# : chemistry qs: write the equillibrim constant for the reverse reaction

1. Nov 29, 2010

### crosbykins

URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

1. The problem statement, all variables and given/known data

2. The equilibrium constant for the following reaction is 2.0 10^ 4

2HBr(g)  H2(g)  Br2(g)

2. Relevant equations

b) What is the equilibrium constant for the reverse reaction?

3. The attempt at a solution

I don't think this is right but

2.0 x 10^4 = (Br2)(H2)/(HBr)^2
2.0 x 10^4 = (x)(x)/(2x)^2
2.0 x 10^4 = 1/4

so for the reverse reaction K = 4 * 2.0 x 10^4.

2. Nov 29, 2010

### zhermes

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

You're right, it's not :) but this is right:

If that is the forward reaction, what would be the equation for the reverse reaction? (it might be easier to to just think of it as k = (A)(B)/(C) for C --> A + B )

3. Nov 29, 2010

### p21bass

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

In equilibrium the forward and reverse reaction rates are the same.

Do you know what the equations for the reaction rates for each side are?

4. Nov 29, 2010

### Staff: Mentor

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

While it can be done through reaction rates - it is unnecessary complication. It can be done much faster and much easier just by writing the equation for the reverse reaction (see zhermes post).

5. Nov 29, 2010

### crosbykins

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

ok it would be K = [HBr]^2/[H2][I2]

but how does that give the the value of K?

6. Nov 29, 2010

### crosbykins

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

ok so the equation for the reverse reaction is 2HBr -> H2 + I2 and K = [HBr]^2/[H2][I2]

...how can i get the value of K from this

7. Nov 29, 2010

### Staff: Mentor

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

$$K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4$$

$$K_2 = \frac {[HBr]^2} {[H_2][Br_2]}$$

If you still don't see the answer, try to calculate K1xK2.

Edit: note, that most of your equations are incorrect as they have HBr on one side of the reaction, I2 on the second.

8. Nov 29, 2010

### crosbykins

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

so K1 * K2 =1

that means K 2 must equal 1/2*10^ 4

Last edited: Nov 29, 2010
9. Nov 29, 2010

### Staff: Mentor

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

Wasn't that hard.

10. Nov 30, 2010

### sjb-2812

Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

Not quite, try multiplying K1 and K2 to check.