How can I calculate the equilibrium constant for this reaction? Silver(II) oxide partially decomposes into solid silver and oxygen gas according to:
$$\ce{2 AgO(s) &lt;=&gt; 4 Ag(s) + O2(g)}$$
with enthalpy $H^\circ = \pu{62.0 kJ}$ and entropy $S^\circ = \pu{133.6 J K-1}.$
a) Calculate the change in standard Gibbs free energy. (Answer: $\pu{22.2 kJ}$)
b) Calculate the equilibrium constant for this reaction.

This is one of the exam questions I got, but I don't understand how I'm supposed to answer question b without the concentration. I also didn't get a mass or volume so I can't calculate it using the molar mass.
I tried using the formula $G^\circ=-RT \ln K,$ but I don't get the right answer with this formula. Could someone help me find the right way and formula to answer this question?
$$\frac{ΔG^\circ}{-RT} = \frac{-RT}{-RT}\ln K$$
$$\ln K = \frac{ΔG^\circ}{-RT}$$
$$
\begin{align}
K &amp;= e^{-ΔG^\circ/(RT)} \\
  &amp;= e^{-22.2/(8.3145\cdot 298)}\\
  &amp;= \pu{8.18545e-9}
\end{align}
$$
The answer I'm supposed to get is $\pu{1.30e-4}.$ The answer I get according to my calculations is $\pu{8.1854e-9}.$
You might want to check your units.

The energy should be in J/mol, according to the values you put in, and not in kJ/mol.

$e^{-22200/(8.314*298)} = 0.000128397 \approx 1.3*10^{-4}$