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JEE Advanced 2020 · Paper 1, Chemistry question 14

Consider the reaction A⇌B\ce{A <=> B} at 1000 K. At time t′t', the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at 1 bar. Given below is the plot of the partial pressure of B with time. What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?

Partial pressure of B (bar) against time. It is steady at 10 bar until time t′, then rises sharply, briefly overshooting to about 115 bar, and settles at a steady 100 bar.
Answer and worked solution

Official answer: 0.25

With pA=1p_A=1 bar, K=pB/pAK=p_B/p_A can be read off the graph.

  • 1000 K: K1=10K_1=10.
  • 2000 K: K2=100K_2=100 (the brief overshoot to 115 is not equilibrium).

ΔG1000∘ΔG2000∘=−R(1000)ln⁡10−R(2000)ln⁡100=1000ln⁡104000ln⁡10=0.25.\frac{\Delta G^\circ_{1000}}{\Delta G^\circ_{2000}}=\frac{-R(1000)\ln10}{-R(2000)\ln100}=\frac{1000\ln10}{4000\ln10}=0.25.
Answer: 0.25

Topics:Equilibrium and Gibbs energy
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