Question

Standard entropy of ${\mathrm{X}}_2,{\mathrm{Y}}_2\mathrm{and}{\mathrm{XY}}_3$ are $60,40\mathrm{and}50{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$, respectively. For the reaction, $\frac{1}{2}{\mathrm{X}}_2+\frac{3}{2}{\mathrm{Y}}_2\to {\mathrm{XY}}_3,\Delta \mathrm{H}=-30\mathrm{kJ}$ to be at equilibrium. What will be the temperature?

• A. $500\mathrm{K}$
• B. $750\mathrm{K}$
• C. $1000\mathrm{K}$
• D. $1250\mathrm{K}$

Answer 1

The correct option is B

$750\mathrm{K}$

Explanation for correct option:

Option (B) $750\mathrm{K}$

Step 1: Analyse the given reaction

The given reaction is as follows:

$\frac{1}{2}{\mathrm{X}}_2+\frac{3}{2}{\mathrm{Y}}_2\to {\mathrm{XY}}_3,\Delta \mathrm{H}=-30\mathrm{kJ}$

If we multiply the whole reaction by the factor $2$, it becomes:

${\mathrm{X}}_2+3{\mathrm{Y}}_2\to 2{\mathrm{XY}}_3∆\mathrm{H}=-60\mathrm{kJ}$

Step 2: Find entropy change of reaction.

The change in entropy of the reaction is given by:

$\Delta {\mathrm{S}}_{\text{reaction}}=\sum \Delta {\mathrm{S}}_{\text{product}}-\sum \Delta {\mathrm{S}}_{\text{reactant}}$

By substituting the values, we get:

$$\begin{gathered} \Delta {\mathrm{S}}_{\mathrm{reaction}}=(2\times 50)-(3\times 40)-(1\times 60) \\ \Delta {\mathrm{S}}_{\mathrm{reaction}}=100-120-60=-80{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 3: Find the required temperature

We know that:$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

So, $∆\mathrm{H}=\mathrm{T}∆\mathrm{S}$

By substituting the values, we get:

$$\begin{gathered} 1000\times (-60)=-80\times \mathrm{T} \\ \mathrm{T}=750\mathrm{K} \end{gathered}$$

So, the required temperature is $750\mathrm{K}$

Hence, option (b) $750\mathrm{K}$ is the correct option.