if volume/pressure of a system changes, then how can equilibrium shift without changing Keq?

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tdod

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for example, if the pressure decreases then the following reaction will shift left:

1HCl(g) + 1CO(g) --><--- 1HCOCl(g)

however, Keq will remain the same since temperature doesn't change. but how is it that the Keq doesn't change if the denominator increases and the numerator decreases??
 
Saying the "equilibrium shifts" is a little misleading. You are correct that the equilibrium constant (Kp in this case) will be constant at the same temperature. What we mean by a "shift" is that, in response to a change in pressure (volume), the reaction ceases to act as a dynamic equilibrium and experiences net progress in one particular direction. This is because the stress upon the system (changed pressure) changes the reaction quotient Q such that Q ≠ K, and thus net shift in the relative amounts of reactants and products occurs until Q = K and the reaction is again in dynamic equilibrium.

For example, let's saw your reaction has a Kp = 1. The reaction will initially be at equilibrium where P = 1 atm for each species: Q = (1 atm HCOCl) / (1 atm HCl)(1 atm CO) = 1 = K.

If we double the volume of the reaction vessel, we see the P drops to 0.5 atm for each species. Now, Q = (0.5 atm HCOCl) / (0.5 atm HCl)(0.5 atm CO) = 2 ≠ K. Now Q > K, and the reaction must shift to reestablish equilibrium.

Solving with the quadratic we can find that after the shift the P_HCOCl = 0.382 atm and P_reactants = 0.618 atm for both HCl and CO. Now we see:

Q = (0.382 atm HCOCl) / (0.618 atm HCl)(0.618 atm CO) = 1 = K. The reaction is again at equilibrium, but now P_product ≠ P_reactants.
 
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