Chapter 18, Problem 45QP

(a)

Calculate $\Delta G°$ and $K_{\text{p}}$ for the following equilibrium reaction at $\text{25°C}$:

(b)

Calculate $\Delta G$ for the reaction if the partial pressures of the initial mixture are $P_{{\text{PCl}}_{\text{3}}}\text{ = 0}\text{.0029 atm, }{P}_{{\text{PCl}}_{\text{3}}}\text{ = 0}\text{.27 atm, and }{P}_{{\text{cl}}_{\text{2}}}\text{ = 0}\text{.40 atm}\text{.}$

Interpretation Introduction

Interpretation:

The standard free energy change and the equilibrium constant, at a constant pressure ${\text{(K}}_{\text{P}})$ and at a temperature of ${\text{25}}^{\circ }\text{C}$, for the given equilibrium reaction are to be determined.

Concept introduction:

For a general chemical reaction, $\text{aA + bB }\underset{}{\overset{}{\rightleftarrows }}\text{cC+dD}$, the reaction quotient will be represented as

$Q=\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

The reaction quotient in terms of partial pressure can be represented as $Q_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

The general formula for writing equilibrium expression for the reaction is given as follows:

$K_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

The standard free energy change of the reaction is the difference between the sums of the standard free energy change of the products and the standard free energy change of the reactants.

$\Delta {\text{G}}_{reaction}^{\circ }\text{=}{\displaystyle \sum \Delta {\text{G}}_{product}^{\circ }}-{\displaystyle \sum \Delta {\text{G}}_{reac\tan t}^{\circ }}$

The relation between free energy change and standard free energy change is as follows:

$\Delta G=\Delta G^{\circ }+RT\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$ is the gas constant, and $T$ is the temperature.

Answer to Problem 45QP

Solution:

a)

$\Delta G_{rxn}^o=35.4\text{kJ/mol}$

${\text{K}}_{\text{P}}=6.2\times {10}^{-7}$

b)

$\Delta G=\text{44}\text{.6 kJ/mol}$

Explanation of Solution

a) Value of $\Delta {\text{G}}^{\text{o}}$ and ${\text{K}}_{\text{P}}$ for equilibrium reaction ${\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_3{\text{(g) + Cl}}_2\text{(g)}$

The given reaction, for which $\Delta {\text{G}}^{\text{o}}$ is to be calculated, is as follows:

${\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_3{\text{(g) + Cl}}_2\text{(g)}$

The standard free energy changes for the formation of ${\text{PCl}}_{\text{5}}\text{(g)}$, ${\text{PCl}}_3\text{(g)}$, and ${\text{Cl}}_2\text{(g)}$ are as follows:

$\Delta {\text{G}}_f^{\circ }[{\text{PCl}}_{\text{5}}\text{(g)}]=-305.0kJ/mol$

$\Delta {\text{G}}_f^{\circ }[{\text{PCl}}_3\text{(g)}]=-269.6kJ/mol$

$\Delta {\text{G}}_f^{\circ }[C{l}_2(\text{g})]=0.00k\text{J/m}ol$

The Gibbs free energy of the reaction is calculated as

$\Delta {\text{G}}_{reaction}^{\circ }\text{=}{\displaystyle \sum \Delta {\text{G}}_{product}^{\circ }}-{\displaystyle \sum \Delta {\text{G}}_{reac\tan t}^{\circ }}$

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}=\Delta {\text{G}}_f^{\circ }[{\text{PCl}}_3\text{(g)}]\text{+}{\text{ΔG}}_{\text{f}}^{\text{o}}{\text{[Cl}}_{\text{2}}\text{(g)]}-\Delta {\text{G}}_f^{\circ }[{\text{PCl}}_5\text{(g)}]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=\text{(}-269.6kJ/mol\text{)}\text{+}\text{(1)(0)}-\text{(}-305.0kJ/mol\text{)}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}=35.4\text{kJ/mol}\end{array}$

The reaction quotient for the above reaction is represented as

$Q_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

${\text{Q}}_{\text{P}}\text{=}\frac{{\text{P}}_{PC{l}_3}{\text{P}}_{C{l}_2}}{{\text{P}}_{PC{l}_5}}$

The relation between free energy change and standard free energy change is as

$\Delta G=\Delta G^{\circ }+\text{R }T\ln Q$

The value of $\Delta \text{G}$ for a reaction at equilibrium is taken to be zero.

