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 is the equilibrium mixture is ________ millimoles. (Round off to the Nearest Integer).
[$$R = 8.314$$ J mol$$^{-1}$$ K$$^{-1}$$; $$\ln 10 = 2.303$$]

We are given the reaction $$A(g) \rightleftharpoons B(g)$$ at 495 K with $$\Delta_r G^\circ = -9.478$$ kJ/mol. We start with 22 millimoles of A in a closed container and need to find the amount of B at equilibrium.

First, we find the equilibrium constant using $$\Delta_r G^\circ = -RT \ln K$$. Rearranging: $$\ln K = \frac{-\Delta_r G^\circ}{RT} = \frac{9478}{8.314 \times 495} = \frac{9478}{4115.43} = 2.3026$$.

Since $$\ln 10 = 2.303$$, we get $$K = 10$$.

For the reaction $$A \rightleftharpoons B$$, if we start with 22 millimoles of A and let $$x$$ millimoles react, at equilibrium we have $$(22 - x)$$ millimoles of A and $$x$$ millimoles of B.

Since both A and B are gases in the same container, the equilibrium constant in terms of mole fractions (which equals $$K_p$$ for this equal-moles reaction) gives us $$K = \frac{x}{22 - x}$$. For a reaction with equal total moles of gas on both sides, $$K_p = K$$ in terms of the ratio of partial pressures, which equals the ratio of moles.

So $$10 = \frac{x}{22 - x}$$. Solving: $$10(22 - x) = x$$, giving $$220 - 10x = x$$, so $$11x = 220$$, hence $$x = 20$$.

The amount of B at equilibrium is $$\mathbf{20}$$ millimoles.