The enthalpy and entropy change for the reaction: $\mathrm{Br}_{2(\mathrm{l})}+\mathrm{Cl}_{2(g)}…

The enthalpy and entropy change for the reaction: $\mathrm{Br}_{2(\mathrm{l})}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{BrCl}_{(g)}$ are $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature which the reaction will be in equilibrium is:
  1. $300 \mathrm{~K}$
  2. $285.5 \mathrm{~K}$
  3. $273 \mathrm{~K}$
  4. $450 \mathrm{~K}$

Solution

For the reaction $\begin{gathered} \mathrm{Br}_2(l)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{BrCl}(g) \\ \Delta \mathrm{H}=30 \mathrm{~kJ} / \mathrm{mol} \\ \Delta \mathrm{S}=105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{gathered}$ For at equilibrium $\Delta \mathrm{G}=0$ $\begin{aligned} & \therefore \quad \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\ & \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\ & \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{30 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1}}{105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ & =285.7 \mathrm{~K} \\ & \end{aligned}$

Asked in: NEET 2006

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