Question

Standard entropy of ${{\text{X}}_{\text{2}}}$, ${{\text{Y}}_{\text{2}}}$ and ${\text{X}}{{\text{Y}}_{\text{3}}}$ are ${\text{60}}$, ${\text{40}}$ and $50{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ respectively. For the reaction, $\dfrac{1}{2}{{\text{X}}_2} + \dfrac{3}{2}{{\text{Y}}_2} \to {\text{X}}{{\text{Y}}_3}{\Delta {\text{H}}} = - 30{\text{ kJ}}$. To be at equilibrium the temperature will be:
A) $500{\text{ K}}$
B) $750{\text{ K}}$
C) $1000{\text{ K}}$
D) $1250{\text{ K}}$

Answer 1

The measure of randomness or disordered distribution is known as entropy. The randomness is always higher in a gaseous state. More the number of gaseous molecules higher is the entropy. To solve this we must know the expression that gives the relation between free energy, entropy and enthalpy.

Formula Used:
${{\Delta }}{{\text{S}}_{{\text{reaction}}}} = {{\Delta }}{{\text{S}}_{{\text{products}}}} - {{\Delta }}{{\text{S}}_{{\text{reactants}}}}$
${{\Delta G}} = {{\Delta H}} - {{T\Delta S}}$

Complete step by step answer:
We are given the reaction as follows:
$\dfrac{1}{2}{{\text{X}}_2} + \dfrac{3}{2}{{\text{Y}}_2} \to {\text{X}}{{\text{Y}}_3}$
We know that the measure of randomness or disordered distribution is known as entropy.
We will first calculate the change in entropy of the reaction using the equation as follows:
${{\Delta }}{{\text{S}}_{{\text{reaction}}}} = {{\Delta }}{{\text{S}}_{{\text{products}}}} - {{\Delta }}{{\text{S}}_{{\text{reactants}}}}$
Where, ${{\Delta }}{{\text{S}}_{{\text{reaction}}}}$ is the change in entropy of the reaction.
We are given that the standard entropies of ${{\text{X}}_{\text{2}}}$, ${{\text{Y}}_{\text{2}}}$ and ${\text{X}}{{\text{Y}}_{\text{3}}}$ are ${\text{60}}$, ${\text{40}}$ and $50{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ respectively. Thus,
${{\Delta }}{{\text{S}}_{{\text{reaction}}}} = {\text{50}} - \left( {\dfrac{3}{2} \times 40 + \dfrac{1}{2} \times 60} \right)$
${{\Delta }}{{\text{S}}_{{\text{reaction}}}} = - 40{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
Thus, the change in entropy of the reaction is $- 40{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$.

We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
${{\Delta G}} = {{\Delta H}} - {{T\Delta S}}$
Where, ${{\Delta G}}$ is the change in Gibbs free energy,
${{\Delta H}}$ is the change in enthalpy,
$T$ is the temperature,
${{\Delta S}}$ is the change in entropy.
At equilibrium, ${\Delta G} = 0$. Thus,
${{\Delta H}} = {{T\Delta S}}$
${\text{T}} = \dfrac{{{{\Delta H}}}}{{{{\Delta S}}}}$

We are given that for the reaction $\Delta {\text{H}} = - 30{\text{ kJ}} = - 30 \times {10^3}{\text{ J}}$ and the change in entropy of the reaction is $- 40{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$. Thus,
${\text{T}} = \dfrac{{ - 30 \times {{10}^3}{\text{ J}}}}{{ - 40{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}}$
${\text{T}} = 750{\text{ K}}$
Thus, for the reaction to be at equilibrium the temperature will be $750{\text{ K}}$.

Thus, the correct option is (B).

Note: For an endothermic process, the value of $\Delta H$ is positive. In an endothermic process, heat is absorbed and the system feels cold. A negative value of $\Delta H$ indicates that the reaction is an exothermic process. In an exothermic process, heat is released or given out. We are given that for the reaction $\Delta {\text{H}} = - 30{\text{ kJ}}$ and thus, the reaction is exothermic.