Di-hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

$CH_4{_{(g)}}+H_2O_{(g)} \ce{<=>} CO_{(g)}+3H_2{_{(g)}}$

How will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?


1) According to this answer $K_c$ only changes on changing temperature i.e. unaffected by pressure.

2) According to Wikipedia the article "Pressure dependence" says, "...However, the composition at equilibrium will depend appreciably on pressure when:

1)the pressure is changed by compression or expansion of the gaseous reacting system, and
2)the reaction results in the change of the number of moles of gas in the system....."

here wiki is using term $K_p$ to denote equilibrium constant. Now the "composition of equilibrium" means $K_p$ and not $Q_p$ because its taking at equilibrium, so the $K_p$ changes with pressure.

3) I have studied that on increasing pressure the reaction shifts the side with less number of sum of gaseous atoms in the reaction formulae. I know from this question that composition of equilibrium is different from "the side at which the reaction shifts". So it seems my study isn't useful here in deciding the composition of equilibrium

So my question is how will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ? Does one change and the other remains unaffected and isn't the composition of equilibrium directly related to $K_p$ and $K_c$

Are dependencies of $K_p$ and $K_c$ different for change in pressure.

  • $\begingroup$ By increasing the pressure, you produce more $CH_4$ and $H_2O$, and the amount of $CO$ and $H_2$ decreases. This is a well known method for producing $CH_4$ (+ $H_2O$) from the so called "water gas", made of a mixture $ CO$ + $H_2$ which is produced by sending water on some hot or burning charcoal : $C + H_2O$ -> $CO + H_2$ $\endgroup$
    – Maurice
    Commented Mar 11, 2020 at 20:49
  • 1
    $\begingroup$ I think you're looking for $\mathrm{K_p} = \mathrm{K_c}(\mathrm{RT})^{\Delta n}$where $\Delta n$ is the change in the number of moles for the reaction. See for example this explanation. $\endgroup$
    – MaxW
    Commented Mar 12, 2020 at 3:37

1 Answer 1


$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.

As for why $K_p$ won't change with pressure:

$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified $p^⦵$ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.

And as for why $K_c$ won't change with pressure, see the answer here


Your Answer

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

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