Confused about Kp and Kc - help please

Biolover2020

Hi, im a bit confused why temperature affects the equilibrium constants, but changes in concentration and pressure don't? I dont understand the explanation in my textbook, could someone please help with this?

Reply 1

ecoworrier

Equilibrium will shift to counteract any changes in concentration and pressure. Kc and Kp do change with temperature as depending on whether the forward reaction is exo or endothermic, equilibrium will shift. If the yield of products increases, Kc and Kp will increase as there is a greater number at the top of the fraction/expression.
Sorry if that explanation was bad, hope it helps? If not, I like Allery Chemistry on youtube but it depends on your spec. :smile:

Reply 2

Tagsmags

I found this quite confusing at first, and also felt my textbook was far from clear in explaining it.

The equilibrium constant, k, shows the relationship between the concentration of reactants and products at equilibrium.

Lets use this equilibrium for an example.

a ⇌ b + c ,

k= x [c] / [a]

Out of the temperature, the pressure and the concentrations of reactants, only changing the temperature will actually change the value of k, so the ratio of ( x [c]) to [a] actually changes.

By altering the concentration and the pressure (given that there is a different number of moles of products and reactants) you are actually changing concentration of the products and/or reactants themselves, so that x [c] / [a] no longer equals k. The equilibrium will therefore respond by changing the concentrations until [b] x [c] / [a] equals k again.

With concentration this is quite straightforward; by increasing/ decreasing the concentration of one of the reactants/products you are changing the value of x [c] / [a] , therefore the concentrations will change (shifting the position of the equilibrium) until the ratio equals K again.

When you change the pressure of the reaction you are essentially changing the concentration, for example if you half the pressure of a reactant at concentration 1dm^3 the concentration will drop to 0.5dm^3.

If there are equal amounts of moles of both the products and reactants, lets say in the equilibrium

a ⇌ b , k= / [a]

both [a] and will decrease by the same factor therefore the value of / [a] will be unchanged (and still equal k).

However if you have different amount of moles in the products to the reactants (as in our first equilibrium) it will change the value of x [c] / [a]. This can be demonstrated by saying that the values of [a], and [c] are all equal 1 at equilibrium;

1 x 1 / 1 = 1, therefore k = 1.

If we half the concentration by halving the pressure you can see that this will change the value of x [c] / [a] so that it no longer equals k.

0.5 x 0.5 / 0.5 = 0.5, 0.5 ≠ 1.

Therefore the concentration of the products (b and c in this case) will increase until k is restored to 1.

I hope this makes some sense.

(edited 11 months ago)

Reply 3

Biolover2020

[quote="Tagsmags;98377886"]I found this quite confusing at first, and also felt my textbook was far from clear in explaining it.

The equilibrium constant, k, shows the relationship between the concentration of reactants and products at equilibrium.

Lets use this equilibrium for an example.

a ⇌ b + c ,

k= x [c] / [a]

Out of the temperature, the pressure and the concentrations of reactants, only changing the temperature will actually change the value of k, so the ratio of ( x [c]) to [a] actually changes.

By altering the concentration and the pressure (given that there is a different number of moles of products and reactants) you are actually changing concentration of the products and/or reactants themselves, so that x [c] / [a] no longer equals k. The equilibrium will therefore respond by changing the concentrations until x [c] / [a] equals k again.

With concentration this is quite straightforward; by increasing/ decreasing the concentration of one of the reactants/products you are changing the value of x [c] / [a] , therefore the concentrations will change (shifting the position of the equilibrium) until the ratio equals K again.

When you change the pressure of the reaction you are essentially changing the concentration, for example if you half the pressure of a reactant at concentration 1dm^3 the concentration will drop to 0.5dm^3.

If there are equal amounts of moles of both the products and reactants, lets say in the equilibrium

a ⇌ b , k= / [a]

both [a] and will decrease by the same factor therefore the value of / [a] will be unchanged (and still equal k).

However if you have different amount of moles in the products to the reactants (as in our first equilibrium) it will change the value of x [c] / [a]. This can be demonstrated by saying that the values of [a], and [c] are all equal 1 at equilibrium;

1 x 1 / 1 = 1, therefore k = 1.

If we half the concentration by halving the pressure you can see that this will change the value of x [c] / [a] so that it no longer equals k.

0.5 x 0.5 / 0.5 = 0.5, 0.5 ≠ 1.

Therefore the concentration of the products (b and c in this case) will increase until k is restored to 1.

I hope this makes some sense.
Thank you this really helps!