Question

The enthalpy and entropy change for the reaction:${\mathrm{Br}}_2(l)+{\mathrm{Cl}}_2(\mathrm{g})\to 2\mathrm{BrCl}(\mathrm{g})$ are $30\mathrm{kJ}{\mathrm{mol}}^{-1}$ and $105{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ respectively.The temperature at which the reaction will be in equilibrium is:

• (A) $273\mathrm{K}$
• (B) $450\mathrm{K}$
• (C) $300\mathrm{K}$
• (D) $285.7\mathrm{K}$

Answer 1

The Correct Option is D

Explanation:
At equilibrium, Gibbs free energy change $\left(\mathrm{\Delta }{\mathrm{G}}^{\circ }\right)$ is equal to zero. The following thermodynamic relation is used to show the relation of $(\mathrm{\Delta }{\mathrm{G}}^{\circ })$ with the enthalpy change $\left(\mathrm{\Delta }{\mathrm{H}}^{\circ }\right)$ and entropy change $\left(\mathrm{\Delta }{\mathrm{S}}^{\circ }\right)$.$\mathrm{\Delta }{\mathrm{G}}^{\circ }=\mathrm{\Delta }{\mathrm{H}}^o-\mathrm{\Delta }{\mathrm{S}}^o$$0=30\times {10}^3\left(\mathrm{J}{\mathrm{mol}}^{-1}\right)-\mathrm{T}\times 105\left(\mathrm{J}{\mathrm{K}}^{-1}\right){\mathrm{mol}}^{-1}$Therefore, $T=\frac{30\times {10}^3}{105}\mathrm{K}=285.71\mathrm{K}$Hence, the correct option is (D).