The enthalpy and entropy change for the reaction: $ ext{Br}_2 (l) + ext{Cl}_2 (g) ightarrow 2 ext{BrCl}

Question

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

Accepted Answer

At equilibrium, the change in Gibbs free energy ($\Delta G$) is zero. 
We know that $\Delta G = \Delta H - T\Delta S$
Substituting $\Delta G = 0$, we get:
$T = \frac{\Delta H}{\Delta S}$
Given:
$\Delta H = 30 	ext{ kJ mol}^{-1} = 30000 	ext{ J mol}^{-1}$
$\Delta S = 105 	ext{ J K}^{-1}	ext{ mol}^{-1}$
$T = \frac{30000}{105} = 285.71 	ext{ K} \approx 285.7 	ext{ K}$

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