Chapter 6, Problem 6.41NP

(a)

Interpretation Introduction

Interpretation: The value of $K$ needs to be calculated at 298 K.

Concept Introduction: The value of $K$ can be calculated as follows:

  $\ln K\left(T\right)=-\frac{\Delta G_R^o\left(T\right)}{R\times T}$

Here, $\Delta G_R^o$ is change in standard Gibbs free energy of the reaction, R is Universal gas constant and T is temperature.

(b)

Interpretation Introduction

Interpretation: The following equation needs to be derived.

  $\alpha =\frac{1}{2}\sqrt{K_P\left(T\right)\frac{P}{P^o}}$

Concept Introduction: For a general equilibrium reaction as follows:

  $aA+bB\rightleftharpoons cC+dD$

The equilibrium constant expression is represented as follows:

  $K=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

(c)

Interpretation Introduction

Interpretation: The degree of reaction needs to be calculated at 298 K and 5.00 bar.

Concept Introduction: For a general equilibrium reaction as follows:

  $aA+bB\rightleftharpoons cC+dD$

The equilibrium constant expression is represented as follows:

  $K=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

(d)

Interpretation Introduction

Interpretation: The value of $K_X$ needs to be calculated at 298 K and 5.00 bar.

Concept Introduction: The relation between $K_X$ and $K_P$ of the reaction is as follows:

  $K_X=K_P{\left(\frac{P}{P^o}\right)}^{-△n}$

Here, $\Delta n$ is change in number of gaseous species at product and reactant side.