[Rutherford]

If I have a reaction:
A  B ΔH<0
If I increase the temperature, the rate constant of both the forward and the reverse reaction will increase, but the equilibrium will be shifted to the left. Does this mean that the numerical value of the equilibrium constant changes? It wouldn't be logic as the both rate constant increase. Do they maybe increase by a different amount?

[curiouscat]

If I have a reaction:
A  B ΔH<0
If I increase the temperature, the rate constant of both the forward and the reverse reaction will increase, but the equilibrium will be shifted to the left. Does this mean that the numerical value of the equilibrium constant changes?

Yes.

It wouldn't be logic as the both rate constant increase.

Why not?

Do they maybe increase by a different amount?

Yes

[Schrödinger]

Do they maybe increase by a different amount?

Just draw the energy vs reaction coordinate diagram and see for yourself. The activation energies for the forward and reverse reactions : they aren't the same. And you know that activation energy is related to the rate constant...

[Rutherford]

Yes, I skipped that  . So in overall, the equilibrium constant's value changes here (and we say that the equilibrium gets shifted). I suppose that a similar thing happens with pressure changes (now the factor A increases). Some factors change the value some don't (like concentration). Thanks for the help.

[Sophia7X]

Here's the equation to calculate the new Keq value after a temperature change:

Arrhenius equation:
ln(K2/K1) = -Ea/R(1/T1 - 1/T2)

[Rutherford]

Thanks for the formula. I understand the derivation.

These are rate constants for the reaction in one direction. I would need the rate constant for the reaction in the other direction to be able to calculate the K_eq. As their Ea-s are different the logarithms will have different values, it is obvious that the K_eq will change. For the pressure change, I think that it has something to do with the A factor (or Z*ρ) that describes the collisions (their frequency should increase as KE increases).

[curiouscat]

Yes, I skipped that  . So in overall, the equilibrium constant's value changes here (and we say that the equilibrium gets shifted). I suppose that a similar thing happens with pressure changes (now the factor A increases). Some factors change the value some don't (like concentration). Thanks for the help.

No. I think you are wrong. Keq is itself independent of P. It only changes with T.

The equilibrium composition itself does depend on Pressure though (in some cases). But that's not the same as saying K depends on P.
 

[Rutherford]

You mean that it is similar to a concentration change, both of them can change the frequency of collision, but the rates of both the reverse and direct reaction should change by the same amount as the factor A is not in the exponent of e.
I found on wikkipedia this: "In a condensed phase, the pressure dependence of the equilibrium constant is associated with the reaction molar volume". There is a formula, too. I don't understand this. What is a "condensed phase" and why is the equilibrium constant changing its value here?

[curiouscat]

You mean that it is similar to a concentration change, both of them can change the frequency of collision, but the rates of both the reverse and direct reaction should change by the same amount as the factor A is not in the exponent of e.
I found on wikkipedia this: "In a condensed phase, the pressure dependence of the equilibrium constant is associated with the reaction molar volume". There is a formula, too. I don't understand this. What is a "condensed phase" and why is the equilibrium constant changing its value here?

"condensed phase" = liq / solids

[curiouscat]

"In a condensed phase, the pressure dependence of the equilibrium constant is associated with the reaction molar volume". There is a formula, too. I don't understand this.

Note carefully that he's talking about Kx here not K



Kx and K are different. One can depend on P the other doesn't.

[Rutherford]

True thing. So, only temperature affects the K_eq. Thanks for clearing this up.

[curiouscat]

True thing. So, only temperature affects the K_eq. Thanks for clearing this up.

Glad to help.