JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Chemical Thermodynamics

Consider the reaction

A ⇌ B

at 1000 K. At time

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?

Answer: 0.25

Solution

From the graph:

At 1000 K, equilibrium partial pressure of B = 10 bar
At 2000 K, equilibrium partial pressure of B = 100 bar

Equilibrium constants:

$K_{1000} = \frac{P _{B}}{P _{A}} = \frac{10}{1} = 10$
$K_{2000} = \frac{P _{B}}{P _{A}} = \frac{100}{1} = 100$

Using

Δ G^{\circ} = - R T ln K

\frac{Δ G _{1000}^{\circ}}{Δ G _{2000}^{\circ}} = \frac{1000 \times ln 10}{2000 \times ln 100} = \frac{1000 \times ln 10}{2000 \times 2 ln 10} = \frac{1}{4} = 0.25

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Chemical Thermodynamics

Solution

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