Chapter 6, Problem 6.6NP

(a)

Interpretation Introduction

Interpretation:

The value of K_p at $1000\text{ K}$ should be calculated.

Concept Introduction :

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

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

(b)

Interpretation Introduction

Interpretation:

The value of K_P at $298.15\text{ K}$ should be calculated.

Concept Introduction :

Clausius-Clapeyron equation is represented as follows:

  $\ln \frac{K_{P,2}}{K_{P,1}}=-\frac{\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Here,

  $\begin{array}{l}{K}_P\text{ - equilibrium constant}\\ \Delta H\text{ - enthalpy change}\\ \text{R - universal gas constant}\\ \text{T - temperature}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $\Delta G_R^0$ at $298.15\text{ K}$ should be calculated.

Concept Introduction :

The Gibbs free energy change can be calculated as follows:

  $\Delta G_R^0=-RT\ln K$

Here,

  $\begin{array}{l}\Delta G_R^0\text{ - Gibbs free energy change}\\ \text{R - universal gas constant}\\ \text{T - temperature}\\ \text{K - equilibrium constant}\end{array}$