Haber-Bosch ammonia synthesis reaction: $$\ce{3H2 +N2 -> 2NH3}$$

According to the ideal gas law: $pV=nRT$ where constant volume implies $\displaystyle \frac nV=\frac p{RT}$

Then $\displaystyle K_\mathrm c=\frac{[\ce{NH3}]^{2}}{[\ce{H2}]^3\cdot [\ce{N2}]}$

When I substitute $\displaystyle \frac nV=c=\frac {p_{\text{gas}}}{RT}$ I get:

$$ K_\mathrm c=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$

But according to Le Chatelier's principle, increased pressure should result in an increased $K_\mathrm c$, but my $\displaystyle K_\mathrm c\approx \frac 1{p^2}$.

Where is the error? Is is impossible to combine the partial pressures like that because they scale differently with total pressure? Can one derive from the gas law the pressure dependency of this reaction?

Edit after solved:
First mistake was that the formula had $K_c$ (constant) instead of Q, $$ Q=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$ where Q is the current value/ratio of concentrations, so the reaction tries to get Q back to $K_c$ so for greater pressure p, Q is smaller so the reaction goes in the direction of NH3...


How do you know that $p_{\ce{NH_3}}$, $p_{\ce{H_2} } $ and $p_{\ce{N_2}}$ are all same and all equals $p$. You can't just substitute same $p$ for all of them because according to Le-chatterlier's principle, the increase in total pressure of the system, distorts the equilibrium, but your $p$ is not at all total pressure. So, you can't just say anything from your result. By the formula, $$K_p = K_x (P)^{\Delta n}$$ you can say something. Here $P$ is the total pressure of the system. (This is because the formula is derived by using Dalton's law of partial pressure which says, $p_i = x_i\dot P$ where $p_i$ is the partial prssure of the $i^{th}$ gas, $x_i$ is its mole fraction and $P$ is the total pressure of the system.)

Here $\Delta n = -2$, so ,$$K_p = K_x/P^2$$ If you now increase $p$ at constant temperature, $K_p$ remains constant as it depends only on temperature. So, $K_x$ also increases to make $K_p$ remain constant, but $$K_x = x^2_{\ce{NH3}}/x^3_{\ce{H2}}\cdot x_{\ce{N2}}=n^2_{\ce{NH3}}\dot N^2/n^3_{\ce{H2}}\cdot n_{\ce{N2}}$$ where, $N = n_\ce{NH3} + n_\ce{N2} + n_\ce{H2}$

$K_x$ increasing means that the moles of $\ce{NH3}$ increase while the moles of $\ce{H2}$ and $\ce{N2}$ decrease, which is the favour of forward reaction and as $$K_p = K_c (RT)^{\Delta n}$$ here $\Delta n = -2$ and the temperature doesn't change so $$K_c= K_p (RT)^2 -> K_c \propto K_p$$ As $K_p$ increases, $K_c$ also increases.


According to Le Chatelier's principle, a pressure increase should result in the equilibrium shifting to the right, but it should not - at least to first approximation - affect the equilibrium constant.

$$K = \frac{\ce{[NH3]}^2}{\ce{[H2]}^3 \ce{[N2]}}$$

Using partial pressures, this can be rewritten as

$$K = \left(\frac{RT}{P}\right)^2 \frac{y_{\ce{NH3}}^2}{y_{\ce{H2}}^3 y_{\ce{N2}}}$$

Then for $K$ to be constant, an increase in the denominator, given by the pressure for example, has to be followed by an increase in the numerator, which means the reaction working in the direction of formation of ammonia.

  • $\begingroup$ Oh I see, so if I have two containers where one has double the pressure of the other, then at the start of my reaction c(H2) and c(N2) are increased (aswell as c(NH3) but the total fraction of the momentarily reaction "fraction" is changed by 1/4 and so the system tries to get back to equilibium / to Kc by synthesising NH3. Thank you @Vinícius. $\endgroup$ – Beny Benz Feb 21 '18 at 20:50

It is relatively easy to work out what happens in a gas phase reaction when the total pressure is changed by considering the degree of dissociation $\alpha$. $K_p$ may be defined in terms of free energies of the species in their standard states, i.e. 1 bar pressure thus it is independent of pressure.

The changes due to dissociation are shown under the equation for the Haber process

$$3H_2 + \;\;N_2 \leftrightharpoons \;\;2NH_3\\ 3(1-\alpha) \;\; 1-\alpha \qquad 2\alpha$$

As $\displaystyle K_p= \frac{p_A^2}{p_H^3p_N}$, where subscript A is for ammonia. To work out the partial pressures $p_A$ etc. in terms of the total pressure $p$ and $\alpha$ recall that each partial pressure is the extent of dissociation over the total (in this case $3-3\alpha+1-\alpha+2\alpha=4-2\alpha=f$) which are $\displaystyle p_A= \frac{2\alpha}{f}p, \; p_N=\frac{1-\alpha}{f}p,\; p_H=\frac{3-\alpha}{f}p$ which produces the equilibrium constant in terms of $\alpha$ as

$$K_p=\frac{4\alpha^2p^2 (4-2\alpha)^4}{(4-2\alpha)^2\cdot 3^3(1-\alpha)^3p^3(1-\alpha)p}= \frac{4\alpha^2(4-2\alpha)^2}{3^3(1-\alpha)^4}\frac{1}{p^2}$$

To find the pressure dependence, rearrange the equation as $\displaystyle p=\sqrt{ \frac{4\alpha^2(4-2\alpha)^2}{3^3(1-\alpha)^4K_p} }$ and workout what the pressure is at small and large $\alpha$. In this case when $\alpha =0,\; p=0$ so as pressure increases $\alpha$ increases as $\alpha$ cannot be negative. If you plot a graph this is clear also. (Incidentally it does not matter what value $K_p$ has as it only multiples the curve by a constant value). Thus in this case as pressure is increased the extent of dissociation increases, and more product is formed. le-Chatelier's principle is in accord with this conclusion.

  • $\begingroup$ Thank you for your answer @porphyrin , but why is the formula for the degree of dissociation alpha for $3H^2=3-a$ and not $3-3a$ or $3*(1-a)$? Degree of dissociation is an unknown topic for me, so I'm sorry if it is obvious... I found chemistry.stackexchange.com/questions/531/… and if I understand it correctely alpha is the fraction of (reacted substance/substance to start with (constant)). So it should be between 0 and 1... $\endgroup$ – Beny Benz Feb 26 '18 at 0:49
  • $\begingroup$ Thank you for noticing my error. I have corrected the equations. The conclusions are unchanged. $\endgroup$ – porphyrin Feb 26 '18 at 14:52

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