding examples (Na+, Ca2+, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as
$\ce{Al(NO3)3(s) <=> Al^{3+}(aq) +3 NO3^{-}(aq)} \nonumber$
However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is
$\ce{Al(NO3)3(s) +6 H2O(l) <=> Al(H2O)6^{3+}(aq) +3 NO_3^{-}(aq)} \nonumber$
As shown in Figure $1$, the ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:
$\ce{Al(H2O)6^{3+}(aq) + H2O(l) <=> H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad \quad K_{ a }=1.4 \times 10^{-5} \nonumber$
The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:
\begin{align*} \ce{Al(H2O)_6^{3+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_5(OH)^{2+}(aq)} \[4pt] \ce{Al(H2O)_5(OH)^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_4(OH)_2^{+}(aq)} \[4pt] \ce{Al(H2O)_4(OH)_2^{+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Al(H2O)_3(OH)_3(aq)} \end{align*} \nonumber
This is an example of a polyprotic acid, the topic of discussion in a later section of this chapter.
Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:
\begin{align*} \ce{Fe(H2O)_6^{3+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Fe(H2O)_5(OH)^{2+}(aq)} && p K_{ a }=2.74 \[4pt] \ce{Cu(H2O)_6^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Cu(H2O)_5(OH)^{+}(aq)} && p K_{ a }=\sim 6.3 \[4pt] \ce{Zn(H2O)_4^{2+}(aq) + H2O(l) &<=> H3O^{+}(aq) + Zn(H2O)_3(OH)^{+}(aq)} && p K_{ a }=9.6 \end{align*} \nonumber
Example $4$: Hydrolysis of [Al(H2O)6]3+
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion ($\ce{[Al(H2O)6]^{3+}}$) in solution.
Solution
The equation for the reaction and Ka are:
$\ce{Al(H2O)6^{3+}(aq) + H2O(l) <=> H3O^{+}(aq) + Al(H2O)5(OH)^{2+}(aq)} \quad K_{ a }=1.4 \times 10^{-5} \nonumber$
An ICE table with the provided information is
Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:
\begin{align*} K_{ a } &=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{Al(H2O)5(OH)^{2+}} \right]}{\left[ \ce{Al(H2O)6^{3+}} \right]}\[4pt] &=\frac{(x)(x)}{0.10-x}=1.4 \times 10^{-5} \end{align*} \nonumber
Assuming $x \ll 0.10$ and solving the simplified equation gives:
$x=1.2 \times 10^{-3} ~\text{M} \nonumber$
The ICE table defined $x$ as equal to the hydronium ion concentration, and so the pH is calculated to be
$[\ce{H3O^{+}}]=0+x=1.2 \times 10^{-3}~\text{M} \nonumber$
$\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=2.92 \nonumber$
Therefore this is an acidic solution.
Exercise $4$
What is $[Al(H2O)5(OH)^{2+}}]$ in a 0.15-M solution of $\ce{Al(NO3)3}$ that contains enough of the strong acid $\ce{HNO3}$ to bring [$\ce{H3O^{+}}$] to 0.10 M?
Answer
$2.1 \times 10^{−5} M$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.05%3A_Hydrolysis_of_Salt.txt |
Learning Objectives
By the end of this section, you will be able to:
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
Acids are classified by the number of protons per molecule that they can give up in a reaction. Acids such as $\ce{HCl}$, $\ce{HNO3}$, and $\ce{HCN}$ that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:
\begin{align*} \ce{HCl (aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + Cl^{-}(aq) } \[4pt] \ce{HNO_3(aq) + H_2 O (l)} & \ce{\longrightarrow H3O^{+}(aq) + NO_3^{-}(aq) } \[4pt] \ce{HCN (aq) + H_2 O (l)} & \ce{\rightleftharpoons H3O^{+}(aq) + CN^{-}(aq) } \end{align*} \nonumber
Even though it contains four hydrogen atoms, acetic acid, $\ce{CH3CO2H}$, is also monoprotic because only the hydrogen atom from the carboxyl group ($\ce{-COOH}$) reacts with bases:
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
$\tag{ First ionization} \ce{H_2 SO_4 (aq) + H2O (l) \rightleftharpoons H_3O^{+}(aq) + HSO_4^{-}(aq)}$
with $K_{ a 1} > 10^2$.
$\tag{ Second ionization} \ce{HSO_4^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + SO_4^{2-}(aq) }$
with $K_{ a 2}=1.2 \times 10^{-2}$.
This stepwise ionization process occurs for all polyprotic acids. Carbonic acid, $\ce{H2CO3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.
$\tag{First ionization} \ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H_3O^{+}(aq) + HCO_3^{-}(aq) }$
with
$K_{ H_2 CO_3}=\frac{\left[\ce{H3O^{+}} \right]\left[ \ce{HCO_3^{-}} \right]}{\left[ \ce{H_2CO_3} \right]}=4.3 \times 10^{-7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
$\tag{Second ionization} \ce{ HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq) }$
with
$K_{\ce{HCO_3^{-}}}=\frac{\left[ \ce{H_3O^{+}}\right]\left[ \ce{CO_3^{2-}}\right]}{\left[ \ce{HCO_3^{-}}\right]}=4.7 \times 10^{-11} \nonumber$
$K_{\ce{H_2CO_3}}$ is larger than $K_{\ce{HCO_3^{-}}}$ by a factor of $10^{4}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in the solution. This means that little of the formed by the ionization of $\ce{H2CO3}$ ionizes to give hydronium ions (and carbonate ions), and the concentrations of $\ce{H3O^{+}}$ and are practically equal in a pure aqueous solution of $\ce{H2CO3}$.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This approach is demonstrated in the following example and exercise.
Example $1$: Ionization of a Diprotic Acid
“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because $\ce{CO2}$ reacts with water to form carbonic acid, $\ce{H2CO3}$. What are and in a saturated solution of $\ce{CO2}$ with an initial $\ce{[H2CO3] = 0.033 \,M}$?
$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$
$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \quad K_{ a 2}=4.7 \times 10^{-11} \nonumber$
Solution
As indicated by the ionization constants, $\ce{H2CO3}$ is a much stronger acid than so the stepwise ionization reactions may be treated separately.
The first ionization reaction is
$\ce{H_2 CO_3(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HCO_3^{-}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \nonumber$
Using provided information, an ICE table for this first step is prepared:
Substituting the equilibrium concentrations into the equilibrium equation gives
$K_{ H_2 CO_3}=\frac{\left[ H3O^{+}\right]\left[ HCO_3^{-}\right]}{\left[ H_2 CO_3\right]}=\frac{(x)(x)}{0.033-x}=4.3 \times 10^{-7} \nonumber$
Assuming $x \ll 0.033$ and solving the simplified equation yields
$x=1.2 \times 10^{-4} \nonumber$
The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:
$\left[ H_2 CO_3\right]=0.033 M \nonumber$
$\left[ H3O^{+}\right]=\left[ HCO_3^{-}\right]=1.2 \times 10^{-4} M \nonumber$
Using the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation:
$\ce{HCO_3^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + CO_3^{2-}(aq)} \nonumber$
$K_{ HCO_{3^{-}}}=\frac{\left[ H3O^{+}\right]\left[ CO_3^{2-}\right]}{\left[ HCO_3^{-}\right]}=\frac{\left(1.2 \times 10^{-4}\right)\left[ CO_3^{2-}\right]}{1.2 \times 10^{-4}} \nonumber$
$\left[ CO_3^{2-}\right]=\frac{\left(4.7 \times 10^{-11}\right)\left(1.2 \times 10^{-4}\right)}{1.2 \times 10^{-4}}=4.7 \times 10^{-11} M \nonumber$
To summarize: at equilibrium $\ce{[H2CO3]} = 0.033 M$; $\ce{[H3O^{+}]} = 1.2 \times 10^{−4}$; $\ce{[HCO3^{-}]}=1.2 \times 10^{−4}\,M$;
$\ce{[CO32^{−}]}=4.7 \times 10^{−11}\,M$.
Exercise $1$
The concentration of $\ce{H2S}$ in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate, $\ce{H3O^{+}}$, $\ce{[HS^{−}]}$, and $\ce{[S^{2−}]}$ in the solution:
$\ce{H_2 S (aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + HS^{-}(aq)} \quad K_{ a 1}=8.9 \times 10^{-8} \nonumber$
$\ce{HS^{-}(aq) + H_2 O (l) \rightleftharpoons H3O^{+}(aq) + S^{2-}(aq)} \quad K_{ a 2}=1.0 \times 10^{-19} \nonumber$
Answer
$[\ce{H2S}] = 0.1\, M$
$[\ce{H3O^{+}}] = [\ce{HS^{-}}] = 0.000094\, M$
$[\ce{S^{2-}}] = 1 \times 10^{−19}\, M$
A triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:
$\tag{First ionization} \ce{H3PO_4(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + H_2 PO_4^{-}(aq)} \quad K_{ a 1}=7.5 \times 10^{-3}$
$\tag{Second ionization} \ce{H2PO4^{-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + HPO_4^{2-}(aq)} \quad K_{ a 2}=6.2 \times 10^{-8}$
$\tag{Third ionization} \ce{HPO4^{2-}(aq) + H2O (l) \rightleftharpoons H3O^{+}(aq) + PO_4^{3-}(aq)} \quad K_{ a 3}=4.2 \times 10^{-13}$
As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about $10^{5}$ to $10^{6}$.
This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of $\ce{H3PO4}$ complicated. However, because the successive ionization constants differ by a factor of $10^{5}$ to $10^{6}$, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.
Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.
$\ce{H_2 O (l) + CO_3^{2-}(aq) \rightleftharpoons HCO_3^{-}(aq) + OH^{-}(aq)} \quad K_{ b 1}=2.1 \times 10^{-4} \nonumber$
$\ce{H_2 O (l) + HCO_3^{-}(aq) \rightleftharpoons H_2 CO_3(aq) + OH^{-}(aq)} \quad K_{ b 2}=2.3 \times 10^{-8} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.06%3A_Polyprotic_Acids.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the composition and function of acid–base buffers
• Calculate the pH of a buffer before and after the addition of added acid or base
A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure $1$). A solution of acetic acid and sodium acetate (CH3COOH + CH3COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH3(aq) + NH4Cl(aq)).
How Buffers Work
To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):
$\ce{CH_3 CO_2 H (aq) + H2O(l) <=> H3O^{+}(aq) + CH_3 CO_2^{-}(aq)} \nonumber$
Likewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). Figure $2$ provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.
Example 14.20: pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.
1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
2. Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.
3. For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
Solution
(a) Following the ICE approach to this equilibrium calculation yields the following:
Substituting the equilibrium concentration terms into the Ka expression, assuming $x \ll 0.10$, and solving the simplified equation for $x$ yields
$x=1.8 \times 10^{-5}~\text{M} \nonumber$
$\left[ \ce{H3O^{+}} \right]=0+x=1.8 \times 10^{-5}~\text{M} \nonumber$
\begin{align*} pH &=-\log \left[ \ce{H3O^{+}} \right] \[4pt] &=-\log \left(1.8 \times 10^{-5}\right) \[4pt] &= 4.74 \end{align*} \nonumber
(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.
Adding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.
$0.0010~ \cancel{\text{L}} \times \left(\dfrac{0.10 ~\text{mol} ~\ce{NaOH}}{1 ~ \text{L}}\right)=1.0 \times 10^{-4}~\text{mol} ~ \ce{NaOH} \nonumber$
The initial molar amount of acetic acid is
$0.100 ~ \cancel{\text{L}} \times\left(\frac{0.100~ \text{mol} ~ \ce{CH3CO2H} }{1 ~ \cancel{\text{L}} }\right)=1.00 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber$
The amount of acetic acid remaining after some is neutralized by the added base is
$\left(1.0 \times 10^{-2}\right)-\left(0.01 \times 10^{-2}\right)=0.99 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber$
The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of
$\left(1.0 \times 10^{-2}\right) + \left(0.01 \times 10^{-2}\right)=1.01 \times 10^{-2}~ \text{mol} ~ \ce{NaCH3CO2} \nonumber$
Compute molar concentrations for the two buffer components:
\begin{align*} \left[ \ce{CH3CO2H} \right] & =\frac{9.9 \times 10^{-3} ~\text{mol} }{0.101~\text{L} }=0.098 ~\text{M} \[4pt] \left[ \ce{NaCH3CO2} \right] & =\frac{1.01 \times 10^{-2} ~\text{mol} }{0.101 ~\text{L}}=0.100~\text{M} \end{align*} \nonumber
Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).
(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74.
The amount of hydronium ion initially present in the solution is
\begin{align*} \left[ \ce{H3O^{+}} \right] &=10^{-4.74}=1.8 \times 10^{-5} ~ \text{M} \[4pt] \text{mol} ~\ce{H3O^{+}} & =(0.100~\text{L})\left(1.8 \times 10^{-5}~\text{M} \right)=1.8 \times 10^{-6}~ \text{mol}~\ce{H3O^{+}} \end{align*} \nonumber
The amount of hydroxide ion added to the solution is
$mol OH^{-}=(0.0010 L)(0.10 M)=1.0 \times 10^{-4} mol \ce{OH^{-}} \nonumber$
The added hydroxide will neutralize hydronium ion via the reaction
$\ce{H3O^{+}(aq) + OH^{-}(aq) \leftrightharpoons 2 H2O(l)} \nonumber$
The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).
The amount of hydroxide ion remaining is
$1.0 \times 10^{-4} mol -1.8 \times 10^{-6} mol =9.8 \times 10^{-5} mol OH^{-} \nonumber$
corresponding to a hydroxide molarity of
$\dfrac{9.8 \times 10^{-5}~\text{mol}~\ce{OH^{-}}}{0.101~\text{L}}=9.7 \times 10^{-4} M \nonumber$
The pH of the solution is then calculated to be
$pH =14.00- pOH =14.00 - -\log \left(9.7 \times 10^{-4}\right)=10.99 \nonumber$
In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).
Exercise $1$
Show that adding 1.0 mL of $0.10~\text{M}~\ce{HCl}$ changes the pH of 100 mL of a $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$ solution from 4.74 to 3.00.
Answer
Initial pH of $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$;
$\text{pH} = −\log[\ce{H3O^{+}}] = −\log[1.8 \times 10^{−5}] = 4.74 \nonumber$
Moles of H3O+ in 100 mL $1.8 \times 10^{−5} ~\text{M}~\ce{HCl}$; $(1.8 \times 10^{−5}~\text{moles/L})\times (0.100 ~\text{L} = 1.8 \times 10^{−6}~\text{mol}\nonumber$
Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: $(0.10~\text{moles/L}) \times (0.0010~ \text{L}) = 1.0 \times 10^{−4}~\text{mol} \nonumber$
Final pH after addition of 1.0 mL of 0.10 M HCl:
$pH =-\log \left[ \ce{H3O^{+}} \right]=-\log \left(\frac{\text { total moles } \ce{H3O^{+}}}{\text {total volume }}\right)=-\log \left(\frac{1.0 \times 10^{-4} mol +1.8 \times 10^{-6}~\text{mol}}{101~\text{mL} \left(\frac{1~\text{L} }{1000~\text{mL}}\right)}\right) = 3.00 \nonumber$
Buffer Capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure $3$). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.
The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.
Selection of Suitable Buffer Mixtures
There are two useful rules of thumb for selecting buffer mixtures:
1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure $4$ shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, $\ce{H2CO3}$, and the bicarbonate ion, $\ce{HCO3^{-}}$. When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:
$\ce{H3O^{+}(aq) + HCO_3^{-}(aq) \longrightarrow H_2 CO_3(aq) + H2O(l)} \nonumber$
An added hydroxide ion is removed by the reaction:
$\ce{ OH^{-}(aq) + H_2CO_3(aq) \longrightarrow HCO_3^{-}(aq) + H2O(l)} \nonumber$
The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H3O+ is converted to H2CO3 and OH- is converted to HCO3-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.
The Henderson-Hasselbalch Equation
The ionization-constant expression for a solution of a weak acid can be written as:
$K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \nonumber$
Rearranging to solve for [H3O+] yields:
$\left[ \ce{H3O^{+}} \right]=K_{ a } \times \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber$
Taking the negative logarithm of both sides of this equation gives
$-\log \left[ \ce{H3O^{+}} \right]=-\log K_{ a }-\log \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber$
which can be written as
$pH = p K_{ a }+\log \frac{\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \nonumber$
where pKa is the negative of the logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.
Portrait of a Chemist: Lawrence Joseph Henderson and Karl Albert Hasselbalch
Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.
In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.
How Sciences Interconnect: Medicine: The Buffer System in Blood
The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:
$\ce{CO_2(g) + 2 H2O(l) <=> H2CO3(aq) <=> HCO_3^{-}(aq) + H3O^{+}(aq)} \nonumber$
The concentration of carbonic acid, $\ce{H2CO3}$ is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood:
$pH = pK_{a} + \log \dfrac{[\text { base }]}{[\text { acid }]} = 6.4 + \log \frac{0.024}{0.0012} = 7.7 \nonumber$
The fact that the $\ce{H2CO3}$ concentration is significantly lower than that of the $\ce{HCO3^{-}}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.
Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH.
View information on the buffer system encountered in natural waters. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.07%3A_Buffers.txt |
Learning Objectives
By the end of this section, you will be able to:
• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration
• Explain the function of acid-base indicators
As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.
Titration Curves
A titration curve is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration’s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.
Example $1$: Calculating pH for Titration Solutions: Strong Acid/Strong Base
A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH (the titration curve is shown in Figure $1$). Calculate the pH at these volumes of added base solution:
1. 0.00 mL
2. 12.50 mL
3. 25.00 mL
4. 37.50 mL
Solution
(a) Titrant volume = 0 mL. The solution pH is due to the acid ionization of HCl. Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 M. The pH of the solution is then
$\text{pH} =-\log (0.100)=1.000 \nonumber$
(b) Titrant volume = 12.50 mL. Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the initial amount and then dividing by the solution volume:
$\left[ \ce{H3O^{+}} \right]=\frac{ n \left( \ce{H^{+}} \right)}{V}=\frac{0.002500~\text{mol} \times\left(\frac{1000 ~mL }{1 ~L }\right)-0.100 ~M \times 12.50 ~mL }{25.00~ mL +12.50~ mL }=0.0333~ M \nonumber$
(c) Titrant volume = 25.00 mL. This titrant addition involves a stoichiometric amount of base (the equivalence point), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autoprotolysis of water. The solution is neutral, having a pH = 7.00.
(d) Titrant volume = 37.50 mL. This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:
$n \left( OH^{-}\right)_0> n \left( H^{+}\right)_0 \nonumber$
$\left[ \ce{OH^{-}} \right]=\frac{ n \left( \ce{OH^{-}} \right)}{\text{V}}=\dfrac{0.100~\text{M} \times 37.50~\text{mL} - 0.002500 ~\text{mol} \times\left(\frac{1000~\text{mL}}{1~\text{L}}\right)}{25.00~\text{mL} + 37.50~\text{mL}}=0.0200~\text{M} \nonumber$
$\text{pH} = 14 − \text{pOH} = 14 + \log([\ce{OH^{-}}]) = 14 + \log(0.0200) = 12.30 \nonumber$
Exercise $1$
Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO3(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.
Answer
0.00: 1.000; 15.0: 1.5111; 25.0: 7; 40.0: 12.523
Example $2$: Titration of a Weak Acid with a Strong Base
Consider the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:
$\ce{CH3CO2H + OH^{-} -> CH3CO2^{-} + H2O} \nonumber$
Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:
1. 0.00 mL
2. 25.00 mL
3. 12.50 mL
4. 37.50 mL
Solution
(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:
$K_{ a }=\frac{\left[ H3O^{+}\right]\left[ CH_3 CO_2^{-}\right]}{\left[ CH_3 CO_2 H \right]} \approx \frac{\left[ H3O^{+}\right]^2}{\left[ CH_3 CO_2 H \right]_0} \nonumber$
and
$\left[ H3O^{+}\right]=\sqrt{K_a \times\left[ CH_3 CO_2 H \right]}=\sqrt{1.8 \times 10^{-5} \times 0.100}=1.3 \times 10^{-3} \nonumber$
$pH =-\log \left(1.3 \times 10^{-3}\right)=2.87 \nonumber$
(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of
$\frac{0.00250~\text{mol} }{0.0500~\text{L}}=0.0500~\text{M}~\ce{CH3CO2^{-}} \nonumber$
Base ionization of acetate is represented by the equation
$\ce{CH3CO2^{-}(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-}(aq)} \nonumber$
$K_{ b }=\frac{\left[ H^{+}\right]\left[ OH^{-}\right]}{K_{ a }}=\frac{K_{ w }}{K_{ a }}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.6 \times 10^{-10} \nonumber$
Assuming $x \ll 0.0500$, the pH may be calculated via the usual ICE approach:
$K_{ b }=\frac{x^2}{0.0500 M} \nonumber$
$x=\left[ OH^{-}\right]=5.3 \times 10^{-6} \nonumber$
\begin{align*} \text{pOH} &=-\log \left(5.3 \times 10^{-6}\right)=5.28 \ \text{pH} &=14.00-5.28=8.72 \end{align*}
Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.
(c) Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:
$pH =p K_{ a }+\log \frac{[\text { Base }]}{[\text { Acid }]}=-\log \left(K_{ a }\right)+\log \frac{\left[ \ce{CH3CO2^{-}} \right]}{\left[ \ce{CH3CO2H} \right]}=-\log \left(1.8 \times 10^{-5}\right)+\log (1) \nonumber$
$pH =-\log \left(1.8 \times 10^{-5}\right)=4.74 \nonumber$
(pH = pKa at the half-equivalence point in a titration of a weak acid)
(d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:
$\left[ \ce{OH^{-}} \right]=\frac{(0.003750 mol -0.00250 mol )}{0.06250 L }=2.00 \times 10^{-2} M \nonumber$
\begin{align*} \text{pOH} &=-\log \left(2.00 \times 10^{-2}\right)=1.70 \[4pt] \text{pH} &=14.00-1.70=12.30 \end{align*}
Exercise $1$
Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.
Answer
0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.097
Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH/volume data pairs for the strong and weak acid titrations is provided in Table $1$ and plotted as titration curves in Figure $1$. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:
• initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH
• pre-equivalence point (0 mL < V < 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base
• equivalence point (V = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid
• postequivalence point (V > 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.
Table $1$: pH Values in the Titrations of a Strong Acid and of a Weak Acid
Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M CH3CO2H2
0.0 0.0 1.00 2.87
5.0 0.00050 1.18 4.14
10.0 0.00100 1.37 4.57
15.0 0.00150 1.60 4.92
20.0 0.00200 1.95 5.35
22.0 0.00220 2.20 5.61
24.0 0.00240 2.69 6.13
24.5 0.00245 3.00 6.44
24.9 0.00249 3.70 7.14
25.0 0.00250 7.00 8.72
25.1 0.00251 10.30 10.30
25.5 0.00255 11.00 11.00
26.0 0.00260 11.29 11.29
28.0 0.00280 11.75 11.75
30.0 0.00300 11.96 11.96
35.0 0.00350 12.22 12.22
40.0 0.00400 12.36 12.36
45.0 0.00450 12.46 12.46
50.0 0.00500 12.52 12.52
Acid-Base Indicators
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than $5.0 \times 10^{−9} ~\text{M}$ (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than $5.0 \times 10^{−9} ~\text{M}$ (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid-base indicators are either weak organic acids or weak organic bases.
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:
$\underset{\text { red }}{ \ce{HIn(aq) + H2O(l)} \ce{ <=> } \underset{\text { yellow }}{ \ce{H3O^{+}(aq)} + In^{-}(aq)}} \nonumber$
$K_a=\dfrac{\left[ \ce{H3O^{+}}\right]\left[ \ce{In^{-}} \right]}{[ \ce{HIn} ]}=4.0 \times 10^{-4} \nonumber$
The anion of methyl orange, $\ce{In^{−}}$, is yellow, and the nonionized form, $\ce{HIn}$, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.
The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species $\ce{In^{−}}$ and $\ce{HIn}$. If most of the indicator (typically about 60−90% or more) is present as $\ce{In^{−}}$, the perceived color of the solution is yellow. If most is present as $\ce{HIn}$, then the solution color appears red. The Henderson-Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):
$pH = p K a +\log \left(\frac{\left[ \ce{In^{-}} \right]}{[ \ce{HIn} ]}\right) \nonumber$
In solutions where pH > pKa, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH < pKa, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately pKa ± 1.
There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure $2$ presents several indicators, their colors, and their color-change intervals.
The titration curves shown in Figure $3$ illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.
The weak acid titration curve in Figure $3$ shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to-yellow color change over a relatively large volume interval (0–6 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7–8 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.
Footnotes
• 1Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH.
• 2Titration of 25.00 mL of 0.100 M CH3CO2H (0.00250 mol of CH3CO2H) with 0.100 M NaOH. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.08%3A_Acid-Base_Titrations.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
acid ionizationreaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
acid ionization constant (Ka)equilibrium constant for an acid ionization reaction
acid-base indicatorweak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH
acidica solution in which [H3O+] > [OH]
amphiproticspecies that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction
amphotericspecies that can act as either an acid or a base
autoionizationreaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionizationreaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
base ionization constant (Kb)equilibrium constant for a base ionization reaction
basica solution in which [H3O+] < [OH]
Brønsted-Lowry acidproton donor
Brønsted-Lowry baseproton acceptor
buffermixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added
buffer capacityamount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)
color-change intervalrange in pH over which the color change of an indicator is observed
conjugate acidsubstance formed when a base gains a proton
conjugate basesubstance formed when an acid loses a proton
diprotic acidacid containing two ionizable hydrogen atoms per molecule
diprotic basebase capable of accepting two protons
Henderson-Hasselbalch equationlogarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions
ion-product constant for water (Kw)equilibrium constant for the autoionization of water
leveling effectobservation that acid-base strength of solutes in a given solvent is limited to that of the solvent’s characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)
monoprotic acidacid containing one ionizable hydrogen atom per molecule
neutraldescribes a solution in which [H3O+] = [OH]
oxyacidternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom
percent ionizationratio of the concentration of ionized acid to initial acid concentration expressed as a percentage
pHlogarithmic measure of the concentration of hydronium ions in a solution
pOHlogarithmic measure of the concentration of hydroxide ions in a solution
stepwise ionizationprocess in which a polyprotic acid is ionized by losing protons sequentially
titration curveplot of some sample property (such as pH) versus volume of added titrant
triprotic acidacid that contains three ionizable hydrogen atoms per molecule
14.10: Key Equations
Kw = [H3O+][OH] = 1.0 \times 10^{−14} (at 25 °C)
pOH = −log[OH]
[H3O+] = \times 10^{−pH}
[OH] = \times 10^{−pOH}
pH + pOH = pKw = 14.00 at 25 °C
Ka Kb = 1.0 \times 10^{−14} = Kw
pKa = −log Ka
pKb = −log Kb | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.09%3A_Key_Terms.txt |
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, OH when it undergoes autoionization:
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([H3O+] > [OH]), basic ([H3O+] < [OH]), or neutral ([H3O+] = [OH]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.
The relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists between the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids.
An acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.
Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.
The titration curve for an acid-base titration is typically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.11%3A_Summary.txt |
1.
Write equations that show NH3 as both a conjugate acid and a conjugate base.
2.
Write equations that show $H2PO4−H2PO4−$ acting both as an acid and as a base.
3.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
(a) $H3O+H3O+$
(b) HCl
(c) NH3
(d) CH3CO2H
(e) $NH4+NH4+$
(f) $HSO4−HSO4−$
4.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
(a) HNO3
(b) $PH4+PH4+$
(c) H2S
(d) CH3CH2COOH
(e) $H2PO4−H2PO4−$
(f) HS
5.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
(a) H2O
(b) OH
(c) NH3
(d) CN
(e) S2−
(f) $H2PO4−H2PO4−$
6.
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
(a) HS
(b) $PO43−PO43−$
(c) $NH2−NH2−$
(d) C2H5OH
(e) O2−
(f) $H2PO4−H2PO4−$
7.
What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) OH
(b) H2O
(c) $HCO3−HCO3−$
(d) NH3
(e) $HSO4−HSO4−$
(f) H2O2
(g) HS
(h) $H5N2+H5N2+$
8.
What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) H2S
(b) $H2PO4−H2PO4−$
(c) PH3
(d) HS
(e) $HSO3−HSO3−$
(f) $H3O2+H3O2+$
(g) H4N2
(h) CH3OH
9.
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) $HNO3+H2O⟶H3O++NO3−HNO3+H2O⟶H3O++NO3−$
(b) $CN−+H2O⟶HCN+OH−CN−+H2O⟶HCN+OH−$
(c) $H2SO4+Cl−⟶HCl+HSO4−H2SO4+Cl−⟶HCl+HSO4−$
(d) $HSO4−+OH−⟶SO42−+H2OHSO4−+OH−⟶SO42−+H2O$
(e) $O2−+H2O⟶2OH−O2−+H2O⟶2OH−$
(f) $[Cu(H2O)3(OH)]++[Al(H2O)6]3+⟶[Cu(H2O)4]2++[Al(H2O)5(OH)]2+[Cu(H2O)3(OH)]++[Al(H2O)6]3+⟶[Cu(H2O)4]2++[Al(H2O)5(OH)]2+$
(g) $H2S+NH2−⟶HS−+NH3H2S+NH2−⟶HS−+NH3$
10.
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) $NO2−+H2O⟶HNO2+OH−NO2−+H2O⟶HNO2+OH−$
(b) $HBr+H2O⟶H3O++Br−HBr+H2O⟶H3O++Br−$
(c) $HS−+H2O⟶H2S+OH−HS−+H2O⟶H2S+OH−$
(d) $H2PO4−+OH−⟶HPO42−+H2OH2PO4−+OH−⟶HPO42−+H2O$
(e) $H2PO4−+HCl⟶H3PO4+Cl−H2PO4−+HCl⟶H3PO4+Cl−$
(f) $[Fe(H2O)5(OH)]2++[Al(H2O)6]3+⟶[Fe(H2O)6]3++[Al(H2O)5(OH)]2+[Fe(H2O)5(OH)]2++[Al(H2O)6]3+⟶[Fe(H2O)6]3++[Al(H2O)5(OH)]2+$
(g) $CH3OH+H−⟶CH3O−+H2CH3OH+H−⟶CH3O−+H2$
11.
What are amphiprotic species? Illustrate with suitable equations.
12.
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
(a) H2O
(b) $H2PO4−H2PO4−$
(c) S2−
(d) $CO32−CO32−$
(e) $HSO4−HSO4−$
13.
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
(a) NH3
(b) $HPO4−HPO4−$
(c) Br
(d) $NH4+NH4+$
(e) $ASO43−ASO43−$
14.
Is the self-ionization of water endothermic or exothermic? The ionization constant for water (Kw) is 2.9 $××$ 10−14 at 40 °C and 9.3 $××$ 10−14 at 60 °C.
15.
Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 $××$ 10−7 M. Kw is 2.9 $××$ 10−14 at 40 °C.
16.
The ionization constant for water (Kw) is 2.9 $××$ 10−14 at 40 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 40 °C.
17.
The ionization constant for water (Kw) is 9.311 $××$ 10−14 at 60 °C. Calculate [H3O+], [OH], pH, and pOH for pure water at 60 °C.
18.
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
(a) 0.200 M HCl
(b) 0.0143 M NaOH
(c) 3.0 M HNO3
(d) 0.0031 M Ca(OH)2
19.
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
(a) 0.000259 M HClO4
(b) 0.21 M NaOH
(c) 0.000071 M Ba(OH)2
(d) 2.5 M KOH
20.
What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
21.
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
22.
Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 14.2 for useful information.
23.
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 14.2 for useful information.
24.
The hydronium ion concentration in a sample of rainwater is found to be 1.7 $××$ 10−6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?
25.
The hydroxide ion concentration in household ammonia is 3.2 $××$ 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
26.
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
27.
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
28.
Use this list of important industrial compounds (and Figure 14.8) to answer the following questions regarding: Ca(OH)2, CH3CO2H, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.
(a) Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
(b) Identify the compounds that can behave as Brønsted-Lowry acids with strengths lying between those of H3O+ and H2O.
(c) Identify the compounds that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH.
29.
The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
30.
Household ammonia is a solution of the weak base NH3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.
31.
Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
32.
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
33.
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
34.
Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.
35.
What is the ionization constant at 25 °C for the weak acid $CH3NH3+,CH3NH3+,$ the conjugate acid of the weak base CH3NH2, Kb = 4.4 $××$ 10−4.
36.
What is the ionization constant at 25 °C for the weak acid $(CH3)2NH2+,(CH3)2NH2+,$ the conjugate acid of the weak base (CH3)2NH, Kb = 5.9 $××$ 10−4?
37.
Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, $(CH3)2NH2+(CH3)2NH2+$ or $CH3NH3+CH3NH3+$, is the stronger acid?
38.
Which is the stronger acid, $NH4+NH4+$ or HBrO?
39.
Which is the stronger base, (CH3)3N or $H2BO3−?H2BO3−?$
40.
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
(a) H2O or HF
(b) B(OH)3 or Al(OH)3
(c) $HSO3−HSO3−$ or $HSO4−HSO4−$
(d) NH3 or H2S
(e) H2O or H2Te
41.
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
(a) $HSO4−HSO4−$ or $HSeO4−HSeO4−$
(b) NH3 or H2O
(c) PH3 or HI
(d) NH3 or PH3
(e) H2S or HBr
42.
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
(a) acidity: HCl, HBr, HI
(b) basicity: H2O, OH, H, Cl
(c) basicity: Mg(OH)2, Si(OH)4, ClO3(OH) (Hint: Formula could also be written as HClO4.)
(d) acidity: HF, H2O, NH3, CH4
43.
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
(a) acidity: NaHSO3, NaHSeO3, NaHSO4
(b) basicity: $BrO2−,BrO2−,$ $ClO2−,ClO2−,$ $IO2−IO2−$
(c) acidity: HOCl, HOBr, HOI
(d) acidity: HOCl, HOClO, HOClO2, HOClO3
(e) basicity: $NH2−,NH2−,$ HS, HTe, $PH2−PH2−$
(f) basicity: BrO, $BrO2−,BrO2−,$ $BrO3−,BrO3−,$ $BrO4−BrO4−$
44.
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base?
45.
The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).
46.
Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
47.
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?
48.
Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water?
(a) addition of NaOH
(b) addition of HCl
(c) addition of NH4Cl
49.
Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?
(a) addition of NaOH
(b) addition of HCl
(c) addition of NaF
50.
What is the effect on the concentrations of $NO2−,NO2−,$ HNO2, and OH when the following are added to a solution of KNO2 in water:
(a) HCl
(b) HNO2
(c) NaOH
(d) NaCl
(e) KNO
51.
What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
(a) HCl
(b) KF
(c) NaCl
(d) KOH
(e) HF
52.
Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
53.
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
(a) CH3CO2H: $[H3O+][H3O+]$ = 1.34 $××$ 10−3 M;
$[CH3CO2−][CH3CO2−]$ = 1.34 $××$ 10−3 M;
[CH3CO2H] = 9.866 $××$ 10−2 M;
(b) ClO: [OH] = 4.0 $××$ 10−4 M;
[HClO] = 2.38 $××$ 10−4 M;
[ClO] = 0.273 M;
(c) HCO2H: [HCO2H] = 0.524 M;
$[H3O+][H3O+]$ = 9.8 $××$ 10−3 M;
$[HCO2−][HCO2−]$ = 9.8 $××$ 10−3 M;
(d) $C6H5NH3+:C6H5NH3+:$ $[C6H5NH3+][C6H5NH3+]$ = 0.233 M;
[C6H5NH2] = 2.3 $××$ 10−3 M;
$[H3O+][H3O+]$ = 2.3 $××$ 10−3 M
54.
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
(a) NH3: [OH] = 3.1 $××$ 10−3 M;
$[NH4+][NH4+]$ = 3.1 $××$ 10−3 M;
[NH3] = 0.533 M;
(b) HNO2: $[H3O+][H3O+]$ = 0.011 M;
$[NO2−][NO2−]$ = 0.0438 M;
[HNO2] = 1.07 M;
(c) (CH3)3N: [(CH3)3N] = 0.25 M;
[(CH3)3NH+] = 4.3 $××$ 10−3 M;
[OH] = 3.7 $××$ 10−3 M;
(d) $NH4+:NH4+:$ $[NH4+][NH4+]$ = 0.100 M;
[NH3] = 7.5 $××$ 10−6 M;
[H3O+] = 7.5 $××$ 10−6 M
55.
Determine Kb for the nitrite ion, $NO2−.NO2−.$ In a 0.10-M solution this base is 0.0015% ionized.
56.
Determine Ka for hydrogen sulfate ion, $HSO4−.HSO4−.$ In a 0.10-M solution the acid is 29% ionized.
57.
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) F
(b) $NH4+NH4+$
(c) $AsO43−AsO43−$
(d) $(CH3)2NH2+(CH3)2NH2+$
(e) $NO2−NO2−$
(f) $HC2O4−HC2O4−$ (as a base)
58.
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) HTe (as a base)
(b) $(CH3)3NH+(CH3)3NH+$
(c) $HAsO42–HAsO42–$ (as a base)
(d) $HO2−HO2−$ (as a base)
(e) $C6H5NH3+C6H5NH3+$
(f) $HSO3−HSO3−$ (as a base)
59.
Using the Ka value of 1.4 $×Figure 14.7.$
60.
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
(a) 0.0092 M HClO, a weak acid
(b) 0.0784 M C6H5NH2, a weak base
(c) 0.0810 M HCN, a weak acid
(d) 0.11 M (CH3)3N, a weak base
(e) 0.120 M $Fe (H2O)6 2+Fe (H2O)6 2+$ a weak acid, Ka = 1.6 $××$ 10−7
61.
Propionic acid, C2H5CO2H (Ka = 1.34 $××$ 10−5), is used in the manufacture of calcium propionate, a food preservative. What is the pH of a 0.698-M solution of C2H5CO2H?
62.
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3, what is the pH?
63.
The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is 1.36 $××$ 10−4. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
64.
Nicotine, C10H14N2, is a base that will accept two protons (Kb1 = 7 $××$ 10−7, Kb2 = 1.4 $××$ 10−11). What is the concentration of each species present in a 0.050-M solution of nicotine?
65.
The pH of a 0.23-M solution of HF is 1.92. Determine Ka for HF from these data.
66.
The pH of a 0.15-M solution of $HSO4−HSO4−$ is 1.43. Determine Ka for $HSO4−HSO4−$ from these data.
67.
The pH of a 0.10-M solution of caffeine is 11.70. Determine Kb for caffeine from these data:
$C8H10N4O2(aq)+H2O(l)⇌C8H10N4O2H+(aq)+OH−(aq)C8H10N4O2(aq)+H2O(l)⇌C8H10N4O2H+(aq)+OH−(aq)$
68.
The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.
69.
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) Al(NO3)3
(b) RbI
(c) KHCO2
(d) CH3NH3Br
70.
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) FeCl3
(b) K2CO3
(c) NH4Br
(d) KClO4
71.
Novocaine, C13H21O2N2Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 $××$ 10−6. Is a solution of novocaine acidic or basic? What are [H3O+], [OH], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.
72.
Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: $[H3O+],[H3O+],$ [OH], [H2CO3], $[HCO3−],[HCO3−],$ $[CO32−]?[CO32−]?$ No calculations are needed to answer this question.
73.
Calculate the concentration of each species present in a 0.050-M solution of H2S.
74.
Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
$C6H4(CO2H)2(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2H)(CO2)−(aq)Ka=1.1×10−3C6H4(CO2H)(CO2)(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2)22−(aq)Ka=3.9×10−6C6H4(CO2H)2(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2H)(CO2)−(aq)Ka=1.1×10−3C6H4(CO2H)(CO2)(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2)22−(aq)Ka=3.9×10−6$
75.
Salicylic acid, HOC6H4CO2H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
(a) Both functional groups of salicylic acid ionize in water, with Ka = 1.0 $××$ 10−3 for the—CO2H group and 4.2 $××$ 10−13 for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).
(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH3CO2C6H4CO2H. The −CO2H functional group is still present, but its acidity is reduced, Ka = 3.0 $××$ 10−4. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
76.
The ion HTe is an amphiprotic species; it can act as either an acid or a base.
(a) What is Ka for the acid reaction of HTe with H2O?
(b) What is Kb for the reaction in which HTe functions as a base in water?
(c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe] in a 0.10 M solution of H2Te.
77.
Explain why a buffer can be prepared from a mixture of NH4Cl and NaOH but not from NH3 and NaOH.
78.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.
79.
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH3 and a salt of its conjugate acid NH4Cl.
80.
What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2?
$CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)Ka=1.8×10−5CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)Ka=1.8×10−5$
81.
What is [H3O+] in a solution of 0.075 M HNO2 and 0.030 M NaNO2?
$HNO2(aq)+H2O(l)⇌H3O+(aq)+NO2−(aq)Ka=4.5×10−5HNO2(aq)+H2O(l)⇌H3O+(aq)+NO2−(aq)Ka=4.5×10−5$
82.
What is [OH] in a solution of 0.125 M CH3NH2 and 0.130 M CH3NH3Cl?
$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)Kb=4.4×10−4CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)Kb=4.4×10−4$
83.
What is [OH] in a solution of 1.25 M NH3 and 0.78 M NH4NO3?
$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kb=1.8×10−5NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kb=1.8×10−5$
84.
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
(a) HCl
(b) KCH3CO2
(c) NaCl
(d) KOH
(e) CH3CO2H
85.
What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
(a) KI
(b) NH3
(c) HI
(d) NaOH
(e) NH4Cl
86.
What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?
87.
Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.
88.
How much solid NaCH3CO2•3H2O must be added to 0.300 L of a 0.50-M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
89.
What mass of NH4Cl must be added to 0.750 L of a 0.100-M solution of NH3 to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)
90.
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 $××$ 10−5 as Ka for acetic acid.
(a) What is the pH of the solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
91.
A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution
diluted to 0.100 L.
(a) What is the pH of this buffer solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
92.
Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
93.
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
94.
Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 $××$ 10−5) with 0.100 M KOH.
(a) no KOH added
(b) 20 mL of KOH solution added
(c) 39 mL of KOH solution added
(d) 40 mL of KOH solution added
(e) 41 mL of KOH solution added
95.
The indicator dinitrophenol is an acid with a Ka of 1.1 $××$ 10−4. In a 1.0 $××$ 10−4-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow). | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/14%3A_Acid-Base_Equilibria/14.12%3A_Exercises.txt |
We previously learned about aqueous solutions and their importance, as well as about solubility rules. While this gives us a picture of solubility, that picture is not complete if we look at the rules alone. Solubility equilibrium, which we will explore in this chapter, is a more complex topic that allows us to determine the extent to which a slightly soluble ionic solid will dissolve, and the conditions under which precipitation.
15: Equilibria of Other Reaction Classes
The mineral fluorite, CaF2 Figure \(1\): , is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble CaF2 in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH, and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions.
15.02: Precipitation and Dissolution
Learning Objectives
By the end of this section, you will be able to:
• Write chemical equations and equilibrium expressions representing solubility equilibria
• Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.
The Solubility Product
Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.
$\ce{AgCl(s) <=>[\text{dissolution}][\text{precipitation}] Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous $\ce{Ag^{+}}$ and $\ce{Cl^{-}}$ ions at the same rate that these aqueous ions combine and precipitate to form solid $\ce{AgCl}$ (Figure $1$). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.
The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, Ksp, in this case
$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \quad \quad K_{ sp }=\left[\ce{Ag^{+}(aq)}\right]\left[\ce{Cl^{-}(aq)}\right] \nonumber$
Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.
Example $1$: Writing Equations and Solubility Products
Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:
1. AgI, silver iodide, a solid with antiseptic properties
2. CaCO3, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
3. Mg(OH)2, magnesium hydroxide, the active ingredient in Milk of Magnesia
4. Mg(NH4)PO4, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium
5. Ca5(PO4)3OH, the mineral apatite, a source of phosphate for fertilizers
Solution
1. $\ce{AgI(s) <=> Ag^{+}(aq) + I^{-}(aq)}$ $K_{ sp }=\left[\ce{Ag^{+}}\right]\left[\ce{I^{-}}\right] \nonumber$
2. $\ce{CaCO_3(s) <=> Ca^{2+}(aq) + CO3^{2-}(aq)}$ $K_{\text {sp}}=\left[\ce{Ca^{2+}} \right]\left[\ce{CO3^{2-}}\right] \nonumber$
3. $\ce{Mg(OH)_2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)}$ $K_{\text {sp}}=\left[\ce{Mg^{2+}}\right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
4. $\ce{Mg(NH4)PO4(s) <=> Mg^{2+}(aq) + NH4^{+}(aq) + PO4^{3-}(aq)}$ $K_{\text{sp}}=\left[\ce{Mg^{2+}}\right]\left[\ce{NH4^{+}}\right]\left[\ce{ PO4^{3-}}\right] \nonumber$
5. $\ce{Ca5(PO4) 3OH(s) <=> 5 Ca^{2+}(aq) + 3 PO4^{3-}(aq) + OH^{-}(aq)}$ $K_{ sp }=\left[\ce{Ca^{2+}}\right]^5\left[\ce{PO4^{3-}}\right]^3[\ce{OH^{-}}] \nonumber$
Exercise $1$
Write the dissolution equation and the solubility product for each of the following slightly soluble compounds:
1. BaSO4
2. Ag2SO4
3. Al(OH)3
4. Pb(OH)Cl
Answer
1. $\ce{BaSO4(s) <=> Ba^{2+}(aq) + SO4^{2-}(aq)}$ $K_{ sp }=\left[ \ce{Ba^{2+}}\right]\left[ \ce{SO4^{2-}}\right] \nonumber$
2. $\ce{Ag2SO4(s) <=> 2 Ag^{+}(aq) + SO4^{2-}(aq)}$ $K_{ sp }=\left[ \ce{Ag^{+}}\right]^2\left[ \ce{SO4^{2-}}\right] \nonumber$
3. $\ce{Al(OH)3(s) <=> Al^{3+}(aq) + 3 OH^{-}(aq)}$ $K_{ sp }=\left[ \ce{Al^{3+}}\right]\left[ \ce{OH^{-}}\right]^3 \nonumber$
4. $\ce{Pb(OH)Cl(s) <=> Pb^{2+}(aq) + OH^{-}(aq) + Cl^{-}(aq)}$ $K_{ sp }=\left[ \ce{Pb^{2+}}\right]\left[ \ce{OH^{-}} \right]\left[ \ce{Cl^{-}}\right] \nonumber$
Ksp and Solubility
The Ksp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:
$\ce{M_{p}X_{q}(s) <=> p~M^{m+}(aq) + q~X^{n-}(aq)} \nonumber$
For cases such as these, one may derive Ksp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.
Example $2$: Calculation of Ksp from Equilibrium Concentrations
Fluorite, $\ce{CaF2}$, is a slightly soluble solid that dissolves according to the equation:
$\ce{CaF_2(s) <=> Ca^{2+}(aq) + 2 F^{-}(aq)} \nonumber$
The concentration of $\ce{Ca^{2+}}$ in a saturated solution of $\ce{CaF2}$ is $2.15 \times 10^{–4} ~\text{M}$. What is the solubility product of fluorite?
Solution
According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a $\ce{CaF2}$ solution is equal to twice its calcium ion molarity:
$\left[ F^{-}\right]=\left(2~\text{mol}~\ce{F^{-}} / 1~\text{mol}~\ce{Ca^{2+}}\right)=(2)\left(2.15 \times 10^{-4} M\right)=4.30 \times 10^{-4} M \nonumber$
Substituting the ion concentrations into the Ksp expression gives
$K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{F^{-}}\right]^2=\left(2.15 \times 10^{-4}\right)\left(4.30 \times 10^{-4}\right)^2=3.98 \times 10^{-11} \nonumber$
Exercise $2$
In a saturated solution of Mg(OH)2, the concentration of Mg2+ is $1.31 \times 10^{–4} ~\text{M}$. What is the solubility product for Mg(OH)2?
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \nonumber$
Answer
$8.99 \times 10^{–12}$
Example $3$: Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide, CuBr, is $6.3 \times 10^{–9}$. Calculate the molar solubility of copper bromide.
Solution
The dissolution equation and solubility product expression are
$\ce{CuBr(s) <=> Cu^{+}(aq)+ Br^{-}(aq)} \nonumber$
with
$K_{ sp }=\left[ \ce{Cu^{+}}\right]\left[ \ce{Br^{-}}\right] \nonumber$
Following the ICE approach to this calculation yields the table
Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields
\begin{align*} K_{ sp }&=\left[ \ce{Cu^{+}}\right]\left[ \ce{Br^{-}}\right] \[4pt] 6.3 \times 10^{-9} &=(x)(x)=x^2 \[4pt] x&=\sqrt{6.3 \times 10^{-9}}\[4pt] &=7.9 \times 10^{-5}\,\text{M} \end{align*} \nonumber
Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of $\ce{Br}$ dissolved, the molar solubility of $\ce{CuBr}$ is $7.9 \times 10^{–5} ~\text{M}$.
Exercise $3$
The Ksp of AgI is $1.5 \times 10^{–16}$. Calculate the molar solubility of silver iodide.
Answer
$1.2 \times 10^{–8}~ \text{M}$
Example $4$: Determination of Molar Solubility from Ksp
The Ksp of calcium hydroxide, Ca(OH)2, is $1.3 \times 10^{–6}$. Calculate the molar solubility of calcium hydroxide.
Solution
The dissolution equation and solubility product expression are
$\ce{Ca(OH)2(s) <=> Ca^{2+}(aq) +2 OH^{-}(aq)} \nonumber$
with
$K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
The ICE table for this system is
Substituting terms for the equilibrium concentrations into the solubility product expression and solving for $x$ gives
\begin{aligned} K_{ sp } &=\left[ \ce{Ca^{2+}}\right]\left[ \ce{OH^{-}}\right]^2\[4pt] 1.3 \times 10^{-6}&=(x)(2 x)^2=(x)\left(4 x^2\right)=4 x^3\[4pt] x&=\sqrt[3]{\frac{1.3 \times 10^{-6}}{4}}\[4pt] &=6.9 \times 10^{-3}\,\text{M} \end{aligned} \nonumber
As defined in the ICE table, $x$ is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)2 is $6.9 \times 10^{–3}~\text{M}$.
Exercise $4$
The Ksp of PbI2 is $1.4 \times 10^{–8}$. Calculate the molar solubility of lead(II) iodide.
Answer
$1.5 \times 10^{–3}~ \text{M}$
Example $5$: Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure $2$) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is $4.6 \times 10^{–6}~\text{g/L}$. Determine the solubility product for PbCrO4.
Solution
Before calculating the solubility product, the provided solubility must be converted to molarity:
\begin{align*} [\ce{PbCrO4}] &= \dfrac{4.6 \times 10^{-6}~\text{g}~\text{PbCrO4}}{1~\text{L} } \times \dfrac{1~ \text{mol}~ \ce{PbCrO4}}{323.2~ \text{g} ~ \ce{PbCrO4}} \[4pt] &= \frac{1.4 \times 10^{-8}~\text{mol} ~ \ce{PbCrO4}}{1\,\text{L} } \[4pt] &= 1.4 \times 10^{-8}\,\text{M} \end{align*} \nonumber
The dissolution equation for this compound is
$\ce{PbCrO4(s) <=> Pb^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both $[\ce{Pb^{2+}}]$ and $[\ce{CrO4^{2-}}]$ are equal to the molar solubility of $\ce{PbCrO4}$:
$[\ce{Pb^{2+}}]= [\ce{CrO4^{2-}}] =1.4 \times 10^{-8}~ \text{M} \nonumber$
Therefore
$K_{sp} = [\ce{Pb^{2+}}][\ce{CrO4^{2-}}] = (1.4 \times 10^{–8})(1.4 \times 10^{–8}) = 2.0 \times 10^{–16} \nonumber$
Exercise $5$
The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20 °C. What is its solubility product?
Answer
$1.69 \times 10^{–4}$
Example $6$: Calculating the Solubility of Hg2Cl2
Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), and chloride ions, Cl. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small Ksp:
$\ce{Hg2Cl2(s) <=> Hg2^{2+}(aq) + 2 Cl^{-}(aq)} \quad K_{ sp }=1.1 \times 10^{-18} \nonumber$
Calculate the molar solubility of $\ce{Hg2Cl2}$.
Solution
The dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg2Cl2 is equal to the concentration of ions
Following the ICE approach results in
Substituting the equilibrium concentration terms into the solubility product expression and solving for x gives
\begin{align*}
K_{ sp } &=\left[ \ce{Hg2^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \4pt] 1.1 \times 10^{-18} &=(x)(2 x)^2 \[4pt] 4 x^3& =1.1 \times 10^{-18} \[4pt] x&=\sqrt[3]{\left(\frac{1.1 \times 10^{-18}}{4}\right)} \[4pt] &=6.5 \times 10^{-7}~ \text{M} \end{align*} Therefore \begin{align*} \left[ \ce{Hg2^{2+}}\right] &=6.5 \times 10^{-7}~\text{M} \[4pt]&=6.5 \times 10^{-7}~ \text{M} \[4pt] \left[ \ce{Cl^{-}}\right] &=2 x=2\left(6.5 \times 10^{-7} ~\text{M}\right) \[4pt] &=1.3 \times 10^{-6} M \end{align*} The dissolution stoichiometry shows the molar solubility of $\ce{Hg2Cl2}$ is equal to $[ \ce{Hg2^{2+}}]$ or $6.5 \times 10^{–7}~ \text{M}$. Exercise $6$ Determine the molar solubility of MgF2 from its solubility product: $K_{sp} = 6.4 \times 10^{–9}$. Answer $1.2 \times 10^{–3} ~\text{M}$ How Sciences Interconnect: Using Barium Sulfate for Medical Imaging Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the $K_{sp}$ of barium sulfate is $2.3 \times 10^{–8}$, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure $3$). Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions. Visit this website for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose. Predicting Precipitation The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is: \[\ce{CaCO3(s) <=> Ca^{2+}(aq) + CO3^{2-}(aq)} \nonumber
with
$K_{s p}=\left[ \ce{Ca^{2+}} \right]\left[ \ce{CO3^{2-}} \right]=8.7 \times 10^{-9} \nonumber$
It is important to realize that this equilibrium is established in any aqueous solution containing Ca2+ and CO32 ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:
• Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)
• Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)
This predictive strategy and related calculations are demonstrated in the next few example exercises.
Example $7$: Precipitation of Mg(OH)2
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH ion:
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \quad K_{ sp }=8.9 \times 10^{-12} \nonumber$
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH] of 0.0010 M?
Solution
Calculation of the reaction quotient under these conditions is shown here:
$Q=\left[ \ce{Mg^{2+}}\right]\left[ \ce{OH^{-}}\right]^2=(0.0537)(0.0010)^2=5.4 \times 10^{-8} \nonumber$
Because Q is greater than Ksp ($Q = 5.4 \times 10^{–8}$ is larger than $K_{sp} = 8.9 \times 10^{–12}$), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that $Q_{sp} = K_{sp}$.
Exercise $7$
Predict whether $\ce{CaHPO4}$ will precipitate from a solution with [Ca2+] = 0.0001 M and = 0.001 M.
Answer
No precipitation of $\ce{CaHPO4}$; $Q = 1 \times 10^{–7}$, which is less than Ksp ($7 \times 10^{–7}$)
Example $8$: Precipitation of AgCl
Does silver chloride precipitate when equal volumes of a $2.0 \times 10^{–4}$-M solution of AgNO3 and a $2.0 \times 10^{–4}$-M solution of NaCl are mixed?
Solution
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
$\ce{AgCl (s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
The solubility product is $1.6 \times 10^{-10}$ (see appendix J).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. Because the volume doubles when equal volumes of AgNO3 and NaCl solutions are mixed, each concentration is reduced to half its initial value
$\frac{1}{2}\left(2.0 \times 10^{-4}\right) M=1.0 \times 10^{-4}~\text{M} \nonumber$
The reaction quotient, Q, is greater than Ksp for AgCl, so a supersaturated solution is formed:
$Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]=\left(1.0 \times 10^{-4}\right)\left(1.0 \times 10^{-4}\right)=1.0 \times 10^{-8}>K_{ sp } \nonumber$
AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to Ksp.
Exercise $8$
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of $\ce{ClO4^{-}}$? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)
Answer
No, $Q = 4.0 \times 10^{–3}$, which is less than $K_{sp} = 1.05 \times 10^{–2}$
Example $9$: Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, $\ce{C2O4^{2−}}$, for this purpose (Figure $4$). At sufficiently high concentrations, the calcium and oxalate ions form solid, $\ce{CaC2O4 \cdot H2O}$ (calcium oxalate monohydrate). The concentration of $\ce{Ca^{2+}}$ in a sample of blood serum is $2.2 \times 10^{–3} M$. What concentration of $\ce{C2O4^{2−}}$ ion must be established before $\ce{CaC2O4 \cdot H2O}$ begins to precipitate?
Solution
The equilibrium expression is:
$\ce{CaC2O4(s) <=> Ca^{2+}(aq) + C2O4^{2-}(aq)} \nonumber$
For this reaction:
$K_{\text {sp }}=\left[ \ce{Ca^{2+}} \right]\left[ \ce{C2O4^{2-}} \right]=1.96 \times 10^{-9} \nonumber$
(see Appendix J)
Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:
\begin{aligned}
Q=K_{ sp }=\left[ \ce{Ca^{2+}}\right]\left[ \ce{C2O4^{2-}}\right]&=1.96 \times 10^{-9}\4pt] \left(2.2 \times 10^{-3}\right)\left[ C_2 O_4{ }^{2-}\right] &=1.96 \times 10^{-9}\[4pt] \left[ \ce{C2O4^{2-}}\right] &=\frac{1.96 \times 10^{-9}}{2.2 \times 10^{-3}} \[4pt]&=8.9 \times 10^{-7}~\text{M} \end{aligned} A concentration of $[\ce{C2O4^{2-}}] = 8.9 \times 10^{-7}~ \text{M}$ is necessary to initiate the precipitation of $\ce{CaC2O4}$ under these conditions. Exercise $9$ If a solution contains 0.0020 mol of $\ce{CrO4^{2−}}$ per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate. Answer $6.7 \time 10^{–5}~\text{M}$ Example $10$: Concentrations Following Precipitation Clothing washed in water that has a manganese $[\ce{Mn^{2+}(aq)}]$ concentration exceeding 0.1 mg/L ($1.8 \times 10^{–6} ~\text{M}$) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be decreased by adding a base to precipitate Mn(OH)2. What pH is required to keep [Mn2+] equal to $1.8 \times 10^{–6}~ \text{M}$? Solution The dissolution of Mn(OH)2 is described by the equation: \[\ce{Mn(OH)2(s) <=> Mn^{2+}(aq) + 2 OH^{-}(aq)} \quad \quad K_{ sp }=2 \times 10^{-13} \nonumber
At equilibrium:
$K_{ sp }=\left[ \ce{Mn^{2+}} \right]\left[ \ce{OH^{-}} \right]^2 \nonumber$
or
$\left(1.8 \times 10^{-6}\right)\left[ \ce{OH^{-}} \right]^2=2 \times 10^{-13} \nonumber$
so
$\left[ \ce{OH^{-}} \right]=3.3 \times 10^{-4} M \nonumber$
Calculate the pH from the pOH:
\begin{align*} \text{pOH} &=-\log \left[ \ce{OH^{-}}\right] \[4pt] &=-\log (3.3 \times 10^{-4})=3.48 \[4pt] \text{pH} &=14.00 - \text{pOH} \[4pt] &=14.00-3.48 \[4pt] &=10.52\end{align*} \nonumber
(final result rounded to one significant digit, limited by the certainty of the Ksp)
Exercise $10$
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is $5.37 \times 10^{–2}~\text{M}$. Calculate the pH at which [Mg2+] is decreased to $1.0 \times 10^{–5}~\text{M}$
Answer
10.97
In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.
Chemistry in Everyday Life: The Role of Precipitation in Wastewater Treatment
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure $5$). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.
One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:
$\ce{5 Ca^{2+} + 3 PO4^{3-} + OH^{-} <=> Ca5 (PO4)3 \cdot OH(s)} \nonumber$
Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.
View this site for more information on how phosphorus is removed from wastewater.
Example $11$: Precipitation of Silver Halides
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
Solution
The two equilibria involved are:
\begin{align*} \ce{AgCl(s) & <=> Ag^{+}(aq) + Cl^{-}(aq)} & K_{ sp } & =1.6 \times 10^{-10} \[4pt] \ce{AgBr(s) & <=> Ag^{+}(aq) + Br^{-}(aq)} & K_{ sp } & =5.0 \times 10^{-13} \end{align*} \nonumber
If the solution contained about equal concentrations of Cl and Br, then the silver salt with the smaller Ksp (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag+] precipitates first.
AgBr precipitates when Q equals Ksp for AgBr
\begin{align*} Q_{ sp }=K_{ sp }=\left[ \ce{Ag^{+}} \right]\left[ \ce{Br^{-}}\right]=\left[ \ce{Ag^{+}}\right](0.00010)&=5.0 \times 10^{-13} \[4pt] \left[ \ce{Ag^{+}}\right] &=\frac{5.0 \times 10^{-13}}{0.00010} \[4pt] &=5.0 \times 10^{-9}~\text{M} \end{align*} \nonumber
AgBr begins to precipitate when [Ag+] is $5.0 \times 10^{–9} ~\text{M}$.
For AgCl: AgCl precipitates when Q equals Ksp for AgCl ($1.6 \times 10^{–10}$). When [Cl] = 0.10 M:
\begin{align*} Q_{ sp }=K_{ sp }=\left[ \ce{Ag^{+}}\right]\left[ \ce{Cl^{-}}\right]=\left[ \ce{Ag^{+}}\right](0.10)&=1.6 \times 10^{-10} \[4pt] \left[ \ce{Ag^{+}}\right] &=\frac{1.6 \times 10^{-10}}{0.10} \[4pt] &=1.6 \times 10^{-9}~\text{M} \end{align*} \nonumber
AgCl begins to precipitate when [Ag+] is $1.6 \times 10^{–9} ~\text{M}$.
AgCl begins to precipitate at a lower [Ag+] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a Ksp greater than that of silver bromide.
Exercise $11$
If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?
Answer
$[\ce{Ag^{+}}] = 1.0 \times 10^{-11}~ \text{M}$; $\ce{AgBr}$ precipitates first
Common Ion Effect
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le Chatelier’s principle. Consider the dissolution of silver iodide:
$\ce{AgI(s) <=> Ag^{+}(aq) + I^{-}(aq)} \nonumber$
This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag+ and I. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.
This effect may also be explained in terms of mass action as represented in the solubility product expression:
$K_{ sp }=\left[ \ce{Ag^{+}} \right]\left[ \ce{I^{-}}\right] \nonumber$
The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.
View this simulation to explore various aspects of the common ion effect.
Example $12$: Common Ion Effect on Solubility
What is the effect on the amount of solid Mg(OH)2 and the concentrations of Mg2+ and OH when each of the following are added to a saturated solution of Mg(OH)2?
1. MgCl2
2. KOH
3. NaNO3
4. Mg(OH)2
Solution
The solubility equilibrium is
$\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2 OH^{-}(aq)} \nonumber$
1. Adding a common ion, $\ce{Mg^{2+}}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.
2. Adding a common ion, $\ce{OH^{-}}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.
3. The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
4. Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same. $Q=\left[ \ce{Mg^{2+}} \right]\left[ \ce{OH^{-}}\right]^2 \nonumber$
Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of $Q$, and no shift is required to restore $Q$ to the value of the equilibrium constant.
Exercise $12$
What is the effect on the amount of solid $\ce{NiCO3}$ and the concentrations of $\ce{Ni^{2+}}$ and $\ce{CO3^{2-}}$ when each of the following are added to a saturated solution of $\ce{NiCO3}$
1. Ni(NO3)2
2. KClO4
3. NiCO3
4. K2CO3
Answer
(a) mass of NiCO3(s) increases, [Ni2+] increases, $\ce{CO3^{2-}}$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO3; (d) mass of NiCO3(s) increases, [Ni2+] decreases, $\ce{CO3^{2-}}$ increases;
Example $13$: Common Ion Effect
Calculate the molar solubility of cadmium sulfide ($\ce{CdS}$) in a 0.010-M solution of cadmium bromide ($\ce{CdBr2}$). The Ksp of $\ce{CdS}$ is $1.0 \times 10^{–28}$.
Solution
This calculation can be performed using the ICE approach:
$\ce{CdS(s) <=> Cd^{2+}(aq) + S^{2-}(aq)} \nonumber$
\begin{align*} K_{\text {sp }}=\left[ \ce{Cd^{2+}} \right]\left[ \ce{S^{2-}}\right] &=1.0 \times 10^{-28} \[4pt] (0.010+x)(x) &=1.0 \times 10^{-28} \end{align*} \nonumber
Because Ksp is very small, assume $x \ll 0.010$ and solve the simplified equation for $x$:
\begin{align*} (0.010)(x) &=1.0 \times 10^{-28}\[4pt] x&=1.0 \times 10^{-26}~ \text{M} \end{align*} \nonumber
The molar solubility of $\ce{CdS}$ in this solution is $1.0 \times 10^{–26} ~\text{M}$.
Exercise $13$
Calculate the molar solubility of aluminum hydroxide, Al(OH)3, in a 0.015-M solution of aluminum nitrate, Al(NO3)3. The Ksp of Al(OH)3 is $2 \times 10^{–32}$.
Answer
$4 \times 10^{–11} ~\text{M}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the Lewis model of acid-base chemistry
• Write equations for the formation of adducts and complex ions
In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.
A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.
Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is H+. A few examples involving other Lewis acids and bases are described below.
The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. Being short of the preferred octet, BF3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:
In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:
Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:
Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:
Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands. These ligands can be neutral molecules like H2O or NH3, or ions such as CN or OH. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry—the topic of another chapter in this text.
The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant (Kf) (sometimes called a stability constant). For example, the complex ion
is produced by the reaction
$\ce{Cu^{+}(aq) + 2 CN^{-}(aq) <=> Cu(CN)2^{-}(aq)} \nonumber$
The formation constant for this reaction is
$K_{ f }=\frac{\left[ \ce{Cu(CN)2^{-}} \right]}{\left[ \ce{Cu^{+}} \right]\left[ \ce{CN^{-}}\right]^2} \nonumber$
Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant (Kd). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, Kd = Kf–1. A tabulation of formation constants is provided in Appendix K.
As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ($[\ce{Ag^{+}}] = 1.3 \times 10^{–5} ~\text{M}$):
$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber$
However, if NH3 is present in the water, the complex ion, $\ce{Ag(NH3)2^{+}}$ can form according to the equation:
$\ce{Ag^{+}(aq) + 2 NH3(aq) <=> Ag(NH3)2^{+}(aq)} \nonumber$
with
$K_{ f }=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][\ce{NH3}]^2}=1.7 \times 10^7 \nonumber$
The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH 3 to form $\ce{Ag(NH3)2^{+}}$. As a consequence, the concentration of silver ions, $\ce{[Ag^{+}]}$, is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag + ][Cl ], falls below the solubility product of AgCl:
$Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]<K_{ sp } \nonumber$
More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.
Example $1$: Dissociation of a Complex Ion
Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to $\ce{Ag(NH3)2^{+}}$).
Solution
Applying the standard ICE approach to this reaction yields the following:
Substituting these equilibrium concentration terms into the Kf expression gives
\begin{align*} K_{ f } &=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][ \ce{NH3} ]^2} \[4pt] 1.7 \times 10^7 &=\frac{0.10-x}{(x)(2 x)^2} \end{align*} \nonumber
The very large equilibrium constant means the amount of the complex ion that will dissociate, x, will be very small. Assuming $x \ll 0.1$ permits simplifying the above equation:
\begin{align*} 1.7 \times 10^7 &=\frac{0.10}{(x)(2 x)^2} \[4pt] x^3&=\frac{0.10}{4\left(1.7 \times 10^7\right)}=1.5 \times 10^{-9} \[4pt] x&=\sqrt[3]{1.5 \times 10^{-9}}=1.1 \times 10^{-3} \end{align*} \nonumber
Because only 1.1% of the $\ce{Ag(NH3)2^{+}}$) dissociates into $\ce{Ag^{+}}$ and $\ce{NH3}$, the assumption that x is small is justified.
Using this value of x and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations:
\begin{align*} [ \ce{Ag^{+}} ] &=0+x=1.1 \times 10^{-3}~\text{M} \[4pt] [ \ce{NH3}] &=0+2 x=2.2 \times 10^{-3} ~\text{M} \[4pt] [ \ce{Ag(NH3)2^{+}}] &=0.10-x=0.10-0.0011 \[4pt] &=0.099 ~\text{M} \end{align*} \nonumber
The concentration of free silver ion in the solution is 0.0011 M.
Exercise $1$
Calculate the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of $\ce{AgNO3}$ and 10.0 g of $\ce{KCN}$ in sufficient water to make 1.00 L of solution. (Hint: Because Kf is very large, assume the reaction goes to completion then calculate the [Ag+] produced by dissociation of the complex.)
Answer
$2.9 \times 10^{–22} ~\text{M}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.03%3A_Lewis_Acids_and_Bases.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe examples of systems involving two (or more) coupled chemical equilibria
• Calculate reactant and product concentrations for coupled equilibrium systems
As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.
An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is
$\ce{CaCO_3(s) <=> Ca^{2+}(aq) + CO_3^{-2}(aq)} \quad K_{\text {sp }}=8.7 \times 10^{-9} \nonumber$
Rising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:
\begin{align*} \ce{CO_2(g) &<=> CO_2(aq)} \[4pt] \ce{CO_2(aq) + H2O(l) &<=> H_2 CO_3(aq)} \[4pt] \ce{H_2 CO_3(aq) + H2O(l) &<=> HCO_3^{-}(aq) + H3O^{+}(aq)} \quad K_{ a 1}=4.3 \times 10^{-7} \[4pt] \ce{HCO_3^{-}(aq) + H2O(l) &<=> CO_3^{2-}(aq) + H3O^{+}(aq)} \quad K_{ a 2}=4.7 \times 10^{-11} \end{align*}
Inspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields
$\ce{CaCO_3(s) + H3O^{+}(aq) <=> Ca^{2+}(aq) + HCO_3^{-}(aq) + H2O(l)} \quad \quad K=K_{\text {sp }} / K_{ a 2}=180 \nonumber$
The equilibrium constant for this net reaction is much greater than the Ksp for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy (Figure $1$).
Learn more about ocean acidification and how it affects other marine creatures.
This site has detailed information about how ocean acidification specifically affects coral reefs.
The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure $2$), a sparingly soluble ionic compound whose dissolution equilibrium is
$\ce{Ca5(PO4)3OH(s) <=> 5 Ca^{2+}(aq) +3 PO4^{3-}(aq) + OH^{-}(aq)} \nonumber$
This compound dissolved to yield two different basic ions: triprotic phosphate ions
\begin{align*} \ce{PO_4^{3-}(aq) + H3O^{+}(aq) &-> H_2 PO_4^{2-}(aq) + H2O(l)} \[4pt] \ce{H_2 PO_4^{2-}(aq) + H3O^{+}(aq) &-> H_2 PO_4^{-}(aq) + H2O(l)} \[4pt] \ce{H_2 PO_4^{-}(aq) + H3O^{+}(aq) &-> H_3 PO_4(aq) + H2O(l)} \end{align*}
and monoprotic hydroxide ions:
$\ce{OH^{-}(aq) + H3O^{+} -> 2 H2O} \nonumber$
Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF2 that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:
$\ce{NaF + Ca5(PO4)3OH <=> Ca5(PO4)3F + Na^{+} + OH^{-}} \nonumber$
The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.
Chemistry in Everyday Life: Role of Fluoride in Preventing Tooth Decay
As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca5(PO4)3F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure $3$).
Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.
The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion
The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)3.
\begin{align*} \ce{Al ( OH )_3(s) &<=> Al^{3+}(aq) +3 OH^{-}(aq)} & K_{ sp }=2 \times 10^{-32} \[4pt] \ce{Al^{3+}(aq) +4 OH^{-}(aq) &<=> Al ( OH )_4^{-}(aq) }& K_{ f }=1.1 \times 10^{33} \[4pt] \text { Net: } \ce{Al ( OH )_3(s) + OH^{-}(aq) &<=> Al ( OH )_4^{-}(aq)} & K=K_{ sp } K_{ f }=22 \end{align*} \nonumber
Example $1$: Increased Solubility in Acidic Solutions
Compute and compare the molar solubilities for aluminum hydroxide, Al(OH)3, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.
Solution
(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:
$\ce{Al ( OH )_3(s) <=> Al^{3+}(aq) +3 OH^{-}(aq)} \quad \quad K_{ sp }=2 \times 10^{-32} \nonumber] \[\text{molar solubility in water} =\left[ \ce{Al^{3+}} \right]= \left(\dfrac{2 \times 10^{-32}}{27}\right)^{1/4} =5 \times 10^{-9} ~\text{M} \nonumber$
(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:
\begin{align*} \text{pH} &= pK_{ a }+\log \dfrac{\left[ \ce{CH3COO^{-}} \right] }{ \left[\ce{CH3COOH} \right]} \[4pt] &=4.74+\log \left(\dfrac{0.100}{0.100}\right)=4.74 \end{align*} \nonumber
At this pH, the concentration of hydroxide ion is
\begin{align*} & \text{pOH} =14.00-4.74=9.26 \[4pt] & {\left[ \ce{OH^{-}} \right]=10^{-9.26}=5.5 \times 10^{-10}} \end{align*} \nonumber
The solubility of $\ce{Al(OH)3}$ in this buffer is then calculated from its solubility product expression:
$K_{ sp }=\left[ \ce{Al^{3+}}\right]\left[ \ce{OH^{-}}\right]^3 \nonumber$
Via
\begin{align*} \text { molar solubility in buffer }&=\left[ \ce{Al^{3+}} \right] \[4pt] &= \dfrac{K_{ sp } }{\left[ \ce{OH^{-}} \right]^3} \[4pt] &=\dfrac{2 \times 10^{-32}}{(5.5 \times 10^{-10})^3} \[4pt] &= 1.2 \times 10^{-4}~\text{M} \end{align*}
Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).
Exercise $1$
What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?
Answer
0.1 M
Example $2$: Multiple Equilibria
Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion $\ce{Ag(S2O3)2^{3-}}$ with $K_f = 4.7 \times 10^{13}$.
What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $\ce{Ag(S2O3)2^{3-}}$
Solution
Two equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the ion:
• dissolution: $\ce{AgBr(s) <=> Ag^{+}(aq) + Br^{-}(aq)} \quad \quad K_{ sp }=5.0 \times 10^{-13} \nonumber$
• complexation: $\nonumber \ce{Ag^{+}(aq) +2 S2O3^{2-}(aq) <=> Ag ( S2O3)_2^{3-}(aq)}\quad \quad K_{ f }=4.7 \times 10^{13} \nonumber$
Combining these two equilibrium equations yields
$\ce{AgBr(s) +2S2O3^{2-}(aq) <=> Ag(S2O3)2^{3-}(aq) + Br^{-}(aq)} \nonumber$
with
$K=\frac{\left[ \ce{Ag(S2O3)2^3} \right][ \ce{Br^{-}} ]}{\left[ \ce{S2O3^{-2}} \right]^2}=K_{ sp } K_{ f }=24 \nonumber$
The concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is
$\left[ Br^{-}\right]=\frac{1.00 ~\text{g} ~\ce{AgBr} \times \dfrac{1 ~ \text{mol} ~\ce{AgBr} }{187.77\, \text{g / mol} } \times \dfrac{1 ~ \text{mol} ~\ce{Br^{-}}}{1~\text{mol} ~\ce{AgBr^{-}}}}{1.00~\text{L} }=0.00532~\text{M} \nonumber$
The stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M, and the very large value of ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion:
$\left[ \ce{Ag(S2O3)2^{3-}}\right]=0.00532 ~\text{M} \nonumber$
Rearranging the K expression for the combined equilibrium equations and solving for the concentration of thiosulfate ion yields
$\left[ \ce{S2O3^{2-}} \right]=\frac{\left[ \ce{Ag(S2O3)2^{3-}} \right]\left[ \ce{Br^{-}} \right]}{K}=\frac{(0.00532 M)(0.00532~\text{M})}{24}=0.0011 ~ \text{M} \nonumber$
Finally, the total mass of $\ce{Na2S2O3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.
Mass of $\ce{Na2S2O3}$ required to yield 0.00532 M $\ce{Ag(2S2O3)2^{3-}}$
$0.00532 \dfrac{ \text{mol} ~\ce{Ag(S2O3)2^{3-}}}{1.00~\text{L}} \times \dfrac{2~\text{mol}~\ce{S2O3^{2-}}}{1~\text{mol} ~\ce{Ag(S2O3)2^{3-}}} \times \dfrac{1~\text{mol}~\ce{Na2S2O3}}{1 ~\text{mol}~\ce{S2O3^{2-}}} \times \dfrac{158.1 ~ \text{g}~\ce{Na2S2O3}}{1 ~\text{mol}~\ce{Na2S2O3}}=1.68~\text{g} \nonumber$
Mass of $\ce{Na2S2O3}$ required to yield 0.00110 M $\ce{S2O3^{2-}}$
$0.0011 \frac{ \text{mol} ~\ce{S2O3^{2-}}}{1.00~\text{L} } \times \frac{1~\text{mol} ~ \ce{S2O3^{2-}}}{1 ~ \text{mol} ~\ce{Na2S2O3}} \times \frac{158.1 ~\text{g}~\ce{Na2S2O3}}{1~\text{mol} ~\ce{Na2S2O3}}=0.17~\text{g} \nonumber$
The mass of required to dissolve 1.00 g of AgBr in 1.00 L of water is thus 1.68 g + 0.17 g = 1.85 g
Exercise $2$
AgCl(s), silver chloride, has a very low solubility:
$\ce{AgCl (s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \quad\quad K_{ so }=1.6 \times 10^{-10} \nonumber$
Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:
$\ce{Ag^{+}(aq) +2 NH_3(aq) <=> Ag(NH3)2^{+}(aq)} \quad\quad K_{ f }=1.7 \times 10^7 \nonumber$
What mass of $\ce{NH3}$ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $\ce{Ag(NH3)^{2+}}$?
Answer
1.00 L of a solution prepared with 4.81 g NH3 dissolves 2.0 g of $\ce{AgCl}$. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.04%3A_Coupled_Equilibria.txt |
Example and Directions
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Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
common ion effecteffect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
complex ionion consisting of a central atom surrounding molecules or ions called ligands via coordinate covalent bonds
coordinate covalent bond(also, dative bond) covalent bond in which both electrons originated from the same atom
coupled equilibriasystem characterized the simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product
dissociation constant(Kd) equilibrium constant for the decomposition of a complex ion into its components
formation constant(Kf) (also, stability constant) equilibrium constant for the formation of a complex ion from its components
Lewis acidany species that can accept a pair of electrons and form a coordinate covalent bond
Lewis acid-base adductcompound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base
Lewis acid-base chemistryreactions involving the formation of coordinate covalent bonds
Lewis baseany species that can donate a pair of electrons and form a coordinate covalent bond
ligandmolecule or ion acting as a Lewis base in complex ion formation; bonds to the central atom of the complex
molar solubilitysolubility of a compound expressed in units of moles per liter (mol/L)
selective precipitationprocess in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product constant (Ksp)equilibrium constant for the dissolution of an ionic compound
15.07: Summary
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid MpXq and its ions Mm+ and Xn–:
the solubility product expression is:
The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its Ksp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.
A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.
A Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts and comprise central metal atoms or ions acting as Lewis acids bonded to molecules or ions called ligands that act as Lewis bases. The equilibrium constant for the reaction between a metal ion and ligands produces a complex ion called a formation constant; for the reverse reaction, it is called a dissociation constant.
Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.05%3A_Key_Terms.txt |
1.
Complete the changes in concentrations for each of the following reactions:
(a)
$AgI(s)⟶ Ag+(aq)+ I−(aq)x_____AgI(s)⟶ Ag+(aq)+ I−(aq)x_____$
(b)
$CaCO3(s)⟶ Ca2+(aq)+CO3 2−(aq)____xCaCO3(s)⟶ Ca2+(aq)+CO3 2−(aq)____x$
(c)
$Mg(OH)2(s)⟶ Mg2+(aq)+2 OH−(aq)x_____Mg(OH)2(s)⟶ Mg2+(aq)+2 OH−(aq)x_____$
(d)
$Mg3(PO4)2(s)⟶ 3Mg2+(aq)+ 2PO43−(aq) x _____ Mg3(PO4)2(s)⟶ 3Mg2+(aq)+ 2PO43−(aq) x _____$
(e)
$Ca5(PO4)3OH(s)⟶5Ca2+(aq)+3PO43−(aq)+OH−(aq)__________xCa5(PO4)3OH(s)⟶5Ca2+(aq)+3PO43−(aq)+OH−(aq)__________x$
2.
