Chapter 13, Problem 116SAE

Interpretation Introduction

(a)

Interpretation:

Whether the given forward reaction is endothermic or exothermic needs to be determined.

Concept introduction:

$\text{ΔH}°$ is calculated using formula:

$\text{ΔrH° = ΔH°(product) - ΔH°(reactant) }$

The positive sign of $\text{ΔH}°$ means that reaction is endothermic whereas the negative sign signifies that reaction is exothermic in nature.

Interpretation Introduction

(b)

Interpretation:

The value of $\text{ΔrG}°$ for the given reaction at 298 K should be determined.

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with Temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

The value of $\text{ΔrH}°$ and $\text{ΔrS}°$ for a chemical reaction can be calculated with the help of standard $\text{ΔH}°$ and $\text{ΔS}°$ values of reactant and product molecules.

$\begin{array}{l}\text{ΔrH° = ΔH°(product) - ΔH°(reactant) }\\ \text{ΔrS° = ΔS°(product) - ΔS°(reactant) }\end{array}$

Interpretation Introduction

(c)

Interpretation:

The value of K for the given reaction should be determined.

Concept introduction:

The mathematical expression relating $\text{ΔrG}°$, R, T and log K is:

$\text{ΔrG}°\text{ = -2}\text{.303 }\times \text{ RT log K}$

Interpretation Introduction

(d)

Interpretation:

Whether the reaction occurs spontaneously below or above 298 K should be determined.

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with Temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔrG}°\text{ = ΔrH}°\text{ - TΔrS}°$

The value of $\text{ΔrH}°$ and $\text{ΔrS}°$ for a chemical reaction can be calculated with the help of standard $\text{ΔH}°$ and $\text{ΔS}°$ values of reactant and product molecules.

$\begin{array}{l}\text{ΔrH° = ΔH°(product) - ΔH°(reactant) }\\ \text{ΔrS° = ΔS°(product) - ΔS°(reactant) }\end{array}$