8
$\begingroup$

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 ($K_p$)?

|improve this question
$\endgroup$
  • $\begingroup$ 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 '14 at 11:50
  • $\begingroup$ Second equation you posted looks like the Clausius-Clapyeron equation, not the van't Hoff equation. $\endgroup$ – Dissenter May 15 '14 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 $\endgroup$ – Kjetil Sonerud May 28 '14 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? $\endgroup$ – Martin - マーチン♦ Jul 16 '14 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 '14 at 9:44
5
$\begingroup$

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.

|improve this answer
$\endgroup$
0
$\begingroup$

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!

|improve this answer
$\endgroup$
  • 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 '14 at 10:43
  • $\begingroup$ Exactly. Kc does not necessarily equal Kp. $\endgroup$ – Dissenter May 15 '14 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.