Complete the changes in concentrations for each of the following reactions:
(a)
$BaSO4(s)⟶Ba2+(aq)+SO42−(aq)x_____BaSO4(s)⟶Ba2+(aq)+SO42−(aq)x_____$
(b
$Ag2SO4(s)⟶2Ag+(aq)+SO42−(aq)_____xAg2SO4(s)⟶2Ag+(aq)+SO42−(aq)_____x$
(c)
$Al(OH)3(s)⟶Al3+(aq)+3OH−(aq)x_____Al(OH)3(s)⟶Al3+(aq)+3OH−(aq)x_____$
(d)
$Pb(OH)Cl(s)⟶Pb2+(aq)+OH−(aq)+Cl−(aq)_____x_____Pb(OH)Cl(s)⟶Pb2+(aq)+OH−(aq)+Cl−(aq)_____x_____$
(e)
$Ca3(AsO4)2(s)⟶3Ca2+(aq)+2AsO43−(aq)3x_____Ca3(AsO4)2(s)⟶3Ca2+(aq)+2AsO43−(aq)3x_____$
3.
How do the concentrations of Ag+ and $CrO42−CrO42−$ in a saturated solution above 1.0 g of solid Ag2CrO4 change when 100 g of solid Ag2CrO4 is added to the system? Explain.
4.
How do the concentrations of Pb2+ and S2– change when K2S is added to a saturated solution of PbS?
5.
What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?
6.
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO3, CuI, PbCO3, PbCl2, Tl2S, KClO4?
7.
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO4, CaF2, Hg2I2, MnCO3, and ZnS?
8.
Write the ionic equation for dissolution and the solubility product (Ksp) expression for each of the following slightly soluble ionic compounds:
(a) PbCl2
(b) Ag2S
(c) Sr3(PO4)2
(d) SrSO4
9.
Write the ionic equation for the dissolution and the Ksp expression for each of the following slightly soluble ionic compounds:
(a) LaF3
(b) CaCO3
(c) Ag2SO4
(d) Pb(OH)2
10.
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSiF6, 0.026 g/100 mL (contains $SiF62−SiF62−$ ions)
(b) Ce(IO3)4, 1.5 $××$ 10–2 g/100 mL
(c) Gd2(SO4)3, 3.98 g/100 mL
(d) (NH4)2PtBr6, 0.59 g/100 mL (contains $PtBr62−PtBr62−$ ions)
11.
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSeO4, 0.0118 g/100 mL
(b) Ba(BrO3)2·H2O, 0.30 g/100 mL
(c) NH4MgAsO4·6H2O, 0.038 g/100 mL
(d) La2(MoO4)3, 0.00179 g/100 mL
12.
Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF2, Hg2Cl2, PbI2, or Sn(OH)2.
13.
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
(a) KHC4H4O6
(b) PbI2
(c) Ag4[Fe(CN)6], a salt containing the $Fe(CN)64–Fe(CN)64–$ ion
(d) Hg2I2
14.
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
(a) Ag2SO4
(b) PbBr2
(c) AgI
(d) CaC2O4·H2O
15.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
(a) AgCl(s) in 0.025 M NaCl
(b) CaF2(s) in 0.00133 M KF
(c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4
(d) Zn(OH)2(s) in a solution buffered at a pH of 11.45
16.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
(a) TlCl(s) in 1.250 M HCl
(b) PbI2(s) in 0.0355 M CaI2
(c) Ag2CrO4(s) in 0.225 L of a solution containing 0.856 g of K2CrO4
(d) Cd(OH)2(s) in a solution buffered at a pH of 10.995
17.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
(a) TlCl(s) in 0.025 M TlNO3
(b) BaF2(s) in 0.0313 M KF
(c) MgC2O4 in 2.250 L of a solution containing 8.156 g of Mg(NO3)2
(d) Ca(OH)2(s) in an unbuffered solution initially with a pH of 12.700
18.
Explain why the changes in concentrations of the common ions in Exercise 15.17 can be neglected.
19.
Explain why the changes in concentrations of the common ions in Exercise 15.18 cannot be neglected.
20.
Calculate the solubility of aluminum hydroxide, Al(OH)3, in a solution buffered at pH 11.00.
21.
Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.
22.
Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of BaSO4 is possible because of its low solubility. Calculate the molar solubility of BaSO4 and the mass of barium present in 1.00 L of water saturated with BaSO4.
23.
Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 $××$ 10–3 M) of $SO42−SO42−$ because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO4 (“gyp” water) as a result or passing through soil containing gypsum, CaSO4·2H2O, meet these standards? What is the concentration of $SO42−SO42−$ in such water?
24.
Perform the following calculations:
(a) Calculate [Ag+] in a saturated aqueous solution of AgBr.
(b) What will [Ag+] be when enough KBr has been added to make [Br] = 0.050 M?
(c) What will [Br] be when enough AgNO3 has been added to make [Ag+] = 0.020 M?
25.
The solubility product of CaSO4·2H2O is 2.4 $××$ 10–5. What mass of this salt will dissolve in 1.0 L of 0.010 M $SO42−?SO42−?$
26.
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).
(a) TlCl
(b) BaF2
(c) Ag2CrO4
(d) CaC2O4·H2O
(e) the mineral anglesite, PbSO4
27.
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):
(a) AgI
(b) Ag2SO4
(c) Mn(OH)2
(d) Sr(OH)2·8H2O
(e) the mineral brucite, Mg(OH)2
28.
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
(a) AgBr: [Ag+] = 5.7 $××$ 10–7 M, [Br] = 5.7 $××$ 10–7 M
(b) CaCO3: [Ca2+] = 5.3 $××$ 10–3 M, $[CO32−][CO32−]$ = 9.0 $××$ 10–7 M
(c) PbF2: [Pb2+] = 2.1 $××$ 10–3 M, [F] = 4.2 $××$ 10–3 M
(d) Ag2CrO4: [Ag+] = 5.3 $××$ 10–5 M, 3.2 $××$ 10–3 M
(e) InF3: [In3+] = 2.3 $××$ 10–3 M, [F] = 7.0 $××$ 10–3 M
29.
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:
(a) TlCl: [Tl+] = 1.21 $××$ 10–2 M, [Cl] = 1.2 $××$ 10–2 M
(b) Ce(IO3)4: [Ce4+] = 1.8 $××$ 10–4 M, $[IO3−][IO3−]$ = 2.6 $××$ 10–13 M
(c) Gd2(SO4)3: [Gd3+] = 0.132 M, $[SO42−][SO42−]$ = 0.198 M
(d) Ag2SO4: [Ag+] = 2.40 $××$ 10–2 M, $[SO42−][SO42−]$ = 2.05 $××$ 10–2 M
(e) BaSO4: [Ba2+] = 0.500 M, $[SO42−][SO42−]$ = 4.6 $××$ 10−8 M
30.
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.)
(a) KClO4: [K+] = 0.01 M, $[ClO4−][ClO4−]$ = 0.01 M
(b) K2PtCl6: [K+] = 0.01 M, $[PtCl62−][PtCl62−]$ = 0.01 M
(c) PbI2: [Pb2+] = 0.003 M, [I] = 1.3 $××$ 10–3 M
(d) Ag2S: [Ag+] = 1 $××$ 10–10 M, [S2–] = 1 $××$ 10–13 M
31.
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.)
(a) CaCO3: [Ca2+] = 0.003 M, $[CO32−][CO32−]$ = 0.003 M
(b) Co(OH)2: [Co2+] = 0.01 M, [OH] = 1 $××$ 10–7 M
(c) CaHPO4: [Ca2+] = 0.01 M, $[HPO42−][HPO42−]$ = 2 $××$ 10–6 M
(d) Pb3(PO4)2: [Pb2+] = 0.01 M, $[PO43−][PO43−]$= 1 $××$ 10–13 M
32.
Calculate the concentration of Tl+ when TlCl just begins to precipitate from a solution that is 0.0250 M in Cl.
33.
Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
34.
Calculate the concentration of Sr2+ when SrCrO4 starts to precipitate from a solution that is 0.0025 M in CrO42–.
35.
Calculate the concentration of $PO43−PO43−$ when Ag3PO4 starts to precipitate from a solution that is 0.0125 M in Ag+.
36.
Calculate the concentration of F required to begin precipitation of CaF2 in a solution that is 0.010 M in Ca2+.
37.
Calculate the concentration of Ag+ required to begin precipitation of Ag2CO3 in a solution that is 2.50 $××$ 10–6 M in $CO32−.CO32−.$
38.
What [Ag+] is required to reduce $[CO32−][CO32−]$ to 8.2 $××$ 10–4 M by precipitation of Ag2CO3?
39.
What [F] is required to reduce [Ca2+] to 1.0 $××$ 10–4 M by precipitation of CaF2?
40.
A volume of 0.800 L of a 2 $××$ 10–4-M Ba(NO3)2 solution is added to 0.200 L of 5 $××$ 10–4 M Li2SO4. Does BaSO4 precipitate? Explain your answer.
41.
Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing NiCO3 be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO3 (Ksp = 1.36 $××$ 10–7).
(b) If the NiCO3 were a contaminant in a sample of CoCO3 (Ksp = 1.0 $××$ 10–12), what mass of CoCO3 would have been lost? Keep in mind that both NiCO3 and CoCO3 dissolve in the same solution.
42.
Iron concentrations greater than 5.4 $××$ 10–6 M in water used for laundry purposes can cause staining. What [OH] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2?
43.
A solution is 0.010 M in both Cu2+ and Cd2+. What percentage of Cd2+ remains in the solution when 99.9% of the Cu2+ has been precipitated as CuS by adding sulfide?
44.
A solution is 0.15 M in both Pb2+ and Ag+. If Cl is added to this solution, what is [Ag+] when PbCl2 begins to precipitate?
45.
What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the Ksp values given in Appendix J.)
(a) $Hg22+Hg22+$ and Cu2+
(b) $SO42−SO42−$ and Cl
(c) Hg2+ and Co2+
(d) Zn2+ and Sr2+
(e) Ba2+ and Mg2+
(f) $CO32−CO32−$ and OH
46.
A solution contains 1.0 $××$ 10–5 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?
47.
A solution contains 1.0 $××$ 10–2 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?
48.
The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, K2C2O4, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC2O4·H2O. It is necessary to remove all but 1.0% of the Ca2+ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca2+ per 100 mL of serum, what mass of K2C2O4 is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the Ksp value for CaC2O4 in serum is the same as in water.)
49.
About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca3(PO4)2. The normal mid range calcium content excreted in the urine is 0.10 g of Ca2+ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?
50.
The pH of normal urine is 6.30, and the total phosphate concentration $([PO43−]Exercise 15.49 for additional information.)$
51.
Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:
$Mg 2+ ( a q ) + Ca(OH) 2 ( a q ) ⟶ Mg(OH) 2 ( s ) + Ca 2+ ( a q ) Mg 2+ ( a q ) + Ca(OH) 2 ( a q ) ⟶ Mg(OH) 2 ( s ) + Ca 2+ ( a q )$
$Mg(OH) 2 ( s ) + 2HCl( a q ) ⟶ MgCl 2 ( s ) + 2H 2 O ( l ) Mg(OH) 2 ( s ) + 2HCl( a q ) ⟶ MgCl 2 ( s ) + 2H 2 O ( l )$
$MgCl 2 ( l ) → electrolysis Mg ( s ) + Cl 2 ( g ) MgCl 2 ( l ) → electrolysis Mg ( s ) + Cl 2 ( g )$
Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium as Mg2+(aq) by mass. What mass, in kilograms, of Ca(OH)2 is required to precipitate 99.9% of the magnesium in 1.00 $××$ 103 L of sea water?
52.
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb2+ and Fe2+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H2S ([H2S] = 0.10 M). What concentrations of Pb2+ and Fe2+ remain in the solution? For a saturated solution of H2S we can use the equilibrium:
$H 2 S ( a q ) + 2H 2 O ( l ) ⇌ 2H 3 O + ( a q ) + S 2− ( a q ) K = 1.0 × 1 0 − 26 H 2 S ( a q ) + 2H 2 O ( l ) ⇌ 2H 3 O + ( a q ) + S 2− ( a q ) K = 1.0 × 1 0 − 26$
(Hint: The $[H3O+][H3O+]$ changes as metal sulfides precipitate.)
53.
Perform the following calculations involving concentrations of iodate ions:
(a) The iodate ion concentration of a saturated solution of La(IO3)3 was found to be 3.1 $××$ 10–3 mol/L. Find the Ksp.
(b) Find the concentration of iodate ions in a saturated solution of Cu(IO3)2 (Ksp = 7.4 $××$ 10–8).
54.
Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 $××$ 10–13).
55.
How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 $××$ 10–15)?
56.
Use the simulation from the earlier Link to Learning to complete the following exercise. Using 0.01 g CaF2, give the Ksp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.
57.
How many grams of Milk of Magnesia, Mg(OH)2 (s) (58.3 g/mol), would be soluble in 200 mL of water. Ksp = 7.1 $××$ 10–12. Include the ionic reaction and the expression for Ksp in your answer. (Kw = 1 $××$ 10–14 = [H3O+][OH])
58.
Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 $××$ 10–5, what is the Ksp value for LQ?
59.
The carbonate ion concentration is gradually increased in a solution containing equal concentrations of the divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will precipitate first? Which will precipitate last? Explain.
(a) $MgCO3 •3H2OKsp=1×10−5MgCO3 •3H2OKsp=1×10−5$
(b) $CaCO3Ksp=8.7×10−9CaCO3Ksp=8.7×10−9$
(c) $SrCO3Ksp=7×10−10SrCO3Ksp=7×10−10$
(d) $BaCO3Ksp=1.6×10−9BaCO3Ksp=1.6×10−9$
(e) $MnCO3Ksp=8.8×10−11MnCO3Ksp=8.8×10−11$
60.
How many grams of Zn(CN)2(s) (117.44 g/mol) would be soluble in 100 mL of H2O? Include the balanced reaction and the expression for Ksp in your answer. The Ksp value for Zn(CN)2(s) is 3.0 $××$ 10–16.
61.
Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2?
62.
Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
63.
Explain why the addition of NH3 or HNO3 to a saturated solution of Ag2CO3 in contact with solid Ag2CO3 increases the solubility of the solid.
64.
Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 0.150 L of 0.100 NH3(aq).
65.
Explain why addition of NH3 or HNO3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH)2 increases the solubility of the solid.
66.
Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion $AlF63−AlF63−$ the dissociation reaction is:
$AlF63−⇌Al3++6F−AlF63−⇌Al3++6F−$ and $Kd=[Al3+][F−]6[AlF63−]=2×10−24Kd=[Al3+][F−]6[AlF63−]=2×10−24$
Calculate the value of the formation constant, Kf, for $AlF63−.AlF63−.$
67.
Using the value of the formation constant for the complex ion $Co(NH3)62+,Co(NH3)62+,$ calculate the dissociation constant.
68.
Using the dissociation constant, Kd = 7.8 $××$ 10–18, calculate the equilibrium concentrations of Cd2+ and CN in a 0.250-M solution of $Cd(CN)42−.Cd(CN)42−.$
69.
Using the dissociation constant, Kd = 3.4 $××$ 10–15, calculate the equilibrium concentrations of Zn2+ and OH in a 0.0465-M solution of $Zn(OH)42−.Zn(OH)42−.$
70.
Using the dissociation constant, Kd = 2.2 $××$ 10–34, calculate the equilibrium concentrations of Co3+ and NH3 in a 0.500-M solution of $Co(NH3)63+.Co(NH3)63+.$
71.
Using the dissociation constant, Kd = 1 $××$ 10–44, calculate the equilibrium concentrations of Fe3+ and CN in a 0.333 M solution of $Fe(CN)63−.Fe(CN)63−.$
72.
Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 $××$ 10–2 mol of silver cyanide, AgCN.
73.
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 $××$ 10–3 mol of silver bromide.
74.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na2S2O3·5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $Ag(S2O3)23−Ag(S2O3)23−$ (Kf = 4.7 $××$ 1013)?
75.
We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
76.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) $CO2+OH−⟶HCO3−CO2+OH−⟶HCO3−$
(b) $B(OH)3+OH−⟶B(OH)4−B(OH)3+OH−⟶B(OH)4−$
(c) $I−+I2⟶I3−I−+I2⟶I3−$
(d) $AlCl3+Cl−⟶AlCl4−AlCl3+Cl−⟶AlCl4−$ (use Al-Cl single bonds)
(e) $O2−+SO3⟶SO42−O2−+SO3⟶SO42−$
77.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) $CS2+SH−⟶HCS3−CS2+SH−⟶HCS3−$
(b) $BF3+F−⟶BF4−BF3+F−⟶BF4−$
(c) $I−+SnI2⟶SnI3−I−+SnI2⟶SnI3−$
(d) $Al(OH)3+OH−⟶Al(OH)4−Al(OH)3+OH−⟶Al(OH)4−$
(e) $F−+SO3⟶SFO3−F−+SO3⟶SFO3−$
78.
Using Lewis structures, write balanced equations for the following reactions:
(a) $HCl(g)+PH3(g)⟶HCl(g)+PH3(g)⟶$
(b) $H3O++CH3−⟶H3O++CH3−⟶$
(c) $CaO+SO3⟶CaO+SO3⟶$
(d) $NH4++C2H5O−⟶NH4++C2H5O−⟶$
79.
Calculate $[HgCl42−][HgCl42−]$ in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl2 solution.
80.
In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN goes to completion, producing the $Ag(CN)2−Ag(CN)2−$ complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of $Ag(CN)2−.Ag(CN)2−.$ How many grams of NaCN were in the original sample?
81.
What are the concentrations of Ag+, CN, and $Ag(CN)2−Ag(CN)2−$ in a saturated solution of AgCN?
82.
In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF3 or SbF5). Write balanced chemical equations for the reaction of pure HNO3 with pure HF and of pure HF with BF3.
83.
The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH2, and a carboxylic acid group, –CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3.
(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.
(b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH2 and $−CO2−−CO2−$ groups.)
84.
Boric acid, H3BO3, is not a Brønsted-Lowry acid but a Lewis acid.
(a) Write an equation for its reaction with water.
(b) Predict the shape of the anion thus formed.
(c) What is the hybridization on the boron consistent with the shape you have predicted?
85.
A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
86.
Calculate the equilibrium concentration of Ni2+ in a 1.0-M solution [Ni(NH3)6](NO3)2.
87.
Calculate the equilibrium concentration of Zn2+ in a 0.30-M solution of $Zn(CN)42−.Zn(CN)42−.$
88.
Calculate the equilibrium concentration of Cu2+ in a solution initially with 0.050 M Cu2+ and 1.00 M NH3.
89.
Calculate the equilibrium concentration of Zn2+ in a solution initially with 0.150 M Zn2+ and 2.50 M CN.
90.
Calculate the Fe3+ equilibrium concentration when 0.0888 mole of K3[Fe(CN)6] is added to a solution with 0.0.00010 M CN.
91.
Calculate the Co2+ equilibrium concentration when 0.010 mole of [Co(NH3)6](NO3)2 is added to a solution with 0.25 M NH3. Assume the volume is 1.00 L.
92.
Calculate the molar solubility of Sn(OH)2 in a buffer solution containing equal concentrations of NH3 and $NH4+.NH4+.$
93.
Calculate the molar solubility of Al(OH)3 in a buffer solution with 0.100 M NH3 and 0.400 M $NH4+.NH4+.$
94.
What is the molar solubility of CaF2 in a 0.100-M solution of HF? Ka for HF = 6.4 $××$ 10–4.
95.
What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for $HSO4−HSO4−$ = 1.2 $××$ 10–2.
96.
What is the molar solubility of Tl(OH)3 in a 0.10-M solution of NH3?
97.
What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?
98.
A solution of 0.075 M CoBr2 is saturated with H2S ([H2S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?
$CoS(s)⇌ Co2+(aq) +S2−(aq) Ksp=2.3× 1027CoS(s)⇌ Co2+(aq) +S2−(aq) Ksp=2.3× 1027$
$H2S(aq)+2 H2O(l)⇌2H3O+ (aq) +S2− (aq) K=8.9×10−27H2S(aq)+2 H2O(l)⇌2H3O+ (aq) +S2− (aq) K=8.9×10−27$
99.
A 0.125-M solution of Mn(NO3)2 is saturated with H2S ([H2S] = 0.10 M). At what pH does MnS begin to precipitate?
$MnS(s)⇌Mn2+(aq)+S2−(aq) Ksp=2.3×10−13MnS(s)⇌Mn2+(aq)+S2−(aq) Ksp=2.3×10−13$
$H 2 S ( a q ) + 2 H 2 O ( l ) ⇌ 2 H 3 O + ( a q ) + S 2− ( a q ) K = 1.0 × 10 −26 H 2 S ( a q ) + 2 H 2 O ( l ) ⇌ 2 H 3 O + ( a q ) + S 2− ( a q ) K = 1.0 × 10 −26$
100.
Both AgCl and AgI dissolve in NH3.
(a) What mass of AgI dissolves in 1.0 L of 1.0 M NH3?
(b) What mass of AgCl dissolves in 1.0 L of 1.0 M NH3?
101.
The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Solve the following problem:
$MgF 2 ( s ) ⇌ Mg 2+ ( a q ) + 2 F − ( a q ) MgF 2 ( s ) ⇌ Mg 2+ ( a q ) + 2 F − ( a q )$
In a saturated solution of MgF2 at 18 °C, the concentration of Mg2+ is 1.21 $××$ 10–3 M. The equilibrium is represented by the preceding equation.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18 °C.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2 solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 $××$ 10–3-M solution of Mg(NO3)2 is mixed with 200.0 mL of a 2.00 $××$ 10–3-M solution of NaF at 18 °C. Show the calculations to support your prediction.
(d) At 27 °C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 $××$ 10–3 M. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.
102.
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: CuCl, CaCO3, MnS, PbBr2, CaF2? Explain your answer.
103.
Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: AgBr, BaF2, Ca3(PO4)2, ZnS, PbI2? Explain your answer.
104.
What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH when each of the following are added to a mixture of solid Mg(OH)2 and water at equilibrium?
(a) MgCl2
(b) KOH
(c) HClO4
(d) NaNO3
(e) Mg(OH)2
105.
What is the effect on the amount of CaHPO4 that dissolves and the concentrations of Ca2+ and $HPO42−HPO42−$ when each of the following are added to a mixture of solid CaHPO4 and water at equilibrium?
(a) CaCl2
(b) HCl
(c) KClO4
(d) NaOH
(e) CaHPO4
106.
Identify all chemical species present in an aqueous solution of Ca3(PO4)2 and list these species in decreasing order of their concentrations. (Hint: Remember that the $PO43−PO43−$ ion is a weak base.) | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/15%3A_Equilibria_of_Other_Reaction_Classes/15.08%3A_Exercises.txt |
Electrochemistry deals with chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer, and so they are oxidation-reduction (or redox) reactions. Many metals may be purified or electroplated using electrochemical methods. Devices such as automobiles, smartphones, electronic tablets, watches, pacemakers, and many others use batteries for power. Batteries use chemical reactions that produce electricity spontaneously and that can be converted into useful work. All electrochemical systems involve the transfer of electrons in a reacting system. In many systems, the reactions occur in a region known as the cell, where the transfer of electrons occurs at electrodes.
16: Electrochemistry
Another chapter in this text introduced the chemistry of reduction-oxidation (redox) reactions. This important reaction class is defined by changes in oxidation states for one or more reactant elements, and it includes a subset of reactions involving the transfer of electrons between reactant species. Around the turn of the nineteenth century, chemists began exploring ways these electrons could be transferred indirectly via an external circuit rather than directly via intimate contact of redox reactants. In the two centuries since, the field of electrochemistry has evolved to yield significant insights on the fundamental aspects of redox chemistry as well as a wealth of technologies ranging from industrial-scale metallurgical processes to robust, rechargeable batteries for electric vehicles (Figure \(1\)). In this chapter, the essential concepts of electrochemistry will be addressed.
16.02: Review of Redox Chemistry
Learning Objectives
By the end of this section, you will be able to:
• Describe defining traits of redox chemistry
• Identify the oxidant and reductant of a redox reaction
• Balance chemical equations for redox reactions using the half-reaction method
Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier text chapter (see chapter on reaction stoichiometry). Readers wishing additional review are referred to the text chapter on reaction stoichiometry.
Oxidation Numbers
By definition, a redox reaction is one that entails changes in oxidation number (or oxidation state) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description, the oxidation number of an atom in an element is equal to zero. For an atom in a compound, the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic. Consequential to these rules, the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.
Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements’ oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl, is comprised of Na+ cations and Cl anions, and so oxidation numbers for sodium and chlorine are, +1 and −1, respectively. Calcium fluoride, CaF2, is comprised of Ca2+ cations and F anions, and so oxidation numbers for calcium and fluorine are, +2 and −1, respectively.
Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent O−H bonds. The shared electrons comprising an O−H bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral.
Imagine that the polarization of shared electrons within the O−H bonds of water were 100% complete—the result would be transfer of electrons from H to O, and water would be an ionic compound comprised of O2− anions and H+ cations. And so, the oxidations numbers for oxygen and hydrogen in water are −2 and +1, respectively. Applying this same logic to carbon tetrachloride, CCl4, yields oxidation numbers of +4 for carbon and −1 for chlorine. In the nitrate ion, $\ce{NO3^{-}}$, the oxidation number for nitrogen is +5 and that for oxygen is −2, summing to equal the 1− charge on the molecule:
$(1 N \text { atom })\left(\frac{+5}{ N \text { atom }}\right)+(3 O \text { atoms })\left(\frac{-2}{ O \text { atom }}\right)=+5+-6=-1 \nonumber$
Balancing Redox Equations
The unbalanced equation below describes the decomposition of molten sodium chloride:
$\ce{NaCl(l) -> Na(l) + Cl2(g)} \tag{unbalanced}$
This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes reduction) and that for Cl is increased from −1 to 0 (it undergoes oxidation). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na:
$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)} \tag{balanced}$
Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, H+(aq) and OH(aq), as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the half-reaction method is helpful.
Balancing REDOX reactions via the half-reaction Method
This approach involves the following steps:
1. Write skeletal equations for the oxidation and reduction half-reactions.
2. Balance each half-reaction for all elements except H and O.
3. Balance each half-reaction for O by adding H2O.
4. Balance each half-reaction for H by adding H+.
5. Balance each half-reaction for charge by adding electrons.
6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
7. Add the two half-reactions and simplify.
8. If the reaction takes place in a basic medium, add OH ions the equation obtained in step 7 to neutralize the H+ ions (add in equal numbers to both sides of the equation) and simplify.
The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.
Example $1$: Balancing Equations for Redox Reactions in Acidic Solutions
Write the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.
Solution
Following the steps of the half-reaction method:
Step 1: Write skeletal equations for the oxidation and reduction half-reactions.
\begin{align*} &\text { oxidation: } \quad \ce{Cu (s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{HNO3(aq) -> NO(g)} \end{align*} \nonumber
Step 2: Balance each half-reaction for all elements except $\ce{H}$ and $\ce{O}$.
\begin{align*} &\text { oxidation: } \quad\ce{Cu(s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad\ce{HNO3(aq) -> NO(g)} \end{align*} \nonumber
Step 3: Balance each half-reaction for $\ce{O}$ by adding $\ce{H2O}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cu (s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{HNO3(aq) -> NO(g) + \mathbf{2 H2O(l)}} \end{align*} \nonumber
Step 4: Balance each half-reaction for $\ce{H}$ by adding $\ce{H^{+}}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cu(s) -> Cu^{2+}(aq)} \[4pt] &\text { reduction: } \quad \ce{\mathbf{3 H^{+}(aq)} + HNO3(aq) -> NO(g) + 2 H2O(l)} \end{align*} \nonumber
Step 5: Balance each half-reaction for charge by adding electrons.
\begin{align*} &\text { oxidation: } \quad \ce{Cu(s) -> Cu^{2+}(aq) + \mathbf{ 2 e^{-}}} \[4pt] &\text {reduction: } \quad \ce{\mathbf{3 e^{-}} + 3 H^{+}(aq) + HNO3(aq) -> NO(g) + 2 H2O(l)} \end{align*} \nonumber
Step 6: If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
\begin{align*} &\text { oxidation }(\times 3): \quad \ce{\mathbf{3} Cu(s) -> \mathbf{3} Cu^{2+}(aq) + \mathbf{6} \cancel{2} e^{-}} \[4pt] &\text {reduction }(\times 2): \quad \ce{\mathbf{6} \cancel{3} e^{-} + \mathbf{6} \cancel{3} H^{+}(aq) + \mathbf{2} HNO3(aq) -> \mathbf{2} NO(g) + \mathbf{4} \cancel{2} H2O (l)} \end{align*} \nonumber
Step 7: Add the two half-reactions and simplify.
\begin{align*} \ce{3 Cu(s) + \cancel{6 e^{-}} + 6 H^{+}(aq) +2 HNO3(aq) &-> 3 Cu^{2+}(aq) + \cancel{6 e^{-}} + 2 NO(g) + 4 H2O (l)} \[4pt] \ce{3 Cu(s) + 6 H^{+}(aq) + 2 HNO3(aq) &-> 3 Cu^{2+}(aq) + 2 NO(g) + 4 H2O (l)} \end{align*} \nonumber
Step 8: If the reaction takes place in a basic medium, add $\ce{OH^{−}}$ ions the equation obtained in step 7 to neutralize the $\ce{H^{+}}$ ions (add in equal numbers to both sides of the equation) and simplify.
This step not necessary since the solution is stipulated to be acidic.
The balanced equation for the reaction in an acidic solution is then
$\ce{3 Cu(s) + 6 H^{+}(aq) + 2 HNO3(aq) -> 3 Cu^{2+}(aq) + 2 NO(g) + 4 H2O(l)} \nonumber$
Exercise $1$
The reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.
Answer
$\ce{Cu(s) + 2 H^{+}(aq) + 2 HNO3(aq) -> Cu^{2+}(aq) + 2 NO2(g) + 2 H2O(l)} \nonumber$
Example $2$: Balancing Equations for Redox Reactions in Basic Solutions
Write the balanced equation representing reaction between aqueous permanganate ion, , and solid chromium(III) hydroxide, Cr(OH)3, to yield solid manganese(IV) oxide, MnO2, and aqueous chromate ion, The reaction takes place in a basic solution.
Solution
Step 1: Write skeletal equations for the oxidation and reduction half-reactions.
\begin{align*} &\text { oxidation: } \quad \ce{Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s)} \end{align*} \nonumber
Step 2: Balance each half-reaction for all elements except $\ce{H}$ and $\ce{O}$.
\begin{align*} &\text { oxidation: } \quad \ce{Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s)} \end{align*} \nonumber
Step 3: Balance each half-reaction for $\ce{O}$ by adding $\ce{H2O}$.
\begin{align*} &\text { oxidation: } \quad \ce{\mathbf{H2O(l)} + Cr(OH)3(s) -> CrO4^{2-}(aq)} \[4pt] &\text { reduction: } \quad \ce{MnO4^{-}(aq) -> MnO2(s) + \mathbf{2 H2O(l)}} \end{align*} \nonumber
Step 4: Balance each half-reaction for $\ce{H}$ by adding $\ce{H^{+}}$.
\begin{align*} &\text { oxidation: } \quad \ce{H2O(l) + Cr(OH)3(s) -> CrO4^{2-}(aq) + \mathbf{5H^{+}(aq)} } \[4pt] &\text { reduction: } \quad \ce{\mathbf{4H^{+}(aq)} + MnO4^{-}(aq) -> MnO2(s) + 2 H2O(l)} \end{align*} \nonumber
Step 5: Balance each half-reaction for charge by adding electrons.
\begin{align*} &\text { oxidation: } \quad \ce{H2O(l) + Cr(OH)3(s) -> CrO4^{2-}(aq) + 5H^{+}(aq) + \mathbf{3 e^{-}}} \[4pt] &\text { reduction: } \quad \ce{\mathbf{3 e^{-}} + 4H^{+}(aq) + MnO4^{-}(aq) -> MnO2(s) + 2 H2O(l)} \end{align*} \nonumber
Step 6: If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
This step is not necessary since the number of electrons is already in balance.
Step 7: Add the two half-reactions and simplify.
\begin{align*} \ce{\cancel{H2O(l)} + Cr(OH)3(s) &-> CrO4^{2-}(aq) + \cancelto{1}{5}H^{+}(aq) + \bcancel{3 e^{-}}} \[4pt] + \quad \quad \ce{\bcancel{3 e^{-}} + \cancel{4H^{+}(aq)} + MnO4^{-}(aq) &-> MnO2(s) + \cancelto{1}{2} H2O(l)} \[4pt] \hline \ce{Cr(OH)3(s) + MnO4^{-}(aq) &-> CrO4^{2-}(aq) + MnO2(s) + H^{+}(aq) + H2O(l)} \end{align*} \nonumber
Step 8: If the reaction takes place in a basic medium, add $\ce{OH^{−}}$ ions the equation obtained in step 7 to neutralize the $\ce{H^{+}}$ ions (add in equal numbers to both sides of the equation) and simplify.
$\ce{OH^{-}(aq) + Cr(OH)3(s) + MnO4^{-}(aq) -> CrO4^{2-}(aq) + MnO2(s) + \cancel{H^{+}(aq)} + H2O(l) + \cancel{OH^{-}(aq)}} \nonumber$
after neutralizing the $\ce{H^{+}}$, we get the balanced equation for the reaction in a basic solution
$\ce{OH^{-}(aq) + Cr(OH)3(s) + MnO4^{-}(aq) -> CrO4^{2-}(aq) + MnO2(s) + 2H2O(l)} \nonumber$
Exercise $2$
Aqueous permanganate ion may also be reduced using aqueous bromide ion, $\ce{Br^{-}}$, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, $\ce{BrO3^{-}}$. Write the balanced equation for this reaction occurring in a basic medium.
Answer
$\ce{H2O(l) + 2 MnO4^{-}(aq) + Br^{-}(aq) -> 2 MnO2(s) + BrO3^{-}(aq) + 2 OH^{-}(aq)} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the function of a galvanic cell and its components
• Use cell notation to symbolize the composition and construction of galvanic cells
As demonstration of spontaneous chemical change, Figure \(1\): shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.
These observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, Cu2+(aq), which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:
Consider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells, and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells).
A galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure \(2\). The cell is comprised of two half-cells, each containing the redox conjugate pair (“couple”) of a single reactant. The half-cell shown at the left contains the Cu(0)/Cu(II) couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the Ag(I)/Ag(0) couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode. By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell’s reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge. The spontaneous reaction in this cell produces Cu2+ cations in the anode half-cell and consumes Ag+ ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of Cu2+ in the anode half-cell are balanced by an influx of NO3 from the salt bridge, while a flow of Na+ into the cathode half-cell compensates for the decreasing Ag+ concentration.
Cell Notation
Abbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics, and they are written following a few guidelines:
• The relevant components of each half-cell are represented by their chemical formulas or element symbols
• All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas
• By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode
A verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure \(2\): consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:
Consider a different galvanic cell (see Figure \(3\)) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:
In this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II) chloride, immersed in which is a platinum cathode. The cell schematic is then written as
Notice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance (Pt) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell’s redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes.
Example \(1\): Writing Galvanic Cell Schematics
A galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a 1 M CrCl3 solution and another in which a copper wire is immersed in 1 M CuCl2. Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode half-reaction, the cathode half-reaction, and the overall cell reaction.
Solution
Since the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:
The half-reactions for this cell are
Multiplying to make the number of electrons lost by Cr and gained by Cu2+ equal yields
Adding the half-reaction equations and simplifying yields an equation for the cell reaction:
Exercise \(1\)
Omitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below.
Answer | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.03%3A_Galvanic_Cells.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe and relate the definitions of electrode and cell potentials
• Interpret electrode potentials in terms of relative oxidant and reductant strengths
• Calculate cell potentials and predict redox spontaneity using standard electrode potentials
Unlike the spontaneous oxidation of copper by aqueous silver(I) ions described previously, immersing a copper wire in an aqueous solution of lead(II) ions yields no reaction. The two species, Ag+(aq) and Pb2+(aq), thus show a distinct difference in their redox activity towards copper: the silver ion spontaneously oxidized copper, but the lead ion did not. Electrochemical cells permit this relative redox activity to be quantified by an easily measured property, potential. This property is more commonly called voltage when referenced in regard to electrical applications, and it is a measure of energy accompanying the transfer of charge. Potentials are measured in the volt unit, defined as one joule of energy per one coulomb of charge, V = J/C.
When measured for purposes of electrochemistry, a potential reflects the driving force for a specific type of charge transfer process, namely, the transfer of electrons between redox reactants. Considering the nature of potential in this context, it is clear that the potential of a single half-cell or a single electrode can’t be measured; “transfer” of electrons requires both a donor and recipient, in this case a reductant and an oxidant, respectively. Instead, a half-cell potential may only be assessed relative to that of another half-cell. It is only the difference in potential between two half-cells that may be measured, and these measured potentials are called cell potentials, Ecell, defined as
$E_{\text {cell }}= E_{\text {cathode }}- E_{\text {anode }} \nonumber$
where $E_{cathode}$ and $E_{anode}$ are the potentials of two different half-cells functioning as specified in the subscripts. As for other thermodynamic quantities, the standard cell potential, E°cell, is a cell potential measured when both half-cells are under standard-state conditions (1 M concentrations, 1 bar pressures, 298 K):
$E_{\text {cell }}^{\circ}= E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \nonumber$
To simplify the collection and sharing of potential data for half-reactions, the scientific community has designated one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of exactly 0 V. This half-cell is the standard hydrogen electrode (SHE) and it is based on half-reaction below:
$\ce{2 H^{+}(aq) + 2 e^{-} -> H2(g)} \nonumber$
A typical SHE contains an inert platinum electrode immersed in precisely 1 M aqueous H+ and a stream of bubbling H2 gas at 1 bar pressure, all maintained at a temperature of 298 K (see Figure $1$).
The assigned potential of the SHE permits the definition of a conveniently measured potential for a single half-cell. The electrode potential (EX) for a half-cell X is defined as the potential measured for a cell comprised of X acting as cathode and the SHE acting as anode:
\begin{aligned} & E_{\text {cell }}= E_{ X }- E_{ SHE } \[4pt] & E_{ SHE }=0 V \text { (defined) } \[4pt] & E_{\text {cell }}= E_{ X } \end{aligned} \nonumber
When the half-cell X is under standard-state conditions, its potential is the standard electrode potential, E°X. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials.
This approach to measuring electrode potentials is illustrated in Figure $2$: , which depicts a cell comprised of an SHE connected to a copper(II)/copper(0) half-cell under standard-state conditions. A voltmeter in the external circuit allows measurement of the potential difference between the two half-cells. Since the Cu half-cell is designated as the cathode in the definition of cell potential, it is connected to the red (positive) input of the voltmeter, while the designated SHE anode is connected to the black (negative) input. These connections insure that the sign of the measured potential will be consistent with the sign conventions of electrochemistry per the various definitions discussed above. A cell potential of +0.337 V is measured, and so
$E_{\text {cell }}^{\circ}= E_{\ce{Cu}}^{\circ}=+0.337 V \nonumber$
Tabulations of E° values for other half-cells measured in a similar fashion are available as reference literature to permit calculations of cell potentials and the prediction of the spontaneity of redox processes.
Table $1$ provides a listing of standard electrode potentials for a selection of half-reactions in numerical order, and a more extensive alphabetical listing is given in Appendix L.
Table $1$: Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
$\ce{F2(g) + 2e^{-}-> 2F^{-}(aq)}$ +2.866
$\ce{PbO2(s) + SO4^{2-}(aq) + 4H^{+}(aq) + 2e^{-}-> PbSO4(s) + 2H2O(l)}$ +1.69
$\ce{MnO4^{-}(aq) + 8H^{+}(aq) + 5e^{-}-> Mn^{2+}(aq) + 4H2O(l)}$ +1.507
$\ce{Au^{3+}(aq) + 3e^{-}-> Au(s)}$ +1.498
$\ce{Cl2(g) + 2e^{-}-> 2Cl^{-}(aq)}$ +1.35827
$\ce{O2(g) + 4H^{+}(aq) + 4e^{-}-> 2H2O(l)}$ +1.229
$\ce{Pt^{2+}(aq) + 2e^{-}-> Pt(s)}$ +1.20
$\ce{Br2(aq) + 2e^{-}-> 2Br^{-}(aq)}$ +1.0873
$\ce{Ag^{+}(aq) + e^{-}-> Ag(s)}$ +0.7996
$\ce{Hg2^{2+}(aq) + 2e^{-}-> 2Hg(l)}$ +0.7973
$\ce{Fe^{3+}(aq) + e^{-} -> Fe^{2+}(aq)}$ +0.771
$\ce{MnO4^{-}(aq) + 2H2O(l) + 3e^{-}-> MnO2(s) + 4OH^{-}(aq)}$ +0.558
$\ce{I2(s) + 2e^{-}-> 2I^{-}(aq)}$ +0.5355
$\ce{NiO2(s) + 2H2O(l) + 2e^{-}-> Ni(OH)2(s) + 2OH^{-}(aq)}$ +0.49
$\ce{Cu^{2+}(aq) + 2e^{-}-> Cu(s)}$ +0.34
$\ce{Hg2Cl2(s) + 2e^{-}-> 2Hg(l) + 2Cl^{-}(aq)}$ +0.26808
$\ce{AgCl(s) + e^{-}-> Ag(s) + Cl^{-}(aq)}$ +0.22233
$\ce{Sn^{4+}(aq) + 2e^{-}-> Sn^{2+}(aq)}$ +0.151
$\ce{2H^{+}(aq) + 2e^{-}-> H2(g)}$ 0.00
$\ce{Pb^{2+}(aq) + 2e^{-}-> Pb(s)}$ -0.1262
$\ce{Sn^{2+}(aq) + 2e^{-}-> Sn(s)}$ -0.1375
$\ce{Ni^{2+}(aq) + 2e^{-}-> Ni(s)}$ -0.257
$\ce{Co^{2+}(aq) + 2e^{-}-> Co(s)}$ -0.28
$\ce{PbSO4(s) + 2e^{-}-> Pb(s) + SO4^{2-}(aq)}$ -0.3505
$\ce{Cd^{2+}(aq) + 2e^{-}-> Cd(s)}$ -0.4030
$\ce{Fe^{2+}(aq) + 2e^{-}-> Fe(s)}$ -0.447
$\ce{Cr^{3+}(aq) + 3e^{-}-> Cr(s)}$ -0.744
$\ce{Mn^{2+}(aq) + 2e^{-}-> Mn(s)}$ -1.185
$\ce{Zn(OH)2(s) + 2e^{-}-> Zn(s) + 2OH^{-}(aq)}$ -1.245
$\ce{Zn^{2+}(aq) + 2e^{-}-> Zn(s)}$ -0.7618
$\ce{Al^{3+}(aq) + 3e^{-}-> Al(s)}$ -1.662
$\ce{Mg2(aq) + 2e^{-}-> Mg(s)}$ -2.372
$\ce{Na^{+}(aq) + e^{-}-> Na(s)}$ -2.71
$\ce{Ca^{2+}(aq) + 2e^{-}-> Ca(s)}$ -2.868
$\ce{Ba^{2+}(aq) + 2e^{-}-> Ba(s)}$ -2.912
$\ce{K^{+}(aq) + e^{-}-> K(s)}$ -2.931
$\ce{K^{+}(aq) + e^{-}-> K(s)}$ -3.04
Example $1$: Calculating Standard Cell Potentials
What is the standard potential of the galvanic cell shown in Figure 17.3?
Solution
The cell in Figure 17.3 is galvanic, the spontaneous cell reaction involving oxidation of its copper anode and reduction of silver(I) ions at its silver cathode:
\begin{align*} & \text {anode: } \quad && \ce{Cu(s) -> Cu^{2+} + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{2Ag^{+}(aq) + 2 e^{-} -> 2Ag(s)}\[4pt] \hline &\text { cell: } \quad && \ce{Cu(s) +2Ag^{+}(aq) -> Cu^{2+} + 2Ag(s) } && E^o_{\text {cell }} = 0\,\text{V} \end{align*} \nonumber
The standard cell potential computed as
\begin{aligned} E_{\text {cell }}^{\circ} & = E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \ & = E_{ Ag }^{\circ}- E_{ Cu }^{\circ} \ & =0.7996 V -0.34 V \ & =+0.46 V \end{aligned} \nonumber
Exercise $1$
What is the standard cell potential expected if the silver cathode half-cell in Figure 17.3 is replaced with a lead half-cell:
$\ce{Pb^{2+}(aq) + 2 e^{-} -> Pb(s)} \nonumber$
Answer
−0. 47 V
Intrepreting Electrode and Cell Potentials
Thinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations in Example $1$ have just shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness (“strength”) with which aqueous Ag+ and Pb2+ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, E°. Since by convention electrode potentials are for reduction processes, an increased value of $E^o$ corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.
Applying this logic to the numerically ordered listing of standard electrode potentials in Table $1$ shows this listing to be likewise in order of the oxidizing strength of the half-reaction’s reactant species, decreasing from strongest oxidant (most positive E°) to weakest oxidant (most negative E°). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, E°cell is positive when $E^o_{\text{cathode}} > E^o_{\text{anode}}$, and so any redox reaction in which the oxidant’s entry is above the reductant’s entry is predicted to be spontaneous.
Reconsideration of the two redox reactions in Example $1$ provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by Ag+ is predicted to be spontaneous (E°cathode > E°anode and so E°cell > 0). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by Pb2+ is nonspontaneous (E°cathode < E°anode and so E°cell < 0).
Recalling the chapter on thermodynamics, the spontaneities of the forward and reverse reactions of a reversible process show a reciprocal relationship: if a process is spontaneous in one direction, it is non-spontaneous in the opposite direction. As an indicator of spontaneity for redox reactions, the potential of a cell reaction shows a consequential relationship in its arithmetic sign. The spontaneous oxidation of copper by lead(II) ions is not observed,
$\ce{Cu(s) + Pb^2+(aq) -> Cu^2+(aq) + Pb(s)} \hspace{20px} E^\circ_\mathrm{forward}=\mathrm{−0.47\: V (negative, non-spontaneous)} \nonumber$
and so the reverse reaction, the oxidation of lead by copper(II) ions, is predicted to occur spontaneously:
$\ce{Pb(s) + Cu^2+(aq) -> Pb^2+(aq) + Cu(s)} \hspace{20px} E^\circ_\mathrm{forward}=\mathrm{+0.47\: V (positive, spontaneous)} \nonumber$
Note that reversing the direction of a redox reaction effectively interchanges the identities of the cathode and anode half-reactions, and so the cell potential is calculated from electrode potentials in the reverse subtraction order than that for the forward reaction. In practice, a voltmeter would report a potential of −0.47 V with its red and black inputs connected to the Pb and Cu electrodes, respectively. If the inputs were swapped, the reported voltage would be +0.47 V.
Example $2$: Predicting Redox Spontaneity
Are aqueous iron(II) ions predicted to spontaneously oxidize elemental chromium under standard state conditions? Assume the half-reactions to be those available in Table $1$.
Solution
Referring to the tabulated half-reactions, the redox reaction in question can be represented by the equations below:
$\ce{Cr(s) + Fe^{2+}(aq) -> Cr^{3+}(aq) + Fe(s)} \nonumber$
The entry for the putative oxidant, Fe2+, appears above the entry for the reductant, Cr, and so a spontaneous reaction is predicted per the quick approach described above. Supporting this predication by calculating the standard cell potential for this reaction gives
\begin{aligned} E_{\text {cell }}^{\circ} & = E_{\text {cathode }}^{\circ}- E_{\text {anode }}^{\circ} \ & = E_{\text {Fe(II) }}^{\circ}- E_{\text {Cr }}^{\circ} \ & =-0.447 V --0.744 V =+0.297 V \end{aligned} \nonumber
The positive value for the standard cell potential indicates the process is spontaneous under standard state conditions.
Exercise $2$
Use the data in Table $1$ to predict the spontaneity of the oxidation of bromide ion by molecular iodine under standard state conditions, supporting the prediction by calculating the standard cell potential for the reaction. Repeat for the oxidation of iodide ion by molecular bromine.
Answer
\begin{align*} \ce{I2(s) + 2 Br^{-}(aq) &-> 2 I^{-}(aq) + Br2(l)} && E_{\text {cell }}=-0.5518\,\text{V} \text { (nonspontaneous) } \ \ce{Br2(s) + 2 I^{-}(aq) &-> 2 Br^{-}(aq) + I2(l)} && E_{\text {cell }}^{\circ}=+0.5518\,\text{V} \text { (spontaneous) } \end{align*} \nonumber | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.04%3A_Electrode_and_Cell_Potentials.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the relations between potential, free energy change, and equilibrium constants
• Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium
• Use the Nernst equation to determine cell potentials under nonstandard conditions
So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties ΔG and K.
E° and ΔG°
The standard free energy change of a process, $ΔG^°$, was defined in a previous chapter as the maximum work that could be performed by a system, $w_{max}$. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, $w_{elec}$:
$\Delta G^{\circ}=w_{\max }=w_{\text {elec }} \nonumber$
The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:
$\Delta G^{\circ}=w_{\text {elec }}=-n F E_{\text {cell }}^{\circ} \nonumber$
where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant, and $E^°_{cell}$ is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.
E° and K
Combining a previously derived relation between ΔG° and K (see the chapter on thermodynamics) and the equation above relating ΔG° and E°cell yields the following:
$\Delta G^{\circ}=-R T \ln K=-n F E_{\text {cell }}^{\circ} \label{eq1}$
with
$E_{\text {coll }}^{\circ}=\left(\frac{R T}{n F}\right) \ln K \nonumber$
This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between E°, ΔG° and K is depicted in Figure $1$, and a table correlating reaction spontaneity to values of these properties is provided in Table $1$.
Table $1$
K ΔG° E°cell Comment
> 1 < 0 > 0
Reaction is spontaneous under standard conditions
Products more abundant at equilibrium
< 1 > 0 < 0
Reaction is non-spontaneous under standard conditions
Reactants more abundant at equilibrium
= 1 = 0 = 0
Reaction is at equilibrium under standard conditions
Reactants and products equally abundant
Example $1$: Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes
Use data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at 25 °C. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products.
$\ce{2 Ag^{+}(aq) + Fe(s) <=> 2 Ag(s) + Fe^{2+}(aq)} \nonumber$
Solution
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L.
\begin{align*} & \text { anode: } \quad && \ce{Fe(s) -> Fe^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Fe^{2+}/Fe}} =-0.447\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{2 \times (Ag^{+}(aq) + e^{-} -> Ag(s))} &&E^o_{\ce{Ag^{+}/Ag}} = +0.7996\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{Fe(s) + 2Ag^{+}(aq) -> Fe^{2+}(aq) + 2Ag(s)} && E^o_{\text {cell }} = +1.2247\,\text{V} \end{align*} \nonumber
with
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}= E^o_{\ce{Fe^{2+}/Fe}} - E^o_{\ce{Ag^{+}/Ag}} = +1.2247\,\text{V} \nonumber$
With $n = 2$, the equilibrium constant is then
\begin{aligned} E_{\text {cell }}^{\circ} &=\frac{0.0592 V }{n} \log K \[4pt] K&=10^{n \times E_{\text {cal }}^o / 0.0592 V } \[4pt] &=10^{2 \times 1.247 V / 0.0592 V } \[4pt] &=10^{42.128} \[4pt] &=1.3 \times 10^{42} \end{aligned} \nonumber
The standard free energy is then Equation \ref{eq1}
\begin{aligned} & \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ} \ & =-2 \times 96,485 \frac{ C }{ mol } \times 1.247 \frac{ J }{ C }=-240.6 \frac{ kJ }{ mol } \end{aligned} \nonumber
The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.
Exercise $1$
What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?
$\ce{Sn(s) + 2 Cu^{2+}(aq) <=> Sn^{2+}(aq) + 2 Cu^{+}(aq)} \nonumber$
Answer
Spontaneous
$n = 2$
$E_{\text {cell }}^{\circ}=+0.291 V$
$[\Delta G^{\circ}=-56.2 \frac{ kJ }{ mol }$
K = 6.8 109.
Potentials at Nonstandard Conditions: The Nernst Equation
Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.
$\Delta G=\Delta G^{\circ}+R T \ln Q \nonumber$
Notice the reaction quotient, Q, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:
\begin{align*} -n F E_{\text {cell }}&=-n F E_{\text {cell }}^{\circ}+R T \ln Q \[4pt] E_{\text {cell }}&=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q \end{align*} \nonumber
This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, n, the temperature, T, and the reaction mixture composition as reflected in Q. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0592 V }{n} \log Q \label{Nernst at room temperature}$
Example $2$: Predicting Redox Spontaneity Under Nonstandard Conditions
Use the Nernst equation to predict the spontaneity of the redox reaction shown below at room temperature.
$\ce{Co(s)} + \ce{Fe^{2+}} \text{(aq, 1.94 M)} \longrightarrow \ce{Co^{2+}}\text{(aq, 0.15 M)} + \ce{Fe (s)} \nonumber$
Solution
Collecting information from Appendix L and the problem,
\begin{align*} & \text { anode: } \quad && \ce{Co(s) -> Co^{2+}(aq) + 2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.28\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{Fe^{2+}(aq) + 2 e^{-} -> Fe(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0.447\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{Co(s) + Fe^{2+}(Aq) -> Fe(s) + Co^{2+}(aq)} && E^o_{\text {cell }} = -0.17\,\text{V} \end{align*} \nonumber
with
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=-0.447 V -(-0.28 V )=-0.17 V \nonumber$
Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions requires calculating the reaction quotient $Q$
$Q =\frac{\left[\ce{Co^{2+}}\right]}{\left[\ce{Fe^{2+}}\right]}=\dfrac{0.15 M}{1.94 M}=0.077 \nonumber$
them $Q$ and $n$ are substituted into Equation \red{Nernst at room temperature}
\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592 \,\text{V} }{n} \log Q \[4pt] & =-0.17 \,\text{V} -\frac{0.0592 \,\text{V} }{2} \log 0.077 \[4pt] & =-0.17 \,\text{V} +0.033 \,\text{V} \[4pt] &=-0.14\,\text{V} \end{aligned} \nonumber
The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.
Exercise $2$
For the cell schematic below at room temperature, identify values for n and Q, and calculate the cell potential, Ecell.
$\ce{Al(s)} | \ce{Al^{3+}}\text{(aq, 0.15 M)} | \ce{Cu^{2+}}\text{(aq, 0.025 M)} | \ce{Cu(s)} \nonumber$
Answer
n = 6; Q = 1440; Ecell = +1.97 V, spontaneous.
A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.
Example $3$: Concentration Cells
What is the cell potential of the concentration cell operating at room temperature described by
$\ce{Zn(s)} | \ce{Zn^{2+}}\text{(aq, 0.10 M)} | \ce{Zn^{2+}} \text{(aq, 0.50 M)} | \ce{Zn(s)} \nonumber$
Solution
From the information given:
\begin{align*} & \text { anode: } \quad && \ce{Zn(s) -> Zn^{2+}}\text{(aq, 0.10 M)} + \ce{2 e^{-}} &&E^o_{\ce{Co^{2+}/Co}} =-0.7618\,\text{V}\[4pt] & \text {cathode: } \quad && \ce{Zn^{2+}}\text{(aq, 0.50 M)} + \ce{2 e^{-} -> Zn(s)} &&E^o_{\ce{Fe^{+}/Fe}} = -0,7618\,\text{V}\[4pt] \hline &\text { cell: } \quad && \ce{\cancel{Zn(s)}} + \ce{Zn^{2+}} \text{(aq, 0.50 M)} \ce{-> Zn^{2+}}\text{(aq, 0.10 M)} + \cancel{\ce{Zn(s)}} && E^o_{\text {cell }} = 0\,\text{V} \end{align*} \nonumber
Substituting into the Nernst equation (Equation \ref{Nernst at room temperature}),
$E_{\text {cell }}=0.000\,\text{V} -\frac{0.0592\,\text{V} }{2} \log \dfrac{0.10}{0.50} = +0.021\, \text{V} \nonumber$
The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (Ecathode > Eanode).
Exercise $1$
The concentration cell above was allowed to operate until the cell reaction reached equilibrium at room temperature. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?
Answer
Ecell = 0.000 V; [Zn2+]cathode = [Zn2+]anode = 0.30 M | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.05%3A_Potential_Free_Energy_and_Equilibrium.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the electrochemistry associated with several common batteries
• Distinguish the operation of a fuel cell from that of a battery
There are many technological products associated with the past two centuries of electrochemistry research, none more immediately obvious than the battery. A battery is a galvanic cell that has been specially designed and constructed in a way that best suits its intended use a source of electrical power for specific applications. Among the first successful batteries was the Daniell cell, which relied on the spontaneous oxidation of zinc by copper(II) ions (Figure $1$):
$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)} \nonumber$
Modern batteries exist in a multitude of forms to accommodate various applications, from tiny button batteries that provide the modest power needs of a wristwatch to the very large batteries used to supply backup energy to municipal power grids. Some batteries are designed for single-use applications and cannot be recharged (primary cells), while others are based on conveniently reversible cell reactions that allow recharging by an external power source (secondary cells). This section will provide a summary of the basic electrochemical aspects of several batteries familiar to most consumers, and will introduce a related electrochemical device called a fuel cell that can offer improved performance in certain applications.
Visit this site to learn more about batteries.
Single-Use Batteries
A common primary battery is the dry cell, which uses a zinc can as both container and anode (“–” terminal) and a graphite rod as the cathode (“+” terminal). The Zn can is filled with an electrolyte paste containing manganese(IV) oxide, zinc(II) chloride, ammonium chloride, and water. A graphite rod is immersed in the electrolyte paste to complete the cell. The spontaneous cell reaction involves the oxidation of zinc:
$\ce{Zn(s) -> Zn^{2+}(aq) +2 e^{-}} \tag{anode}$
and the reduction of manganese(IV)
$\ce{2 MnO2(s) + 2 NH4Cl(aq) + 2 e^{-} -> Mn2O3(s) + 2 NH3(aq) + H2O(l) + 2 Cl^{-}} \tag{cathode}$
which together yield the cell reaction:
$\ce{2 MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Zn^{2+}(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l) + 2 Cl^{-}} \tag{cell}$
The voltage (cell potential) of a dry cell is approximately 1.5 V. Dry cells are available in various sizes (e.g., D, C, AA, AAA). All sizes of dry cells comprise the same components, and so they exhibit the same voltage, but larger cells contain greater amounts of the redox reactants and therefore are capable of transferring correspondingly greater amounts of charge. Like other galvanic cells, dry cells may be connected in series to yield batteries with greater voltage outputs, if needed.
Visit this site to learn more about zinc-carbon batteries.
Alkaline batteries (Figure $3$) were developed in the 1950s to improve on the performance of the dry cell, and they were designed around the same redox couples. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are
\begin{align*}
& \text { anode: } \quad && \ce{Zn(s) + 2OH^{-}(aq) -> ZnO(s) + H2O(l) + 2e^{-}} \4pt] & \text {cathode: } \quad && \ce{2 MnO2(s) + H2O(l) + 2 e^{-} -> Mn2O3(s) + 2 OH^{-}(aq)} \[4pt] \hline &\text { cell: } \quad && \ce{Zn(s) + 2 MnO2(s) -> ZnO(s) + Mn2O3(s)} \quad \quad \quad E_{\text {cell }}=+1.43\,\text{V} \end{align*} An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so they should 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. Visit this site to learn more about alkaline batteries. Rechargeable (Secondary) Batteries Nickel-cadmium, or NiCd, batteries (Figure $4$) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a “jelly-roll” design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are \[\begin{align*} & \text { anode: } \quad && \ce{Cd(s) + 2OH^{-}(aq) -> Cd(OH)2(s) + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{NiO2(s) + 2 H2O(l) + 2 e^{-} -> Ni(OH)2(s) + 2OH^{-}(aq)} \[4pt] \hline &\text { cell: } \quad && \ce{Cd(s) + NiO2(s) + 2 H2O(l) -> Cd(OH)2(s) + Ni(OH)2(s)} \quad \quad \quad E_{\text {cell }}\approx+1.2 \,\text{V} \end{align*} \nonumber
When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be ruptured or incinerated, and they should be disposed of in accordance with relevant toxic waste guidelines.
Visit this site for more information about nickel cadmium rechargeable batteries.
Lithium ion batteries (Figure $5$) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are
\begin{align*} & \text { anode: } \quad && \ce{LiCoO2 -> Li_{1-x}CoO2 + x~ Li^{+} + x~ e^{-}} \[4pt] & \text {cathode: } \quad && \ce{x~ Li^{+} + x~e^{-} + x~ C6 -> x~ LiC6} \[4pt] \hline &\text { cell: } \quad && \ce{LiCoO2 + x~ C6 -> Li_{1-x}CoO2 + x~ LiC6} \quad \quad \quad E_{\text {cell }}\approx+3.7\,\text{V} \end{align*} \nonumber
The variable stoichiometry of the cell reaction leads to variation in cell voltages, but for typical conditions, x is usually no more than 0.5 and the cell voltage is approximately 3.7 V. Lithium batteries are popular because they can provide a large amount 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.
Visit this site for more information about lithium ion batteries.
The lead acid battery (Figure $6$) is the type of secondary battery commonly used in automobiles. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are
\begin{align*} & \text { anode: } \quad && \ce{Pb(s) + HSO4^{-}(aq) -> PbSO4(s) + H^{+}(aq) + 2 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{PbO2(s) + HSO4^{-}(aq) + 3 H^{+}(aq) + 2 e^{-} -> PbSO4(s) + 2 H2O(l)} \[4pt] \hline &\text { cell: } \quad && \ce{Pb(s) + PbO2(s) + 2 H2SO4(aq) -> 2 PbSO4(s) + 2 H2O (l)} \quad \quad \quad E_{\text {cell }}\approx +2\,\text{V} \end{align*} \nonumber
Each cell produces 2 V, so six cells are connected in series to produce a 12-V car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, H2SO4(aq), but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.
Visit this site for more information about lead acid batteries.
Fuel Cells
A fuel cell is a galvanic cell that uses traditional combustive fuels, most often hydrogen or methane, that are continuously fed into the cell along with an oxidant. (An alternative, but not very popular, name for a fuel cell is a flow battery.) Within the cell, fuel and oxidant undergo the same redox chemistry as when they are combusted, but via a catalyzed electrochemical that is significantly more efficient. For example, a typical hydrogen fuel cell uses graphite electrodes embedded with platinum-based catalysts to accelerate the two half-cell reactions:
\begin{align*} & \text { anode: } \quad && \ce{2 H2(g) -> 4 H^{+}(aq) + 4 e^{-}} \[4pt] & \text {cathode: } \quad && \ce{O2(g) + 4 H^{+}(aq) + 4 e^{-} -> 2 H2O (g)} \[4pt] \hline &\text { cell: } \quad && \ce{2 H2(g) + O2(g) -> 2 H2O (g)} \quad \quad \quad E_{\text {cell }}\approx+1.2\,\text{V} \end{align*} \nonumber
These types of fuel cells generally produce voltages of approximately 1.2 V. Compared to an internal combustion engine, the energy efficiency of a fuel cell using the same redox reaction is typically more than double (~20%–25% for an engine versus ~50%–75% for a fuel cell). Hydrogen fuel cells are commonly used on extended space missions, and prototypes for personal vehicles have been developed, though the technology remains relatively immature.
Check out this link to learn more about fuel cells. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.06%3A_Batteries_and_Fuel_Cells.txt |
Learning Objectives
By the end of this section, you will be able to:
• Define corrosion
• List some of the methods used to prevent or slow corrosion
Corrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year.
Chemistry in Everyday Life: Statue of Liberty: Changing Colors
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure $1$). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide (Cu2O), which is red, and then to copper(II) oxide, which is black
\begin{align} \ce{2 Cu(s) + 1/2 O2(g) -> Cu_2 O (s)} \tag{red} \[4pt] \ce{2 Cu2O(s) + 1/2 O2(g) -> 2CuO (s)} \tag{black} \end{align}
Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the $\ce{CuO}$:
\begin{align*} \ce{2 CuO(s) + CO2(g) + H2O (l) &-> Cu2CO3(OH)2(s)} \tag{green}\[4pt] \ce{3 CuO(s) +2 CO2(g) + H2O (l) &-> Cu2(CO3)2(OH)2(s)} \tag{blue} \[4pt] \ce{4 CuO(s) + SO3(g) + 3 H2O (l) &-> Cu4SO4(OH)6(s)} \tag{green} \end{align*} \nonumber
These three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation, a phenomenon discussed further in another chapter of this text.
Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure $1$. The relevant redox reactions are described by the following equations:
\begin{align*} &\text { anode: } & \ce{Fe(s) \longrightarrow Fe^{2+}(aq) + 2 e^{-}} && E_{\ce{Fe^{2+} / Fe} }^{\circ}=-0.44\,\text{V} \[4pt] &\text { cathode: } & \ce{O2(g) + 4 H^{+}(aq) + 4 e^{-} -> 2 H2O (l)} && E_{\ce{O2 /O^{2-}}}^{\circ}=+1.23\,\text{V} \[4pt] \hline &\text { overall: } & \ce{2 Fe(s) + O2(g) + 4 H^{+}(aq) -> 2 Fe^{2+}(aq) + 2 H2O (l)} && E_{\text {cell }}^{\circ}=+1.67\,\text{V} \end{align*} \nonumber
Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:
$\ce{4 Fe^{2+}(aq) + O2(g) + (4+2 x) ~H2O(l) -> 2 Fe2O3 \cdot x~ H2O(s) + 8 H^{+}(aq)} \nonumber$
The stoichiometry of the hydrate varies, as indicated by the use of x in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.
One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.
Other strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron.
Iron and other metals may also be protected from corrosion by galvanization, a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph.
Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure $3$). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.07%3A_Corrosion.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the process of electrolysis
• Compare the operation of electrolytic cells with that of galvanic cells
• Perform stoichiometric calculations for electrolytic processes
Electrochemical cells in which spontaneous redox reactions take place (galvanic cells) have been the topic of discussion so far in this chapter. In these cells, electrical work is done by a redox system on its surroundings as electrons produced by the redox reaction are transferred through an external circuit. This final section of the chapter will address an alternative scenario in which an external circuit does work on a redox system by imposing a voltage sufficient to drive an otherwise nonspontaneous reaction, a process known as electrolysis. A familiar example of electrolysis is recharging a battery, which involves use of an external power source to drive the spontaneous (discharge) cell reaction in the reverse direction, restoring to some extent the composition of the half-cells and the voltage of the battery. Perhaps less familiar is the use of electrolysis in the refinement of metallic ores, the manufacture of commodity chemicals, and the electroplating of metallic coatings on various products (e.g., jewelry, utensils, auto parts). To illustrate the essential concepts of electrolysis, a few specific processes will be considered.
The Electrolysis of Molten Sodium Chloride
Metallic sodium, Na, and chlorine gas, Cl2, are used in numerous applications, and their industrial production relies on the large-scale electrolysis of molten sodium chloride, NaCl(l). The industrial process typically uses a Downs cell similar to the simplified illustration shown in Figure $1$. The reactions associated with this process are:
\begin{aligned} &\text { oxidation: } &&\ce{2Cl^{-}(l) &&-> Cl2(g} + 2e^{-}} \[4pt] + \quad &\text { reduction: } && \ce{Na^{+}(l) + e^{-} &&-> Na(l)} \[4pt] \hline &\text{cell:} &&\ce{2Cl^{-}(l) + 2Na^{+}(l) &&-> 2Na(l) + Cl2(g} + 2e^{-}} \end{aligned} \nonumber
The cell potential for the above process is negative, indicating the reaction as written (decomposition of liquid NaCl) is not spontaneous. To force this reaction, a positive potential of magnitude greater than the negative cell potential must be applied to the cell.
The Electrolysis of Water
Water may be electrolytically decomposed in a cell similar to the one illustrated in Figure $2$. To improve electrical conductivity without introducing a different redox species, the hydrogen ion concentration of the water is typically increased by addition of a strong acid. The redox processes associated with this cell are
\begin{aligned} &\text { oxidation: } &&\ce{2H2(l) &&-> O2(g) + 4H^{+} + 4e^{-}} &&E^o_{anode} = +1.299\,\text{V} \[4pt] + \quad &\text { reduction: } && \ce{2H^{+}(aq) + 2e^{-} &&-> H2(g)} &&E^o_{anode} = 0\,\text{V} \[4pt] \hline &\text{cell:} &&\ce{2H2O(l) &&-> 2H2(g) + O2(g}} &&E^o_{cell} = -1.299\,\text{V} \end{aligned} \nonumber
Again, the cell potential as written is negative, indicating a nonspontaneous cell reaction that must be driven by imposing a cell voltage greater than +1.229 V. Keep in mind that standard electrode potentials are used to inform thermodynamic predictions here, though the cell is not operating under standard state conditions. Therefore, at best, calculated cell potentials should be considered ballpark estimates.
The Electrolysis of Aqueous Sodium Chloride
When aqueous solutions of ionic compounds are electrolyzed, the anode and cathode half-reactions may involve the electrolysis of either water species (H2O, H+, OH-) or solute species (the cations and anions of the compound). As an example, the electrolysis of aqueous sodium chloride could involve either of these two anode reactions:
$\ce{2 Cl^{-}(aq) -> Cl2(g) + 2 e^{-}} \quad E_{\text {anode}}^{\circ}=+1.35827\,\text{V} \tag{i}$
or
$\ce{2 H2O(l) -> O2(g) + 4 H^{+}(aq) + 4 e^{-}} \quad E_{\text {anode }}^{\circ}=+1.229\,\text{V} \tag{ii}$
The standard electrode (reduction) potentials of these two half-reactions indicate water may be oxidized at a less negative/more positive potential (–1.229 V) than chloride ion (–1.358 V). Thermodynamics thus predicts that water would be more readily oxidized, though in practice it is observed that both water and chloride ion are oxidized under typical conditions, producing a mixture of oxygen and chlorine gas.
Turning attention to the cathode, the possibilities for reduction are:
$\ce{2 H^{+}(aq) +2 e^{-} -> H2(g)} \quad E_{\text {anode}}^{\circ}=0\,\text{V} \tag{iii}$
or
$\ce{2 H2O(l) +2 e^{-} -> H2(g) + 4 OH^{-}(aq)} \quad E_{\text {anode }}^{\circ}=-0.8277\,\text{V} \tag{iv}$
or
$\ce{Na^{+}(aq) + e^{-} -> Na(s)} \quad E_{\text {anode }}^{\circ}=-2.71\,\text{V} \tag{v}$
Comparison of these standard half-reaction potentials suggests the reduction of hydrogen ion is thermodynamically favored. However, in a neutral aqueous sodium chloride solution, the concentration of hydrogen ion is far below the standard state value of 1 M (approximately 10-7 M), and so the observed cathode reaction is actually reduction of water. The net cell reaction in this case is then
$\ce{ 2 H2O(l) + 2 Cl^{-}(aq) -> H2(g) + Cl2(g) +2 OH^{-}(aq)} \quad E_{\text {cell }}^{\circ}=-2.186\,\text{V} \nonumber$
This electrolysis reaction is part of the chlor-alkali process used by industry to produce chlorine and sodium hydroxide (lye).
Chemistry in Everyday Life: Electroplating
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. The silver plating of eating utensils is used here to illustrate the process. (Figure $3$).
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. Applying a sufficient potential results in the oxidation of the silver anode
$\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \tag{anode}$
and reduction of silver ion at the (spoon) cathode:
$\ce{Ag^{+}(aq) + e^{-} -> Ag(s)} \tag{cathode}$
The net result is the transfer of silver metal from the anode to the cathode. Several experimental factors must be carefully controlled to obtain high-quality silver coatings, including the exact composition of the electrolyte solution, the cell voltage applied, and the rate of the electrolysis reaction (electrical current).
Quantitative Aspects of Electrolysis
Electrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second (A = 1 C/s). The charge transferred, Q, by passage of a constant current, I, over a specified time interval, t, is then given by the simple mathematical product
$Q=I t \nonumber$
When electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process
$\ce{M^{n+}(aq) + n\,e^{-} -> M(s)} \nonumber$
involves the transfer of $n$ mole of electrons. The total charge transferred is, therefore,
$Q=n F \nonumber$
where $F$ is Faraday’s constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction.
Example $1$: Converting Current to Moles of Electrons
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?
Solution
Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
$n=\frac{Q}{F}=\frac{\frac{10.23 C }{ s } \times 1 hr \times \frac{60 min }{ hr } \times \frac{60 s }{\min }}{96,485 C / mol e^{-}}=\frac{36,830 C }{96,485 C / mol e^{-}}=0.3817 mol \textrm {e }^ { - } \nonumber$
From the problem, the solution contains AgNO3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver
$\ce{Ag^{+}(aq) + e^{-} -> Ag(s)} \tag{cathode}$
The atomic mass of silver is 107.9 g/mol, so
$\text{mass}\, Ag =0.3817\,\text{mol e}^{-} \times \dfrac{1\, \text{mol}\,\ce{Ag}}{1\,\text{mol}\,\ce{e}^{-}} \times \frac{107.9\,\text{g}\,\ce{Ag}}{1\,\text{mol}\,\ce{Ag}}=41.19\,\text{g}\,\ce{Ag} \nonumber$
Exercise $1$
Aluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?
Answer
$Al^{3+}(aq) +3 e^{-} \longrightarrow Al (s) \nonumber$
0.0777 mol Al = 2.10 g Al.
Example $2$: Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm3.
Solution
First, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):
$\text { volume }=\left(0.010 mm \times \frac{1 cm }{10 mm }\right) \times\left(3.3 m^2 \times\left(\frac{10,000 cm^2}{1 m^2}\right)\right)=33 cm^3 \nonumber$
Use the computed volume and the provided density to calculate the molar amount of chromium required:
$\text { mass }=\text { volume } \times \text { density }=33 cm^3 \times \frac{7.19 g }{ cm^3}=237 g Cr \nonumber$
$mol Cr =237 g Cr \times \frac{1 mol Cr }{52.00 g Cr }=4.56 mol Cr \nonumber$
The stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:
$Q=4.56 mol Cr \times \frac{3 mol e^{-}}{1 mol Cr } \times \frac{96485 C }{ mol e^{-}}=1.32 \times 10^6 C \nonumber$
Finally, if this charge is passed at a rate of 33.46 C/s, the required time is:
$t=\frac{Q}{I}=\frac{1.32 \times 10^6 C }{33.46 C / s }=3.95 \times 10^4 s =11.0 hr \nonumber$
Exercise $2$
What mass of zinc is required to galvanize the top of a 3.00 m 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3.
Answer
11.8 kg Zn requires 382 hours. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.08%3A_Electrolysis.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
active electrodeelectrode that participates as a reactant or product in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction
alkaline batteryprimary battery similar to a dry cell that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an improved replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell
anodeelectrode in an electrochemical cell at which oxidation occurs
batterysingle or series of galvanic cells designed for use as a source of electrical power
cathodeelectrode in an electrochemical cell at which reduction occurs
cathodic protectionapproach to preventing corrosion of a metal object by connecting it to a sacrificial anode composed of a more readily oxidized metal
cell notation (schematic)symbolic representation of the components and reactions in an electrochemical cell
cell potential (Ecell)difference in potential of the cathode and anode half-cells
concentration cellgalvanic cell comprising half-cells of identical composition but for the concentration of one redox reactant or product
corrosiondegradation of metal via a natural electrochemical process
dry cellprimary battery, also called a zinc-carbon battery, based on the spontaneous oxidation of zinc by manganese(IV)
electrode potential (EX)the potential of a cell in which the half-cell of interest acts as a cathode when connected to the standard hydrogen electrode
electrolysisprocess using electrical energy to cause a nonspontaneous process to occur
electrolytic cellelectrochemical cell in which an external source of electrical power is used to drive an otherwise nonspontaneous process
Faraday’s constant (F)charge on 1 mol of electrons; F = 96,485 C/mol e
fuel celldevices similar to galvanic cells that require a continuous feed of redox reactants; also called a flow battery
galvanic (voltaic) cellelectrochemical cell in which a spontaneous redox reaction takes place; also called a voltaic cell
galvanizationmethod of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.
half cellcomponent of a cell that contains the redox conjugate pair (“couple”) of a single reactant
inert electrodeelectrode that conducts electrons to and from the reactants in a half-cell but that is not itself oxidized or reduced
lead acid batteryrechargeable battery commonly used in automobiles; it typically comprises six galvanic cells based on Pb half-reactions in acidic solution
lithium ion batterywidely used rechargeable battery commonly used in portable electronic devices, based on lithium ion transfer between the anode and cathode
Nernst equationrelating the potential of a redox system to its composition
nickel-cadmium batteryrechargeable battery based on Ni/Cd half-cells with applications similar to those of lithium ion batteries
primary cellnonrechargeable battery, suitable for single use only
sacrificial anodeelectrode constructed from an easily oxidized metal, often magnesium or zinc, used to prevent corrosion of metal objects via cathodic protection
salt bridgetube filled with inert electrolyte solution
secondary cellbattery designed to allow recharging
standard cell potential the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K
standard electrode potential ()electrode potential measured under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K
standard hydrogen electrode (SHE)half-cell based on hydrogen ion production, assigned a potential of exactly 0 V under standard state conditions, used as the universal reference for measuring electrode potential
16.10: Key Equations
ΔG = −nFEcell
Q = I t = n F | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.09%3A_Key_Terms.txt |
Redox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, H+ and OH, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.
Galvanic cells are devices in which a spontaneous redox reaction occurs indirectly, with the oxidant and reductant redox couples contained in separate half-cells. Electrons are transferred from the reductant (in the anode half-cell) to the oxidant (in the cathode half-cell) through an external circuit, and inert solution phase ions are transferred between half-cells, through a salt bridge, to maintain charge neutrality. The construction and composition of a galvanic cell may be succinctly represented using chemical formulas and others symbols in the form of a cell schematic (cell notation).
The property of potential, E, is the energy associated with the separation/transfer of charge. In electrochemistry, the potentials of cells and half-cells are thermodynamic quantities that reflect the driving force or the spontaneity of their redox processes. The cell potential of an electrochemical cell is the difference in between its cathode and anode. To permit easy sharing of half-cell potential data, the standard hydrogen electrode (SHE) is assigned a potential of exactly 0 V and used to define a single electrode potential for any given half-cell. The electrode potential of a half-cell, EX, is the cell potential of said half-cell acting as a cathode when connected to a SHE acting as an anode. When the half-cell is operating under standard state conditions, its potential is the standard electrode potential, E°X. Standard electrode potentials reflect the relative oxidizing strength of the half-reaction’s reactant, with stronger oxidants exhibiting larger (more positive) X values. Tabulations of standard electrode potentials may be used to compute standard cell potentials, cell, for many redox reactions. The arithmetic sign of a cell potential indicates the spontaneity of the cell reaction, with positive values for spontaneous reactions and negative values for nonspontaneous reactions (spontaneous in the reverse direction).
Potential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction’s cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of K). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient, Q, as described by the Nernst equation.
Galvanic cells designed specifically to function as electrical power supplies are called batteries. A variety of both single-use batteries (primary cells) and rechargeable batteries (secondary cells) are commercially available to serve a variety of applications, with important specifications including voltage, size, and lifetime. Fuel cells, sometimes called flow batteries, are devices that harness the energy of spontaneous redox reactions normally associated with combustion processes. Like batteries, fuel cells enable the reaction’s electron transfer via an external circuit, but they require continuous input of the redox reactants (fuel and oxidant) from an external reservoir. Fuel cells are typically much more efficient in converting the energy released by the reaction to useful work in comparison to internal combustion engines.
Spontaneous oxidation of metals by natural electrochemical processes is called corrosion, familiar examples including the rusting of iron and the tarnishing of silver. Corrosion process involve the creation of a galvanic cell in which different sites on the metal object function as anode and cathode, with the corrosion taking place at the anodic site. Approaches to preventing corrosion of metals include use of a protective coating of zinc (galvanization) and the use of sacrificial anodes connected to the metal object (cathodic protection).
