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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2024 · 6 April · Evening Shift · Chemistry · Thermodynamics (C)

For the reaction at 298 ~K, 2 ~A+B arrow C . Δ H =400 ~kJ ~mol^ -1...

For the reaction at 298 K,2 A+B→C.ΔH298 \mathrm{~K}, 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C} . \Delta \mathrm{H} =400 kJ mol−1=400 \mathrm{~kJ} \mathrm{~mol}^{-1} and ΔS=0.2 kJ mol−1 K−1\Delta \mathrm{S}=0.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}. The reaction will become spontaneous above \qquad K.

Correct answer

2000

Step-by-step solution

ΔG=0 T=ΔHΔS=4000.2=2000 K\begin{aligned} & \Delta \mathrm{G}=0 \\ & \mathrm{~T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{400}{0.2}=2000 \mathrm{~K}\end{aligned}
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