# Chemical Equilibrium - equilibrium constant

Lol, I just took a test on the Chemical Kinetics chpater I been asking help on and I got a 70.. :( ..after all that studying. For some reason, the last couple of chapter tests have not been too good for me, not sure why, maybe because we are starting to get towards the end of the book.

Anyways, the question I am having trouble with:

Suppose that a gas-phase reactions A --> B and B --> are both elementary processes with rate constants of 3.8 x 10^-2 s- and 3.1 x 10^-1 s-, respectively. (a) What is the value of the equilibrium constant for the equilibrium reaction A(g) B(g) ? (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Explain.

I've tried to use K(forward) / K(reverse) = a equilibrium constant. But that doesnt give me the right answer..help.

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That should get the right answer. That's what I did. Could you have made a miscalculation?

The rate expression for the forward reaction must be:

Ratef = kf x [A]

and the rate expression for the revrese reaction must be:

Rate r= kr x [B]

At equilibrium the Rate f = Rate r so

kf x [A] = kr x [B]

Rearranging yields

kf/kr = [B]/[A] = Keq

Therefore

Keq = 3.8 x 10-2 / 3.1 x 10-1 = .123

Since the value of Keq is less than 1, the reactants are favored and the partial pressure of A will be greater than the partial pressure of B (Based on the ideal gas equation, pressure is directly related to Molarity - P = n/v xRT)

spock

Wed, 2008-02-27 20:48

No, because the answer given in the back of the book is:

(a) K(p) = K(c) = 2.8 x 10^-2

(b) Since K(f)

Sunil

Wed, 2008-02-27 20:52

So basicly we agree with the book with the exception of the numeric value for the equilibrium constant. If you came up with the same value as I did, I think the book is wrong.

spock

Wed, 2008-02-27 21:06

Yes I did come up with the same value as you, so I am also thinking the book is wrong, because there is really no other way to do this problem.

Sunil

Wed, 2008-02-27 21:08