# Why is there no impact of pressure on a equilibrium condition when the number of molecule of gas reactant and the number of molecule of gas product are same? What will the theoretical explanation?

May 6, 2018

(Previous ${K}_{p}$ explanation was replaced because it was too confusing. Huge thanks to @Truong-Son N. for clearing up my understanding!)

Let's take a sample gaseous equilibrium:

$2 C \left(g\right) + 2 D \left(g\right) r i g h t \le f t h a r p \infty n s A \left(g\right) + 3 B \left(g\right)$

At equilibrium, ${K}_{c} = {Q}_{c}$:

${K}_{c} = \frac{\left[A\right] \times {\left[B\right]}^{3}}{{\left[C\right]}^{2} \times {\left[D\right]}^{2}} = {Q}_{c}$

When pressure is changed, you might think that ${Q}_{c}$ would change away from ${K}_{c}$ (because pressure changes are often caused by changes in volume, which factors into concentration), so the reaction position will shift to favour one side temporarily.

This doesn't happen, however!

When the volume is changed to cause a change in pressure, yes, the concentration will change.

$\left[A\right]$, $\left[B\right]$, $\left[C\right]$, and $\left[D\right]$ will all change, but here's the thing—they will all be changing by the same factor.

And, since there's an equal number of moles on each side, these changes will cancel out (since you can factor out a constant raised to $1 + 3 - \left(2 + 2\right) = 0$) to result in the same ${Q}_{c}$ so that it remains that ${Q}_{c} = {K}_{c}$.

It's unaffected, so the system is still at equilibrium and the position does not change.