The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$. For a sample reaction:

$$ \ce{A \underset{k_{-1}}{\overset{k_1}{<=>}} P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A} $$

Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:

$$ k_i = A_i e^{-E_{a,i}\over RT} $$

Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:

So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?


1 Answer 1


TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria. The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$ {P_i\over RT} = {n_i\over V} = C_i $$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$ K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p $$

Thus, for this reaction:

$$ K_p = RT\,K_c \tag{1}\label{Kp} $$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in the sum of the stoichiometric coefficients between the reactants and the products.

Since $K_c = f(T)$ and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$. Again assuming ideal behavior, the mole fraction of a species is defined as:

$$ P_i = x_i P $$

Substituting this into the $K_P$ expression:

$$ K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x $$

Thus, for this reaction:

$$ K_x = {1\over P}K_p \tag{2}\label{Kx} $$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in the sums of the stoichiometric coefficients between the reactants and the products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$ V = {nRT \over P} $$

and, equivalently:

$$ P = {nRT \over V}\tag{3}\label{igl-P} $$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.

  • $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$ Jun 11, 2018 at 15:34
  • $\begingroup$ @Martin-マーチン Aww, nuts, I did it again... old habits die hard. Better? $\endgroup$
    – hBy2Py
    Jun 11, 2018 at 15:57
  • $\begingroup$ Do you say the total kilograms, or would you say the total mass? What would be the problem with 'the total amount of substance'? $\endgroup$ Jun 11, 2018 at 16:21
  • $\begingroup$ @Martin-マーチン Because 'total amount of substance' implies an overall mass balance to my ear, not the overall molecularity of the reaction. (And "molecularity" isn't even the right word here, if this is correct.) The 'total amount of substance' never changes in any reaction -- it's just reordered. $\endgroup$
    – hBy2Py
    Jun 11, 2018 at 16:23
  • $\begingroup$ I don't understand your point, does the total mass change in any reaction? The added amounts of the substances of the components can be referred to as the total amount of substance of the components, if you really mean that. If you want to say the power of RT depends on the molecularity of the reaction, then say that. Or you could say the power of RT depends on the difference of the amounts of substances of the reactants and the amounts of substances of the products. In any case, when you say 'moles', you are referring to a quantity by its unit. It only makes an inaccurate statement worse. $\endgroup$ Jun 11, 2018 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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