Nonspontaneous redox processes may be forced to occur in electrochemical cells by the application of an appropriate potential using an external power source—a process known as electrolysis. Electrolysis is the basis for certain ore refining processes, the industrial production of many chemical commodities, and the electroplating of metal coatings on various products. Measurement of the current flow during electrolysis permits stoichiometric calculations. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.11%3A_Summary.txt |
1.
Identify each half-reaction below as either oxidation or reduction.
(a) $Fe3++3e−⟶FeFe3++3e−⟶Fe$
(b) $Cr⟶Cr3++3e−Cr⟶Cr3++3e−$
(c) $MnO42−⟶MnO4−+e−MnO42−⟶MnO4−+e−$
(d) $Li++e−⟶LiLi++e−⟶Li$
2.
Identify each half-reaction below as either oxidation or reduction.
(a) $Cl−⟶Cl2Cl−⟶Cl2$
(b) $Mn2+⟶MnO2Mn2+⟶MnO2$
(c) $H2⟶H+H2⟶H+$
(d) $NO3−⟶NONO3−⟶NO$
3.
Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.
(a) $Ca⟶Ca2++2e−,Ca⟶Ca2++2e−,$ $F2+2e−⟶2F−F2+2e−⟶2F−$
(b) $Li⟶Li++e−,Li⟶Li++e−,$ $Cl2+2e−⟶2Cl−Cl2+2e−⟶2Cl−$
(c) $Fe⟶Fe3++3e−,Fe⟶Fe3++3e−,$ $Br2+2e−⟶2Br−Br2+2e−⟶2Br−$
(d) $Ag⟶Ag++e−,Ag⟶Ag++e−,$ $MnO4−+4H++3e−⟶MnO2+2H2OMnO4−+4H++3e−⟶MnO2+2H2O$
4.
Balance the equations below assuming they occur in an acidic solution.
(a) $H2O2+Sn2+⟶H2O+Sn4+H2O2+Sn2+⟶H2O+Sn4+$
(b) $PbO2+Hg⟶Hg22++Pb2+PbO2+Hg⟶Hg22++Pb2+$
(c) $Al+Cr2O72−⟶Al3++Cr3+Al+Cr2O72−⟶Al3++Cr3+$
5.
Identify the oxidant and reductant of each reaction of the previous exercise.
6.
Balance the equations below assuming they occur in a basic solution.
(a) $SO32−(aq)+Cu(OH)2(s)⟶SO42−(aq)+Cu(OH)(s)SO32−(aq)+Cu(OH)2(s)⟶SO42−(aq)+Cu(OH)(s)$
(b) $O2(g)+Mn(OH)2(s)⟶MnO2(s)O2(g)+Mn(OH)2(s)⟶MnO2(s)$
(c) $NO3−(aq)+H2(g)⟶NO(g)NO3−(aq)+H2(g)⟶NO(g)$
(d) $Al(s)+CrO42−(aq)⟶Al(OH)3(s)+Cr(OH)4−(aq)Al(s)+CrO42−(aq)⟶Al(OH)3(s)+Cr(OH)4−(aq)$
7.
Identify the oxidant and reductant of each reaction of the previous exercise.
8.
Why don’t hydroxide ions appear in equations for half-reactions occurring in acidic solution?
9.
Why don’t hydrogen ions appear in equations for half-reactions occurring in basic solution?
10.
Why must the charge balance in oxidation-reduction reactions?
11.
Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.
(a) $Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)$
(b) $2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)$
(c) $Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)$
(d) $3CuNO3(aq)+Au(NO3)3(aq)⟶3Cu(NO3)2(aq)+Au(s)3CuNO3(aq)+Au(NO3)3(aq)⟶3Cu(NO3)2(aq)+Au(s)$
12.
Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.
(a) $Mg(s)│Mg2+(aq)║Cu2+(aq)│Cu(s)Mg(s)│Mg2+(aq)║Cu2+(aq)│Cu(s)$
(b) $Ni(s)│Ni2+(aq)║Ag+(aq)│Ag(s)Ni(s)│Ni2+(aq)║Ag+(aq)│Ag(s)$
13.
Write a balanced equation for the cell reaction of each cell in the previous exercise.
14.
Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.
(a) $Al(s)+Zr4+(aq)⟶Al3+(aq)+Zr(s)Al(s)+Zr4+(aq)⟶Al3+(aq)+Zr(s)$
(b) $Ag+(aq)+NO(g)⟶Ag(s)+NO3−(aq)(acidic solution)Ag+(aq)+NO(g)⟶Ag(s)+NO3−(aq)(acidic solution)$
(c) $SiO32−(aq)+Mg(s)⟶Si(s)+Mg(OH)2(s)(basic solution)SiO32−(aq)+Mg(s)⟶Si(s)+Mg(OH)2(s)(basic solution)$
(d) $ClO3−(aq)+MnO2(s)⟶Cl−(aq)+MnO4−(aq)(basic solution)ClO3−(aq)+MnO2(s)⟶Cl−(aq)+MnO4−(aq)(basic solution)$
15.
Identify the oxidant and reductant in each reaction of the previous exercise.
16.
From the information provided, use cell notation to describe the following systems:
(a) In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO3)2 solution with all solute concentrations 1 M.
(b) The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution.
(c) One half-cell consists of a silver electrode in a 1 M AgNO3 solution, and in the other half-cell, a copper electrode in 1 M Cu(NO3)2 is oxidized.
17.
Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?
18.
An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
19.
An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.
20.
The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode.
21.
Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
(a) $Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)$
(b) $2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)$
(c) $Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)$
(d) $3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)$
22.
Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.
(a) $Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)$
(b) $3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+3Cu(s)3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+3Cu(s)$
(c) $Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)$
(d) $Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)$
23.
Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
$Cu ( s ) │ Cu 2+ ( a q ) ║ Au 3+ ( a q ) │ Au ( s ) Cu ( s ) │ Cu 2+ ( a q ) ║ Au 3+ ( a q ) │ Au ( s )$
24.
Determine the cell reaction and standard cell potential at 25 °C for a cell made from a cathode half-cell consisting of a silver electrode in 1 M silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?
25.
Determine the cell reaction and standard cell potential at 25 °C for a cell made from an anode half-cell containing a cadmium electrode in 1 M cadmium nitrate and a cathode half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?
26.
Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.
$Pt(s)│H2(g)│H+(aq)║Br2(aq),Br−(aq)│Pt(s)Pt(s)│H2(g)│H+(aq)║Br2(aq),Br−(aq)│Pt(s)$
27.
For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).
(a) 0.000 V, n = 2
(b) +0.434 V, n = 2
(c) −2.439 V, n = 1
28.
For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.
(a) 12 kJ/mol, n = 3
(b) −45 kJ/mol, n = 1
29.
Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.
(a) $Hg(l)+S2−(aq, 0.10M)+2Ag+(aq, 0.25M)⟶2Ag(s)+HgS(s)Hg(l)+S2−(aq, 0.10M)+2Ag+(aq, 0.25M)⟶2Ag(s)+HgS(s)$
(b) The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
(c) The cell comprised of a half-cell in which aqueous bromide ion (1.0 M) is being oxidized to aqueous bromine (0.11 M) and a half-cell in which Al3+ (0.023 M) is being reduced to aluminum metal.
30.
Determine ΔG and ΔG° for each of the reactions in the previous problem.
31.
Use the data in Appendix L to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.
(a) $AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s)⇌Ag+(aq)+Cl−(aq)$
(b) $CdS(s)⇌Cd2+(aq)+S2−(aq)at 377 KCdS(s)⇌Cd2+(aq)+S2−(aq)at 377 K$
(c) $Hg2+(aq)+4Br−(aq)⇌[HgBr4]2−(aq)Hg2+(aq)+4Br−(aq)⇌[HgBr4]2−(aq)$
(d) $H2O(l)⇌H+(aq)+OH−(aq)at 25°CH2O(l)⇌H+(aq)+OH−(aq)at 25°C$
32.
Consider a battery made from one half-cell that consists of a copper electrode in 1 M CuSO4 solution and another half-cell that consists of a lead electrode in 1 M Pb(NO3)2 solution.
(a) What is the standard cell potential for the battery?
(b) What are the reactions at the anode, cathode, and the overall reaction?
(c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.
(d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some PbSO4(s) forms. Would the cell potential increase, decrease, or remain the same?
33.
Consider a battery with the overall reaction: $Cu(s)+2Ag+(aq)⟶2Ag(s)+Cu2+(aq).Cu(s)+2Ag+(aq)⟶2Ag(s)+Cu2+(aq).$
(a) What is the reaction at the anode and cathode?
(b) A battery is “dead” when its cell potential is zero. What is the value of Q when this battery is dead?
(c) If a particular dead battery was found to have [Cu2+] = 0.11 M, what was the concentration of silver ion?
34.
Why do batteries go dead, but fuel cells do not?
35.
Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.
36.
Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.
37.
Which member of each pair of metals is more likely to corrode (oxidize)?
(a) Mg or Ca
(b) Au or Hg
(c) Fe or Zn
(d) Ag or Pt
38.
Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use −0.447 V as the standard reduction potential for steel.
39.
Aluminum $(EAl3+/Al°=−2.07 V)(EAl3+/Al°=−2.07 V)$ is more easily oxidized than iron $(EFe3+/Fe°=−0.477 V),(EFe3+/Fe°=−0.477 V),$ and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. What might explain this observation?
40.
If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.
41.
Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?
42.
Why would a sacrificial anode made of lithium metal be a bad choice
43.
If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?
44.
For the scenario in the previous question, how many electrons moved through the circuit?
45.
Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.
(a) CaCl2
(b) LiH
(c) AlCl3
(d) CrBr3
46.
What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 $××$ 105 C passes through each cell?
47.
How long would it take to reduce 1 mole of each of the following ions using the current indicated?
(a) Al3+, 1.234 A
(b) Ca2+, 22.2 A
(c) Cr5+, 37.45 A
(d) Au3+, 3.57 A
48.
A current of 2.345 A passes through the cell shown in Figure 16.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? (Hint: Is hydrogen the only gas present above the water?)
49.
An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO3)2 solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm3. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/16%3A_Electrochemistry/16.12%3A_Exercises.txt |
Chemical kinetics is the study of rates of chemical processes and includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.
17: Kinetics
The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun’s rays is critical to the lizard’s survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators.
From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: “Will the reaction produce the desired products in useful quantities?” The second question is: “How rapidly will the reaction occur?” A third question is often asked when investigating reactions in greater detail: “What specific molecular-level processes take place as the reaction occurs?” Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.
The study of chemical kinetics concerns the second and third questions—that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur.
17.02: Chemical Reaction Rates
Learning Objectives
By the end of this section, you will be able to:
• Define chemical reaction rate
• Derive rate expressions from the balanced equation for a given chemical reaction
• Calculate reaction rates from experimental data
A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.
The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.
For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, H2O2, in an aqueous solution changes slowly over time as it decomposes according to the equation:
$2 H_2 O_2(aq) \longrightarrow 2 H_2 O (l) + O_2(g) \nonumber$
The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:
\text { rate of decomposition of } \begin{align*} H_2O_2 & =-\dfrac{\text { change in concentration of reactant }}{\text { time interval }} \[4pt] & =-\dfrac{\left[ H_2 O_2\right]_{t_2}-\left[ H_2 O_2\right]_{t_1}}{t_2-t_1} \[4pt] & =-\dfrac{\Delta\left[ H_2 O_2\right]}{\Delta t} \end{align*}
This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta ($Δ$) indicates “change in.” Thus, $[\ce{H2O2}]_{t_1}}$ represents the molar concentration of hydrogen peroxide at some time $t_1$; likewise, $[\ce{H2O2}]_{t_2}}$ represents the molar concentration of hydrogen peroxide at a later time t2; and Δ[H2O2] represents the change in molar concentration of hydrogen peroxide during the time interval $Δt$ (that is, $t_2 − t_1$). Since the reactant concentration decreases as the reaction proceeds, $Δ[\ce{H2O2}]$ is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by −1. Figure $1$ provides an example of data collected during the decomposition of $\ce{H2O2}$.
To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:
$\frac{-\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{-(0.500 mol / L -1.000 mol / L )}{(6.00 h -0.00 h )}=0.0833 mol L^{-1} h^{-1} \nonumber$
Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:
$\frac{-\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{-(0.0625 mol / L -0.125 mol / L )}{(24.00 h -18.00 h )}=0.010 mol L^{-1} h^{-1} \nonumber$
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t0). A few moments later, the instantaneous rate at a specific moment—call it t1—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.
The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\ce{H2O2}$ at any time $t$ is given by the slope of a straight line that is tangent to the curve at that time (Figure $2$). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.
Chemistry in Everyday Life: Reaction Rates in Analysis: Test Strips for Urinalysis
Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure $3$). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.
The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:
$\begin{gathered} C_6 H_{12} O_6+ O_2 \stackrel{\text { catalyst }}{\longrightarrow} C_6 H_{10} O_6+ H_2 O_2 \[4pt] 2 H_2 O_2+2 I^{-} \stackrel{\text { catalyst }}{\longrightarrow} I_2+2 H_2 O + O_2 \end{gathered} \nonumber$
The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.
The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.
Relative Rates of Reaction
The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction
$aA \longrightarrow bB] can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction: \[\text { rate }=-\left(\frac{1}{ a }\right)\left(\frac{\Delta A }{\Delta t}\right)=\left(\frac{1}{ b }\right)\left(\frac{\Delta B }{\Delta t}\right) \nonumber$
Consider the reaction represented by the following equation:
$2 NH_3(g) \longrightarrow N_2(g) + 3 H_2(g) \nonumber$
The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:
$-\frac{\Delta mol NH_3}{\Delta t} \times \frac{1 mol N_2}{2 mol NH_3}=\frac{\Delta mol N_2}{\Delta t} \nonumber$
This may be represented in an abbreviated format by omitting the units of the stoichiometric factor:
$-\frac{1}{2} \frac{\Delta mol NH_3}{\Delta t}=\frac{\Delta mol N_2}{\Delta t} \nonumber$
Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:
$-\frac{1}{2} \frac{\Delta\left[ NH_3\right]}{\Delta t}=\frac{\Delta\left[ N_2\right]}{\Delta t} \nonumber$
Similarly, the rate of formation of H2 is three times the rate of formation of N2 because three moles of H2 are produced for each mole of N2 produced.
$\frac{1}{3} \frac{\Delta\left[ H_2\right]}{\Delta t}=\frac{\Delta\left[ N_2\right]}{\Delta t} \nonumber$
Figure $4$: illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. Slopes of the tangent lines at t = 500 s show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:
$\frac{2.91 \times 10^{-6} M / s }{9.70 \times 10^{-7} M / s } \approx 3 \nonumber$
Example $1$: Expressions for Relative Reaction Rates
The first step in the production of nitric acid is the combustion of ammonia:
Exercise $1$
The rate of formation of Br2 is 6.0 10−6 mol/L/s in a reaction described by the following net ionic equation:
$5 Br^{-}+ BrO_3^{-}+6 H^{+} \longrightarrow 3 Br_2+3 H_2 O \nonumber$
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Answer
$-\frac{1}{5} \frac{\Delta\left[ Br^{-}\right]}{\Delta t}=-\frac{\Delta\left[ BrO_3^{-}\right]}{\Delta t}=-\frac{1}{6} \frac{\Delta\left[ H^{+}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[ Br_2\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[ H_2 O \right]}{\Delta t} \nonumber$
Example $2$: Reaction Rate Expressions for Decomposition of H2O2
The graph in Figure $2$: shows the rate of the decomposition of H2O2 over time:
$2 H_2 O_2 \longrightarrow 2 H_2 O + O_2 \nonumber$
Based on these data, the instantaneous rate of decomposition of H2O2 at t = 11.1 h is determined to be
3.20 10−2 mol/L/h, that is:
$-\frac{\Delta\left[ H_2 O_2\right]}{\Delta t}=3.20 \times 10^{-2} mol L^{-1} h^{-1} \nonumber$
What is the instantaneous rate of production of $\ce{H2O}$ and $\ce{O2}$?
Solution
The reaction stoichiometry shows that
$-\frac{1}{2} \frac{\Delta\left[ H_2 O_2\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[ H_2 O \right]}{\Delta t}=\frac{\Delta\left[ O_2\right]}{\Delta t}] Therefore: \[\frac{1}{2} \times 3.20 \times 10^{-2} mol L^{-1} h^{-1}=\frac{\Delta\left[ O_2\right]}{\Delta t} \nonumber$
and
$\frac{\Delta\left[ O_2\right]}{\Delta t}=1.60 \times 10^{-2} mol L^{-1} h^{-1} \nonumber$
Exercise $2$
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 10−6 mol/L/s, what is the rate of production of nitrogen and hydrogen?
Answer
1.05 10−6 mol/L/s, N2 and 3.15 10−6 mol/L/s, H2. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates
The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.
The Chemical Nature of the Reacting Substances
The rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air overnight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.
The Physical States of the Reactants
A chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder (Figure $1$). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.
Watch this video to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.
Temperature of the Reactants
Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by 10 °C.
Concentrations of the Reactants
The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate (CaCO3) deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure $2$). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:
$\ce{SO_2(g) + H_2O(g) \longrightarrow H_2SO_3(aq)} \nonumber$
Calcium carbonate reacts with sulfurous acid as follows:
$\ce{CaCO_3(s) + H_2SO_3(aq) \longrightarrow CaSO_3(aq) + CO_2(g) + H_2O(l)} \nonumber$
In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about 20% oxygen.
Phosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this video to see an example.
The Presence of a Catalyst
Relatively dilute aqueous solutions of hydrogen peroxide, $\ce{H2O2}$, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:
$\ce{2 H_2O_2(aq) \longrightarrow 2 H_2O(l) + O_2(g)} \nonumber$
Under typical conditions, this decomposition occurs very slowly. When dilute $\ce{H2O2(aq)}$ is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound’s exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter.
Chemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions & Rates interactive to explore how temperature, concentration, and the nature of the reactants affect reaction rates. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.03%3A_Factors_Affecting_Reaction_Rates.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the form and function of a rate law
• Use rate laws to calculate reaction rates
• Use rate and concentration data to identify reaction orders and derive rate laws
As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. Rate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation
$a A+b B \longrightarrow \text { products } \nonumber$
where $a$ and $b$ are stoichiometric coefficients. The rate law for this reaction is written as:
$\text { rate }=k[A]^m[B]^n \nonumber$
in which [A] and [B] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m and n are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant k and the reaction orders m and n must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the reactant concentrations, but it does vary with temperature.
The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is m order with respect to A and n order with respect to B. For example, if m = 1 and n = 2, the reaction is first order in A and second order in B. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.
The rate law:
$\text { rate }=k\left[ H_2 O_2\right] \nonumber$
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
$\text { rate }=k\left[ \ce{C4H6} \right]^2 \nonumber$
describes a reaction that is second order in $\ce{C4H6}$ and second order overall. The rate law:
$\text { rate }=k\left[ \ce{H^{+}} \right]\left[ \ce{OH^{-}} \right] \nonumber$
describes a reaction that is first order in $\ce{H^{+}}$, first order in $\ce{OH^{−}}$, and second order overall.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2(g) + CO(g) -> NO(g) + CO2(g)} \nonumber$
is second order in $\ce{NO2}$ and zero order in $\ce{CO}$ at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\text { rate }=k\left[ \ce{NO_2} \right]^m[ \ce{CO} ]^n \nonumber$
The reaction is second order in $\ce{NO2}$; thus $m = 2$. The reaction is zero order in $\ce{CO}$; thus $n = 0$. The rate law is:
$\text { rate }=k\left[ \ce{NO_2} \right]^2[ \ce{CO} ]^0=k\left[ \ce{NO2} \right]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why the $\ce{CO}$ concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of $\ce{NO2}$. A later chapter section on reaction mechanisms will explain how a reactant’s concentration can have no effect on a reaction rate despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2(g) + 2 NO(g) -> N2O(g) + H2O(g)} \nonumber$
has been determined to be rate = k[NO]2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
order in NO = 2; order in H2 = 1; overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol ($\ce{CH3OH}$) and ethyl acetate ($\ce{CH3CH2OCOCH3}$) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 -> CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:
$\text { rate }=k\left[ \ce{CH3OH} \right] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
order in CH3OH = 1; order in CH3CH2OCOCH3 = 0; overall order = 1
A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.
Example $2$: Determining a Rate Law from Initial Rates
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure $1$). One such reaction is the combination of nitric oxide, NO, with ozone, O3:
$\ce{NO(g) + O3(g) -> NO2(g) + O2(g)} \nonumber$
This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.
Trial [NO] (mol/L) [O3] (mol/L) $\dfrac{\Delta[ \ce{NO2}]}{\Delta t} (\text{mol}\text{L}^{-1} \text{s}^{-1})$
1 1.00 × 10−6 3.00 × 10−6 6.60 × 10−5
2 1.00 × 10−6 6.00 × 10−6 1.32 × 10−4
3 1.00 × 10−6 9.00 × 10−6 1.98 × 10−4
4 2.00 × 10−6 9.00 × 10−6 3.96 × 10−4
5 3.00 × 10−6 9.00 × 10−6 5.94 × 10−4
Determine the rate law and the rate constant for the reaction at 25 °C.
Solution
The rate law will have the form:
$\text { rate }=k[ NO ]^m\left[ O_3\right]^n \nonumber$
Determine the values of m, n, and k from the experimental data using the following three-part process:
Step 1:
Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
Step 2:
Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:
$\text { rate }=k[ NO ]^1\left[ O_3\right]^1=k[ NO ]\left[ O_3\right] \nonumber$
Step 3:
Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:
\begin{align*} k & =\dfrac{\text { rate }}{[ \ce{NO} ]\left[ \ce{O3} \right]} \[4pt] & =\dfrac{6.60 \times 10^{-5} ~\text{mol L}^{-1}\text{s}^{-1}}{\left(1.00 \times 10^{-6} ~\text{mol L}^{-1}\right)\left(3.00 \times 10^{-6}~\text{mol L}^{-1}\right)} \[4pt] & =2.20 \times 10^7~\text{L mol}^{-1}\text{s}^{-1} \end{align*}
Exercise $2$
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
$\ce{CH3CHO(g) -> CH4(g) + CO(g)} \nonumber$
Determine the rate law and the rate constant for the reaction from the following experimental data:
Trial [CH3CHO] (mol/L) $-\dfrac{\Delta [\ce{CH3CHO}]}{\Delta t} (\text{mol}\,\text{L}^{-1}\text{s}^{-1})$
1 1.75 × 10−3 2.06 × 10−11
2 3.50 × 10−3 8.24 × 10−11
3 7.00 × 10−3 3.30 × 10−10
Answer
$\text { rate }=k\left[ \ce{CH3CHO} \right]^2$ with k = 6.7310−6 L/mol/s
Example $3$: Determining Rate Laws from Initial Rates
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
$2 NO (g) + Cl_2(g) \longrightarrow 2 NOCl (g) \nonumber$
Trial [NO] (mol/L) [Cl2] (mol/L) $-\dfrac{\Delta [\ce{CO}]}{\Delta t} (\text{mol} ~L^{-1} \text{s}^{-1})$
1 0.10 0.10 0.00300
2 0.10 0.15 0.00450
3 0.15 0.10 0.00675
Solution
The rate law for this reaction will have the form:
$\text { rate }=k[ NO ]^m\left[ Cl_2\right]^n \nonumber$
As in Example $2$, approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of m and n:
Step 1
Determine the value of m from the data in which $[\ce{NO}]$ varies and $[\ce{Cl2}]$ is constant. Write the ratios with the subscripts x and y to indicate data from two different trials:
$\dfrac{\text { rate }_x}{\text { rate }_y}=\dfrac{k[ NO ]_x^m\left[ Cl_2\right]_x^n}{k[ NO ]_y^m\left[ Cl_2\right]_y^n} \nonumber$
Using the third trial and the first trial, in which $[\ce{Cl2}]$ does not vary, gives:
$\dfrac{\text { rate } 3}{\text { rate } 1}=\dfrac{0.00675}{0.00300}=\dfrac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n} \nonumber$
Canceling equivalent terms in the numerator and denominator leaves:
$\dfrac{0.00675}{0.00300}=\dfrac{(0.15)^m}{(0.10)^m} \nonumber$
which simplifies to:
$2.25=(1.5)^m \nonumber$
Use logarithms to determine the value of the exponent m:
\begin{align*} \ln (2.25) & =m \ln (1.5) \[4pt] \dfrac{\ln (2.25)}{\ln (1.5)} & =m \[4pt] 2 & =m \end{align*}
Confirm the result
$1.5^2=2.25 \nonumber$
Step 2
Determine the value of $n$ from data in which $[\ce{Cl2}]$ varies and $[\ce{NO}]$ is constant.
$\dfrac{\text { rate } 2}{\text { rate } 1}=\dfrac{0.00450}{0.00300}=\dfrac{k0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n} \nonumber$
Cancelation gives:
$\dfrac{0.0045}{0.0030}=\dfrac{(0.15)^n}{(0.10)^n} \nonumber$
which simplifies to:
$1.5=(1.5)^n \nonumber$
Thus $n$ must be 1, and the form of the rate law is:
$\text { rate }=k[ NO ]^m\left[ Cl_2\right]^n=k[ NO ]^2\left[ Cl_2\right] \nonumber$
Step 3.
Determine the numerical value of the constant with appropriate units. The units for the rate of a reaction are mol/L/s. The units for are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol L/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:
\begin{align*} 0.00300~\text{mol L}^{-1} \text{s}^{-1} & =k\left(0.10~\text{mol L}^{-1}\right)^2\left(0.10 ~\text{mol L}^{-1}\right)^1 \[4pt] k & =3.0 ~\text{mol}^{-2}\text{L}^2 \text{s}^{-1} \end{align*}
Exercise $3$
Use the provided initial rate data to derive the rate law for the reaction whose equation is:
$\ce{OCl^{-}(aq) + I^{-}(aq) -> OI^{-}(aq) + Cl^{-}(aq)} \nonumber$
Trial [OCl] (mol/L) [I] (mol/L) Initial Rate (mol/L/s)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
Answer
$\dfrac{\text { rate } 2}{\text { rate 3 }}=\dfrac{0.00092}{0.00046}=\dfrac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y} \nonumber$
\begin{align*} 2.00 &=2.00^y \[4pt] y&=1 \end{align*}
$\dfrac{\text { rate } 1}{\text { rate } 2}=\dfrac{0.00184}{0.00092}=\dfrac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y} \nonumber$
\begin{align*} 2.00 & =\dfrac{2^x}{2^y} \[4pt] 2.00 & =\dfrac{2^x}{2^1} \[4pt] 4.00 & =2^x \[4pt] x & =2 \end{align*}
Substituting the concentration data from trial 1 and solving for $k$ yields:
\begin{align*} \text { rate } & =k\left[ OCl^{-}\right]^2\left[ I^{-}\right]^1 \[4pt] 0.00184 & =k(0.0040)^2(0.0020)^1 \[4pt] k & =5.75 \times 10^4 mol^{-2} L^2 s^{-1} \end{align*}
Reaction Order and Rate Constant Units
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.
Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
\begin{align*} \ce{NO_2 + CO & -> NO + CO_2} && \text { rate }=k\left[ NO_2\right]^2 \[4pt] \ce{CH_3 CHO & -> CH_4 + CO} && \text { rate }=k\left[ CH_3 CHO \right]^2 \[4pt] \ce{2 N_2 O_5 & -> NO_2 + O_2} && \text { rate }=k\left[ N_2 O_5\right] \[4pt] \ce{2 NO_2 + F_2 & -> 2 NO_2 F} && \text { rate }=k\left[ NO_2\right]\left[ F_2\right] \[4pt] \ce{2 NO_2 Cl & -> 2 NO_2 + Cl_2} && \text { rate }=k\left[ NO_2 Cl \right] \end{align*} \nonumber
Note
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example $2$ was determined to be $k$ was derived to be $\text{L}\,\text{mol}^{−1}\text{s}^{−1}$. For the third-order reaction, $k$ was derived to be $\text{L}^{2}\text{mol}^{−2}\text{s}^{−1}$. Dimensional analysis requires the rate constant unit for a reaction whose overall order is x to be $\text{L}^{x-1}\,\text{mol}^{1-x}\text{s}^{−1}$. Table $1$ summarizes the rate constant units for common reaction orders.
Table $1$: Rate Constant Units for Common Reaction Orders
Overall Reaction Order (x) Rate Constant Unit (Lx−1 mol1x s−1)
0 (zero) mol L−1 s−1
1 (first) s−1
2 (second) L mol−1 s−1
3 (third) L2 mol−2 s−1
Note that the units in this table were derived using specific units for concentration (mol/L) and time (s), though any valid units for these two properties may be used. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.04%3A_Rate_Laws.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the form and function of an integrated rate law
• Perform integrated rate law calculations for zero-, first-, and second-order reactions
• Define half-life and carry out related calculations
• Identify the order of a reaction from concentration/time data
The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.
First-Order Reactions
Integration of the rate law for a simple first-order reaction (rate = k[A]) results in an equation describing how the reactant concentration varies with time:
$[A]_t=[A]_0 e^{-k t} \nonumber$
where [A]t is the concentration of A at any time t, [A]0 is the initial concentration of A, and k is the first-order rate constant.
For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:
$\ln \left(\frac{[A]_t}{[A]_0}\right)=-k t \nonumber$
or
$\ln \left(\frac{[A]_0}{[A]_t}\right)=k t \nonumber$
and a format showing a linear dependence of concentration in time:
$\ln [A]_t=\ln [A]_0-k t \nonumber$
Example $1$: The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, $\ce{C4H8}$ at 500 °C is 9.2 10−3 s−1:
$\ce{C_4H_8 \longrightarrow 2 C_2H_4} \nonumber$
How long will it take for 80.0% of a sample of $\ce{C4H8}$ to decompose?
Solution
Since the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:
$\ln \left(\frac{[A]_0}{[A]_t}\right)=k t \nonumber$
The initial concentration of $\ce{C4H8}$, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:
\begin{align*} t & =\ln \frac{[x]}{[0.200 x]} \times \frac{1}{k} \[4pt] & =\ln 5 \times \frac{1}{9.2 \times 10^{-3} ~\text{s}^{-1}} \[4pt] & =1.609 \times \frac{1}{9.2 \times 10^{-3} ~\text{s}^{-1}} \[4pt] & =1.7 \times 10^2 ~\text{s} \end{align*}
Exercise $1$
Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:
$\text{I-131} \longrightarrow \text{Xe-131} + \text{electron } \nonumber$
The decay is first-order with a rate constant of 0.138 d−1. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?
Answer
16.7 days
In the next example exercise, a linear format for the integrated rate law will be convenient:
\begin{align*} \ln [A]_t & =(-k)(t) + \ln [A]_0 \[4pt] y & =m x+b \end{align*}
A plot of ln[A]t versus t for a first-order reaction is a straight line with a slope of −k and a y-intercept of ln[A]0. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.
Example $2$: Graphical Determination of Reaction Order and Rate Constant
Show that the data in Figure $1$ can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the decomposition of H2O2 from these data.
Solution
The data from Figure $1$ are tabulated below, and a plot of ln[H2O2] is shown in Figure $1$.
Time (h) [H2O2] (M) ln[H2O2]
0.00 1.000 0.000
6.00 0.500 −0.693
12.00 0.250 −1.386
18.00 0.125 −2.079
24.00 0.0625 −2.772
The plot of ln[H2O2] versus time is linear, indicating that the reaction may be described by a first-order rate law.
According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot’s slope.
$\text { slope }=\frac{\text { change in } y}{\text { change in } x}=\frac{\Delta y}{\Delta x}=\frac{\Delta \ln \left[ H_2 O_2\right]}{\Delta t} \nonumber$
The slope of this line may be derived from two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 0.00 h is 0.000; the value when t = 24.00 h is −2.772
\begin{align*} \text { slope } & =\frac{-2.772-0.000}{24.00-0.00 h } \[4pt] & =\frac{-2.772}{24.00 h } \[4pt] & =-0.116 h^{-1} \[4pt] k & =- \text { slope }=-\left(-0.116 h^{-1}\right)=0.116 h^{-1} \end{align*}
Exercise $2$
Graph the following data to determine whether the reaction $A \longrightarrow B+C$ is first order.
Time (s) [A]
4.0 0.220
8.0 0.144
12.0 0.110
16.0 0.088
20.0 0.074
Answer
The plot of ln[A]t vs. t is not linear, indicating the reaction is not first order:
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:
$\text { rate }=k[A]^2 \nonumber$
For these second-order reactions, the integrated rate law is:
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
where the terms in the equation have their usual meanings as defined earlier.
Example $3$: The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas (C4H6) to yield C8H12 gas is described by the equation:
$2 C_4 H_6(g) \longrightarrow C_8 H_{12}(g) \nonumber$
This “dimerization” reaction is second order with a rate constant equal to 5.76 10−2 L mol−1 min−1 under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration after 10.0 min?
Solution
For a second-order reaction, the integrated rate law is written
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:
\begin{align*} \frac{1}{[A]_t} & =\left(5.76 \times 10^{-2} L mol^{-1} min^{-1}\right)(10 min ) + \dfrac{1}{0.200 mol^{-1}} \[4pt] & =\left(5.76 \times 10^{-1} L mol^{-1}\right) + 5.00 L mol^{-1} \[4pt] & =5.58 L mol^{-1} \[4pt] [A]_t & =1.79 \times 10^{-1} mol L^{-1} \end{align*}
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
Exercise $3$
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?
Answer
0.0195 mol/L
The integrated rate law for second-order reactions has the form of the equation of a straight line:
\begin{align*} \frac{1}{[A]_t} & =k t+\frac{1}{[A]_0} \[4pt] y & =m x+b \end{align*} \nonumber
A plot of $\frac{1}{[A]_t}$ versus $t$ for a second-order reaction is a straight line with a slope of k and a y-intercept of $\frac{1}{[A]_0}$. If the plot is not a straight line, then the reaction is not second order.
Example $4$: Graphical Determination of Reaction Order and Rate Constant
The data below are for the same reaction described in Example $3$. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.
Solution
Time (s) [C4H6] (M)
0 1.00 × 10−2
1600 5.04 × 10−3
3200 3.37 × 10−3
4800 2.53 × 10−3
6200 2.08 × 10−3
In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of ln[C4H6]t versus t and compare it to a plot of $\frac{1}{[\ce{C4H6}]_t}$ versus $t$. The values needed for these plots follow.
Time (s) $\frac{1}{[ \ce{C4H6}]} (\text{M}^{-1})$ $\ln[\ce{C4H6}]$
0 100 −4.605
1600 198 −5.289
3200 296 −5.692
4800 395 −5.978
6200 481 −6.175
The plots are shown in Figure $2$, which clearly shows the plot of $\ln[\ce{C4H6}]_t$ versus $t$ is not linear, therefore the reaction is not first order. The plot of $\frac{1}{[\ce{C4H6}]_t}$ versus $t$ is linear, indicating that the reaction is second order.
According to the second-order integrated rate law, the rate constant is equal to the slope of the versus t plot. Using the data for t = 0 s and t = 6200 s, the rate constant is estimated as follows:
$k=\text { slope }=\frac{\left(481 M^{-1}-100 M^{-1}\right)}{(6200 s -0 s )}=0.0614 M^{-1} s^{-1} \nonumber$
Exercise $4$
Do the following data fit a second-order rate law?
Time (s) [A] (M)
5 0.952
10 0.625
15 0.465
20 0.370
25 0.308
35 0.230
Answer
Yes. The plot of $\frac{1}{[A]_t}$ vs. $t$ is linear:
Zero-Order Reactions
For zero-order reactions, the differential rate law is:
$\text { rate }=k \nonumber$
A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can’t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren’t met, and for this reason the more prudent term pseudo-zero-order is sometimes used.
The integrated rate law for a zero-order reaction is a linear function:
\begin{align*} {[A]_t } & =-k t+[A]_0 \[4pt] y & =m x+b \end{align*}
A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and a y-intercept of [A]0. Figure $3$: shows a plot of [NH3] versus t for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz (SiO2) surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.
Example $5$: Graphical Determination of Zero-Order Rate Constant
Use the data plot in Figure $3$ to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.
Solution
The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [A]t, versus time, t, with a slope equal to the negative of the rate constant, −k. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at t = 0 and t = 1000 s:
$k=- \text { slope }=-\frac{\left(0.0015 mol L^{-1}-0.0028 mol L^{-1}\right)}{(1000 s -0 s )}=1.3 \times 10^{-6} mol L^{-1} s^{-1} \nonumber$
Exercise $5$
The zero-order plot in Figure $3$ shows an initial ammonia concentration of 0.0028 mol L−1 decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, at what time (min) will the concentration reach 0.0001 mol L−1?
Answer
35 min
The Half-Life of a Reaction
The half-life of a reaction (t1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure $1$) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H2O2 decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H2O2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.
First-Order Reactions
An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:
\begin{align*} \ln \frac{[A]_0}{[A]_t} & =k t \[4pt] t & =\ln \frac{[A]_0}{[A]_t} \times \frac{1}{k} \end{align*}
Invoking the definition of half-life, symbolized requires that the concentration of A at this point is one-half its initial concentration: $t=t_{1 / 2}$, $[A]_t=\frac{1}{2}[A]_0$
Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:
\begin{align*} t_{1 / 2} & =\ln \frac{[A]_0}{\frac{1}{2}[A]_0} \times \frac{1}{k} \[4pt] & =\ln 2 \times \frac{1}{k}=0.693 \times \frac{1}{k} \[4pt] t_{1 / 2} & =\frac{0.693}{k} \end{align*}
This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.
Example $6$: Calculation of a First-order Rate Constant using Half-Life
Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure $4$.
Solution
Inspecting the concentration/time data in Figure $4$ shows the half-life for the decomposition of H2O2 is 2.16 × 104 s:
\begin{align*} t_{1 / 2} & =\frac{0.693}{k} \[4pt] k & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.16 \times 10^4 s }=3.21 \times 10^{-5} s^{-1} \end{align*}
Exercise $1$
The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d−1. What is the half-life for this decay?
Answer
5.02 d.
Second-Order Reactions
Following the same approach as used for first-order reactions, an equation relating the half-life of a second-order reaction to its rate constant and initial concentration may be derived from its integrated rate law:
$\frac{1}{[A]_t}=k t+\frac{1}{[A]_0} \nonumber$
or
$\frac{1}{[A]}-\frac{1}{[A]_0}=k t \nonumber$
Restrict t to t1/2
$t=t_{1 / 2} \nonumber$
define [A]t as one-half [A]0
$[A]_t=\frac{1}{2}[A]_0 \nonumber$
and then substitute into the integrated rate law and simplify:
\begin{align*} \frac{1}{\frac{1}{2}[A]_0}-\frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] \frac{2}{[A]_0}-\frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] \frac{1}{[A]_0} & =k t_{1 / 2} \[4pt] t_{1 / 2} & =\frac{1}{k[A]_0} \end{align*}
For a second-order reaction, $t_{1/2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zero-Order Reactions
As for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law:
$[A]=-k t+[A]_0 \nonumber$
Restricting the time and concentrations to those defined by half-life: and Substituting these terms into the zero-order integrated rate law yields:
\begin{align*} \frac{[ A ]_0}{2} & =-k t_{1 / 2}+[ A ]_0 \[4pt] k t_{1 / 2} & =\frac{[ A ]_0}{2} \[4pt] t_{1 / 2} & =\frac{[A]_0}{2 k} \end{align*}
As for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases.
Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table $1$.
Table $1$: Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k[A] rate = k[A]2
units of rate constant M s−1 s−1 M−1 s−1
integrated rate law $[A]=-k t+[A]_0$ $\ln [A]=-k t+\ln [A]_0$ $\frac{1}{[A]}=k t+\left(\frac{1}{[A]_0}\right)$
plot needed for linear fit of rate data [A] vs. t ln[A] vs. t vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = slope
half-life $t_{1 / 2}=\frac{[A]_0}{2 k}$ $t_{1 / 2}=\frac{0.693}{k}$ $t_{1 / 2}=\frac{1}{[A]_0 k}$
Example $7$: Half-Life for Zero-Order and Second-Order Reactions
What is the half-life for the butadiene dimerization reaction described in Example $3$?
Solution
The reaction in question is second order, is initiated with a 0.200 mol L−1 reactant solution, and exhibits a rate constant of 0.0576 L mol−1 min−1. Substituting these quantities into the second-order half-life equation:
$t_{1 / 2}=\frac{1}{\left[\left(0.0576 L mol^{-1} min^{-1}\right)\left(0.200 mol L^{-1}\right)\right]}=86.8~\text{min} \nonumber$
Exercise $7$
What is the half-life (min) for the thermal decomposition of ammonia on tungsten (see Figure $3$)?
Answer
87 min | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.05%3A_Integrated_Rate_Laws.txt |
Learning Objectives
By the end of this section, you will be able to:
• Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
• Define the concepts of activation energy and transition state
• Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.
Collision theory is based on the following postulates:
1. The rate of a reaction is proportional to the rate of reactant collisions: $\text { reaction rate } \propto \frac{\# \text { collisions }}{\text { time }} \nonumber$
2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).
We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:
$\ce{2 CO (g) + O_2(g) \longrightarrow 2 CO_2(g)} \nonumber$
Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure.
The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:
$\ce{CO (g) + O_2(g) \longrightarrow CO_2(g) + O (g)} \nonumber$
Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure $1$. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms (\ce{O=C=O}\). This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.
If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.
Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.
Activation Energy and the Arrhenius Equation
The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.
Figure $2$ shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation
$\ce{A+B \longrightarrow C+D} \nonumber$
These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, $\ce{A + B}$. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, C + D. The diagram depicts the reaction's activation energy, Ea, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, ΔH, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic (ΔH < 0) since it yields a decrease in system enthalpy.
The Arrhenius equation relates the activation energy and the rate constant, k, for many chemical reactions:
$k=A e^{-E_{ a } / R T} \nonumber$
In this equation, $R$ is the ideal gas constant, which has a value 8.314 J/mol/K, $T$ is temperature on the Kelvin scale, $E_a$ is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.
Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, A, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates.
The exponential term, e−Ea/RT, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure $\PageIndex{3a}$. Two shaded areas under the curve represent the numbers of molecules possessing adequate energy (RT) to overcome the activation barriers (Ea). A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.
The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy (RT) to overcome the activation barrier (Ea), as shown in Figure $\PageIndex{3b}$. This yields a greater value for the rate constant and a correspondingly faster reaction rate.
A convenient approach for determining $E_a$ for a reaction involves the measurement of k at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation
\begin{align*} \ln k & =\left(\frac{-E_{ a }}{R}\right)\left(\frac{1}{T}\right) + \ln A \[4pt] y & =m x+b \end{align*}
A plot of $\ln k$ versus $\frac{1}{T}$ is linear with a slope equal to $\frac{-E_a}{R}$ and a y-intercept equal to $\ln A$.
Example $1$: Determination of Ea
The variation of the rate constant with temperature for the decomposition of $\ce{HI(g)}$ to $\ce{H2(g)}$ and $\ce{I2(g)}$ is given here. What is the activation energy for the reaction?
$\ce{2 HI (g) \longrightarrow H_2(g) + I_2(g)} \nonumber$
T (K) k (L/mol/s)
555 3.52 × 10−7
575 1.22 × 10−6
645 8.59 × 10−5
700 1.16 × 10−3
781 3.95 × 10−2
Solution
Use the provided data to derive values of $\frac{1}{T}$ and $\ln k$:
$\frac{1}{T}$ $\ln k$
1.80 × 10−3 −14.860
1.74 × 10−3 −13.617
1.55 × 10−3 −9.362
1.43 × 10−3 −6.759
1.28 × 10−3 −3.231
Figure $4$ is a graph of $\ln k$ versus $\frac{1}{T}$. In practice, the equation of the line (slope and y-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.
\begin{align*} \text { Slope }= & \frac{\Delta(\ln k)}{\Delta\left(\frac{1}{T}\right)} \[4pt] = & \frac{(-14.860)-(-3.231)}{\left(1.80 \times 10^{-3} K^{-1}\right)-\left(1.28 \times 10^{-3} K^{-1}\right)} \[4pt] = & \frac{-11.629}{0.52 \times 10^{-3} K^{-1}}=-2.2 \times 10^4 K \[4pt] = & -\frac{E_{ a }}{R} \[4pt] E_{ a }= & - \text { slope } \times R=-\left(-2.2 \times 10^4 K \times 8.314 J mol^{-1} K^{-1}\right) \[4pt] & 1.8 \times 10^5 J mol^{-1} \text { or } 180 kJ mol^{-1} \end{align*}
Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:
$\ln \frac{k_1}{k_2}=\frac{E_{ a }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \nonumber$
Rearranging this equation to isolate activation energy yields:
$E_{ a }=-R\left(\frac{\ln k_2-\ln k_1}{\left(\frac{1}{T_2}\right)-\left(\frac{1}{T_1}\right)}\right) \nonumber$
Any two data pairs may be substituted into this equation—for example, the first and last entries from the above data table:
$E_{ a }=-8.314 J mol^{-1} K^{-1}\left(\frac{-3.231-(-14.860)}{1.28 \times 10^{-3} K^{-1}-1.80 \times 10^{-3} K^{-1}}\right) \nonumber$
and the result is Ea = 1.8 105 J mol−1 or 180 kJ mol−1
This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.
Exercise $1$
The rate constant for the rate of decomposition of $\ce{N2O5}$ to $\ce{NO}$ and $\ce{O2}$ in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:
$\ce{2 N2O5(g) -> 4 NO(g) + 3 O2(g)} \nonumber$
Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.
Answer
1.1 105 J mol−1 or 110 kJ mol−1 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.06%3A_Collision_Theory.txt |
Learning Objectives
By the end of this section, you will be able to:
• Distinguish net reactions from elementary reactions (steps)
• Identify the molecularity of elementary reactions
• Write a balanced chemical equation for a process given its reaction mechanism
• Derive the rate law consistent with a given reaction mechanism
Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.
The decomposition of ozone, for example, appears to follow a mechanism with two steps:
\begin{align*} & O_3(g) \longrightarrow O_2(g) + O \[4pt] & O + O_3(g) \longrightarrow 2 O_2(g) \end{align*}
Each of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:
$2 O_3(g) \longrightarrow 3 O_2(g) \nonumber$
Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.
While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one O3 decomposes to yield O2 and an oxygen atom, and a second O3 molecule subsequently reacts with the oxygen atom to yield two additional O2 molecules.
Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction’s equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.
Unimolecular Elementary Reactions
The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product:
$A \longrightarrow \text { products } \nonumber$
The rate law for a unimolecular reaction is first order:
$\text { rate }=k[A] \nonumber$
A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:
$O_3 \longrightarrow O_2 + O \nonumber$
illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an “overall” reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C4H8, to ethylene, C2H4, is represented by the following chemical equation:
This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first-order behavior:
$\text { rate }=-\frac{\Delta\left[ C_4 H_8\right]}{\Delta t}=k\left[ C_4 H_8\right] \nonumber$
This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.
Bimolecular Elementary Reactions
A bimolecular reaction involves two reactant species, for example:
$A+B \longrightarrow \text { products } \nonumber$
and
$2 A \longrightarrow \text { products } \nonumber$
For the first type, in which the two reactant molecules are different, the rate law is first-order in A and first order in B (second-order overall):
$\text { rate }=k[A][B] \nonumber$
For the second type, in which two identical molecules collide and react, the rate law is second order in A:
$\text { rate }=k[A][A]=k[A]^2 \nonumber$
Some chemical reactions occur by mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:
$NO_2(g) + CO (g) \longrightarrow NO (g) + CO_2(g) \nonumber$
(see Figure $1$)
Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:
$O (g) + O_3(g) \longrightarrow 2 O_2(g) \nonumber$
Termolecular Elementary Reactions
An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:
\begin{align*} & 2 NO + O_2 \longrightarrow 2 NO_2 \[4pt] & \text { rate }=k[ NO ]^2\left[ O_2\right] \end{align*}
Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:
\begin{align*} & 2 NO + Cl_2 \longrightarrow 2 NOCl \[4pt] & \text { rate }=k[ NO ]^2\left[ Cl_2\right] \end{align*}
Relating Reaction Mechanisms to Rate Laws
It’s often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure $2$.
As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example:
$NO_2(g) + CO (g) \longrightarrow CO_2(g) + NO (g) \nonumber$
For temperatures above 225 °C, the rate law has been found to be:
$\text { rate }=k\left[ NO_2\right][ CO ] \nonumber$
The reaction is first order with respect to NO2 and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.
At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to NO2:
$\text { rate }=k\left[ NO_2\right]^2 \nonumber$
This rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:
\begin{align*} & NO_2(g) + NO_2(g) \longrightarrow NO_3(g) + NO (g) \text { (slow) } \[4pt] & NO_3(g) + CO (g) \longrightarrow NO_2(g) + CO_2(g) \text { (fast) } \end{align*}
The rate-determining (slower) step gives a rate law showing second-order dependence on the NO2 concentration, and the sum of the two equations gives the net overall reaction.
In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive.
As discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species N2O2.
$NO + NO \rightleftharpoons N_2 O_2 \nonumber$
When this reaction is at equilibrium:
\begin{align*} & \text { rate }_{\text {forward }}=\text { rate }_{\text {reverse }} \[4pt] & k_1[ NO ]^2=k_{-1}\left[ N_2 O_2\right] \end{align*}
This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:
$\left(\frac{ k_1[ NO ]^2}{ k_{-1}}\right)=\left[ N_2 O_2\right] \nonumber$
Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.
Example $1$: Deriving a Rate Law from a Reaction Mechanism
The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:
\begin{align*} &\text{Step 1: } \quad \ce{NO(g) + Cl2(g) <=> NOCl2(g)} \tag{fast} \[4pt] &\text{Step 2: } \quad \ce{NOCl2(g) + NO(g) <=> 2 NOCl(g)} \tag{slow} \end{align*}
Use this mechanism to derive the equation and predicted rate law for the overall reaction.
Solution
The equation for the overall reaction is obtained by adding the two elementary reactions:
$2 NO (g) + Cl_2(g) \longrightarrow 2 NOCl (g) \nonumber$
To derive a rate law from this mechanism, first write rates laws for each of the two steps.
\begin{align*} & \text { rate }_1=k_1[ NO ]\left[ Cl_2\right] \text { for the forward reaction of step } 1 \[4pt] & \text { rate }_{-1}=k_{-1}\left[ NOCl_2\right] \text { for the reverse reaction of step } 1 \[4pt] & \text { rate }_2=k_2\left[ NOCl_2\right][ NO ] \text { for step } 2 \end{align*}
Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [NOCl2]. To remedy this, use the first step’s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.
Assuming step 1 is at equilibrium:
\begin{align*} \text { rate }_1 & =\text { rate }_{-1} \[4pt] k_1[ NO ]\left[ Cl_2\right] & =k_{-1}\left[ NOCl_2\right] \[4pt] {\left[ NOCl_2\right] } & =\left(\frac{k_1}{k_{-1}}\right)[ NO ]\left[ Cl_2\right] \end{align*}
Substituting this expression into the rate law for step 2 yields:
$\text { rate }_2=\text { rate }_{\text {overall }}=\left(\frac{k_2 k_1}{k_{-1}}\right)[ NO ]^2\left[ Cl_2\right] \nonumber$
Exercise $1$
The first step of a proposed multistep mechanism is:
$\ce{F2(g) <=> 2 F(g) \tag{fast}$
Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.
Answer
$[\ce{F}]=\left(\frac{k_1\left[ F_2\right]}{k_{-1}}\right)^{1 / 2} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.07%3A_Reaction_Mechanisms.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams
• List examples of catalysis in natural and industrial processes
Catalysts Do Not Affect Equilibrium
A catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.
The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation
$N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year.
Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the 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.
Fritz Haber
In the early 20th century, German chemist Fritz Haber (Figure 17.19) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb.
$N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (N2) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).
Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.
Figure 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.
In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.
Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.
It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.
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 in which N2, H2, and NH3 are at equilibrium or are coming to equilibrium.
$N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.
Although increasing the pressure of a mixture of N2, H2, and NH3 will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of N2 and H2, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:
$N2(g)+3H2(g)⟶2NH3(g)ΔH=−92.2kJN2(g)+3H2(g)⟶2NH3(g)ΔH=−92.2kJ$
Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.
Part of the rate of formation lost 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.
In the commercial production of ammonia, conditions of about 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 17.20).
Figure 17.20 Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.
Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function.
Figure 17.21 shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn’t necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower Ea).
Figure 17.21 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states).
Reaction Diagrams for Catalyzed Reactions
The two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst.
Solution
Activation energies are calculated by subtracting the reactant energy from the transition state energy. $diagram (a):Ea=32kJ−6kJ=26kJdiagram (a):Ea=32kJ−6kJ=26kJ$
$diagram (b):Ea=20kJ−6kJ=14kJdiagram (b):Ea=20kJ−6kJ=14kJ$
The catalyzed reaction is the one with lesser activation energy, in this case represented by diagram (b).
Check Your Learning
Reaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare?
Answer:
For the first step, Ea = 80 kJ for (a) and 70 kJ for (b), so diagram (b) depicts the catalyzed reaction. Activation energies for the second steps of both mechanisms are the same, 20 kJ.
Homogeneous Catalysts
A homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product.
As an important illustration of homogeneous catalysis, consider the earth’s ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:
$3O2(g)→hv2O3(g)3O2(g)→hv2O3(g)$
Ozone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following two-step mechanism:
$O3⟶O2+OO+O3⟶2O2O3⟶O2+OO+O3⟶2O2$
A number of substances can catalyze the decomposition of ozone. For example, the nitric oxide -catalyzed decomposition of ozone is believed to occur via the following three-step mechanism:
$NO(g)+O3(g)⟶NO2(g)+O2(g)O3(g)⟶O2(g)+O(g)NO2(g)+O(g)⟶NO(g)+O2(g)NO(g)+O3(g)⟶NO2(g)+O2(g)O3(g)⟶O2(g)+O(g)NO2(g)+O(g)⟶NO(g)+O2(g)$
As required, the overall reaction is the same for both the two-step uncatalyzed mechanism and the three-step NO-catalyzed mechanism:
$2O3(g)⟶3O2(g)2O3(g)⟶3O2(g)$
Notice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction.
Mario J. Molina
The 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure 17.22), and F. Sherwood Rowland “for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone.”2 Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).
Figure 17.22 (a) Mexican chemist Mario Molina (1943 –) shared the Nobel Prize in Chemistry in 1995 for his research on (b) the Antarctic ozone hole. (credit a: courtesy of Mario Molina; credit b: modification of work by NASA)
In 1974, Molina and Rowland published a paper in the journal Nature detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth’s upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable “hole” forms above Antarctica, and an increase in the amount of solar ultraviolet radiation— strongly linked to the prevalence of skin cancers—reaches earth’s surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction.
Molina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons—once widely used as refrigerants and propellants—are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:
$CH3Cl+OH⟶Cl+other productsCH3Cl+OH⟶Cl+other products$
Chlorine radicals break down ozone and are regenerated by the following catalytic cycle:
$Cl+O3⟶ClO+O2ClO+O⟶Cl+O2overall Reaction:O3+O⟶2O2Cl+O3⟶ClO+O2ClO+O⟶Cl+O2overall Reaction:O3+O⟶2O2$
A single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms Cl2 and ClONO2.
Since receiving his portion of the Nobel Prize, Molina has continued his work in atmospheric chemistry at MIT.
Glucose-6-Phosphate Dehydrogenase Deficiency
Enzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure 17.23, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (Figure 17.24).
Figure 17.23 Glucose-6-phosphate dehydrogenase is a rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells.
A disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.
Figure 17.24 In the mechanism for the pentose phosphate pathway, G6PD catalyzes the reaction that regulates NADPH, a co-enzyme that regulates glutathione, an antioxidant that protects red blood cells and other cells from oxidative damage.
Heterogeneous Catalysts
A heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.
Heterogeneous catalysis typically involves the following processes:
1. Adsorption of the reactant(s) onto the surface of the catalyst
2. Activation of the adsorbed reactant(s)
3. Reaction of the adsorbed reactant(s)
4. Desorption of product(s) from the surface of the catalyst
Figure 17.25 illustrates the steps of a mechanism for the reaction of compounds containing a carbon–carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon–carbon double bonds) to produce saturated fats and oils (which contain only carbon–carbon single bonds).
Figure 17.25 Mechanism for the Ni-catalyzed reaction $C2H4+H2⟶C2H6.C2H4+H2⟶C2H6.$ (a) Hydrogen is adsorbed on the surface, breaking the H–H bonds and forming Ni–H bonds. (b) Ethylene is adsorbed on the surface, breaking the C-C π-bond and forming Ni–C bonds. (c) Atoms diffuse across the surface and form new C–H bonds when they collide. (d) C2H6 molecules desorb from the Ni surface.
Many important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure 17.26).
Automobile Catalytic Converters
Scientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure 17.26).
Figure 17.26 A catalytic converter allows for the combustion of all carbon-containing compounds to carbon dioxide, while at the same time reducing the output of nitrogen oxide and other pollutants in emissions from gasoline-burning engines.
Most modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion of nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:
$2NO2(g)⟶N2(g)+2O2(g)2CO(g)+O2(g)⟶2CO2(g)2C8H18(g)+25O2(g)⟶16CO2(g)+18H2O(g)2NO2(g)⟶N2(g)+2O2(g)2CO(g)+O2(g)⟶2CO2(g)2C8H18(g)+25O2(g)⟶16CO2(g)+18H2O(g)$
In order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures.
Link to Learning
The University of California at Davis’ “ChemWiki” provides a thorough explanation of how catalytic converters work.
Enzyme Structure and Function
The study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table 17.3.
Classes of Enzymes and Their Functions
Class Function
oxidoreductases redox reactions
transferases transfer of functional groups
hydrolases hydrolysis reactions
lyases group elimination to form double bonds
isomerases isomerization
ligases bond formation with ATP hydrolysis
Table 17.3
Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme’s active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure 17.27).
Figure 17.27 (a) According to the lock-and-key model, the shape of an enzyme’s active site is a perfect fit for the substrate. (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate.
Link to Learning
The Royal Society of Chemistry provides an excellent introduction to enzymes for students and teachers.
The connection between the rate of a reaction and its equilibrium constant is one we can easily determine with just a bit of algebraic substitution. For a reaction where substance A forms B (and the reverse)
$A⇌BA⇌B$
The rate of the forward reaction is
$Rate(f)=k(f)[A]Rate(f)=k(f)[A]$
And the rate of the reverse reaction is
$Rate(r)=k(r)[B]Rate(r)=k(r)[B]$
Once equilibrium is established, the rates of the forward and reverse reactions are equal:
$Rate(f)=Rate(r)=k(f)[A]=k(r)[B]Rate(f)=Rate(r)=k(f)[A]=k(r)[B]$
Rearranging a bit, we get
$Ratef=Ratersok(f)[A]=k(r)[B]Ratef=Ratersok(f)[A]=k(r)[B]$
Also recall that the equilibrium constant is simply the ratio of product to reactant concentration at equilibrium:
$k(f)k(r)=[B][A]k(f)k(r)=[B][A]$ $K=[B][A]K=[B][A]$
So the equilibrium constant turns out to be the ratio of the forward to the reverse rate constants. This relationship also helps cement our understanding of the nature of a catalyst. That is, a catalyst does not change the fundamental equilibrium (or the underlying thermodynamics) of a reaction. Rather, what it does is alter the rate constant for the reaction – that is, both rate constants, forward and reverse, equally. In doing so, catalysts usually speed up the rate at which reactions attain equilibrium (though they can be used to slow down the rate of reaction as well!).
Footnotes
• 1Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.
• 2“The Nobel Prize in Chemistry 1995,” Nobel Prize.org, accessed February 18, 2015, http://www.nobelprize.org/nobel_priz...aureates/1995/. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.08%3A_Catalysis.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
activated complex(also, transition state) unstable combination of reactant species formed during a chemical reaction
activation energy (Ea)minimum energy necessary in order for a reaction to take place
Arrhenius equationmathematical relationship between a reaction’s rate constant, activation energy, and temperature
average raterate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred
bimolecular reactionelementary reaction involving two reactant species
catalystsubstance that increases the rate of a reaction without itself being consumed by the reaction
collision theorymodel that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics
elementary reactionreaction that takes place in a single step, precisely as depicted in its chemical equation
frequency factor (A)proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation
half-life of a reaction (tl/2)time required for half of a given amount of reactant to be consumed
heterogeneous catalystcatalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur
homogeneous catalystcatalyst present in the same phase as the reactants
initial rateinstantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)
instantaneous raterate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time
integrated rate lawequation that relates the concentration of a reactant to elapsed time of reaction
intermediatespecies produced in one step of a reaction mechanism and consumed in a subsequent step
method of initial ratescommon experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations
molecularitynumber of reactant species involved in an elementary reaction
overall reaction ordersum of the reaction orders for each substance represented in the rate law
rate constant (k)proportionality constant in a rate law
rate expressionmathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time
rate law(also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
rate of reactionmeasure of the speed at which a chemical reaction takes place
rate-determining step(also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction
reaction diagramused in chemical kinetics to illustrate various properties of a reaction
reaction mechanismstepwise sequence of elementary reactions by which a chemical change takes place
reaction ordervalue of an exponent in a rate law (for example, zero order for 0, first order for 1, second order for 2, and so on)
termolecular reactionelementary reaction involving three reactant species
unimolecular reactionelementary reaction involving a single reactant species | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.09%3A_Key_Terms.txt |
integrated rate law for zero-order reactions:
half-life for a zero-order reaction
integrated rate law for first-order reactions:
half-life for a first-order reaction
integrated rate law for second-order reactions:
half-life for a second-order reaction
17.11: Summary
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy.
Rate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.
The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction’s half-life varies with rate constant and, for some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
Chemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant, activation energy, temperature, and dependence on collision orientation.
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.
Catalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy. Catalysts can be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than the reactants). | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.10%3A_Key_Equations.txt |
1.
What is the difference between average rate, initial rate, and instantaneous rate?
2.
Ozone decomposes to oxygen according to the equation $2O3(g)⟶3O2(g).2O3(g)⟶3O2(g).$ Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
3.
In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction $Cl2(g)+3F2(g)⟶2ClF3(g).Cl2(g)+3F2(g)⟶2ClF3(g).$ Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.
4.
A study of the rate of dimerization of C4H6 gave the data shown in the table:
$2C4H6⟶C8H122C4H6⟶C8H12$
Time (s) 0 1600 3200 4800 6200
[C4H6] (M) 1.00 $××$ 10−2 5.04 $××$ 10−3 3.37 $××$ 10−3 2.53 $××$ 10−3 2.08 $××$ 10−3
(a) Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
(b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate?
(c) Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).
5.
A study of the rate of the reaction represented as $2A⟶B2A⟶B$ gave the following data:
Time (s) 0.0 5.0 10.0 15.0 20.0 25.0 35.0
[A] (M) 1.00 0.775 0.625 0.465 0.360 0.285 0.230
(a) Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
(b) Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?
(c) Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.
6.
Consider the following reaction in aqueous solution:
$5Br−(aq)+BrO3−(aq)+6H+(aq)⟶3Br2(aq)+3H2O(l)5Br−(aq)+BrO3−(aq)+6H+(aq)⟶3Br2(aq)+3H2O(l)$
If the rate of disappearance of Br(aq) at a particular moment during the reaction is 3.5 $××$ 10−4 mol L−1 s−1, what is the rate of appearance of Br2(aq) at that moment?
7.
Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.
8.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
9.
Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.
(a) What happens when the angle of the collision is changed?
(b) Explain how this is relevant to rate of reaction.
10.
In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature?
11.
In the PhET Reactions & Rates interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.
(a) Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
(b) Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.
12.
How do the rate of a reaction and its rate constant differ?
13.
Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:
(a) What is the order of the reaction with respect to that reactant?
(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
14.
Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:
(a) What is the order of the reaction with respect to that reactant?
(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold. What is the order of the reaction with respect to that reactant?
15.
How will the rate of reaction change for the process: $CO(g)+NO2(g)⟶CO2(g)+NO(g)CO(g)+NO2(g)⟶CO2(g)+NO(g)$ if the rate law for the reaction is $rate=k[NO2]2?rate=k[NO2]2?$
(a) Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm.
(b) Increasing the concentration of CO from 0.01 M to 0.03 M.
16.
How will each of the following affect the rate of the reaction: $CO(g)+NO2(g)⟶CO2(g)+NO(g)CO(g)+NO2(g)⟶CO2(g)+NO(g)$ if the rate law for the reaction is $rate=k[NO2][CO]rate=k[NO2][CO]$?
(a) Increasing the pressure of NO2 from 0.1 atm to 0.3 atm
(b) Increasing the concentration of CO from 0.02 M to 0.06 M.
17.
Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $NO+O3⟶NO2+O2NO+O3⟶NO2+O2$ is first order with respect to both NO and O3 with a rate constant of 2.20 $××$ 107 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 $××$ 10−6 M and [O3] = 5.9 $××$ 10−7 M?
18.
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:
$1532P⟶ 1632S+e−1532P⟶ 1632S+e−$
rate = 4.85 $××$ 10−2 $day−1[32P]day−1[32P]$
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?
19.
The rate constant for the radioactive decay of 14C is 1.21 $××$ 10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles):
$614C⟶714N+e−614C⟶714N+e−$
$rate=k[614C]rate=k[614C]$
What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 $××$ 10−9 M?
20.
The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 $××$ 10−8 L mol−1 s−1. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of 5.55 $××$ 10−4 M?
21.
Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:
[C2H5OH] (M) 4.4 $××$ 10−2 3.3 $××$ 10−2 2.2 $××$ 10−2
Rate (mol L−1 h−1) 2.0 $××$ 10−2 2.0 $××$ 10−2 2.0 $××$ 10−2
Determine the rate law, the rate constant, and the overall order for this reaction.
22.
Under certain conditions the decomposition of ammonia on a metal surface gives the following data:
[NH3] (M) 1.0 $××$ 10−3 2.0 $××$ 10−3 3.0 $××$ 10−3
Rate (mol L−1 h−1) 1.5 $××$ 10−6 1.5 $××$ 10−6 1.5 $××$ 10−6
Determine the rate law, the rate constant, and the overall order for this reaction.
23.
Nitrosyl chloride, NOCl, decomposes to NO and Cl2.
$2NOCl(g)⟶2NO(g)+Cl2(g)2NOCl(g)⟶2NO(g)+Cl2(g)$
Determine the rate law, the rate constant, and the overall order for this reaction from the following data:
[NOCl] (M) 0.10 0.20 0.30
Rate (mol L−1 h−1) 8.0 $××$ 10−10 3.2 $××$ 10−9 7.2 $××$ 10−9
24.
From the following data, determine the rate law, the rate constant, and the order with respect to A for the reaction $A⟶2C.A⟶2C.$
[A] (M) 1.33 $××$ 10−2 2.66 $××$ 10−2 3.99 $××$ 10−2
Rate (mol L−1 h−1) 3.80 $××$ 10−7 1.52 $××$ 10−6 3.42 $××$ 10−6
25.
Nitrogen monoxide reacts with chlorine according to the equation:
$2NO(g)+Cl2(g)⟶2NOCl(g)2NO(g)+Cl2(g)⟶2NOCl(g)$
The following initial rates of reaction have been observed for certain reactant concentrations:
[NO] (mol/L) [Cl2] (mol/L) Rate (mol L−1 h−1)
0.50 0.50 1.14
1.00 0.50 4.56
1.00 1.00 9.12
What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?
26.
Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: $H2(g)+2NO(g)⟶N2O(g)+H2O(g)H2(g)+2NO(g)⟶N2O(g)+H2O(g)$
Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:
[NO] (M) 0.30 0.60 0.60
[H2] (M) 0.35 0.35 0.70
Rate (mol L−1 s−1) 2.835 $××$ 10−3 1.134 $××$ 10−2 2.268 $××$ 10−2
27.
For the reaction $A⟶B+C,A⟶B+C,$ the following data were obtained at 30 °C:
[A] (M) 0.230 0.356 0.557
Rate (mol L−1 s−1) 4.17 $××$ 10−4 9.99 $××$ 10−4 2.44 $××$ 10−3
(a) What is the order of the reaction with respect to [A], and what is the rate law?
(b) What is the rate constant?
28.
For the reaction $Q⟶W+X,Q⟶W+X,$ the following data were obtained at 30 °C:
[Q]initial (M) 0.170 0.212 0.357
Rate (mol L−1 s−1) 6.68 $××$ 10−3 1.04 $××$ 10−2 2.94 $××$ 10−2
(a) What is the order of the reaction with respect to [Q], and what is the rate law?
(b) What is the rate constant?
29.
The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6.2 $××$ 10−4 min−1.
$2N2O5⟶4NO2+O22N2O5⟶4NO2+O2$
What is the rate of the reaction when [N2O5] = 0.40 M?
30.
The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.
(a) $4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)$
(b) $2NO(g)+O2(g)⟶2NO2(g)2NO(g)+O2(g)⟶2NO2(g)$
(c) $3NO2(g)+H2O(l)⟶2HNO3(aq)+NO(g)3NO2(g)+H2O(l)⟶2HNO3(aq)+NO(g)$
The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 $××$ 10−6 L2 mol−2 s−1.
31.
The following data have been determined for the reaction:
$I−+OCl−⟶IO−+Cl−I−+OCl−⟶IO−+Cl−$
1 2 3
$[I−]initial[I−]initial$ (M) 0.10 0.20 0.30
$[OCl−]initial[OCl−]initial$ (M) 0.050 0.050 0.010
Rate (mol L−1 s−1) 3.05 $××$ 10−4 6.20 $××$ 10−4 1.83 $××$ 10−4
Determine the rate law and the rate constant for this reaction.
32.
Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of A at varying times.
33.
Use the data provided to graphically determine the order and rate constant of the following reaction: $SO2Cl2⟶SO2+Cl2SO2Cl2⟶SO2+Cl2$
Time (s) 0 5.00 $××$ 103 1.00 $××$ 104 1.50 $××$ 104
[SO2Cl2] (M) 0.100 0.0896 0.0802 0.0719
Time (s) 2.50 $××$ 104 3.00 $××$ 104 4.00 $××$ 104
[SO2Cl2] (M) 0.0577 0.0517 0.0415
34.
Pure ozone decomposes slowly to oxygen, $2O3(g)⟶3O2(g).2O3(g)⟶3O2(g).$ Use the data provided in a graphical method and determine the order and rate constant of the reaction.
Time (h) 0 2.0 $××$ 103 7.6 $××$ 103 1.00 $××$ 104
[O3] (M) 1.00 $××$ 10−5 4.98 $××$ 10−6 2.07 $××$ 10−6 1.66 $××$ 10−6
Time (h) 1.23 $××$ 104 1.43 $××$ 104 1.70 $××$ 104
[O3] (M) 1.39 $××$ 10−6 1.22 $××$ 10−6 1.05 $××$ 10−6
35.
From the given data, use a graphical method to determine the order and rate constant of the following reaction:
$2X⟶Y+Z2X⟶Y+Z$
Time (s) 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0
[X] (M) 0.0990 0.0497 0.0332 0.0249 0.0200 0.0166 0.0143 0.0125
36.
What is the half-life for the first-order decay of phosphorus-32? $( 1532P ⟶ 1632S +e−)(1532P⟶1632S+e−)$ The rate constant for the decay is 4.85 $××$ 10−2 day−1.
37.
What is the half-life for the first-order decay of carbon-14? $( 614C ⟶ 714N +e−)(614C⟶714N+e−)$ The rate constant for the decay is 1.21 $××$ 10−4 year−1.
38.
What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 $××$ 10−8 L mol−1 s−1.
39.
What is the half-life for the decomposition of O3 when the concentration of O3 is 2.35 $××$ 10−6 M? The rate constant for this second-order reaction is 50.4 L mol−1 h−1.
40.
The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L mol−1 s−1. If the initial concentration is 0.500 mol/L, what is the value of t1/2?
41.
The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 M. How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to A or (b) second order with respect to A?
42.
Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 $××$ 104 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 $××$ 10−6 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.
[Penicillin] (M) Rate (mol L−1 min−1)
2.0 $××$ 10−6 1.0 $××$ 10−10
3.0 $××$ 10−6 1.5 $××$ 10−10
4.0 $××$ 10−6 2.0 $××$ 10−10
43.
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
44.
There are two molecules with the formula C3H6. Propene, $CH3CH=CH2,CH3CH=CH2,$ is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:
When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of
5.95 $××$ 10−4 s−1. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?
45.
Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is $918F ⟶ 818O ++10e)918F⟶818O++10e)$ Physicians use 18F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.
(a) What is the rate constant for the decomposition of fluorine-18?
(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?
(c) How long does it take for 99.99% of the 18F to decay?
46.
Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for $164164$ of the initial dose to remain in the athlete’s body?
47.
Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.
48.
Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:
Initial [C3H5N3O9] (M) 4.88 3.52 2.29 1.81 5.33 4.05 2.95 1.72
t (s) 300 300 300 300 180 180 180 180
% Decomposed 52.0 52.9 53.2 53.9 34.6 35.9 36.0 35.4
49.
For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene $(CH2=CH–CH=CH2)(CH2=CH–CH=CH2)$ has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:
The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 $××$ 10−4 s−1 at 150 °C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 °C with an initial pressure of 55 torr.
50.
Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
51.
When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
52.
What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
53.
Account for the relationship between the rate of a reaction and its activation energy.
54.
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
55.
How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
56.
The rate of a certain reaction doubles for every 10 °C rise in temperature.
(a) How much faster does the reaction proceed at 45 °C than at 25 °C?
(b) How much faster does the reaction proceed at 95 °C than at 25 °C?
57.
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher? (Hint: Assume the rate doubles for each 10 °C rise in temperature.)
58.
The rate constant at 325 °C for the decomposition reaction $C4H8⟶2C2H4C4H8⟶2C2H4$ is 6.1 $××$ 10−8 s−1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
59.
The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 $××$ 10−2 L mol−1 s−1 at 703 K and 4.95 L mol−1 s−1 at 865 K. Determine the activation energy for this decomposition.
60.
An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
61.
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
(a) the change in free energy per second
(b) the change in temperature per second
(c) the number of collisions per second
(d) the number of product molecules
62.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
Temperature (K) k (L mol−1 s−1)
555 6.23 $××$ 10−7
575 2.42 $××$ 10−6
645 1.44 $××$ 10−4
700 2.01 $××$ 10−3
What is the value of the activation energy (in kJ/mol) for this reaction?
63.
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
T (K) k (s−1)
293 0.054
298 0.100
64.
The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
$C12H22O11+H2O⟶C6H12O6+C6H12O6C12H22O11+H2O⟶C6H12O6+C6H12O6$
follows a first-order rate law for the disappearance of sucrose: rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
(a) In neutral solution, k = 2.1 $××$ 10−11 s−1 at 27 °C and 8.5 $××$ 10−11 s−1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
(b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 $××$ 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
65.
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+CA+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
66.
Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first $A+BC⟶AB+CA+BC⟶AB+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
67.
Why are elementary reactions involving three or more reactants very uncommon?
68.
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction $A+B⟶CA+B⟶C$? Can we predict the effect if the reaction is known to be an elementary reaction?
69.
Define these terms:
(a) unimolecular reaction
(b) bimolecular reaction
(c) elementary reaction
(d) overall reaction
70.
What is the rate law for the elementary termolecular reaction $A+2B⟶products?A+2B⟶products?$ For $3A⟶products?3A⟶products?$
71.
Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?
$(a) Cl 2 + CO ⟶ Cl 2 CO rate = k [ Cl 2 ] 3/2 [ CO ] (a) Cl 2 + CO ⟶ Cl 2 CO rate = k [ Cl 2 ] 3/2 [ CO ]$
$(b) PCl 3 + Cl 2 ⟶ PCl 5 rate = k [ PCl 3 ] [ Cl 2 ] (b) PCl 3 + Cl 2 ⟶ PCl 5 rate = k [ PCl 3 ] [ Cl 2 ]$
$(c) 2 NO + H 2 ⟶ N 2 + H 2 O 2 rate = k [ NO ] [ H 2 ] (c) 2 NO + H 2 ⟶ N 2 + H 2 O 2 rate = k [ NO ] [ H 2 ]$
$(d) 2 NO + O 2 ⟶ 2 NO 2 rate = k [NO] 2 [ O 2 ] (d) 2 NO + O 2 ⟶ 2 NO 2 rate = k [NO] 2 [ O 2 ]$
$(e) NO + O 3 ⟶ NO 2 + O 2 rate = k [ NO ] [ O 3 ] (e) NO + O 3 ⟶ NO 2 + O 2 rate = k [ NO ] [ O 3 ]$
72.
Write the rate law for each of the following elementary reactions:
(a) $O3→sunlightO2+OO3→sunlightO2+O$
(b) $O3+Cl⟶O2+ClOO3+Cl⟶O2+ClO$
(c) $ClO+O⟶Cl+O2ClO+O⟶Cl+O2$
(d) $O3+NO⟶NO2+O2O3+NO⟶NO2+O2$
(e) $NO2+O⟶NO+O2NO2+O⟶NO+O2$
73.
Nitrogen monoxide, NO, reacts with hydrogen, H2, according to the following equation:
$2NO+2H2⟶N2+2H2O2NO+2H2⟶N2+2H2O$
What would the rate law be if the mechanism for this reaction were:
$2NO+H2⟶N2+H2O2(slow)H2O2+H2⟶2H2O(fast)2NO+H2⟶N2+H2O2(slow)H2O2+H2⟶2H2O(fast)$
74.
Experiments were conducted to study the rate of the reaction represented by this equation.3
$2NO(g)+2H2(g)⟶N2(g)+2H2O(g)2NO(g)+2H2(g)⟶N2(g)+2H2O(g)$
Initial concentrations and rates of reaction are given here.
Experiment Initial Concentration [NO] (mol L−1) Initial Concentration, [H2] (mol L−1 min−1) Initial Rate of Formation of N2 (mol L−1 min−1)
1 0.0060 0.0010 1.8 $××$ 10−4
2 0.0060 0.0020 3.6 $××$ 10−4
3 0.0010 0.0060 0.30 $××$ 10−4
4 0.0020 0.0060 1.2 $××$ 10−4
Consider the following questions:
(a) Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning.
(b) Write the overall rate law for the reaction.
(c) Calculate the value of the rate constant, k, for the reaction. Include units.
(d) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H2 had been consumed.
(e) The following sequence of elementary steps is a proposed mechanism for the reaction.
Step 1: $NO+NO⇌N2O2NO+NO⇌N2O2$
Step 2: $N2O2+H2⇌H2O+N2ON2O2+H2⇌H2O+N2O$
Step 3: $N2O+H2⇌N2+H2ON2O+H2⇌N2+H2O$
Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.
75.
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:
$Cl2(g)⇌2Cl(g)Cl2(g)⇌2Cl(g)$ (fast, k1 represents the forward rate constant, k−1 the reverse rate constant)
$CO(g)+Cl(g)⟶COCl(g)CO(g)+Cl(g)⟶COCl(g)$ (slow, k2 the rate constant)
$COCl(g)+Cl(g)⟶COCl2(g)COCl(g)+Cl(g)⟶COCl2(g)$ (fast, k3 the rate constant)
(a) Write the overall reaction.
(b) Identify all intermediates.
(c) Write the rate law for each elementary reaction.
(d) Write the overall rate law expression.
76.
Account for the increase in reaction rate brought about by a catalyst.
77.
Compare the functions of homogeneous and heterogeneous catalysts.
78.
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:
$O3→sunlightO2+OO3+Cl⟶O2+ClOClO+O⟶Cl+O2O3→sunlightO2+OO3+Cl⟶O2+ClOClO+O⟶Cl+O2$
(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:
$2O3⟶3O22O3⟶3O2$
(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism:
$O3→sunlightO2+OO3+NO⟶NO2+O2NO2+O⟶NO+O2O3→sunlightO2+OO3+NO⟶NO2+O2NO2+O⟶NO+O2$
Is NO a catalyst for the decomposition? Explain your answer.
79.
Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: $H2O(g)+C(s)⇌H2(g)+CO(g).H2O(g)+C(s)⇌H2(g)+CO(g).$ Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst. What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
80.
Nitrogen and oxygen react at high temperatures. What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
81.
For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:
(a)
(b)
82.
For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:
(a)
(b)
83.
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
84.
For each of the following reaction diagrams, estimate the activation energy (Ea) of the reaction:
(a)
(b)
85.
Assuming the diagrams in Exercise 17.83 represent different mechanisms for the same reaction, which of the reactions has the faster rate?
86.
Consider the similarities and differences in the two reaction diagrams shown in Exercise 17.84. Do these diagrams represent two different overall reactions, or do they represent the same overall reaction taking place by two different mechanisms? Explain your answer.
