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

[crosbykins]

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

Homework Statement

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

2HBr(g)  H2(g)  Br2(g)

Homework Equations

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

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]

I don't think this is right but

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

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 )

 

[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]

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).

 

[crosbykins]

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/

[I2]

but how does that give the the value of K?

 

[crosbykins]

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/

[I2]

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

 

[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.

 

[crosbykins]

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

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

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

 

[Borek]

Wasn't that hard.

 

[sjb-2812]

so K1 * K2 =1

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

Not quite, try multiplying K1 and K2 to check.