Question

In a reaction (at equilibrium) that makes more moles of gas than it consumes, what is the effect of increasing the pressure?

Answer 1

Well, if you make more mols of gas...

`aA(g) -> bB(g)`

where `b > a` and these are stoichiometric coefficients.

Write the equilibrium constant.

`K_P = P_B^b/(P_A^a)`

As `b > a`, let `b = a + c`, where `c` is any positive constant. Then:

`K_P = P_B^(a+c)/(P_A^a)`

`= (P_B/P_A)^a P_B^c`

If you increase the total pressure, you must increase the partial pressures.

Even if the mols of `B` are equal to the mols of `A` (instead of larger), even though the partial pressures increase by the same amount, the `P_B^c` out front causes the `K_P` to increase overall.

Therefore, the equilibrium is skewed towards the products, and by Le Chatelier's principle, it wishes to undo the imbalance and shift towards the reactants.