Footnotes
• 3This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/17%3A_Kinetics/17.12%3A_Exercises.txt |
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table.
18: Representative Metals Metalloids and Nonmetals
The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table.
18.02: Periodicity
Learning Objectives
By the end of this section, you will be able to:
• Classify elements
• Make predictions about the periodicity properties of the representative elements
We begin this section by examining the behaviors of representative metals in relation to their positions in the periodic table. The primary focus of this section will be the application of periodicity to the representative metals.
It is possible to divide elements into groups according to their electron configurations. The representative elements are elements where the s and p orbitals are filling. The transition elements are elements where the d orbitals (groups 3–11 on the periodic table) are filling, and the inner transition metals are the elements where the f orbitals are filling. The d orbitals fill with the elements in group 11; therefore, the elements in group 12 qualify as representative elements because the last electron enters an s orbital. Metals among the representative elements are the representative metals. Metallic character results from an element’s ability to lose its outer valence electrons and results in high thermal and electrical conductivity, among other physical and chemical properties. There are 20 nonradioactive representative metals in groups 1, 2, 3, 12, 13, 14, and 15 of the periodic table (the elements shaded in yellow in Figure $1$). The radioactive elements copernicium, flerovium, polonium, and livermorium are also metals but are beyond the scope of this chapter.
In addition to the representative metals, some of the representative elements are metalloids. A metalloid is an element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors.
The remaining representative elements are nonmetals. Unlike metals, which typically form cations and ionic compounds (containing ionic bonds), nonmetals tend to form anions or molecular compounds. In general, the combination of a metal and a nonmetal produces a salt. A salt is an ionic compound consisting of cations and anions.
Most of the representative metals do not occur naturally in an uncombined state because they readily react with water and oxygen in the air. However, it is possible to isolate elemental beryllium, magnesium, zinc, cadmium, mercury, aluminum, tin, and lead from their naturally occurring minerals and use them because they react very slowly with air. Part of the reason why these elements react slowly is that these elements react with air to form a protective coating. The formation of this protective coating is passivation. The coating is a nonreactive film of oxide or some other compound. Elemental magnesium, aluminum, zinc, and tin are important in the fabrication of many familiar items, including wire, cookware, foil, and many household and personal objects. Although beryllium, cadmium, mercury, and lead are readily available, there are limitations in their use because of their toxicity.
Group 1: The Alkali Metals
The alkali metals lithium, sodium, potassium, rubidium, cesium, and francium constitute group 1 of the periodic table. Although hydrogen is in group 1 (and also in group 17), it is a nonmetal and deserves separate consideration later in this chapter. The name alkali metal is in reference to the fact that these metals and their oxides react with water to form very basic (alkaline) solutions.
The properties of the alkali metals are similar to each other as expected for elements in the same family. The alkali metals have the largest atomic radii and the lowest first ionization energy in their periods. This combination makes it very easy to remove the single electron in the outermost (valence) shell of each. The easy loss of this valence electron means that these metals readily form stable cations with a charge of 1+. Their reactivity increases with increasing atomic number due to the ease of losing the lone valence electron (decreasing ionization energy). Since oxidation is so easy, the reverse, reduction, is difficult, which explains why it is hard to isolate the elements. The solid alkali metals are very soft; lithium, shown in Figure $2$, has the lowest density of any metal (0.5 g/cm3).
The alkali metals all react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. This means they are easier to oxidize than is hydrogen. As an example, the reaction of lithium with water is:
$\ce{2 Li(s) + 2 H2O(l) \longrightarrow 2 LiOH(aq) + H_2(g)} \nonumber$
Alkali metals react directly with all the nonmetals (except the noble gases) to yield binary ionic compounds containing 1+ metal ions. These metals are so reactive that it is necessary to avoid contact with both moisture and oxygen in the air. Therefore, they are stored in sealed containers under mineral oil, as shown in Figure $3$, to prevent contact with air and moisture. The pure metals never exist free (uncombined) in nature due to their high reactivity. In addition, this high reactivity makes it necessary to prepare the metals by electrolysis of alkali metal compounds.
Unlike many other metals, the reactivity and softness of the alkali metals make these metals unsuitable for structural applications. However, there are applications where the reactivity of the alkali metals is an advantage. For example, the production of metals such as titanium and zirconium relies, in part, on the ability of sodium to reduce compounds of these metals. The manufacture of many organic compounds, including certain dyes, drugs, and perfumes, utilizes reduction by lithium or sodium.
Sodium and its compounds impart a bright yellow color to a flame, as seen in Figure $4$. Passing an electrical discharge through sodium vapor also produces this color. In both cases, this is an example of an emission spectrum as discussed in the chapter on electronic structure. Streetlights sometime employ sodium vapor lights because the sodium vapor penetrates fog better than most other light. This is because the fog does not scatter yellow light as much as it scatters white light. The other alkali metals and their salts also impart color to a flame. Lithium creates a bright, crimson color, whereas the others create a pale, violet color.
This video demonstrates the reactions of the alkali metals with water.
Group 2: The Alkaline Earth Metals
The alkaline earth metals (beryllium, magnesium, calcium, strontium, barium, and radium) constitute group 2 of the periodic table. The name alkaline metal comes from the fact that the oxides of the heavier members of the group react with water to form alkaline solutions. The nuclear charge increases when going from group 1 to group 2. Because of this charge increase, the atoms of the alkaline earth metals are smaller and have higher first ionization energies than the alkali metals within the same period. The higher ionization energy makes the alkaline earth metals less reactive than the alkali metals; however, they are still very reactive elements. Their reactivity increases, as expected, with increasing size and decreasing ionization energy. In chemical reactions, these metals readily lose both valence electrons to form compounds in which they exhibit an oxidation state of 2+. Due to their high reactivity, it is common to produce the alkaline earth metals, like the alkali metals, by electrolysis. Even though the ionization energies are low, the two metals with the highest ionization energies (beryllium and magnesium) do form compounds that exhibit some covalent characters. Like the alkali metals, the heavier alkaline earth metals impart color to a flame. As in the case of the alkali metals, this is part of the emission spectrum of these elements. Calcium and strontium produce shades of red, whereas barium produces a green color.
Magnesium is a silver-white metal that is malleable and ductile at high temperatures. Passivation decreases the reactivity of magnesium metal. Upon exposure to air, a tightly adhering layer of magnesium oxycarbonate forms on the surface of the metal and inhibits further reaction. (The carbonate comes from the reaction of carbon dioxide in the atmosphere.) Magnesium is the lightest of the widely used structural metals, which is why most magnesium production is for lightweight alloys.
Magnesium (shown in Figure $5$), calcium, strontium, and barium react with water and air. At room temperature, barium shows the most vigorous reaction. The products of the reaction with water are hydrogen and the metal hydroxide. The formation of hydrogen gas indicates that the heavier alkaline earth metals are better reducing agents (more easily oxidized) than is hydrogen. As expected, these metals react with both acids and nonmetals to form ionic compounds. Unlike most salts of the alkali metals, many of the common salts of the alkaline earth metals are insoluble in water because of the high lattice energies of these compounds, containing a divalent metal ion.
The potent reducing power of hot magnesium is useful in preparing some metals from their oxides. Indeed, magnesium’s affinity for oxygen is so great that burning magnesium reacts with carbon dioxide, producing elemental carbon:
$\ce{2 Mg(s) + CO_2(g) \longrightarrow 2 MgO(s) + C(s)} \nonumber$
For this reason, a CO2 fire extinguisher will not extinguish a magnesium fire. Additionally, the brilliant white light emitted by burning magnesium makes it useful in flares and fireworks.
Group 12
The elements in group 12 are transition elements; however, the last electron added is not a d electron, but an s electron. Since the last electron added is an s electron, these elements qualify as representative metals, or post-transition metals. The group 12 elements behave more like the alkaline earth metals than transition metals. Group 12 contains the four elements zinc, cadmium, mercury, and copernicium. Each of these elements has two electrons in its outer shell (ns2). When atoms of these metals form cations with a charge of 2+, where the two outer electrons are lost, they have pseudo-noble gas electron configurations. Mercury is sometimes an exception because it also exhibits an oxidation state of 1+ in compounds that contain a diatomic ion. In their elemental forms and in compounds, cadmium and mercury are both toxic.
Zinc is the most reactive in group 12, and mercury is the least reactive. (This is the reverse of the reactivity trend of the metals of groups 1 and 2, in which reactivity increases down a group. The increase in reactivity with increasing atomic number only occurs for the metals in groups 1 and 2.) The decreasing reactivity is due to the formation of ions with a pseudo-noble gas configuration and to other factors that are beyond the scope of this discussion. The chemical behaviors of zinc and cadmium are quite similar to each other but differ from that of mercury.
Zinc and cadmium have lower reduction potentials than hydrogen, and, like the alkali metals and alkaline earth metals, they will produce hydrogen gas when they react with acids. The reaction of zinc with hydrochloric acid, shown in Figure $6$, is:
$\ce{Zn(s) + 2 H3O^{+}(aq) + 2 Cl^{-}(aq) \longrightarrow H_2(g) + Zn^{2+}(aq) + 2 Cl^{-}(aq) + 2 H2O(l)} \nonumber$
Zinc is a silvery metal that quickly tarnishes to a blue-gray appearance. This change in color is due to an adherent coating of a basic carbonate, Zn2(OH)2CO3, which passivates the metal to inhibit further corrosion. Dry cell and alkaline batteries contain a zinc anode. Brass (Cu and Zn) and some bronze (Cu, Sn, and sometimes Zn) are important zinc alloys. About half of zinc production serves to protect iron and other metals from corrosion. This protection may take the form of a sacrificial anode (also known as a galvanic anode, which is a means of providing cathodic protection for various metals) or as a thin coating on the protected metal. Galvanized steel is steel with a protective coating of zinc.
Chemistry in Everyday Life: Sacrificial Anodes
A sacrificial anode, or galvanic anode, is a means of providing cathodic protection of various metals. Cathodic protection refers to the prevention of corrosion by converting the corroding metal into a cathode. As a cathode, the metal resists corrosion, which is an oxidation process. Corrosion occurs at the sacrificial anode instead of at the cathode.
The construction of such a system begins with the attachment of a more active metal (more negative reduction potential) to the metal needing protection. Attachment may be direct or via a wire. To complete the circuit, a salt bridge is necessary. This salt bridge is often seawater or ground water. Once the circuit is complete, oxidation (corrosion) occurs at the anode and not the cathode.
The commonly used sacrificial anodes are magnesium, aluminum, and zinc. Magnesium has the most negative reduction potential of the three and serves best when the salt bridge is less efficient due to a low electrolyte concentration such as in freshwater. Zinc and aluminum work better in saltwater than does magnesium. Aluminum is lighter than zinc and has a higher capacity; however, an oxide coating may passivate the aluminum. In special cases, other materials are useful. For example, iron will protect copper.
Mercury is very different from zinc and cadmium. Mercury is the only metal that is liquid at 25 °C. Many metals dissolve in mercury, forming solutions called amalgams (see the feature on Amalgams), which are alloys of mercury with one or more other metals. Mercury, shown in Figure $7$, is a nonreactive element that is more difficult to oxidize than hydrogen. Thus, it does not displace hydrogen from acids; however, it will react with strong oxidizing acids, such as nitric acid:
\begin{align*} \ce{Hg(l) + HCl(aq) &->}~\text{no reaction} \[4pt] \ce{3 Hg(l) + 8 HNO_3(aq) &-> 3 Hg(NO3)2(aq) + 4 H2O(l) + 2 NO(g)} \end{align*}
The clear NO initially formed quickly undergoes further oxidation to the reddish brown NO2.
Most mercury compounds decompose when heated. Most mercury compounds contain mercury with a 2+-oxidation state. When there is a large excess of mercury, it is possible to form compounds containing the ion. All mercury compounds are toxic, and it is necessary to exercise great care in their synthesis.
Chemistry in Everyday Life: Amalgams
An amalgam is an alloy of mercury with one or more other metals. This is similar to considering steel to be an alloy of iron with other metals. Most metals will form an amalgam with mercury, with the main exceptions being iron, platinum, tungsten, and tantalum.
Due to toxicity issues with mercury, there has been a significant decrease in the use of amalgams. Historically, amalgams were important in electrolytic cells and in the extraction of gold. Amalgams of the alkali metals still find use because they are strong reducing agents and easier to handle than the pure alkali metals.
Prospectors had a problem when they found finely divided gold. They learned that adding mercury to their pans collected the gold into the mercury to form an amalgam for easier collection. Unfortunately, losses of small amounts of mercury over the years left many streams in California polluted with mercury.
Dentists use amalgams containing silver and other metals to fill cavities. There are several reasons to use an amalgam including low cost, ease of manipulation, and longevity compared to alternate materials. Dental amalgams are approximately 50% mercury by weight, which, in recent years, has become a concern due to the toxicity of mercury.
After reviewing the best available data, the Food and Drug Administration (FDA) considers amalgam-based fillings to be safe for adults and children over six years of age. Even with multiple fillings, the mercury levels in the patients remain far below the lowest levels associated with harm. Clinical studies have found no link between dental amalgams and health problems. Health issues may not be the same in cases of children under six or pregnant women. The FDA conclusions are in line with the opinions of the Environmental Protection Agency (EPA) and Centers for Disease Control (CDC). The only health consideration noted is that some people are allergic to the amalgam or one of its components.
Group 13
Group 13 contains the metalloid boron and the metals aluminum, gallium, indium, and thallium. The lightest element, boron, is semiconducting, and its binary compounds tend to be covalent and not ionic. The remaining elements of the group are metals, but their oxides and hydroxides change characters. The oxides and hydroxides of aluminum and gallium exhibit both acidic and basic behaviors. A substance, such as these two, that will react with both acids and bases is amphoteric. This characteristic illustrates the combination of nonmetallic and metallic behaviors of these two elements. Indium and thallium oxides and hydroxides exhibit only basic behavior, in accordance with the clearly metallic character of these two elements. The melting point of gallium is unusually low (about 30 °C) and will melt in your hand.
Aluminum is amphoteric because it will react with both acids and bases. A typical reaction with an acid is:
$\ce{2 Al(s) + 6 HCl(aq) \longrightarrow 2 AlCl_3(aq) + 3 H_2(g)} \nonumber$
The products of the reaction of aluminum with a base depend upon the reaction conditions, with the following being one possibility:
$\ce{2 Al(s) + 2 NaOH(aq) + 6 H2O(l) -> 2 Na[Al(OH)4](aq) + 3 H2(g)} \nonumber$
With both acids and bases, the reaction with aluminum generates hydrogen gas.
The group 13 elements have a valence shell electron configuration of ns2np1. Aluminum normally uses all of its valence electrons when it reacts, giving compounds in which it has an oxidation state of 3+. Although many of these compounds are covalent, others, such as AlF3 and Al2(SO4)3, are ionic. Aqueous solutions of aluminum salts contain the cation $\ce{[Al(H2O)6]^{3+}}$ abbreviated as Al3+(aq). Gallium, indium, and thallium also form ionic compounds containing M3+ ions. These three elements exhibit not only the expected oxidation state of 3+ from the three valence electrons but also an oxidation state (in this case, 1+) that is two below the expected value. This phenomenon, the inert pair effect, refers to the formation of a stable ion with an oxidation state two lower than expected for the group. The pair of electrons is the valence s orbital for those elements. In general, the inert pair effect is important for the lower p-block elements. In an aqueous solution, the Tl+(aq) ion is more stable than is Tl3+(aq). In general, these metals will react with air and water to form 3+ ions; however, thallium reacts to give thallium(I) derivatives. The metals of group 13 all react directly with nonmetals such as sulfur, phosphorus, and the halogens, forming binary compounds.
The metals of group 13 (Al, Ga, In, and Tl) are all reactive. However, passivation occurs as a tough, hard, thin film of the metal oxide forms upon exposure to air. Disruption of this film may counter the passivation, allowing the metal to react. One way to disrupt the film is to expose the passivated metal to mercury. Some of the metal dissolves in the mercury to form an amalgam, which sheds the protective oxide layer to expose the metal to further reaction. The formation of an amalgam allows the metal to react with air and water.
Although easily oxidized, the passivation of aluminum makes it very useful as a strong, lightweight building material. Because of the formation of an amalgam, mercury is corrosive to structural materials made of aluminum. This video demonstrates how the integrity of an aluminum beam can be destroyed by the addition of a small amount of elemental mercury.
The most important uses of aluminum are in the construction and transportation industries, and in the manufacture of aluminum cans and aluminum foil. These uses depend on the lightness, toughness, and strength of the metal, as well as its resistance to corrosion. Because aluminum is an excellent conductor of heat and resists corrosion, it is useful in the manufacture of cooking utensils.
Aluminum is a very good reducing agent and may replace other reducing agents in the isolation of certain metals from their oxides. Although more expensive than reduction by carbon, aluminum is important in the isolation of Mo, W, and Cr from their oxides.
Group 14
The metallic members of group 14 are tin, lead, and flerovium. Carbon is a typical nonmetal. The remaining elements of the group, silicon and germanium, are examples of semimetals or metalloids. Tin and lead form the stable divalent cations, Sn2+ and Pb2+, with oxidation states two below the group oxidation state of 4+. The stability of this oxidation state is a consequence of the inert pair effect. Tin and lead also form covalent compounds with a formal 4+-oxidation state. For example, SnCl4 and PbCl4 are low-boiling covalent liquids.
Tin reacts readily with nonmetals and acids to form tin(II) compounds (indicating that it is more easily oxidized than hydrogen) and with nonmetals to form either tin(II) or tin(IV) compounds (shown in Figure $8$), depending on the stoichiometry and reaction conditions. Lead is less reactive. It is only slightly easier to oxidize than hydrogen, and oxidation normally requires a hot concentrated acid.
Many of these elements exist as allotropes. Allotropes are two or more forms of the same element in the same physical state with different chemical and physical properties. There are two common allotropes of tin. These allotropes are grey (brittle) tin and white tin. As with other allotropes, the difference between these forms of tin is in the arrangement of the atoms. White tin is stable above 13.2 °C and is malleable like other metals. At low temperatures, gray tin is the more stable form. Gray tin is brittle and tends to break down to a powder. Consequently, articles made of tin will disintegrate in cold weather, particularly if the cold spell is lengthy. The change progresses slowly from the spot of origin, and the gray tin that is first formed catalyzes further change. In a way, this effect is similar to the spread of an infection in a plant or animal body, leading people to call this process tin disease or tin pest.
The principal use of tin is in the coating of steel to form tin plate-sheet iron, which constitutes the tin in tin cans. Important tin alloys are bronze (Cu and Sn) and solder (Sn and Pb). Lead is important in the lead storage batteries in automobiles.
Group 15
Bismuth, the heaviest member of group 15, is a less reactive metal than the other representative metals. It readily gives up three of its five valence electrons to active nonmetals to form the tri-positive ion, Bi3+. It forms compounds with the group oxidation state of 5+ only when treated with strong oxidizing agents. The stability of the 3+-oxidation state is another example of the inert pair effect. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify natural sources of representative metals
• Describe electrolytic and chemical reduction processes used to prepare these elements from natural sources
Because of their reactivity, we do not find most representative metals as free elements in nature. However, compounds that contain ions of most representative metals are abundant. In this section, we will consider the two common techniques used to isolate the metals from these compounds—electrolysis and chemical reduction.
These metals primarily occur in minerals, with lithium found in silicate or phosphate minerals, and sodium and potassium found in salt deposits from evaporation of ancient seas and in silicates. The alkaline earth metals occur as silicates and, with the exception of beryllium, as carbonates and sulfates. Beryllium occurs as the mineral beryl, Be3Al2Si6O18, which, with certain impurities, may be either the gemstone emerald or aquamarine. Magnesium is in seawater and, along with the heavier alkaline earth metals, occurs as silicates, carbonates, and sulfates. Aluminum occurs abundantly in many types of clay and in bauxite, an impure aluminum oxide hydroxide. The principle tin ore is the oxide cassiterite, SnO2, and the principle lead and thallium ores are the sulfides or the products of weathering of the sulfides. The remaining representative metals occur as impurities in zinc or aluminum ores.
Electrolysis
Ions of metals in of groups 1 and 2, along with aluminum, are very difficult to reduce; therefore, it is necessary to prepare these elements by electrolysis, an important process discussed in the chapter on electrochemistry. Briefly, electrolysis involves using electrical energy to drive unfavorable chemical reactions to completion; it is useful in the isolation of reactive metals in their pure forms. Sodium, aluminum, and magnesium are typical examples.
The Preparation of Sodium
The most important method for the production of sodium is the electrolysis of molten sodium chloride; the set-up is a Downs cell, shown in Figure $1$. The reaction involved in this process is:
$\ce{2 NaCl(l) ->[\text { electrolysis }][600^{\circ} C] 2 Na(l) + Cl_2(g)} \nonumber$
The electrolysis cell contains molten sodium chloride (melting point 801 °C), to which calcium chloride has been added to lower the melting point to 600 °C (a colligative effect). The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Chloride ions migrate to the positively charged anode, lose electrons, and undergo oxidation to chlorine gas. The overall cell reaction comes from adding the following reactions:
\begin{align*} &\text{at the cathode:} && \ce{2 Na^{+} + 2 e^{-} -> 2 Na(l)} \[4pt] &\text{at the anode:} && \ce{2 Cl^{-} -> Cl_2(g) + 2 e^{-}} \[4pt] \hline &\text{overall change:} && \ce{ 2 Na^{+} + 2 Cl^{-} -> 2 Na(l) + Cl_2(g)} \end{align*}
Separation of the molten sodium and chlorine prevents recombination. The liquid sodium, which is less dense than molten sodium chloride, floats to the surface and flows into a collector. The gaseous chlorine goes to storage tanks. Chlorine is also a valuable product.
The Preparation of Aluminum
The preparation of aluminum utilizes a process invented in 1886 by Charles M. Hall, who began to work on the problem while a student at Oberlin College in Ohio. Paul L. T. Héroult discovered the process independently a month or two later in France. In honor to the two inventors, this electrolysis cell is known as the Hall–Héroult cell. The Hall–Héroult cell is an electrolysis cell for the production of aluminum. Figure $2$: illustrates the Hall–Héroult cell.
The production of aluminum begins with the purification of bauxite, the most common source of aluminum. The reaction of bauxite, AlO(OH), with hot sodium hydroxide forms soluble sodium aluminate, while clay and other impurities remain undissolved:
$\ce{AlO(OH)(s) + NaOH(aq) + H2O(l) -> Na[Al(OH)4](aq)} \nonumber$
After the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:
$\ce{Na[Al(OH)4](aq) + H3O^{+}(aq) -> Al(OH)3(s) + Na^{+}(aq) + 2 H2O(l)} \nonumber$
The next step is to remove the precipitated aluminum hydroxide by filtration. Heating the hydroxide produces aluminum oxide, Al2O3, which dissolves in a molten mixture of cryolite, Na3AlF6, and calcium fluoride, CaF2. Electrolysis of this solution takes place in a cell like that shown in Figure $2$. Reduction of aluminum ions to the metal occurs at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode.
The Preparation of Magnesium
Magnesium is the other metal that is isolated in large quantities by electrolysis. Seawater, which contains approximately 0.5% magnesium chloride, serves as the major source of magnesium. Addition of calcium hydroxide to seawater precipitates magnesium hydroxide. The addition of hydrochloric acid to magnesium hydroxide, followed by evaporation of the resultant aqueous solution, leaves pure magnesium chloride. The electrolysis of molten magnesium chloride forms liquid magnesium and chlorine gas:
\begin{align*} \ce{MgCl2(aq) + Ca(OH)2(aq) &-> Mg(OH)2(s) + CaCl2(aq)} \[4pt] \ce{Mg(OH)2(s) + 2 HCl(aq) &-> MgCl_2(aq) + 2 H2O(l)} \[4pt] \ce{MgCl2(l) &-> Mg(l) + Cl_2(g)} \end{align*}
Some production facilities have moved away from electrolysis completely. In the next section, we will see how the Pidgeon process leads to the chemical reduction of magnesium.
Chemical Reduction
It is possible to isolate many of the representative metals by chemical reduction using other elements as reducing agents. In general, chemical reduction is much less expensive than electrolysis, and for this reason, chemical reduction is the method of choice for the isolation of these elements. For example, it is possible to produce potassium, rubidium, and cesium by chemical reduction, as it is possible to reduce the molten chlorides of these metals with sodium metal. This may be surprising given that these metals are more reactive than sodium; however, the metals formed are more volatile than sodium and can be distilled for collection. The removal of the metal vapor leads to a shift in the equilibrium to produce more metal (see how reactions can be driven in the discussions of Le Chatelier’s principle in the chapter on fundamental equilibrium concepts).
The production of magnesium, zinc, and tin provide additional examples of chemical reduction.
The Preparation of Magnesium
The Pidgeon process involves the reaction of magnesium oxide with elemental silicon at high temperatures to form pure magnesium:
$\ce{Si(s) + 2 MgO(s) ->[\Delta] SiO_2(s) + 2 Mg(g)} \nonumber$
Although this reaction is unfavorable in terms of thermodynamics, the removal of the magnesium vapor produced takes advantage of Le Chatelier’s principle to continue the forward progress of the reaction. Over 75% of the world’s production of magnesium, primarily in China, comes from this process.
The Preparation of Zinc
Zinc ores usually contain zinc sulfide, zinc oxide, or zinc carbonate. After separation of these compounds from the ores, heating in air converts the ore to zinc oxide by one of the following reactions:
\begin{align*} \ce{2 ZnS(s) + 3 O_2(g) &->[\Delta] 2 ZnO(s) + 2 SO_2(g)} \[4pt] \ce{ZnCO_3(s) &->[\Delta] ZnO(s) + CO_2(g)} \end{align*}
Carbon, in the form of coal, reduces the zinc oxide to form zinc vapor:
$\ce{ZnO(s) + C(s) \longrightarrow Zn(g) + CO(g)} \nonumber$
The zinc can be distilled (boiling point 907 °C) and condensed. This zinc contains impurities of cadmium (767 °C), iron (2862 °C), lead (1750 °C), and arsenic (613 °C). Careful redistillation produces pure zinc. Arsenic and cadmium are distilled from the zinc because they have lower boiling points. At higher temperatures, the zinc is distilled from the other impurities, mainly lead and iron.
The Preparation of Tin
The ready reduction of tin(IV) oxide by the hot coals of a campfire accounts for the knowledge of tin in the ancient world. In the modern process, the roasting of tin ores containing SnO2 removes contaminants such as arsenic and sulfur as volatile oxides. Treatment of the remaining material with hydrochloric acid removes the oxides of other metals. Heating the purified ore with carbon at temperature above 1000 °C produces tin:
$\ce{SnO_2(s) + 2 C(s) ->[\Delta] Sn(s) + 2 CO(g)} \nonumber$
The molten tin collects at the bottom of the furnace and is drawn off and cast into blocks. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.03%3A_Occurrence_and_Preparation_of_the_Representative_Metals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the general preparation, properties, and uses of the metalloids
• Describe the preparation, properties, and compounds of boron and silicon
A series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements—boron and silicon—in more detail.
The metalloid boron exhibits many similarities to its neighbor carbon and its diagonal neighbor silicon. All three elements form covalent compounds. However, boron has one distinct difference in that its 2s22p1 outer electron structure gives it one less valence electron than it has valence orbitals. Although boron exhibits an oxidation state of 3+ in most of its stable compounds, this electron deficiency provides boron with the ability to form other, sometimes fractional, oxidation states, which occur, for example, in the boron hydrides.
Silicon has the valence shell electron configuration 3s23p2, and it commonly forms tetrahedral structures in which it is sp3 hybridized with a formal oxidation state of 4+. The major differences between the chemistry of carbon and silicon result from the relative strength of the carbon-carbon bond, carbon’s ability to form stable bonds to itself, and the presence of the empty 3d valence-shell orbitals in silicon. Silicon’s empty d orbitals and boron’s empty p orbital enable tetrahedral silicon compounds and trigonal planar boron compounds to act as Lewis acids. Carbon, on the other hand, has no available valence shell orbitals; tetrahedral carbon compounds cannot act as Lewis acids. Germanium is very similar to silicon in its chemical behavior.
Arsenic and antimony generally form compounds in which an oxidation state of 3+ or 5+ is exhibited; however, arsenic can form arsenides with an oxidation state of 3−. These elements tarnish only slightly in dry air but readily oxidize when warmed.
Tellurium combines directly with most elements. The most stable tellurium compounds are the tellurides—salts of Te2 formed with active metals and lanthanides—and compounds with oxygen, fluorine, and chlorine, in which tellurium normally exhibits an oxidation state 2+ or 4+. Although tellurium(VI) compounds are known (for example, TeF6), there is a marked resistance to oxidation to this maximum group oxidation state.
Structures of the Metalloids
Covalent bonding is the key to the crystal structures of the metalloids. In this regard, these elements resemble nonmetals in their behavior.
Elemental silicon, germanium, arsenic, antimony, and tellurium are lustrous, metallic-looking solids. Silicon and germanium crystallize with a diamond structure. Each atom within the crystal has covalent bonds to four neighboring atoms at the corners of a regular tetrahedron. Single crystals of silicon and germanium are giant, three-dimensional molecules. There are several allotropes of arsenic with the most stable being layer like and containing puckered sheets of arsenic atoms. Each arsenic atom forms covalent bonds to three other atoms within the sheet. The crystal structure of antimony is similar to that of arsenic, both shown in Figure $1$. The structures of arsenic and antimony are similar to the structure of graphite, covered later in this chapter. Tellurium forms crystals that contain infinite spiral chains of tellurium atoms. Each atom in the chain bonds to two other atoms.
Explore a cubic diamond crystal structure.
Pure crystalline boron is transparent. The crystals consist of icosahedra, as shown in Figure $2$, with a boron atom at each corner. In the most common form of boron, the icosahedra pack together in a manner similar to the cubic closest packing of spheres. All boron-boron bonds within each icosahedron are identical and are approximately 176 pm in length. In the different forms of boron, there are different arrangements and connections between the icosahedra.
The name silicon is derived from the Latin word for flint, silex. The metalloid silicon readily forms compounds containing Si-O-Si bonds, which are of prime importance in the mineral world. This bonding capability is in contrast to the nonmetal carbon, whose ability to form carbon-carbon bonds gives it prime importance in the plant and animal worlds.
Occurrence, Preparation, and Compounds of Boron and Silicon
Boron constitutes less than 0.001% by weight of the earth’s crust. In nature, it only occurs in compounds with oxygen. Boron is widely distributed in volcanic regions as boric acid, B(OH)3, and in dry lake regions, including the desert areas of California, as borates and salts of boron oxyacids, such as borax, Na2B4O7⋅10H2O.
Elemental boron is chemically inert at room temperature, reacting with only fluorine and oxygen to form boron trifluoride, BF3, and boric oxide, B2O3, respectively. At higher temperatures, boron reacts with all nonmetals, except tellurium and the noble gases, and with nearly all metals; it oxidizes to B2O3 when heated with concentrated nitric or sulfuric acid. Boron does not react with nonoxidizing acids. Many boron compounds react readily with water to give boric acid, B(OH)3 (sometimes written as H3BO3).
Reduction of boric oxide with magnesium powder forms boron (95–98.5% pure) as a brown, amorphous powder:
$\ce{B_2 O_3(s) + 3 Mg(s) \longrightarrow 2 B(s) + 3 MgO(s)} \nonumber$
An amorphous substance is a material that appears to be a solid, but does not have a long-range order like a true solid. Treatment with hydrochloric acid removes the magnesium oxide. Further purification of the boron begins with conversion of the impure boron into boron trichloride. The next step is to heat a mixture of boron trichloride and hydrogen:
$\ce{2 BCl_3(g) + 3 H_2(g) \stackrel{1500^{\circ} C }{\longrightarrow} 2 B(s) + 6 HCl(g)} \quad \quad \Delta H^{\circ}=253.7 kJ \nonumber$
Silicon makes up nearly one-fourth of the mass of the earth’s crust—second in abundance only to oxygen. The crust is composed almost entirely of minerals in which the silicon atoms are at the center of the silicon-oxygen tetrahedron, which connect in a variety of ways to produce, among other things, chains, layers, and three-dimensional frameworks. These minerals constitute the bulk of most common rocks, soil, and clays. In addition, materials such as bricks, ceramics, and glasses contain silicon compounds.
It is possible to produce silicon by the high-temperature reduction of silicon dioxide with strong reducing agents, such as carbon and magnesium:
\begin{align*} \ce{SiO_2(s) + 2 C(s) ->[\Delta] Si(s) + 2 CO(g)} \[4pt] \ce{SiO_2(s) + 2 Mg(s) ->[\Delta] Si(s) + 2 MgO(s)} \end{align*}
Extremely pure silicon is necessary for the manufacture of semiconductor electronic devices. This process begins with the conversion of impure silicon into silicon tetrahalides, or silane (SiH4), followed by decomposition at high temperatures. Zone refining, illustrated in Figure $3$, completes the purification. In this method, a rod of silicon is heated at one end by a heat source that produces a thin cross-section of molten silicon. Slowly lowering the rod through the heat source moves the molten zone from one end of the rod to other. As this thin, molten region moves, impurities in the silicon dissolve in the liquid silicon and move with the molten region. Ultimately, the impurities move to one end of the rod, which is then cut off.
This highly purified silicon, containing no more than one part impurity per million parts of silicon, is the most important element in the computer industry. Pure silicon is necessary in semiconductor electronic devices such as transistors, computer chips, and solar cells.
Like some metals, passivation of silicon occurs due the formation of a very thin film of oxide (primarily silicon dioxide, SiO2). Silicon dioxide is soluble in hot aqueous base; thus, strong bases destroy the passivation. Removal of the passivation layer allows the base to dissolve the silicon, forming hydrogen gas and silicate anions. For example:
$\ce{Si(s) + 4 OH^{-}(aq) \longrightarrow SiO_4^{4-}(aq) + 2 H_2(g)} \nonumber$
Silicon reacts with halogens at high temperatures, forming volatile tetrahalides, such as SiF4.
Unlike carbon, silicon does not readily form double or triple bonds. Silicon compounds of the general formula SiX4, where X is a highly electronegative group, can act as Lewis acids to form six-coordinate silicon. For example, silicon tetrafluoride, SiF4, reacts with sodium fluoride to yield Na2[SiF6], which contains the octahedral ion in which silicon is sp3d2 hybridized:
$\ce{2 NaF(s) + SiF_4(g) \longrightarrow Na_2 SiF_6(s)} \nonumber$
Antimony reacts readily with stoichiometric amounts of fluorine, chlorine, bromine, or iodine, yielding trihalides or, with excess fluorine or chlorine, forming the pentahalides SbF5 and SbCl5. Depending on the stoichiometry, it forms antimony(III) sulfide, Sb2S3, or antimony(V) sulfide when heated with sulfur. As expected, the metallic nature of the element is greater than that of arsenic, which lies immediately above it in group 15.
Boron and Silicon Halides
Boron trihalides—BF3, BCl3, BBr3, and BI3—can be prepared by the direct reaction of the elements. These nonpolar molecules contain boron with sp2 hybridization and a trigonal planar molecular geometry. The fluoride and chloride compounds are colorless gasses, the bromide is a liquid, and the iodide is a white crystalline solid.
Except for boron trifluoride, the boron trihalides readily hydrolyze in water to form boric acid and the corresponding hydrohalic acid. Boron trichloride reacts according to the equation:
$\ce{BCl_3(g) + 3 H2O(l) \longrightarrow B(OH)_3(aq) + 3 HCl(aq)} \nonumber$
Boron trifluoride reacts with hydrofluoric acid, to yield a solution of fluoroboric acid, HBF4:
$\ce{BF_3(aq) + HF(aq) + H2O(l) \longrightarrow H3O^{+}(aq) + BF_4^{-}(aq)} \nonumber$
In this reaction, the BF3 molecule acts as the Lewis acid (electron pair acceptor) and accepts a pair of electrons from a fluoride ion:
All the tetrahalides of silicon, SiX4, have been prepared. Silicon tetrachloride can be prepared by direct chlorination at elevated temperatures or by heating silicon dioxide with chlorine and carbon:
$\ce{SiO_2(s) + 2 C(s) + 2 Cl_2(g) \stackrel{\Delta}{\longrightarrow} SiCl_4(g) + 2 CO(g)}$
Silicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling (57 °C), colorless liquid.
It is possible to prepare silicon tetrafluoride by the reaction of silicon dioxide with hydrofluoric acid:
$\ce{SiO_2(s) + 4 HF(g) \longrightarrow SiF_4(g) + 2 H2O(l)} \quad \quad \Delta H^{\circ}=-191.2~\text{kJ} \nonumber$
Hydrofluoric acid is the only common acid that will react with silicon dioxide or silicates. This reaction occurs because the silicon-fluorine bond is the only bond that silicon forms that is stronger than the silicon-oxygen bond. For this reason, it is possible to store all common acids, other than hydrofluoric acid, in glass containers.
Except for silicon tetrafluoride, silicon halides are extremely sensitive to water. Upon exposure to water, SiCl4 reacts rapidly with hydroxide groups, replacing all four chlorine atoms to produce unstable orthosilicic acid, Si(OH)4 or H4SiO4, which slowly decomposes into SiO2.
Boron and Silicon Oxides and Derivatives
Boron burns at 700 °C in oxygen, forming boric oxide, B2O3. Boric oxide is necessary for the production of heat-resistant borosilicate glass, like that shown in Figure $4$ and certain optical glasses. Boric oxide dissolves in hot water to form boric acid, B(OH)3:
$\ce{B2O3(s) + 3 H2O(l) -> 2 B(OH)3(aq)} \nonumber$
The boron atom in B(OH)3 is sp2 hybridized and is located at the center of an equilateral triangle with oxygen atoms at the corners. In solid B(OH)3, hydrogen bonding holds these triangular units together. Boric acid, shown in Figure $5$, is a very weak acid that does not act as a proton donor but rather as a Lewis acid, accepting an unshared pair of electrons from the Lewis base OH:
$\ce{B(OH)3(aq) + 2 H2O(l) <=> B(OH)4^{-}(aq) + H3O^{+}(aq)} \quad \quad K_{ a }=5.8 \times 10^{-10} \nonumber$
Heating boric acid to 100 °C causes molecules of water to split out between pairs of adjacent –OH groups to form metaboric acid, HBO2. At about 150 °C, additional B-O-B linkages form, connecting the BO3 groups together with shared oxygen atoms to form tetraboric acid, H2B4O7. Complete water loss, at still higher temperatures, results in boric oxide.
Borates are salts of the oxyacids of boron. Borates result from the reactions of a base with an oxyacid or from the fusion of boric acid or boric oxide with a metal oxide or hydroxide. Borate anions range from the simple trigonal planar 2[B4O5(OH)4]⋅8H2O, which is an important component of some laundry detergents. Most of the supply of borax comes directly from dry lakes, such as Searles Lake in California, or is prepared from kernite, Na2B4O7⋅4H2O.
Silicon dioxide, silica, occurs in both crystalline and amorphous forms. The usual crystalline form of silicon dioxide is quartz, a hard, brittle, clear, colorless solid. It is useful in many ways—for architectural decorations, semiprecious jewels, and frequency control in radio transmitters. Silica takes many crystalline forms, or polymorphs, in nature. Trace amounts of Fe3+ in quartz give amethyst its characteristic purple color. The term quartz is also used for articles such as tubing and lenses that are manufactured from amorphous silica. Opal is a naturally occurring form of amorphous silica.
The contrast in structure and physical properties between silicon dioxide and carbon dioxide is interesting, as illustrated in Figure $7$. Solid carbon dioxide (dry ice) contains single CO2 molecules with each of the two oxygen atoms attached to the carbon atom by double bonds. Very weak intermolecular forces hold the molecules together in the crystal. The volatility of dry ice reflect these weak forces between molecules. In contrast, silicon dioxide is a covalent network solid. In silicon dioxide, each silicon atom links to four oxygen atoms by single bonds directed toward the corners of a regular tetrahedron, and SiO4 tetrahedra share oxygen atoms. This arrangement gives a three dimensional, continuous, silicon-oxygen network. A quartz crystal is a macromolecule of silicon dioxide. The difference between these two compounds is the ability of the group 14 elements to form strong π bonds. Second-period elements, such as carbon, form very strong π bonds, which is why carbon dioxide forms small molecules with strong double bonds. Elements below the second period, such as silicon, do not form π bonds as readily as second-period elements, and when they do form, the π bonds are weaker than those formed by second-period elements. For this reason, silicon dioxide does not contain π bonds but only σ bonds.
At 1600 °C, quartz melts to yield a viscous liquid. When the liquid cools, it does not crystallize readily but usually supercools and forms a glass, also called silica. The SiO4 tetrahedra in glassy silica have a random arrangement characteristic of supercooled liquids, and the glass has some very useful properties. Silica is highly transparent to both visible and ultraviolet light. For this reason, it is important in the manufacture of lamps that give radiation rich in ultraviolet light and in certain optical instruments that operate with ultraviolet light. The coefficient of expansion of silica glass is very low; therefore, rapid temperature changes do not cause it to fracture. CorningWare and other ceramic cookware contain amorphous silica.
Silicates are salts containing anions composed of silicon and oxygen. In nearly all silicates, sp3-hybridized silicon atoms occur at the centers of tetrahedra with oxygen at the corners. There is a variation in the silicon-to-oxygen ratio that occurs because silicon-oxygen tetrahedra may exist as discrete, independent units or may share oxygen atoms at corners in a variety of ways. In addition, the presence of a variety of cations gives rise to the large number of silicate minerals.
Many ceramics are composed of silicates. By including small amounts of other compounds, it is possible to modify the physical properties of the silicate materials to produce ceramics with useful characteristics. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.04%3A_Structure_and_General_Properties_of_the_Metalloids.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe structure and properties of nonmetals
The nonmetals are elements located in the upper right portion of the periodic table. Their properties and behavior are quite different from those of metals on the left side. Under normal conditions, more than half of the nonmetals are gases, one is a liquid, and the rest include some of the softest and hardest of solids. The nonmetals exhibit a rich variety of chemical behaviors. They include the most reactive and least reactive of elements, and they form many different ionic and covalent compounds. This section presents an overview of the properties and chemical behaviors of the nonmetals, as well as the chemistry of specific elements. Many of these nonmetals are important in biological systems.
In many cases, trends in electronegativity enable us to predict the type of bonding and the physical states in compounds involving the nonmetals. We know that electronegativity decreases as we move down a given group and increases as we move from left to right across a period. The nonmetals have higher electronegativities than do metals, and compounds formed between metals and nonmetals are generally ionic in nature because of the large differences in electronegativity between them. The metals form cations, the nonmetals form anions, and the resulting compounds are solids under normal conditions. On the other hand, compounds formed between two or more nonmetals have small differences in electronegativity between the atoms, and covalent bonding—sharing of electrons—results. These substances tend to be molecular in nature and are gases, liquids, or volatile solids at room temperature and pressure.
In normal chemical processes, nonmetals do not form monatomic positive ions (cations) because their ionization energies are too high. All monatomic nonmetal ions are anions; examples include the chloride ion, Cl, the nitride ion, N3−, and the selenide ion, Se2.
The common oxidation states that the nonmetals exhibit in their ionic and covalent compounds are shown in Figure $1$. Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.
The first member of each nonmetal group exhibits different behaviors, in many respects, from the other group members. The reasons for this include smaller size, greater ionization energy, and (most important) the fact that the first member of each group has only four valence orbitals (one 2s and three 2p) available for bonding, whereas other group members have empty d orbitals in their valence shells, making possible five, six, or even more bonds around the central atom. For example, nitrogen forms only NF3, whereas phosphorus forms both PF3 and PF5.
Another difference between the first group member and subsequent members is the greater ability of the first member to form π bonds. This is primarily a function of the smaller size of the first member of each group, which allows better overlap of atomic orbitals. Nonmetals, other than the first member of each group, rarely form π bonds to nonmetals that are the first member of a group. For example, sulfur-oxygen π bonds are well known, whereas sulfur does not normally form stable π bonds to itself.
The variety of oxidation states displayed by most of the nonmetals means that many of their chemical reactions involve changes in oxidation state through oxidation-reduction reactions. There are five general aspects of the oxidation-reduction chemistry:
1. Nonmetals oxidize most metals. The oxidation state of the metal becomes positive as it undergoes oxidation and that of the nonmetal becomes negative as it undergoes reduction. For example: $4 \overset{\large 0}{\text{Fe}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +3}{\text{Fe}}_2 \overset{\large -2}{\text{O}}_3(s) \nonumber$
2. With the exception of nitrogen and carbon, which are poor oxidizing agents, a more electronegative nonmetal oxidizes a less electronegative nonmetal or the anion of the nonmetal: $\overset{\large 0}{\text{S}}(s) + 3 \overset{\large 0}{\text{O}}_2(g) \longrightarrow 2 \overset{\large +4}{\text{S}} \overset{\large -2}{\text{O}}_2(s) \nonumber$\overset{\large 0}{\text{Cl}}_2\text{(s)} + 2 \text{I}^{-}\text{(aq)} \longrightarrow \overset{\large 0}{\text{I}_2}\text{(s)} + 2 \overset{\large }{\text{Cl}^{-}} \text{(aq)} \nonumber$
3. Fluorine and oxygen are the strongest oxidizing agents within their respective groups; each oxidizes all the elements that lie below it in the group. Within any period, the strongest oxidizing agent is in group 17. A nonmetal often oxidizes an element that lies to its left in the same period. For example: $2 \overset{\large 0}{\text{As}}(s) + 3 \overset{\large 0}{\text{Br}}_2(l) \longrightarrow 2 \overset{\large +3}{\text{As}}_2 \overset{\large -1}{\text{Br}}_3(s) \nonumber$
4. The stronger a nonmetal is as an oxidizing agent, the more difficult it is to oxidize the anion formed by the nonmetal. This means that the most stable negative ions are formed by elements at the top of the group or in group 17 of the period.
5. Fluorine and oxygen are the strongest oxidizing elements known. Fluorine does not form compounds in which it exhibits positive oxidation states; oxygen exhibits a positive oxidation state only when combined with fluorine. For example: $2 \overset{\large 0}{\text{F}}_2\text{(s)} + 2 \text{OH}^{-}\text{(aq)} \longrightarrow \overset{\large +2}{\text{O}} \text{F}_2 \text{(s)} + 2 \overset{\large -1}{\text{F}^{-}} \text{(aq)} + \text{H}_2\text{O(l)} \nonumber$
With the exception of most of the noble gases, all nonmetals form compounds with oxygen, yielding covalent oxides. Most of these oxides are acidic, that is, they react with water to form oxyacids. Recall from the acid-base chapter that an oxyacid is an acid consisting of hydrogen, oxygen, and some other element. Notable exceptions are carbon monoxide, CO, nitrous oxide, N2O, and nitric oxide, NO. There are three characteristics of these acidic oxides:
1. Oxides such as SO2 and N2O5, in which the nonmetal exhibits one of its common oxidation states, are acid anhydrides and react with water to form acids with no change in oxidation state. The product is an oxyacid. For example:
2. Those oxides such as NO2 and ClO2, in which the nonmetal does not exhibit one of its common oxidation states, also react with water. In these reactions, the nonmetal is both oxidized and reduced. For example:
Reactions in which the same element is both oxidized and reduced are called disproportionation reactions.
3. The acid strength increases as the electronegativity of the central atom increases. To learn more, see the discussion in the chapter on acid-base chemistry.
The binary hydrogen compounds of the nonmetals also exhibit an acidic behavior in water, although only HCl, HBr, and HI are strong acids. The acid strength of the nonmetal hydrogen compounds increases from left to right across a period and down a group. For example, ammonia, NH3, is a weaker acid than is water, H2O, which is weaker than is hydrogen fluoride, HF. Water, H2O, is also a weaker acid than is hydrogen sulfide, H2S, which is weaker than is hydrogen selenide, H2Se. Weaker acidic character implies greater basic character.
Structures of the Nonmetals
The structures of the nonmetals differ dramatically from those of metals. Metals crystallize in closely packed arrays that do not contain molecules or covalent bonds. Nonmetal structures contain covalent bonds, and many nonmetals consist of individual molecules. The electrons in nonmetals are localized in covalent bonds, whereas in a metal, there is delocalization of the electrons throughout the solid.
The noble gases are all monatomic, whereas the other nonmetal gases—hydrogen, nitrogen, oxygen, fluorine, and chlorine—normally exist as the diatomic molecules H2, N2, O2, F2, and Cl2. The other halogens are also diatomic; Br2 is a liquid and I2 exists as a solid under normal conditions. The changes in state as one moves down the halogen family offer excellent examples of the increasing strength of intermolecular London forces with increasing molecular mass and increasing polarizability.
Oxygen has two allotropes: O2, dioxygen, and O3, ozone. Phosphorus has three common allotropes, commonly referred to by their colors: white, red, and black. Sulfur has several allotropes. There are also many carbon allotropes. Most people know of diamond, graphite, and charcoal, but fewer people know of the recent discovery of fullerenes, carbon nanotubes, and graphene.
Descriptions of the physical properties of three nonmetals that are characteristic of molecular solids follow.
Carbon
Carbon occurs in the uncombined (elemental) state in many forms, such as diamond, graphite, charcoal, coke, carbon black, graphene, and fullerene.
Diamond, shown in Figure $2$, is a very hard crystalline material that is colorless and transparent when pure. Each atom forms four single bonds to four other atoms at the corners of a tetrahedron (sp3 hybridization); this makes the diamond a giant molecule. Carbon-carbon single bonds are very strong, and, because they extend throughout the crystal to form a three-dimensional network, the crystals are very hard and have high melting points (~4400 °C).
Graphite, also shown in Figure $2$, is a soft, slippery, grayish-black solid that conducts electricity. These properties relate to its structure, which consists of layers of carbon atoms, with each atom surrounded by three other carbon atoms in a trigonal planar arrangement. Each carbon atom in graphite forms three σ bonds, one to each of its nearest neighbors, by means of sp2-hybrid orbitals. The unhybridized p orbital on each carbon atom will overlap unhybridized orbitals on adjacent carbon atoms in the same layer to form π bonds. Many resonance forms are necessary to describe the electronic structure of a graphite layer; Figure $3$: illustrates two of these forms.
Atoms within a graphite layer are bonded together tightly by the σ and π bonds; however, the forces between layers are weak. London dispersion forces hold the layers together. To learn more, see the discussion of these weak forces in the chapter on liquids and solids. The weak forces between layers give graphite the soft, flaky character that makes it useful as the so-called “lead” in pencils and the slippery character that makes it useful as a lubricant. The loosely held electrons in the resonating π bonds can move throughout the solid and are responsible for the electrical conductivity of graphite.
Other forms of elemental carbon include carbon black, charcoal, and coke. Carbon black is an amorphous form of carbon prepared by the incomplete combustion of natural gas, CH4. It is possible to produce charcoal and coke by heating wood and coal, respectively, at high temperatures in the absence of air.
Recently, new forms of elemental carbon molecules have been identified in the soot generated by a smoky flame and in the vapor produced when graphite is heated to very high temperatures in a vacuum or in helium. One of these new forms, first isolated by Professor Richard Smalley and coworkers at Rice University, consists of icosahedral (soccer-ball-shaped) molecules that contain 60 carbon atoms, C60. This is buckminsterfullerene (often called bucky balls) after the architect Buckminster Fuller, who designed domed structures, which have a similar appearance (Figure $4$).
Chemistry in Everyday Life: Nanotubes and Graphene
Graphene and carbon nanotubes are two recently discovered allotropes of carbon. Both of the forms bear some relationship to graphite. Graphene is a single layer of graphite (one atom thick), as illustrated in Figure $5$, whereas carbon nanotubes roll the layer into a small tube, as illustrated in Figure $5$.
Graphene is a very strong, lightweight, and efficient conductor of heat and electricity discovered in 2003. As in graphite, the carbon atoms form a layer of six-membered rings with sp2-hybridized carbon atoms at the corners. Resonance stabilizes the system and leads to its conductivity. Unlike graphite, there is no stacking of the layers to give a three-dimensional structure. Andre Geim and Kostya Novoselov at the University of Manchester won the 2010 Nobel Prize in Physics for their pioneering work characterizing graphene.
The simplest procedure for preparing graphene is to use a piece of adhesive tape to remove a single layer of graphene from the surface of a piece of graphite. This method works because there are only weak London dispersion forces between the layers in graphite. Alternative methods are to deposit a single layer of carbon atoms on the surface of some other material (ruthenium, iridium, or copper) or to synthesize it at the surface of silicon carbide via the sublimation of silicon.
There currently are no commercial applications of graphene. However, its unusual properties, such as high electron mobility and thermal conductivity, should make it suitable for the manufacture of many advanced electronic devices and for thermal management applications.
Carbon nanotubes are carbon allotropes, which have a cylindrical structure. Like graphite and graphene, nanotubes consist of rings of sp2-hybridized carbon atoms. Unlike graphite and graphene, which occur in layers, the layers wrap into a tube and bond together to produce a stable structure. The walls of the tube may be one atom or multiple atoms thick.
Carbon nanotubes are extremely strong materials that are harder than diamond. Depending upon the shape of the nanotube, it may be a conductor or semiconductor. For some applications, the conducting form is preferable, whereas other applications utilize the semiconducting form.
The basis for the synthesis of carbon nanotubes is the generation of carbon atoms in a vacuum. It is possible to produce carbon atoms by an electrical discharge through graphite, vaporization of graphite with a laser, and the decomposition of a carbon compound.
The strength of carbon nanotubes will eventually lead to some of their most exciting applications, as a thread produced from several nanotubes will support enormous weight. However, the current applications only employ bulk nanotubes. The addition of nanotubes to polymers improves the mechanical, thermal, and electrical properties of the bulk material. There are currently nanotubes in some bicycle parts, skis, baseball bats, fishing rods, and surfboards.
Phosphorus
The name phosphorus comes from the Greek words meaning light bringing. When phosphorus was first isolated, scientists noted that it glowed in the dark and burned when exposed to air. Phosphorus is the only member of its group that does not occur in the uncombined state in nature; it exists in many allotropic forms. We will consider two of those forms: white phosphorus and red phosphorus.
White phosphorus is a white, waxy solid that melts at 44.2 °C and boils at 280 °C. It is insoluble in water (in which it is stored—see Figure $6$), is very soluble in carbon disulfide, and bursts into flame in air. As a solid, as a liquid, as a gas, and in solution, white phosphorus exists as P4 molecules with four phosphorus atoms at the corners of a regular tetrahedron, as illustrated in Figure $6$. Each phosphorus atom covalently bonds to the other three atoms in the molecule by single covalent bonds. White phosphorus is the most reactive allotrope and is very toxic.
Heating white phosphorus to 270–300 °C in the absence of air yields red phosphorus. Red phosphorus (shown in Figure $6$) is denser, has a higher melting point (~600 °C), is much less reactive, is essentially nontoxic, and is easier and safer to handle than is white phosphorus. Its structure is highly polymeric and appears to contain three-dimensional networks of P4 tetrahedra joined by P-P single bonds. Red phosphorus is insoluble in solvents that dissolve white phosphorus. When red phosphorus is heated, P4 molecules sublime from the solid.
Sulfur
The allotropy of sulfur is far greater and more complex than that of any other element. Sulfur is the brimstone referred to in the Bible and other places, and references to sulfur occur throughout recorded history—right up to the relatively recent discovery that it is a component of the atmospheres of Venus and of Io, a moon of Jupiter. The most common and most stable allotrope of sulfur is yellow, rhombic sulfur, so named because of the shape of its crystals. Rhombic sulfur is the form to which all other allotropes revert at room temperature. Crystals of rhombic sulfur melt at 113 °C. Cooling this liquid gives long needles of monoclinic sulfur. This form is stable from 96 °C to the melting point, 119 °C. At room temperature, it gradually reverts to the rhombic form.
Both rhombic sulfur and monoclinic sulfur contain S8 molecules in which atoms form eight-membered, puckered rings that resemble crowns, as illustrated in Figure $7$. Each sulfur atom is bonded to each of its two neighbors in the ring by covalent S-S single bonds.
When rhombic sulfur melts, the straw-colored liquid is quite mobile; its viscosity is low because S8 molecules are essentially spherical and offer relatively little resistance as they move past each other. As the temperature rises, S-S bonds in the rings break, and polymeric chains of sulfur atoms result. These chains combine end to end, forming still longer chains that tangle with one another. The liquid gradually darkens in color and becomes so viscous that finally (at about 230 °C) it does not pour easily. The dangling atoms at the ends of the chains of sulfur atoms are responsible for the dark red color because their electronic structure differs from those of sulfur atoms that have bonds to two adjacent sulfur atoms. This causes them to absorb light differently and results in a different visible color. Cooling the liquid rapidly produces a rubberlike amorphous mass, called plastic sulfur.
Sulfur boils at 445 °C and forms a vapor consisting of S2, S6, and S8 molecules; at about 1000 °C, the vapor density corresponds to the formula S2, which is a paramagnetic molecule like O2 with a similar electronic structure and a weak sulfur-sulfur double bond.
As seen in this discussion, an important feature of the structural behavior of the nonmetals is that the elements usually occur with eight electrons in their valence shells. If necessary, the elements form enough covalent bonds to supplement the electrons already present to possess an octet. For example, members of group 15 have five valence electrons and require only three additional electrons to fill their valence shells. These elements form three covalent bonds in their free state: triple bonds in the N2 molecule or single bonds to three different atoms in arsenic and phosphorus. The elements of group 16 require only two additional electrons. Oxygen forms a double bond in the O2 molecule, and sulfur, selenium, and tellurium form two single bonds in various rings and chains. The halogens form diatomic molecules in which each atom is involved in only one bond. This provides the electron required necessary to complete the octet on the halogen atom. The noble gases do not form covalent bonds to other noble gas atoms because they already have a filled outer shell. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.05%3A_Structure_and_General_Properties_of_the_Nonmetals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and compounds of hydrogen
Hydrogen is the most abundant element in the universe. The sun and other stars are composed largely of hydrogen. Astronomers estimate that 90% of the atoms in the universe are hydrogen atoms. Hydrogen is a component of more compounds than any other element. Water is the most abundant compound of hydrogen found on earth. Hydrogen is an important part of petroleum, many minerals, cellulose and starch, sugar, fats, oils, alcohols, acids, and thousands of other substances.
At ordinary temperatures, hydrogen is a colorless, odorless, tasteless, and nonpoisonous gas consisting of the diatomic molecule H2. Hydrogen is composed of three isotopes, and unlike other elements, these isotopes have different names and chemical symbols: protium, 1H, deuterium, 2H(or “D”), and tritium 3H(or “T”). In a naturally occurring sample of hydrogen, there is one atom of deuterium for every 7000 H atoms and one atom of radioactive tritium for every 1018 H atoms. The chemical properties of the different isotopes are very similar because they have identical electron structures, but they differ in some physical properties because of their differing atomic masses. Elemental deuterium and tritium have lower vapor pressure than ordinary hydrogen. Consequently, when liquid hydrogen evaporates, the heavier isotopes are concentrated in the last portions to evaporate. Electrolysis of heavy water, D2O, yields deuterium. Most tritium originates from nuclear reactions.
Preparation of Hydrogen
Elemental hydrogen must be prepared from compounds by breaking chemical bonds. The most common methods of preparing hydrogen follow.
From Steam and Carbon or Hydrocarbons
Water is the cheapest and most abundant source of hydrogen. Passing steam over coke(an impure form of elemental carbon) at 1000 °C produces a mixture of carbon monoxide and hydrogen known as water gas:
$\ce{C(s) + H2O(g) ->[1000^{\circ} ~\text{C}] \underset{\text{water gas}}{\ce{CO(g) + H_2(g)}} } \nonumber$
Water gas is as an industrial fuel. It is possible to produce additional hydrogen by mixing the water gas with steam in the presence of a catalyst to convert the CO to CO2. This reaction is the water gas shift reaction.
It is also possible to prepare a mixture of hydrogen and carbon monoxide by passing hydrocarbons from natural gas or petroleum and steam over a nickel-based catalyst. Propane is an example of a hydrocarbon reactant:
$\ce{C3H8(g) + 3H2O(g) ->[900^{\circ} ~\text{C}][\text{catalyst}] 3CO(g) + 7H2(g) } \nonumber$
Electrolysis
Hydrogen forms when direct current electricity passes through water containing an electrolyte such as H2SO4, as illustrated in Figure $1$. Bubbles of hydrogen form at the cathode, and oxygen evolves at the anode. The net reaction is:
$\ce{2 H2O(l) + electrical energy \longrightarrow 2 H2(g) + O2(g)} \nonumber$
Reaction of Metals with Acids
This is the most convenient laboratory method of producing hydrogen. Metals with lower reduction potentials reduce the hydrogen ion in dilute acids to produce hydrogen gas and metal salts. For example, as shown in Figure $2$, iron in dilute hydrochloric acid produces hydrogen gas and iron(II) chloride:
$\ce{Fe(s) + 2 H3O^{+}(aq) + 2 Cl^{-}(aq) \longrightarrow Fe^{2+}(aq) + 2 Cl^{-}(aq) + H_2(g) + 2 H2O(l)} \nonumber$
Reaction of Ionic Metal Hydrides with Water
It is possible to produce hydrogen from the reaction of hydrides of the active metals, which contain the very strongly basic H anion, with water:
$\ce{CaH_2(s) + 2 H2O(l) \longrightarrow Ca^{2+}(aq) + 2 OH^{-}(aq) + 2 H_2(g)} \nonumber$
Metal hydrides are expensive but convenient sources of hydrogen, especially where space and weight are important factors. They are important in the inflation of life jackets, life rafts, and military balloons.
Reactions
Under normal conditions, hydrogen is relatively inactive chemically, but when heated, it enters into many chemical reactions.
Two thirds of the world’s hydrogen production is devoted to the manufacture of ammonia, which is a fertilizer and used in the manufacture of nitric acid. Large quantities of hydrogen are also important in the process of hydrogenation, discussed in the chapter on organic chemistry.
It is possible to use hydrogen as a nonpolluting fuel. The reaction of hydrogen with oxygen is a very exothermic reaction, releasing 286 kJ of energy per mole of water formed. Hydrogen burns without explosion under controlled conditions. The oxygen-hydrogen torch, because of the high heat of combustion of hydrogen, can achieve temperatures up to 2800 °C. The hot flame of this torch is useful in cutting thick sheets of many metals. Liquid hydrogen is also an important rocket fuel (Figure $3$).
An uncombined hydrogen atom consists of a nucleus and one valence electron in the 1s orbital. The n = 1 valence shell has a capacity for two electrons, and hydrogen can rightfully occupy two locations in the periodic table. It is possible to consider hydrogen a group 1 element because hydrogen can lose an electron to form the cation, H+. It is also possible to consider hydrogen to be a group 17 element because it needs only one electron to fill its valence orbital to form a hydride ion, H, or it can share an electron to form a single, covalent bond. In reality, hydrogen is a unique element that almost deserves its own location in the periodic table.
Reactions with Elements
When heated, hydrogen reacts with the metals of group 1 and with Ca, Sr, and Ba(the more active metals in group 2). The compounds formed are crystalline, ionic hydrides that contain the hydride anion, H, a strong reducing agent and a strong base, which reacts vigorously with water and other acids to form hydrogen gas.
The reactions of hydrogen with nonmetals generally produce acidic hydrogen compounds with hydrogen in the 1+ oxidation state. The reactions become more exothermic and vigorous as the electronegativity of the nonmetal increases. Hydrogen reacts with nitrogen and sulfur only when heated, but it reacts explosively with fluorine(forming HF) and, under some conditions, with chlorine(forming HCl). A mixture of hydrogen and oxygen explodes if ignited. Because of the explosive nature of the reaction, it is necessary to exercise caution when handling hydrogen(or any other combustible gas) to avoid the formation of an explosive mixture in a confined space. Although most hydrides of the nonmetals are acidic, ammonia and phosphine(PH3) are very, very weak acids and generally function as bases. There is a summary of these reactions of hydrogen with the elements in Table $1$.
Table $1$: Chemical Reactions of Hydrogen with Other Elements
General Equation Comments
$\ce{MH} \text { or } \ce{MH2 \longrightarrow MOH} \text { or } \ce{M(OH)2 + H 2}$ ionic hydrides with group 1 and Ca, Sr, and Ba
$\ce{H2 + C \longrightarrow(no reaction)}$
$\ce{3 H2 + N2 \longrightarrow 2 NH3}$ requires high pressure and temperature; low yield
$\ce{2 H2 + O2 \longrightarrow 2 H2O}$ exothermic and potentially explosive
Hydrogen Compounds
Other than the noble gases, each of the nonmetals forms compounds with hydrogen. For brevity, we will discuss only a few hydrogen compounds of the nonmetals here.
Nitrogen Hydrogen Compounds
Ammonia, NH3, forms naturally when any nitrogen-containing organic material decomposes in the absence of air. The laboratory preparation of ammonia is by the reaction of an ammonium salt with a strong base such as sodium hydroxide. The acid-base reaction with the weakly acidic ammonium ion gives ammonia, illustrated in Figure $4$. Ammonia also forms when ionic nitrides react with water. The nitride ion is a much stronger base than the hydroxide ion:
$\ce{Mg3N2(s) + 6 H2O(l) \longrightarrow 3 Mg(OH)2(s) + 2 NH3(g)} \nonumber$
The commercial production of ammonia is by the direct combination of the elements in the Haber process:
$\ce{N2(g) + 3 H2(g) ->[\text{catalyst}] 2 NH3(g)} \quad \quad \quad \quad \Delta H^{\circ}=-92 ~\text{kJ} \nonumber$
Ammonia is a colorless gas with a sharp, pungent odor. Smelling salts utilize this powerful odor. Gaseous ammonia readily liquefies to give a colorless liquid that boils at −33 °C. Due to intermolecular hydrogen bonding, the enthalpy of vaporization of liquid ammonia is higher than that of any other liquid except water, so ammonia is useful as a refrigerant. Ammonia is quite soluble in water(658 L at STP dissolves in 1 L H2O).
The chemical properties of ammonia are as follows:
1. Ammonia acts as a Brønsted base, as discussed in the chapter on acid-base chemistry. The ammonium ion is similar in size to the potassium ion; compounds of the two ions exhibit many similarities in their structures and solubilities.
2. Ammonia can display acidic behavior, although it is a much weaker acid than water. Like other acids, ammonia reacts with metals, although it is so weak that high temperatures are necessary. Hydrogen and(depending on the stoichiometry) amides(salts of $\ce{NH2^{-}}$), imides(salts of NH2), or nitrides(salts of N3−) form.
3. The nitrogen atom in ammonia has its lowest possible oxidation state(3−) and thus is not susceptible to reduction. However, it can be oxidized. Ammonia burns in air, giving NO and water. Hot ammonia and the ammonium ion are active reducing agents. Of particular interest are the oxidations of ammonium ion by nitrite ion, to yield pure nitrogen and by nitrate ion to yield nitrous oxide, N2O.
4. There are a number of compounds that we can consider derivatives of ammonia through the replacement of one or more hydrogen atoms with some other atom or group of atoms. Inorganic derivations include chloramine, NH2Cl, and hydrazine, N2H4:
Chloramine, NH2Cl, results from the reaction of sodium hypochlorite, NaOCl, with ammonia in basic solution. In the presence of a large excess of ammonia at low temperature, the chloramine reacts further to produce hydrazine, N2H4:
\begin{align*} \ce{NH3(aq) + OCl^{-}(aq) & -> NH2Cl(aq) + OH^{-}(aq)} \[4pt][4pt] \ce{NH2Cl(aq) + NH3(aq) + OH^{-}(aq) &-> N2H4(aq) + Cl^{-}(aq) + H2O(l)} \end{align*}
Anhydrous hydrazine is relatively stable in spite of its positive free energy of formation:
$\ce{N2(g) + 2 H2(g) \longrightarrow N2H4(l)} \quad \quad \Delta G_{ f }^{\circ}=149.2 kJ mol^{-1} \nonumber$
Hydrazine is a fuming, colorless liquid that has some physical properties remarkably similar to those of H2O(it melts at 2 °C, boils at 113.5 °C, and has a density at 25 °C of 1.00 g/mL). It burns rapidly and completely in air with substantial evolution of heat:
$\ce{N2H4(l) + O2(g) \longrightarrow N2(g) +2 H2O(l)} \quad \quad \Delta H^{\circ}=-621.5 kJ mol^{-1} \nonumber$
Like ammonia, hydrazine is both a Brønsted base and a Lewis base, although it is weaker than ammonia. It reacts with strong acids and forms two series of salts that contain the and ions, respectively. Some rockets use hydrazine as a fuel.
Phosphorus Hydrogen Compounds
The most important hydride of phosphorus is phosphine, PH3, a gaseous analog of ammonia in terms of both formula and structure. Unlike ammonia, it is not possible to form phosphine by direct union of the elements. There are two methods for the preparation of phosphine. One method is by the action of an acid on an ionic phosphide. The other method is the disproportionation of white phosphorus with hot concentrated base to produce phosphine and the hydrogen phosphite ion:
\begin{align*} \ce{AlP(s) + 3 H3O^{+}(aq) \longrightarrow PH3(g) + Al^{3+}(aq) + 3 H2O(l)} \[4pt][4pt] \ce{P4(s) + 4 OH^{-}(aq) + 2 H2O(l) \longrightarrow 2 HPO3^{2-}(aq) + 2 PH3(g)} \end{align*}
Phosphine is a colorless, very poisonous gas, which has an odor like that of decaying fish. Heat easily decomposes phosphine ($\ce{4 PH3 -> P4 + 6H2}$) and the compound burns in air. The major uses of phosphine are as a fumigant for grains and in semiconductor processing. Like ammonia, gaseous phosphine unites with gaseous hydrogen halides, forming phosphonium compounds like PH4Cl and PH4I. Phosphine is a much weaker base than ammonia; therefore, these compounds decompose in water, and the insoluble PH3 escapes from solution.
Sulfur Hydrogen Compounds
Hydrogen sulfide, H2S, is a colorless gas that is responsible for the offensive odor of rotten eggs and of many hot springs. Hydrogen sulfide is as toxic as hydrogen cyanide; therefore, it is necessary to exercise great care in handling it. Hydrogen sulfide is particularly deceptive because it paralyzes the olfactory nerves; after a short exposure, one does not smell it.
The production of hydrogen sulfide by the direct reaction of the elements(H2 + S) is unsatisfactory because the yield is low. A more effective preparation method is the reaction of a metal sulfide with a dilute acid. For example:
$\ce{FeS(s) + 2 H3O^{+}(aq) \longrightarrow Fe^{2+}(aq) + H_2 S(g) + 2 H2O(l)} \nonumber$
It is easy to oxidize the sulfur in metal sulfides and in hydrogen sulfide, making metal sulfides and H2S good reducing agents. In acidic solutions, hydrogen sulfide reduces Fe3+ to Fe2+, to Mn2+, to Cr3+, and HNO3 to NO2. The sulfur in H2S usually oxidizes to elemental sulfur, unless a large excess of the oxidizing agent is present. In which case, the sulfide may oxidize to or(or to SO2 or SO3 in the absence of water):
$\ce{2 H_2 S(g) + O_2(g) \longrightarrow 2 S(s) + 2 H2O(l)} \nonumber$
This oxidation process leads to the removal of the hydrogen sulfide found in many sources of natural gas. The deposits of sulfur in volcanic regions may be the result of the oxidation of H2S present in volcanic gases.
Hydrogen sulfide is a weak diprotic acid that dissolves in water to form hydrosulfuric acid. The acid ionizes in two stages, yielding hydrogen sulfide ions, HS, in the first stage and sulfide ions, S2−, in the second. Since hydrogen sulfide is a weak acid, aqueous solutions of soluble sulfides and hydrogen sulfides are basic:
\begin{align*} \ce{S^{2-}(aq) + H2O(l) &\rightleftharpoons HS^{-}(aq) + OH^{-}(aq)} \[4pt][4pt] \ce{HS^{-}(aq) + H2O(l) &\rightleftharpoons H2S(g) + OH^{-}(aq)} \end{align*}
Halogen Hydrogen Compounds
Binary compounds containing only hydrogen and a halogen are hydrogen halides. At room temperature, the pure hydrogen halides HF, HCl, HBr, and HI are gases.
In general, it is possible to prepare the halides by the general techniques used to prepare other acids. Fluorine, chlorine, and bromine react directly with hydrogen to form the respective hydrogen halide. This is a commercially important reaction for preparing hydrogen chloride and hydrogen bromide.
The acid-base reaction between a nonvolatile strong acid and a metal halide will yield a hydrogen halide. The escape of the gaseous hydrogen halide drives the reaction to completion. For example, the usual method of preparing hydrogen fluoride is by heating a mixture of calcium fluoride, CaF2, and concentrated sulfuric acid:
$\ce{CaF2(s) + H2SO4(aq) \longrightarrow CaSO4(s) + 2 HF(g)} \nonumber$
Gaseous hydrogen fluoride is also a by-product in the preparation of phosphate fertilizers by the reaction of fluoroapatite, Ca5(PO4)3F, with sulfuric acid. The reaction of concentrated sulfuric acid with a chloride salt produces hydrogen chloride both commercially and in the laboratory.
In most cases, sodium chloride is the chloride of choice because it is the least expensive chloride. Hydrogen bromide and hydrogen iodide cannot be prepared using sulfuric acid because this acid is an oxidizing agent capable of oxidizing both bromide and iodide. However, it is possible to prepare both hydrogen bromide and hydrogen iodide using an acid such as phosphoric acid because it is a weaker oxidizing agent. For example:
$\ce{H3PO4(l) + Br^{-}(aq) \longrightarrow HBr(g) + H2PO4^{-}(aq)} \nonumber$
All of the hydrogen halides are very soluble in water, forming hydrohalic acids. With the exception of hydrogen fluoride, which has a strong hydrogen-fluoride bond, they are strong acids. Reactions of hydrohalic acids with metals, metal hydroxides, oxides, or carbonates produce salts of the halides. Most chloride salts are soluble in water. AgCl, PbCl2, and Hg2Cl2 are the commonly encountered exceptions.
The halide ions give the substances the properties associated with X(aq). The heavier halide ions(Cl, Br, and I) can act as reducing agents, and the lighter halogens or other oxidizing agents will oxidize them:
\begin{align*} \ce{Cl_2(aq) + 2 e^{-} &<=> 2 Cl^{-}(aq)} & E^{\circ}=1.36 V \[4pt][4pt] \ce{Br2(aq) + 2 e^{-} &<=> 2 Br^{-}(aq)} & E^{\circ}=1.09 V \[4pt][4pt] \ce{I2(aq) + 2 e^{-} &<=> 2 I^{-}(aq)} & E^{\circ}=0.54 V \end{align*}
For example, bromine oxidizes iodine:
$\ce{Br2(aq) + 2 HI(aq) \longrightarrow 2 HBr(aq) + I2(aq)} \quad \quad E^{\circ}=0.55 V \nonumber$
Hydrofluoric acid is unique in its reactions with sand(silicon dioxide) and with glass, which is a mixture of silicates:
\begin{align*} \ce{SiO2(s) + 4 HF(aq) &\longrightarrow SiF4(g) + 2 H2O(l)} \[4pt][4pt] \ce{CaSiO3(s) + 6 HF(aq) &\longrightarrow CaF2(s) + SiF4(g) + 3 H2O}(l) \end{align*} \nonumber
The volatile silicon tetrafluoride escapes from these reactions. Because hydrogen fluoride attacks glass, it can frost or etch glass and is used to etch markings on thermometers, burets, and other glassware.
The largest use for hydrogen fluoride is in production of hydrochlorofluorocarbons for refrigerants, in plastics, and in propellants. The second largest use is in the manufacture of cryolite, Na3AlF6, which is important in the production of aluminum. The acid is also important in the production of other inorganic fluorides(such as BF3), which serve as catalysts in the industrial synthesis of certain organic compounds.
Hydrochloric acid is relatively inexpensive. It is an important and versatile acid in industry and is important for the manufacture of metal chlorides, dyes, glue, glucose, and various other chemicals. A considerable amount is also important for the activation of oil wells and as pickle liquor—an acid used to remove oxide coating from iron or steel that is to be galvanized, tinned, or enameled. The amounts of hydrobromic acid and hydroiodic acid used commercially are insignificant by comparison. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Compounds_of_Hydrogen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the preparation, properties, and uses of some representative metal carbonates
The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates—compounds that contain the carbonate anions, The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates—compounds that contain the hydrogen carbonate anion, $\ce{HCO3^{−}}$, also known as the bicarbonate anion.
With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include:
\begin{align*} \ce{Na_2 O(s) + CO_2(g) &-> Na_2CO_3(s) } \[4pt][4pt] \ce{Ca( OH )_2(s) + CO_2(g) &-> CaCO_3(s) + H2O(l)} \end{align*}
The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are:
\begin{align*} \ce{Ca^{2+}(aq) + CO_3^{2-}(aq) &-> CaCO_3(s)} \[4pt][4pt] \ce{Pb^{2+}(aq) + CO_3^{2-}(aq) &-> PbCO_3(s)} \end{align*}
Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al3+ or Sn4+ behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate.
Alkali metal hydrogen carbonates such as NaHCO3 and CsHCO3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide:
$\ce{OH^{-}(aq) + CO_2(aq) -> HCO_3^{-}(aq)} \nonumber$
It is possible to isolate the solids by evaporation of the water from the solution.
Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO3 dissolves in water containing dissolved carbon dioxide:
$\ce{CaCO_3(s) + CO_2(aq) + H2O(l) -> Ca^{2+}(aq) + 2 HCO_3^{-}(aq)} \nonumber$
Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure $1$: , form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate.
The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na3(CO3)(HCO3)(H2O)2. Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na2CO3:
$\ce{2 Na_3(CO3)(HCO3)(H2O)2(s) -> 3 Na_2CO_3(s) + 5 H2O(l) + CO_2(g)} \nonumber$
Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction:
$\ce{CO_3^{2-}(aq) + H2O(l) <=> HCO_3^{-}(aq) + OH^{-}(aq)} \nonumber$
Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid(stomach acid), as shown in Figure $2$, illustrates the reaction:
$\ce{CaCO_3(s) + 2 HCl(aq) -> CaCl_2(aq) + CO_2(g) + H2O(l)} \nonumber$
Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides.
Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water:
$\ce{KHCO_3(aq) + KOH(aq) -> K_2 CO_3(aq) + H2O(l)} \nonumber$
With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda(bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate(cream of tartar), KHC4H4O6. As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide:
$\ce{HC_4 H_4 O_6^{-}(aq) + HCO_3^{-}(aq) -> C_4 H_4 O_6^{2-}(aq) + CO_2(g) + H2O(l)} \nonumber$
Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.07%3A_Occurrence_Preparation_and_Properties_of_Carbonates.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of nitrogen
Most pure nitrogen comes from the fractional distillation of liquid air. The atmosphere consists of 78% nitrogen by volume. This means there are more than 20 million tons of nitrogen over every square mile of the earth’s surface. Nitrogen is a component of proteins and of the genetic material (DNA/RNA) of all plants and animals.
Under ordinary conditions, nitrogen is a colorless, odorless, and tasteless gas. It boils at 77 K and freezes at 63 K. Liquid nitrogen is a useful coolant because it is inexpensive and has a low boiling point. Nitrogen is very unreactive because of the very strong triple bond between the nitrogen atoms. The only common reactions at room temperature occur with lithium to form Li3N, with certain transition metal complexes, and with hydrogen or oxygen in nitrogen-fixing bacteria. The general lack of reactivity of nitrogen makes the remarkable ability of some bacteria to synthesize nitrogen compounds using atmospheric nitrogen gas as the source one of the most exciting chemical events on our planet. This process is one type of nitrogen fixation. In this case, nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. Nitrogen fixation also occurs when lightning passes through air, causing molecular nitrogen to react with oxygen to form nitrogen oxides, which are then carried down to the soil.
Chemistry in Everyday Life: Nitrogen Fixation
All living organisms require nitrogen compounds for survival. Unfortunately, most of these organisms cannot absorb nitrogen from its most abundant source—the atmosphere. Atmospheric nitrogen consists of N2 molecules, which are very unreactive due to the strong nitrogen-nitrogen triple bond. However, a few organisms can overcome this problem through a process known as nitrogen fixation, illustrated in Figure $1$.
Nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the best-known example being the presence of rhizobia in the root nodules of legumes.
