The gas phase reaction
$$2 A(g) \rightleftharpoons A_2(g)$$
at 400 K has $$\Delta G^\circ = +25.2$$ kJ mol$$^{-1}$$.
The equilibrium constant $$K_C$$ for this reaction is ___ $$\times 10^{-2}$$. (Round off to the Nearest integer)
Use : $$R = 8.3$$ J mol$$^{-1}$$ K$$^{-1}$$, ln 10 = 2.3, $$\log_{10} 2 = 0.30$$, 1 atm = 1 bar
antilog(-0.3) = 0.501

We are given $$\Delta G^\circ = +25.2 \text{ kJ mol}^{-1} = 25200 \text{ J mol}^{-1}$$ at $$T = 400 \text{ K}$$, and need to find $$K_C$$. We use the relation $$\Delta G^\circ = -RT \ln K_p$$, where $$R = 8.3 \text{ J mol}^{-1} \text{K}^{-1}$$ (as given in the problem).

Rearranging: $$\ln K_p = -\frac{\Delta G^\circ}{RT} = -\frac{25200}{8.3 \times 400} = -\frac{25200}{3320} = -7.59$$.

Converting to $$\log_{10}$$: since $$\ln x = 2.3 \times \log_{10} x$$, we get $$\log_{10} K_p = \frac{\ln K_p}{2.3} = \frac{-7.59}{2.3} = -3.3$$. Writing this as $$\log_{10} K_p = -3 - 0.3$$, we get $$K_p = 10^{-3} \times 10^{-0.3}$$. Since $$\log_{10} 2 = 0.30$$, we have $$10^{0.3} = 2$$, so $$10^{-0.3} = \frac{1}{2} = 0.5$$. Therefore $$K_p = 0.5 \times 10^{-3} = 5 \times 10^{-4}$$.

Now we convert $$K_p$$ to $$K_C$$ using the relation $$K_p = K_C \times (RT)^{\Delta n}$$. For the reaction $$2A(g) \rightleftharpoons A_2(g)$$, the change in moles of gas is $$\Delta n = 1 - 2 = -1$$. So $$K_p = K_C \times (RT)^{-1}$$, which gives $$K_C = K_p \times (RT)^{1} = K_p \times RT$$.

Here we need $$RT$$ in units of $$\text{L atm mol}^{-1}$$ (or equivalently $$\text{L bar mol}^{-1}$$, since the problem states 1 atm = 1 bar). We use $$R = 0.0831 \text{ L bar mol}^{-1} \text{K}^{-1}$$, so $$RT = 0.0831 \times 400 = 33.24 \text{ L bar mol}^{-1}$$. Note that this is a different numerical value of $$R$$ from the one used earlier because the units are different: $$8.3 \text{ J mol}^{-1}\text{K}^{-1}$$ is used for energy calculations, while $$0.0831 \text{ L bar mol}^{-1}\text{K}^{-1}$$ is used for the $$K_p$$-$$K_C$$ conversion involving volume units.

Therefore $$K_C = 5 \times 10^{-4} \times 33.24 = 1.662 \times 10^{-2} \approx 2 \times 10^{-2}$$.

The answer is $$2$$ (i.e., $$K_C = 2 \times 10^{-2}$$).