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
  • 1
    $\begingroup$ 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

3 Answers 3


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.


Here is a simple and detailed explanation.

For a general reaction $aA+bB ⇌cC+dD$, $K_{eq}$ is defined as:

$$K_{eq} = \frac{(a_{\ce{C}})^c(a_{\ce{D}})^d}{(a_{\ce{A}})^a(a_{\ce{B}})^b}$$ Without going into detail, the activity "$a_X$" is basically :

  • magnitude of concentration in molarity(if solution)
  • magnitude of partial pressure in bar(if gas)
  • $1$ (if solid or pure liquid)

Observe, the repetitive use of "magnitude". This is to emphasize that activity is a dimensionless quantity. It trivially follows that $K_{eq}$ is also a dimensionless quantity.

The above expression for $K_{eq}$ holds for any general reaction, no matter how many or which phases are involved. For example, for $aA(s)+bB(aq)+cC(g)⇌dD(l)+eE(g)+fF(aq)+gG(aq)$, the value of $K_{eq}$ is:- $$K_{eq} = \frac{(P_{\ce{E}})^e[\ce{F}]^f[\ce{G}]^g}{[\ce{B}]^b(P_{\ce{C}})^c}$$ Note here that I have taken the magnitudes of partial pressures and concentrations.

So, this in itself answers your question as to whether to take $K_p$ or $K_c$. The answer is you take neither-you simply take $K_{eq}$ which can be calculated for any reaction as explained above. This way, you'll never need to worry about taking $K_p$ or $K_c$.

Addendum: So, what's all the fuss about $K_p$ and $K_c$? Well, it so happens that sometimes you may be given the value of $K_p$ and $K_c$ for a reaction which you may need to use. Although, this is never useful over simply giving $K_{eq}$ (in my experience), it happens to be the way things are done sometimes. So, it's worth knowing a few important things about $K_p$ and $K_c$. I'll proceed by breaking down the concept into a number of small points:

  1. Like $K_{eq}$ , $K_p$ and $K_c$ are defined in terms of activities and are dimensionless.
  2. $K_p$ is typically only used when the reactants and products have only solids, pure liquids and gases(i.e. no solutions). Thus, we can say that $K_p$ is another term for $K_{eq}$ for reactions not containing solutions.
  3. $K_c$ can be used for any general reaction, including those with all of solids, pure liquids, gases and solutions. $K_c$ is defined as-

$$K_{c} = \frac{(a_{\ce{C}})^c(a_{\ce{D}})^d}{(a_{\ce{A}})^a(a_{\ce{B}})^b}$$ where "$a_X$"is :

  • magnitude of concentration in molarity(if solution )
  • magnitude of concentration in molarity(if gas)
  • $1$ (if solid or pure liquid)

Observe, unlike $K_{eq}$, here "$a_X$" for gases is not magnitude of partial pressure in bar and is instead the same definition as solutions. This is also why I didn't take $a_X$ to refer to activity here, because technically "magnitude of concentration in molarity" is not the activity of a gas. This "re-definition" has the effect that $K_c \neq K_{eq}$(in general). Now, what do we mean by "magnitude of concentration in molarity" for a gas? Well, rewriting the ideal gas equation $P_XV=n_XRT$, we get $\frac{P_X}{RT}=\frac{n_X}{V}$=[X]. so "magnitude of concentration in molarity" is $=\frac{P_X}{RT}$. So, now instead of $a_X=P_X$(in $K_{eq}$), we have $a_X=\frac{P_X}{RT}$ (in $K_{c}$). For example, for $aA(s)+bB(aq)+cC(g)⇌dD(l)+eE(g)+fF(aq)+gG(aq)$,the value of $K_{c}$ is:- $$K_{c} = \frac{(\frac{P_E}{RT})^e[F]^f[G]^g}{[B]^b(\frac{P_C}{RT})^c}=K_{eq}(RT)^{c-e}$$ Note here, I have taken the magnitudes of partial pressures and concentrations and also the magnitude of temperature.

  1. We can extend the last result to $K_c=K_{eq}(RT)^{ - \Delta n_g}$ or $K_{eq}=K_c(RT)^{ \Delta n_g}$. Several sources cite that $K_p=K_c(RT)^{ \Delta n_g}$ which is true, but only when $K_p$ is valid i.e. for reactions not involving solutions. So, it doesn't show the whole picture. $K_{eq}=K_c(RT)^{ \Delta n_g}$ is true for any general reaction.We, can also see from this result that $K_c=K_{eq}$ only when $\Delta n_g=0$ which is usually seen in reactions only involving solutions but may also be seen in some gaseous reactions where the amount of gas is same on both sides.

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!

  • 2
    $\begingroup$ 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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