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# Kc and Kp help. Watch

1. Have I got this right?

The only factor that alters the equilibrium constant is temperature. For exothermic reactions, an increase in temperature results in lower Kc values. For endothermic reactions, an increase in temperature results in higher Kc values. An increase in temperature always results in an increase in the rate of reaction.

I was wondering if this applies to Kp as well? Does increasing pressure or the addition of a catalyst alter the value of Kp?
2. K = Ae-Ea/RT

Thus is K effected by both Temperature and Activation Energy. And the effect of a catalyst is.......
3. (Original post by gingerbreadman85)
K = Ae-Ea/RT

Thus is K effected by both Temperature and Activation Energy. And the effect of a catalyst is.......
So a catalyst lowers Ea and so would lower K for exothermic reactions and increase K for endothermic reactions?

Is this K, Kc or Kp or both as we haven't been taught the formula and it's not in the text book...?

Thumbs up given.

Edit: ...but that K is the rate constant K not the equilibrium constant Kc or Partial pressure constant Kp.
4. (Original post by chaz1992)
So a catalyst lowers Ea and so would lower K for exothermic reactions and increase K for endothermic reactions?

Is this K, Kc or Kp or both as we haven't been taught the formula and it's not in the text book...?

Thumbs up given.

Edit: ...but that K is the rate constant K not the equilibrium constant Kc or Partial pressure constant Kp.
Yes, your edit is correct.

Catalyst changes the rate of attainment of equilibrium, but not the position. Hence Kc, Kp stay the same.
5. (Original post by charco)
Catalyst changes the rate of attainment of equilibrium, but not the position. Hence Kc, Kp stay the same.
Indeed, i'm getting my Ks mixed up.

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