1. Use the relation between $\Delta_r G^\circ$ and equilibrium constant

For the reaction $A(g) \rightleftharpoons B(g)$ we use $\Delta_r G^\circ = -RT\ln K$

Given:

• $\Delta_r G^\circ = -9.478\ \text{kJ mol}^{-1} = -9478\ \text{J mol}^{-1}$
• $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
• $T = 495\ \text{K}$

So, $-9478 = -(8.314)(495)\ln K$ $\ln K = \frac{9478}{8.314\times 495}$

Now, $8.314\times 495 = 4115.43$ $\ln K = \frac{9478}{4115.43} \approx 2.303$

Since $\ln 10 = 2.303$, $K = 10$

2. Set up equilibrium expression

Initially, only $A$ is present.

Initial moles:

• $A = 22$ mmol
• $B = 0$ mmol

Let $x$ mmol of $A$ convert to $B$.

At equilibrium:

• $A = 22-x$
• $B = x$

For $A(g) \rightleftharpoons B(g)$

$K = \frac{[B]}{[A]}$

Since stoichiometric coefficients are equal and total moles remain same, we can use moles directly: $K = \frac{x}{22-x}$

Given $K=10$, $\frac{x}{22-x} = 10$

3. Solve for $x$

$x = 10(22-x)$ $x = 220 - 10x$ $11x = 220$ $x = 20$

Thus, the amount of $B$ at equilibrium is $\boxed{20\ \text{mmol}}$

4. Comparison with stored answer

Stored correct answer = 20

Our derived answer = 20

So, the answer agrees with the stored correct answer.