Large volumes of atmospheric nitrogen are necessary for making ammonia—the principal starting material used for preparation of large quantities of other nitrogen-containing compounds. Most other uses for elemental nitrogen depend on its inactivity. It is helpful when a chemical process requires an inert atmosphere. Canned foods and luncheon meats cannot oxidize in a pure nitrogen atmosphere, so they retain a better flavor and color, and spoil less rapidly, when sealed in nitrogen instead of air. This technology allows fresh produce to be available year-round, regardless of growing season.
There are compounds with nitrogen in all of its oxidation states from 3− to 5+. Much of the chemistry of nitrogen involves oxidation-reduction reactions. Some active metals (such as alkali metals and alkaline earth metals) can reduce nitrogen to form metal nitrides. In the remainder of this section, we will examine nitrogen-oxygen chemistry.
There are well-characterized nitrogen oxides in which nitrogen exhibits each of its positive oxidation numbers from 1+ to 5+. When ammonium nitrate is carefully heated, nitrous oxide (dinitrogen oxide) and water vapor form. Stronger heating generates nitrogen gas, oxygen gas, and water vapor. No one should ever attempt this reaction—it can be very explosive. In 1947, there was a major ammonium nitrate explosion in Texas City, Texas, and, in 2013, there was another major explosion in West, Texas. In the last 100 years, there were nearly 30 similar disasters worldwide, resulting in the loss of numerous lives. In this oxidation-reduction reaction, the nitrogen in the nitrate ion oxidizes the nitrogen in the ammonium ion. Nitrous oxide, shown in Figure $2$: , is a colorless gas possessing a mild, pleasing odor and a sweet taste. It finds application as an anesthetic for minor operations, especially in dentistry, under the name “laughing gas.”
Low yields of nitric oxide, NO, form when heating nitrogen and oxygen together. NO also forms when lightning passes through air during thunderstorms. Burning ammonia is the commercial method of preparing nitric oxide. In the laboratory, the reduction of nitric acid is the best method for preparing nitric oxide. When copper reacts with dilute nitric acid, nitric oxide is the principal reduction product:
$\ce{3 Cu(s) + 8 HNO_3(aq) -> 2 NO(g) + 3 Cu(NO_3)_2(aq) + 4 H2O(l)} \nonumber$
Gaseous nitric oxide is the most thermally stable of the nitrogen oxides and is the simplest known thermally stable molecule with an unpaired electron. It is one of the air pollutants generated by internal combustion engines, resulting from the reaction of atmospheric nitrogen and oxygen during the combustion process.
At room temperature, nitric oxide is a colorless gas consisting of diatomic molecules. As is often the case with molecules that contain an unpaired electron, two molecules combine to form a dimer by pairing their unpaired electrons to form a bond. Liquid and solid NO both contain N2O2 dimers, like that shown in Figure $3$. Most substances with unpaired electrons exhibit color by absorbing visible light; however, NO is colorless because the absorption of light is not in the visible region of the spectrum.
Cooling a mixture of equal parts nitric oxide and nitrogen dioxide to −21 °C produces dinitrogen trioxide, a blue liquid consisting of N2O3 molecules (shown in Figure $4$). Dinitrogen trioxide exists only in the liquid and solid states. When heated, it reverts to a mixture of NO and NO2.
It is possible to prepare nitrogen dioxide in the laboratory by heating the nitrate of a heavy metal, or by the reduction of concentrated nitric acid with copper metal, as shown in Figure $5$. Commercially, it is possible to prepare nitrogen dioxide by oxidizing nitric oxide with air.
The nitrogen dioxide molecule (illustrated in Figure $6$) contains an unpaired electron, which is responsible for its color and paramagnetism. It is also responsible for the dimerization of NO2. At low pressures or at high temperatures, nitrogen dioxide has a deep brown color that is due to the presence of the NO2 molecule. At low temperatures, the color almost entirely disappears as dinitrogen tetraoxide, N2O4, forms. At room temperature, an equilibrium exists:
$\ce{2 NO_2(g) <=> N2O4(g)} \quad \quad \quad \quad K_P=6.86 \nonumber$
Dinitrogen pentaoxide, N2O5 (illustrated in Figure $7$), is a white solid that is formed by the dehydration of nitric acid by phosphorus(V) oxide (tetraphosphorus decoxide):
$\ce{P4O10(s) + 4 HNO3(l) \longrightarrow 4 HPO3(s) + 2 N2O5(s)} \nonumber$
It is unstable above room temperature, decomposing to N2O4 and O2.
The oxides of nitrogen(III), nitrogen(IV), and nitrogen(V) react with water and form nitrogen-containing oxyacids. Nitrogen(III) oxide, N2O3, is the anhydride of nitrous acid; HNO2 forms when N2O3 reacts with water. There are no stable oxyacids containing nitrogen with an oxidation state of 4+; therefore, nitrogen(IV) oxide, NO2, disproportionates in one of two ways when it reacts with water. In cold water, a mixture of HNO2 and HNO3 forms. At higher temperatures, HNO3 and NO will form. Nitrogen(V) oxide, N2O5, is the anhydride of nitric acid; HNO3 is produced when N2O5 reacts with water:
$\ce{N2O5(s) + H2O(l) \longrightarrow 2 HNO3(aq)} \nonumber$
The nitrogen oxides exhibit extensive oxidation-reduction behavior. Nitrous oxide resembles oxygen in its behavior when heated with combustible substances. N2O is a strong oxidizing agent that decomposes when heated to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air (one-fifth oxygen). A glowing splinter bursts into flame when thrust into a bottle of this gas. Nitric oxide acts both as an oxidizing agent and as a reducing agent. For example:
\begin{align*} &\text{oxidizing agent:} \quad \quad \ce{P_4(s) + 6 NO(g) -> P4O6(s) + 3 N2(g)} \[4pt][4pt] &\text{reducing agent:} \quad \quad \ce{Cl_2(g) + 2 NO(g) -> 2 ClNO(g)} \end{align*} \nonumber
Nitrogen dioxide (or dinitrogen tetraoxide) is a good oxidizing agent. For example:
\begin{align*} \ce{NO_2(g) + CO(g) &\longrightarrow NO(g) + CO_2(g)} \[4pt][4pt] \ce{NO_2(g) + 2 HCl(aq) &\longrightarrow NO(g) + Cl_2(g) + H2O(l)} \end{align*} \nonumber | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.08%3A_Occurrence_Preparation_and_Properties_of_Nitrogen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of phosphorus
The industrial preparation of phosphorus is by heating calcium phosphate, obtained from phosphate rock, with sand and coke:
$\ce{2 Ca_3(PO4)2(s) + 6 SiO2(s) + 10 C(s) ->[\Delta] 6 CaSiO3(l) + 10 CO(g) + P4(g)} \nonumber$
The phosphorus distills out of the furnace and is condensed into a solid or burned to form P4O10. The preparation of many other phosphorus compounds begins with P4O10. The acids and phosphates are useful as fertilizers and in the chemical industry. Other uses are in the manufacture of special alloys such as ferrophosphorus and phosphor bronze. Phosphorus is important in making pesticides, matches, and some plastics. Phosphorus is an active nonmetal. In compounds, phosphorus usually occurs in oxidation states of 3−, 3+, and 5+. Phosphorus exhibits oxidation numbers that are unusual for a group 15 element in compounds that contain phosphorus-phosphorus bonds; examples include diphosphorus tetrahydride, H2P-PH2, and tetraphosphorus trisulfide, P4S3, illustrated in Figure $1$.
Phosphorus Oxygen Compounds
Phosphorus forms two common oxides, phosphorus(III) oxide (or tetraphosphorus hexaoxide), P4O6, and phosphorus(V) oxide (or tetraphosphorus decaoxide), P4O10, both shown in Figure $2$. Phosphorus(III) oxide is a white crystalline solid with a garlic-like odor. Its vapor is very poisonous. It oxidizes slowly in air and inflames when heated to 70 °C, forming P4O10. Phosphorus(III) oxide dissolves slowly in cold water to form phosphorous acid, H3PO3.
Phosphorus(V) oxide, P4O10, is a white powder that is prepared by burning phosphorus in excess oxygen. Its enthalpy of formation is very high (−2984 kJ), and it is quite stable and a very poor oxidizing agent. Dropping P4O10 into water produces a hissing sound, heat, and orthophosphoric acid:
$\ce{P_4 O_{10}(s) + 6 H2O(l) \longrightarrow 4 H_3 PO_4(aq)} \nonumber$
Because of its great affinity for water, phosphorus(V) oxide is an excellent drying agent for gases and solvents, and for removing water from many compounds.
Phosphorus Halogen Compounds
Phosphorus will react directly with the halogens, forming trihalides, PX3, and pentahalides, PX5. The trihalides are much more stable than the corresponding nitrogen trihalides; nitrogen pentahalides do not form because of nitrogen’s inability to form more than four bonds.
The chlorides PCl3 and PCl5, both shown in Figure $3$: , are the most important halides of phosphorus. Phosphorus trichloride is a colorless liquid that is prepared by passing chlorine over molten phosphorus. Phosphorus pentachloride is an off-white solid that is prepared by oxidizing the trichloride with excess chlorine. The pentachloride sublimes when warmed and forms an equilibrium with the trichloride and chlorine when heated.
Like most other nonmetal halides, both phosphorus chlorides react with an excess of water and yield hydrogen chloride and an oxyacid: PCl3 yields phosphorous acid H3PO3 and PCl5 yields phosphoric acid, H3PO4.
The pentahalides of phosphorus are Lewis acids because of the empty valence d orbitals of phosphorus. These compounds readily react with halide ions (Lewis bases) to give the anion $\ce{PX6^{−}}$. Whereas phosphorus pentafluoride is a molecular compound in all states, X-ray studies show that solid phosphorus pentachloride is an ionic compound, $\ce{[PCl4^{+}][PCl6^{-}]}$ as are phosphorus pentabromide, $\ce{[PBr4^{+}][Br^{-}]}$, and phosphorus pentaiodide, [$\ce{[PI6^{+}][I^{-}]}$. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.09%3A_Occurrence_Preparation_and_Properties_of_Phosphorus.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and compounds of oxygen
• Describe the preparation, properties, and uses of some representative metal oxides, peroxides, and hydroxides
Oxygen is the most abundant element on the earth’s crust. The earth’s surface is composed of the crust, atmosphere, and hydrosphere. About 50% of the mass of the earth’s crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as O2 molecules and, to a limited extent, as O3 (ozone) molecules in air. It forms about 20% of the mass of the air. About 89% of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals.
Oxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water (49 mL of gas dissolves in 1 L at STP), oxygen’s solubility is very important to aquatic life.
Most of the oxygen isolated commercially comes from air and the remainder from the electrolysis of water. The separation of oxygen from air begins with cooling and compressing the air until it liquefies. As liquid air warms, oxygen with its higher boiling point (90 K) separates from nitrogen, which has a lower boiling point (77 K). It is possible to separate the other components of air at the same time based on differences in their boiling points.
Oxygen is essential in combustion processes such as the burning of fuels. Plants and animals use the oxygen from the air in respiration. The administration of oxygen-enriched air is an important medical practice when a patient is receiving an inadequate supply of oxygen because of shock, pneumonia, or some other illness.
The chemical industry employs oxygen for oxidizing many substances. A significant amount of oxygen produced commercially is important in the removal of carbon from iron during steel production. Large quantities of pure oxygen are also necessary in metal fabrication and in the cutting and welding of metals with oxyhydrogen and oxyacetylene torches.
Liquid oxygen is important to the space industry. It is an oxidizing agent in rocket engines. It is also the source of gaseous oxygen for life support in space.
As we know, oxygen is very important to life. The energy required for the maintenance of normal body functions in human beings and in other organisms comes from the slow oxidation of chemical compounds. Oxygen is the final oxidizing agent in these reactions. In humans, oxygen passes from the lungs into the blood, where it combines with hemoglobin, producing oxyhemoglobin. In this form, blood transports the oxygen to tissues, where it is transferred to the tissues. The ultimate products are carbon dioxide and water. The blood carries the carbon dioxide through the veins to the lungs, where the blood releases the carbon dioxide and collects another supply of oxygen. Digestion and assimilation of food regenerate the materials consumed by oxidation in the body; the energy liberated is the same as if the food burned outside the body.
Green plants continually replenish the oxygen in the atmosphere by a process called photosynthesis. The products of photosynthesis may vary, but, in general, the process converts carbon dioxide and water into glucose (a sugar) and oxygen using the energy of light:
$\underset{\text{carbon} \[4pt][4pt]\text{dioxide}}{\ce{6 CO2(g)}} + \underset{\text{water}}{\ce{6 H2O(l)}} \ce{->[\text{chlorophyll}][\text{light}]} \underset{\text{glucose}}{\ce{C6H12O6(aq)}} + \underset{\text{oxygen}}{\ce{6O2(g)}} \nonumber$
Thus, the oxygen that became carbon dioxide and water by the metabolic processes in plants and animals returns to the atmosphere by photosynthesis.
When dry oxygen is passed between two electrically charged plates, ozone (O3, illustrated in Figure $1$), an allotrope of oxygen possessing a distinctive odor, forms. The formation of ozone from oxygen is an endothermic reaction, in which the energy comes from an electrical discharge, heat, or ultraviolet light:
$\ce{3 O2(g) ->[\text { electric discharge }] O3(g)} \quad \quad \Delta H^{\circ}=287~\text{kJ} \nonumber$
The sharp odor associated with sparking electrical equipment is due, in part, to ozone.
Ozone forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth’s surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction:
$\ce{O3(g) ->[\text{ultraviolet light}] O(g) + O2(g)} \nonumber$
The reactive oxygen atoms recombine with molecular oxygen to complete the ozone cycle. The presence of stratospheric ozone decreases the frequency of skin cancer and other damaging effects of ultraviolet radiation. It has been clearly demonstrated that chlorofluorocarbons, CFCs (known commercially as Freons), which were present as aerosol propellants in spray cans and as refrigerants, caused depletion of ozone in the stratosphere. This occurred because ultraviolet light also causes CFCs to decompose, producing atomic chlorine. The chlorine atoms react with ozone molecules, resulting in a net removal of O3 molecules from stratosphere. This process is explored in detail in our coverage of chemical kinetics. There is a worldwide effort to reduce the amount of CFCs used commercially, and the ozone hole is already beginning to decrease in size as atmospheric concentrations of atomic chlorine decrease. While ozone in the stratosphere helps protect us, ozone in the troposphere is a problem. This ozone is a toxic component of photochemical smog.
The uses of ozone depend on its reactivity with other substances. It can be used as a bleaching agent for oils, waxes, fabrics, and starch: It oxidizes the colored compounds in these substances to colorless compounds. It is an alternative to chlorine as a disinfectant for water.
Reactions
Elemental oxygen is a strong oxidizing agent. It reacts with most other elements and many compounds.
Reaction with Elements
Oxygen reacts directly at room temperature or at elevated temperatures with all other elements except the noble gases, the halogens, and few second- and third-row transition metals of low reactivity (those with higher reduction potentials than copper). Rust is an example of the reaction of oxygen with iron. The more active metals form peroxides or superoxides. Less active metals and the nonmetals give oxides. Two examples of these reactions are:
\begin{align*} \ce{2 Mg(s) + O_2(g) &\longrightarrow 2 MgO(s)} \[4pt][4pt] \ce{P_4(s) + 5 O_2(g) &\longrightarrow P4O10(s)} \end{align*}
The oxides of halogens, at least one of the noble gases, and metals with higher reduction potentials than copper do not form by the direct action of the elements with oxygen.
Reaction with Compounds
Elemental oxygen also reacts with some compounds. If it is possible to oxidize any of the elements in a given compound, further oxidation by oxygen can occur. For example, hydrogen sulfide, H2S, contains sulfur with an oxidation state of 2−. Because the sulfur does not exhibit its maximum oxidation state, we would expect H2S to react with oxygen. It does, yielding water and sulfur dioxide. The reaction is:
$\ce{2 H2S(g) + 3O2(g) \longrightarrow 2 H2O(l) + 2 SO2(g)} \nonumber$
It is also possible to oxidize oxides such as CO and P4O6 that contain an element with a lower oxidation state. The ease with which elemental oxygen picks up electrons is mirrored by the difficulty of removing electrons from oxygen in most oxides. Of the elements, only the very reactive fluorine can oxidize oxides to form oxygen gas.
Oxides, Peroxides, and Hydroxides
Compounds of the representative metals with oxygen fall into three categories:
1. oxides containing oxide ions, $\ce{O^{2−}}$,
2. peroxides containing peroxides ions, $\ce{O_2^{2−}}$ with oxygen-oxygen covalent single bonds and a very limited number of superoxides, containing superoxide ions, $\ce{O_2^{−}}$ with oxygen-oxygen covalent bonds that have a bond order of $\frac{3}{2}$,
3. hydroxides containing hydroxide ions, $\ce{OH^{−}}$. All representative metals form oxides. Some of the metals of group 2 also form peroxides, $\ce{MO2}$, and the metals of group 1 also form peroxides, $\ce{M2O2}$, and superoxides, $\ce{MO2}$.
Oxides
It is possible to produce the oxides of most representative metals by heating the corresponding hydroxides (forming the oxide and gaseous water) or carbonates (forming the oxide and gaseous CO2). Equations for example reactions are:
\begin{align*} \ce{2 Al(OH)3(s) ->[\Delta] Al2O3(s) + 3 H2O(g)} \[4pt][4pt] \ce{CaCO3(s) ->[\Delta] CaO(s) + CO2(g)} \end{align*} \nonumber
However, alkali metal salts generally are very stable and do not decompose easily when heated. Alkali metal oxides result from the oxidation-reduction reactions created by heating nitrates or hydroxides with the metals. Equations for sample reactions are:
\begin{align*} \ce{2 KNO3(s) + 10 K(s) ->[\Delta] 6 K_2O(s) + N2(g)} \[4pt][4pt] \ce{2 LiOH(s) + 2 Li(s) ->[\Delta] 2 Li2O(s) + H2(g)} \end{align*} \nonumber
With the exception of mercury(II) oxide, it is possible to produce the oxides of the metals of groups 2–15 by burning the corresponding metal in air. The heaviest member of each group, the member for which the inert pair effect is most pronounced, forms an oxide in which the oxidation state of the metal ion is two less than the group oxidation state (inert pair effect). Thus, Tl2O, PbO, and Bi2O3 form when burning thallium, lead, and bismuth, respectively. The oxides of the lighter members of each group exhibit the group oxidation state. For example, SnO2 forms from burning tin. Mercury(II) oxide, HgO, forms slowly when mercury is warmed below 500 °C; it decomposes at higher temperatures.
Burning the members of groups 1 and 2 in air is not a suitable way to form the oxides of these elements. These metals are reactive enough to combine with nitrogen in the air, so they form mixtures of oxides and ionic nitrides. Several also form peroxides or superoxides when heated in air.
Ionic oxides all contain the oxide ion, a very powerful hydrogen ion acceptor. With the exception of the very insoluble aluminum oxide, Al2O3, tin(IV), SnO2, and lead(IV), PbO2, the oxides of the representative metals react with acids to form salts. Some equations for these reactions are:
\begin{align*} \ce{Na2O + 2 HNO3(aq) &-> 2 NaNO3(aq) + H2O(l)} \[4pt][4pt] \ce{CaO(s) + 2 HCl(aq) & -> CaCl2(aq) + H2O(l)} \[4pt][4pt] \ce{SnO(s) + 2 HClO4(aq) & -> Sn(ClO4)2(aq) + H2O(l)} \end{align*} \nonumber
The oxides of the metals of groups 1 and 2 and of thallium(I) oxide react with water and form hydroxides. Examples of such reactions are:
\begin{align*} \ce{Na2O(s) + H2O(l) \longrightarrow NaOH(aq)} \[4pt][4pt] \ce{CaO(s) + H2O(l) \longrightarrow Ca(OH)2(aq)} \[4pt][4pt] \ce{Tl2O(s) + H2O(aq) \longrightarrow 2 TlOH(aq)} \end{align*} \nonumber
The oxides of the alkali metals have little industrial utility, unlike magnesium oxide, calcium oxide, and aluminum oxide. Magnesium oxide is important in making firebrick, crucibles, furnace linings, and thermal insulation—applications that require chemical and thermal stability. Calcium oxide, sometimes called quicklime or lime in the industrial market, is very reactive, and its principal uses reflect its reactivity. Pure calcium oxide emits an intense white light when heated to a high temperature (as illustrated in Figure $2$). Blocks of calcium oxide heated by gas flames were the stage lights in theaters before electricity was available. This is the source of the phrase “in the limelight.”
Calcium oxide and calcium hydroxide are inexpensive bases used extensively in chemical processing, although most of the useful products prepared from them do not contain calcium. Calcium oxide, CaO, is made by heating calcium carbonate, CaCO3, which is widely and inexpensively available as limestone or oyster shells:
$\ce{CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)} \nonumber$
Although this decomposition reaction is reversible, it is possible to obtain a 100% yield of CaO by allowing the CO2 to escape. It is possible to prepare calcium hydroxide by the familiar acid-base reaction of a soluble metal oxide with water:
$\ce{CaO(s) + H2O(l) \longrightarrow Ca(OH)_2(s)} \nonumber$
Both CaO and Ca(OH)2 are useful as bases; they accept protons and neutralize acids.
Alumina (Al2O3) occurs in nature as the mineral corundum, a very hard substance used as an abrasive for grinding and polishing. Corundum is important to the jewelry trade as ruby and sapphire. The color of ruby is due to the presence of a small amount of chromium; other impurities produce the wide variety of colors possible for sapphires. Artificial rubies and sapphires are now manufactured by melting aluminum oxide (melting point = 2050 °C) with small amounts of oxides to produce the desired colors and cooling the melt in such a way as to produce large crystals. Ruby lasers use synthetic ruby crystals.
Zinc oxide, ZnO, was a useful white paint pigment; however, pollutants tend to discolor the compound. The compound is also important in the manufacture of automobile tires and other rubber goods, and in the preparation of medicinal ointments. For example, zinc-oxide-based sunscreens, as shown in Figure $3$, help prevent sunburn. The zinc oxide in these sunscreens is present in the form of very small grains known as nanoparticles. Lead dioxide is a constituent of charged lead storage batteries. Lead(IV) tends to revert to the more stable lead(II) ion by gaining two electrons, so lead dioxide is a powerful oxidizing agent.
Peroxides and Superoxides
Peroxides and superoxides are strong oxidizers and are important in chemical processes. Hydrogen peroxide, H2O2, prepared from metal peroxides, is an important bleach and disinfectant. Peroxides and superoxides form when the metal or metal oxides of groups 1 and 2 react with pure oxygen at elevated temperatures. Sodium peroxide and the peroxides of calcium, strontium, and barium form by heating the corresponding metal or metal oxide in pure oxygen:
\begin{align*} \ce{2 Na(s) + O_2(g) ->[\Delta] Na2O2(s)} \[4pt][4pt] \ce{2 Na2O(s) + O_2(g) ->[\Delta] 2 Na2O2(s)} \[4pt][4pt] \ce{2 SrO(s) + O2(g) ->[\Delta] 2 SrO2(s)} \end{align*}
The peroxides of potassium, rubidium, and cesium can be prepared by heating the metal or its oxide in a carefully controlled amount of oxygen:
$\ce{2 K(s) + O_2(g) \longrightarrow K2O2(s)} \quad \quad (2~\text{mol}~\ce{K} \text { per mol} ~\ce{O2}) \nonumber$
With an excess of oxygen, the superoxides KO2, RbO2, and CsO2 form. For example:
$\ce{K(s) + O2(g) \longrightarrow KO2(s)} \quad \quad (1 ~\text{mol} ~\ce{K} \text { per mol} ~\ce{O2}) \nonumber$
The stability of the peroxides and superoxides of the alkali metals increases as the size of the cation increases.
Hydroxides
Hydroxides are compounds that contain the OH ion. It is possible to prepare these compounds by two general types of reactions. Soluble metal hydroxides can be produced by the reaction of the metal or metal oxide with water. Insoluble metal hydroxides form when a solution of a soluble salt of the metal combines with a solution containing hydroxide ions.
With the exception of beryllium and magnesium, the metals of groups 1 and 2 react with water to form hydroxides and hydrogen gas. Examples of such reactions include:
\begin{align*} \ce{2 Li(s) + 2 H2O(l) \longrightarrow 2 LiOH(aq) + H_2(g)} \[4pt][4pt] \ce{Ca(s) + 2 H2O(l) \longrightarrow Ca ( OH )_2(aq) + H_2(g)} \end{align*}
However, these reactions can be violent and dangerous; therefore, it is preferable to produce soluble metal hydroxides by the reaction of the respective oxide with water:
\begin{align*} \ce{Li_2O(s) + H2O(l) &\longrightarrow 2 LiOH(aq)} \[4pt][4pt] \ce{CaO(s) + H2O(l) &\longrightarrow Ca(OH)2(aq)} \end{align*}
Most metal oxides are base anhydrides. This is obvious for the soluble oxides because they form metal hydroxides. Most other metal oxides are insoluble and do not form hydroxides in water; however, they are still base anhydrides because they will react with acids.
It is possible to prepare the insoluble hydroxides of beryllium, magnesium, and other representative metals by the addition of sodium hydroxide to a solution of a salt of the respective metal. The net ionic equations for the reactions involving a magnesium salt, an aluminum salt, and a zinc salt are:
\begin{align*} \ce{Mg^{2+}(aq) + 2 OH^{-}(aq) &\longrightarrow Mg(OH)2(s)} \[4pt][4pt] \ce{Al^{3+}(aq) + 3 OH^{-}(aq) &\longrightarrow Al(OH)3(s)} \[4pt][4pt] \ce{Zn^{2+}(aq) + 2 OH^{-}(aq) &\longrightarrow Zn(OH)2(s)} \end{align*}
An excess of hydroxide must be avoided when preparing aluminum, gallium, zinc, and tin(II) hydroxides, or the hydroxides will dissolve with the formation of the corresponding complex ions: $\ce{ Al(OH)4^{−}}$, $\ce{Ga(OH)4^{−}}$, $\ce{Zn(OH)4^{2−}}$, and $\ce{Sn(OH)3^{−}}$ (Figure $4$). The important aspect of complex ions for this chapter is that they form by a Lewis acid-base reaction with the metal being the Lewis acid.
Industry uses large quantities of sodium hydroxide as a cheap, strong base. Sodium chloride is the starting material for the production of NaOH because NaCl is a less expensive starting material than the oxide. Sodium hydroxide is among the top 10 chemicals in production in the United States, and this production was almost entirely by electrolysis of solutions of sodium chloride. This process is the chlor-alkali process, and it is the primary method for producing chlorine.
Sodium hydroxide is an ionic compound and melts without decomposition. It is very soluble in water, giving off a great deal of heat and forming very basic solutions: 40 grams of sodium hydroxide dissolves in only 60 grams of water at 25 °C. Sodium hydroxide is employed in the production of other sodium compounds and is used to neutralize acidic solutions during the production of other chemicals such as petrochemicals and polymers.
Many of the applications of hydroxides are for the neutralization of acids (such as the antacid shown in Figure $5$) and for the preparation of oxides by thermal decomposition. An aqueous suspension of magnesium hydroxide constitutes the antacid milk of magnesia. Because of its ready availability (from the reaction of water with calcium oxide prepared by the decomposition of limestone, CaCO3), low cost, and activity, calcium hydroxide is used extensively in commercial applications needing a cheap, strong base. The reaction of hydroxides with appropriate acids is also used to prepare salts.
Chemistry in Everyday Life: The Chlor-Alkali Process
Although they are very different chemically, there is a link between chlorine and sodium hydroxide because there is an important electrochemical process that produces the two chemicals simultaneously. The process known as the chlor-alkali process, utilizes sodium chloride, which occurs in large deposits in many parts of the world. This is an electrochemical process to oxidize chloride ion to chlorine and generate sodium hydroxide.
Passing a direct current of electricity through a solution of NaCl causes the chloride ions to migrate to the positive electrode where oxidation to gaseous chlorine occurs when the ion gives up an electron to the electrode:
$\ce{2 Cl^{-}(aq) -> Cl_2(g) + 2 e^{-}} \quad \quad \text { (at the positive electrode) } \nonumber$
The electrons produced travel through the outside electrical circuit to the negative electrode. Although the positive sodium ions migrate toward this negative electrode, metallic sodium does not form because sodium ions are too difficult to reduce under the conditions used. (Recall that metallic sodium is active enough to react with water and hence, even if produced, would immediately react with water to produce sodium ions again.) Instead, water molecules pick up electrons from the electrode and undergo reduction to form hydrogen gas and hydroxide ions:
$\ce{2 H2O(l)} + \ce{2 e^{-}} (\text{from the negative electrode}) \ce{-> H_2(g) + 2 OH^{-}(aq)} \nonumber$
The overall result is the conversion of the aqueous solution of NaCl to an aqueous solution of NaOH, gaseous Cl2, and gaseous H2:
$\ce{2 Na^{+}(aq) + 2 Cl^{-}(aq) + 2 H2O(l) ->[\text {electrolysis}] 2 Na^{+}(aq) + 2 OH^{-}(aq) + Cl_2(g) + H_2(g)} \nonumber$
Nonmetal Oxygen Compounds
Most nonmetals react with oxygen to form nonmetal oxides. Depending on the available oxidation states for the element, a variety of oxides might form. Fluorine will combine with oxygen to form fluorides such as OF2, where the oxygen has a 2+-oxidation state.
Sulfur Oxygen Compounds
The two common oxides of sulfur are sulfur dioxide, SO2, and sulfur trioxide, SO3. The odor of burning sulfur comes from sulfur dioxide. Sulfur dioxide, shown in Figure $6$, occurs in volcanic gases and in the atmosphere near industrial plants that burn fuel containing sulfur compounds.
Commercial production of sulfur dioxide is from either burning sulfur or roasting sulfide ores such as ZnS, FeS2, and Cu2S in air. (Roasting, which forms the metal oxide, is the first step in the separation of many metals from their ores.) A convenient method for preparing sulfur dioxide in the laboratory is by the action of a strong acid on either sulfite salts containing the $\ce{SO3^{2-}}$ ion or hydrogen sulfite salts containing $\ce{HSO3^{-}}$. Sulfurous acid, H2SO3, forms first, but quickly decomposes into sulfur dioxide and water. Sulfur dioxide also forms when many reducing agents react with hot, concentrated sulfuric acid. Sulfur trioxide forms slowly when heating sulfur dioxide and oxygen together, and the reaction is exothermic:
$\ce{2 SO2(g) + O2(g) \longrightarrow 2 SO3(g)} \quad \quad \quad \quad \Delta H^{\circ}=-197.8 kJ \nonumber$
Sulfur dioxide is a gas at room temperature, and the SO2 molecule is bent. Sulfur trioxide melts at 17 °C and boils at 43 °C. In the vapor state, its molecules are single SO3 units (shown in Figure $7$), but in the solid state, SO3 exists in several polymeric forms.
The sulfur oxides react as Lewis acids with many oxides and hydroxides in Lewis acid-base reactions, with the formation of sulfites or hydrogen sulfites, and sulfates or hydrogen sulfates, respectively.
Halogen Oxygen Compounds
The halogens do not react directly with oxygen, but it is possible to prepare binary oxygen-halogen compounds by the reactions of the halogens with oxygen-containing compounds. Oxygen compounds with chlorine, bromine, and iodine are oxides because oxygen is the more electronegative element in these compounds. On the other hand, fluorine compounds with oxygen are fluorides because fluorine is the more electronegative element.
As a class, the oxides are extremely reactive and unstable, and their chemistry has little practical importance. Dichlorine oxide, formally called dichlorine monoxide, and chlorine dioxide, both shown in Figure $8$, are the only commercially important compounds. They are important as bleaching agents (for use with pulp and flour) and for water treatment.
Nonmetal Oxyacids and Their Salts
Nonmetal oxides form acids when allowed to react with water; these are acid anhydrides. The resulting oxyanions can form salts with various metal ions.
Nitrogen Oxyacids and Salts
Nitrogen pentaoxide, N2O5, and NO2 react with water to form nitric acid, HNO3. Alchemists, as early as the eighth century, knew nitric acid (shown in Figure $9$) as aqua fortis (meaning "strong water"). The acid was useful in the separation of gold from silver because it dissolves silver but not gold. Traces of nitric acid occur in the atmosphere after thunderstorms, and its salts are widely distributed in nature. There are tremendous deposits of Chile saltpeter, NaNO3, in the desert region near the boundary of Chile and Peru. Bengal saltpeter, KNO3, occurs in India and in other countries of the Far East.
In the laboratory, it is possible to produce nitric acid by heating a nitrate salt (such as sodium or potassium nitrate) with concentrated sulfuric acid:
$\ce{NaNO3(s) + H2SO4(l) ->[\Delta] NaHSO4(s) + HNO3(g)} \nonumber$
The Ostwald process is the commercial method for producing nitric acid. This process involves the oxidation of ammonia to nitric oxide, NO; oxidation of nitric oxide to nitrogen dioxide, NO2; and further oxidation and hydration of nitrogen dioxide to form nitric acid:
\begin{align*} \ce{4 NH3(g) + 5 O2(g) &\longrightarrow 4 NO(g) + 6 H2O(g)} \[4pt][4pt] \ce{2 NO(g) + O2(g) &\longrightarrow 2 NO2(g)} \[4pt][4pt] \ce{3 NO2(g) + H2O(l) &\longrightarrow 2 HNO3(aq) + NO(g)} \end{align*}
Or
$\ce{4 NO_2(g) + O_2(g) + 2 H2O(g) \longrightarrow 4 HNO_3(l)} \nonumber$
Pure nitric acid is a colorless liquid. However, it is often yellow or brown in color because NO2 forms as the acid decomposes. Nitric acid is stable in aqueous solution; solutions containing 68% of the acid are commercially available concentrated nitric acid. It is both a strong oxidizing agent and a strong acid.
The action of nitric acid on a metal rarely produces H2 (by reduction of H+) in more than small amounts. Instead, the reduction of nitrogen occurs. The products formed depend on the concentration of the acid, the activity of the metal, and the temperature. Normally, a mixture of nitrates, nitrogen oxides, and various reduction products form. Less active metals such as copper, silver, and lead reduce concentrated nitric acid primarily to nitrogen dioxide. The reaction of dilute nitric acid with copper produces NO. In each case, the nitrate salts of the metals crystallize upon evaporation of the resultant solutions.
Nonmetallic elements, such as sulfur, carbon, iodine, and phosphorus, undergo oxidation by concentrated nitric acid to their oxides or oxyacids, with the formation of NO2:
\begin{align*} \ce{S(s) + 6 HNO3(aq) &\longrightarrow H2SO4(aq) + 6 NO2(g) + 2 H2O(l)} \[4pt][4pt] \ce{C(s) + 4 HNO3(aq) &\longrightarrow CO2(g) + 4 NO2(g) + 2 H2O(l)} \end{align*}
Nitric acid oxidizes many compounds; for example, concentrated nitric acid readily oxidizes hydrochloric acid to chlorine and chlorine dioxide. A mixture of one part concentrated nitric acid and three parts concentrated hydrochloric acid (called aqua regia, which means royal water) reacts vigorously with metals. This mixture is particularly useful in dissolving gold, platinum, and other metals that are more difficult to oxidize than hydrogen. A simplified equation to represent the action of aqua regia on gold is:
$\ce{Au(s) + 4 HCl(aq) + 3 HNO3(aq) \longrightarrow HAuCl4(aq) + 3 NO2(g) + 3 H2O(l)} \nonumber$
Although gold is generally unreactive, you can watch a video of the complex mixture of compounds present in aqua regia dissolving it into solution.
Nitrates, salts of nitric acid, form when metals, oxides, hydroxides, or carbonates react with nitric acid. Most nitrates are soluble in water; indeed, one of the significant uses of nitric acid is to prepare soluble metal nitrates.
Nitric acid finds extensive use in the laboratory and in chemical industries as a strong acid and strong oxidizing agent. It is important in the manufacture of explosives, dyes, plastics, and drugs. Salts of nitric acid (nitrates) are valuable as fertilizers. Gunpowder is a mixture of potassium nitrate, sulfur, and charcoal.
The reaction of N2O3 with water gives a pale blue solution of nitrous acid, HNO2. However, HNO2 (shown in Figure $10$) is easier to prepare by the addition of an acid to a solution of nitrite; nitrous acid is a weak acid, so the nitrite ion is basic in aqueous solution:
$\ce{NO_2^{-}(aq) + H3O^{+}(aq) \longrightarrow HNO_2(aq) + H2O(l)} \nonumber$
Nitrous acid is very unstable and exists only in solution. It disproportionates slowly at room temperature (rapidly when heated) into nitric acid and nitric oxide. Nitrous acid is an active oxidizing agent with strong reducing agents, and strong oxidizing agents oxidize it to nitric acid.
Sodium nitrite, NaNO2, is an additive to meats such as hot dogs and cold cuts. The nitrite ion has two functions. It limits the growth of bacteria that can cause food poisoning, and it prolongs the meat’s retention of its red color. The addition of sodium nitrite to meat products is controversial because nitrous acid reacts with certain organic compounds to form a class of compounds known as nitrosamines. Nitrosamines produce cancer in laboratory animals. This has prompted the FDA to limit the amount of NaNO2 in foods.
The nitrites are much more stable than the acid, but nitrites, like nitrates, can explode. Nitrites, like nitrates, are also soluble in water (AgNO2 is only slightly soluble).
Phosphorus Oxyacids and Salts
Pure orthophosphoric acid, H3PO4 (shown in Figure $11$), forms colorless, deliquescent crystals that melt at 42 °C. The common name of this compound is phosphoric acid, and is commercially available as a viscous 82% solution known as syrupy phosphoric acid. One use of phosphoric acid is as an additive to many soft drinks.
One commercial method of preparing orthophosphoric acid is to treat calcium phosphate rock with concentrated sulfuric acid:
$\ce{Ca_3(PO_4)2(s) + 3 H2SO4(aq) \longrightarrow 2 H3PO4(aq) + 3 CaSO4(s)} \nonumber$
Dilution of the products with water, followed by filtration to remove calcium sulfate, gives a dilute acid solution contaminated with calcium dihydrogen phosphate, Ca(H2PO4)2, and other compounds associated with calcium phosphate rock. It is possible to prepare pure orthophosphoric acid by dissolving P4O10 in water.
The action of water on P4O6, PCl3, PBr3, or PI3 forms phosphorous acid, H3PO3 (shown in Figure $12$). The best method for preparing pure phosphorous acid is by hydrolyzing phosphorus trichloride:
$\ce{PCl_3(l) + 3 H2O(l) \longrightarrow H_3 PO_3(aq) + 3 HCl(g)} \nonumber$
Heating the resulting solution expels the hydrogen chloride and leads to the evaporation of water. When sufficient water evaporates, white crystals of phosphorous acid will appear upon cooling. The crystals are deliquescent, very soluble in water, and have an odor like that of garlic. The solid melts at 70.1 °C and decomposes at about 200 °C by disproportionation into phosphine and orthophosphoric acid:
$\ce{4 H_3PO_3(l) \longrightarrow PH_3(g) + 3 H_3 PO_4(l)} \nonumber$
Phosphorous acid forms only two series of salts, which contain the dihydrogen phosphite ion, $\ce{H2PO3^{-}}$, or the hydrogen phosphate ion, $\ce{HPO3^{2-}}$ respectively. It is not possible to replace the third atom of hydrogen because it is not very acidic, as it is not easy to ionize the P-H bond.
Sulfur Oxyacids and Salts
The preparation of sulfuric acid, H2SO4 (shown in Figure $13$), begins with the oxidation of sulfur to sulfur trioxide and then converting the trioxide to sulfuric acid. Pure sulfuric acid is a colorless, oily liquid that freezes at 10.5 °C. It fumes when heated because the acid decomposes to water and sulfur trioxide. The heating process causes the loss of more sulfur trioxide than water, until reaching a concentration of 98.33% acid. Acid of this concentration boils at 338 °C without further change in concentration (a constant boiling solution) and is commercially concentrated H2SO4. The amount of sulfuric acid used in industry exceeds that of any other manufactured compound.
The strong affinity of concentrated sulfuric acid for water makes it a good dehydrating agent. It is possible to dry gases and immiscible liquids that do not react with the acid by passing them through the acid.
Sulfuric acid is a strong diprotic acid that ionizes in two stages. In aqueous solution, the first stage is essentially complete. The secondary ionization is not nearly so complete, and $\ce{HSO4^{-}}$ is a moderately strong acid (about 25% ionized in solution of a $\ce{HSO4^{-}}$ salt ($K_a = 1.2 \times 10^{−2}}$).
Being a diprotic acid, sulfuric acid forms both sulfates, such as Na2SO4, and hydrogen sulfates, such as NaHSO4. Most sulfates are soluble in water; however, the sulfates of barium, strontium, calcium, and lead are only slightly soluble in water.
Among the important sulfates are Na2SO4⋅10H2O and Epsom salts, MgSO4⋅7H2O. Because the$\ce{HSO4^{-}}$ ion is an acid, hydrogen sulfates, such as NaHSO4, exhibit acidic behavior, and this compound is the primary ingredient in some household cleansers.
Hot, concentrated sulfuric acid is an oxidizing agent. Depending on its concentration, the temperature, and the strength of the reducing agent, sulfuric acid oxidizes many compounds and, in the process, undergoes reduction to SO2, $\ce{HSO3^{-}}$, $\ce{SO3^{2-}}$, S, H2S, or S2−.
Sulfur dioxide dissolves in water to form a solution of sulfurous acid, as expected for the oxide of a nonmetal. Sulfurous acid is unstable, and it is not possible to isolate anhydrous H2SO3. Heating a solution of sulfurous acid expels the sulfur dioxide. Like other diprotic acids, sulfurous acid ionizes in two steps: The hydrogen sulfite ion, $\ce{HSO3^{-}}$ and the sulfite ion, $\ce{SO3^{2-}}$, form. Sulfurous acid is a moderately strong acid. Ionization is about 25% in the first stage, but it is much less in the second ($K_{a1} = 1.2 \times 10^{−2}$ and $K_{a2} = 6.2 \times 10^{−8}$).
In order to prepare solid sulfite and hydrogen sulfite salts, it is necessary to add a stoichiometric amount of a base to a sulfurous acid solution and then evaporate the water. These salts also form from the reaction of SO2 with oxides and hydroxides. Heating solid sodium hydrogen sulfite forms sodium sulfite, sulfur dioxide, and water:
$\ce{2 NaHSO_3(s) ->[\Delta] Na_2SO_3(s) + SO_2(g) + H2O(l)} \nonumber$
Strong oxidizing agents can oxidize sulfurous acid. Oxygen in the air oxidizes it slowly to the more stable sulfuric acid:
$\ce{2 H_2 SO_3(aq) + O_2(g) + 2 H2O(l) ->[\Delta] 2 H3O^{+}(aq) + 2 HSO_4^{-}(aq)} \nonumber$
Solutions of sulfites are also very susceptible to air oxidation to produce sulfates. Thus, solutions of sulfites always contain sulfates after exposure to air.
Halogen Oxyacids and Their Salts
The compounds HXO, HXO2, HXO3, and HXO4, where X represents Cl, Br, or I, are the hypohalous, halous, halic, and perhalic acids, respectively. The strengths of these acids increase from the hypohalous acids, which are very weak acids, to the perhalic acids, which are very strong. Table $1$ lists the known acids, and, where known, their pKa values are given in parentheses.
Table $1$: Oxyacids of the Halogens
Name Fluorine Chlorine Bromine Iodine
hypohalous HOF HOCl (7.5) HOBr (8.7) HOI (11)
halous HClO2 (2.0)
halic HClO3 HBrO3 HIO3 (0.8)
perhalic HClO4 HBrO4 HIO4 (1.6)
paraperhalic H5IO6 (1.6)
The only known oxyacid of fluorine is the very unstable hypofluorous acid, HOF, which is prepared by the reaction of gaseous fluorine with ice:
$\ce{F_2(g) + H2O(s) \longrightarrow HOF(g) + HF(g)} \nonumber$
The compound is very unstable and decomposes above −40 °C. This compound does not ionize in water, and there are no known salts. It is uncertain whether the name hypofluorous acid is even appropriate for HOF; a more appropriate name might be hydrogen hypofluorite.
The reactions of chlorine and bromine with water are analogous to that of fluorine with ice, but these reactions do not go to completion, and mixtures of the halogen and the respective hypohalous and hydrohalic acids result. Other than HOF, the hypohalous acids only exist in solution. The hypohalous acids are all very weak acids; however, HOCl is a stronger acid than HOBr, which, in turn, is stronger than HOI.
The addition of base to solutions of the hypohalous acids produces solutions of salts containing the basic hypohalite ions, OX. It is possible to isolate these salts as solids. All of the hypohalites are unstable with respect to disproportionation in solution, but the reaction is slow for hypochlorite. Hypobromite and hypoiodite disproportionate rapidly, even in the cold:
$\ce{3 XO^{-}(aq) \longrightarrow 2 X^{-}(aq) + XO_3^{-}(aq)} \nonumber$
Sodium hypochlorite is an inexpensive bleach (Clorox) and germicide. The commercial preparation involves the electrolysis of cold, dilute, aqueous sodium chloride solutions under conditions where the resulting chlorine and hydroxide ion can react. The net reaction is:
$\ce{Cl^{-}(aq) + H2O(l) \stackrel{\text { electrical energy }}{\longrightarrow} ClO^{-}(aq) + H_2(g)} \nonumber$
The only definitely known halous acid is chlorous acid, HClO2, obtained by the reaction of barium chlorite with dilute sulfuric acid:
$\ce{Ba \left( ClO_2\right)_2(aq) + H_2 SO_4(aq) \longrightarrow BaSO_4(s) + 2 HClO_2(aq)} \nonumber$
Filtering the insoluble barium sulfate leaves a solution of HClO2. Chlorous acid is not stable; it slowly decomposes in solution to yield chlorine dioxide, hydrochloric acid, and water. Chlorous acid reacts with bases to give salts containing the chlorite ion (shown in Figure $14$). Sodium chlorite finds an extensive application in the bleaching of paper because it is a strong oxidizing agent and does not damage the paper.
Chloric acid, HClO3, and bromic acid, HBrO3, are stable only in solution. The reaction of iodine with concentrated nitric acid produces stable white iodic acid, HIO3:
$\ce{I_2(s) + 10 HNO_3(aq) \longrightarrow 2 HIO_3(s) + 10 NO_2(g) + 4 H2O(l)} \nonumber$
It is possible to obtain the lighter halic acids from their barium salts by reaction with dilute sulfuric acid. The reaction is analogous to that used to prepare chlorous acid. All of the halic acids are strong acids and very active oxidizing agents. The acids react with bases to form salts containing chlorate ions (shown in Figure $15$). Another preparative method is the electrochemical oxidation of a hot solution of a metal halide to form the appropriate metal chlorates. Sodium chlorate is a weed killer; potassium chlorate is used as an oxidizing agent.
Perchloric acid, HClO4, forms when treating a perchlorate, such as potassium perchlorate, with sulfuric acid under reduced pressure. The HClO4 can be distilled from the mixture:
$\ce{KClO_4(s) + H_2 SO_4(aq) \longrightarrow HClO_4(g) + KHSO_4(s)} \nonumber$
Dilute aqueous solutions of perchloric acid are quite stable thermally, but concentrations above 60% are unstable and dangerous. Perchloric acid and its salts are powerful oxidizing agents, as the very electronegative chlorine is more stable in a lower oxidation state than 7+. Serious explosions have occurred when heating concentrated solutions with easily oxidized substances. However, its reactions as an oxidizing agent are slow when perchloric acid is cold and dilute. The acid is among the strongest of all acids. Most salts containing the perchlorate ion (shown in Figure $16$) are soluble. It is possible to prepare them from reactions of bases with perchloric acid and, commercially, by the electrolysis of hot solutions of their chlorides.
Perbromate salts are difficult to prepare, and the best syntheses currently involve the oxidation of bromates in basic solution with fluorine gas followed by acidification. There are few, if any, commercial uses of this acid or its salts.
There are several different acids containing iodine in the 7+-oxidation state; they include metaperiodic acid, HIO4, and paraperiodic acid, H5IO6. These acids are strong oxidizing agents and react with bases to form the appropriate salts. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.10%3A_Occurrence_Preparation_and_Compounds_of_Oxygen.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of sulfur
Sulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure $1$. Sulfur is a constituent of many proteins and is essential for life.
The Frasch process, illustrated in Figure $2$: , is important in the mining of free sulfur from enormous underground deposits in Texas and Louisiana. Superheated water (170 °C and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is 99.5% to 99.9% pure and requires no purification for most uses.
Larger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.
Sulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is S8. However, chemists commonly use S to simplify the coefficients in chemical equations; we will follow this practice in this book.
Like oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2−). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:
$\ce{S(s) + 6 HNO_3(aq) \longrightarrow 2 H3O^{+}(aq) + SO_4^{2-}(aq) + 6 NO_2(g)} \nonumber$
The chemistry of sulfur with an oxidation state of 2− is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states.
18.12: Occurrence Preparation and Properties of Halogens
Learning Objectives
By the end of this section, you will be able to:
• Describe the preparation, properties, and uses of halogens
• Describe the properties, preparation, and uses of halogen compounds
The elements in group 17 are the halogens. These are the elements fluorine, chlorine, bromine, iodine, and astatine. These elements are too reactive to occur freely in nature, but their compounds are widely distributed. Chlorides are the most abundant; although fluorides, bromides, and iodides are less common, they are reasonably available. In this section, we will examine the occurrence, preparation, and properties of halogens. Next, we will examine halogen compounds with the representative metals followed by an examination of the interhalogens. This section will conclude with some applications of halogens.
Occurrence and Preparation
All of the halogens occur in seawater as halide ions. The concentration of the chloride ion is 0.54 M; that of the other halides is less than 10–4 M. Fluoride also occurs in minerals such as CaF2, Ca(PO4)3F, and Na3AlF6. Chloride also occurs in the Great Salt Lake and the Dead Sea, and in extensive salt beds that contain NaCl, KCl, or MgCl2. Part of the chlorine in your body is present as hydrochloric acid, which is a component of stomach acid. Bromine compounds occur in the Dead Sea and underground brines. Iodine compounds are found in small quantities in Chile saltpeter, underground brines, and sea kelp. Iodine is essential to the function of the thyroid gland.
The best sources of halogens (except iodine) are halide salts. It is possible to oxidize the halide ions to free diatomic halogen molecules by various methods, depending on the ease of oxidation of the halide ion. Fluoride is the most difficult to oxidize, whereas iodide is the easiest.
The major method for preparing fluorine is electrolytic oxidation. The most common electrolysis procedure is to use a molten mixture of potassium hydrogen fluoride, KHF2, and anhydrous hydrogen fluoride. Electrolysis causes HF to decompose, forming fluorine gas at the anode and hydrogen at the cathode. It is necessary to keep the two gases separated to prevent their explosive recombination to reform hydrogen fluoride.
Most commercial chlorine comes from the electrolysis of the chloride ion in aqueous solutions of sodium chloride; this is the chlor-alkali process discussed previously. Chlorine is also a product of the electrolytic production of metals such as sodium, calcium, and magnesium from their fused chlorides. It is also possible to prepare chlorine by the chemical oxidation of the chloride ion in acid solution with strong oxidizing agents such as manganese dioxide (MnO2) or sodium dichromate (Na2Cr2O7). The reaction with manganese dioxide is:
$\ce{MnO_2(s) + 2 Cl^{-}(aq) + 4 H3O^{+}(aq) \longrightarrow Mn^{2+}(aq) + Cl_2(g) + 6 H2O(l)} \nonumber$
The commercial preparation of bromine involves the oxidation of bromide ion by chlorine:
$\ce{2 Br^{-}(aq) + Cl_2(g) \longrightarrow Br_2(l) + 2 Cl^{-}(aq)} \nonumber$
Chlorine is a stronger oxidizing agent than bromine. This method is important for the production of essentially all domestic bromine.
Some iodine comes from the oxidation of iodine chloride, ICl, or iodic acid, HlO3. The commercial preparation of iodine utilizes the reduction of sodium iodate, NaIO3, an impurity in deposits of Chile saltpeter, with sodium hydrogen sulfite:
$\ce{2 IO_3^{-}(aq) + 5 HSO_3^{-}(aq) \longrightarrow 3 HSO_4^{-}(aq) + 2 SO_4^{2-}(aq) + H2O(l) + I_2(s)} \nonumber$
Properties of the Halogens
Fluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a deep reddish-brown liquid, and iodine is a grayish-black crystalline solid. Liquid bromine has a high vapor pressure, and the reddish vapor is readily visible in Figure $1$. Iodine crystals have a noticeable vapor pressure. When gently heated, these crystals sublime and form a beautiful deep violet vapor.
Bromine is only slightly soluble in water, but it is miscible in all proportions in less polar (or nonpolar) solvents such as chloroform, carbon tetrachloride, and carbon disulfide, forming solutions that vary from yellow to reddish-brown, depending on the concentration.
Iodine is soluble in chloroform, carbon tetrachloride, carbon disulfide, and many hydrocarbons, giving violet solutions of I2 molecules. Iodine dissolves only slightly in water, giving brown solutions. It is quite soluble in aqueous solutions of iodides, with which it forms brown solutions. These brown solutions result because iodine molecules have empty valence d orbitals and can act as weak Lewis acids towards the iodide ion. The equation for the reversible reaction of iodine (Lewis acid) with the iodide ion (Lewis base) to form triiodide ion, is:
$\ce{I_2(s) + I^{-}(aq) \longrightarrow I_3^{-}(aq)} \nonumber$
The easier it is to oxidize the halide ion, the more difficult it is for the halogen to act as an oxidizing agent. Fluorine generally oxidizes an element to its highest oxidation state, whereas the heavier halogens may not. For example, when excess fluorine reacts with sulfur, SF6 forms. Chlorine gives SCl2 and bromine, S2Br2. Iodine does not react with sulfur.
Fluorine is the most powerful oxidizing agent of the known elements. It spontaneously oxidizes most other elements; therefore, the reverse reaction, the oxidation of fluorides, is very difficult to accomplish. Fluorine reacts directly and forms binary fluorides with all of the elements except the lighter noble gases (He, Ne, and Ar). Fluorine is such a strong oxidizing agent that many substances ignite on contact with it. Drops of water inflame in fluorine and form O2, OF2, H2O2, O3, and HF. Wood and asbestos ignite and burn in fluorine gas. Most hot metals burn vigorously in fluorine. However, it is possible to handle fluorine in copper, iron, or nickel containers because an adherent film of the fluoride salt passivates their surfaces. Fluorine is the only element that reacts directly with the noble gas xenon.
Although it is a strong oxidizing agent, chlorine is less active than fluorine. Mixing chlorine and hydrogen in the dark makes the reaction between them to be imperceptibly slow. Exposure of the mixture to light causes the two to react explosively. Chlorine is also less active towards metals than fluorine, and oxidation reactions usually require higher temperatures. Molten sodium ignites in chlorine. Chlorine attacks most nonmetals (C, N2, and O2 are notable exceptions), forming covalent molecular compounds. Chlorine generally reacts with compounds that contain only carbon and hydrogen (hydrocarbons) by adding to multiple bonds or by substitution.
In cold water, chlorine undergoes a disproportionation reaction:
$\ce{Cl_2(aq) + 2 H2O(l) \longrightarrow HOCl(aq) + H3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
Half the chlorine atoms oxidize to the 1+ oxidation state (hypochlorous acid), and the other half reduce to the 1− oxidation state (chloride ion). This disproportionation is incomplete, so chlorine water is an equilibrium mixture of chlorine molecules, hypochlorous acid molecules, hydronium ions, and chloride ions. When exposed to light, this solution undergoes a photochemical decomposition:
$\ce{2 HOCl(aq) + 2 H2O(l) ->[\text{sunlight}] 2 H3O^{+}(aq) + 2 Cl^{-}(aq) + O2(g)} \nonumber$
The nonmetal chlorine is more electronegative than any other element except fluorine, oxygen, and nitrogen. In general, very electronegative elements are good oxidizing agents; therefore, we would expect elemental chlorine to oxidize all of the other elements except for these three (and the nonreactive noble gases). Its oxidizing property, in fact, is responsible for its principal use. For example, phosphorus(V) chloride, an important intermediate in the preparation of insecticides and chemical weapons, is manufactured by oxidizing the phosphorus with chlorine:
$\ce{P_4(s) + 10 Cl_2(g) \longrightarrow 4 PCl_5(l)} \nonumber$
A great deal of chlorine is also used to oxidize, and thus to destroy, organic or biological materials in water purification and in bleaching.
The chemical properties of bromine are similar to those of chlorine, although bromine is the weaker oxidizing agent and its reactivity is less than that of chlorine.
Iodine is the least reactive of the halogens. It is the weakest oxidizing agent, and the iodide ion is the most easily oxidized halide ion. Iodine reacts with metals, but heating is often required. It does not oxidize other halide ions.
Compared with the other halogens, iodine reacts only slightly with water. Traces of iodine in water react with a mixture of starch and iodide ion, forming a deep blue color. This reaction is a very sensitive test for the presence of iodine in water.
Halides of the Representative Metals
Thousands of salts of the representative metals have been prepared. The binary halides are an important subclass of salts. A salt is an ionic compound composed of cations and anions, other than hydroxide or oxide ions. In general, it is possible to prepare these salts from the metals or from oxides, hydroxides, or carbonates. We will illustrate the general types of reactions for preparing salts through reactions used to prepare binary halides.
The binary compounds of a metal with the halogens are the halides. Most binary halides are ionic. However, mercury, the elements of group 13 with oxidation states of 3+, tin(IV), and lead(IV) form covalent binary halides.
The direct reaction of a metal and a halogen produce the halide of the metal. Examples of these oxidation-reduction reactions include:
\begin{align*} \ce{Cd(s) + Cl_2(g) & \longrightarrow CdCl_2(s)} \[4pt][4pt] \ce{2 Ga(l) + 3 Br_2(l) & \longrightarrow 2 GaBr_3(s)} \end{align*}
Reactions of the alkali metals with elemental halogens are very exothermic and often quite violent. Under controlled conditions, they provide exciting demonstrations for budding students of chemistry. You can view the initial heating of the sodium that removes the coating of sodium hydroxide, sodium peroxide, and residual mineral oil to expose the reactive surface. The reaction with chlorine gas then proceeds very nicely.
If a metal can exhibit two oxidation states, it may be necessary to control the stoichiometry in order to obtain the halide with the lower oxidation state. For example, preparation of tin(II) chloride requires a 1:1 ratio of Sn to Cl2, whereas preparation of tin(IV) chloride requires a 1:2 ratio:
\begin{align*} \ce{Sn(s) + Cl_2(g) &\longrightarrow SnCl_2(s) }\[4pt][4pt] \ce{Sn(s) + 2 Cl_2(g) &\longrightarrow SnCl_4(l)} \end{align*}
The active representative metals—those that are easier to oxidize than hydrogen—react with gaseous hydrogen halides to produce metal halides and hydrogen. The reaction of zinc with hydrogen fluoride is:
$\ce{Zn(s) + 2 HF(g) \longrightarrow ZnF_2(s) + H_2(g)} \nonumber$
The active representative metals also react with solutions of hydrogen halides to form hydrogen and solutions of the corresponding halides. Examples of such reactions include:
\begin{align*} \ce{Cd(s) + 2 HBr(aq) &\longrightarrow CdBr_2(aq) + H_2(g)} \[4pt][4pt] \ce{Sn(s) + 2 HI(aq) &\longrightarrow SnI_2(aq) + H_2(g)} \end{align*}
Hydroxides, carbonates, and some oxides react with solutions of the hydrogen halides to form solutions of halide salts. It is possible to prepare additional salts by the reaction of these hydroxides, carbonates, and oxides with aqueous solution of other acids:
\begin{align*} \ce{CaCo_3(s) + 2 HCl(aq) &\longrightarrow CaCl_2(aq) + CO_2(g) + H2O(l)} \[4pt][4pt] \ce{TlOH(aq) + HF(aq) &\longrightarrow TlF(aq) + H2O(l)} \end{align*}
A few halides and many of the other salts of the representative metals are insoluble. It is possible to prepare these soluble salts by metathesis reactions that occur when solutions of soluble salts are mixed (see Figure $2$). Metathesis reactions are examined in the chapter on the stoichiometry of chemical reactions.
Several halides occur in large quantities in nature. The ocean and underground brines contain many halides. For example, magnesium chloride in the ocean is the source of magnesium ions used in the production of magnesium. Large underground deposits of sodium chloride, like the salt mine shown in Figure $3$, occur in many parts of the world. These deposits serve as the source of sodium and chlorine in almost all other compounds containing these elements. The chlor-alkali process is one example.
Interhalogens
Compounds formed from two or more different halogens are interhalogens. Interhalogen molecules consist of one atom of the heavier halogen bonded by single bonds to an odd number of atoms of the lighter halogen. The structures of IF3, IF5, and IF7 are illustrated in Figure $4$. Formulas for other interhalogens, each of which comes from the reaction of the respective halogens, are in Table $1$.
Note from Table $1$ that fluorine is able to oxidize iodine to its maximum oxidation state, 7+, whereas bromine and chlorine, which are more difficult to oxidize, achieve only the 5+-oxidation state. A 7+-oxidation state is the limit for the halogens. Because smaller halogens are grouped about a larger one, the maximum number of smaller atoms possible increases as the radius of the larger atom increases. Many of these compounds are unstable, and most are extremely reactive. The interhalogens react like their component halides; halogen fluorides, for example, are stronger oxidizing agents than are halogen chlorides.
The ionic polyhalides of the alkali metals, such as KI3, KICl2, KICl4, CsIBr2, and CsBrCl2, which contain an anion composed of at least three halogen atoms, are closely related to the interhalogens. As seen previously, the formation of the polyhalide $\ce{I3^{-}}$ anion is responsible for the solubility of iodine in aqueous solutions containing an iodide ion.
Table $1$: Interhalogens
YX YX3 YX5 YX7
ClF(g) ClF3(g) ClF5(g)
BrF(g) BrF3(l) BrF5(l)
BrCl(g)
IF(s) IF3(s) IF5(l) IF7(g)
ICl(l) ICl3(s)
IBr(s)
Applications
The fluoride ion and fluorine compounds have many important uses. Compounds of carbon, hydrogen, and fluorine are replacing Freons (compounds of carbon, chlorine, and fluorine) as refrigerants. Teflon is a polymer composed of –CF2CF2– units. Fluoride ion is added to water supplies and to some toothpastes as SnF2 or NaF to fight tooth decay. Fluoride partially converts teeth from Ca5(PO4)3(OH) into Ca5(PO4)3F.
Chlorine is important to bleach wood pulp and cotton cloth. The chlorine reacts with water to form hypochlorous acid, which oxidizes colored substances to colorless ones. Large quantities of chlorine are important in chlorinating hydrocarbons (replacing hydrogen with chlorine) to produce compounds such as tetrachloride (CCl4), chloroform (CHCl3), and ethyl chloride (C2H5Cl), and in the production of polyvinyl chloride (PVC) and other polymers. Chlorine is also important to kill the bacteria in community water supplies.
Bromine is important in the production of certain dyes, and sodium and potassium bromides are used as sedatives. At one time, light-sensitive silver bromide was a component of photographic film.
Iodine in alcohol solution with potassium iodide is an antiseptic (tincture of iodine). Iodide salts are essential for the proper functioning of the thyroid gland; an iodine deficiency may lead to the development of a goiter. Iodized table salt contains 0.023% potassium iodide. Silver iodide is useful in the seeding of clouds to induce rain; it was important in the production of photographic film and iodoform, CHI3, is an antiseptic. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.11%3A_Occurrence_Preparation_and_Properties_of_Sulfur.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the properties, preparation, and uses of the noble gases
The elements in group 18 are the noble gases (helium, neon, argon, krypton, xenon, and radon). They earned the name “noble” because they were assumed to be nonreactive since they have filled valence shells. In 1962, Dr. Neil Bartlett at the University of British Columbia proved this assumption to be false.
These elements are present in the atmosphere in small amounts. Some natural gas contains 1–2% helium by mass. Helium is isolated from natural gas by liquefying the condensable components, leaving only helium as a gas. The United States possesses most of the world’s commercial supply of this element in its helium-bearing gas fields. Argon, neon, krypton, and xenon come from the fractional distillation of liquid air. Radon comes from other radioactive elements. More recently, it was observed that this radioactive gas is present in very small amounts in soils and minerals. Its accumulation in well-insulated, tightly sealed buildings, however, constitutes a health hazard, primarily lung cancer.
The boiling points and melting points of the noble gases are extremely low relative to those of other substances of comparable atomic or molecular masses. This is because only weak London dispersion forces are present, and these forces can hold the atoms together only when molecular motion is very slight, as it is at very low temperatures. Helium is the only substance known that does not solidify on cooling at normal pressure. It remains liquid close to absolute zero (0.001 K) at ordinary pressures, but it solidifies under elevated pressure.
Helium is used for filling balloons and lighter-than-air craft because it does not burn, making it safer to use than hydrogen. Helium at high pressures is not a narcotic like nitrogen. Thus, mixtures of oxygen and helium are important for divers working under high pressures. Using a helium-oxygen mixture avoids the disoriented mental state known as nitrogen narcosis, the so-called rapture of the deep. Helium is important as an inert atmosphere for the melting and welding of easily oxidizable metals and for many chemical processes that are sensitive to air.
Liquid helium (boiling point, 4.2 K) is an important coolant to reach the low temperatures necessary for cryogenic research, and it is essential for achieving the low temperatures necessary to produce superconduction in traditional superconducting materials used in powerful magnets and other devices. This cooling ability is necessary for the magnets used for magnetic resonance imaging, a common medical diagnostic procedure. The other common coolant is liquid nitrogen (boiling point, 77 K), which is significantly cheaper.
Neon is a component of neon lamps and signs. Passing an electric spark through a tube containing neon at low pressure generates the familiar red glow of neon. It is possible to change the color of the light by mixing argon or mercury vapor with the neon or by utilizing glass tubes of a special color.
Argon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air.
Krypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only of a second. Krypton forms a difluoride, KrF2, which is thermally unstable at room temperature.
Stable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, XeF2, forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, XeF4, and xenon hexafluoride, XeF6, are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen.
When XeF6 reacts with water, a solution of XeO3 results and the xenon remains in the 6+-oxidation state:
$\ce{XeF6(s) + 3 H2O(l) \longrightarrow XeO3(aq) + 6 HF(aq)} \nonumber$
Dry, solid xenon trioxide, XeO3, is extremely explosive—it will spontaneously detonate. Both XeF6 and XeO3 disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, in which xenon reaches its maximum oxidation sate of 8+.
Radon apparently forms RnF2—evidence of this compound comes from radiochemical tracer techniques.
Unstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.13%3A_Occurrence_Preparation_and_Properties_of_the_Noble_Gases.txt |
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acid anhydride
compound that reacts with water to form an acid or acidic solution
alkaline earth metal
any of the metals (beryllium, magnesium, calcium, strontium, barium, and radium) occupying group 2 of the periodic table; they are reactive, divalent metals that form basic oxides
allotropes
two or more forms of the same element, in the same physical state, with different chemical structures
amorphous
solid material such as a glass that does not have a regular repeating component to its three-dimensional structure; a solid but not a crystal
base anhydride
metal oxide that behaves as a base towards acids
bicarbonate anion
salt of the hydrogen carbonate ion,
bismuth
heaviest member of group 15; a less reactive metal than other representative metals
borate
compound containing boron-oxygen bonds, typically with clusters or chains as a part of the chemical structure
carbonate
salt of the anion often formed by the reaction of carbon dioxide with bases
chemical reduction
method of preparing a representative metal using a reducing agent
chlor-alkali process
electrolysis process for the synthesis of chlorine and sodium hydroxide
disproportionation reaction
chemical reaction where a single reactant is simultaneously reduced and oxidized; it is both the reducing agent and the oxidizing agent
Downs cell
electrochemical cell used for the commercial preparation of metallic sodium (and chlorine) from molten sodium chloride
Frasch process
important in the mining of free sulfur from enormous underground deposits
Haber process
main industrial process used to produce ammonia from nitrogen and hydrogen; involves the use of an iron catalyst and elevated temperatures and pressures
halide
compound containing an anion of a group 17 element in the 1− oxidation state (fluoride, F; chloride, Cl; bromide, Br; and iodide, I)
Hall–Héroult cell
electrolysis apparatus used to isolate pure aluminum metal from a solution of alumina in molten cryolite
hydrogen carbonate
salt of carbonic acid, H2CO3 (containing the anion in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion
hydrogen halide
binary compound formed between hydrogen and the halogens: HF, HCl, HBr, and HI
hydrogen sulfate
ion
hydrogen sulfite
ion
hydrogenation
addition of hydrogen (H2) to reduce a compound
hydroxide
compound of a metal with the hydroxide ion OH or the group −OH
interhalogen
compound formed from two or more different halogens
metal (representative)
atoms of the metallic elements of groups 1, 2, 12, 13, 14, 15, and 16, which form ionic compounds by losing electrons from their outer s or p orbitals
metalloid
element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors
nitrate
ion; salt of nitric acid
nitrogen fixation
formation of nitrogen compounds from molecular nitrogen
Ostwald process
industrial process used to convert ammonia into nitric acid
oxide
binary compound of oxygen with another element or group, typically containing O2− ions or the group –O– or =O
ozone
allotrope of oxygen; O3
passivation
metals with a protective nonreactive film of oxide or other compound that creates a barrier for chemical reactions; physical or chemical removal of the passivating film allows the metals to demonstrate their expected chemical reactivity
peroxide
molecule containing two oxygen atoms bonded together or as the anion,
photosynthesis
process whereby light energy promotes the reaction of water and carbon dioxide to form carbohydrates and oxygen; this allows photosynthetic organisms to store energy
Pidgeon process
chemical reduction process used to produce magnesium through the thermal reaction of magnesium oxide with silicon
polymorph
variation in crystalline structure that results in different physical properties for the resulting compound
representative element
element where the s and p orbitals are filling
representative metal
metal among the representative elements
silicate
compound containing silicon-oxygen bonds, with silicate tetrahedra connected in rings, sheets, or three-dimensional networks, depending on the other elements involved in the formation of the compounds
sulfate
ion
sulfite
ion
superoxide
oxide containing the anion | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.14%3A_Key_Terms.txt |
This section focuses on the periodicity of the representative elements. These are the elements where the electrons are entering the s and p orbitals. The representative elements occur in groups 1, 2, and 12–18. These elements are representative metals, metalloids, and nonmetals. The alkali metals (group 1) are very reactive, readily form ions with a charge of 1+ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. The outermost electrons of the alkaline earth metals (group 2) are more difficult to remove than the outer electron of the alkali metals, leading to the group 2 metals being less reactive than those in group 1. These elements easily form compounds in which the metals exhibit an oxidation state of 2+. Zinc, cadmium, and mercury (group 12) commonly exhibit the group oxidation state of 2+ (although mercury also exhibits an oxidation state of 1+ in compounds that contain Aluminum, gallium, indium, and thallium (group 13) are easier to oxidize than is hydrogen. Aluminum, gallium, and indium occur with an oxidation state 3+ (however, thallium also commonly occurs as the Tl+ ion). Tin and lead form stable divalent cations and covalent compounds in which the metals exhibit the 4+-oxidation state.
Because of their chemical reactivity, it is necessary to produce the representative metals in their pure forms by reduction from naturally occurring compounds. Electrolysis is important in the production of sodium, potassium, and aluminum. Chemical reduction is the primary method for the isolation of magnesium, zinc, and tin. Similar procedures are important for the other representative metals.
The elements boron, silicon, germanium, arsenic, antimony, and tellurium separate the metals from the nonmetals in the periodic table. These elements, called metalloids or sometimes semimetals, exhibit properties characteristic of both metals and nonmetals. The structures of these elements are similar in many ways to those of nonmetals, but the elements are electrical semiconductors.
Nonmetals have structures that are very different from those of the metals, primarily because they have greater electronegativity and electrons that are more tightly bound to individual atoms. Most nonmetal oxides are acid anhydrides, meaning that they react with water to form acidic solutions. Molecular structures are common for most of the nonmetals, and several have multiple allotropes with varying physical properties.
Hydrogen is the most abundant element in the universe and its chemistry is truly unique. Although it has some chemical reactivity that is similar to that of the alkali metals, hydrogen has many of the same chemical properties of a nonmetal with a relatively low electronegativity. It forms ionic hydrides with active metals, covalent compounds in which it has an oxidation state of 1− with less electronegative elements, and covalent compounds in which it has an oxidation state of 1+ with more electronegative nonmetals. It reacts explosively with oxygen, fluorine, and chlorine, less readily with bromine, and much less readily with iodine, sulfur, and nitrogen. Hydrogen reduces the oxides of metals with lower reduction potentials than chromium to form the metal and water. The hydrogen halides are all acidic when dissolved in water.
The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO3), the antacid Tums (CaCO3), and baking soda (NaHCO3) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating.
Nitrogen exhibits oxidation states ranging from 3− to 5+. Because of the stability of the N≡N triple bond, it requires a great deal of energy to make compounds from molecular nitrogen. Active metals such as the alkali metals and alkaline earth metals can reduce nitrogen to form metal nitrides. Nitrogen oxides and nitrogen hydrides are also important substances.
Phosphorus (group 15) commonly exhibits oxidation states of 3− with active metals and of 3+ and 5+ with more electronegative nonmetals. The halogens and oxygen will oxidize phosphorus. The oxides are phosphorus(V) oxide, P4O10, and phosphorus(III) oxide, P4O6. The two common methods for preparing orthophosphoric acid, H3PO4, are either the reaction of a phosphate with sulfuric acid or the reaction of water with phosphorus(V) oxide. Orthophosphoric acid is a triprotic acid that forms three types of salts.
Oxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood.
Compounds of the representative metals with oxygen exist in three categories (1) oxides, (2) peroxides and superoxides, and (3) hydroxides. Heating the corresponding hydroxides, nitrates, or carbonates is the most common method for producing oxides. Heating the metal or metal oxide in oxygen may lead to the formation of peroxides and superoxides. The soluble oxides dissolve in water to form solutions of hydroxides. Most metals oxides are base anhydrides and react with acids. The hydroxides of the representative metals react with acids in acid-base reactions to form salts and water. The hydroxides have many commercial uses.
All nonmetals except fluorine form multiple oxides. Nearly all of the nonmetal oxides are acid anhydrides. The acidity of oxyacids requires that the hydrogen atoms bond to the oxygen atoms in the molecule rather than to the other nonmetal atom. Generally, the strength of the oxyacid increases with the number of oxygen atoms bonded to the nonmetal atom and not to a hydrogen.
Sulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, S2−, in which it has as oxidation state of 2−. Sulfur reacts with most nonmetals.
The halogens form halides with less electronegative elements. Halides of the metals vary from ionic to covalent; halides of nonmetals are covalent. Interhalogens form by the combination of two or more different halogens.
All of the representative metals react directly with elemental halogens or with solutions of the hydrohalic acids (HF, HCl, HBr, and HI) to produce representative metal halides. Other laboratory preparations involve the addition of aqueous hydrohalic acids to compounds that contain such basic anions, such as hydroxides, oxides, or carbonates.
The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.15%3A_Summary.txt |
1.
How do alkali metals differ from alkaline earth metals in atomic structure and general properties?
2.
Why does the reactivity of the alkali metals decrease from cesium to lithium?
3.
Predict the formulas for the nine compounds that may form when each species in column 1 of the table reacts with each species in column 2.
1 2
Na I
Sr Se
Al O
4.
Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples.
(a) the most metallic of the elements Al, Be, and Ba
(b) the most covalent of the compounds NaCl, CaCl2, and BeCl2
(c) the lowest first ionization energy among the elements Rb, K, and Li
(d) the smallest among Al, Al+, and Al3+
(e) the largest among Cs+, Ba2+, and Xe
5.
Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other?
6.
The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH)2, which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic:
$CaO(s)+H2O(l)⟶Ca(OH)2(s)ΔH=−350 kJmol−1CaO(s)+H2O(l)⟶Ca(OH)2(s)ΔH=−350 kJmol−1$
(a) What is the enthalpy of reaction per gram of quicklime that reacts?
(b) How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime?
7.
Write a balanced equation for the reaction of elemental strontium with each of the following:
(a) oxygen
(b) hydrogen bromide
(c) hydrogen
(d) phosphorus
(e) water
8.
How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?
9.
What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl2.) How many pounds of fish is this?
10.
The elements sodium, aluminum, and chlorine are in the same period.
(a) Which has the greatest electronegativity?
(b) Which of the atoms is smallest?
(c) Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine.
(d) Will the oxide of each element be acidic, basic, or amphoteric?
11.
Does metallic tin react with HCl?
12.
What is tin pest, also known as tin disease?
13.
Compare the nature of the bonds in PbCl2 to that of the bonds in PbCl4.
14.
Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare?
15.
Write an equation for the reduction of cesium chloride by elemental calcium at high temperature.
16.
Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal?
17.
Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples.
18.
The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine.
Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples.
19.
What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water?
20.
How many grams of oxygen gas are necessary to react completely with 3.01 $××$ 1021 atoms of magnesium to yield magnesium oxide?
21.
Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction?
22.
Why is it possible for an active metal like aluminum to be useful as a structural metal?
23.
Describe the production of metallic aluminum by electrolytic reduction.
24.
What is the common ore of tin and how is tin separated from it?
25.
A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H2SnO3, is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample?
26.
Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield.
(a) How long will it take to produce the 100 kg of sodium metal?
(b) What volume of chlorine at 25 °C and 1.00 atm forms?
27.
What mass of magnesium forms when 100,000 A is passed through a MgCl2 melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield?
28.
Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples.
(a) GeH4
(b) SbF3
(c) Te(OH)6
(d) H2Te
(e) GeF2
(f) TeCl4
(g) $SiF62−SiF62−$
(h) SbCl5
(i) TeF6
29.
Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.
(a) H3BPH3
(b) $BF4−BF4−$
(c) BBr3
(d) B(CH3)3
(e) B(OH)3
30.
Describe the hybridization of boron and the molecular structure about the boron in each of the following:
(a) H3BPH3
(b) $BF4−BF4−$
(c) BBr3
(d) B(CH3)3
(e) B(OH)3
31.
Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure.
32.
Write a Lewis structure for each of the following molecules and ions:
(a) (CH3)3SiH
(b) $SiO44−SiO44−$
(c) Si2H6
(d) Si(OH)4
(e) $SiF62−SiF62−$
33.
Describe the hybridization of silicon and the molecular structure of the following molecules and ions:
(a) (CH3)3SiH
(b) $SiO44−SiO44−$
(c) Si2H6
(d) Si(OH)4
(e) $SiF62−SiF62−$
34.
Describe the hybridization and the bonding of a silicon atom in elemental silicon.
35.
Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding.
(a) SiH4
(b) Si2H6
(c) SiCl3H
(d) SiF4
(e) SiCl2F2
36.
Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide.
37.
Name each of the following compounds:
(a) TeO2
(b) Sb2S3
(c) GeF4
(d) SiH4
(e) GeH4
38.
Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature):
(a) F2
(b) O2
(c) S
(d) Se
(e) Br2
39.
Why is boron limited to a maximum coordination number of four in its compounds?
40.
Write a formula for each of the following compounds:
(a) silicon dioxide
(b) silicon tetraiodide
(c) silane
(d) silicon carbide
(e) magnesium silicide
41.
From the data given in Appendix G, determine the standard enthalpy change and the standard free energy change for each of the following reactions:
(a) $BF3(g)+3H2O(l)⟶B(OH)3(s)+3HF(g)BF3(g)+3H2O(l)⟶B(OH)3(s)+3HF(g)$
(b) $BCl3(g)+3H2O(l)⟶B(OH)3(s)+3HCl(g)BCl3(g)+3H2O(l)⟶B(OH)3(s)+3HCl(g)$
(c) $B2H6(g)+6H2O(l)⟶2B(OH)3(s)+6H2(g)B2H6(g)+6H2O(l)⟶2B(OH)3(s)+6H2(g)$
42.
A hydride of silicon prepared by the reaction of Mg2Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride?
43.
Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?
44.
Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure?
45.
Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P4 molecules instead of P2 molecules?
46.
Write balanced chemical equations for the reaction of the following acid anhydrides with water:
(a) SO3
(b) N2O3
(c) Cl2O7
(d) P4O10
(e) NO2
47.
Determine the oxidation number of each element in each of the following compounds:
(a) HCN
(b) OF2
(c) AsCl3
48.
Determine the oxidation state of sulfur in each of the following:
(a) SO3
(b) SO2
(c) $SO32−SO32−$
49.
Arrange the following in order of increasing electronegativity: F; Cl; O; and S.
50.
