We know that the Gibbs free energy is related to the equilibrium constant by the following equation: $$\Delta_\mathrm{r}G^\circ=-RT\ln K$$ We also know the Van't Hoff equation: $$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$

My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?

  • $\begingroup$ You use the equilibrium constant $K$ that is formulated in terms of activities. This are relations connecting this $K$ to $K_{c}$ and $K_{p}$ and in the limit of ideal behaviour $K = K_{c}$. Have a look at this answer of mine. It explains the derivation of $K$ and gives the relation to $K_{c}$ ($K_{p}$ is not included but I could add it if that would be helpful). $\endgroup$
    – Philipp
    Apr 17, 2014 at 11:50
  • $\begingroup$ Second equation you posted looks like the Clausius-Clapyeron equation, not the van't Hoff equation. $\endgroup$
    – Dissenter
    May 15, 2014 at 17:42
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    $\begingroup$ I would disagree to the last comment, @Dissenter. As far as I can see, the posted equation is indeed the Van't Hoff equation $\endgroup$ May 28, 2014 at 9:23
  • $\begingroup$ @Philipp As you use van't Hoff for determining the equilibrium for $\ce{aA + bB <=> cC + dD}$ shouldn't it be all the same, since you would use the same definition for $K^\circ_{T_1} = \frac{a^{c}(\ce{C})\cdot{}a^{d}(\ce{D})}{a^{a}(\ce{A})\cdot{}a^{b}(\ce{B})}$ and possible constants would cancel each other in the quotient? $\endgroup$ Jul 16, 2014 at 6:55
  • $\begingroup$ @Martin I'm not 100 % sure but looking at the relation between $K$ and for example $K_{c}$, \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ , \end{equation} I would say that the quotients $\frac{K(T_{1})}{K(T_{2})}$ and $\frac{K_{c}(T_{1})}{K_{c}(T_{2})}$ will be equal only for small temperature differences since $\left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}}$ contains $T$ directly and the fugacity $\varphi_{i}$ is temperature-dependent as well. $\endgroup$
    – Philipp
    Jul 16, 2014 at 9:44

2 Answers 2


The first equation actually contains the definition of the standard equilibrium constant: $$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$ With this definition the equilibrium constant is dimensionless.

Under standard conditions the van't Hoff equation is $$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$ and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant. $$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) = \frac{{\ - \Delta H^\circ }}{R} \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$

Now the ordinary equilibrium constant may be defined in various forms: $$K_x = \prod_B x_B^{\nu_B}.$$

Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction, $$\ce{\nu_{A}A + \nu_{B}B -> \nu_{C}C + \nu_{D}D},$$ this resolves in $$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$

The concentration is connected to the activity via $$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$ where the standard concentration is $c^\circ = \pu{1 mol/L}$. At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as $$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$ See also a very detailed answer of Philipp.

The partial pressure is connected to the activity via $$a(\ce{A}) = \frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$ with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$. For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since \begin{align} \lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &=1, & p(\ce{A}) &= y_{\ce{A}}\cdot{}p. \end{align}

Of course concentrations and partial pressures are connected via the ideal gas \begin{aligned} pV\ &=nRT\\ p\ &\propto \text{const.} \cdot c, \end{aligned} and therefore it is valid to write: $$K_c\propto \text{const.} \cdot K_p.$$ It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.

While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.


Strictly speaking these equations should contain some correction of activity (concentration quotients) or non-ideal behaviour of gases (pressure quotients). If you don't need to deal with those and you can treat your system as ideal, the concentration of a gas and pressure are equivalent. Be careful with the units, though!

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    $\begingroup$ Even with ideal gases, $K_p$ and $K_c$ are not always equivalent - they are related, but one might be scaled relative to the other by powers of $(RT)$, since $C=\frac{n}{V}=\frac{P}{RT}$. $\endgroup$
    – Ben Norris
    Apr 16, 2014 at 10:43
  • $\begingroup$ Exactly. Kc does not necessarily equal Kp. $\endgroup$
    – Dissenter
    May 15, 2014 at 17:42

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