So, the above expression is reduced to the following expression:

$\begin{array}{c}0=\Delta G^{\circ }+\text{R}T\ln Q\\ \Delta G^{\circ }=\text{R }T\ln Q\end{array}$

Substitute all the values in the above expression

$\begin{array}{c}\Delta G^o\text{=}-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K×}\text{ln}\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}\\ 35.4\text{kJ/mol = }\left(-\text{0}\text{.008314}\text{kJ/mol}\text{.K}\text{×}\text{298}\text{.15}\text{K×}\ln {\text{K}}_{\text{P}}\right)\\ \ln {\text{K}}_{\text{P}}=\left(\frac{-35.4\cancel{\text{kJ/mol}}}{\text{0}\text{.008314}\cancel{\text{kJ/mol}\text{.K}}\text{×}\text{298}\text{.15}\cancel{\text{K}}}\right)\\ {\text{K}}_{\text{P}}=e^{\left(\frac{-35.4}{\text{0}\text{.008314}\text{×}\text{298}\text{.15}}\right)}\end{array}$

Again, solving further, we get

$\begin{array}{l}{\text{K}}_{\text{P}}=e^{-14.2809}\\ {\text{K}}_{\text{P}}=6.2\times {10}^{-7}\end{array}$

Thus, the $\Delta G^o$ value for the given equilibrium reaction at a temperature of ${\text{25}}^{\circ }\text{C}$ is $35.4\text{kJ/mol}$, and the equilibrium constant at a constant pressure ${\text{(K}}_{\text{P}})$ for the given equilibrium reaction at a temperature of ${\text{25}}^{\circ }\text{C}$ is $6.2\times {10}^{-7}$.

b) Value of $\Delta {\text{G}}^{\text{o}}$ if the partial pressures of initial mixture ${\text{PCl}}_{\text{5}}\text{(g)}$, ${\text{PCl}}_3\text{(g)}$, and ${\text{Cl}}_2\text{(g)}$ are $\text{0}\text{.0029}\text{atm}$, $\text{0}\text{.27}\text{atm}$, and $\text{0}\text{.40}\text{atm}$, respectively.

The given equilibrium reaction is as follows:

${\text{PCl}}_{\text{5}}\text{(g)}\rightleftharpoons {\text{PCl}}_3{\text{(g) + Cl}}_2\text{(g)}$

The reaction quotient for the above reaction is represented as

$Q_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

${\text{Q}}_{\text{P}}\text{=}\frac{{\text{P}}_{PC{l}_3}{\text{P}}_{C{l}_2}}{{\text{P}}_{PC{l}_5}}$

The partial pressures of ${\text{PCl}}_{\text{5}}\text{(g)}$, ${\text{PCl}}_3\text{(g)}$, and ${\text{Cl}}_2\text{(g)}$ are $\text{0}\text{.0029}\text{atm}$, $\text{0}\text{.27}\text{atm}$, and $\text{0}\text{.40}\text{atm}$, respectively.

Substitute all the values of partial pressure in the above expression

$\begin{array}{c}{\text{Q}}_{\text{P}}\text{=}\frac{\text{(0}\text{.27)(0}\text{.40)}}{\text{(0}\text{.0029)}}\\ \text{=}\text{37}\end{array}$

The relation between free energy change and standard free energy change is as follows:

$\begin{array}{l}\Delta G=\Delta G^o\text{+}\text{R }T\ln Q_P\\ \Delta G=\text{ (35}\text{.4 kJ/mol) + (0}\text{.008314 kJ/K}\text{.mol)}\left(\text{298 K}\right)\text{ln}\left(\text{37}\right)\\ \Delta G=\text{ 44}\text{.6 kJ/mol}\end{array}$

Thus, the Gibbs free energy for the given equilibrium reaction at a temperature of ${\text{25}}^{\circ }\text{C}$ is $\text{44}\text{.6 kJ/mol}$.