Why does white phosphorus consist of tetrahedral P4 molecules while nitrogen consists of diatomic N2 molecules?
51.
Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals?
52.
The reaction of calcium hydride, CaH2, with water can be characterized as a Lewis acid-base reaction:
$CaH2(s)+2H2O(l)⟶Ca(OH)2(aq)+2H2(g)CaH2(s)+2H2O(l)⟶Ca(OH)2(aq)+2H2(g)$
Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction.
53.
In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why?
54.
What mass of CaH2 is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is:
$CaH2(s)+2H2O(l)⟶Ca(OH)2(aq)+2H2(g)CaH2(s)+2H2O(l)⟶Ca(OH)2(aq)+2H2(g)$
55.
What mass of hydrogen gas results from the reaction of 8.5 g of KH with water?
$KH+H2O⟶KOH+H2KH+H2O⟶KOH+H2$
56.
Carbon forms the $CO32−CO32−$ ion, yet silicon does not form an analogous $SiO32−SiO32−$ ion. Why?
57.
Complete and balance the following chemical equations:
(a) hardening of plaster containing slaked lime
$Ca(OH)2+CO2⟶Ca(OH)2+CO2⟶$
(b) removal of sulfur dioxide from the flue gas of power plants
$CaO+SO2⟶CaO+SO2⟶$
(c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise
$NaHCO3+NaH2PO4⟶NaHCO3+NaH2PO4⟶$
58.
Heating a sample of Na2CO3xH2O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na2CO3. What is the formula of the hydrated compound?
59.
Write the Lewis structures for each of the following:
(a) NH2−
(b) N2F4
(c) $NH2−NH2−$
(d) NF3
(e) $N3−N3−$
60.
For each of the following, indicate the hybridization of the nitrogen atom (for $N3−,N3−,$ the central nitrogen).
(a) N2F4
(b) $NH2−NH2−$
(c) NF3
(d) $N3−N3−$
61.
Explain how ammonia can function both as a Brønsted base and as a Lewis base.
62.
Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples.
(a) NCl3
(b) ClNO
(c) N2O5
(d) N2O3
(e) $NO2−NO2−$
(f) N2O4
(g) N2O
(h) $NO3−NO3−$
(i) HNO2
(j) HNO3
63.
For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples.
(a) NO2
(b) $NO2−NO2−$
(c) $NO2+NO2+$
64.
How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce?
65.
Although PF5 and AsF5 are stable, nitrogen does not form NF5 molecules. Explain this difference among members of the same group.
66.
The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 M HNO3 is at 35.27 mL. What is the concentration of the CsOH solution?
67.
Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
(a) PH3
(b) $PH4+PH4+$
(c) P2H4
(d) $PO43−PO43−$
(e) PF5
68.
Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry.
(a) PH3
(b) $PH4+PH4+$
(c) P2H4
(d) $PO43−PO43−$
69.
Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.)
(a) $P4+Al⟶P4+Al⟶$
(b) $P4+Na⟶P4+Na⟶$
(c) $P4+F2⟶P4+F2⟶$
(d) $P4+Cl2⟶P4+Cl2⟶$
(e) $P4+O2⟶P4+O2⟶$
(f) $P4O6+O2⟶P4O6+O2⟶$
70.
Describe the hybridization of phosphorus in each of the following compounds: P4O10, P4O6, PH4I (an ionic compound), PBr3, H3PO4, H3PO3, PH3, and P2H4. You may wish to review the chapter on advanced theories of covalent bonding.
71.
What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl3 is an excess of water? Note that when H3PO3 is titrated under these conditions, only one proton of the acid molecule reacts.
72.
How much POCl3 can form from 25.0 g of PCl5 and the appropriate amount of H2O?
73.
How many tons of Ca3(PO4)2 are necessary to prepare 5.0 tons of phosphorus if the yield is 90%?
74.
Write equations showing the stepwise ionization of phosphorous acid.
75.
Draw the Lewis structures and describe the geometry for the following:
(a) $PF4+PF4+$
(b) PF5
(c) $PF6−PF6−$
(d) POF3
76.
Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms?
77.
Assign an oxidation state to phosphorus in each of the following:
(a) NaH2PO3
(b) PF5
(c) P4O6
(d) K3PO4
(e) Na3P
(f) Na4P2O7
78.
Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus.
(a) Write the empirical formula of phosphorus(V) oxide.
(b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280.
(c) Write balanced equations for the production of phosphorus(V) oxide and phosphoric acid.
(d) Determine the mass of phosphorus required to make 1.00 $××$ 104 kg of phosphoric acid, assuming a yield of 98.85%.
79.
Predict the product of burning francium in air.
80.
Using equations, describe the reaction of water with potassium and with potassium oxide.
81.
Write balanced chemical equations for the following reactions:
(a) zinc metal heated in a stream of oxygen gas
(b) zinc carbonate heated until loss of mass stops
(c) zinc carbonate added to a solution of acetic acid, CH3CO2H
(d) zinc added to a solution of hydrobromic acid
82.
Write balanced chemical equations for the following reactions:
(a) cadmium burned in air
(b) elemental cadmium added to a solution of hydrochloric acid
(c) cadmium hydroxide added to a solution of acetic acid, CH3CO2H
83.
Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations.
84.
Write balanced chemical equations for the following reactions:
(a) metallic aluminum burned in air
(b) elemental aluminum heated in an atmosphere of chlorine
(c) aluminum heated in hydrogen bromide gas
(d) aluminum hydroxide added to a solution of nitric acid
85.
Write balanced chemical equations for the following reactions:
(a) sodium oxide added to water
(b) cesium carbonate added to an excess of an aqueous solution of HF
(c) aluminum oxide added to an aqueous solution of HClO4
(d) a solution of sodium carbonate added to solution of barium nitrate
(e) titanium metal produced from the reaction of titanium tetrachloride with elemental sodium
86.
What volume of 0.250 M H2SO4 solution is required to neutralize a solution that contains 5.00 g of CaCO3?
87.
Which is the stronger acid, HClO4 or HBrO4? Why?
88.
Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state.
(a) Mg
(b) Rb
(c) Ga
(d) C2H2
(e) CO
89.
Which is the stronger acid, H2SO4 or H2SeO4? Why? You may wish to review the chapter on acid-base equilibria.
90.
Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid.
91.
Give the hybridization and oxidation state for sulfur in SO2, in SO3, and in H2SO4.
92.
Which is the stronger acid, NaHSO3 or NaHSO4?
93.
Determine the oxidation state of sulfur in SF6, SO2F2, and KHS.
94.
Which is a stronger acid, sulfurous acid or sulfuric acid? Why?
95.
Oxygen forms double bonds in O2, but sulfur forms single bonds in S8. Why?
96.
Give the Lewis structure of each of the following:
(a) SF4
(b) K2SO4
(c) SO2Cl2
(d) H2SO3
(e) SO3
97.
Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent.
98.
Explain why sulfuric acid, H2SO4, which is a covalent molecule, dissolves in water and produces a solution that contains ions.
99.
How many grams of Epsom salts (MgSO4⋅7H2O) will form from 5.0 kg of magnesium?
100.
What does it mean to say that mercury(II) halides are weak electrolytes?
101.
Why is SnCl4 not classified as a salt?
102.
The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions:
(a) reaction of a weak base and a strong acid
$NH3+HClO4⟶NH3+HClO4⟶$
(b) preparation of a soluble silver salt for silver plating
$Ag2CO3+HNO3⟶Ag2CO3+HNO3⟶$
(c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride
$SrCl2(aq)+H2O(l)→electrolysisSrCl2(aq)+H2O(l)→electrolysis$
103.
Which is the stronger acid, HClO3 or HBrO3? Why?
104.
What is the hybridization of iodine in IF3 and IF5?
105.
Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.
(a) IF5
(b) $I3−I3−$
(c) PCl5
(d) SeF4
(e) ClF3
106.
Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties?
107.
Name each of the following compounds:
(a) BrF3
(b) NaBrO3
(c) PBr5
(d) NaClO4
(e) KClO
108.
Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid.
109.
What is the oxidation state of the halogen in each of the following?
(a) H5IO6
(b) $IO4−IO4−$
(c) ClO2
(d) ICl3
(e) F2
110.
Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 M. A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought?
111.
Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.
(a) XeF2
(b) XeF4
(c) XeO3
(d) XeO4
(e) XeOF4
112.
What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry.
(a) XeF2
(b) XeF4
(c) XeO3
(d) XeO4
(e) XeOF4
113.
Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.
(a) XeF2
(b) XeF4
(c) XeO3
(d) XeO4
(e) XeOF4
114.
What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry.
(a) XeO2F2
(b) KrF2
(c) $XeF3+XeF3+$
(d) $XeO64−XeO64−$
(e) XeO3
115.
A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.
116.
Basic solutions of Na4XeO6 are powerful oxidants. What mass of Mn(NO3)2•6H2O reacts with 125.0 mL of a 0.1717 M basic solution of Na4XeO6 that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.16%3A_Exercises.txt |
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. These include the d-block (groups 3–11) and f-block element elements. The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
19: Transition Metals and Coordination Chemistry
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.
In addition to being used in their pure elemental forms, many compounds containing transition metals have numerous other applications. Silver nitrate is used to create mirrors, zirconium silicate provides friction in automotive brakes, and many important cancer-fighting agents, like the drug cisplatin and related species, are platinum compounds.
The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
19.02: Occurrence Preparation and Properties of Transition Metals and Their Compounds
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure $1$, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.
The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.
The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series.
Example $1$: Valence Electrons in Transition Metals
Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration:
1. cerium(III)
2. lead(II)
3. Ti2+
4. Am3+
5. Pd2+
For the examples that are transition metals, determine to which series they belong.
Solution
For ions, the s-valence electrons are lost prior to the d or f electrons.
1. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series.
2. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element.
3. titanium(II) [Ar]3d2; first transition series
4. americium(III) [Rn]5f6; actinide
5. palladium(II) [Kr]4d8; second transition series
Exercise $1$
Give an example of an ion from the first transition series with no d electrons.
Answer
V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+.
Chemistry in Everyday Life: Uses of Lanthanides in Devices
Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver ($4.5 \times 10^{−5}~ \%$ versus $0.79 \times 10^{−5}~\%$ by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.
The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure $2$). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.
As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to$470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials.
The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series.
Properties of the Transition Elements
Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (see Appendix H), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.
Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as and
Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.
Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts.
Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure $3$. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).
For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air.
The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+.
Example $2$: Activity of the Transition Metals
Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?
Solution
First, we need to look up the reduction half reactions (in Appendix L) for each oxide in the specified oxidation state:
$\begin{array}{cc} Cr_2 O_7^{2-}+14 H^{+}+6 e^{-} \longrightarrow 2 Cr^{3+}+7 H_2 O & +1.33 V \ MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+ H_2 O & +1.51 V \ TiO_2+4 H^{+}+2 e^{-} \longrightarrow Ti^{2+}+2 H_2 O & -0.50 V \end{array} \nonumber$
A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set.
Exercise $2$
Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from Appendix L.
Answer
$\ce{Co(s) + 2 HCl \longrightarrow H2 + CoCl2(aq)} \nonumber$
no reaction because Pt(s) will not be oxidized by H+
Preparation of the Transition Elements
Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure $4$). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.
Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.
In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.
We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining.
1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.
2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal.
3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals.
Isolation of Iron
The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.
The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure $5$) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.
Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:
$\ce{CO2(g) + C(s) \longrightarrow 2 CO(g)} \nonumber$
The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure $5$.
The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:
$\ce{CaO(s) + SiO2(s) \longrightarrow CaSiO3(l)} \nonumber$
Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure $6$).
Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle.
You can watch an animation of steelmaking that walks you through the process.
Isolation of Copper
The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:
\begin{align*} \ce{CaCO3(s) + SiO2(s) &\longrightarrow CaSiO3(l) + CO2(g)} \ \ce{FeO(s) + SiO2(s) &\longrightarrow FeSiO3(l)} \end{align*}
In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).
Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:
\begin{align*} \ce{2 Cu2S(l) + 3 O2(g) & \longrightarrow 2 Cu2O(l) + 2 SO2(g)} \ \ce{2 Cu2O(l) + Cu2S(l) & \longrightarrow 6 Cu(l) + SO2(g)} \end{align*}
The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure $7$). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).
Isolation of Silver
Silver sometimes occurs in large nuggets (Figure $8$) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores. Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are:
\begin{align*} \ce{4 Ag(s) + 8 CN^{-}(aq) + O2(g) + 2 H2O (l) \longrightarrow 4\left[ Ag(CN)2 \right]^{-}(aq) + 4 OH^{-}(aq)} \ \ce{2 Ag2S(s) + 8 CN^{-}(aq) + O2(g) + 2 H2O(l) \longrightarrow 4\left[ Ag(CN)2 \right]^{-}(aq) + 2 S(s) + 4 OH^{-}(aq)} \ \ce{AgCl(s) + 2 CN^{-}(aq) \longrightarrow \left[ Ag(CN)2 \right]^{-}(aq) + Cl^{-}(aq)} \end{align*}
The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:
$\ce{ 2\left[ Ag(CN)2\right]^{-}(aq) + Zn(s) \longrightarrow 2 Ag(s) + \left[ Zn(CN)4\right]^{2-}(aq)} \nonumber$
Example $3$: Refining Redox
One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:
$\ce{4 Ag(s) + 8 CN^{-}(aq) + O_2(g) + 2 H_2O(l) \longrightarrow 4 \left[ Ag(CN)2 \right]^{-}(aq) + 4 OH^{-}(aq)} \nonumber$
Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:
$\ce{4 Ag(s) + 8 CN^{-}(aq) \longrightarrow 4\left[ Ag(CN)2\right]^{-}(aq)} ? \nonumber$
Solution
The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state.
Exercise $3$
During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?
Answer
The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0).
Transition Metal Compounds
The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows.
Halides
Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:
$\ce{2 Fe(s) + 3 Cl2(g) \longrightarrow 2 FeCl3(s)} \nonumber$
Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state:
$\ce{Fe(s) + 2 FeCl3(s) \longrightarrow 3 FeCl2(s)} \nonumber$
The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.
In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:
\begin{align*} \ce{NiCO3(s) + 2 HF(aq) \longrightarrow NiF2(aq) + H2O(l) + CO2(g)} \ \ce{Co(OH)2(s) + 2 HBr(aq) \longrightarrow CoBr2(aq) + 2 H2O(l)} \end{align*}
Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:
$\ce{Cr(s) + 2 HCl(aq) \longrightarrow CrCl2(aq) + H2(g)} \nonumber$
The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics.
The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:
\begin{aligned} \ce{SiCl4(l) + 2 H2O(l) &\longrightarrow SiO2(s) + 4 HCl(aq)}\ \ce{TiCl4(l) + 2 H2O(l) &\longrightarrow TiO2(s) + 4 HCl(aq)} \end{aligned} \nonumber
Oxides
As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.
The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO.
Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:
$\begin{gathered} \ce{FeC2O4(s) \longrightarrow FeO(s) + CO(g) + CO2(g)} \ \ce{Co(OH)2(s) \longrightarrow CoO(s) + H2O (g)} \end{gathered} \nonumber$
With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:
\begin{align*} \ce{CoO(s) + 2 HNO3(aq) &\longrightarrow Co(NO3)2(aq) + H2O(l)} \ \ce{Sc2O3(s) + 6 HCl(aq) &\longrightarrow 2 ScCl3(aq) + 3 H2O(l)} \end{align*}] The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions and $\ce{VO4^{3−}$, $\ce{CrO4^{2−}$, and $\ce{MnO4−}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by: \[\ce{CrO3(s) + 2 Na^{+}(aq) + 2 OH^{-}(aq) \longrightarrow 2 Na^{+}(aq) + CrO4^{2-}(aq) + H2O(l)} \nonumber
Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively.
Hydroxides
When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:
$Co^{2+}(aq) +2 OH^{-}(aq) \longrightarrow Co ( OH )_2(s) \nonumber$
In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:
$4 Fe^{3+}(aq) +6 OH^{-}(aq) + n H_2 O (l) \longrightarrow 2 Fe_2 O_3 \cdot( n +3) H_2 O (s) \nonumber$
These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal.
Carbonates
Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:
$Ni^{2+}(aq) + CO_3{ }^{2-} \longrightarrow NiCO_3(s) \nonumber$
The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides.
Other Salts
In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.
A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:
$2 Sc (s)+6 HBr (aq) \longrightarrow 2 ScBr_3(aq) +3 H_2(g) \nonumber$
The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:
$Ni ( OH )_2(s)+2 H_3 O^{+}(aq) +2 ClO_4^{-}(aq) \longrightarrow Ni^{2+}(aq) +2 ClO_4^{-}(aq) +4 H_2 O (l) \nonumber$
Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example:
$Ba^{2+}(aq) +2 Cl^{-}(aq) +2 K^{+}(aq) + CrO_4^{2-}(aq) \longrightarrow BaCrO_4(s)+2 K^{+}(aq) +2 Cl^{-}(aq) \nonumber$
In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements.
How Sciences Interconnect: High Temperature Superconductors
A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.
Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.
One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7.
The new materials become superconducting at temperatures close to 90 K (Figure $9$), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.
Further advances during the same period included materials that became superconducting at even higher temperatures and with a wider array of materials. The DuPont team led by Uma Chowdry and Arthur Sleight identified Bismouth-Strontium-Copper-Oxides that became superconducting at temperatures as high as 110 K and, importantly, did not contain rare earth elements. Advances continued through the subsequent decades until, in 2020, a team led by Ranga Dias at University of Rochester announced the development of a room-temperature superconductor, opening doors to widespread applications. More research and development is needed to realize the potential of these materials, but the possibilities are very promising.
Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.
Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure $10$).
Watch how a high-temperature superconductor levitates around a magnetic racetrack in the video. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• List the defining traits of coordination compounds
• Describe the structures of complexes containing monodentate and polydentate ligands
• Use standard nomenclature rules to name coordination compounds
• Explain and provide examples of geometric and optical isomerism
• Identify several natural and technological occurrences of coordination compounds
The hemoglobin in your blood, the chlorophyll in green plants, vitamin B-12, and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(1\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH4. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(2\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure \(3\)). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H2NCH2CH2NH2) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(4\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en)3]3+, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.
Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or “many teeth”) because it can bite into the metal center with more than one bond. The term chelate (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(4\): showed one example of a chelate. The heme complex in hemoglobin is another important example (Figure \(5\)). It contains a polydentate ligand with four donor atoms that coordinate to iron.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as \(\ce{NH3}\), \(\ce{Cl^{−}}\), and \(\ce{H2O}\), are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, \(\cE{H2NCH2CH2NH2}\), and the anion of the acid glycine, \(\ce{NH2CH2CO2^{−}}\) (Figure \(6\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The ligand in heme (Figure \(5\)) is a tetradentate ligand.
The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group. For examples, see Table \(1\). For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua (H2O), ammine (NH3), carbonyl (CO), and nitrosyl (NO). For example, name [Pt(NH3)2Cl4] as diamminetetrachloroplatinum(IV).
Table \(1\): Examples of Anionic Ligands
Anionic Ligand Name
F fluoro
Cl chloro
Br bromo
I iodo
CN cyano
nitrato
OH hydroxo
O2– oxo
oxalato
carbonato
1. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri- (for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris- (for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups (C5H4N).
When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Table \(2\) and Table \(3\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state (Table \(4\)). Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H2O)4Cl2]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + x, so the oxidation state (x) is equal to 3+.
Table \(2\): Examples in Which the Complex Is a Cation
[Co(NH3)6]Cl3 hexaamminecobalt(III) chloride
[Pt(NH3)4Cl2]2+ tetraamminedichloroplatinum(IV) ion
[Ag(NH3)2]+ diamminesilver(I) ion
[Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride
[Co(H2NCH2CH2NH2)3]2(SO4)3 tris(ethylenediamine)cobalt(III) sulfate
Table \(3\): Examples in Which the Complex Is Neutral
[Pt(NH3)2Cl4] diamminetetrachloroplatinum(IV)
[Ni(H2NCH2CH2NH2)2Cl2] dichlorobis(ethylenediamine)nickel(II)
Table \(4\): Examples in Which the Complex Is an Anion
[PtCl6]2− hexachloroplatinate(IV) ion
Na2[SnCl6] sodium hexachlorostannate(IV)
Do you think you understand naming coordination complexes? You can look over more examples and test yourself with online quizzes at the University of Sydney’s site.
Example \(1\): Coordination Numbers and Oxidation States
Determine the name of the following complexes and give the coordination number of the central metal atom.
1. \(\ce{Na2[PtCl6]}\)
2. \(\ce{K3[Fe(C2O4)3]}\)
3. \(\ce{[Co(NH3)5Cl]Cl2}\)
Solution
(a) There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]2−. There are six anionic chloride ligands, so −2 = −6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
(b) The coordination sphere has a charge of 3− (based on the potassium) and the oxalate ligands each have a charge of 2−, so the metal oxidation state is given by −3 = −6 + x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
(c) In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1−. The oxidation state is found by +2 = −1 + x and is 3+, so the complex is pentaamminechlorocobalt(III) chloride and the coordination number is six.
Exercise \(1\)
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.
Answer
K[Ag(CN)2]; coordination number two
The Structures of Complexes
For transition metal complexes, the coordination number determines the geometry around the central metal ion. The most common structures of the complexes in coordination compounds are square planar, tetrahedral, and octahedral, corresponding to coordination numbers of four, four, and six, respectively. Coordination numbers greater than six are less common and yield a variety of structures (see Figure \(7\) and Table \(5\)):
Table \(5\): Coordination Numbers and Molecular Geometry
Coordination Number Molecular Geometry Example
2 linear [Ag(NH3)2]+
3 trigonal planar [Cu(CN)3]2−
4 tetrahedral(d0 or d10), low oxidation states for M [Ni(CO)4]
4 square planar (d8) [Ni(CN)4]2−
5 trigonal bipyramidal [CoCl5]2−
5 square pyramidal [VO(CN)4]2−
6 octahedral [CoCl6]3−
7 pentagonal bipyramid [ZrF7]3−
8 square antiprism [ReF8]2−
8 dodecahedron [Mo(CN)8]4−
9 and above more complicated structures [ReH9]2−
Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding d-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(8\). The chloride and nitrate anions in [Co(H2O)6]Cl2 and [Cr(en)3](NO3)3, and the potassium cations in K2[PtCl6], are outside the brackets and are not bonded to the metal ion.
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN)4]2− (Figure \(9\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH3)2Cl2], each ligand has two other ligands at 90° angles (called the cis positions) and one additional ligand at an 180° angle, in the trans position.
Isomerism in Complexes
Isomers are different chemical species that have the same chemical formula. Transition metal complexes often exist as geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral [Co(NH3)4Cl2]+ ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure \(10\)). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH3)4Cl2]NO3 isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.
Example \(2\): Geometric Isomers
Identify which geometric isomer of [Pt(NH3)2Cl2] is shown in Figure \(9\). Draw the other geometric isomer and give its full name.
Solution
In the Figure \(9\), the two chlorine ligands occupy cis positions. The other form is shown in Figure \(11\). When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diamminedichloroplatinum(II).
Exercise \(1\)
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).
Answer
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(12\). These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other.
The [Co(en)2Cl2]+ ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure \(13\)).
Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH3)5SCN]2+ or [Co(NH3)5NCS]2+).
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6][Br] and [CoCl5Br][Cl].
Coordination Complexes in Nature and Technology
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(14\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.
Chemistry in Everyday Life: Transition Metal Catalysts
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (see Figure \(15\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.
Portrait of a Chemist: Deanna D’Alessandro
Dr. Deanna D’Alessandro develops new metal-containing materials that demonstrate unique electronic, optical, and magnetic properties. Her research combines the fields of fundamental inorganic and physical chemistry with materials engineering. She is working on many different projects that rely on transition metals. For example, one type of compound she is developing captures carbon dioxide waste from power plants and catalytically converts it into useful products (see Figure \(16\)).
Another project involves the development of porous, sponge-like materials that are “photoactive.” The absorption of light causes the pores of the sponge to change size, allowing gas diffusion to be controlled. This has many potential useful applications, from powering cars with hydrogen fuel cells to making better electronics components. Although not a complex, self-darkening sunglasses are an example of a photoactive substance.
Watch this video to learn more about this research and listen to Dr. D’Alessandro (shown in Figure \(17\)) describe what it is like being a research chemist.
Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(14\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.
The structure of heme (Figure \(18\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe2+; oxidation of the iron to Fe3+ prevents oxygen transport.
Complexing agents often are used for water softening because they tie up such ions as Ca2+, Mg2+, and Fe2+, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, (HO2CCH2)2NCH2CH2N(CH2CO2H)2, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure \(19\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH2CH(SH)CH2OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(20\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.
Example \(3\): Chelation Therapy
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(21\). Identify which atoms in this molecule could act as donor atoms.
Exercise \(1\)
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations.1 Identify at least two biologically important metals that could be disrupted by chelation therapy.
Answer
Ca, Fe, Zn, and Cu
Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN)2] and [Au(CN)2] are used extensively in the electroplating industry.
In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diamminedichloroplatinum(II), [Pt(NH3)2(Cl)2], and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diammine (NH3)2 portion is retained with other groups, replacing the dichloro [(Cl)2] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.
Footnotes
• 1National Council against Health Fraud, NCAHF Policy Statement on Chelation Therapy, (Peabody, MA, 2002). | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.03%3A_Coordination_Chemistry_of_Transition_Metals.txt |
Learning Objectives
By the end of this section, you will be able to:
• Outline the basic premise of crystal field theory (CFT)
• Identify molecular geometries associated with various d-orbital splitting patterns
• Predict electron configurations of split d orbitals for selected transition metal atoms or ions
• Explain spectral and magnetic properties in terms of CFT concepts
The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.
Crystal Field Theory
To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.
CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.
All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure $1$. In an octahedral complex, the six ligands coordinate along the axes.
In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same.
In octahedral complexes, the lobes in two of the five d orbitals, the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals, point toward the ligands (Figure $1$). These two orbitals are called the $e_g$ orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals, have lobes that point between the ligands and are called the $t_{2g}$ orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the d orbitals of the metal ion. However, the repulsions between the electrons in the eg orbitals (the $d_{z^2}$ and $d_{x^2−y^2}$ orbitals) and the ligands are greater than the repulsions between the electrons in the $t_{2g}$ orbitals (the $d_{zy}$, $d_{xz}$, and $d_{yz}$ orbitals) and the ligands. This is because the lobes of the eg orbitals point directly at the ligands, whereas the lobes of the $t_{2g}$ orbitals point between them. Thus, electrons in the eg orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the $t_{2g}$ orbitals. The difference in energy may be represented as shown in Figure $2$.
The difference in energy between the eg and the t2g orbitals is called the crystal field splitting and is symbolized by $Δ_{oct}$, where oct stands for octahedral.
The magnitude of Δoct depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3d, 4d, or 5d orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:
$\Large \xrightarrow[\text{a few ligands of the spectrochemical series, in order of increasing field strength of the ligand}]{\ce{I^{-}< Br^{-}< Cl^{-}< F^{-} < H_2O < NH_3 < en < NO_2^{-} < CN^{-}}} \nonumber$
In this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the Δoct value for an octahedral complex with iodide ligands (I) is much smaller than the Δoct value for the same metal with cyanide ligands (CN).
Electrons in the d orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy (P). Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δoct. When electrons fill the d orbitals, the relative magnitudes of Δoct and P determine which orbitals will be occupied.
In [Fe(CN)6]4−, the strong field of six cyanide ligands produces a large Δoct. Under these conditions, the electrons require less energy to pair than they require to be excited to the eg orbitals (Δoct > P). The six 3d electrons of the Fe2+ ion pair in the three t2g orbitals (Figure $3$). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.
In [Fe(H2O)6]2+, on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δoct < P). Because it requires less energy for the electrons to occupy the eg orbitals than to pair together, there will be an electron in each of the five 3d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H2O)6]2+, there will be one pair of electrons and four unpaired electrons (Figure $3$). Complexes such as the [Fe(H2O)6]2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.
A similar line of reasoning shows why the [Fe(CN)6]3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H2O)6]3+ and [FeF6]3− ions are high-spin complexes with five unpaired electrons.
Example $1$: High- and Low-Spin Complexes
Predict the number of unpaired electrons.
1. $\ce{K3[CrI6]}$
2. $\ce{[Cu(en)2(H2O)2]Cl2}$
3. $\ce{Na3[Co(NO2)6]}$
Solution
The complexes are octahedral.
1. Cr3+ has a d3 configuration. These electrons will all be unpaired.
2. Cu2+ is d9, so there will be one unpaired electron.
3. Co3+ has d6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t2g orbitals, leaving 0 unpaired.
Exercise $1$
The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which d-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?
Answer
d4, d5, d6, and d7
Example $2$: CFT for Other Geometries
CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the eg set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes (Figure $4$). None of the orbitals points directly at the tetrahedral ligands. However, the eg set (along the Cartesian axes) overlaps with the ligands less than does the t2g set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral eg set becomes a tetrahedral e set, and the octahedral t2g set becomes a t2 set.
Solution
Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so $Δ_{tet}$ is usually small $\left(\Delta_{\text {tet}}=\frac{4}{9} \Delta_{oct}\right)$
Exercise $2$
Explain how many unpaired electrons a tetrahedral d4 ion will have.
Answer
4; because $Δ_{tet}$ is small, all tetrahedral complexes are high spin and the electrons go into the t2 orbitals before pairing
The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the z-axis. This changes the distribution of the d orbitals, as orbitals on or near the z-axis become more stable, and those on or near the x- or y-axes become less stable. This results in the octahedral t2g and the eg sets splitting and gives a more complicated pattern, as depicted below:
Magnetic Moments of Molecules and Ions
Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N2 and ions such as Na+ and [Fe(CN)6]4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.
When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d6 [Fe(CN)6]4− confirms that iron is diamagnetic, whereas high-spin d6 [Fe(H2O)6]2+ has four unpaired electrons with a magnetic moment that confirms this arrangement.
Colors of Transition Metal Complexes
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed.
The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH3)4]2+ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure $5$).
Example $3$: Colors of Complexes
The octahedral complex [Ti(H2O)6]3+ has a single d electron. To excite this electron from the ground state t2g orbital to the eg orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δoct and occurs at 499 nm. Calculate the value of Δoct in Joules and predict what color the solution will appear.
Solution
Using Planck's equation (refer to the section on electromagnetic energy), we calculate:
$\nu =\frac{c}{\lambda} \text { so } \frac{3.00 \times 10^8 m / s }{\frac{499 nm \times 1 m }{10^9 nm }}=6.01 \times 10^{14} Hz \nonumber$
$E=h \nu \text { so } 6.63 \times 10^{-34} J \cdot s \times 6.01 \times 10^{14} Hz =3.99 \times 10^{-19} \text { Joules/ion } \nonumber$
Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.
Exercise $3$
A complex that appears green, absorbs photons of what wavelengths?
Answer
red, 620–800 nm
Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure $6$, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.
The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H2O)6]SO4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure $7$). In contrast, the low-spin iron(II) complex K4[Fe(CN)6] appears pale yellow because it absorbs higher-energy violet photons.
Watch this video of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.
In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δoct). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.
A coordination compound of the Cu+ ion has a d10 configuration, and all the eg orbitals are filled. To excite an electron to a higher level, such as the 4p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN)2], for example, is colorless. On the other hand, octahedral Cu2+ complexes have a vacancy in the eg orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure $8$). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.04%3A_Spectroscopic_and_Magnetic_Properties_of_Coordination_Compounds.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
actinide series(also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103
bidentate ligandligand that coordinates to one central metal through coordinate bonds from two different atoms
central metalion or atom to which one or more ligands is attached through coordinate covalent bonds
chelatecomplex formed from a polydentate ligand attached to a central metal
chelating ligandligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configurationconfiguration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compoundstable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
coordination compoundsubstance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination numbernumber of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination spherecentral metal atom or ion plus the attached ligands of a complex
crystal field splitting (Δoct)difference in energy between the t2g and eg sets or t and e sets of orbitals
crystal field theorymodel that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding
d-block elementone of the elements in groups 3–11 with valence electrons in d orbitals
donor atomatom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
eg orbitalsset of two d orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the t2g orbitals
f-block element(also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table
first transition seriestransition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29
fourth transition seriestransition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111
geometric isomersisomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties
high-spin complexcomplex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals
hydrometallurgyprocess in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
ionization isomer(or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
lanthanide series(also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71
ligandion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomercoordination compound that possesses a ligand that can bind to the transition metal in two different ways (CN vs. NC)
low-spin complexcomplex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals
monodentateligand that attaches to a central metal through just one coordinate covalent bond
optical isomer(also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
pairing energy (P)energy required to place two electrons with opposite spins into a single orbital
platinum metalsgroup of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
polydentate ligandligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
rare earth elementcollection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition seriestransition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47
smeltingprocess of extracting a pure metal from a molten ore
spectrochemical seriesranking of ligands according to the magnitude of the crystal field splitting they induce
steelmaterial made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
strong-field ligandligand that causes larger crystal field splittings
superconductormaterial that conducts electricity with no resistance
t2g orbitalsset of three d orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the eg orbitals according to CFT
third transition seriestransition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79
trans configurationconfiguration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule
weak-field ligandligand that causes small crystal field splittings
19.06: Summary
The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.
Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.
The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δoct) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t2g orbitals are completely filled before any electrons occupy the eg orbitals. Weak-field ligands favor formation of high-spin complexes. The t2g and the eg orbitals are singly occupied before any are doubly occupied. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.05%3A_Key_Terms.txt |
1.
Write the electron configurations for each of the following elements:
(a) Sc
(b) Ti
(c) Cr
(d) Fe
(e) Ru
2.
Write the electron configurations for each of the following elements and its ions:
(a) Ti
(b) Ti2+
(c) Ti3+
(d) Ti4+
3.
Write the electron configurations for each of the following elements and its 3+ ions:
(a) La
(b) Sm
(c) Lu
4.
Why are the lanthanoid elements not found in nature in their elemental forms?
5.
Which of the following elements is most likely to be used to prepare La by the reduction of La2O3: Al, C, or Fe? Why?
6.
Which of the following is the strongest oxidizing agent: $VO43,VO43,$ $CrO42−,CrO42−,$ or $MnO4−?MnO4−?$
7.
Which of the following elements is most likely to form an oxide with the formula MO3: Zr, Nb, or Mo?
8.
The following reactions all occur in a blast furnace. Which of these are redox reactions?
(a) $3Fe2O3(s)+CO(g)⟶2Fe3O4(s)+CO2(g)3Fe2O3(s)+CO(g)⟶2Fe3O4(s)+CO2(g)$
(b) $Fe3O4(s)+CO(g)⟶3FeO(s)+CO2(g)Fe3O4(s)+CO(g)⟶3FeO(s)+CO2(g)$
(c) $FeO(s)+CO(g)⟶Fe(l)+CO2(g)FeO(s)+CO(g)⟶Fe(l)+CO2(g)$
(d) $C(s)+O2(g)⟶CO2(g)C(s)+O2(g)⟶CO2(g)$
(e) $C(s)+CO2(g)⟶2CO(g)C(s)+CO2(g)⟶2CO(g)$
(f) $CaCO3(s)⟶CaO(s)+CO2(g)CaCO3(s)⟶CaO(s)+CO2(g)$
(g) $CaO(s)+SiO2(s)⟶CaSiO3(l)CaO(s)+SiO2(s)⟶CaSiO3(l)$
9.
Why is the formation of slag useful during the smelting of iron?
10.
Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer.
11.
Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 M Na2Cr2O7 is required in the titration. What percentage of the ore sample was iron?
12.
How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe2O3 to convert that Fe2O3 into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume.
13.
Find the potentials of the following electrochemical cell:
Cd | Cd2+, M = 0.10 ‖ Ni2+, M = 0.50 | Ni
14.
A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
15.
The standard reduction potential for the reaction $[Co(H2O)6]3+(aq)+e−⟶[Co(H2O)6]2+(aq)[Co(H2O)6]3+(aq)+e−⟶[Co(H2O)6]2+(aq)$ is about 1.8 V. The reduction potential for the reaction $[Co(NH3)6]3+(aq)+e−⟶[Co(NH3)6]2+(aq)[Co(NH3)6]3+(aq)+e−⟶[Co(NH3)6]2+(aq)$ is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H2O)6]2+ and/or [Co(NH3)6]2+, can be oxidized to the corresponding cobalt(III) complex by oxygen.
16.
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
(a) $MnCO3(s)+HI(aq)⟶MnCO3(s)+HI(aq)⟶$
(b) $CoO(s)+O2(g)⟶CoO(s)+O2(g)⟶$
(c) $La(s)+O2(g)⟶La(s)+O2(g)⟶$
(d) $V(s)+VCl4(s)⟶V(s)+VCl4(s)⟶$
(e) $Co(s)+xsF2(g)⟶Co(s)+xsF2(g)⟶$
(f) $CrO3(s)+CsOH(aq)⟶CrO3(s)+CsOH(aq)⟶$
17.
Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
(a) $Fe(s)+H2SO4(aq)⟶Fe(s)+H2SO4(aq)⟶$
(b) $FeCl3(aq)+NaOH(aq)⟶FeCl3(aq)+NaOH(aq)⟶$
(c) $Mn(OH)2(s)+HBr(aq)⟶Mn(OH)2(s)+HBr(aq)⟶$
(d) $Cr(s)+O2(g)⟶Cr(s)+O2(g)⟶$
(e) $Mn2O3(s)+HCl(aq)⟶Mn2O3(s)+HCl(aq)⟶$
(f) $Ti(s)+xsF2(g)⟶Ti(s)+xsF2(g)⟶$
18.
Describe the electrolytic process for refining copper.
19.
Predict the products of the following reactions and balance the equations.
(a) Zn is added to a solution of Cr2(SO4)3 in acid.
(b) FeCl2 is added to a solution containing an excess of $Cr2O72−Cr2O72−$ in hydrochloric acid.
(c) Cr2+ is added to $Cr2O72−Cr2O72−$ in acid solution.
(d) Mn is heated with CrO3.
(e) CrO is added to 2HNO3 in water.
(f) FeCl3 is added to an aqueous solution of NaOH.
20.
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
21.
Predict the products of each of the following reactions and then balance the chemical equations.
(a) Fe is heated in an atmosphere of steam.
(b) NaOH is added to a solution of Fe(NO3)3.
(c) FeSO4 is added to an acidic solution of KMnO4.
(d) Fe is added to a dilute solution of H2SO4.
(e) A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
(f) FeCO3 is added to a solution of HClO4.
(g) Fe is heated in air.
22.
Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$Co(NO3)2(s)⟶Co2O3(s)+NO2(g)+O2(g)Co(NO3)2(s)⟶Co2O3(s)+NO2(g)+O2(g)$
23.
Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
24.
Predict which will be more stable, [CrO4]2− or [WO4]2−, and explain.
25.
Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M3O4 are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M2O3, to permit estimation of the metal’s two oxidation states.)
(a) Sc2O3
(b) TiO2
(c) V2O5
(d) CrO3
(e) MnO2
(f) Fe3O4
(g) Co3O4
(h) NiO
(i) Cu2O
26.
Indicate the coordination number for the central metal atom in each of the following coordination compounds:
(a) [Pt(H2O)2Br2]
(b) [Pt(NH3)(py)(Cl)(Br)] (py = pyridine, C5H5N)
(c) [Zn(NH3)2Cl2]
(d) [Zn(NH3)(py)(Cl)(Br)]
(e) [Ni(H2O)4Cl2]
(f) [Fe(en)2(CN)2]+ (en = ethylenediamine, C2H8N2)
27.
Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
(a) tetrahydroxozincate(II) ion (tetrahedral)
(b) hexacyanopalladate(IV) ion
(c) dichloroaurate(I) ion (note that aurum is Latin for "gold")
(d) diamminedichloroplatinum(II)
(e) potassium diamminetetrachlorochromate(III)
(f) hexaamminecobalt(III) hexacyanochromate(III)
(g) dibromobis(ethylenediamine) cobalt(III) nitrate
28.
Give the coordination number for each metal ion in the following compounds:
(a) [Co(CO3)3]3− (note that CO32− is bidentate in this complex)
(b) [Cu(NH3)4]2+
(c) [Co(NH3)4Br2]2(SO4)
(d) [Pt(NH3)4][PtCl4]
(e) [Cr(en)3](NO3)3
(f) [Pd(NH3)2Br2] (square planar)
(g) K3[Cu(Cl)5]
(h) [Zn(NH3)2Cl2]
29.
Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.
(a) [Pt(H2O)2Br2] (square planar)
(b) [Pt(NH3)(py)(Cl)(Br)] (square planar, py = pyridine, C5H5N)
(c) [Zn(NH3)3Cl]+ (tetrahedral)
(d) [Pt(NH3)3Cl]+ (square planar)
(e) [Ni(H2O)4Cl2]
(f) [Co(C2O4)2Cl2]3− (note that $C2O42−C2O42−$ is the bidentate oxalate ion, $−O2CCO2−)−O2CCO2−)$
30.
Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):
(a) [Co(en)2(NO2)Cl]+
(b) [Co(en)2Cl2]+
(c) [Pt(NH3)2Cl4]
(d) [Cr(en)3]3+
(e) [Pt(NH3)2Cl2]
31.
Name each of the compounds or ions given in Exercise 19.28, including the oxidation state of the metal.
32.
Name each of the compounds or ions given in Exercise 19.30.
33.
Specify whether the following complexes have isomers.
(a) tetrahedral [Ni(CO)2(Cl)2]
(b) trigonal bipyramidal [Mn(CO)4NO]
(c) [Pt(en)2Cl2]Cl2
34.
Predict whether the carbonate ligand $CO32−CO32−$will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.
35.
Draw the geometric, linkage, and ionization isomers for [CoCl5CN][CN].
36.
Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
37.
Draw the crystal field diagrams for [Fe(NO2)6]4− and [FeF6]3−. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δoct to P for each complex.
38.
Give the oxidation state of the metal, number of d electrons, and the number of unpaired electrons predicted for [Co(NH3)6]Cl3.
39.
The solid anhydrous solid CoCl2 is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
40.
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
41.
How many unpaired electrons are present in each of the following?
(a) [CoF6]3− (high spin)
(b) [Mn(CN)6]3− (low spin)
(c) [Mn(CN)6]4− (low spin)
(d) [MnCl6]4− (high spin)
(e) [RhCl6]3− (low spin)
42.
Explain how the diphosphate ion, [O3P−O−PO3]4−, can function as a water softener that prevents the precipitation of Fe2+ as an insoluble iron salt.
43.
For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?
(a) [Fe(H2O)6]2+ or [Fe(CN)6]4−
(b) [Co(NH3)6]3+ or [CoF6]3−
(c) [Mn(CN)6]4− or [MnCl6]4−
44.
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
45.
Would you expect the complex [Co(en)3]Cl3 to have any unpaired electrons? Any isomers?
46.
Would you expect the Mg3[Cr(CN)6]2 to be diamagnetic or paramagnetic? Explain your reasoning.
47.
Would you expect salts of the gold(I) ion, Au+, to be colored? Explain.
48.
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.07%3A_Exercises.txt |
The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions. This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas.
• 20.1: Introduction
• 20.2: Nuclear Structure and Stability
An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2.
• 20.3: Nuclear Equations
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation).
• 20.4: Radioactive Decay
Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life.
• 20.5: Transmutation and Nuclear Energy
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way. Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons).
• 20.6: Uses of Radioisotopes
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US.
• 20.7: Biological Effects of Radiation
We are constantly exposed to radiation from naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays are most penetrating.
• 20.8: Key Terms
• 20.9: Key Equations
• 20.10: Summary
• 20.11: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax.
20: Nuclear Chemistry
The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions.
This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe nuclear structure in terms of protons, neutrons, and electrons
• Calculate mass defect and binding energy for nuclei
• Explain trends in the relative stability of nuclei
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^{1}_{1}H}$.
Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation $\ce{_{A}^{Z}X}$, where $X$ is the symbol for the element, $A$ is the mass number, and $Z$ is the atomic number (for example, $\ce{^{14}_6C}$). Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about $10^{−15}$ meters, a nucleus is quite small compared to the radius of the entire atom, which is about $10^{−10}$ meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger). Example $1$ demonstrates just how great nuclear densities can be in the natural world.
Example $1$: Density of a Neutron Star
Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star’s outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses (1 solar mass = $M_{\odot}$ = mass of the sun = $1.99 \times 10^{30}\, kg$) and a diameter of 26 km.
1. What is the density of this neutron star?
2. How does this neutron star’s density compare to the density of a uranium nucleus, which has a diameter of about 15 fm (1 fm = 10–15 m)?
Solution
We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:
$d=\frac{m}{V} \quad \text { with } \quad V=\frac{4}{3} \pi r^3 \nonumber$
1. The radius of the neutron star is so the density of the neutron star is:
$d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^3}=\frac{2.4\left(1.99 \times 10^{30} \, \text{kg} \right)}{\frac{4}{3} \pi\left(1.3 \times 10^4 m \right)^3}=5.2 \times 10^{17} \, \text{kg}/ \text{m}^3 \nonumber$
2. The radius of the U-235 nucleus is so the density of the U-235 nucleus is:
$d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^3}=\frac{235 \, \text{amu} \left(\dfrac{1.66 \times 10^{-27} \, \text{kg}}{1\, \text{amu}}\right)}{\frac{4}{3} \pi\left(7.5 \times 10^{-15}\, \text{m}\right)^3}=2.2 \times 10^{17}\, \text{kg} / \text{m}^3 \nonumber$
These values are fairly similar (same order of magnitude), but the neutron star is more than twice as dense as the U-235 nucleus.
Exercise $1$
Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km, and compare it to the density of a hydrogen nucleus, which has a diameter of $1.75\, \text{fm}$ ($1\, \text{fm} = 1 \times 10^{–15}\, \text{m}$).
Answer
The density of the neutron star is $3.4 \times 10^{18} \, \text{kg}/\text{m}^3$. The density of a hydrogen nucleus is $6.0 \times 10^{17}\, \text{kg}/\text{m}^3}$. The neutron star is 5.7 times denser than the hydrogen nucleus.
To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than 10−15 meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.
Visit this website for more information about the four fundamental forces.
Nuclear Binding Energy
As a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:
$\underbrace{(2 \times 1.0073 \, \text{amu})}_{\text{protons}} + \underbrace{(2 \times 1.0087\, \text{amu})}_{\text{neutrons}} + \underbrace{(2 \times 0.00055\, \text{amu})}_{\text{electrons}} =4.0331\, \text{amu} \nonumber$
However, mass spectrometric measurements reveal that the mass of an atom is 4.0026 amu, less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a “loss” in mass of 4.0331 amu – 4.0026 amu = 0.0305 amu. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms’ nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions.
The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:
$E=m c^2 \label{mass-energy}$
where $E$ is energy, $m$ is mass of the matter being converted, and $c$ is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this mass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example $2$. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making 1 eV = 1.602 10–19 J.
Example $2$: Calculation of Nuclear Binding Energy
Determine the binding energy for the nuclide in:
1. joules per mole of nuclei
2. joules per nucleus
3. MeV per nucleus
Solution
The mass defect for a $\ce{_{2}^{4}He}$ nucleus is 0.0305 amu, as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that 1 J = 1 kg m2/s2).
(a) First, express the mass defect in g/mol. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass (g/mol) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore 0.0305 g/mol. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg, since 1 J = 1 kg m2/s2. Converting grams into kilograms yields a mass defect of $3.05 \times 10^{–5}\, \text{kg/mol}$. Substituting this quantity into the mass-energy equivalence equation yields:
\begin{aligned} E &=m c^2=\frac{3.05 \times 10^{-5} kg }{ mol } \times\left(\frac{2.998 \times 10^8 m }{ s }\right)^2=2.74 \times 10^{12} \, \text{kg m}^2 \text{s}^{-2}\text{mol}^{-1} \[4pt] & =2.74 \times 10^{12}\text{J mol}^{-1}=2.74 \,\text{TJ} \, \text{mol}^{-1} \end{aligned} \nonumber
Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg, roughly the mass of typical drop of water).
(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro’s number:
\begin{align*} E &= 2.74 \times 10^{12} \, \text{J mol}^{-1} \times \dfrac{1\, \text{mol}}{6.022 \times 10^{23} \text { nuclei }} \[4pt] &= 4.55 \times 10^{-12}\, \text{J} =4.55 \, \text{pJ} \end{align*} \nonumber
(c) Recall that $1\, \text{eV} = 1.602 \times 10^{–19}\,\text{J}$. Using the binding energy computed in part (b):
$E=4.55 \times 10^{-12}\, \text{J} \times \frac{1\, \text{eV}}{1.602 \times 10^{-19}\,\text{J}} =2.84 \times 10^7\, \text{eV} =28.4\, \text{MeV} \nonumber$
Exercise $1$
What is the binding energy for the $\ce{_{9}^{19}F}$ nuclide (atomic mass: $18.9984\, \text{amu}$) in MeV per nucleus?
Answer
148.4 MeV
Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit enthalpies on the order of thousands of kJ/mol, which is equivalent to mass differences in the nanogram range (10–9 g). On the other hand, nuclear binding energies are typically on the order of billions of kJ/mol, corresponding to mass differences in the milligram range (10–3 g).
Nuclear Stability
A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure $1$: represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.
The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.
Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table $1$). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as and are called “double magic” and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.
Table $1$: Stable Nuclear Isotopes
Number of Stable Isotopes Proton Number Neutron Number
157 even even
53 even odd
50 odd even
5 odd odd
The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example $2$ that the binding energy for a $\ce{_{2}^{4}He}$ nucleus is 28.4 MeV. The binding energy per nucleon for a $\ce{_{2}^{4}He}$ nucleus is therefore:
In Example $3$, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure $2$.
Example $3$: Calculation of Binding Energy per Nucleon
The iron nuclide atom $\ce{_{26}^{56}Fe}$ lies near the top of the binding energy curve (Figure $2$) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide $\ce{_{26}^{56}Fe}$ (atomic mass of 55.9349 amu)?
Solution
As in Example $2$, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an $\ce{_{26}^{56}Fe}$ atom:
\begin{aligned} \text { Mass defect } & = [(26 \times 1.0073 \, \text{amu})+(30 \times 1.0087 \, \text{amu})+(26 \times 0.00055 \, \text{amu})]-55.9349 amu \[4pt] & =56.4651\, \text{amu} - 55.9349 \, \text{amu}\[4pt] & =0.5302 \, \text{amu} \end{aligned} \nonumber
We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:
\begin{aligned} E &=m c^2=0.5302 \, \text{amu} \times \left( \dfrac{1.6605 \times 10^{-27} \,\text{kg} }{1 \, \text{amu} } \right) \times \left(2.998 \times 10^8\, \text{m} / \text{s} \right)^2 \[4pt] & =7.913 \times 10^{-11} \, \text{kg} \cdot \text{m} /\text{s}^2 \[4pt] & =7.913 \times 10^{-11} \, \text{J} \end{aligned} \nonumber
We then convert the binding energy in joules per nucleus into units of MeV per nuclide:
$7.913 \times 10^{-11}\, \text{J} \times \frac{1 \, \text{MeV}}{1.602 \times 10^{-13} \, \text{J} }=493.9\, \text{MeV} \nonumber$
Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:
Note that this is almost 25% larger than the binding energy per nucleon for $\ce{_{2}^{4}He}$.
$\text { Binding energy per nucleon }=\frac{493.9 \, \text{MeV} }{56}=8.820\, \text{MeV} / \text { nucleon } \nonumber$
(Note also that this is the same process as in Example $1$, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)
Exercise $1$
What is the binding energy per nucleon in $\ce{_{9}^{19}F}$ (atomic mass, 18.9984 amu)?
Answer
$7.810\, \text{MeV}/\text{nucleon}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.02%3A_Nuclear_Structure_and_Stability.txt |
Learning Objectives
By the end of this section, you will be able to:
• Identify common particles and energies involved in nuclear reactions
• Write and balance nuclear equations
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Types of Particles in Nuclear Reactions
Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure $1$. Protons ($\ce{^1_1p}$, also represented by the symbol $\ce{^1_1H}$) and neutrons ($\ce{_0^1n}$) are the constituents of atomic nuclei, and have been described previously. Alpha particles ($\ce{_2^4He}$, also represented by the symbol $\ce{_2^4\alpha}$) are high-energy helium nuclei. Beta particles ($\ce{_{−1}^0β}$, also represented by the symbol $\ce{_1^0e}$) are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons ($\ce{_{+1}^0e}$, also represented by the symbol $\ce{_1^0β}$) are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus ($\ce{He}$) with a charge of +2 and a mass number of 4, so it is symbolized $\ce{_2^4He}$. This works because, in general, the ion charge is not important in the balancing of nuclear equations.
Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays ($\gamma$)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation $E = mc^2$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:
$\ce{_{-1}^0 e +_{+1}^0 e -> \gamma+\gamma} \nonumber$
As seen in the chapter discussing light and electromagnetic radiation, gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays that can behave as particles in the wave-particle duality sense. Gamma rays are a type of high energy electromagnetic radiation produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions.
Balancing Nuclear Reactions
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of nucleons (subatomic particles within the atoms’ nuclei) rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, were one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an $α$ particle with magnesium-25 ($\ce{_{12}^{25}Mg}$) produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{_{12}^{25}Mg +_2^4He ->_1^1H +_{Z}^{A}X} \nonumber$
where $A$ is the mass number and $Z$ is the atomic number of the new nuclide, $X$. Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$25+4= A +1 \nonumber$
or $A =28$.
Similarly, the charges must balance, so:
$12+2= Z +1 \nonumber$
and $Z =13$.
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{_{13}^{28}Al}$
Exercise $1$
The nuclide $\ce{_{53}^{125}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{_{53}^{125}I +_{−1}^{0}e ->_{52}^{125}Te} \nonumber$
Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:
• The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting α particles: $\ce{^{212}_{84}Po ->^{208}_{82}Pb +^{4}_{2}He} \nonumber$
• The first nuclide to be prepared by artificial means was an isotope of oxygen, 17O. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with α particles: $\ce{^{14}_7N +^{4}_{2}He ->^{17}_8O +^{1}_{1}H} \nonumber$
• James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with 12C by the nuclear reaction between 9Be and 4He: $\ce{^{9}_4Be +^{4}_2He ->^{12}_6C +^{1}_0n} \nonumber$
• The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, $\ce{^{2}_1H}$), by Emilio Segre and Carlo Perrier in 1937: $\ce{^{2}_1H +^{97}_{42}Mo -> 2^{1}_0n +^{97}_{43}Tc} \nonumber$
• The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was: $\ce{^{235}_{92}U +^{1}_{0} n ->^{87}_{35}Br +^{146}_{57}La + 3^{1}_{0}n} \nonumber$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.03%3A_Nuclear_Equations.txt |
Learning Objectives
By the end of this section, you will be able to:
• Recognize common modes of radioactive decay
• Identify common particles and energies involved in nuclear decay reactions
• Write and balance nuclear decay equations
• Calculate kinetic parameters for decay processes, including half-life
• Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure $1$).
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure $2$) helped him determine that one type of radiation consisted of positively charged and relatively massive $α$ particles; a second type was made up of negatively charged and much less massive $β$ particles; and a third was uncharged electromagnetic waves, γ rays. We now know that $α$ particles are high-energy helium nuclei, $β$ particles are high-energy electrons, and γ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Alpha ($α$) decay is the emission of an $α$ particle from the nucleus. For example, polonium-210 undergoes $α$ decay:
$\ce{_{84}^{210} Po ->_{2}^{4} He +_{82}^{206} Pb} \nonumber$
or
$\ce{_{84}^{210} Po ->_{2}^{4} \alpha +_{82}^{206} Pb} \nonumber$
Alpha decay occurs primarily in heavy nuclei ($A > 200$, $Z > 83$). Because the loss of an $α$ particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing $α$ decay lies below the band of stability (refer to Figure 21.2), the daughter nuclide will lie closer to the band.
Beta ($β$) decay is the emission of an electron (i.e., a $\beta$ particle) from a nucleus. Iodine-131 is an example of a nuclide that undergoes $β$ decay:
$\ce{_{53}^{131} I ->_{-1}^0 e +_{54}^{131} Xe} \nonumber$
or
$\ce{_{53}^{131} I ->_{-1}^0 \beta +_{54}^{131} Xe} \nonumber$
Beta decay, which can be thought of as the conversion of a neutron into a proton and a $β$ particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission ($γ$ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a $γ$ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits $γ$ radiation and is used in many applications including cancer treatment:
$\ce{_{27}^{60} Co^{*} ->_0^0 \gamma +_{27}^{60} Co} \nonumber$
There is no change in mass number or atomic number during the emission of a $γ$ ray unless the $γ$ emission accompanies one of the other modes of decay.
Positron emission ($β^{+}$ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
$\ce{_{8}^{15} O ->_{+1}^0 e +_7^{15} N} \nonumber$
or
$\ce{_{8}^{15} O ->_{+1}^0 \beta +_7^{15} N} \nonumber$
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
$\ce{_{19}^{40} K +_{-1}^0 e ->_{18}^{40} Ar} \nonumber$
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur.
Figure $3$ summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
Chemistry in Everyday Life: PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure $4$). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
For example, F-18 is produced by proton bombardment of $\ce{^{18}O}$
$\ce{_{8}^{18} O +_{1}^{1} p ->_{9}^{18} F +_{0}^{1} n} \nonumber$
and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The $\ce{^{18}F}$ emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure $5$). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life ($t_{1/2}$), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, coba source, since half of the nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, $N$, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
$\text{decay rate} = λN \nonumber$
with $λ$ = the decay constant for the particular radioisotope.
The decay constant, $λ$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, $t_{1/2}$:
$\lambda=\dfrac{\ln 2}{t_{1 / 2}}=\dfrac{0.693}{t_{1 / 2}} \nonumber$
or
$t_{1 / 2}=\dfrac{\ln 2}{\lambda}=\dfrac{0.693}{\lambda} \nonumber$
The first-order equations relating amount, $N$, and time are:
$N_t=N_0 e^{-\lambda t} \label{firstorder}$
or
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \nonumber$
Example $1$: Rates of Radioactive Decay
$\ce{^{60}_{27}Co}$ decays with a half-life of 5.27 years to produce $\ce{^{60}_{28}Co}$
1. What is the decay constant for the radioactive disintegration of cobalt-60?
2. Calculate the fraction of a sample of the $\ce{^{60}_{27}Co}$ isotope that will remain after 15 years.
3. How long does it take for a sample of $\ce{^{60}_{27}Co}$ to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{5.27 y }=0.132 y^{-1} \nonumber$
(b) The fraction of $\ce{^{60}_{27}Co}$ that is left after time $t$ is given by $\frac{N_t}{N_0}$. Rearranging the first-order relationship in Equation \ref{firstorder} ($N_t = N_0e^{–λt}$ to solve for this ratio yields:
$\frac{N_t}{N_0}=e^{-\lambda t}=e^{-(0.132 / \text{y} )(15 \, \text{y} )}=0.138 \nonumber$
The fraction of $\ce{^{60}_{27}Co}$ that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the original $\ce{^{60}_{27}Co}$ present will remain after 15 years.
(c) 2.00% of the original amount of $\ce{^{60}_{27}Co}$ is equal to 0.0200 N0. Substituting this into the equation for time for first-order kinetics, we have:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right)=-\frac{1}{0.132\, \text{y}^{-1}} \ln \left(\frac{0.0200 \times N_0}{N_0}\right)=29.6 y \nonumber$
Exercise $1$
Radon-222, $\ce{^{222}_{86}Rn}$ has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of $\ce{^{209}_{83}Bi}$ is $1.9 \times 10^{19}$ years; $\ce{^{239}_{94}Ra}$ is 24,000 years; $\ce{^{222}_{86}Rn}$ is 3.82 days; and element-111 (Rg for roentgenium) is $1.5 \times 10^{–3}$ seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table $1$, and others are listed in Appendix M.
Table $1$: Half-lives of Radioactive Isotopes Important to Medicine
Type1 Decay Mode Half-Life Uses
F-18 β+ decay 110. minutes PET scans
Co-60 β decay, γ decay 5.27 years cancer treatment
Tc-99m γ decay 8.01 hours scans of brain, lung, heart, bone
I-131 β decay 8.02 days thyroid scans and treatment
Tl-201 electron capture 73 hours heart and arteries scans; cardiac stress tests
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: $\ce{^{12}_{6}C}$ which constitutes about 99% of the carbon on earth; $\ce{^{13}_6C}$, about 1% of the total; and trace amounts of $\ce{^{14}_6C}$. Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
$\ce{_{7}^{14} N +_{0}^{1}n ->_{6}^{14}C + _{1}^{1} H} \nonumber$
All isotopes of carbon react with oxygen to produce $\ce{CO2}$ molecules. The ratio of $\ce{^{14}_{6}CO2}$ to $\ce{^{12}_{6}CO2}$ depends on the ratio of $\ce{^{14}_{6}CO}$ to $\ce{^{12}_{6}CO}$ in the atmosphere. The natural abundance of $\ce{^{14}_{6}CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of $\ce{^{14}_{6}CO2}$ and $\ce{^{14}_{6}CO2}$ into plants is a regular part of the photosynthesis process, which means that the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio found in a living plant is the same as the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because $\ce{^{12}_{6}C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by $β$ emission with a half-life of 5730 years:
$\ce{_{6}^{14} C ->_{7}^{14} N +_{-1}^0 e} \nonumber$
Thus, the ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure $7$: visually depicts this process.
For example, with the half-life of being 5730 years, if the ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Visit this website to perform simulations of radiometric dating.
Example $2$: Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, $N$, in the relationship:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \longrightarrow t=-\frac{1}{\lambda} \ln \left(\frac{\text { Rate }_t}{\text { Rate }_0}\right) \nonumber$
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{5730 \,\text{y}}=1.21 \times 10^{-4}\, \text{y}^{-1} \nonumber$
Substituting and solving, we have:
$t=-\frac{1}{\lambda} \ln \left(\frac{\text { Rate }_t}{\text { Rate }_0}\right) = -\frac{1}{1.21 \times 10^{-4}\, \text{y}^{-1}} \ln \left(\frac{10.8 \, \cancel{ \text{dis} / \text{min} / \text{g}\, \ce{C}} }{13.6 \, \cancel{ \text{dis} / \text{min} / \text{g}\, \ce{C}} }\right)=1910\, \text{y} \nonumber$
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure $8$).
Exercise $1$
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio. The accuracy of a straightforward application of this technique depends on the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of $\ce{CO2}$ molecules (largely $\ce{^{12}_{6}C}$) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the $\ce{^{14}_{6}C}$ has decayed), the ratio of $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ in the atmosphere may be changing. This manmade increase in $\ce{^{12}_{6}C}$ in the atmosphere causes the $\ce{^{14}_{6}C} : \ce{^{12}_{6}C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example $3$: Radioactive Dating of Rocks
An igneous rock contains $9.58 \times 10^{–5} \, \text{g}$ of U-238 and $2.51 \times 10^{–5} \, \text{g}$ of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
$9.58 \times 10^{-5}\, \cancel{\text{g}\, \ce{U}} \times\left(\frac{1\, \text{mol}\, \ce{U} }{238\, \cancel{\text{g}\, \ce{U}}}\right)=4.03 \times 10^{-7} \, \text{mol}\, \ce{U} \nonumber$
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
$2.51 \times 10^{-5}\, \cancel{ \text{g}\, \ce{Pb}} \times\left(\frac{1\, \bcancel{\text{mol}\, \ce{Pb}} }{206\, \cancel{ \text{g}\, \ce{Pb}} }\right) \times\left(\frac{1\, \text{mol}\, \ce{U}}{1\, \bcancel{\text{mol}\, \ce{Pb}} }\right)=1.22 \times 10^{-7}\, \text{mol}\, \ce{U} \nonumber$
The total amount of U-238 originally present in the rock is therefore:
$4.03 \times 10^{-7}\, \text{mol}\, \ce{U} + 1.22 \times 10^{-7}\, \text{mol}\, \ce{U} = 5.25 \times 10^{-7}\, \text{mol}\, \ce{U} \nonumber$
The amount of time that has passed since the formation of the rock is given by:
$t=-\frac{1}{\lambda} \ln \left(\frac{N_t}{N_0}\right) \nonumber$
with $N_0$ representing the original amount of U-238 and $N_t$ representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of $4.5 \times 10^9\, \text{y}$, so the decay constant $λ$ is:
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{0.693}{4.5 \times 10^9\, \text{y} }=1.54 \times 10^{-10} \, \text{y}^{-1} \nonumber$
Substituting and solving, we have:
$t=-\frac{1}{1.54 \times 10^{-10}\, \text{y}^{-1}} \ln \left(\frac{4.03 \times 10^{-7} \, \cancel{\text{mol}\, \ce{U}}}{5.25 \times 10^{-7} \, \cancel{\text{mol}\, \ce{U}}}\right)=1.7 \times 10^9 \, \text{y} \nonumber$
Therefore, the rock is approximately 1.7 billion years old.
Exercise $1$
A sample of rock contains $6.14 \times 10^{–4}$ g of Rb-87 and $3.51 \times 10^{–5} g$ of Sr-87. Calculate the age of the rock. (The half-life of the $β$ decay of Rb-87 is $4.7 \times 10^{10}\, \text{y}$.)
Answer
3.7 109 y
Footnotes
• 1The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit γ radiation to rid themselves of excess energy and become (more) stable. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.04%3A_Radioactive_Decay.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the synthesis of transuranium nuclides
• Explain nuclear fission and fusion processes
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission and fusion reactors
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Scientists learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace.
Synthesis of Nuclides
Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:
The and nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news.
Chemistry in Everyday Life: CERN Particle Accelerator
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure \(1\)). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2013 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Famous physicist Brian Cox talks about his work on the Large Hadron Collider at CERN, providing an entertaining and engaging tour of this massive project and the physics behind it.
View a short video from CERN, describing the basics of how its particle accelerators work.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
Plutonium is now mostly formed in nuclear reactors as a byproduct during the fission of U-235. Additional neutrons are released during this fission process (see the next section), some of which combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. These processes are summarized in the equation:
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table 21.3.
Table 21.3: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95
curium Cm 96
californium Cf 98
einsteinium Es 99
mendelevium Md 101
nobelium No 102
rutherfordium Rf 104
seaborgium Sg 106
meitnerium Mt 107
Nuclear Fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56 (see Figure 21.3). Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(2\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(3\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
View this link to see a simulation of nuclear fission.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (see Figure \(4\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(5\)).
An atomic bomb (Figure \(6\)) contains several pounds of fissionable material, or a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(7\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (refer back to Figure \(5\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Chemistry in Everyday Life: Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(9\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(10\)).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed.
Explore the information in this link to learn about the approaches to nuclear waste management.
Nuclear Fusion and Fusion Reactors
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 1011 kJ of energy per mole of produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, and a triton, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 109 kilojoules per mole of formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure \(11\)). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.05%3A_Transmutation_and_Nuclear_Energy.txt |
Learning Objectives
By the end of this section, you will be able to:
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioimmunossays (RIA), for example, rely on radioisotopes to detect the presence and/or concentration of certain antigens. Developed by Rosalyn Sussman Yalow and Solomon Berson in the 1950s, the technique is known for extreme sensitivity, meaning that it can detect and measure very small quantities of a substance. Prior to its discovery, most similar detection relied on large enough quantities to produce visible outcomes. RIA revolutionized and expanded entire fields of study, most notably endocrinology, and is commonly used in narcotics detection, blood bank screening, early cancer screening, hormone measurement, and allergy diagnosis. Based on her significant contribution to medicine, Yalow received a Nobel Prize, making her the second woman to be awarded the prize for medicine.
Radioisotopes have revolutionized medical practice (see Appendix M), where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 4399Tc)(4399Tc), thallium-201 81201Tl)(81201Tl), iodine-131 53131I)(53131I), and sodium-24 1124Na)(1124Na) . Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure \(1\)) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure \(2\)). The parent nuclide Mo-99 is part of a molybdate ion, MoO42;MoO42; when it decays, it forms the pertechnetate ion, TcO4.TcO4. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure \(3\)). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Coba , which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
n 2760 Co 2860 Ni+ −10 β+2 00 γ2759 Co+ 01 n 2760 Co 2860 Ni+ −10 β+2 00 γ
The overall decay scheme for this is shown graphically in Figure \(4\).
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
6CO2(g)+6H2O(l)C6H12O6(s)+6O2(g),6CO2(g)+6H2O(l)C6H12O6(s)+6O2(g),
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of 614C614C. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure \(5\)). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure \(6\)). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.06%3A_Uses_of_Radioisotopes.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the biological impact of ionizing radiation
• Define units for measuring radiation exposure
• Explain the operation of common tools for detecting radioactivity
• List common sources of radiation exposure in the US
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure \(1\)).
Ionizing and Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure \(2\)).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Because the hydroxyl radical has an unpaired electron, it is highly reactive. (This is true of any substance with unpaired electrons, known as a free radical.) This hydroxyl radical can react with all kinds of biological molecules (DNA, proteins, enzymes, and so on), causing damage to the molecules and disrupting physiological processes. Examples of direct and indirect damage are shown in Figure 21.32.
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure \(3\)). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
Chemistry in Everyday Life: Radon Exposure
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238 (Figure 21.9), which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure \(4\)).
Radon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure \(5\)). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Figure \(6\)). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = 3.7 1010 disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose. The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (100 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy) along with a biological factor referred to as the RBE (for relative biological effectiveness) that is an approximate measure of the relative damage done by the radiation. These are related by:
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Units of Radiation Measurement
Table 21.4 summarizes the units used for measuring radiation.
Table 21.4: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes 3.7 1010 decays/second
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE Gy
roentgen equivalent for man (rem) Rem = RBE rad
Example 21.8: Amount of Radiation
Coba is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
And to convert this to decays per second:
(a) Since 1 Bq = the activity in Becquerel (Bq) is:
(b) Since 1 Ci = the activity in curie (Ci) is:
Exercise \(1\)
Tritium is a radioactive isotope of hydrogen (t1/2 = 12.32 y) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer
(a) 3.56 1011 Bq; (b) 0.962 Ci
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure \(7\): , the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table 21.5.
Table 21.5: Health Effects of Radiation2
Exposure (rem) Health Effect Time to Onset (without treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness; minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Footnotes
• 2Source: US Environmental Protection Agency | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.07%3A_Biological_Effects_of_Radiation.txt |
Example and Directions
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(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
alpha (α) decayloss of an alpha particle during radioactive decay
alpha particle(α or or high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons
antimatterparticles with the same mass but opposite properties (such as charge) of ordinary particles
band of stability(also, belt of stability, zone of stability, or valley of stability) region of graph of number of protons versus number of neutrons containing stable (nonradioactive) nuclides
becquerel (Bq)SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s
beta (β) decaybreakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle
beta particle or or high-energy electron
binding energy per nucleontotal binding energy for the nucleus divided by the number of nucleons in the nucleus
chain reactionrepeated fission caused when the neutrons released in fission bombard other atoms
chemotherapysimilar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
containment system(also, shield) a three-part structure of materials that protects the exterior of a nuclear fission reactor and operating personnel from the high temperatures, pressures, and radiation levels inside the reactor
control rodmaterial inserted into the fuel assembly that absorbs neutrons and can be raised or lowered to adjust the rate of a fission reaction
critical massamount of fissionable material that will support a self-sustaining (nuclear fission) chain reaction
curie (Ci)larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 1010 disintegrations/s
daughter nuclidenuclide produced by the radioactive decay of another nuclide; may be stable or may decay further
electron capturecombination of a core electron with a proton to yield a neutron within the nucleus
electron volt (eV)measurement unit of nuclear binding energies, with 1 eV equaling the amount energy due to the moving an electron across an electric potential difference of 1 volt
external beam radiation therapyradiation delivered by a machine outside the body
fissile (or fissionable)when a material is capable of sustaining a nuclear fission reaction
fissionsplitting of a heavier nucleus into two or more lighter nuclei, usually accompanied by the conversion of mass into large amounts of energy
fusioncombination of very light nuclei into heavier nuclei, accompanied by the conversion of mass into large amounts of energy
fusion reactornuclear reactor in which fusion reactions of light nuclei are controlled
gamma (γ) emissiondecay of an excited-state nuclide accompanied by emission of a gamma ray
gamma ray(γ or short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality
Geiger counterinstrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube
gray (Gy)SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue
half-life (t1/2)time required for half of the atoms in a radioactive sample to decay
internal radiation therapy(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
ionizing radiationradiation that can cause a molecule to lose an electron and form an ion
magic numbernuclei with specific numbers of nucleons that are within the band of stability
mass defectdifference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass “lost” when nucleons are brought together to form a nucleus)
mass-energy equivalence equationAlbert Einstein’s relationship showing that mass and energy are equivalent
millicurie (mCi)larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 1010 disintegrations/s
nonionizing radiationradiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules
nuclear binding energyenergy lost when an atom’s nucleons are bound together (or the energy needed to break a nucleus into its constituent protons and neutrons)
nuclear chemistrystudy of the structure of atomic nuclei and processes that change nuclear structure
nuclear fuelfissionable isotope present in sufficient quantities to provide a self-sustaining chain reaction in a nuclear reactor
nuclear moderatorsubstance that slows neutrons to a speed low enough to cause fission
nuclear reactionchange to a nucleus resulting in changes in the atomic number, mass number, or energy state
nuclear reactorenvironment that produces energy via nuclear fission in which the chain reaction is controlled and sustained without explosion
nuclear transmutationconversion of one nuclide into another nuclide
nucleoncollective term for protons and neutrons in a nucleus
nuclidenucleus of a particular isotope
parent nuclideunstable nuclide that changes spontaneously into another (daughter) nuclide
particle acceleratordevice that uses electric and magnetic fields to increase the kinetic energy of nuclei used in transmutation reactions
positron or antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge
positron emission(also, β+ decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted
radiation absorbed dose (rad)SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy
radiation dosimeterdevice that measures ionizing radiation and is used to determine personal radiation exposure
radiation therapyuse of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive decayspontaneous decay of an unstable nuclide into another nuclide
radioactive decay serieschains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product
radioactive tracer(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions
radioactivityphenomenon exhibited by an unstable nucleon that spontaneously undergoes change into a nucleon that is more stable; an unstable nucleon is said to be radioactive
radiocarbon datinghighly accurate means of dating objects 30,000–50,000 years old that were derived from once-living matter; achieved by calculating the ratio of in the object vs. the ratio of in the present-day atmosphere
radioisotopeisotope that is unstable and undergoes conversion into a different, more stable isotope
radiometric datinguse of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations
reactor coolantassembly used to carry the heat produced by fission in a reactor to an external boiler and turbine where it is transformed into electricity
relative biological effectiveness (RBE)measure of the relative damage done by radiation
roentgen equivalent man (rem)unit for radiation damage, frequently used in medicine; 100 rem = 1 Sv
scintillation counterinstrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation
sievert (Sv)SI unit measuring tissue damage caused by radiation; takes into account energy and biological effects of radiation
strong nuclear forceforce of attraction between nucleons that holds a nucleus together
subcritical massamount of fissionable material that cannot sustain a chain reaction; less than a critical mass
supercritical massamount of material in which there is an increasing rate of fission
transmutation reactionbombardment of one type of nuclei with other nuclei or neutrons
transuranium elementelement with an atomic number greater than 92; these elements do not occur in nature | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.08%3A_Key_Terms.txt |
E = mc2
decay rate = λN
rem = RBE rad
Sv = RBE Gy
20.10: Summary
An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2. Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56; these are the most stable nuclei.
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.
Nuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some substances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more.
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way.
Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production.
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, and including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source, and limiting time of exposure. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.09%3A_Key_Equations.txt |
1.
Write the following isotopes in hyphenated form (e.g., “carbon-14”)
2.
Write the following isotopes in nuclide notation (e.g.,
1. oxygen-14
2. copper-70
3. tantalum-175
4. francium-217
3.
For the following isotopes that have missing information, fill in the missing information to complete the notation
4.
For each of the isotopes in Exercise 21.1, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.
5.
Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:
1. 25 protons, 20 neutrons, 24 electrons
2. 45 protons, 24 neutrons, 43 electrons
3. 53 protons, 89 neutrons, 54 electrons
4. 97 protons, 146 neutrons, 97 electrons
6.
Calculate the density of the nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 10–13 cm and is spherical in shape.
7.
What are the two principal differences between nuclear reactions and ordinary chemical changes?
8.
The mass of the atom is 22.9898 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
9.
Which of the following nuclei lie within the band of stability shown in Figure 21.2?
1. chlorine-37
2. calcium-40
3. 204Bi
4. 56Fe
5. 206Pb
6. 211Pb
7. 222Rn
8. carbon-14
10.
Which of the following nuclei lie within the band of stability shown in Figure 21.2?
1. argon-40
2. oxygen-16
3. 122Ba
4. 58Ni
5. 205Tl
6. 210Tl
7. 226Ra
8. magnesium-24
11.
Write a brief description or definition of each of the following:
1. nucleon
2. α particle
3. β particle
4. positron
5. γ ray
6. nuclide
7. mass number
8. atomic number
12.
Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei?
13.
Complete each of the following equations by adding the missing species:
14.
Complete each of the following equations:
15.
Write a balanced equation for each of the following nuclear reactions:
1. the production of 17O from 14N by α particle bombardment
2. the production of 14C from 14N by neutron bombardment
3. the production of 233Th from 232Th by neutron bombardment
4. the production of 239U from 238U by bombardment
16.
Technetium-99 is prepared from 98Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as 99Tc*. This excited nucleus relaxes to the ground state, represented as 99Tc, by emitting a γ ray. The ground state of 99Tc then emits a β particle. Write the equations for each of these nuclear reactions.
17.
The mass of the atom is 18.99840 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
18.
For the reaction if 100.0 g of carbon reacts, what volume of nitrogen gas (N2) is produced at 273K and 1 atm?
19.
What are the types of radiation emitted by the nuclei of radioactive elements?
20.
What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?
1. an α particle is emitted
2. a β particle is emitted
3. γ radiation is emitted
4. a positron is emitted
5. an electron is captured
21.
What is the change in the nucleus that results from the following decay scenarios?
1. emission of a β particle
2. emission of a β+ particle
3. capture of an electron
22.
Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles.
23.
Why is electron capture accompanied by the emission of an X-ray?
24.
Explain, in terms of Figure 21.2, how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.
25.
Which of the following nuclei is most likely to decay by positron emission? Explain your choice.
1. chromium-53
2. manganese-51
3. iron-59
26.
The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.
27.
The following nuclei do not lie in the band of stability. How would they be expected to decay?
28.
Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:
1. (f) 129Ba
(g) 237Pu
29.
Write a nuclear reaction for each step in the formation of from which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α particles, in that order.
30.
Write a nuclear reaction for each step in the formation of from which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.
31.
Define the term half-life and illustrate it with an example.
32.
A 1.00 10–6-g sample of nobelium, has a half-life of 55 seconds after it is formed. What is the percentage of remaining at the following times?
1. 5.0 min after it forms
2. 1.0 h after it forms
33.
239Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y?
34.
The isotope 208Tl undergoes β decay with a half-life of 3.1 min.
1. What isotope is produced by the decay?
2. How long will it take for 99.0% of a sample of pure 208Tl to decay?
3. What percentage of a sample of pure 208Tl remains un-decayed after 1.0 h?
35.
If 1.000 g of produces 0.0001 mL of the gas at STP (standard temperature and pressure) in 24 h, what is the half-life of 226Ra in years?
36.
The isotope is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.
37.
Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of
38.
What is the age of mummified primate skin that contains 8.25% of the original quantity of 14C?
39.
A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87.
1. Calculate the age of the rock if the half-life of the decay of rubidium by β emission is 4.7 1010 y.
2. If some was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.
40.
A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of and 2.52 mg of Calculate the age of the ore. The half-life of is 4.5 109 yr.
41.
Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 109 years ago?
42.
A atom (mass = 7.0169 amu) decays into a atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction?
43.
A atom (mass = 8.0246 amu) decays into a atom (mass = 8.0053 amu) by loss of a β+ particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?
44.
Isotopes such as 26Al (half-life: 7.2 105 years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.
1. 26Al decays by β+ emission or electron capture. Write the equations for these two nuclear transformations.
2. The earth was formed about 4.7 109 (4.7 billion) years ago. How old was the earth when 99.999999% of the 26Al originally present had decayed?
45.
Write a balanced equation for each of the following nuclear reactions:
1. bismuth-212 decays into polonium-212
2. beryllium-8 and a positron are produced by the decay of an unstable nucleus
3. neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239
4. strontium-90 decays into yttrium-90
46.
