Silver(II) oxide partially decomposes into solid silver and oxygen gas according to:

$$\ce{2 AgO(s) <=> 4 Ag(s) + O2(g)}$$

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?

$$\frac{ΔG^\circ}{-RT} = \frac{-RT}{-RT}\ln K$$

$$\ln K = \frac{ΔG^\circ}{-RT}$$

$$ \begin{align} K &= e^{-ΔG^\circ/(RT)} \\ &= e^{-22.2/(8.3145\cdot 298)}\\ &= \pu{8.18545e-9} \end{align} $$

The answer I'm supposed to get is $\pu{1.30e-4}.$ The answer I get according to my calculations is $\pu{8.1854e-9}.$

  • $\begingroup$ Please show us your calculation. $\endgroup$ Commented Aug 5, 2019 at 0:52
  • $\begingroup$ You need to give more information about what you did. Have you consider using dG= dH - TdS? and then replacing the value for dG in dG = -RTlnKp? Is any other data provided to solve the problem? Also notice that in this eq system you have to consider only O2, since the other reagents are solid. Also you need to define which constant are you going to calculate. Kc, Kp... $\endgroup$ Commented Aug 5, 2019 at 1:11
  • $\begingroup$ What is the temperature? What is "the right answer" for b? $\endgroup$
    – Karsten
    Commented Aug 5, 2019 at 5:38
  • $\begingroup$ @Rose For calculating $\Delta G^0$ use the relation $\Delta G^0 = \Delta H^0 - T\Delta S^0$, and put $T = 298$ K, as it is in the standard state (Note it should be change in enthalpy that is given in the question not the enthalpy. Change in enthalpy is the ( $\sum_{products} H_f^0 - \sum_{reactants} H_f^0$) which is actually meant in the question). Similarly for the next part use $\mathrm{\Delta G^0 = -RT lnK}$ and put $T = 298 $K, and use the result of $\Delta G^0$ obtained in the previous result. $\endgroup$
    – Soumik Das
    Commented Aug 6, 2019 at 12:05
  • 1
    $\begingroup$ For the record, this is why we propagate units through the calculation. You could have spotted the error yourself if you had done that in your calculations here. $\endgroup$
    – Zhe
    Commented Aug 6, 2019 at 17:05

1 Answer 1


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.

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


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