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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?

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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.

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

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