Why does Kc only change with temperature?

Like why doesn't Kc change with pressure?
ok lets take the Haber process:

N₂ + 3H₂ ⇌ 2NH₃

If pressure is increased the the position of equilibrium will shift to the right to favour the products as its the side with fewer moles of gas. Therefore we get a higher yield of the product.

As Kc = [products]/[reactants] wouldn't that mean that the value of Kc should be larger at a higher pressure for the Haber process.

Btw im doing A level chem (AQA) so im not sure if they would ask a question like this, its just that im curious about this.

Cheers

(edited 7 years ago)

Reply 1

Pigster

Original post by chem revision

Kc = [products]²/[reactants]⁴

Increasing P, increases the conc of all species.

The bottom term grows faster than the top term,

The ratio no longer is equal to Kc.

The concs on the bottom get smaller and the concs on the top get bigger.

Until the ratio returns to the value of Kc.

Reply 2

chem revision

Original post by Pigster

ok so the bottom terms grows faster because its raised to the power of 4 rather than the power of 2?

So conc. of reactants will initially increase (and at a faster rate than products) due to the increase in pressure. (So at this point are the conc. of reactants greater than conc. of products?). At this point in time is it true to say the value of Kc has changed?

but then conc. of reactants decrease since the equilibrium will favour the products, leading conc. of products to increase as well. And this returns kc to its original value.

Reply 3

Pigster

Original post by chem revision

At this point in time is it true to say the value of Kc has changed?

Absolutely not. Kc is only changed by changes to T.

Like I said before, the ratio is no longer equal to Kc, i.e. the reaction is NOT in dynamic equilibrium.

Original post by chem revision

but then conc. of reactants decrease since the equilibrium will favour the products, leading conc. of products to increase as well. And this returns kc to its original value.

If you had the conc of all species = 1, then Kc = 1²/1⁴ = 1
Then you double the pressure, all concs double:
2²/2⁴ = 0.25 i.e. not Kc
So the forward reaction is favoured until:
2.71²/1.64⁴ = 1 = Kc once again.

(N.B. at the higher P, both concs are higher than at initial P)

Reply 4

chem revision

Original post by Pigster

Absolutely not. Kc is only changed by changes to T.

Like I said before, the ratio is no longer equal to Kc, i.e. the reaction is NOT in dynamic equilibrium.

If you had the conc of all species = 1, then Kc = 1²/1⁴ = 1
Then you double the pressure, all concs double:
2²/2⁴ = 0.25 i.e. not Kc
So the forward reaction is favoured until:
2.71²/1.64⁴ = 1 = Kc once again.

(N.B. at the higher P, both concs are higher than at initial P)