Another Equilibrium Constant question

[ilzmastr]

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Somehow, I can't get this from the previous posts:

If equilibrium constant K is a ratio of the forward reaction's rate constant kf to the reverse reaction's rate constant kr, then K=kf/kr

Why does changing temperature not affect kf and kr symmetrically, ie why does changing Temperature change K?

Why does changing pressure affect kf and kr symmetrically, ie changing pressure does not change K?

 

[ilzmastr]

Hmmmm...could it be that from the Arrhenius equation, k=zpe^(-Ea\RT), pressure dependence of a rate constant is negligible?

 

[Night Hawk]

Hmmmm...could it be that from the Arrhenius equation, k=zpe^(-Ea\RT), pressure dependence of a rate constant is negligible?

This, unless you're dealing with gasses, in which the R can be replaced with: PV/nT. Normally chemical rxn's are at atmospheric pressure and we know liquids to be incompressable but this isn't always true of gasseous reactions.

 

[Godric]

Changing the temperature changes the numerical value for Keq because temperature supply's energy to the reaction so at different temps you are supplying differ amounts of energy. So the reason Kf and Kr aren't changed equally when energy is added (heat) is because it's not a linear relationship, all depends on the system.

When you change the pressure or add something at equilibrium the system will try to undo the stress you placed on it, however this will not change Keq at that temp. The relative amounts may change to readjust the system back to equilibrium.

You stated that if you change pressure that "Why does changing pressure affect kf and kr symmetrically, ie changing pressure does not change K?"

Ans: It only changes equally only if you have the same amount of moles on both side of the reaction, if not then it changes the Kf and Kr unequally