Write a balanced equation for each of the following nuclear reactions:
1. mercury-180 decays into platinum-176
2. zirconium-90 and an electron are produced by the decay of an unstable nucleus
3. thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium-228 by beta decay
4. neon-19 decays into fluorine-19
47.
Write the balanced nuclear equation for the production of the following transuranium elements:
1. berkelium-244, made by the reaction of Am-241 and He-4
2. fermium-254, made by the reaction of Pu-239 with a large number of neutrons
3. lawrencium-257, made by the reaction of Cf-250 and B-11
4. dubnium-260, made by the reaction of Cf-249 and N-15
48.
How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic?
49.
Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?
50.
Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.
51.
Describe the components of a nuclear reactor.
52.
In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.
53.
Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.
54.
The mass of a hydrogen atom is 1.007825 amu; that of a tritium atom is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of produced is released by the following fusion reaction:
55.
How can a radioactive nuclide be used to show that the equilibrium:
is a dynamic equilibrium?
56.
Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
57.
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β emission.
1. Write an equation for the decay.
2. How long will it take for 95.0% of a dose of I-131 to decay?
58.
If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:
1. cobalt-60 (a strong γ emitter used for irradiation)
2. molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)
59.
Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous?
60.
Given specimens uranium-232 (t1/2 = 68.9 y) and uranium-233 (t1/2 = 159,200 y) of equal mass, which one would have greater activity and why?
61.
A scientist is studying a 2.234 g sample of thorium-229 (t1/2 = 7340 y) in a laboratory.
1. What is its activity in Bq?
2. What is its activity in Ci?
62.
Given specimens neon-24 (t1/2 = 3.38 min) and bismuth-211 (t1/2 = 2.14 min) of equal mass, which one would have greater activity and why? | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/20%3A_Nuclear_Chemistry/20.11%3A_Exercises.txt |
Organic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms. Study of structure includes many physical and chemical methods to determine the chemical composition and the chemical constitution of organic compounds and materials. Study of properties includes both physical properties and chemical properties, and uses similar methods as well as methods to evaluate chemical reactivity, with the aim to understand the behavior of the organic matter.
• 21.1: Introduction
Early chemists regarded substances isolated from organisms as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. Thr defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
• 21.2: Hydrocarbons
Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring of delocalized π electrons.
• 21.3: Alcohols and Ethers
Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
• 21.4: Aldehydes, Ketones, Carboxylic Acids, and Esters
The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom.
• 21.5: Amines and Amides
The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides.
• 21.6: Key Terms
• 21.7: Summary
• 21.8: Exercises
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. 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.
21: Organic Chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and CO2. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.01%3A_Introduction.txt |
Learning Objectives
By the end of this section, you will be able to:
• Explain the importance of hydrocarbons and the reason for their diversity
• Name saturated and unsaturated hydrocarbons, and molecules derived from them
• Describe the reactions characteristic of saturated and unsaturated hydrocarbons
• Identify structural and geometric isomers of hydrocarbons
The largest database1 of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated2 at 1060—an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities.
The simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals.
Alkanes
Alkanes, or saturated hydrocarbons, contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has sp3 hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure \(1\). Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the sp3 hybridization, the bond angles in carbon chains are close to 109.5°, giving such chains in an alkane a zigzag shape.
The structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, condensed formulas). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure \(1\): , and several additional examples are provided in the exercises at the end of this chapter.
A common method used by organic chemists to simplify the drawings of larger molecules is to use a skeletal structure (also called a line-angle structure). In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. Figure \(2\): shows three different ways to draw the same structure.
Example \(1\): Drawing Skeletal Structures
Draw the skeletal structures for these two molecules:
Solution
Each carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):
Exercise \(1\)
Draw the skeletal structures for these two molecules:
Answer
Example \(2\): Interpreting Skeletal Structures
Identify the chemical formula of the molecule represented here:
Solution
There are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of C8H16.
Location of the hydrogen atoms:
Exercise \(1\)
Identify the chemical formula of the molecule represented here:
Answer
C9H20
All alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of CnH2n+2. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table 20.1) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change.
Table 20.1: Properties of Some Alkanes3
Alkane Molecular Formula Melting Point (°C) Boiling Point (°C) Phase at STP4 Number of Structural Isomers
methane CH4 –182.5 –161.5 gas 1
ethane C2H6 –183.3 –88.6 gas 1
propane C3H8 –187.7 –42.1 gas 1
butane C4H10 –138.3 –0.5 gas 2
pentane C5H12 –129.7 36.1 liquid 3
hexane C6H14 –95.3 68.7 liquid 5
heptane C7H16 –90.6 98.4 liquid 9
octane C8H18 –56.8 125.7 liquid 18
nonane C9H20 –53.6 150.8 liquid 35
decane C10H22 –29.7 174.0 liquid 75
tetradecane C14H30 5.9 253.5 solid 1858
octadecane C18H38 28.2 316.1 solid 60,523
Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula C4H10: They are called n-butane and 2-methylpropane (or isobutane), and have the following Lewis structures:
The compounds n-butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The n-butane molecule contains an unbranched chain, meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal, or the prefix n, to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms)
Identifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in Figure \(3\): all represent the same molecule, n-butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.
The Basics of Organic Nomenclature: Naming 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). These prefixes can be seen in the names of the alkanes described in Table 20.1.
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 \(3\): 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.
Exercise \(1\)
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 \(4\): 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.
Exercise \(1\)
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. (It may be easier to see the equivalency in the ball and stick models in Figure \(1\). Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent (Figure \(1\)) 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 \(4\).
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:
Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:
Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (see Figure \(5\)). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.
In a substitution reaction, another typical reaction of alkanes, one or more of the alkane’s hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and the hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:
The C–Cl portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter.
Want more practice naming alkanes? Watch this brief video tutorial to review the nomenclature process.
Alkenes
Organic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one σ bond and one π bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.
Ethene, C2H4, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure \(6\)); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.
Ethylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.
Chemistry in Everyday Life: Recycling Plastics
Polymers (from Greek words poly meaning “many” and mer meaning “parts”) are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter).
An example of a polymerization reaction is shown in Figure \(7\). The monomer ethylene (C2H4) is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of –CH2– units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).
Polyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials.
Plastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (see Figure \(8\)). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today.
The name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:
Isomers of Alkenes
Molecules of 1-butene and 2-butene are structural isomers; the arrangement of the atoms in these two molecules differs. As an example of arrangement differences, the first carbon atom in 1-butene is bonded to two hydrogen atoms; the first carbon atom in 2-butene is bonded to three hydrogen atoms.
The compound 2-butene and some other alkenes also form a second type of isomer called a geometric isomer. In a set of geometric isomers, the same types of atoms are attached to each other in the same order, but the geometries of the two molecules differ. Geometric isomers of alkenes differ in the orientation of the groups on either side of a bond.
Carbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid. This makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with 120° bond angles around the sp2-hybridized carbon atoms participating in the double bond, the isomers are apparent. The 2-butene isomer in which the two methyl groups are on the same side is called a cis-isomer; the one in which the two methyl groups are on opposite sides is called a trans-isomer (Figure \(9\)). The different geometries produce different physical properties, such as boiling point, that may make separation of the isomers possible:
Alkenes are much more reactive than alkanes because the moiety is a reactive functional group. A π bond, being a weaker bond, is disrupted much more easily than a σ bond. Thus, alkenes undergo a characteristic reaction in which the π bond is broken and replaced by two σ bonds. This reaction is called an addition reaction. The hybridization of the carbon atoms in the double bond in an alkene changes from sp2 to sp3 during an addition reaction. For example, halogens add to the double bond in an alkene instead of replacing hydrogen, as occurs in an alkane:
Example \(5\): Alkene Reactivity and Naming
Provide the IUPAC names for the reactant and product of the halogenation reaction shown here:
Solution
The reactant is a five-carbon chain that contains a carbon-carbon double bond, so the base name will be pentene. We begin counting at the end of the chain closest to the double bond—in this case, from the left—the double bond spans carbons 2 and 3, so the name becomes 2-pentene. Since there are two carbon-containing groups attached to the two carbon atoms in the double bond—and they are on the same side of the double bond—this molecule is the cis-isomer, making the name of the starting alkene cis-2-pentene. The product of the halogenation reaction will have two chlorine atoms attached to the carbon atoms that were a part of the carbon-carbon double bond:
This molecule is now a substituted alkane and will be named as such. The base of the name will be pentane. We will count from the end that numbers the carbon atoms where the chlorine atoms are attached as 2 and 3, making the name of the product 2,3-dichloropentane.
Exercise \(1\)
Provide names for the reactant and product of the reaction shown:
Answer
reactant: cis-3-hexene product: 3,4-dichlorohexane
Alkynes
Hydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The sp-hybridized carbons involved in the triple bond have bond angles of 180°, giving these types of bonds a linear, rod-like shape.
The simplest member of the alkyne series is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:
The IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, is called 1-butyne.
Example \(6\): Structure of Alkynes
Describe the geometry and hybridization of the carbon atoms in the following molecule:
Solution
Carbon atoms 1 and 4 have four single bonds and are thus tetrahedral with sp3 hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as sp hybrids.
Exercise \(1\)
Identify the hybridization and bond angles at the carbon atoms in the molecule shown:
Answer
carbon 1: sp, 180°; carbon 2: sp, 180°; carbon 3: sp2, 120°; carbon 4: sp2, 120°; carbon 5: sp3, 109.5°
Chemically, the alkynes are similar to the alkenes. Since the functional group has two π bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:
Acetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene.
Aromatic Hydrocarbons
Benzene, C6H6, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C6H6, are:
Valence bond theory describes the benzene molecule and other planar aromatic hydrocarbon molecules as hexagonal rings of sp2-hybridized carbon atoms with the unhybridized p orbital of each carbon atom perpendicular to the plane of the ring. Three valence electrons in the sp2 hybrid orbitals of each carbon atom and the valence electron of each hydrogen atom form the framework of σ bonds in the benzene molecule. The fourth valence electron of each carbon atom is shared with an adjacent carbon atom in their unhybridized p orbitals to yield the π bonds. Benzene does not, however, exhibit the characteristics typical of an alkene. Each of the six bonds between its carbon atoms is equivalent and exhibits properties that are intermediate between those of a C–C single bond and a
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.
Example \(7\): Structure of Aromatic Hydrocarbons
One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:
Solution
Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:
Exercise \(1\)
Draw three isomers of a six-membered aromatic ring compound substituted with two bromines.
Answer
Footnotes
• 1This is the Beilstein database, now available through the Reaxys site (www.elsevier.com/online-tools/reaxys).
• 2Peplow, Mark. “Organic Synthesis: The Robo-Chemist,” Nature 512 (2014): 20–2.
• 3Physical properties for C4H10 and heavier molecules are those of the normal isomer, n-butane, n-pentane, etc.
• 4STP indicates a temperature of 0 °C and a pressure of 1 atm. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.02%3A_Hydrocarbons.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of alcohols
• Describe the structure and properties of ethers
• Name and draw structures for alcohols and ethers
In this section, we will learn about alcohols and ethers.
Alcohols
Incorporation of an oxygen atom into carbon- and hydrogen-containing molecules leads to new functional groups and new families of compounds. When the oxygen atom is attached by single bonds, the molecule is either an alcohol or ether.
Alcohols are derivatives of hydrocarbons in which an –OH group has replaced a hydrogen atom. Although all alcohols have one or more hydroxyl (–OH) functional groups, they do not behave like bases such as NaOH and KOH. NaOH and KOH are ionic compounds that contain OH ions. Alcohols are covalent molecules; the –OH group in an alcohol molecule is attached to a carbon atom by a covalent bond.
Ethanol, CH3CH2OH, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars:
Large quantities of ethanol are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:
Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):
Naming 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.5
Example \(1\): Naming Alcohols
Consider the following example. How should it be named?
Solution
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.
Exercise \(1\)
Name the following molecule:
Answer
2-methyl-2-pentanol
Ethers
Ethers are compounds that contain the functional group –O–. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named “methoxy.” The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by “ether.” The common name for the compound shown in Example \(2\) is ethylmethyl ether:
Example \(2\): Naming Ethers
Provide the IUPAC and common name for the ether shown here:
Solution
IUPAC: The molecule is made up of an ethoxy group attached to an ethane chain, so the IUPAC name would be ethoxyethane.
Common: The groups attached to the oxygen atom are both ethyl groups, so the common name would be diethyl ether.
Exercise \(1\)
Provide the IUPAC and common name for the ether shown:
Answer
IUPAC: 2-methoxypropane; common: isopropylmethyl ether
Ethers can be obtained from alcohols by the elimination of a molecule of water from two molecules of the alcohol. For example, when ethanol is treated with a limited amount of sulfuric acid and heated to 140 °C, diethyl ether and water are formed:
In the general formula for ethers, R—O—R, the hydrocarbon groups (R) may be the same or different. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Tertiary-butyl methyl ether, C4H9OCH3 (abbreviated MTBE—italicized portions of names are not counted when ranking the groups alphabetically—so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline.
Want more practice naming ethers? This brief video review summarizes the nomenclature for ethers.
Chemistry in Everyday Life: Carbohydrates and Diabetes
Carbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name “carbohydrate” comes from the formula of the molecules, which can be described by the general formula Cm(H2O)n, which shows that they are in a sense “carbon and water” or “hydrates of carbon.” In many cases, m and n have the same value, but they can be different. The smaller carbohydrates are generally referred to as “sugars,” the biochemical term for this group of molecules is “saccharide” from the Greek word for sugar (Figure 20.12). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars—polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix -ose at the end of the name (for instance, fruit sugar is a monosaccharide called “fructose” and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether.
Organisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of RNA and DNA, respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles.
Diabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure \(1\)). Diabetes may be caused by insufficient insulin production by the pancreas or by the body’s cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy. The long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure.
In 2013, it was estimated that approximately 3.3% of the world’s population (~380 million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin.
Even after treatment protocols were introduced, the need to continually monitor their glucose levels posed a challenge for people with diabetes. The first tests required a doctor or lab, and therefore limited access and frequency. Eventually, researchers developed small tablets that would react to the presence of glucose in urine, but these still required a relatively complex process. Chemist Helen Free, who was working on improvements to the tablets, conceived a simpler device: a small test strip. With her husband and research partner, Alfred Free, she produced the first such product for measuring glucose; soon after, she expanded the technology to provide test strips for other compounds and conditions. While very recent advances (such as breath tests, discussed earlier in the text) have shown promise in replacing test strips, they have been widely used for decades and remain a primary method today.
Footnotes
• 5The IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an “infix” rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.03%3A_Alcohols_and_Ethers.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters
Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section.
Aldehydes and Ketones
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:
In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:
As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.
In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.
Like the
The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:
Example \(1\): Oxidation and Reduction in Organic Chemistry
Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide:
What are the oxidation numbers for the carbon atoms in the molecules shown here?
Solution
In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described). For CH4, the carbon atom carries a –4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of –4). For the alcohol (in this case, methanol), the carbon atom has an oxidation number of –2 (the oxygen atom is assigned –2, the four hydrogen atoms each are assigned +1, and the carbon atom balances the sum by having an oxidation number of –2; note that compared to the carbon atom in CH4, this carbon atom has lost two electrons so it was oxidized); for the aldehyde, the carbon atom’s oxidation number is 0 (–2 for the oxygen atom and +1 for each hydrogen atom already balances to 0, so the oxidation number for the carbon atom is 0); for the carboxylic acid, the carbon atom’s oxidation number is +2 (two oxygen atoms each at –2 and two hydrogen atoms at +1); and for carbon dioxide, the carbon atom’s oxidation number is +4 (here, the carbon atom needs to balance the –4 sum from the two oxygen atoms).
Exercise \(1\)
Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol:
There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms:
Answer
(a) reduced (bond to oxygen atom replaced by bond to hydrogen atom); (b) oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); (c) oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom)
Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms:
An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation.
Formaldehyde, an aldehyde with the formula HCHO, is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.
Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.
Carboxylic Acids and Esters
The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (see Figure \(2\)).
Both carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples):
The functional groups for an acid and for an ester are shown in red in these formulas.
The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt:
Carboxylic acids are weak acids (see the chapter on acids and bases), meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.
We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:
Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH3CO2CH2CH3, is formed when acetic acid reacts with ethanol:
The simplest carboxylic acid is formic acid, HCO2H, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.
Acetic acid, CH3CO2H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon.
The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure \(3\)). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14CO2H, stearic acid, CH3(CH2)16CO2H, and oleic acid, Oleic acid is an unsaturated acid; it contains a double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.04%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure and properties of an amine
• Describe the structure and properties of an amide
Amines are molecules that contain carbon-nitrogen bonds. The nitrogen atom in an amine has a lone pair of electrons and three bonds to other atoms, either carbon or hydrogen. Various nomenclatures are used to derive names for amines, but all involve the class-identifying suffix –ine as illustrated here for a few simple examples:
In some amines, the nitrogen atom replaces a carbon atom in an aromatic hydrocarbon. Pyridine (Figure \(1\)) is one such heterocyclic amine. A heterocyclic compound contains atoms of two or more different elements in its ring structure.
How Sciences Interconnect: DNA in Forensics and Paternity
The genetic material for all living things is a polymer of four different molecules, which are themselves a combination of three subunits. The genetic information, the code for developing an organism, is contained in the specific sequence of the four molecules, similar to the way the letters of the alphabet can be sequenced to form words that convey information. The information in a DNA sequence is used to form two other types of polymers, one of which are proteins. The proteins interact to form a specific type of organism with individual characteristics.
A genetic molecule is called DNA, which stands for deoxyribonucleic acid. The four molecules that make up DNA are called nucleotides. Each nucleotide consists of a single- or double-ringed molecule containing nitrogen, carbon, oxygen, and hydrogen called a nitrogenous base. Each base is bonded to a five-carbon sugar called deoxyribose. The sugar is in turn bonded to a phosphate group PO43)Figure \(2\)).
It probably makes sense that the sequence of nucleotides in the DNA of a cat differs from those of a dog. But it is also true that the sequences of the DNA in the cells of two individual pugs differ. Likewise, the sequences of DNA in you and a sibling differ (unless your sibling is an identical twin), as do those between you and an unrelated individual. However, the DNA sequences of two related individuals are more similar than the sequences of two unrelated individuals, and these similarities in sequence can be observed in various ways. This is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals.
Using similarities in sequences, technicians can determine whether a man is the father of a child (the identity of the mother is rarely in doubt, except in the case of an adopted child and a potential birth mother). Likewise, forensic geneticists can determine whether a crime scene sample of human tissue, such as blood or skin cells, contains DNA that matches exactly the DNA of a suspect.
Watch this video animation of how DNA is packaged for a visual lesson in its structure.
Like ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms:
The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA.
How Sciences Interconnect: Addictive Alkaloids
Since ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H3O+ in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant:
R3N+H3O++Cl[R3NH+]Cl+H2OR3N+H3O++Cl[R3NH+]Cl+H2O
The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base:
R3NH+]Cl+OHR3N+H2O+Cl[R3NH+]Cl+OHR3N+H2O+Cl
The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:
In these diagrams, as is common in representing structures of large organic compounds, carbon atoms in the rings and the hydrogen atoms bonded to them have been omitted for clarity. The solid wedges indicate bonds that extend out of the page. The dashed wedges indicate bonds that extend into the page. Notice that small changes to a part of the molecule change the properties of morphine, codeine, and heroin. Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure \(3\)).
Amides are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group. Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific suffix -amide:
Amides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol discussed in the previous section):
The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins.
How Sciences Interconnect: Proteins and Enzymes
Proteins are large biological molecules made up of long chains of smaller molecules called amino acids. Organisms rely on proteins for a variety of functions—proteins transport molecules across cell membranes, replicate DNA, and catalyze metabolic reactions, to name only a few of their functions. The properties of proteins are functions of the combination of amino acids that compose them and can vary greatly. Interactions between amino acid sequences in the chains of proteins result in the folding of the chain into specific, three-dimensional structures that determine the protein’s activity.
Amino acids are organic molecules that contain an amine functional group (–NH2), a carboxylic acid functional group (–COOH), and a side chain (that is specific to each individual amino acid). Most living things build proteins from the same 20 different amino acids. Amino acids connect by the formation of a peptide bond, which is a covalent bond formed between two amino acids when the carboxylic acid group of one amino acid reacts with the amine group of the other amino acid. The formation of the bond results in the production of a molecule of water (in general, reactions that result in the production of water when two other molecules combine are referred to as condensation reactions). The resulting bond—between the carbonyl group carbon atom and the amine nitrogen atom is called a peptide link or peptide bond. Since each of the original amino acids has an unreacted group (one has an unreacted amine and the other an unreacted carboxylic acid), more peptide bonds can form to other amino acids, extending the structure. (Figure \(4\)) A chain of connected amino acids is called a polypeptide. Proteins contain at least one long polypeptide chain.
Enzymes are large biological molecules, mostly composed of proteins, which are responsible for the thousands of metabolic processes that occur in living organisms. Enzymes are highly specific catalysts; they speed up the rates of certain reactions. Enzymes function by lowering the activation energy of the reaction they are catalyzing, which can dramatically increase the rate of the reaction. Most reactions catalyzed by enzymes have rates that are millions of times faster than the noncatalyzed version. Like all catalysts, enzymes are not consumed during the reactions that they catalyze. Enzymes do differ from other catalysts in how specific they are for their substrates (the molecules that an enzyme will convert into a different product). Each enzyme is only capable of speeding up one or a few very specific reactions or types of reactions. Since the function of enzymes is so specific, the lack or malfunctioning of an enzyme can lead to serious health consequences. One disease that is the result of an enzyme malfunction is phenylketonuria. In this disease, the enzyme that catalyzes the first step in the degradation of the amino acid phenylalanine is not functional (Figure \(5\)). Untreated, this can lead to an accumulation of phenylalanine, which can lead to intellectual disabilities.
Chemistry in Everyday Life: Kevlar
Kevlar (Figure \(6\)) is a synthetic polymer made from two monomers 1,4-phenylene-diamine and terephthaloyl chloride (Kevlar is a registered trademark of DuPont). The material was developed by Susan Kwolek while she worked to find a replacement for steel in tires. Kwolek's work involved synthesizing polyamides and dissolving them in solvents, then spinning the resulting solution into fibers. One of her solutions proved to be quite different in initial appearance and structure. And once spun, the resulting fibers were particularly strong. From this initial discovery, Kevlar was created. The material has a high tensile strength-to-weight ratio (it is about 5 times stronger than an equal weight of steel), making it useful for many applications from bicycle tires to sails to body armor.
The material owes much of its strength to hydrogen bonds between polymer chains (refer back to the chapter on intermolecular interactions). These bonds form between the carbonyl group oxygen atom (which has a partial negative charge due to oxygen’s electronegativity) on one monomer and the partially positively charged hydrogen atom in the N–H bond of an adjacent monomer in the polymer structure (see dashed lines in Figure \(7\)). There is additional strength derived from the interaction between the unhybridized p orbitals in the six-membered rings, called aromatic stacking.
Kevlar may be best known as a component of body armor, combat helmets, and face masks. Since the 1980s, the US military has used Kevlar as a component of the PASGT (personal armor system for ground troops) helmet and vest. Kevlar is also used to protect armored fighting vehicles and aircraft carriers. Civilian applications include protective gear for emergency service personnel such as body armor for police officers and heat-resistant clothing for fire fighters. Kevlar based clothing is considerably lighter and thinner than equivalent gear made from other materials (Figure \(8\)). Beyond Kevlar, Susan Kwolek was instrumental in the development of Nomex, a fireproof material, and was also involved in the creation of Lycra. She became just the fourth woman inducted into the National Inventors Hall of Fame, and received a number of other awards for her significant contributions to science and society.
In addition to its better-known uses, Kevlar is also often used in cryogenics for its very low thermal conductivity (along with its high strength). Kevlar maintains its high strength when cooled to the temperature of liquid nitrogen (–196 °C).
The table here summarizes the structures discussed in this chapter: | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.05%3A_Amines_and_Amides.txt |
Example and Directions
Words (or words that have the same definition)The definition is case sensitive(Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages](Optional) Caption for Image(Optional) External or Internal Link(Optional) Source for Definition
(Eg. "Genetic, Hereditary, DNA ...")(Eg. "Relating to genes or heredity")The infamous double helix https://bio.libretexts.org/CC-BY-SA; Delmar Larsen
Glossary Entries
Word(s)DefinitionImageCaptionLinkSource
addition reactionreaction in which a double carbon-carbon bond forms a single carbon-carbon bond by the addition of a reactant. Typical reaction for an alkene.
alcoholorganic compound with a hydroxyl group (–OH) bonded to a carbon atom
aldehydeorganic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent
alkanemolecule consisting of only carbon and hydrogen atoms connected by single (σ) bonds
alkenemolecule consisting of carbon and hydrogen containing at least one carbon-carbon double bond
alkyl groupsubstituent, consisting of an alkane missing one hydrogen atom, attached to a larger structure
alkynemolecule consisting of carbon and hydrogen containing at least one carbon-carbon triple bond
amideorganic molecule that features a nitrogen atom connected to the carbon atom in a carbonyl group
amineorganic molecule in which a nitrogen atom is bonded to one or more alkyl group
aromatic hydrocarboncyclic molecule consisting of carbon and hydrogen with delocalized alternating carbon-carbon single and double bonds, resulting in enhanced stability
carbonyl groupcarbon atom double bonded to an oxygen atom
carboxylic acidorganic compound containing a carbonyl group with an attached hydroxyl group
esterorganic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent
etherorganic compound with an oxygen atom that is bonded to two carbon atoms
functional grouppart of an organic molecule that imparts a specific chemical reactivity to the molecule
ketoneorganic compound containing a carbonyl group with two carbon substituents attached to it
organic compoundnatural or synthetic compound that contains carbon
saturated hydrocarbonmolecule containing carbon and hydrogen that has only single bonds between carbon atoms
skeletal structureshorthand method of drawing organic molecules in which carbon atoms are represented by the ends of lines and bends in between lines, and hydrogen atoms attached to the carbon atoms are not shown (but are understood to be present by the context of the structure)
substituentbranch or functional group that replaces hydrogen atoms in a larger hydrocarbon chain
substitution reactionreaction in which one atom replaces another in a molecule
21.07: Summary
Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. The chemistry of these compounds is called organic chemistry. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO2H group of a carboxylic acid, and the –CO2R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group.
The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.06%3A_Key_Terms.txt |
1.
Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms:
1. an alkane
2. an alkene
3. an alkyne
2.
What is the difference between the hybridization of carbon atoms’ valence orbitals in saturated and unsaturated hydrocarbons?
3.
On a microscopic level, how does the reaction of bromine with a saturated hydrocarbon differ from its reaction with an unsaturated hydrocarbon? How are they similar?
4.
On a microscopic level, how does the reaction of bromine with an alkene differ from its reaction with an alkyne? How are they similar?
5.
Explain why unbranched alkenes can form geometric isomers while unbranched alkanes cannot. Does this explanation involve the macroscopic domain or the microscopic domain?
6.
Explain why these two molecules are not isomers:
7.
Explain why these two molecules are not isomers:
8.
How does the carbon-atom hybridization change when polyethylene is prepared from ethylene?
9.
Write the Lewis structure and molecular formula for each of the following hydrocarbons:
1. (f) 4-methyl-2-pentyne
10.
Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons:
1. (f) 3,4-dimethyl-1-pentyne
11.
Give the complete IUPAC name for each of the following compounds:
1. (c)
(d)
(e)
(f)
(g)
12.
Give the complete IUPAC name for each of the following compounds:
1. (c)
(d)
(e)
(f)
13.
Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane.
14.
Write Lewis structures and name the five structural isomers of hexane.
15.
Write Lewis structures for the cis–trans isomers of
16.
Write structures for the three isomers of the aromatic hydrocarbon xylene, C6H4(CH3)2.
17.
Isooctane is the common name of the isomer of C8H18 used as the standard of 100 for the gasoline octane rating:
1. What is the IUPAC name for the compound?
2. Name the other isomers that contain a five-carbon chain with three methyl substituents.
18.
Write Lewis structures and IUPAC names for the alkyne isomers of C4H6.
19.
Write Lewis structures and IUPAC names for all isomers of C4H9Cl.
20.
Name and write the structures of all isomers of the propyl and butyl alkyl groups.
21.
Write the structures for all the isomers of the –C5H11 alkyl group.
22.
Write Lewis structures and describe the molecular geometry at each carbon atom in the following compounds:
1. cis-3-hexene
2. cis-1-chloro-2-bromoethene
3. 2-pentyne
4. trans-6-ethyl-7-methyl-2-octene
23.
Benzene is one of the compounds used as an octane enhancer in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene:
Draw Lewis structures for these compounds, with resonance structures as appropriate, and determine the hybridization of the carbon atoms in each.
24.
Teflon is prepared by the polymerization of tetrafluoroethylene. Write the equation that describes the polymerization using Lewis symbols.
25.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 1 mol of 1-butyne reacts with 2 mol of iodine.
2. Pentane is burned in air.
26.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 2-butene reacts with chlorine.
2. benzene burns in air.
27.
What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.
28.
Acetylene is a very weak acid; however, it will react with moist silver(I) oxide and form water and a compound composed of silver and carbon. Addition of a solution of HCl to a 0.2352-g sample of the compound of silver and carbon produced acetylene and 0.2822 g of AgCl.
1. What is the empirical formula of the compound of silver and carbon?
2. The production of acetylene on addition of HCl to the compound of silver and carbon suggests that the carbon is present as the acetylide ion, . Write the formula of the compound showing the acetylide ion.
29.
Ethylene can be produced by the pyrolysis of ethane:
How many kilograms of ethylene is produced by the pyrolysis of 1.000 103 kg of ethane, assuming a 100.0% yield?
30.
Why do the compounds hexane, hexanol, and hexene have such similar names?
31.
Write condensed formulas and provide IUPAC names for the following compounds:
1. ethyl alcohol (in beverages)
2. methyl alcohol (used as a solvent, for example, in shellac)
3. ethylene glycol (antifreeze)
4. isopropyl alcohol (used in rubbing alcohol)
5. glycerine
32.
Give the complete IUPAC name for each of the following compounds:
(a)
(b)
(c)
33.
Give the complete IUPAC name and the common name for each of the following compounds:
(a)
(b)
(c)
34.
Write the condensed structures of both isomers with the formula C2H6O. Label the functional group of each isomer.
35.
Write the condensed structures of all isomers with the formula C2H6O2. Label the functional group (or groups) of each isomer.
36.
Draw the condensed formulas for each of the following compounds:
1. dipropyl ether
2. 2,2-dimethyl-3-hexanol
3. 2-ethoxybutane
37.
MTBE, Methyl tert-butyl ether, CH3OC(CH3)3, is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2-methylpropene with methanol.
1. Using Lewis structures, write the chemical equation representing the reaction.
2. What volume of methanol, density 0.7915 g/mL, is required to produce exactly 1000 kg of MTBE, assuming a 100% yield?
38.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. propanol is converted to dipropyl ether
2. propene is treated with water in dilute acid.
39.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 2-butene is treated with water in dilute acid
2. ethanol is dehydrated to yield ethene
40.
Order the following molecules from least to most oxidized, based on the marked carbon atom:
41.
Predict the products of oxidizing the molecules shown in this problem. In each case, identify the product that will result from the minimal increase in oxidation state for the highlighted carbon atom:
(a)
(b)
(c)
42.
Predict the products of reducing the following molecules. In each case, identify the product that will result from the minimal decrease in oxidation state for the highlighted carbon atom:
(a)
(b)
(c)
43.
Explain why it is not possible to prepare a ketone that contains only two carbon atoms.
44.
How does hybridization of the substituted carbon atom change when an alcohol is converted into an aldehyde? An aldehyde to a carboxylic acid?
45.
Fatty acids are carboxylic acids that have long hydrocarbon chains attached to a carboxylate group. How does a saturated fatty acid differ from an unsaturated fatty acid? How are they similar?
46.
Write a condensed structural formula, such as CH3CH3, and describe the molecular geometry at each carbon atom.
1. (f) formaldehyde
47.
Write a condensed structural formula, such as CH3CH3, and describe the molecular geometry at each carbon atom.
1. 2-propanol
2. acetone
3. dimethyl ether
4. acetic acid
5. 3-methyl-1-hexene
48.
The foul odor of rancid butter is caused by butyric acid, CH3CH2CH2CO2H.
1. Draw the Lewis structure and determine the oxidation number and hybridization for each carbon atom in the molecule.
2. The esters formed from butyric acid are pleasant-smelling compounds found in fruits and used in perfumes. Draw the Lewis structure for the ester formed from the reaction of butyric acid with 2-propanol.
49.
Write the two-resonance structures for the acetate ion.
50.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures:
1. ethanol reacts with propionic acid
2. benzoic acid, C6H5CO2H, is added to a solution of sodium hydroxide
51.
Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
1. 1-butanol reacts with acetic acid
2. propionic acid is poured onto solid calcium carbonate
52.
Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from 10.0 g of CH3CO2H?
53.
Alcohols A, B, and C all have the composition C4H10O. Molecules of alcohol A contain a branched carbon chain and can be oxidized to an aldehyde; molecules of alcohol B contain a linear carbon chain and can be oxidized to a ketone; and molecules of alcohol C can be oxidized to neither an aldehyde nor a ketone. Write the Lewis structures of these molecules.
54.
Write the Lewis structures of both isomers with the formula C2H7N.
55.
What is the molecular structure about the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion, (CH3)3NH+? What is the hybridization of the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion?
56.
Write the two resonance structures for the pyridinium ion, C5H5NH+.
57.
Draw Lewis structures for pyridine and its conjugate acid, the pyridinium ion, C5H5NH+. What are the hybridizations, electron domain geometries, and molecular geometries about the nitrogen atoms in pyridine and in the pyridinium ion?
58.
Write the Lewis structures of all isomers with the formula C3H7ON that contain an amide linkage.
59.
Write two complete balanced equations for the following reaction, one using condensed formulas and one using Lewis structures.
Methyl amine is added to a solution of HCl.
60.
Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.
Ethylammonium chloride is added to a solution of sodium hydroxide.
61.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.26.
62.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.39.
63.
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.51. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/21%3A_Organic_Chemistry/21.08%3A_Exercises.txt |
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22: Appendices
Exponential Arithmetic
Exponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:
\begin{align} 1000&=1×10^3\ 100&=1×10^2\ 10&=1×10^1\ 1&=1×10^0\ 0.1&=1×10^{−1}\ 0.001&=1×10^{−3}\ 2386&=2.386×1000=2.386×10^3\ 0.123&=1.23×0.1=1.23×10^{−1} \end{align}
The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 × 109, and 0.00000000036 × 10−10.
Addition of Exponentials
Convert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.
Adding Exponentials Add 5.00 × 10−5 and 3.00 × 10−3.
Solution
\begin{align} 3.00×10^{−3}&=300×10^{−5}\ (5.00×10^{−5})+(300×10^{−5})&=305×10^{−5}=3.05×10^{−3} \end{align}
Subtraction of Exponentials
Convert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.
Subtracting Exponentials Subtract 4.0 × 10−7 from 5.0 × 10−6.
Solution
$4.0×10^{−7}=0.40×10^{−6}\ (5.0×10^{−6})−(0.40×10^{−6})=4.6×10^{−6}$
Multiplication of Exponentials
Multiply the digit terms in the usual way and add the exponents of the exponential terms.
Multiplying Exponentials Multiply 4.2 × 10−8 by 2.0 × 103.
Solution
$(4.2×10^{−8})×(2.0×10^3)=(4.2×2.0)×10^{(−8)+(+3)}=8.4×10^{−5}$
Division of Exponentials
Divide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.
Dividing Exponentials Divide 3.6 × 105 by 6.0 × 10−4.
Solution
$\dfrac{3.6×10^{−5}}{6.0×10^{−4}}=\left(\dfrac{3.6}{6.0}\right)×10^{(−5)−(−4)}=0.60×10^{−1}=6.0×10^{−2}$
Squaring of Exponentials
Square the digit term in the usual way and multiply the exponent of the exponential term by 2.
Squaring Exponentials Square the number 4.0 × 10−6.
Solution
$(4.0×10^{−6})^2=4×4×10^{2×(−6)}=16×10^{−12}=1.6×10^{−11}$
Cubing of Exponentials
Cube the digit term in the usual way and multiply the exponent of the exponential term by 3.
Cubing Exponentials Cube the number 2 × 104.
Solution
$(2×10^4)^3=2×2×2×10^{3×4}=8×10^{12}$
Taking Square Roots of Exponentials
If necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.
Finding the Square Root of Exponentials Find the square root of 1.6 × 10−7.
Solution
\begin{align} 1.6×10^{−7}&=16×10^{−8}\ \sqrt{16×10^{−8}}=\sqrt{16}×\sqrt{10^{−8}}&=\sqrt{16}×10^{−\large{\frac{8}{2}}}=4.0×10^{−4} \end{align}
Significant Figures
A beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.
The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).
Addition and Subtraction with Significant Figures Add 4.383 g and 0.0023 g.
Solution
\begin{align} &\mathrm{4.38\underline{3}\:g}\ &\mathrm{\underline{0.002\underline{3}\:g}}\ &\mathrm{4.38\underline{5}\:g} \end{align}
In multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures.
Multiplication and Division with Significant Figures Multiply 0.6238 by 6.6.
Solution
$0.623\underline{8}×6.\underline{6}=4.\underline{1}$
When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).
The Use of Logarithms and Exponential Numbers
The common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2, because 10 must be raised to the second power to equal 100. Additional examples follow.
Logarithms and Exponential Numbers
Number Number Expressed Exponentially Common Logarithm
1000 103 3
10 101 1
1 100 0
0.1 10−1 −1
0.001 10−3 −3
What is the common logarithm of 60? Because 60 lies between 10 and 100, which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782; that is,
$60=10^{1.7782}$
The common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is −1.4069, or
$0.03918=10^{-1.4069}=\dfrac{1}{10^{1.4069}}$
To obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate 10x (where x is the logarithm of the number).
The natural logarithm of a number (ln) is the power to which e must be raised to equal the number; e is the constant 2.7182818. For example, the natural logarithm of 10 is 2.303; that is,
$10=e^{2.303}=2.7182818^{2.303}$
To obtain the natural logarithm of a number, use the ln button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate ex (where x is the natural logarithm of the number).
Logarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving exponents.
1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers.
$\log xy= \log x + \log y, \textrm{ and }\ln xy=\ln x + \ln y$
2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers.
$\log\dfrac{x}{y}=\log x-\log y,\textrm{ and } \ln\dfrac{x}{y}=\ln x-\ln y$
3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number.
$\log x^n=n\log x \textrm{ and }\ln x^n=n\ln x$
The Solution of Quadratic Equations
Mathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions.
$ax^2+bx+c=0$
The solution or roots for any quadratic equation can be calculated using the following formula:
$x=\dfrac{-b±\sqrt{b^2−4ac}}{2a}$
Solving Quadratic Equations Solve the quadratic equation 3x2 + 13x − 10 = 0.
Solution Substituting the values a = 3, b = 13, c = −10 in the formula, we obtain
$x=\dfrac{−13±\sqrt{(13)^2−4×3×(−10)}}{2×3}$
$x=\dfrac{−13±\sqrt{169+120}}{6}=\dfrac{−13±\sqrt{289}}{6}=\dfrac{−13±17}{6}$
The two roots are therefore
$x=\dfrac{−13+17}{6}=\dfrac{2}{3}\textrm{ and }x=\dfrac{−13−17}{6}=−5$
Quadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance.
Two-Dimensional (x-y) Graphing
The relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled (x), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured (y).
When the value of y is changing as a function of x (that is, different values of x correspond to different values of y), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for (x,y) data pairs.
Graphing the Dependence of y on x
x y
1 5
2 10
3 7
4 14
This table contains the following points: (1,5), (2,10), (3,7), and (4,14). Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of y on x.
If the function that describes the dependence of y on x is known, it may be used to compute x,y data pairs that may subsequently be plotted.
Plotting Data Pairs If we know that y = x2 + 2, we can produce a table of a few (x,y) values and then plot the line based on the data shown here.
x y = x2 + 2
1 3
2 6
3 11
4 18 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.02%3A_Essential_Mathematics.txt |
Units of Length
meter (m) = 39.37 inches (in.) = 1.094 yards (yd) angstrom (Å) = 10–8 cm (exact, definition) = 10–10 m (exact, definition)
centimeter (cm) = 0.01 m (exact, definition) yard (yd) = 0.9144 m
millimeter (mm) = 0.001 m (exact, definition) inch (in.) = 2.54 cm (exact, definition)
kilometer (km) = 1000 m (exact, definition) mile (US) = 1.60934 km
Units of Volume
liter (L) = 0.001 m3 (exact, definition) = 1000 cm3 (exact, definition) = 1.057 (US) quarts liquid quart (US) = 32 (US) liquid ounces (exact, definition) = 0.25 (US) gallon (exact, definition) = 0.9463 L
milliliter (mL) = 0.001 L (exact, definition) = 1 cm3 (exact, definition) dry quart = 1.1012 L
microliter (μL) = 10–6 L (exact, definition) = 10–3 cm3 (exact, definition) cubic foot (US) = 28.316 L
Units of Mass
gram (g) = 0.001 kg (exact, definition) ounce (oz) (avoirdupois) = 28.35 g
milligram (mg) = 0.001 g (exact, definition) pound (lb) (avoirdupois) = 0.4535924 kg
kilogram (kg) = 1000 g (exact, definition) = 2.205 lb ton (short) =2000 lb (exact, definition) = 907.185 kg
ton (metric) =1000 kg (exact, definition) = 2204.62 lb ton (long) = 2240 lb (exact, definition) = 1.016 metric ton
Units of Energy
4.184 joule (J) = 1 thermochemical calorie (cal)
1 thermochemical calorie (cal) = 4.184 × 107 erg
erg = 10–7 J (exact, definition)
electron-volt (eV) = 1.60218 × 10−19 J = 23.061 kcal mol−1
liter∙atmosphere = 24.217 cal = 101.325 J (exact, definition)
nutritional calorie (Cal) = 1000 cal (exact, definition) = 4184 J
British thermal unit (BTU) = 1054.804 J1
Units of Pressure
torr = 1 mm Hg (exact, definition)
pascal (Pa) = N m–2 (exact, definition) = kg m–1 s–2 (exact, definition)
atmosphere (atm) = 760 mm Hg (exact, definition) = 760 torr (exact, definition) = 101,325 N m–2 (exact, definition) = 101,325 Pa (exact, definition)
bar = 105 Pa (exact, definition) = 105 kg m–1 s–2 (exact, definition)
Footnotes
1. 1 BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. 59 °F (15 °C) is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table.
22.04: Fundamental Physical Constants
Fundamental Physical Constants
Name and Symbol Value
atomic mass unit (amu) 1.6605402 × 10−27 kg
Avogadro’s number 6.0221367 × 1023 mol−1
Boltzmann’s constant (k) 1.380658 × 10−23 J K−1
charge-to-mass ratio for electron (e/me) 1.75881962 × 1011 C kg−1
electron charge (e) 1.60217733 × 10−19 C
electron rest mass (me) 9.1093897 × 10−31 kg
Faraday’s constant (F) 9.6485309 × 104 C mol−1
gas constant (R) 8.205784 × 10−2 L atm mol−1 K−1 = 8.314510 J mol−1 K−1
molar volume of an ideal gas, 1 atm, 0 °C 22.41409 L mol–1
molar volume of an ideal gas, 1 bar, 0 °C 22.71108 L mol–1
neutron rest mass (mn) 1.6749274 × 10−27 kg
Planck’s constant (h) 6.6260755 × 10−34 J s
proton rest mass (mp) 1.6726231 × 10−27 kg
Rydberg constant (R) 1.0973731534 × 107 m−1 = 2.1798736 × 10−18 J
speed of light (in vacuum) (c) 2.99792458 × 108 m s−1
22.05: Water Properties
Water Density (kg/m3) at Different Temperatures (°C)
Temperature1 Density
0 999.8395
4 999.9720 (density maximum)
10 999.7026
15 999.1026
20 998.2071
22 997.7735
25 997.0479
30 995.6502
40 992.2
60 983.2
80 971.8
100 958.4
Water Vapor Pressure at Different Temperatures (°C)
Temperature Vapor Pressure (torr) Vapor Pressure (Pa)
0 4.6 613.2812
4 6.1 813.2642
10 9.2 1226.562
15 12.8 1706.522
20 17.5 2333.135
22 19.8 2639.776
25 23.8 3173.064
30 31.8 4239.64
35 42.2 5626.188
40 55.3 7372.707
45 71.9 9585.852
50 92.5 12332.29
55 118.0 15732
60 149.4 19918.31
65 187.5 24997.88
70 233.7 31157.35
75 289.1 38543.39
80 355.1 47342.64
85 433.6 57808.42
90 525.8 70100.71
95 633.9 84512.82
100 760.0 101324.7
Water Kw and pKw at Different Temperatures (°C)
Temperature Kw 10–14 pKw2
0 0.112 14.95
5 0.182 14.74
10 0.288 14.54
15 0.465 14.33
20 0.671 14.17
25 0.991 14.00
30 1.432 13.84
35 2.042 13.69
40 2.851 13.55
45 3.917 13.41
50 5.297 13.28
55 7.080 13.15
60 9.311 13.03
75 19.95 12.70
100 56.23 12.25
Specific Heat Capacity for Water
C°(H2O(l)) = 4184 J∙K−1∙kg−1 = 4.184 J∙g-1∙°C-1
C°(H2O(s)) = 1864 J∙K−1∙kg−1
C°(H2O(g)) = 2093 J∙K−1∙kg−1
Standard Water Melting and Boiling Temperatures and Enthalpies of the Transitions
Temperature (K) ΔH (kJ/mol)
melting 273.15 6.088
boiling 373.15 40.656 (44.016 at 298 K)
Water Cryoscopic (Freezing Point Depression) and Ebullioscopic (Boiling Point Elevation) Constants
Kf = 1.86°C∙kg∙mol−1 (cryoscopic constant)
Kb = 0.51°C∙kg∙mol−1 (ebullioscopic constant)
<figure class="ui-has-child-figcaption" id="CNX_Chem_00_EE_LiqWatAbso"> <figcaption>Water full-range spectral absorption curve. This curve shows the full-range spectral absorption for water. The y-axis signifies the absorption in 1/cm. If we divide 1 by this value, we will obtain the length of the path (in cm) after which the intensity of a light beam passing through water decays by a factor of the base of the natural logarithm e (e = 2.718281828).</figcaption> </figure>
Footnotes
1. 1 Data for t < 0 °C are for supercooled water
2. 2 pKw = –log10(Kw)
22.06: Composition of Commercial Acids and Bases
Composition of Commercial Acids and Bases
Acid or Base1 Density (g/mL)2 Percentage by Mass Molarity
acetic acid, glacial 1.05 99.5% 17.4
aqueous ammonia3 0.90 28% 14.8
hydrochloric acid 1.18 36% 11.6
nitric acid 1.42 71% 16.0
perchloric acid 1.67 70% 11.65
phosphoric acid 1.70 85% 14.7
sodium hydroxide 1.53 50% 19.1
sulfuric acid 1.84 96% 18.0
Footnotes
1. 1 Acids and bases are commercially available as aqueous solutions. This table lists properties (densities and concentrations) of common acid and base solutions. Nominal values are provided in cases where the manufacturer cites a range of concentrations and densities.
2. 2 This column contains specific gravity data. In the case of this table, specific gravity is the ratio of density of a substance to the density of pure water at the same conditions. Specific gravity is often cited on commercial labels.
3. 3 This solution is sometimes called “ammonium hydroxide,” although this term is not chemically accurate. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.03%3A_Units_and_Conversion_Factors.txt |
Standard Thermodynamic Properties for Selected Substances
Substance \(ΔH^∘_\ce{f}\) (kJ mol) \(ΔG^∘_\ce{f}\) (kJ mol–1) \(S^∘_{298}\) (J K–1 mol–1)
aluminum
Al(s) 0 0 28.3
Al(g) 324.4 285.7 164.54
Al2O3(s) –1676 –1582 50.92
AlF3(s) –1510.4 –1425 66.5
AlCl3(s) –704.2 –628.8 110.67
AlCl3·6H2O(s) –2691.57 –2269.40 376.56
Al2S3(s) –724.0 –492.4 116.9
Al2(SO4)3(s) –3445.06 –3506.61 239.32
antimony
Sb(s) 0 0 45.69
Sb(g) 262.34 222.17 180.16
Sb4O6(s) –1440.55 –1268.17 220.92
SbCl3(g) –313.8 –301.2 337.80
SbCl5(g) –394.34 –334.29 401.94
Sb2S3(s) –174.89 –173.64 182.00
SbCl3(s) –382.17 –323.72 184.10
SbOCl(s) –374.0
arsenic
As(s) 0 0 35.1
As(g) 302.5 261.0 174.21
As4(g) 143.9 92.4 314
As4O6(s) –1313.94 –1152.52 214.22
As2O5(s) –924.87 –782.41 105.44
AsCl3(g) –261.50 –248.95 327.06
As2S3(s) –169.03 –168.62 163.59
AsH3(g) 66.44 68.93 222.78
H3AsO4(s) –906.3
barium
Ba(s) 0 0 62.5
Ba(g) 180 146 170.24
BaO(s) –548.0 –520.3 72.1
BaCl2(s) –855.0 –806.7 123.7
BaSO4(s) –1473.2 –1362.3 132.2
beryllium
Be(s) 0 0 9.50
Be(g) 324.3 286.6 136.27
BeO(s) –609.4 –580.1 13.8
bismuth
Bi(s) 0 0 56.74
Bi(g) 207.1 168.2 187.00
Bi2O3(s) –573.88 –493.7 151.5
BiCl3(s) –379.07 –315.06 176.98
Bi2S3(s) –143.1 –140.6 200.4
boron
B(s) 0 0 5.86
B(g) 565.0 521.0 153.4
B2O3(s) –1273.5 –1194.3 53.97
B2H6(g) 36.4 87.6 232.1
H3BO3(s) –1094.33 –968.92 88.83
BF3(g) –1136.0 –1119.4 254.4
BCl3(g) –403.8 –388.7 290.1
B3N3H6(l) –540.99 –392.79 199.58
HBO2(s) –794.25 –723.41 37.66
bromine
Br2(l) 0 0 152.23
Br2(g) 30.91 3.142 245.5
Br(g) 111.88 82.429 175.0
BrF3(g) –255.60 –229.45 292.42
HBr(g) –36.3 –53.43 198.7
cadmium
Cd(s) 0 0 51.76
Cd(g) 112.01 77.41 167.75
CdO(s) –258.2 –228.4 54.8
CdCl2(s) –391.5 –343.9 115.3
CdSO4(s) –933.3 –822.7 123.0
CdS(s) –161.9 –156.5 64.9
calcium
Ca(s) 0 0 41.6
Ca(g) 178.2 144.3 154.88
CaO(s) –634.9 –603.3 38.1
Ca(OH)2(s) –985.2 –897.5 83.4
CaSO4(s) –1434.5 –1322.0 106.5
CaSO4·2H2O(s) –2022.63 –1797.45 194.14
CaCO3(s) (calcite) –1220.0 –1081.4 110.0
CaSO3·H2O(s) –1752.68 –1555.19 184.10
carbon
C(s) (graphite) 0 0 5.740
C(s) (diamond) 1.89 2.90 2.38
C(g) 716.681 671.2 158.1
CO(g) –110.52 –137.15 197.7
CO2(g) –393.51 –394.36 213.8
CH4(g) –74.6 –50.5 186.3
CH3OH(l) –239.2 –166.6 126.8
CH3OH(g) –201.0 –162.3 239.9
CCl4(l) –128.2 –62.5 214.4
CCl4(g) –95.7 –58.2 309.7
CHCl3(l) –134.1 –73.7 201.7
CHCl3(g) –103.14 –70.34 295.71
CS2(l) 89.70 65.27 151.34
CS2(g) 116.9 66.8 238.0
C2H2(g) 227.4 209.2 200.9
C2H4(g) 52.4 68.4 219.3
C2H6(g) –84.0 –32.0 229.2
CH3CO2H(l) –484.3 –389.9 159.8
CH3CO2H(g) –434.84 –376.69 282.50
C2H5OH(l) –277.6 –174.8 160.7
C2H5OH(g) –234.8 –167.9 281.6
C3H8(g) –103.8 –23.4 270.3
C6H6(g) 82.927 129.66 269.2
C6H6(l) 49.1 124.50 173.4
CH2Cl2(l) –124.2 –63.2 177.8
CH2Cl2(g) –95.4 –65.90 270.2
CH3Cl(g) –81.9 –60.2 234.6
C2H5Cl(l) –136.52 –59.31 190.79
C2H5Cl(g) –112.17 –60.39 276.00
C2N2(g) 308.98 297.36 241.90
HCN(l) 108.9 125.0 112.8
HCN(g) 135.5 124.7 201.8
chlorine
Cl2(g) 0 0 223.1
Cl(g) 121.3 105.70 165.2
ClF(g) –54.48 –55.94 217.78
ClF3(g) –158.99 –118.83 281.50
Cl2O(g) 80.3 97.9 266.2
Cl2O7(l) 238.1
Cl2O7(g) 272.0
HCl(g) –92.307 –95.299 186.9
HClO4(l) –40.58
chromium
Cr(s) 0 0 23.77
Cr(g) 396.6 351.8 174.50
Cr2O3(s) –1139.7 –1058.1 81.2
CrO3(s) –589.5
(NH4)2Cr2O7(s) –1806.7
cobalt
Co(s) 0 0 30.0
CoO(s) –237.9 –214.2 52.97
Co3O4(s) –910.02 –794.98 114.22
Co(NO3)2(s) –420.5
copper
Cu(s) 0 0 33.15
Cu(g) 338.32 298.58 166.38
CuO(s) –157.3 –129.7 42.63
Cu2O(s) –168.6 –146.0 93.14
CuS(s) –53.1 –53.6 66.5
Cu2S(s) –79.5 –86.2 120.9
CuSO4(s) –771.36 –662.2 109.2
Cu(NO3)2(s) –302.9
fluorine
F2(g) 0 0 202.8
F(g) 79.4 62.3 158.8
F2O(g) 24.7 41.9 247.43
HF(g) –273.3 –275.4 173.8
hydrogen
H2(g) 0 0 130.7
H(g) 217.97 203.26 114.7
H2O(l) –285.83 –237.1 70.0
H2O(g) –241.82 –228.59 188.8
H2O2(l) –187.78 –120.35 109.6
H2O2(g) –136.3 –105.6 232.7
HF(g) –273.3 –275.4 173.8
HCl(g) –92.307 –95.299 186.9
HBr(g) –36.3 –53.43 198.7
HI(g) 26.48 1.70 206.59
H2S(g) –20.6 –33.4 205.8
H2Se(g) 29.7 15.9 219.0
iodine
I2(s) 0 0 116.14
I2(g) 62.438 19.3 260.7
I(g) 106.84 70.2 180.8
IF(g) 95.65 –118.49 236.06
ICl(g) 17.78 –5.44 247.44
IBr(g) 40.84 3.72 258.66
IF7(g) –943.91 –818.39 346.44
HI(g) 26.48 1.70 206.59
iron
Fe(s) 0 0 27.3
Fe(g) 416.3 370.7 180.5
Fe2O3(s) –824.2 –742.2 87.40
Fe3O4(s) –1118.4 –1015.4 146.4
Fe(CO)5(l) –774.04 –705.42 338.07
Fe(CO)5(g) –733.87 –697.26 445.18
FeCl2(s) –341.79 –302.30 117.95
FeCl3(s) –399.49 –334.00 142.3
FeO(s) –272.0 –255.2 60.75
Fe(OH)2(s) –569.0 –486.5 88.
Fe(OH)3(s) –823.0 –696.5 106.7
FeS(s) –100.0 –100.4 60.29
Fe3C(s) 25.10 20.08 104.60
lead
Pb(s) 0 0 64.81
Pb(g) 195.2 162. 175.4
PbO(s) (yellow) –217.32 –187.89 68.70
PbO(s) (red) –218.99 –188.93 66.5
Pb(OH)2(s) –515.9
PbS(s) –100.4 –98.7 91.2
Pb(NO3)2(s) –451.9
PbO2(s) –277.4 –217.3 68.6
PbCl2(s) –359.4 –314.1 136.0
lithium
Li(s) 0 0 29.1
Li(g) 159.3 126.6 138.8
LiH(s) –90.5 –68.3 20.0
Li(OH)(s) –487.5 –441.5 42.8
LiF(s) –616.0 –587.5 35.7
Li2CO3(s) –1216.04 –1132.19 90.17
manganese
Mn(s) 0 0 32.0
Mn(g) 280.7 238.5 173.7
MnO(s) –385.2 –362.9 59.71
MnO2(s) –520.03 –465.1 53.05
Mn2O3(s) –958.97 –881.15 110.46
Mn3O4(s) –1378.83 –1283.23 155.64
mercury
Hg(l) 0 0 75.9
Hg(g) 61.4 31.8 175.0
HgO(s) (red) –90.83 –58.5 70.29
HgO(s) (yellow) –90.46 –58.43 71.13
HgCl2(s) –224.3 –178.6 146.0
Hg2Cl2(s) –265.4 –210.7 191.6
HgS(s) (red) –58.16 –50.6 82.4
HgS(s) (black) –53.56 –47.70 88.28
HgSO4(s) –707.51 –594.13 0.00
nitrogen
N2(g) 0 0 191.6
N(g) 472.704 455.5 153.3
NO(g) 90.25 87.6 210.8
NO2(g) 33.2 51.30 240.1
N2O(g) 81.6 103.7 220.0
N2O3(g) 83.72 139.41 312.17
N2O4(g) 11.1 99.8 304.4
N2O5(g) 11.3 115.1 355.7
NH3(g) –45.9 –16.5 192.8
N2H4(l) 50.63 149.43 121.21
N2H4(g) 95.4 159.4 238.5
NH4NO3(s) –365.56 –183.87 151.08
NH4Cl(s) –314.43 –202.87 94.6
NH4Br(s) –270.8 –175.2 113.0
NH4I(s) –201.4 –112.5 117.0
NH4NO2(s) –256.5
HNO3(l) –174.1 –80.7 155.6
HNO3(g) –133.9 –73.5 266.9
oxygen
O2(g) 0 0 205.2
O(g) 249.17 231.7 161.1
O3(g) 142.7 163.2 238.9
phosphorus
P4(s) 0 0 164.4
P4(g) 58.91 24.4 280.0
P(g) 314.64 278.25 163.19
PH3(g) 5.4 13.5 210.2
PCl3(g) –287.0 –267.8 311.78
PCl5(g) –374.9 –305.0 364.4
P4O6(s) –1640.1
P4O10(s) –2984.0 –2697.0 228.86
HPO3(s) –948.5
H3PO2(s) –604.6
H3PO3(s) –964.4
H3PO4(s) –1279.0 –1119.1 110.50
H3PO4(l) –1266.9 –1124.3 110.5
H4P2O7(s) –2241.0
POCl3(l) –597.1 –520.8 222.5
POCl3(g) –558.5 –512.9 325.5
potassium
K(s) 0 0 64.7
K(g) 89.0 60.5 160.3
KF(s) –576.27 –537.75 66.57
KCl(s) –436.5 –408.5 82.6
silicon
Si(s) 0 0 18.8
Si(g) 450.0 405.5 168.0
SiO2(s) –910.7 –856.3 41.5
SiH4(g) 34.3 56.9 204.6
H2SiO3(s) –1188.67 –1092.44 133.89
H4SiO4(s) –1481.14 –1333.02 192.46
SiF4(g) –1615.0 –1572.8 282.8
SiCl4(l) –687.0 –619.8 239.7
SiCl4(g) –662.75 –622.58 330.62
SiC(s, beta cubic) –73.22 –70.71 16.61
SiC(s, alpha hexagonal) –71.55 –69.04 16.48
silver
Ag(s) 0 0 42.55
Ag(g) 284.9 246.0 172.89
Ag2O(s) –31.05 –11.20 121.3
AgCl(s) –127.0 –109.8 96.3
Ag2S(s) –32.6 –40.7 144.0
sodium
Na(s) 0 0 51.3
Na(g) 107.5 77.0 153.7
Na2O(s) –414.2 –375.5 75.1
NaCl(s) –411.2 –384.1 72.1
sulfur
S8(s) (rhombic) 0 0 256.8
S(g) 278.81 238.25 167.82
SO2(g) –296.83 –300.1 248.2
SO3(g) –395.72 –371.06 256.76
H2S(g) –20.6 –33.4 205.8
H2SO4(l) –813.989 690.00 156.90
H2S2O7(s) –1273.6
SF4(g) –728.43 –684.84 291.12
SF6(g) –1220.5 –1116.5 291.5
SCl2(l) –50
SCl2(g) –19.7
S2Cl2(l) –59.4
S2Cl2(g) –19.50 –29.25 319.45
SOCl2(g) –212.55 –198.32 309.66
SOCl2(l) –245.6
SO2Cl2(l) –394.1
SO2Cl2(g) –354.80 –310.45 311.83
tin
Sn(s) 0 0 51.2
Sn(g) 301.2 266.2 168.5
SnO(s) –285.8 –256.9 56.5
SnO2(s) –577.6 –515.8 49.0
SnCl4(l) –511.3 –440.1 258.6
SnCl4(g) –471.5 –432.2 365.8
titanium
Ti(s) 0 0 30.7
Ti(g) 473.0 428.4 180.3
TiO2(s) –944.0 –888.8 50.6
TiCl4(l) –804.2 –737.2 252.4
TiCl4(g) –763.2 –726.3 353.2
tungsten
W(s) 0 0 32.6
W(g) 849.4 807.1 174.0
WO3(s) –842.9 –764.0 75.9
zinc
Zn(s) 0 0 41.6
Zn(g) 130.73 95.14 160.98
ZnO(s) –350.5 –320.5 43.7
ZnCl2(s) –415.1 –369.43 111.5
ZnS(s) –206.0 –201.3 57.7
ZnSO4(s) –982.8 –871.5 110.5
ZnCO3(s) –812.78 –731.57 82.42
complexes
[Co(NH3)4(NO2)2]NO3, cis –898.7
[Co(NH3)4(NO2)2]NO3, trans –896.2
NH4[Co(NH3)2(NO2)4] –837.6
[Co(NH3)6][Co(NH3)2(NO2)4]3 –2733.0
[Co(NH3)4Cl2]Cl, cis –874.9
[Co(NH3)4Cl2]Cl, trans –877.4
[Co(en)2(NO2)2]NO3, cis –689.5
[Co(en)2Cl2]Cl, cis –681.2
[Co(en)2Cl2]Cl, trans –677.4
[Co(en)3](ClO4)3 –762.7
[Co(en)3]Br2 –595.8
[Co(en)3]I2 –475.3
[Co(en)3]I3 –519.2
[Co(NH3)6](ClO4)3 –1034.7 –221.1 615
[Co(NH3)5NO2](NO3)2 –1088.7 –412.9 331
[Co(NH3)6](NO3)3 –1282.0 –524.5 448
[Co(NH3)5Cl]Cl2 –1017.1 –582.5 366.1
[Pt(NH3)4]Cl2 –725.5
[Ni(NH3)6]Cl2 –994.1
[Ni(NH3)6]Br2 –923.8
[Ni(NH3)6]I2 –808.3 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.07%3A_Standard_Thermodynamic_Properties_for_Selected_Substances.txt |
Ionization Constants of Weak Acids
Acid Formula Ka at 25 °C Lewis Structure
acetic CH3CO2H 1.8 × 10−5
arsenic H3AsO4 5.5 × 10−3
1.7 × 10−7
5.1 × 10−12
arsenous H3AsO3 5.1 × 10−10
boric H3BO3 5.4 × 10−10
carbonic H2CO3 4.3 × 10−7
5.6 × 10−11
cyanic HCNO 2 × 10−4
formic HCO2H 1.8 × 10−4
hydrazoic HN3 2.5 × 10−5
hydrocyanic HCN 4.9 × 10−10
hydrofluoric HF 3.5 × 10−4
hydrogen peroxide H2O2 2.4 × 10−12
hydrogen selenide H2Se 1.29 × 10−4
HSe 1 × 10−12
hydrogen sulfate ion 1.2 × 10−2
hydrogen sulfide H2S 8.9 × 10−8
HS 1.0 × 10−19
hydrogen telluride H2Te 2.3 × 10−3
HTe 1.6 × 10−11
hypobromous HBrO 2.8 × 10−9
hypochlorous HClO 2.9 × 10−8
nitrous HNO2 4.6 × 10−4
oxalic H2C2O4 6.0 × 10−2
6.1 × 10−5
phosphoric H3PO4 7.5 × 10−3
6.2 × 10−8
4.2 × 10−13
phosphorous H3PO3 5 × 10−2
2.0 × 10−7
sulfurous H2SO3 1.6 × 10−2
6.4 × 10−8
22.09: Ionization Constants of Weak Bases
Ionization Constants of Weak Bases
Base Lewis Structure Kb at 25 °C
ammonia 1.8 × 10−5
dimethylamine 5.9 × 10−4
methylamine 4.4 × 10−4
phenylamine (aniline) 4.3 × 10−10
trimethylamine 6.3 × 10−5
22.10: Solubility Products
Solubility Products
Substance Ksp at 25 °C
aluminum
Al(OH)3 2 × 10−32
barium
BaCO3 1.6 × 10−9
BaC2O4·2H2O 1.1 × 10−7
BaSO4 2.3 × 10−8
BaCrO4 8.5 × 10−11
BaF2 2.4 × 10−5
Ba(OH)2·8H2O 5.0 × 10−3
Ba3(PO4)2 6 × 10−39
Ba3(AsO4)2 1.1 × 10−13
bismuth
BiO(OH) 4 × 10−10
BiOCl 1.8 × 10−31
Bi2S3 1 × 10−97
cadmium
Cd(OH)2 5.9 × 10−15
CdS 1.0 × 10−28
CdCO3 5.2 × 10−12
calcium
Ca(OH)2 1.3 × 10−6
CaCO3 8.7 × 10−9
CaSO4·2H2O 6.1 × 10−5
CaC2O4·H2O 1.96 × 10−8
Ca3(PO4)2 1.3 × 10−32
CaHPO4 7 × 10−7
CaF2 4.0 × 10−11
chromium
Cr(OH)3 6.7 × 10−31
cobalt
Co(OH)2 2.5 × 10−16
CoS(α) 5 × 10−22
CoS(β) 3 × 10−26
CoCO3 1.4 × 10−13
Co(OH)3 2.5 × 10−43
copper
CuCl 1.2 × 10−6
CuBr 6.27 × 10−9
CuI 1.27 × 10−12
CuSCN 1.6 × 10−11
Cu2S 2.5 × 10−48
Cu(OH)2 2.2 × 10−20
CuS 8.5 × 10−45
CuCO3 2.5 × 10−10
iron
Fe(OH)2 1.8 × 10−15
FeCO3 2.1 × 10−11
FeS 3.7 × 10−19
Fe(OH)3 4 × 10−38
lead
Pb(OH)2 1.2 × 10−15
PbF2 4 × 10−8
PbCl2 1.6 × 10−5
PbBr2 4.6 × 10−6
PbI2 1.4 × 10−8
PbCO3 1.5 × 10−15
PbS 7 × 10−29
PbCrO4 2 × 10−16
PbSO4 1.3 × 10−8
Pb3(PO4)2 1 × 10−54
magnesium
Mg(OH)2 8.9 × 10−12
MgCO3·3H2O ca 1 × 10−5
MgNH4PO4 3 × 10−13
MgF2 6.4 × 10−9
MgC2O4 7 × 10−7
manganese
Mn(OH)2 2 × 10−13
MnCO3 8.8 × 10−11
MnS 2.3 × 10−13
mercury
Hg2O·H2O 3.6 × 10−26
Hg2Cl2 1.1 × 10−18
Hg2Br2 1.3 × 10−22
Hg2I2 4.5 × 10−29
Hg2CO3 9 × 10−15
Hg2SO4 7.4 × 10−7
Hg2S 1.0 × 10−47
Hg2CrO4 2 × 10−9
HgS 1.6 × 10−54
nickel
Ni(OH)2 1.6 × 10−16
NiCO3 1.4 × 10−7
NiS(α) 4 × 10−20
NiS(β) 1.3 × 10−25
potassium
KClO4 1.05 × 10−2
K2PtCl6 7.48 × 10−6
KHC4H4O6 3 × 10−4
silver
\(\ce{^1_2Ag2O(Ag+ + OH- )}\) 2 × 10−8
AgCl 1.6 × 10−10
AgBr 5.0 × 10−13
AgI 1.5 × 10−16
AgCN 1.2 × 10−16
AgSCN 1.0 × 10−12
Ag2S 1.6 × 10−49
Ag2CO3 8.1 × 10−12
Ag2CrO4 9.0 × 10−12
Ag4Fe(CN)6 1.55 × 10−41
Ag2SO4 1.2 × 10−5
Ag3PO4 1.8 × 10−18
strontium
Sr(OH)2·8H2O 3.2 × 10−4
SrCO3 7 × 10−10
SrCrO4 3.6 × 10−5
SrSO4 3.2 × 10−7
SrC2O4·H2O 4 × 10−7
thallium
TlCl 1.7 × 10−4
TlSCN 1.6 × 10−4
Tl2S 6 × 10−22
Tl(OH)3 6.3 × 10−46
tin
Sn(OH)2 3 × 10−27
SnS 1 × 10−26
Sn(OH)4 1.0 × 10−57
zinc
ZnCO3 2 × 10−10
22.11: Formation Constants for Complex Ions
Formation Constants for Complex Ions
Equilibrium Kf
\(\ce{Al^3+ + 6F- ⇌ [AlF6]^3-}\) 7 × 1019
\(\ce{Cd^2+ + 4NH3 ⇌ [Cd(NH3)4]^2+}\) 1.3 × 107
\(\ce{Cd^2+ + 4CN- ⇌ [Cd(CN)4]^2-}\) 3 × 1018
\(\ce{Co^2+ + 6NH3 ⇌ [Co(NH3)6]^2+}\) 1.3 × 105
\(\ce{Co^3+ + 6NH3 ⇌ [Co(NH3)6]^3+}\) 2.3 × 1033
\(\ce{Cu+ + 2CN ⇌ [Cu(CN)2]-}\) 1.0 × 1016
\(\ce{Cu^2+ + 4NH3⇌[Cu(NH3)4]^2+}\) 1.7 × 1013
\(\ce{Fe^2+ + 6CN- ⇌[Fe(CN)6]^4-}\) 1.5 × 1035
\(\ce{Fe^3+ + 6CN- ⇌[Fe(CN)6]^3-}\) 2 × 1043
\(\ce{Fe^3+ + 6SCN- ⇌[Fe(SCN)6]^3-}\) 3.2 × 103
\(\ce{Hg^2+ + 4Cl- ⇌[HgCl4]^2-}\) 1.1 × 1016
\(\ce{Ni^2+ + 6NH3⇌[Ni(NH3)6]^2+}\) 2.0 × 108
\(\ce{Ag+ + 2Cl- ⇌[AgCl2]-}\) 1.8 × 105
\(\ce{Ag+ + 2CN- ⇌[Ag(CN)2]-}\) 1 × 1021
\(\ce{Ag+ + 2NH3⇌[Ag(NH3)2]+}\) 1.7 × 107
\(\ce{Zn^2+ + 4CN- ⇌[Zn(CN)4]^2-}\) 2.1 × 1019
\(\ce{Zn^2+ + 4OH- ⇌[Zn(OH)4]^2-}\) 2 × 1015
\(\ce{Fe^3+ + SCN- ⇌[Fe(SCN)]^2+}\) 8.9 × 102
\(\ce{Ag+ + 4SCN- ⇌[Ag(SCN)4]^3-}\) 1.2 × 1010
\(\ce{Pb^2+ + 4I- ⇌[PbI4]^2-}\) 3.0 × 104
\(\ce{Pt^2+ + 4Cl- ⇌[PtCl4]^2-}\) 1 × 1016
\(\ce{Cu^2+ + 4CN⇌[Cu(CN)4]^2-}\) 1.0 × 1025
\(\ce{Co^2+ + 4SCN- ⇌[Co(SCN)4]^2-}\) 1 × 103 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.08%3A_Ionization_Constants_of_Weak_Acids.txt |
Standard Electrode (Half-Cell) Potentials
Half-Reaction E° (V)
\(\ce{Ag+ + e- ⟶Ag}\) +0.7996
\(\ce{AgCl + e- ⟶Ag + Cl-}\) +0.22233
\(\ce{[Ag(CN)2]- + e- ⟶Ag + 2CN-}\) −0.31
\(\ce{Ag2CrO4 + 2e- ⟶2Ag + CrO4^2-}\) +0.45
\(\ce{[Ag(NH3)2]+ + e- ⟶Ag + 2NH3}\) +0.373
\(\ce{[Ag(S2O3)2]^3+ + e- ⟶Ag + 2S2O3^2-}\) +0.017
\(\ce{[AlF6]^3- + 3e- ⟶Al + 6F-}\) −2.07
\(\ce{Al^3+ + 3e- ⟶Al}\) −1.662
\(\ce{Am^3+ + 3e- ⟶Am}\) −2.048
\(\ce{Au^3+ + 3e- ⟶Au}\) +1.498
\(\ce{Au+ + e- ⟶Au}\) +1.692
\(\ce{Ba^2+ + 2e- ⟶Ba}\) −2.912
\(\ce{Be^2+ + 2e- ⟶Be}\) −1.847
\(\ce{Br2}(aq) + \ce{2e-} ⟶\ce{2Br-}\) +1.0873
\(\ce{Ca^2+ + 2e- ⟶Ca}\) −2.868
\(\ce{Ce^3 + 3e- ⟶Ce}\) −2.483
\(\ce{Ce^4+ + e- ⟶Ce^3+}\) +1.61
\(\ce{Cd^2+ + 2e- ⟶Cd}\) −0.4030
\(\ce{[Cd(CN)4]^2- + 2e- ⟶Cd + 4CN-}\) −1.09
\(\ce{[Cd(NH3)4]^2+ + 2e- ⟶Cd + 4NH3}\) −0.61
\(\ce{CdS + 2e- ⟶Cd + S^2-}\) −1.17
\(\ce{Cl2 + 2e- ⟶2Cl-}\) +1.35827
\(\ce{ClO4- + H2O + 2e- ⟶ClO3- + 2OH-}\) +0.36
\(\ce{ClO3- + H2O + 2e- ⟶ClO2- + 2OH-}\) +0.33
\(\ce{ClO2- + H2O + 2e- ⟶ClO- + 2OH-}\) +0.66
\(\ce{ClO- + H2O + 2e- ⟶Cl- + 2OH-}\) +0.89
\(\ce{ClO4- + 2H3O+ + 2e- ⟶ClO3- + 3H2O}\) +1.189
\(\ce{ClO3- + 3H3O+ + 2e- ⟶HClO2 + 4H2O}\) +1.21
\(\ce{HClO + H3O+ + 2e- ⟶Cl- + 2H2O}\) +1.482
\(\ce{HClO + H3O+ + e- ⟶\dfrac{1}{2}Cl2 + 2H2O}\) +1.611
\(\ce{HClO2 + 2H3O+ + 2e- ⟶HClO + 3H2O}\) +1.628
\(\ce{Co^3+ + e- ⟶Co^2+ \:(2\:mol\,//\,H2SO4)}\) +1.83
\(\ce{Co^2+ + 2e- ⟶Co}\) −0.28
\(\ce{[Co(NH3)6]^3+ + e- ⟶[Co(NH3)6]^2+}\) +0.1
\(\ce{Co(OH)3 + e- ⟶Co(OH)2 + OH-}\) +0.17
\(\ce{Cr^3 + 3e- ⟶Cr}\) −0.744
\(\ce{Cr^3+ + e- ⟶Cr^2+}\) −0.407
\(\ce{Cr^2+ + 2e- ⟶Cr}\) −0.913
\(\ce{[Cu(CN)2]- + e- ⟶Cu + 2CN-}\) −0.43
\(\ce{CrO4^2- + 4H2O + 3e- ⟶ Cr(OH)3 + 5OH-}\) −0.13
\(\ce{Cr2O7^2- + 14H3O+ + 6e- ⟶2Cr^3+ + 21H2O}\) +1.232
\(\ce{[Cr(OH)4]- + 3e- ⟶Cr + 4OH-}\) −1.2
\(\ce{Cr(OH)3 + 3e- ⟶Cr + 3OH-}\) −1.48
\(\ce{Cu^2+ + e- ⟶Cu+}\) +0.153
\(\ce{Cu^2+ + 2e- ⟶Cu}\) +0.34
\(\ce{Cu+ + e- ⟶Cu}\) +0.521
\(\ce{F2 + 2e- ⟶2F-}\) +2.866
\(\ce{Fe^2+ + 2e- ⟶Fe}\) −0.447
\(\ce{Fe^3+ + e- ⟶Fe^2+}\) +0.771
\(\ce{[Fe(CN)6]^3- + e- ⟶[Fe(CN)6]^4-}\) +0.36
\(\ce{Fe(OH)2 + 2e- ⟶Fe + 2OH-}\) −0.88
\(\ce{FeS + 2e- ⟶Fe + S^2-}\) −1.01
\(\ce{Ga^3+ + 3e- ⟶Ga}\) −0.549
\(\ce{Gd^3+ + 3e- ⟶Gd}\) −2.279
\(\ce{\dfrac{1}{2}H2 + e- ⟶H- }\) −2.23
\(\ce{2H2O + 2e- ⟶H2 + 2OH-}\) −0.8277
\(\ce{H2O2 + 2H3O+ + 2e- ⟶4H2O}\) +1.776
\(\ce{2H3O+ + 2e- ⟶H2 + 2H2O}\) 0.00
\(\ce{HO2- + H2O + 2e- ⟶3OH-}\) +0.878
\(\ce{Hf^4+ + 4e- ⟶Hf}\) −1.55
\(\ce{Hg^2+ + 2e- ⟶Hg}\) +0.851
\(\ce{2Hg^2+ + 2e- ⟶Hg2^2+}\) +0.92
\(\ce{Hg2^2+ + 2e- ⟶2Hg}\) +0.7973
\(\ce{[HgBr4]^2- + 2e- ⟶Hg + 4Br-}\) +0.21
\(\ce{Hg2Cl2 + 2e- ⟶2Hg + 2Cl-}\) +0.26808
\(\ce{[Hg(CN)4]^2- + 2e- ⟶Hg + 4CN-}\) −0.37
\(\ce{[HgI4]^2- + 2e- ⟶Hg + 4I-}\) −0.04
\(\ce{HgS + 2e- ⟶Hg + S^2-}\) −0.70
\(\ce{I2 + 2e- ⟶2I-}\) +0.5355
\(\ce{In^3+ + 3e- ⟶In}\) −0.3382
\(\ce{K+ + e- ⟶K}\) −2.931
\(\ce{La^3+ + 3e- ⟶La}\) −2.52
\(\ce{Li+ + e- ⟶Li}\) −3.04
\(\ce{Lu^3+ + 3e- ⟶Lu}\) −2.28
\(\ce{Mg^2+ + 2e- ⟶Mg}\) −2.372
\(\ce{Mn^2+ + 2e- ⟶Mn}\) −1.185
\(\ce{MnO2 + 2H2O + 2e- ⟶Mn(OH)2 + 2OH-}\) −0.05
\(\ce{MnO4- + 2H2O + 3e- ⟶MnO2 + 4OH-}\) +0.558
\(\ce{MnO2 + 4H+ + 2e- ⟶Mn^2+ + 2H2O}\) +1.23
\(\ce{MnO4- + 8H+ + 5e- ⟶Mn^2+ + 4H2O}\) +1.507
\(\ce{Na+ + e- ⟶Na}\) −2.71
\(\ce{Nd^3+ + 3e- ⟶Nd}\) −2.323
\(\ce{Ni^2+ + 2e- ⟶Ni}\) −0.257
\(\ce{[Ni(NH3)6]^2+ + 2e- ⟶Ni + 6NH3}\) −0.49
\(\ce{NiO2 + 4H+ + 2e- ⟶Ni^2+ + 2H2O}\) +1.593
\(\ce{NiO2 + 2H2O + 2e- ⟶Ni(OH)2 + 2OH-}\) +0.49
\(\ce{NiS + 2e- ⟶Ni + S^2-}\) +0.76
\(\ce{NO3- + 4H+ + 3e- ⟶NO + 2H2O}\) +0.957
\(\ce{NO3- + 3H+ + 2e- ⟶HNO2 + H2O}\) +0.92
\(\ce{NO3- + H2O + 2e- ⟶NO2- + 2OH-}\) +0.10
\(\ce{Np^3+ + 3e- ⟶Np}\) −1.856
\(\ce{O2 + 2H2O + 4e- ⟶4OH-}\) +0.401
\(\ce{O2 + 2H+ + 2e- ⟶H2O2}\) +0.695
\(\ce{O2 + 4H+ + 4e- ⟶2H2O}\) +1.229
\(\ce{Pb^2+ + 2e- ⟶Pb}\) −0.1262
\(\ce{PbO2 + SO4^2- + 4H+ + 2e- ⟶PbSO4 + 2H2O}\) +1.69
\(\ce{PbS + 2e- ⟶Pb + S^2-}\) −0.95
\(\ce{PbSO4 + 2e- ⟶Pb + SO4^2-}\) −0.3505
\(\ce{Pd^2+ + 2e- ⟶Pd}\) +0.987
\(\ce{[PdCl4]^2- + 2e- ⟶Pd + 4Cl-}\) +0.591
\(\ce{Pt^2+ + 2e- ⟶Pt}\) +1.20
\(\ce{[PtBr4]^2- + 2e- ⟶Pt + 4Br-}\) +0.58
\(\ce{[PtCl4]^2- + 2e- ⟶Pt + 4Cl-}\) +0.755
\(\ce{[PtCl6]^2- + 2e- ⟶[PtCl4]^2- + 2Cl-}\) +0.68
\(\ce{Pu^3 + 3e- ⟶Pu}\) −2.03
\(\ce{Ra^2+ + 2e- ⟶Ra}\) −2.92
\(\ce{Rb+ + e- ⟶Rb}\) −2.98
\(\ce{[RhCl6]^3- + 3e- ⟶Rh + 6Cl-}\) +0.44
\(\ce{S + 2e- ⟶S^2-}\) −0.47627
\(\ce{S + 2H+ + 2e- ⟶H2S}\) +0.142
\(\ce{Sc^3+ + 3e- ⟶Sc}\) −2.09
\(\ce{Se + 2H+ + 2e- ⟶H2Se}\) −0.399
\(\ce{[SiF6]^2- + 4e- ⟶Si + 6F-}\) −1.2
\(\ce{SiO3^2- + 3H2O + 4e- ⟶Si + 6OH-}\) −1.697
\(\ce{SiO2 + 4H+ + 4e- ⟶Si + 2H2O}\) −0.86
\(\ce{Sm^3+ + 3e- ⟶Sm}\) −2.304
\(\ce{Sn^4+ + 2e- ⟶Sn^2+}\) +0.151
\(\ce{Sn^2+ + 2e- ⟶Sn}\) −0.1375
\(\ce{[SnF6]^2- + 4e- ⟶Sn + 6F-}\) −0.25
\(\ce{SnS + 2e- ⟶Sn + S^2-}\) −0.94
\(\ce{Sr^2+ + 2e- ⟶Sr}\) −2.89
\(\ce{TeO2 + 4H+ + 4e- ⟶Te + 2H2O}\) +0.593
\(\ce{Th^4+ + 4e- ⟶Th}\) −1.90
\(\ce{Ti^2+ + 2e- ⟶Ti}\) −1.630
\(\ce{U^3+ + 3e- ⟶U}\) −1.79
\(\ce{V^2+ + 2e- ⟶V}\) −1.19
\(\ce{Y^3+ + 3e- ⟶Y}\) −2.37
\(\ce{Zn^2+ + 2e- ⟶Zn}\) −0.7618
\(\ce{[Zn(CN)4]^2- + 2e- ⟶Zn + 4CN-}\) −1.26
\(\ce{[Zn(NH3)4]^2+ + 2e- ⟶Zn + 4NH3}\) −1.04
\(\ce{Zn(OH)2 + 2e- ⟶Zn + 2OH-}\) −1.245
\(\ce{[Zn(OH)4]^2 + 2e- ⟶Zn + 4OH-}\) −1.199
\(\ce{ZnS + 2e- ⟶Zn + S^2-}\) −1.40
\(\ce{Zr^4 + 4e- ⟶Zr}\) −1.539 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.12%3A_Standard_Electrode_%28Half-Cell%29_Potentials.txt |
Half-Lives for Several Radioactive Isotopes
Isotope Half-Life1 Type of Emission2 Isotope Half-Life3 Type of Emission4
$614C614C$ 5730 y $(β−)(β−)$ $83210Bi83210Bi$ 5.01 d $(β−)(β−)$
$713N713N$ 9.97 m $(β+)(β+)$ $83212Bi83212Bi$ 60.55 m $(αorβ−)(αorβ−)$
$915F915F$ 4.1 $××$ 10−22 s $(p)(p)$ $84210Po84210Po$ 138.4 d $(α)(α)$
$1124Na1124Na$ 15.00 h $(β−)(β−)$ $84212Po84212Po$ 3 $××$ 10−7 s $(α)(α)$
$1532P1532P$ 14.29 d $(β−)(β−)$ $84216Po84216Po$ 0.15 s $(α)(α)$
$1940K1940K$ 1.27 $××$ 109 y $(βorE.C.)(βorE.C.)$ $84218Po84218Po$ 3.05 m $(α)(α)$
$2649Fe2649Fe$ 0.08 s $(β+)(β+)$ $85215At85215At$ 1.0 $××$ 10−4 s $(α)(α)$
$2660Fe2660Fe$ 2.6 $××$ 106 y $(β−)(β−)$ $85218At85218At$ 1.6 s $(α)(α)$
$2760Co2760Co$ 5.27 y $(β−)(β−)$ $86220Rn86220Rn$ 55.6 s $(α)(α)$
$3787Rb3787Rb$ 4.7 $××$ 1010 y $(β−)(β−)$ $86222Rn86222Rn$ 3.82 d $(α)(α)$
$3890Sr3890Sr$ 29 y $(β−)(β−)$ $88224Ra88224Ra$ 3.66 d $(α)(α)$
$49115In49115In$ 5.1 $××$ 1015 y $(β−)(β−)$ $88226Ra88226Ra$ 1600 y $(α)(α)$
$53131I53131I$ 8.040 d $(β−)(β−)$ $88228Ra88228Ra$ 5.75 y $(β−)(β−)$
$58142Ce58142Ce$ 5 $××$ 1015 y $(α)(α)$ $89228Ac89228Ac$ 6.13 h $(β−)(β−)$
$81208Tl81208Tl$ 3.07 m $(β−)(β−)$ $90228Th90228Th$ 1.913 y $(α)(α)$
$82210Pb82210Pb$ 22.3 y $(β−)(β−)$ $90232Th90232Th$ 1.4 $××$ 1010 y $(α)(α)$
$82212Pb82212Pb$ 10.6 h $(β−)(β−)$ $90233Th90233Th$ 22 m $(β−)(β−)$
$82214Pb82214Pb$ 26.8 m $(β−)(β−)$ $90234Th90234Th$ $24.10 d24.10 d$ $(β−)(β−)$
$83206Bi83206Bi$ 6.243 d $(E.C.)(E.C.)$ $91233Pa91233Pa$ 27 d $(β−)(β−)$
$92233U92233U$ 1.59 $××$ 105 y $(α)(α)$ $96242Cm96242Cm$ 162.8 d $(α)(α)$
$92234U92234U$ 2.45 $××$ 105 y $(α)(α)$ $97243Bk97243Bk$ 4.5 h $(αorE.C.)(αorE.C.)$
$92235U92235U$ 7.03 $××$ 108 y $(α)(α)$ $99253Es99253Es$ 20.47 d $(α)(α)$
$92238U92238U$ 4.47 $××$ 109 y $(α)(α)$ $100254Fm100254Fm$ 3.24 h $(αorS.F.)(αorS.F.)$
$92239U92239U$ 23.54 m $(β−)(β−)$ $100255Fm100255Fm$ 20.1 h $(α)(α)$
$93239Np93239Np$ 2.3 d $(β−)(β−)$ $101256Md101256Md$ 76 m $(αorE.C.)(αorE.C.)$
$94239Pu94239Pu$ 2.407 $××$ 104 y $(α)(α)$ $102254No102254No$ 55 s $(α)(α)$
$94239Pu94239Pu$ 6.54 $××$ 103 y $(α)(α)$ $103257Lr103257Lr$ 0.65 s $(α)(α)$
$94241Pu94241Pu$ 14.4 y $(αorβ−)(αorβ−)$ $105260Ha105260Ha$ 1.5 s $(αorS.F.)(αorS.F.)$
$95241Am95241Am$ 432.2 y $(α)(α)$ $106263Sg106263Sg$ 0.8 s $(αorS.F.)(αorS.F.)$
Table M1
Footnotes
• 1y = years, d = days, h = hours, m = minutes, s = seconds
• 2E.C. = electron capture, S.F. = Spontaneous fission
• 3y = years, d = days, h = hours, m = minutes, s = seconds
• 4E.C. = electron capture, S.F. = Spontaneous fission
22.14.01: Chapter 1
1.
