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NEET CHEMISTRYEasy

The enthalpy and entropy change for the reaction: Br2(l)+Cl2(g)→2BrCl(g)\text{Br}_2 (l) + \text{Cl}_2 (g) \rightarrow 2\text{BrCl} (g) are 30 kJ mol−130 \text{ kJ mol}^{-1} and 105 J K−1 mol−1105 \text{ J K}^{-1}\text{ mol}^{-1} respectively. The temperature at which the reaction will be in equilibrium is :

A

285.7 K

B

273.4 K

C

450.9 K

D

300.1 K

Step-by-Step Solution

At equilibrium, the change in Gibbs free energy (ΔG\Delta G) is zero. We know that ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S Substituting ΔG=0\Delta G = 0, we get: T=ΔHΔST = \frac{\Delta H}{\Delta S} Given: ΔH=30 kJ mol−1=30000 J mol−1\Delta H = 30 \text{ kJ mol}^{-1} = 30000 \text{ J mol}^{-1} ΔS=105 J K−1 mol−1\Delta S = 105 \text{ J K}^{-1}\text{ mol}^{-1} T=30000105=285.71 K≈285.7 KT = \frac{30000}{105} = 285.71 \text{ K} \approx 285.7 \text{ K}

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