Chemistry · Solutions and Colligative Properties

JEE Advanced 2020 — Paper 1 — Question 25

Consider the reaction $A \rightleftharpoons B$ at 1000 K . At time $\mathrm{t}^{\prime}$ , 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 standard Gibbs energy of the reaction at 1000 K to that at 2000 K ?

Answer: 0.25

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{K}_{1}$ at $1000 \mathrm{~K}=\frac{10}{1}=10$

$\mathrm{K}_{2}$ at $2000 \mathrm{~K}=\frac{100}{1}=100$

$\Delta G_{1}^{0}=-R T_{1} \ln 10=-2.303 R T_{1}$

$\Delta G_{2}^{0}=-2.303 R T_{2} \log 100=-2 \times 2.303 R T_{2}$

$\frac{\Delta G_{1}^{\circ}}{\Delta G_{2}^{\circ}}=\frac{1 \times 1000}{2 \times 2000}=\frac{1}{4}=0.25$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)