Place a glass of water outside. It will freeze if the temperature is below 0 °C.
3.
(a) law (states a consistently observed phenomenon, can be used for prediction); (b) theory (a widely accepted explanation of the behavior of matter); (c) hypothesis (a tentative explanation, can be investigated by experimentation)
5.
(a) symbolic, microscopic; (b) macroscopic; (c) symbolic, macroscopic; (d) microscopic
7.
Macroscopic. The heat required is determined from macroscopic properties.
9.
Liquids can change their shape (flow); solids can’t. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not.
11.
The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point.
13.
Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together.
15.
Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate.
17.
(a) element; (b) element; (c) compound; (d) mixture; (e) compound; (f) compound; (g) compound; (h) mixture
19.
In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next.
21.
Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts.
23.
(a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass. (b) 0.9 g
25.
(a) 200.0 g; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g
27.
(a) physical; (b) chemical; (c) chemical; (d) physical; (e) physical
29.
physical
31.
The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered.
33.
Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect “cancel” this dependence on amount, yielding a ratio that is independent of amount (an intensive property).
35.
about a yard
37.
(a) kilograms; (b) meters; (c) meters/second; (d) kilograms/cubic meter; (e) kelvin; (f) square meters; (g) cubic meters
39.
(a) centi-, $××$ 10−2; (b) deci-, $××$ 10−1; (c) Giga-, $××$ 109; (d) kilo-, $××$ 103; (e) milli-, $××$ 10−3; (f) nano-, $××$ 10−9; (g) pico-, $××$ 10−12; (h) tera-, $××$ 1012
41.
(a) m = 18.58 g, V = 5.7 mL. (b) d = 3.3 g/mL (c) dioptase (copper cyclosilicate, d = 3.28—3.31 g/mL); malachite (basic copper carbonate, d = 3.25—4.10 g/mL); Paraiba tourmaline (sodium lithium boron silicate with copper, d = 2.82—3.32 g/mL)
43.
(a) displaced water volume = 2.8 mL; (b) displaced water mass = 2.8 g; (c) The block mass is 2.76 g, essentially equal to the mass of displaced water (2.8 g) and consistent with Archimedes’ principle of buoyancy.
45.
(a) 7.04 $××$ 102; (b) 3.344 $××$ 10−2; (c) 5.479 $××$ 102; (d) 2.2086 $××$ 104; (e) 1.00000 $××$ 103; (f) 6.51 $××$ 10−8; (g) 7.157 $××$ 10−3
47.
(a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain
49.
(a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five
51.
(a) 0.44; (b) 9.0; (c) 27; (d) 140; (e) 1.5 $××$ 10−3; (f) 0.44
53.
(a) 2.15 $××$ 105; (b) 4.2 $××$ 106; (c) 2.08; (d) 0.19; (e) 27,440; (f) 43.0
55.
(a) Archer X; (b) Archer W; (c) Archer Y
57.
(a) $1.0936 yd1 m1.0936 yd1 m$; (b) $0.94635 L1 qt0.94635 L1 qt$; (c) $2.2046 lb1 kg2.2046 lb1 kg$
59.
$2.0 L67.6 fl oz=0.030 L1 fl oz2.0 L67.6 fl oz=0.030 L1 fl oz$
Only two significant figures are justified.
61.
68–71 cm; 400–450 g
63.
355 mL
65.
8 $××$ 10−4 cm
67.
yes; weight = 89.4 kg
69.
5.0 $××$ 10−3 mL
71.
(a) 1.3 $××$ 10−4 kg; (b) 2.32 $××$ 108 kg; (c) 5.23 $××$ 10−12 m; (d) 8.63 $××$ 10−5 kg; (e) 3.76 $××$ 10−1 m; (f) 5.4 $××$ 10−5 m; (g) 1 $××$ 1012 s; (h) 2.7 $××$ 10−11 s; (i) 1.5 $××$ 10−4 K
73.
45.4 L
75.
1.0160 $××$ 103 kg
77.
(a) 394 ft; (b) 5.9634 km; (c) 6.0 $××$ 102; (d) 2.64 L; (e) 5.1 $××$ 1018 kg; (f) 14.5 kg; (g) 324 mg
79.
0.46 m; 1.5 ft/cubit
81.
Yes, the acid’s volume is 123 mL.
83.
62.6 in (about 5 ft 3 in.) and 101 lb
85.
(a) 3.81 cm $××$ 8.89 cm $××$ 2.44 m; (b) 40.6 cm
87.
2.70 g/cm3
89.
(a) 81.6 g; (b) 17.6 g
91.
(a) 5.1 mL; (b) 37 L
93.
5371 °F, 3239 K
95.
−23 °C, 250 K
97.
−33.4 °C, 239.8 K
99.
113 °F | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.13%3A_Half-Lives_for_Several_Radioactive_Isotopes.txt |
1.
The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed.
3.
This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.
5.
Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.
7.
Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.
9.
(a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. (b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. (c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the α particles match the predictions from (a), (b), and (c).
11.
(a) 133Cs+; (b) 127I; (c) 31P3; (d) 57Co3+
13.
(a) Carbon-12, 12C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral 12C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.
15.
(a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is 6Li or $36Li.36Li.$ (b) 6Li+ or $36Li+36Li+$
17.
(a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons
19.
(a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons
21.
Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are 9% Ne-22, 91% Ne-20, and only a trace of Ne-21. The average mass would be 20.18 amu. Checking the nature’s mix of isotopes shows that the abundances are 90.48% Ne-20, 9.25% Ne-22, and 0.27% Ne-21, so our guessed amounts have to be slightly adjusted.
23.
79.90 amu
25.
Turkey source: 20.3% (of 10.0129 amu isotope); US source: 19.1% (of 10.0129 amu isotope)
27.
The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O2, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.
29.
(a) molecular CO2, empirical CO2; (b) molecular C2H2, empirical CH; (c) molecular C2H4, empirical CH2; (d) molecular H2SO4, empirical H2SO4
31.
(a) C4H5N2O; (b) C12H22O11; (c) HO; (d) CH2O; (e) C3H4O3
33.
(a) CH2O; (b) C2H4O
35.
(a) ethanol
(b) methoxymethane, more commonly known as dimethyl ether
(c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.
37.
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
39.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
41.
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 $××$ 1023 molecules.
43.
(a) 256.48 g/mol; (b) 72.150 g mol−1; (c) 378.103 g mol−1; (d) 58.080 g mol−1; (e) 180.158 g mol−1
45.
(a) 197.382 g mol−1; (b) 257.163 g mol−1; (c) 194.193 g mol−1; (d) 60.056 g mol−1; (e) 306.464 g mol−1
47.
(a) 0.819 g;
(b) 307 g;
(c) 0.23 g;
(d) 1.235 $××$ 106 g (1235 kg);
(e) 765 g
49.
(a) 99.41 g;
(b) 2.27 g;
(c) 3.5 g;
(d) 222 kg;
(e) 160.1 g
51.
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
53.
zirconium: 2.038 $××$ 1023 atoms; 30.87 g; silicon: 2.038 $××$ 1023 atoms; 9.504 g; oxygen: 8.151 $××$ 1023 atoms; 21.66 g
55.
AlPO4: 1.000 mol or 26.98 g Al
Al2Cl6: 1.994 mol or 53.74 g Al
Al2S3: 3.00 mol or 80.94 g Al
The Al2S3 sample thus contains the greatest mass of Al.
57.
3.113 $××$ 1025 C atoms
59.
0.865 servings, or about 1 serving.
61.
20.0 g H2O represents the least number of molecules since it has the least number of moles. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.02%3A_Chapter_2.txt |
1.
The spectrum consists of colored lines, at least one of which (probably the brightest) is red.
3.
3.15 m
5.
3.233 $××$ 10−19 J; 2.018 eV
7.
ν = 4.568 $××$ 1014 s−1; λ = 656.3 nm; Energy mol−1 = 1.823 $××$ 105 J mol−1; red
9.
(a) λ = 8.69 $××$ 10−7 m; E = 2.29 $××$ 10−19 J; (b) λ = 4.59 $××$ 10−7 m; E = 4.33 $××$ 10−19 J; The color of (a) is red; (b) is blue.
11.
E = 9.502 $××$ 10−15 J; ν = 1.434 $××$ 1019 s−1
13.
Red: 660 nm; 4.54 $××$ 1014 Hz; 3.01 $××$ 10−19 J. Green: 520 nm; 5.77 $××$ 1014 Hz; 3.82 $××$ 10−19 J. Blue: 440 nm; 6.81 $××$ 1014 Hz; 4.51 $××$ 10−19 J. Somewhat different numbers are also possible.
15.
5.49 $××$ 1014 s−1; no
17.
Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.
19.
$2.856 eV 2.856 eV$
21.
−8.716 $××$ 10−18 J
23.
−3.405 $××$ 10−20 J
25.
33.9 Å
27.
1.471 $××$ 10−17 J
29.
Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature “solar system” with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, n. The orbiting electron in Bohr’s model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such “quantum jumps” will produce discrete spectra, in agreement with observations.
31.
Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, n = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function ψ called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, n (the same one used by Bohr), which specifies shells such that orbitals having the same n all have the same energy and approximately the same spatial extent; the angular momentum quantum number l, which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same l (and n) all have the same energy; and the orientation quantum number m, which is a measure of the z component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [s orbitals] can have zero angular momentum).
33.
n determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. l determines the shape of the orbital. m1 determines the orientation of the orbitals of the same l value with respect to one another. ms determines the spin of an electron.
35.
(a) 2p; (b) 4d; (c) 6s
37.
(a) 3d; (b) 1s; (c) 4f
39.
41.
(a) x. 2, y. 2, z. 2; (b) x. 1, y. 3, z. 0; (c) x. 4 0 0 $12,12,$ y. 2 1 0 $12,12,$ z. 3 2 0 $12;12;$ (d) x. 1, y. 2, z. 3; (e) x. l = 0, ml = 0, y. l = 1, ml = –1, 0, or +1, z. l = 2, ml = –2, –1, 0, +1, +2
43.
12
45.
n l ml s
4 0 0 $+12+12$
4 0 0 $−12−12$
4 1 −1 $+12+12$
4 1 0 $+12+12$
4 1 +1 $+12+12$
4 1 −1 $−12−12$
47.
For example, Na+: 1s22s22p6; Ca2+: 1s22s22p63s23p6; Sn2+: 1s22s22p63s23p63d104s24p64d105s2; F: 1s22s22p6; O2–: 1s22s22p6; Cl: 1s22s22p63s23p6.
49.
(a) 1s22s22p3; (b) 1s22s22p63s23p2; (c) 1s22s22p63s23p64s23d6; (d) 1s22s22p63s23p64s23d104p65s24d105p4; (e) 1s22s22p63s23p64s23d104p65s24d105p66s24f9
51.
The charge on the ion.
53.
(a)
(b)
(c)
(d)
(e)
55.
Zr
57.
Rb+, Se2
59.
Although both (b) and (c) are correct, (e) encompasses both and is the best answer.
61.
K
63.
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10
65.
Co has 27 protons, 27 electrons, and 33 neutrons: 1s22s22p63s23p64s23d7. I has 53 protons, 53 electrons, and 78 neutrons: 1s22s22p63s23p63d104s24p64d105s25p5.
67.
Cl
69.
O
71.
Rb < Li < N < F
73.
15 (5A)
75.
Mg < Ca < Rb < Cs
77.
Si4+ < Al3+ < Ca2+ < K+
79.
Se, As
81.
Mg2+ < K+ < Br < As3–
83.
O, IE1
85.
Ra
87.
(a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element; (d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element
89.
(a) He; (b) Be; (c) Li; (d) O
91.
(a) krypton, Kr; (b) calcium, Ca; (c) fluorine, F; (d) tellurium, Te
93.
(a) $1123Na1123Na$; (b) $54129Xe54129Xe$; (c) $3373As3373As$; (d) $88226Ra88226Ra$
95.
Ionic: KCl, MgCl2; Covalent: NCl3, ICl, PCl5, CCl4
97.
(a) covalent; (b) ionic, Ba2+, O2−; (c) ionic, $NH4+,NH4+,$ $CO32−;CO32−;$ (d) ionic, Sr2+, $H2PO4−;H2PO4−;$ (e) covalent; (f) ionic, Na+, O2−
99.
(a) CaS; (b) (NH4)2SO4; (c) AlBr3; (d) Na2HPO4; (e) Mg3 (PO4)2 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.03%3A_Chapter_3.txt |
1.
The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.
3.
P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.
5.
(a) P3–; (b) Mg2+; (c) Al3+; (d) O2–; (e) Cl; (f) Cs+
7.
(a) [Ar]4s23d104p6; (b) [Kr]4d105s25p6 (c) 1s2 (d) [Kr]4d10; (e) [He]2s22p6; (f) [Ar]3d10; (g) 1s2 (h) [He]2s22p6 (i) [Kr]4d105s2 (j) [Ar]3d7 (k) [Ar]3d6, (l) [Ar]3d104s2
9.
(a) 1s22s22p63s23p1; Al3+: 1s22s22p6; (b) 1s22s22p63s23p63d104s24p5; 1s22s22p63s23p63d104s24p6; (c) 1s22s22p63s23p63d104s24p65s2; Sr2+: 1s22s22p63s23p63d104s24p6; (d) 1s22s1; Li+: 1s2; (e) 1s22s22p63s23p63d104s24p3; 1s22s22p63s23p63d104s24p6; (f) 1s22s22p63s23p4; 1s22s22p63s23p6
11.
NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.
13.
ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)
15.
(a) Cl; (b) O; (c) O; (d) S; (e) N; (f) P; (g) N
17.
(a) H, C, N, O, F; (b) H, I, Br, Cl, F; (c) H, P, S, O, F; (d) Na, Al, H, P, O; (e) Ba, H, As, N, O
19.
N, O, F, and Cl
21.
(a) HF; (b) CO; (c) OH; (d) PCl; (e) NH; (f) PO; (g) CN
23.
(a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide; (f) aluminum fluoride
25.
(a) RbBr; (b) MgSe; (c) Na2O; (d) CaCl2; (e) HF; (f) GaP; (g) AlBr3; (h) (NH4)2SO4
27.
(a) ClO2; (b) N2O4; (c) K3P; (d) Ag2S; (e) AIF3∙3H2O; (f) SiO2
29.
(a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) chloride hexahydrate; (f) molybdenum(IV) sulfide
31.
(a) K3PO4; (b) CuSO4; (c) CaCl2; (d) TiO2; (e) NH4NO3; (f) NaHSO4
33.
(a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide
34.
(a) eight electrons:
(b) eight electrons:
(c) no electrons Be2+
(d) eight electrons:
(e) no electrons Ga3+
(f) no electrons Li+
(g) eight electrons:
36.
(a)
(b)
(c)
(d)
(e)
(f)
38.
40.
(a)
In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
42.
(a) SeF6:
(b) XeF4:
(c) $SeCl3+:SeCl3+:$
(d) Cl2BBCl2:
44.
Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb2+ ion has a 6s2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.
46.
48.
50.
(a)
(b)
(c)
(d)
(e)
52.
54.
Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.
56.
(a)
(b)
(c)
(d)
(e)
58.
60.
(a)
(b)
CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
62.
(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
64.
Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
66.
(a)
(b)
(c)
(d)
68.
HOCl
70.
The structure that gives zero formal charges is consistent with the actual structure:
72.
NF3;
74.
75.
The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.
77.
Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.
79.
As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.
81.
(a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar.
83.
(a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109°)
85.
(a) electron-pair geometry: trigonal planar, molecular structure: bent (120°); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal
87.
All of these molecules and ions contain polar bonds. Only ClF5, $ClO2−,ClO2−,$ PCl3, SeF4, and $PH2−PH2−$ have dipole moments.
89.
SeS2, CCl2F2, PCl3, and ClNO all have dipole moments.
91.
P
93.
nonpolar
95.
(a) tetrahedral; (b) trigonal pyramidal; (c) bent (109°); (d) trigonal planar; (e) bent (109°); (f) bent (109°); (g) CH3CCH tetrahedral, CH3CCH linear; (h) tetrahedral; (i) H2CCCH2 linear; H2CCCH2 trigonal planar
97.
99.
(a)
(b)
(c)
(d) $CS32−CS32−$ includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear
101.
The Lewis structure is made from three units, but the atoms must be rearranged:
103.
The molecular dipole points away from the hydrogen atoms.
105.
The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.04%3A_Chapter_4.txt |
1.
Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).
3.
Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.
5.
No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.
7.
Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.
9.
There are no d orbitals in the valence shell of carbon.
11.
trigonal planar, sp2; trigonal pyramidal (one lone pair on A) sp3; T-shaped (two lone pairs on A sp3d, or (three lone pairs on A) sp3d2
13.
(a) Each S has a bent (109°) geometry, sp3
(b) Bent (120°), sp2
(c) Trigonal planar, sp2
(d) Tetrahedral, sp3
15.
(a) XeF2
(b)
(c) linear (d) sp3d
17.
(a)
(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp3; (d) Oxidation states P +1, S$−113,−113,$ Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1
19.
Phosphorus and nitrogen can form sp3 hybrids to form three bonds and hold one lone pair in PF3 and NF3, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp3d hybrid orbitals to bind five fluorine atoms in NF5. Phosphorus has d orbitals and can bind five fluorine atoms with sp3d hybrid orbitals in PF5.
21.
A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.
23.
(a)
(b) The terminal carbon atom uses sp3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.
25.
(a) sp2; (b) sp; (c) sp2; (d) sp3; (e) sp3; (f) sp3d; (g) sp3
27.
(a) sp2, delocalized; (b) sp, localized; (c) sp2, delocalized; (d) sp3, delocalized
29.
Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.
31.
(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.
33.
An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.
35.
Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.
37.
The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.
39.
(a) H2 bond order = 1, $H2+H2+$ bond order = 0.5, $H2−H2−$ bond order = 0.5, strongest bond is H2; (b) O2 bond order = 2, $O22+O22+$ bond order = 3; $O22−O22−$ bond order = 1, strongest bond is $O22+;O22+;$ (c) Li2 bond order = 1, $Be2+Be2+$ bond order = 0.5, Be2 bond order = 0, strongest bond is $Li2Li2$; (d) F2 bond order = 1, $F2+F2+$ bond order = 1.5, $F2−F2−$ bond order = 0.5, strongest bond is $F2+;F2+;$ (e) N2 bond order = 3, $N2+N2+$ bond order = 2.5, $N2−N2−$ bond order = 2.5, strongest bond is N2
41.
(a) H2; (b) N2; (c) O; (d) C2; (e) B2
43.
Yes, fluorine is a smaller atom than Li, so atoms in the 2s orbital are closer to the nucleus and more stable.
45.
2+
47.
N2 has s-p mixing, so the π orbitals are the last filled in $N22+.N22+.$ O2 does not have s-p mixing, so the σp orbital fills before the π orbitals. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.05%3A_Chapter_5.txt |
1.
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
3.
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
5.
(a) 56.107 amu;
(b) 54.091 amu;
(c) 199.9976 amu;
(d) 97.9950 amu
8.
(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca2+ = 38.76%
10.
% NH3 = 38.2%
12.
(a) CS2
(b) CH2O
14.
C6H6
16.
Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)
18.
C15H15N3
20.
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
22.
(a) 0.679 M; (b) 1.00 M; (c) 0.06998 M; (d) 1.75 M; (e) 0.070 M; (f) 6.6 M
24.
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
26.
(a) 37.0 mol H2SO4, 3.63 $××$ 103 g H2SO4; (b) 3.8 $××$ 10−7 mol NaCN, 1.9 $××$ 10−5 g NaCN; (c) 73.2 mol H2CO, 2.20 kg H2CO; (d) 5.9 $××$ 10−7 mol FeSO4, 8.9 $××$ 10−5 g FeSO4
28.
(a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 $××$ 10−3 M
30.
(a) 5.04 $××$ 10−3 M; (b) 0.499 M; (c) 9.92 M; (d) 1.1 $××$ 10−3 M
32.
0.025 M
34.
0.5000 L
36.
1.9 mL
38.
(a) 0.125 M; (b) 0.04888 M; (c) 0.206 M; (d) 0.0056 M
40.
11.9 M
42.
1.6 L
44.
(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: $%mass1×mass1=%mass2×mass2%mass1×mass1=%mass2×mass2$. This equation can be rearranged to isolate mass1 and the given quantities substituted into this equation. (b) 58.8 g
46.
114 g
48.
1.75 $××$ 10−3 M
50.
95 mg/dL
52.
2.38 $××$ 10−4 mol
54.
0.29 mol
22.14.07: Chapter 7
1.
An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.
3.
(a) $PCl5(s)+H2O(l)⟶POCl3(l)+2HCl(aq);PCl5(s)+H2O(l)⟶POCl3(l)+2HCl(aq);$ (b) $3Cu(s)+8HNO3(aq)⟶3Cu(NO3)2(aq)+4H2O(l)+2NO(g);3Cu(s)+8HNO3(aq)⟶3Cu(NO3)2(aq)+4H2O(l)+2NO(g);$ (c) $H2(g)+I2(s)⟶2HI(s);H2(g)+I2(s)⟶2HI(s);$ (d) $4Fe(s)+3O2(g)⟶2Fe2O3(s);4Fe(s)+3O2(g)⟶2Fe2O3(s);$ (e) $2Na(s)+2H2O(l)⟶2NaOH(aq)+H2(g);2Na(s)+2H2O(l)⟶2NaOH(aq)+H2(g);$ (f) $(NH4 )2Cr2O7(s)⟶Cr2O3(s)+N2(g)+4H2O(g);(NH4 )2Cr2O7(s)⟶Cr2O3(s)+N2(g)+4H2O(g);$ (g) $P4(s)+6Cl2(g)⟶4PCl3(l);P4(s)+6Cl2(g)⟶4PCl3(l);$ (h) $PtCl4(s)⟶Pt(s)+2Cl2(g)PtCl4(s)⟶Pt(s)+2Cl2(g)$
5.
(a) $CaCO3(s)⟶CaO(s)+CO2(g);CaCO3(s)⟶CaO(s)+CO2(g);$ (b) $2C4H10(g)+13O2(g)⟶8CO2(g)+10H2O(g);2C4H10(g)+13O2(g)⟶8CO2(g)+10H2O(g);$ (c) $MgC12(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq);MgC12(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq);$ (d) $2H2O(g)+2Na(s)⟶2NaOH(s)+H2(g)2H2O(g)+2Na(s)⟶2NaOH(s)+H2(g)$
7.
(a) Ba(NO3)2, KClO3; (b) $2KClO3(s)⟶2KCl(s)+3O2(g);2KClO3(s)⟶2KCl(s)+3O2(g);$ (c) $2Ba(NO3)2(s)⟶2BaO(s)+2N2(g)+5O2(g);2Ba(NO3)2(s)⟶2BaO(s)+2N2(g)+5O2(g);$ (d) $2Mg(s)+O2(g)⟶2MgO(s);2Mg(s)+O2(g)⟶2MgO(s);$ $4Al(s)+3O2(g)⟶2Al2O3(s);4Al(s)+3O2(g)⟶2Al2O3(s);$ $4Fe(s)+3O2(g)⟶2Fe2O3(s)4Fe(s)+3O2(g)⟶2Fe2O3(s)$
9.
(a) $4HF(aq)+SiO2(s)⟶SiF4(g)+2H2O(l);4HF(aq)+SiO2(s)⟶SiF4(g)+2H2O(l);$ (b) complete ionic equation: $2Na+(aq)+2F−(aq)+Ca2+(aq)+2Cl−(aq)⟶CaF2(s)+2Na+(aq)+2Cl−(aq),2Na+(aq)+2F−(aq)+Ca2+(aq)+2Cl−(aq)⟶CaF2(s)+2Na+(aq)+2Cl−(aq),$ net ionic equation: $2F−(aq)+Ca2+(aq)⟶CaF2(s)2F−(aq)+Ca2+(aq)⟶CaF2(s)$
11.
(a) $2K+(aq)+C2O42−(aq)+Ba2+(aq)+2OH−(aq)⟶2K+(aq)+2OH−(aq)+BaC2O4(s)(complete)Ba2+(aq)+C2O42−(aq)⟶BaC2O4(s)(net)2K+(aq)+C2O42−(aq)+Ba2+(aq)+2OH−(aq)⟶2K+(aq)+2OH−(aq)+BaC2O4(s)(complete)Ba2+(aq)+C2O42−(aq)⟶BaC2O4(s)(net)$
(b) $Pb2+(aq)+2NO3−(aq)+2H+(aq)+SO42−(aq)⟶PbSO4(s)+2H+(aq)+2NO3−(aq)(complete)Pb2+(aq)+SO42−(aq)⟶PbSO4(s)(net)Pb2+(aq)+2NO3−(aq)+2H+(aq)+SO42−(aq)⟶PbSO4(s)+2H+(aq)+2NO3−(aq)(complete)Pb2+(aq)+SO42−(aq)⟶PbSO4(s)(net)$
(c) $CaCO3(s)+2H+(aq)+SO42−(aq)⟶CaSO4(s)+CO2(g)+H2O(l)(complete)CaCO3(s)+2H+(aq)+SO42−(aq)⟶CaSO4(s)+CO2(g)+H2O(l)(net)CaCO3(s)+2H+(aq)+SO42−(aq)⟶CaSO4(s)+CO2(g)+H2O(l)(complete)CaCO3(s)+2H+(aq)+SO42−(aq)⟶CaSO4(s)+CO2(g)+H2O(l)(net)$
13.
(a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)
15.
It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.
17.
(a) H +1, P +5, O −2; (b) Al +3, H +1, O −2; (c) Se +4, O −2; (d) K +1, N +3, O −2; (e) In +3, S −2; (f) P +3, O −2
19.
(a) acid-base; (b) oxidation-reduction: Na is oxidized, H+ is reduced; (c) oxidation-reduction: Mg is oxidized, Cl2 is reduced; (d) acid-base; (e) oxidation-reduction: P3− is oxidized, O2 is reduced; (f) acid-base
21.
(a) $2HCl(g)+Ca(OH)2(s)⟶CaCl2(s)+2H2O(l);2HCl(g)+Ca(OH)2(s)⟶CaCl2(s)+2H2O(l);$ (b) $Sr(OH)2(aq)+2HNO3(aq)⟶Sr(NO3)2(aq)+2H2O(l)Sr(OH)2(aq)+2HNO3(aq)⟶Sr(NO3)2(aq)+2H2O(l)$
23.
(a) $2Al(s)+3F2(g)⟶2AlF3(s);2Al(s)+3F2(g)⟶2AlF3(s);$ (b) $2Al(s)+3CuBr2(aq)⟶3Cu(s)+2AlBr3(aq);2Al(s)+3CuBr2(aq)⟶3Cu(s)+2AlBr3(aq);$ (c) $P4(s)+5O2(g)⟶P4O10(s);P4(s)+5O2(g)⟶P4O10(s);$ (d) $Ca(s)+2H2O(l)⟶Ca(OH)2(aq)+H2(g)Ca(s)+2H2O(l)⟶Ca(OH)2(aq)+H2(g)$
25.
(a) $Mg(OH)2(s)+2HClO4(aq)⟶Mg2+(aq)+2ClO4−(aq)+2H2O(l);Mg(OH)2(s)+2HClO4(aq)⟶Mg2+(aq)+2ClO4−(aq)+2H2O(l);$ (b) $SO3(g)+2H2O(l)⟶H3O+(aq)+HSO4−(aq),SO3(g)+2H2O(l)⟶H3O+(aq)+HSO4−(aq),$ (a solution of H2SO4); (c) $SrO(s)+H2SO4(l)⟶SrSO4(s)+H2OSrO(s)+H2SO4(l)⟶SrSO4(s)+H2O$
27.
$H 2 ( g ) + F 2 ( g ) ⟶ 2HF ( g ) H 2 ( g ) + F 2 ( g ) ⟶ 2HF ( g )$
29.
$2NaBr ( a q ) + Cl 2 ( g ) ⟶ 2 NaCl ( a q ) + Br 2 ( l ) 2NaBr ( a q ) + Cl 2 ( g ) ⟶ 2 NaCl ( a q ) + Br 2 ( l )$
31.
$2 LiOH ( a q ) + CO 2 ( g ) ⟶ Li 2 CO 3 ( a q ) + H 2 O ( l ) 2 LiOH ( a q ) + CO 2 ( g ) ⟶ Li 2 CO 3 ( a q ) + H 2 O ( l )$
33.
(a) $Ca(OH)2(s)+H2S(g)⟶CaS(s)+2H2O(l);Ca(OH)2(s)+H2S(g)⟶CaS(s)+2H2O(l);$ (b) $Na2CO3(aq)+H2S(g)⟶Na2S(aq)+CO2(g)+H2O(l)Na2CO3(aq)+H2S(g)⟶Na2S(aq)+CO2(g)+H2O(l)$
35.
(a) step 1: $N2(g)+3H2(g)⟶2NH3(g),N2(g)+3H2(g)⟶2NH3(g),$ step 2: $NH3(g)+HNO3(aq)⟶NH4NO3(aq)⟶NH4NO3(s)(after drying);NH3(g)+HNO3(aq)⟶NH4NO3(aq)⟶NH4NO3(s)(after drying);$ (b) $H2(g)+Br2(l)⟶2HBr(g);H2(g)+Br2(l)⟶2HBr(g);$ (c) $Zn(s)+S(s)⟶ZnS(s)Zn(s)+S(s)⟶ZnS(s)$ and $ZnS(s)+2HCl(aq)⟶ZnCl2(aq)+H2S(g)ZnS(s)+2HCl(aq)⟶ZnCl2(aq)+H2S(g)$
37.
(a) $Sn4+(aq)+2e−⟶Sn2+(aq),Sn4+(aq)+2e−⟶Sn2+(aq),$ (b) $[Ag(NH3)2]+(aq)+e−⟶Ag(s)+2NH3(aq);[Ag(NH3)2]+(aq)+e−⟶Ag(s)+2NH3(aq);$ (c) $Hg2Cl2(s)+2e−⟶2Hg(l)+2Cl−(aq);Hg2Cl2(s)+2e−⟶2Hg(l)+2Cl−(aq);$ (d) $2H2O(l)⟶O2(g)+4H+(aq)+4e−;2H2O(l)⟶O2(g)+4H+(aq)+4e−;$ (e) $6H2O(l)+2IO3−(aq)+10e−⟶I2(s)+12OH−(aq);6H2O(l)+2IO3−(aq)+10e−⟶I2(s)+12OH−(aq);$ (f) $H2O(l)+SO32−(aq)⟶SO42−(aq)+2H+(aq)+2e−;H2O(l)+SO32−(aq)⟶SO42−(aq)+2H+(aq)+2e−;$ (g) $8H+(aq)+MnO4−(aq)+5e−⟶Mn2+(aq)+4H2O(l);8H+(aq)+MnO4−(aq)+5e−⟶Mn2+(aq)+4H2O(l);$ (h) $Cl−(aq)+6OH−(aq)⟶ClO3−(aq)+3H2O(l)+6e−Cl−(aq)+6OH−(aq)⟶ClO3−(aq)+3H2O(l)+6e−$
39.
(a) $Sn2+(aq)+2Cu2+(aq)⟶Sn4+(aq)+2Cu+(aq);Sn2+(aq)+2Cu2+(aq)⟶Sn4+(aq)+2Cu+(aq);$ (b) $H2S(g)+Hg22+(aq)+2H2O(l)⟶2Hg(l)+S(s)+2H3O+(aq);H2S(g)+Hg22+(aq)+2H2O(l)⟶2Hg(l)+S(s)+2H3O+(aq);$ (c) $5CN−(aq)+2ClO2(aq)+3H2O(l)⟶5CNO−(aq)+2Cl−(aq)+2H3O+(aq);5CN−(aq)+2ClO2(aq)+3H2O(l)⟶5CNO−(aq)+2Cl−(aq)+2H3O+(aq);$ (d) $Fe2+(aq)+Ce4+(aq)⟶Fe3+(aq)+Ce3+(aq);Fe2+(aq)+Ce4+(aq)⟶Fe3+(aq)+Ce3+(aq);$ (e) $2HBrO(aq)+2H2O(l)⟶2H3O+(aq)+2Br−(aq)+O2(g)2HBrO(aq)+2H2O(l)⟶2H3O+(aq)+2Br−(aq)+O2(g)$
41.
(a) $2MnO4−(aq)+3NO2−(aq)+H2O(l)⟶2MnO2(s)+3NO3−(aq)+2OH−(aq);2MnO4−(aq)+3NO2−(aq)+H2O(l)⟶2MnO2(s)+3NO3−(aq)+2OH−(aq);$ (b) $3MnO42−(aq)+2H2O(l)⟶2MnO4−(aq)+4OH−(aq)+MnO2(s)(in base);3MnO42−(aq)+2H2O(l)⟶2MnO4−(aq)+4OH−(aq)+MnO2(s)(in base);$ (c) $Br2(l)+SO2(g)+2H2O(l)⟶4H+(aq)+2Br−(aq)+SO42−(aq)Br2(l)+SO2(g)+2H2O(l)⟶4H+(aq)+2Br−(aq)+SO42−(aq)$
43.
(a) 0.435 mol Na, 0.217 mol Cl2, 15.4 g Cl2; (b) 0.005780 mol HgO, 2.890 $××$ 10−3 mol O2, 9.248 $××$ 10−2 g O2; (c) 8.00 mol NaNO3, 6.8 $××$ 102 g NaNO3; (d) 1665 mol CO2, 73.3 kg CO2; (e) 18.86 mol CuO, 2.330 kg CuCO3; (f) 0.4580 mol C2H4Br2, 86.05 g C2H4Br2
45.
(a) 0.0686 mol Mg, 1.67 g Mg; (b) 2.701 $××$ 10−3 mol O2, 0.08644 g O2; (c) 6.43 mol MgCO3, 542 g MgCO3 (d) 768 mol H2O, 13.8 kg H2O; (e) 16.31 mol BaO2, 2762 g BaO2; (f) 0.207 mol C2H4, 5.81 g C2H4
47.
(a) $volume HCl solution⟶mol HCl⟶mol GaCl3;volume HCl solution⟶mol HCl⟶mol GaCl3;$ (b) 1.25 mol GaCl3, 2.2 $××$ 102 g GaCl3
49.
(a) 5.337 $××$ 1022 molecules; (b) 10.41 g Zn(CN)2
51.
$SiO2+3C⟶SiC+2CO,SiO2+3C⟶SiC+2CO,$ 4.50 kg SiO2
53.
5.00 $××$ 103 kg
55.
1.28 $××$ 105 g CO2
57.
161.4 mL KI solution
59.
176 g TiO2
61.
The limiting reactant is Cl2.
63.
$Percent yield=31%Percent yield=31%$
65.
$gCCl4⟶molCCl4⟶molCCl2F2⟶gCCl2F2,gCCl4⟶molCCl4⟶molCCl2F2⟶gCCl2F2,$ $percent yield=48.3%percent yield=48.3%$
67.
$percent yield = 91.3% percent yield = 91.3%$
69.
Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6%
71.
The conversion needed is $mol Cr⟶molH3PO4.mol Cr⟶molH3PO4.$ Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.
73.
Na2C2O4 is the limiting reactant. percent yield = 86.56%
75.
Only four molecules can be made.
77.
This amount cannot be weighted by ordinary balances and is worthless.
79.
3.4 $××$ 10−3 M H2SO4
81.
9.6 $××$ 10−3 M Cl
83.
22.4%
85.
The empirical formula is BH3. The molecular formula is B2H6.
87.
49.6 mL
89.
13.64 mL
91.
0.0122 M
93.
34.99 mL KOH
95.
The empirical formula is WCl4. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.06%3A_Chapter_6.txt |
1.
The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.
3.
Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.
5.
0.809 atm; 82.0 kPa
7.
2.2 $××$ 102 kPa
9.
Earth: 14.7 lb in–2; Venus: 1.30 × 103 lb in−2
11.
(a) 101.5 kPa; (b) 51 torr drop
13.
(a) 264 torr; (b) 35,200 Pa; (c) 0.352 bar
15.
(a) 623 mm Hg; (b) 0.820 atm; (c) 83.1 kPa
17.
With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since Pgas = Patm + Pvol liquid.
19.
As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.
21.
(a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure
23.
The curve would be farther to the right and higher up, but the same basic shape.
25.
About 12.5 L
27.
3.40 $××$ 103 torr
29.
12.1 L
31.
217 L
33.
8.190 $××$ 10–2 mol; 5.553 g
35.
(a) 7.24 $××$ 10–2 g; (b) 23.1 g; (c) 1.5 $××$ 10–4 g
37.
5561 L
39.
46.4 g
41.
For a gas exhibiting ideal behavior:
43.
(a) 1.85 L CCl2F2; (b) 4.66 L CH3CH2F
45.
0.644 atm
47.
The pressure decreases by a factor of 3.
49.
4.64 g L−1
51.
38.8 g
53.
72.0 g mol−1
55.
88.1 g mol−1; PF3
57.
141 atm
59.
CH4: 276 kPa; C2H6: 27 kPa; C3H8: 3.4 kPa
61.
Yes
63.
740 torr
65.
(a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of O2 produced by decomposition of this amount of HgO; and determine the volume of O2 from the moles of O2, temperature, and pressure. (b) 0.308 L
67.
(a) Determine the molar mass of CCl2F2. From the balanced equation, calculate the moles of H2 needed for the complete reaction. From the ideal gas law, convert moles of H2 into volume. (b) 3.72 $××$ 103 L
69.
(a) Balance the equation. Determine the grams of CO2 produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) 7.43 $××$ 105 L
71.
42.00 L
73.
(a) 18.0 L; (b) 0.533 atm
75.
10.57 L O2
77.
5.40 $××$ 105 L
79.
XeF4
81.
4.2 hours
83.
Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham’s law states that with a mixture of two gases A and B: $(rate Arate B)=(molar mass of Bmolar mass of A)1/2.(rate Arate B)=(molar mass of Bmolar mass of A)1/2.$ Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:
$KEA=KEB KE=12mv2KEA=KEB KE=12mv2$
Therefore, $12mAvA2=12mBvB212mAvA2=12mBvB2$
$vA2vB2=mBmAvA2vB2=mBmA$
$(vA2vB2)1/2=(mBmA)1/2(vA2vB2)1/2=(mBmA)1/2$
$vAvB=(mBmA)1/2vAvB=(mBmA)1/2$
85.
F2, N2O, Cl2, H2S
87.
1.4; 1.2
89.
51.7 cm
91.
Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average speed of all the molecules is constant at constant temperature.
93.
H2O. Cooling slows the speeds of the He atoms, causing them to behave as though they were heavier.
95.
(a) The pressure of the gas remains constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to $22$ times its initial value; urms is proportional to $KEavg.KEavg.$
97.
(a) equal; (b) less than; (c) 29.48 g mol−1; (d) 1.0966 g L−1; (e) 0.129 g/L; (f) 4.01 $××$ 105 g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min−1
99.
Gases C, E, and F
101.
The gas behavior most like an ideal gas will occur under the conditions that minimize the chances of significant interactions between the gaseous atoms/molecules, namely, low pressures (fewer atoms/molecules per unit volume) and high temperatures (greater kinetic energies of atoms/molecules make them less susceptible to attractive forces). The conditions described in (b), high temperature and low pressure, are therefore most likely to yield ideal gas behavior.
103.
SF6
105.
(a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures, they behave close enough to ideal gases that they are approximated as such; however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation. (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 8.35). (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 8.35. (e) Low temperatures | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.08%3A_Chapter_8.txt |
1.
The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold.
3.
Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one.
5.
(a) 47.6 J/°C; 11.38 cal °C−1; (b) 407 J/°C; 97.3 cal °C−1
7.
1310 J; 313 cal
9.
7.15 °C
11.
(a) 0.390 J/g °C; (b) Copper is a likely candidate.
13.
We assume that the density of water is 1.0 g/cm3(1 g/mL) and that it takes as much energy to keep the water at 85 °F as to heat it from 72 °F to 85 °F. We also assume that only the water is going to be heated. Energy required = 7.47 kWh
15.
lesser; more heat would be lost to the coffee cup and the environment and so ΔT for the water would be lesser and the calculated q would be lesser
17.
greater, since taking the calorimeter’s heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as “surroundings”: qrxn = −(qsolution + qcalorimeter); since both qsolution and qcalorimeter are negative, including the latter term (qrxn) will yield a greater value for the heat of the dissolution
19.
The temperature of the coffee will drop 1 degree.
21.
5.7 $××$ 102 kJ
23.
38.5 °C
25.
−2.2 kJ; The heat produced shows that the reaction is exothermic.
27.
1.4 kJ
29.
22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change.
31.
11.7 kJ
33.
30%
35.
0.24 g
37.
1.4 $××$ 102 Calories
39.
The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH.
41.
25 kJ mol−1
43.
81 kJ mol−1
45.
5204.4 kJ
47.
1.83 $××$ 10−2 mol
49.
–802 kJ mol−1
51.
15.5 kJ/ºC
53.
7.43 g
55.
Yes.
57.
459.6 kJ
59.
−494 kJ/mol
61.
44.01 kJ/mol
63.
−394 kJ
65.
265 kJ
67.
90.3 kJ/mol
69.
(a) −1615.0 kJ mol−1; (b) −484.3 kJ mol−1; (c) 164.2 kJ; (d) −232.1 kJ
71.
−54.04 kJ mol−1
73.
−2660 kJ mol−1
75.
–66.4 kJ
77.
−122.8 kJ
79.
3.7 kg
81.
On the assumption that the best rocket fuel is the one that gives off the most heat, B2H6 is the prime candidate.
83.
−88.2 kJ
85.
(a) $C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l);C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l);$ (b) 330 L air; (c) −104.5 kJ mol−1; (d) 75.4 °C
88.
(a) −114 kJ;
(b) 30 kJ;
(c) −1055 kJ
91.
The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.
92.
The greater bond energy is in the figure on the left. It is the more stable form.
94.
$HCl(g)⟶12H2(g)+12Cl2(g)ΔH1°=−ΔHf[HCl(g)]°12H2(g)⟶H(g)ΔH2°=ΔHf[H(g)]°12Cl2(g)⟶Cl(g)ΔH3°=ΔHf[Cl(g)]°¯HCl(g)⟶H(g)+Cl(g)ΔH298°=ΔH1°+ΔH2°+ΔH3°HCl(g)⟶12H2(g)+12Cl2(g)ΔH1°=−ΔHf[HCl(g)]°12H2(g)⟶H(g)ΔH2°=ΔHf[H(g)]°12Cl2(g)⟶Cl(g)ΔH3°=ΔHf[Cl(g)]°¯HCl(g)⟶H(g)+Cl(g)ΔH298°=ΔH1°+ΔH2°+ΔH3°$
$DHCl=ΔH298°=ΔHf[HCl(g)]°+ΔHf[H(g)]°+ΔHf[Cl(g)]°=−(−92.307kJ)+217.97kJ+121.3kJ=431.6kJDHCl=ΔH298°=ΔHf[HCl(g)]°+ΔHf[H(g)]°+ΔHf[Cl(g)]°=−(−92.307kJ)+217.97kJ+121.3kJ=431.6kJ$
96.
The S–F bond in SF4 is stronger.
98.
The C–C single bonds are longest.
100.
(a) When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. (b) The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4s electron in Ca requires more energy than removal of the 4s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. (d) In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2s electron.
102.
(d)
104.
4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy
106.
(a) Na2O; Na+ has a smaller radius than K+; (b) BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; (d) BaS; S has a larger charge
108.
(e) | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.09%3A_Chapter_9.txt |
1.
Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid.
3.
They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed.
5.
All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature.
7.
(a) Dispersion forces occur as an atom develops a temporary dipole moment when its electrons are distributed asymmetrically about the nucleus. This structure is more prevalent in large atoms such as argon or radon. A second atom can then be distorted by the appearance of the dipole in the first atom. The electrons of the second atom are attracted toward the positive end of the first atom, which sets up a dipole in the second atom. The net result is rapidly fluctuating, temporary dipoles that attract one another (e.g., Ar). (b) A dipole-dipole attraction is a force that results from an electrostatic attraction of the positive end of one polar molecule for the negative end of another polar molecule (e.g., ICI molecules attract one another by dipole-dipole interaction). (c) Hydrogen bonds form whenever a hydrogen atom is bonded to one of the more electronegative atoms, such as a fluorine, oxygen, or nitrogen atom. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom in another molecule gives rise to a strong dipole-dipole interaction called a hydrogen bond (e.g., $HF⋯HF).HF⋯HF).$
9.
The London forces typically increase as the number of electrons increase.
11.
(a) SiH4 < HCl < H2O; (b) F2 < Cl2 < Br2; (c) CH4 < C2H6 < C3H8; (d) N2 < O2 < NO
13.
Only rather small dipole-dipole interactions from C-H bonds are available to hold n-butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction, therefore, is stronger, leading to a higher boiling point.
15.
−85 °C. Water has stronger hydrogen bonds, so it melts at a higher temperature.
17.
The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O.
19.
H-bonding is the principle IMF holding the protein strands together. The H-bonding is between the $N−HN−H$ and $C=O.C=O.$
21.
(a) hydrogen bonding, dipole-dipole attraction, and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces
23.
The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water.
25.
Temperature has an effect on intermolecular forces: The higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid. The lower the temperature, the less the intermolecular forces are overcome, and so the less viscous the liquid.
27.
(a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason.
29.
1.7 $××$ 10−4 m
31.
The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise.
33.
We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids.
35.
The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases.
37.
As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures.
39.
They are equal when the pressure of gas above the liquid is exactly 1 atm.
41.
approximately 95 °C
43.
(a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water.
45.
Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids.
47.
The boiling point of CS2 is higher than that of CO2 partially because of the higher molecular weight of CS2; consequently, the attractive forces are stronger in CS2. It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO2. A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible.
49.
The thermal energy (heat) needed to evaporate the liquid is removed from the skin.
51.
1125 kJ
53.
(a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.
55.
At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values.
57.
(a) gas; (b) gas; (c) gas; (d) gas; (e) solid; (f) gas
59.
61.
Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry.
63.
(a)
(b)
(c)
(d)
(e) liquid phase (f) sublimation
65.
(e) molecular crystals
67.
Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range.
69.
(a) ionic; (b) covalent network; (c) molecular; (d) metallic; (e) covalent network; (f) molecular; (g) molecular; (h) ionic; (i) ionic
71.
X = ionic; Y = metallic; Z = covalent network
73.
(b) metallic solid
75.
The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube.
77.
eight
79.
12
81.
(a) 1.370 Å; (b) 19.26 g/cm
83.
(a) 2.176 Å; (b) 3.595 g/cm3
85.
The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12).
87.
In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS.
89.
Co3O4
91.
In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI.
93.
59.95%; The oxidation number of titanium is +4.
95.
Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si4+ for Al3+.
97.
Mn2O3
99.
1.48 Å
101.
2.874 Å
103.
20.2°
105.
1.74 $××$ 104 eV | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.10%3A_Chapter_10.txt |
1.
A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.
3.
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and $NO3−NO3−$ ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.
5.
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
7.
Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.
9.
Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.
11.
(a) Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and $NO3−NO3−$ ions. Therefore, (z) best represents the solution. (b) $Fe(NO3)3(s)⟶Fe3+(aq)+3NO3−(aq)Fe(NO3)3(s)⟶Fe3+(aq)+3NO3−(aq)$
13.
(a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)
15.
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
17.
The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.
19.
40%
21.
2.8 g
23.
2.9 atm
25.
102 L HCl
27.
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.
29.
Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.
31.
(a) Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components. (b) The mole fraction of HNO3 is 0.378. The mole fraction of H2O is 0.622.
33.
(a) $XNa2CO3=0.0119;XNa2CO3=0.0119;$ $XH2O=0.988;XH2O=0.988;$ (b) $XNH4NO3=0.0928;XNH4NO3=0.0928;$ $XH2O=0.907;XH2O=0.907;$ (c) $XCl2=0.192;XCl2=0.192;$ $XCH2CI2=0.808;XCH2CI2=0.808;$ (d) $XC5H9N=0.00426;XC5H9N=0.00426;$ $XCHCl3=0.997XCHCl3=0.997$
35.
In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
37.
(a) Determine the molar mass of HNO3. Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m
39.
(a) 6.70 $××$ 10−1 m; (b) 5.67 m; (c) 2.8 m; (d) 0.0358 m
41.
1.08 m
43.
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. (b) 100.5 °C
45.
(a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C
47.
(a) Determine the molar mass of Ca(NO3)2; determine the number of moles of Ca(NO3)2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm
49.
(a) Determine the molal concentration from the change in boiling point and Kb; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 $××$ 102 g mol−1
51.
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by ΔTf = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is ΔTf = (1.0 m)(5.14 °C/m) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized.
53.
144 g mol−1
55.
0.870 °C
57.
S8
59.
1.39 $××$ 104 g mol−1
61.
54 g
63.
100.26 °C
65.
(a) $XCH3OH=0.590;XCH3OH=0.590;$ $XC2H5OH=0.410;XC2H5OH=0.410;$ (b) Vapor pressures are: CH3OH: 55 torr; C2H5OH: 18 torr; (c) CH3OH: 0.75; C2H5OH: 0.25
67.
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C.
69.
$Δbp=Kbm=(1.20°C/m)(9.41g×1mol HgCl2271.496g0.03275kg)=1.27°CΔbp=Kbm=(1.20°C/m)(9.41g×1mol HgCl2271.496g0.03275kg)=1.27°C$
The observed change equals the theoretical change; therefore, no dissociation occurs.
71.
Colloidal System Dispersed Phase Dispersion Medium
starch dispersion starch water
smoke solid particles air
fog water air
pearl water calcium carbonate (CaCO3)
whipped cream air cream
floating soap air soap
jelly fruit juice pectin gel
milk butterfat water
ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3)
73.
Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
75.
If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.11%3A_Chapter_11.txt |
1.
A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.
3.
(a) spontaneous; (b) nonspontaneous; (c) spontaneous; (d) nonspontaneous; (e) spontaneous; (f) spontaneous
5.
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
7.
There are four initial microstates and four final microstates.
$ΔS=klnWfWi=1.38×10−23J/K×ln44=0ΔS=klnWfWi=1.38×10−23J/K×ln44=0$
9.
The probability for all the particles to be on one side is $132.132.$ This probability is noticeably lower than the $1818$ result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
11.
There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.
$ΔS=kln(WfWi)=1.38×10−23J/K×ln(41)=1.91×10−23J/KΔS=kln(WfWi)=1.38×10−23J/K×ln(41)=1.91×10−23J/K$
13.
The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I2 is a solid, Br2 is a liquid, and Cl2 is a gas.
15.
(a) C3H7OH(l) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. (b) C2H5OH(g) as it is in the gaseous state. (c) 2H(g), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
17.
(a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. (b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.
19.
$C6H6(l)+7.5O2(g)⟶3H2O(g)+6CO2(g)C6H6(l)+7.5O2(g)⟶3H2O(g)+6CO2(g)$
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.
21.
(a) 107 J/K; (b) −86.4 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K
23.
100.6 J/K
25.
(a) −198.1 J/K; (b) −348.9 J/K
27.
As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
29.
(a) 2.86 J/K; (b) 24.8 J/K; (c) −113.2 J/K; (d) −24.7 J/K; (e) 15.5 J/K; (f) 290.0 J/K
31.
The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.
33.
(a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −291.9 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −154.3 kJ spontaneous
35.
(a) The standard free energy of formation is –1124.3 kJ/mol. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.
37.
(a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.
39.
(a) 1.5 $××$ 102 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ
41.
(a) K = 41; (b) K = 0.053; (c) K = 6.9 $××$ 1013; (d) K = 1.9; (e) K = 0.04
42.
(a) 22.1 kJ;
(b) 98.9 kJ/mol
44.
90 kJ/mol
46.
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) Kp = 0.031; (c) The evaporation of water is spontaneous; (d) $PH2OPH2O$ must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
48.
(a) Nonspontaneous as $ΔG°>0;ΔG°>0;$ (b) $ΔG°=ΔG°+RTlnQ,ΔG°=ΔG°+RTlnQ,$ $ΔG=1.7×103+(8.314×310×ln28120)=−2.1 kJ.ΔG=1.7×103+(8.314×310×ln28120)=−2.1 kJ.$ The forward reaction to produce F6P is spontaneous under these conditions.
50.
ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
52.
(a) Increasing $PO2PO2$ will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(b) Increasing $PO2PO2$ will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(c) Increasing $PO2PO2$ will shift the equilibrium the reactants, which decreases the value of K. ΔG° therefore becomes more positive. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.12%3A_Chapter_12.txt |
1.
The reaction can proceed in both the forward and reverse directions.
3.
When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates.
5.
Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.
7.
Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.
9.
(a) Kc = [Ag+][Cl] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) $Kc=1[Pb2+][Cl−]2Kc=1[Pb2+][Cl−]2$ > 1 because PbCl2 is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).
11.
Since $Kc=[C6H6][C2H2]3,Kc=[C6H6][C2H2]3,$ a value of Kc ≈ 10 means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
13.
Kc > 1
15.
(a) $Qc=[CH3Cl][HCl][CH4][Cl2];Qc=[CH3Cl][HCl][CH4][Cl2];$ (b) $Qc=[NO]2[N2][O2];Qc=[NO]2[N2][O2];$ (c) $Qc=[SO3]2[SO2]2[O2];Qc=[SO3]2[SO2]2[O2];$ (d) Qc = [SO2]; (e) $Qc=1[P4][O2]5;Qc=1[P4][O2]5;$ (f) $Qc=[Br]2[Br2];Qc=[Br]2[Br2];$ (g) $Qc=[CO2][CH4][O2]2;Qc=[CO2][CH4][O2]2;$ (h) Qc = [H2O]5
17.
(a) Qc 25 proceeds left; (b) QP 0.22 proceeds right; (c) Qc undefined proceeds left; (d) QP 1.00 proceeds right; (e) QP 0 proceeds right; (f) Qc 4 proceeds left
19.
The system will shift toward the reactants to reach equilibrium.
21.
(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous
23.
This situation occurs in (a) and (b).
25.
(a) KP = 1.6 $××$ 10−4; (b) KP = 50.2; (c) Kc = 5.34 $××$ 10−39; (d) Kc = 4.60 $××$ 10−3
27.
$K P = P H 2 O = 0.042 . K P = P H 2 O = 0.042 .$
29.
$Q c = [ NH 4 + ] [ OH − ] [ NH 3 ] Q c = [ NH 4 + ] [ OH − ] [ NH 3 ]$
31.
The amount of CaCO3 must be so small that $PCO2PCO2$ is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full $PCO2PCO2$ required for equilibrium.
33.
The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.
34.
No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
36.
Add N2; add H2; decrease the container volume; heat the mixture.
38.
(a) T increase = shift right, V decrease = shift left; (b) T increase = shift right, V = no effect; (c) T increase = shift left, V decrease = shift left; (d) T increase = shift left, V decrease = shift right.
40.
(a) $Kc=[CH3OH][H2]2[CO];Kc=[CH3OH][H2]2[CO];$ (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases
42.
(a) $Kc=[CO][H2][H2O];Kc=[CO][H2][H2O];$ (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.
44.
Only (b)
46.
Add NaCl or some other salt that produces Cl to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).
48.
Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive Ag+ ions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive Ag+.
50.
$Kc=[C]2[A][B]2.Kc=[C]2[A][B]2.$ [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.
52.
Kc = 6.00 $××$ 10−2
54.
Kc = 0.50
56.
KP = 1.9 $××$ 103
58.
KP = 3.06
60.
(a) −2x, +2x; (b) $+43x+43x$, $–23x–23x$, −2x; (c) −2x, 3x; (d) +x, −x, −3x; (e) +x; (f) $–14x–14x$.
62.
Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.
64.
[NH3] = 9.1 $××$ 10−2 M
66.
PBrCl = 4.9 $××$ 10−2 atm
68.
[CO] = 2.04 $××$ 10−4 M
70.
$P H 2 O = 3.64 × 10 −3 atm P H 2 O = 3.64 × 10 −3 atm$
72.
Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium.
74.
(a) [NO2] = 1.17 $××$ 10−3 M
[N2O4] = 0.128 M
(b) The assumption that x is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the N2O4 to its concentration at equilibrium (they differ by just 1 in the least significant digit’s place).
76.
(a) [H2S] = 0.810 atm
[H2] = 0.014 atm
[S2] = 0.0072 atm
(b) The assumption that 2x is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the H2S to its concentration at equilibrium (0.824 atm versus 0.810 atm, a difference of less than 2%).
78.
507 g
80.
330 g
83.
(a) 0.33 mol.
(b) [CO]2 = 0.50 M. Added H2 forms some water as a result of a shift to the left after H2 is added.
85.
(a) $Kc=[NH3]4[O2]7[NO2]4[H2O]6.Kc=[NH3]4[O2]7[NO2]4[H2O]6.$ (b) [NH3] must increase for Qc to reach Kc. (c) The increase in system volume would lower the partial pressures of all reactants (including NO2). (d) $PO2=49torrPO2=49torr$
87.
$P N 2 O 3 = 1.90 atm and P NO = P NO 2 = 1.90 atm P N 2 O 3 = 1.90 atm and P NO = P NO 2 = 1.90 atm$
90.
In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°.
(a) K = 1.07 × 10–13;
(b) K = 2.42 × 10−3;
(c) K = 2.73 × 104;
(d) K = 0.229;
(e) K = 16.1
92.
The standard free energy change is $ΔG298°=−RTlnK=4.84 kJ/mol.ΔG298°=−RTlnK=4.84 kJ/mol.$ When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and $ΔG298ΔG298$ becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.
94.
The reaction will be spontaneous at temperatures greater than 287 K.
96.
K = 5.35 $××$ 1015
The process is exothermic.
98.
1.0 $××$ 10−8 atm. This is the maximum pressure of the gases under the stated conditions.
100.
$x = 1.29 × 10 − 5 atm = P O 2 x = 1.29 × 10 − 5 atm = P O 2$
102.
−0.16 kJ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.13%3A_Chapter_13.txt |
1.
One example for NH3 as a conjugate acid: $NH2−+H+⟶NH3;NH2−+H+⟶NH3;$ as a conjugate base: $NH4+(aq)+OH−(aq)⟶NH3(aq)+H2O(l)NH4+(aq)+OH−(aq)⟶NH3(aq)+H2O(l)$
3.
(a) $H3O+(aq)⟶H+(aq)+H2O(l);H3O+(aq)⟶H+(aq)+H2O(l);$ (b) $HCl(aq)⟶H+(aq)+Cl−(aq);HCl(aq)⟶H+(aq)+Cl−(aq);$ (c) $NH3(aq)⟶H+(aq)+NH2−(aq);NH3(aq)⟶H+(aq)+NH2−(aq);$ (d) $CH3CO2H(aq)⟶H+(aq)+CH3CO2−(aq);CH3CO2H(aq)⟶H+(aq)+CH3CO2−(aq);$ (e) $NH4+(aq)⟶H+(aq)+NH3(aq);NH4+(aq)⟶H+(aq)+NH3(aq);$ (f) $HSO4−(aq)⟶H+(aq)+SO42−(aq)HSO4−(aq)⟶H+(aq)+SO42−(aq)$
5.
(a) $H2O(l)+H+(aq)⟶H3O+(aq);H2O(l)+H+(aq)⟶H3O+(aq);$ (b) $OH−(aq)+H+(aq)⟶H2O(l);OH−(aq)+H+(aq)⟶H2O(l);$ (c) $NH3(aq)+H+(aq)⟶NH4+(aq);NH3(aq)+H+(aq)⟶NH4+(aq);$ (d) $CN−(aq)+H+(aq)⟶HCN(aq);CN−(aq)+H+(aq)⟶HCN(aq);$ (e) $S2−(aq)+H+(aq)⟶HS−(aq);S2−(aq)+H+(aq)⟶HS−(aq);$ (f) $H2PO4−(aq)+H+(aq)⟶H3PO4(aq)H2PO4−(aq)+H+(aq)⟶H3PO4(aq)$
7.
(a) H2O, O2−; (b) H3O+, OH; (c) H2CO3, $CO32−;CO32−;$ (d) $NH4+,NH4+,$ $NH2−;NH2−;$ (e) H2SO4, $SO42−;SO42−;$ (f) $H3O2+,H3O2+,$ $HO2−;HO2−;$ (g) H2S; S2−; (h) $H6N22+,H6N22+,$ H4N2
9.
The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO3(BA), H2O(BB), H3O+(CA), $NO3−(CB);NO3−(CB);$ (b) CN(BB), H2O(BA), HCN(CA), OH(CB); (c) H2SO4(BA), Cl(BB), HCl(CA), $HSO4−(CB);HSO4−(CB);$ (d) $HSO4−(BA),HSO4−(BA),$ OH(BB), $SO42−SO42−$(CB), H2O(CA); (e) O2−(BB), H2O(BA) OH(CB and CA); (f) [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); (g) H2S(BA), $NH2−(BB),NH2−(BB),$ HS(CB), NH3(CA)
11.
Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O. As an acid: $H2O(aq)+NH3(aq)⇌NH4+(aq)+OH−(aq).H2O(aq)+NH3(aq)⇌NH4+(aq)+OH−(aq).$ As a base: $H2O(aq)+HCl(aq)⇌H3O+(aq)+Cl−(aq)H2O(aq)+HCl(aq)⇌H3O+(aq)+Cl−(aq)$
13.
amphiprotic: (a) $NH3+H3O+⟶NH4OH+H2O,NH3+H3O+⟶NH4OH+H2O,$ $NH3+OCH3−⟶NH2−+CH3OH;NH3+OCH3−⟶NH2−+CH3OH;$ (b) $HPO42−+OH−⟶PO43−+H2O,HPO42−+OH−⟶PO43−+H2O,$ $HPO42−+HClO4⟶H2PO4−+ClO4−;HPO42−+HClO4⟶H2PO4−+ClO4−;$ not amphiprotic: (c) Br; (d) $NH4+;NH4+;$ (e) $AsO43−AsO43−$
15.
In a neutral solution [H3O+] = [OH]. At 40 °C, [H3O+] = [OH] = (2.910 × 10−14)1/2 = 1.7 $××$ 10−7.
17.
x = 3.051 $××$ 10−7 M = [H3O+] = [OH]; pH = −log3.051 $××$ 10−7 = −(−6.5156) = 6.5156; pOH = pH = 6.5156
19.
(a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4
21.
[H3O+] = 3.0 $××$ 10−7 M; [OH] = 3.3 $××$ 10−8 M
23.
[H3O+] = 1 $××$ 10−2 M; [OH] = 1 $××$ 10−12 M
25.
[OH] = 3.1 $××$ 10−12 M
27.
The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH, which causes the solution to be basic.
29.
[H2O] > [CH3CO2H] > $[H3O+][H3O+]$[CH3CO2−][CH3CO2−]$ > [OH]
31.
The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.
33.
$Mg ( OH ) 2 ( s ) + 2HCl ( a q ) ⟶ Mg 2+ ( a q ) + 2 Cl − ( a q ) + 2H 2 O ( l ) BB BA CB CA Mg ( OH ) 2 ( s ) + 2HCl ( a q ) ⟶ Mg 2+ ( a q ) + 2 Cl − ( a q ) + 2H 2 O ( l ) BB BA CB CA$
35.
$Ka=2.3×10−11Ka=2.3×10−11$
37.
The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH and $CH3NH3+.CH3NH3+.$
39.
triethylamine
41.
(a) $HSO4−;HSO4−;$ higher electronegativity of the central ion. (b) H2O; NH3 is a base and water is neutral, or decide on the basis of Ka values. (c) HI; PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 is weaker than HI. (d) PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
43.
(a) NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) $ClO2− the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) $HTe− $PH2−PH2−$ and $NH2−NH2−$ are anions of weak bases, so they act as strong bases toward H+. $HTe−HTe−$ and HS are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) $BrO4− with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
45.
$[ H 2 O ] > [ C 6 H 4 OH ( CO 2 H ) ] > [H + ] 0 > [C 6 H 4 OH ( CO 2 ) − ] ≫ [ C 6 H 4 O ( CO 2 H ) − ] > [ OH − ] [ H 2 O ] > [ C 6 H 4 OH ( CO 2 H ) ] > [H + ] 0 > [C 6 H 4 OH ( CO 2 ) − ] ≫ [ C 6 H 4 O ( CO 2 H ) − ] > [ OH − ]$
47.
1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.
48.
(b) The addition of HCl
50.
(a) Adding HCl will add H3O+ ions, which will then react with the OH ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of $NO2−NO2−$ ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of $NO2−NO2−$ ions and decreasing the concentration of OH ions. (c) Adding NaOH adds OH ions, which shifts the equilibrium to the left, increasing the concentration of $NO2−NO2−$ ions and decreasing the concentrations of HNO2. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO2 adds $NO2−NO2−$ ions and shifts the equilibrium to the right, increasing the HNO2 and OH ion concentrations.
52.
This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.
54.
(a) $Kb=1.8×10−5;Kb=1.8×10−5;$ (b) $Ka=4.5×10−4;Ka=4.5×10−4;$ (c) $Kb=6.4×10−5;Kb=6.4×10−5;$ (d) $Ka=5.6×10−10Ka=5.6×10−10$
56.
$K a = 1.2 × 10 −2 K a = 1.2 × 10 −2$
58.
(a) $Kb=4.3×10−12Kb=4.3×10−12$ (b) $Ka=1.6×1010Ka=1.6×1010$ (c) $Kb=5.9×108Kb=5.9×108$ (d) $Kb=4.2×10−3Kb=4.2×10−3$ (e) $Kb=2.3×105Kb=2.3×105$ (f) $Kb=6.3×10−13Kb=6.3×10−13$
60.
(a) $[H3O+] [ClO−][HClO]= (x)(x)(0.0092−x) ≈(x)(x)0.0092=2.9×10−8[H3O+] [ClO−][HClO]= (x)(x)(0.0092−x) ≈(x)(x)0.0092=2.9×10−8$
Solving for x gives 1.63 $××$ 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [ClO] = 1.6 $××$ 10−5 M
[HClO] = 0.0092 M
[OH] = 6.1 $××$ 10−10 M;
(b) $[C6 H5NH3 +][OH−][C6H5NH2]=(x)(x)(0.0784−x)≈ (x)(x)0.0784=4.3×10−10[C6 H5NH3 +][OH−][C6H5NH2]=(x)(x)(0.0784−x)≈ (x)(x)0.0784=4.3×10−10$
Solving for x gives 5.81 $××$ 10−6 M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$[C6H5 NH3+][C6H5 NH3+]$ = [OH] = 5.8 $××$ 10−6 M
[C6H5NH2] = 0.0784 M
[H3O+] = 1.7$××$ 10−9 M;
(c) $[H3O+][CN−][HCN]=(x)(x)(0.0810−x)≈(x)(x)0.0810=4.9×10−10[H3O+][CN−][HCN]=(x)(x)(0.0810−x)≈(x)(x)0.0810=4.9×10−10$
Solving for x gives 6.30 $××$ 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [CN] = 6.3 $××$ 10−6 M
[HCN] = 0.0810 M
[OH] = 1.6 $××$ 10−9 M;
(d) $[(CH3)3NH+][OH−][(CH3)3N]=(x)(x)(0.11−x)≈(x)(x)0.11=6.3×10−5[(CH3)3NH+][OH−][(CH3)3N]=(x)(x)(0.11−x)≈(x)(x)0.11=6.3×10−5$
Solving for x gives 2.63 $××$ 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH3)3NH+] = [OH] = 2.6 $××$ 10−3 M
[(CH3)3N] = 0.11 M
[H3O+] = 3.8 $××$ 10−12 M;
(e) $[Fe(H2O)5(OH)+][H3O+][Fe(H2O)62+]=(x)(x)(0.120−x)≈(x)(x)0.120=1.6×10−7[Fe(H2O)5(OH)+][H3O+][Fe(H2O)62+]=(x)(x)(0.120−x)≈(x)(x)0.120=1.6×10−7$
Solving for x gives 1.39 $××$ 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H2O)5(OH)+] = [H3O+] = 1.4 $××$ 10−4 M
$[Fe(H2O)62+][Fe(H2O)62+]$ = 0.120 M
[OH] = 7.2 $××$ 10−11 M
62.
pH = 2.41
64.
[C10H14N2] = 0.049 M; [C10H14N2H+] = 1.9 $××$ 10−4 M; $[C10H14N2H22+][C10H14N2H22+]$ = 1.4 $××$ 10−11 M; [OH] = 1.9 $××$ 10−4 M; [H3O+] = 5.3 $××$ 10−11 M
66.
$K a = 1.2 × 10 −2 K a = 1.2 × 10 −2$
68.
$K b = 1.77 × 10 −5 K b = 1.77 × 10 −5$
70.
(a) acidic; (b) basic; (c) acidic; (d) neutral
72.
[H3O+] and $[HCO3−][HCO3−]$ are practically equal
74.
[C6H4(CO2H)2] 7.2 $××$ 10−3 M, [C6H4(CO2H)(CO2)] = [H3O+] 2.8 $××$ 10−3 M, $[C6H4(CO2)22−][C6H4(CO2)22−]$3.9 $××$ 10−6 M, [OH] 3.6 $××$ 10−12 M
76.
(a) $Ka2=1.5×10−11;Ka2=1.5×10−11;$
(b) $Kb=4.3×10−12;Kb=4.3×10−12;$
(c) $[Te2−][H3O+][HTe−]=(x)(0.0141+x)(0.0141−x)≈(x)(0.0141)0.0141=1.5×10−11[Te2−][H3O+][HTe−]=(x)(0.0141+x)(0.0141−x)≈(x)(0.0141)0.0141=1.5×10−11$
Solving for x gives 1.5 $××$ 10−11 M. Therefore, compared with 0.014 M, this value is negligible (1.1 $××$ 10−7%).
78.
Excess H3O+ is removed primarily by the reaction: $H3O+(aq)+H2PO4−(aq)⟶H3PO4(aq)+H2O(l)H3O+(aq)+H2PO4−(aq)⟶H3PO4(aq)+H2O(l)$
Excess base is removed by the reaction: $OH−(aq)+H3PO4(aq)⟶H2PO4−(aq)+H2O(l)OH−(aq)+H3PO4(aq)⟶H2PO4−(aq)+H2O(l)$
80.
[H3O+] = 1.5 $××$ 10−4 M
82.
[OH] = 4.2 $××$ 10−4 M
84.
(a) The added HCl will increase the concentration of H3O+ slightly, which will react with $CH3CO2−CH3CO2−$ and produce CH3CO2H in the process. Thus, $[CH3CO2−][CH3CO2−]$ decreases and [CH3CO2H] increases. (b) The added KCH3CO2 will increase the concentration of $[CH3CO2−][CH3CO2−]$ which will react with H3O+ and produce CH3CO2 H in the process. Thus, [H3O+] decreases slightly and [CH3CO2H] increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce OH ions, which will react with the H3O+, thus reducing [H3O+]. Some additional CH3CO2H will dissociate, producing $[CH3CO2−][CH3CO2−]$ ions in the process. Thus, [CH3CO2H] decreases slightly and $[CH3CO2−][CH3CO2−]$ increases. (e) The added CH3CO2H will increase its concentration, causing more of it to dissociate and producing more $[CH3CO2−][CH3CO2−]$ and H3O+ in the process. Thus, [H3O+] increases slightly and $[CH3CO2−][CH3CO2−]$ increases.
86.
pH = 8.95
88.
37 g (0.27 mol)
90.
(a) pH = 5.222; (b) The solution is acidic. (c) pH = 5.220
92.
At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
94.
(a) pH = 2.50; (b) pH = 4.01; (c) pH = 5.60; (d) pH = 8.35; (e) pH = 11.08 | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_2e_(OpenStax)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.14%3A_Chapter_14.txt |
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