Question

For the chemical reaction \(\mathrm{X \rightleftharpoons Y}\), the standard reaction Gibbs energy depends on temperature \(T\) (in K) as \(\Delta_r G^\circ\) (in kJ mol\(^{-1}\)) = \(120 - \frac{3}{8}T\).
The major component of the reaction mixture at \(T\) is :

• A. Y if \(T = 300\) K
• B. Y if \(T = 280\) K
• C. X if \(T = 350\) K
• D. X if \(T = 315\) K

Hint:

Find the temperature at which \(\Delta_r G^\circ = 0\); below it \(\Delta_r G^\circ \gt 0\) so the reactant X dominates, above it the product Y dominates.

Answer 1

The Correct Option is D

\(\Delta_r G^\circ = 120 - \frac{3}{8}T\) kJ mol\(^{-1}\).
At equilibrium composition equal (K = 1): \(\Delta_r G^\circ = 0\) so \(T = \frac{120 \times 8}{3} = 320\) K.
If \(T \lt 320\) K, \(\Delta_r G^\circ \gt 0\), so \(K = e^{-\Delta_r G^\circ/RT} \lt 1\) and the reactant X is the major component. If \(T \gt 320\) K, \(\Delta_r G^\circ \lt 0\), \(K \gt 1\) and Y is major.
Check each option:
(1) 300 K: \(\Delta_r G^\circ = 120 - 112.5 = +7.5\) kJ, so X is major, not Y. False.
(2) 280 K: \(\Delta_r G^\circ = 120 - 105 = +15\) kJ, so X is major, not Y. False.
(3) 350 K: \(\Delta_r G^\circ = 120 - 131.25 = -11.25\) kJ, so Y is major, not X. False.
(4) 315 K: \(\Delta_r G^\circ = 120 - 118.125 = +1.875\) kJ, so \(K = e^{-1875/(8.314 \times 315)} \approx 0.49 \lt 1\); X is major. True.
Answer: (4) X if T = 315 K