Chapter 6, Problem 6.35NP

(a)

Interpretation Introduction

Interpretation:

The following given equation needs to be proved:

  $\ln K_P(T)=\ln K_P(T_0)-\frac{\Delta H_R^0}{R}\left(\frac{1}{T}-\frac{1}{T_0}\right)+\frac{T_0\times \Delta C_P}{R}\left(\frac{1}{T}-\frac{1}{T_0}\right)+\frac{\Delta C_P}{R}\ln \frac{T}{T_0}$

Concept Introduction :

The relation between equilibrium constant and enthalpy change of the reaction is represented as follows:

  ${\displaystyle \underset{K_P(T_2)}{\overset{K_P(T_1)}{\int }}d\ln K_P}=\frac{1}{R}{\displaystyle \underset{T_2}{\overset{T_1}{\int }}\frac{\Delta H_R^0}{T^2}dT}$

  $\begin{array}{l}{K}_P\text{ - equilibrium constant}\\ \Delta H_R^0\text{ - enthalpy change of the reaction}\\ \text{T - temperature}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium pressure of oxygen over copper and CuO(s) at 1275 K should be calculated. The value of K_P value should be determined.

Concept Introduction :

When gases involve in a reaction, partial pressures are used instead of concentrations in the equilibrium expression.

  $\begin{array}{l}A(g)+B(g)\to C(c)+D(d)\\ K_p=\frac{P_C^c\times P_D^d}{P_A^a\times P_B^b}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of equilibrium pressure if $\Delta H_R^0$ were constant at 298.15 K up to 1275 K should be calculated.

Concept Introduction :

When gases involve in a reaction, partial pressures are used instead of concentrations in the equilibrium expression.

  $\begin{array}{l}A(g)+B(g)\to C(c)+D(d)\\ K_p=\frac{P_C^c\times P_D^d}{P_A^a\times P_B^b}\end{array}$