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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 ($K_p$)?

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  • $\begingroup$ 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

3 Answers 3

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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.

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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.
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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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