# How to calculate the standard Gibbs energy at equilibrium?

Given that $$K_c = 1.7 \times 10^{-13}$$, calculate $$\Delta G^{\circ}$$ for this equilibrium mixture at $$\pu{298 K}$$.

$$\ce{N2O (g) + \frac{1}{2}O2 (g) <=> 2 NO (g)}$$

I've calculated:

\begin{align} \Delta G^{\circ} &= -RT \ln K \\ &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln (1.7 \times 10^{-13}) \\ &= \pu{72.8 kJ mol-1} \end{align}

The result I'm expecting is $$\pu{68.9 kJ mol-1}$$. Where have I gone astray?

I think the author of this problem forgot to add units to $$K_c$$ since it's not a dimensionless entity:

$$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$

and should be $$K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$$. In general

$$[K_c] = \mathrm{dim}(c)^{Δn}$$

where square brackets denote the dimensions of the quantity $$K_c$$, $$c$$ is concentration and $$Δn$$ is the difference in the amounts between gaseous products and reactants, e.g. here

$$Δn = 2 - (1 + 0.5) = 0.5$$

On the other hand, the equilibrium constant $$K$$ you use for determining the standard Gibbs energy, must be dimensionless. Since we are dealing with gases only, the easiest way is to use $$K_p$$, which is dimensionless as required (when normalized to the standard state of pressure $$p^\circ = \pu{1 bar}$$):

$$K_p = \frac{\left(\frac{p(\ce{NO})}{p^\circ}\right)^2}{\left(\frac{p(\ce{N2O})}{p^\circ}\right) \left(\frac{p(\ce{O2})}{p^\circ}\right)^{0.5}}$$

so that in general

$$[K_p] = \mathrm{dim}(p)^{Δn}\cdot \mathrm{dim}(p^\circ)^{-Δn}$$

$$K_p$$ and $$K_c$$ are related (via the ideal gas law):

$$K_p = K_c (RT)^{Δn}$$

So, the equilibrium constant is

\begin{align} K &= K_p(p^\circ)^{-Δn}\\ &= K_c(RT)^{Δn}(p^\circ)^{-Δn} \\ &= \pu{1.7e-13 mol^{0.5} L^{-0.5}}\cdot(\pu{8.314e-2 L bar K−1 mol−1}\cdot\pu{298 K})^{0.5}(\pu{1 bar})^{-0.5} \\ &= \pu{8.5e-13} \end{align}

Note that here I used gas constant expressed as $$\pu{8.314e-2 L bar K−1 mol−1}$$ since in this case all dimensions are cancelled out and $$K$$ is left dimensionless. Now we can finally find the standard Gibbs energy:

\begin{align} Δ G^\circ &= -RT\ln K\\ &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln\left(\pu{8.5e-13}\right)\\ &= \pu{68.9 kJ mol-1} \end{align}

Here the product before the logarithm includes $$R = \pu{8.314 J mol-1 K-1}$$ to get answer in $$\pu{kJ mol-1}$$ straight away.

• Yes, that definitely does the trick. It also explains why my method worked for another calculation, where $\Delta n$ happened to be $0$. – Adam Hrankowski Jan 29 at 22:26
• That’s right that expression is written using Kp instead of Kc so you must use the equivalence Kp = Kc(RT)^(np-nr) – Marange Jan 29 at 22:31