[skylar]

(whenever I write ---> I mean inverted arrows as in an equilibrium reaction)

Ok...to express my question, I need to write 2 equilibrium equations first which I just made up:

1.- P ----> 2I
2.- P ----> I + J

For both of them, I put 5 mol dm-3 in a test tube. When equilibrium is reached for both of them, the remaining concentration of P is 3mol dm-3.

So, the calculation for Kc of reaction 1 would be:

(2 x 2)² /(3)
and the answer is Kc = 5.33

On the other hand, the calculation for reaction 2 would be:

(2)x(2) / (3)
giving a Kc of = 1.33, which is 4 times less than Kc of the first reaction.

Are my calculations right? Because if they are, I don’t understand why the equilibrium constant of the first reaction, (which is the same as saying P --- > I + I), is 4 times larger than the constant for P --- > I + J.

I’d have no problems if Kc1 was double Kc2, because in the first one there is only one element as a product while in the second one there are two… but why is there such a big difference?

[Dan]

1.- P ----> 2I

So, the calculation for Kc of reaction 1 would be:

(2 x 2)² /(3)
and the answer is Kc = 5.33

This is incorrect, you have a stray factor of 2. Recall that:

For

So for your equation 1:

K_c = [I ]²/[P]

[Borek]

[I ] = 2*2, that's from the stoichiometry. 2 moles produced for each of 2 moles P that reacted.