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

1 Answer
Truong-Son N.
Feb 9, 2017

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.

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