Question

For the reaction \(A(g) \rightleftharpoons B(g)\) at 495 K, \(\Delta_rG^\circ = -9.478\) kJ mol\(^{-1}\).
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol\(^{-1}\) K\(^{-1}\); ln 10 = 2.303]

Hint:

Use \(\Delta G^\circ = -RT\ln K\) to get K, then \(K = n_B/n_A\) (since \(\Delta n_g = 0\)).

Answer 1

Correct Answer: 20

\(\Delta_rG^\circ = -RT\ln K\), so \(\ln K = \frac{9478}{8.314 \times 495} = \frac{9478}{4115.4} = 2.303 = \ln 10\). Hence \(K = 10\).
For \(A(g) \rightleftharpoons B(g)\), \(\Delta n_g = 0\), so \(K = \frac{n_B}{n_A}\) (volume cancels). Let x millimoles of B form: \(\frac{x}{22 - x} = 10\), so \(x = 220 - 10x\), \(11x = 220\), \(x = 20\).
Answer: 20