## 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.
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 Quote by crosbykins [b] I don't think this is right but
You're right, it's not :) but this is right:

 Quote by crosbykins [b] 2.0 x 10^4 = (Br2)(H2)/(HBr)^2
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 )
 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?

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

 Quote by p21bass 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?
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).

 Quote by zhermes 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 )
ok it would be K = [HBr]^2/[H2][I2]

but how does that give the the value of K?

 Quote by Borek 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).
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
 Admin $$K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4$$ $$K_2 = \frac {[HBr]^2} {[H_2][Br_2]}$$ 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.

 Quote by Borek $$K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4$$ $$K_2 = \frac {[HBr]^2} {[H_2][Br_2]}$$ 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.
so K1 * K2 =1

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