The enthalpy and entropy change for the reaction: Br_2(l)+Cl_2(g) arrow 2 BrCl_(g) are 30 ~kJ ~mol^-1 and 105 ~J ~K^-1 ~mol^-1 respectively. The temperature which the reaction will be in equilibrium is:

Correct answer: (B) 285.5 ~K. For the reaction Br_2(l)+Cl_2(g) arrow 2 BrCl(g) Δ H=30 ~kJ / mol Δ S=105 JK^-1 ~mol^-1 For at equilibrium Δ G=0 ∴ Δ G=Δ H-T Δ S Δ H=T Δ S T=(Δ H)/(Δ S)=frac30 × 1000 ~J ~mol^-1105 JK^-1 ~mol^-1 =285.7 ~K

Q58 / 200 · NEET 2006 · Chemistry

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:

A
$300 \mathrm{~K}$
Correct answer
B
$285.5 \mathrm{~K}$
C
$273 \mathrm{~K}$
D
$450 \mathrm{~K}$

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 $\Delta \mathrm{G}=0$ $$ \begin{aligned} & \therefore \quad \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\ & \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\ & \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{30 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1}}{105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ & =285.7 \mathrm{~K} \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

Solution

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