Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas?
I was trying to explain it in terms of reaction rates and such; more than just 'Le Chatelier's principle states ...', but I can't figure out the underlying reason as to why. I can explain why for increasing the pressure, temperature changes and such, but not for this specific one. If there is more space for the particles to move without collided, this reduces the frequency of successful collisions, reducing the reaction rate of both forward and reverse reactions. But, in $$\ce{A + B<=>3C}$$ there is initially more moles of C than A and B combined. So, isn't it reasonable to think C will collide more frequently than A and B, meaning the reaction rate of the reverse is quicker, rather than the reaction rate of the forward being quicker?
equilibrium pressure
Buck Thorn
EeknormousEeknormous
$\begingroup$ By your reasoning, the equilibrium is never possible at all, because there are still 3C on the right, hence more collisions. But no, it doesn't work quite like that. $\endgroup$ – Ivan Neretin Oct 1 '19 at 9:22
For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.
When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g >0$.
In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.
Similarly, increase in pressure favours the side with lesser number of gaseous moles.
You can relate an equilibrium constant $K_p$ written in terms of partial pressures to one written in terms of mole fractions $K_\chi$, as follows:
$$K_\chi=K_p (p/p_0)^{-\Delta \nu}\tag{1}$$
where $\Delta \nu$ is the difference in the stoichiometric coefficients for a molar amount of reaction. $K_p$ is independent of the pressure $p$ of the system but depends on the temperature.
Now take the logarithm on both sides of Equation (1) followed by the derivative with respect to pressure, keeping temperature constant. Remembering that $K_p$ is not a function of pressure at constant temperature, you can show that
$$\left(\frac{\partial \ln K_\chi}{\partial p}\right)_T=-\frac{\Delta \nu}{p}$$
This equation can be rewritten as
$$d(\ln K_\chi)=-\Delta \nu \times d (\ln p) \tag{constant T}$$
This expression shows that if $\Delta \nu>0$ as in the example you provide, then increasing p (or equivalently ln(p)) will cause $K_\chi$ to decrease, implying that the mole fraction of product will diminish.
This is of course a mathematical explanation and not entirely satisfying, but there are various ways to justify it heuristically, the simplest being that (in your example) by reverting to reagents the number of molecules is reduced, decreasing the total volume, and thereby reducing the amount of work and heat exchanged during the compression.
Buck ThornBuck Thorn
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