Chapter 3, Problem 6P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Calculating and K_eq for Coupled Reactions For the process

A $\rightleftharpoons$B. K_eq (AB) is 0.02 at 37⁰C. For the process B $\rightleftharpoons$C.

K_eq (BC)=1000 at 37⁰C.

1. Determine K_eq (AC), the equilibrium constant for the overall process A $\rightleftharpoons$ C, from K_eq((AB) and (BC).

Determine standard-state free energy changes for all three processes, and use $\Delta G^{\circ }$.

(AC) to determine K_eq (AC). Make sure that ibis value agrees with that determined m part a of this problem.

Interpretation Introduction

(a)

To determine:

K_eq(AC) for the overall process, from K_eq(AB) and K_eq(BC).

Introduction:

  ${\text{K}}_{\text{eq}}\text{=}\frac{\left[\text{Products}\right]}{\left[\text{Reactants}\right]}$

Where K_eqis equilibrium constant, square brackets represents concentration.

Explanation of Solution

  $\begin{array}{l}{K}_{eq}(AB)=\frac{\left[B\right]}{\left[A\right]}\to 1\\ K_{eq}(BC)=\frac{\left[C\right]}{\left[B\right]}\to 2\end{array}$

  $K_{eq}(AC)=\frac{\left[C\right]}{\left[A\right]}\to 3$

When equation 1 multiplied by equation 2

  $K_{eq}(AB)\times K_{eq}(BC)=\frac{\left[B\right]}{\left[A\right]}\times \frac{\left[C\right]}{\left[B\right]}=\frac{\left[C\right]}{\left[A\right]}$

Then we can substitute equation 3 here.

  $K_{eq}(AB)\times K_{eq}(BC)=K_{eq}(AC)$

  $\begin{array}{l}{K}_{eq}(AC)=0.02\times 1000\\ K_{eq}(AC)=20\end{array}$

Interpretation Introduction

(b)

To determine:

Standard-state free energy change for all three processes. Then using $\Delta G^0(AC)$, K_eq(AC) should be determined and should be compared with part (a).

Introduction:

  $\Delta G^0=-RT\ln K_{eq}$

Where K_eqis equilibrium constant, $\Delta G^0$is change in Gibbs free energy at standard conditions, R is Gas constant, T is temperature.

Explanation of Solution

  $\begin{array}{l}\Delta G^0(AB)=-RT\ln K_{eq}(AB)\\ \Delta G^0(AB)=-8.314\times {10}^{-3}\text{ kJ/mol}\text{.K }\times \text{ 310 K }\times \text{ ln (0}\text{.02)}\\ \Delta G^0(AB)=10.1\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0(BC)=-RT\ln K_{eq}(BC)\\ \Delta G^0(BC)=-8.314\times {10}^{-3}\text{ kJ/mol}\text{.K }\times \text{ 310 K }\times \text{ ln (1000)}\\ \Delta G^0(BC)=-17.8\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0(AC)=\Delta G^0(AB)+\Delta G^0(BC)\\ \Delta G^0(AC)=\left(10.1-17.8\right)\text{ kJ/mol}\\ \Delta G^0(AC)=-7.7\text{ kJ/mol}\end{array}$

  $\begin{array}{l}\Delta G^0(AC)=-RT\ln K_{eq}(AC)\\ -7.7\text{ kJ/mol}=-8.314\times {10}^{-3}\text{ kJ/mol}\text{.K }\times \text{ 310 K }\times \text{ ln }{K}_{eq}(AC)\\ \text{ln }{K}_{eq}(AC)=\frac{-7.7\text{ kJ/mol}}{-8.314\times {10}^{-3}\text{ kJ/mol}\text{.K }\times \text{ 310 K}}\\ \text{ln }{K}_{eq}(AC)=2.99\\ K_{eq}(AC)=e^{2.99}\\ K_{eq}(AC)=19.88\simeq 20\end{array}$

Thus,

  $\Delta G^0(AB)=10.1\text{ kJ/mol}$

  $\Delta G^0(BC)=-17.8\text{ kJ/mol}$

  $\Delta G^0(AC)=-7.7\text{ kJ/mol}$

  $K_{eq}(AC)=19.88\simeq 20$