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

crosbykins

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.

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 )

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?

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).

crosbykins

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

but how does that give the the value of K?

crosbykins

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

Borek

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

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

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.

crosbykins

so K1 * K2 =1

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

Borek

Wasn't that hard.

sjb-2812

Not quite, try multiplying K1 and K2 to check.