[maakii]

I don't understand one part about the equilibrium constant K_p. If I am not wrong, K_p tells you about the extent of the reaction at equilibirum, i.e. the proportion of products to reactants. If products >>reactants, then K_p is large. If reactants >>products, k_p is small.

Now i understand that adding a catalyst will not change the k_p, as the ratio of products to reactants remain unchanged. Also, the rates of the forward and backwards reactions are equally increased.

For exothermic reactions, increasing temperature will shift equilibrium to the left, and K_p decreases.

However, why does pressure not affect the equilibrium constant? For a reaction

N₂ + 3H₂ -> 2NH₃

Increasing pressure favours the forward reaction. And when equliibrium is reached again, there should be an increase in the proportion of products to reactants, and looking at the extent of reaction, we should see that it lies furthur to the left, and thus K_p should increase.

But why does it not?

[xiankai]

i.e. the proportion of products to reactants.

in terms of partial pressures.

But why does it not?

the apparent increased value for K_p is offset by the increase in pressure term in the calculation of partial pressure.

[charco]

If you consider a gaseous system at equilibrium then the total number of moles is constant, the temperature is constant and so according to the gas laws if you wish to increase the pressure you MUST change the volume (decreasing it).
Concentration is given by moles/volume so you therefore change (increase) the concentrations as well. If the moles on either side of the equilibrium are unequal then the concentration must increase by unequal amounts temporarily giving proportions that are not now equal to Kc. The system then responds to RESTORE the value of Kc and in doing so moves towards the side of fewer moles.

You can demonstrate this with simple maths.

In the equation: A <==> 2C
let the [A]=a and the [C]=b

Kc = b²/a

if the pressure is then doubled (by halving the volume) then the concentraton of C becomes 2b and the concentration of A becomes 2a

the ratio [products]²/[reactants] now equals 4b²/2a

which is precisely double the original ratio. The system is not at equilibrium as this ratio does not have the value of Kc and so restores the value of Kc by reducing the [products] and increasing the [reactants].

In other words the system moves towards the side of fewer moles to restore the value of Kc.

In all of this I have used Kc but the same logic can be applied to Kp. It's just easier to visualise with Kc that's all.

[maakii]

Thanks, I get it now  you've been a great *delete me*