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 gathered Br_ 2 (l)+ Cl_ 2 (g) → 2 BrCl(g) Δ H=30 ~kJ/ mol Δ S=105 JK^ -1 ~mol^ -1 gathered For at equilibrium Δ G=0 aligned & Δ G=Δ H-T Δ S & Δ H=T Δ S & T= Δ H Δ S = 30 × 1000 ~J ~mol^ -1 105 JK^ -1 ~mol^ -1 & =285.7 ~K & aligned

NEET 2006

Exam: NEET
Year: 2006
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2006 · Chemistry · Thermodynamics (C)

The enthalpy and entropy change for the reaction: Br_ 2(l) +Cl_...

The enthalpy and entropy change for the reaction:

\mathrm{Br}_{2(\mathrm{l})}+\mathrm{Cl}_{2(g)} \rightarrow

\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:

Options

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

Correct answer

B. $285.5 \mathrm{~K}$

Step-by-step solution

For the reaction

\begin{gathered}\mathrm{Br}_{2}(l)+ \mathrm{Cl}_{2}(g) → 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}

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