Chapter 6, Problem 6.4NP

(a)

Interpretation Introduction

Interpretation:

The value of K_p at $700\text{ K and 800 K}$ should be calculated.

Concept Introduction :

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

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

Here, P is partial pressure of respective components at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta G_R^0\text{ and }\Delta H_R^0$ for this reaction at $298.15\text{ K}$ should be calculated using only the data in the problem.

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 $\Delta G_R^0\text{ and }\Delta H_R^0$ for this reaction at $298.15\text{ K}$ should be calculated using the data tables.

Concept Introduction :

The relation between Gibbs free energy change, enthalpy change and entropy change is represented as follows:

  $\Delta G^0=\Delta H^0-T.\Delta S^0$

  $\begin{array}{l}\Delta G^0\text{ - Gibbs free energy change}\\ \Delta H^0\text{ - enthalpy change}\\ T\text{ - temperature}\\ \Delta S^0\text{ - entropy change}\end{array}$