Chapter 16, Problem 51QAP

Interpretation Introduction

Interpretation:

The equilibrium temperature for the given transition should be calculated.

$\text{S(rhombic) }\rightleftharpoons \text{S (monoclinic)}$

Concept introduction:

The mathematical expression for the standard entropy value at room temperature is:

$\Delta S°=\Delta S{°}_{298}=\sum \text{n}\text{S°(products})-\sum \text{p}\text{S°(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The mathematical expression for the standard enthalpy change value at room temperature is:

$\Delta H°=\Delta H{°}_{298}=\sum \text{n}\text{H°(products})-\sum \text{p}\text{H°(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Spontaneity depends upon the temperature and also depends upon the sign of free energy change.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$

If both $\text{ΔH°}$ and $\text{ΔS°}$ is positive, the reaction is endothermic results in increase in entropy.

When the magnitude of $\text{TΔS}$ is more than the $\text{ΔH}$, then $\text{ΔG}$ will be negative, and when the magnitude of $\text{TΔS}$ is less than the $\text{ΔH}$, then $\text{ΔG}$ will be positive. If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

Therefore, reaction is non- spontaneous at low temperature and spontaneous at high temperature.