For the reaction $$A_g \to B_g$$, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $$\Delta G^o$$ for the reaction at 300 K and 1 atm in J mol$$^{-1}$$ is $$-xR$$, where $$x$$ is ______ (Rounded off to the nearest integer) R = 8.31 J mol$$^{-1}$$ K$$^{-1}$$ and ln10 = 2.3

We are given the reaction $$A_g \to B_g$$ with equilibrium constant $$K = 100.0$$ at 300 K and 1 atm. We need to find $$\Delta G^{\circ}$$ in the form $$-xR$$.

The relationship between standard Gibbs free energy and equilibrium constant is:

Substituting the given values:

Now, $$\ln(100) = \ln(10^2) = 2 \ln 10 = 2 \times 2.3 = 4.6$$

So:

Comparing with $$\Delta G^{\circ} = -xR$$, we get:

So, the answer is $$1380$$.