Chapter 6, Problem 6.11P

(a)

Interpretation Introduction

Interpretation:

For given reaction, the equilibrium constant $\left({\text{K}}_{\text{b}}\right)$ at $\text{5°C}$ is $\text{1}{\text{.479×10}}^{\text{-5}}$ and at $\text{10°C}$ is $\text{1}{\text{.570×10}}^{\text{-5}}$

${\text{ΔH}}^{\text{0}}$ has to be find.

Concept introduction:

Equilibrium constant (K): In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.  If value of K is small than 1 the reaction should be move to the left and k is more than 1 the reaction should be move to right.

Relating ${\text{ΔG}}^{\text{0}}$to K:

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{= -RTlnK}\\ \\ {\text{K = e}}^{\left({\text{-ΔG}}^{\text{0}}\text{/RT}\right)}\end{array}$

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{-Change in free energy}\\ \\ \text{K-Equilibrium constant}\\ \\ \text{R-Gas constant (8}\text{.314472)}\\ \\ \text{T-Temperature}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant equation used to form linear graph to calculate ${\text{ΔH}}^{\text{o}}$ when ${\text{ΔH}}^{\text{o}}$ and ${\text{ΔS}}^{\text{o}}$ are constant over some temperature. This is has to be described.