Chapter 6, Problem 6.1CP

Interpretation Introduction

Interpretation: The information updated from the independent nature of ${\text{K}}_{\text{p}}$ on T for a chemical reaction needs to be explained.

Concept Introduction: A reversible chemical reaction can move in either forward or backward direction. The stage of a reversible reaction at which the rate of forward and backward reaction become equal to each other is called an equilibrium stage. The equilibrium constant for a reaction is the ratio of the equilibrium concentration of product and reactant. If the reaction is given in the gaseous state, then the equilibrium pressure must be considered. The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ can be written as:

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{=K}}_{\text{c}}{\text{(R×T)}}^{\text{Δn}}\\ \begin{array}{l}\text{Here:}\hfill \\ {\text{K}}_{\text{p}}{\text{and K}}_{\text{c}}\text{= equilibrium constant}\hfill \\ \text{R = Gas constant}\hfill \\ \text{T = temperature}\hfill \\ \Delta \text{n=number of gaseous product – number of gaseous reactants}\hfill \end{array}\end{array}$

Answer to Problem 6.1CP

  ${\text{ln K}}_{\text{p}}\text{= -}\frac{\text{ΔH° }}{\text{RT}}\text{+}\frac{\text{ΔS°}}{\text{R}}$

Hence for the reaction $\text{ΔH°}$ = 0 ; ${\text{K}}_{\text{p}}$ is independent on T for that chemical reaction.

Explanation of Solution

Given information: ${\text{K}}_{\text{p}}$ is independent on T for a chemical reaction.

  ${\text{ΔG° = - RTlnK}}_{\text{p}}$

Also,

  $\text{ΔG° = ΔH° -TΔS° }$

Thus,

  ${\text{ΔH° -TΔS° = - RT ln K}}_{\text{p}}$

Or,

  ${\text{ln K}}_{\text{p}}\text{= -}\frac{\text{ΔH° }}{\text{RT}}\text{+}\frac{\text{ΔS°}}{\text{R}}$

Hence for the reaction $\text{ΔH°}$ = 0 ; ${\text{K}}_{\text{p}}$ is independent on T for that chemical reaction.

Conclusion

Thus, the value of $\text{ΔH°}$ = 0 for the reaction.