[AhBeng]

In a reaction involving gases, eg. Haber process for manufacture of ammonia : 1N2 + 3H2 --> 2NH3

Increasing pressure shifts equilibrium to the right (since there are 2 moles of gas on RHS compared to 4 moles of gas on LHS; a mole of any gas occupies the same volume at the same temp and pressure).

And yet, as long as the temperature remains constant, equilibrium constant remains unchanged.

For many students, this is puzzling and counter-intuitive, because they imagine that equilibrium constant means the ratio of amount of product to amount of reactants. If this was so, then increasing pressure certainly changes this ratio.

I believe the key to solving this 'riddle', is in the definition of Equilibrium constant, defined mathematically as the molarity of product(s) / molarity of reactant(s), each raised to the power of its stoichiometric coefficient.

Therefore, in a manner of looking at it, it is in order to maintain this mathematical definition of Kc, the equilibrium must (and therefore does) shift when pressure is increased/decreased. That is to say, the ratio of the amount of product to reactant does shift, but the *concentration* ratio or "equilibrium constant" (defined mathematically rather than by a simplistic ratio of "amounts") remains, by its very definition, constant.

Do you agree? Any comments? Thanks!

[Yggdrasil]

I agree with your explanation.  Here it is useful to define a quantity called the reaction quotient, Q, which is equal to the molarity of product(s)/molarity of reactant(s), each raised to the power of its stochiometric coefficient.  When the system is at equilibrium, Q = K_c (this becomes the new definition of K_c).  When Q < K_c, the forward reaction becomes thermodynamically favorable and when Q > K_c, the backward reaction becomes thermodynamically favorable.

So, when the pressure increases for the Haber process, Q decreases (since the change in pressure affects the denominator of Q [i.e. the reactants] more than the numerator of Q [i.e. the products]).  This means Q < K_c and the reaction will go forward the reaction reaches a set of new equilibrium concentrations of products and reactants.

[AhBeng]

Thanks ("snacks" as appreciation), Yggdrasil! It's always great to have confirmation of one's ideas, and a further clarification and discussion of such.

Chemistry (and Biology and every subject!) sure becomes so much more fun (as they should be!) when everything makes perfect sense! 

[Donaldson Tan]

Let total pressure of system be P_o.
Let P_i be the partial pressure of component i
Let X_i be the mole fraction of component i

x_N2 + x_H2 + x_NH3 = 1

K_p = P_NH3²/(P_N2P_H2³)
=> K_pP_o² = x_NH3²/(x_N2x_H2³)

For an isothermal process, K_p is a constant.

According to the above equations, if P_o increases, then x_NH3² must increase, accompanied by the reduction in  x_N2x_H2³