IMAT 2020 Q47 - Equilibrium constant


please why can’t the answer to this question be A?

According to Le Chatelier’s principles, when we have a different number of gases on each side of the equation (here it’s 4:2), an increase in pressure will cause the reaction that produces the least number of gases to be favoured. So it would produce more P (product) and therefore increase the equilibrium constant. Please correct me if I am wrong, thanks a lot!

This is an exothermic reaction, meaning if we decrease temperature the right side will be favoured. If the right side is favoured (the side P is on) then the numerator increases and so K increases.

Also, the value of Kc is not affected by the change in pressure it solely depends on temperature changes and the heat of reaction whether it’s exothermic or endothermic.
Hope it helps:)


Thank you very much, this indeed helps a lot!

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hey, I wonder why doesn’t it depend on both of the factors? it DOES change by the temperature alteration indeed, but why can’t we conclude for the pressure as well ?
regardless of the questions options btw

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration , temperature, activation energy, and how often molecules collide with the correct orientation which together determines the rate constant K.

The equilibrium constant, Kc is the ratio of the rate constants, so only variables that affect the rate constants can affect Kc. Pressure doesn’t show in any of these relationships so it doesn’t have a effect on the kc of the reaction although pressure is involved in rate change but again it has no direct effect on rate constants so Kc doesn’t change either.
Hope this makes things clear:)