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


by crosbykins
Tags: chemistry, constant, equillibrim, reaction, reverse, urgent, write
crosbykins
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#1
Nov29-10, 12:28 PM
P: 53
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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zhermes
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#2
Nov29-10, 02:35 PM
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Quote Quote by crosbykins View Post
[b]
I don't think this is right but
You're right, it's not :) but this is right:

Quote Quote by crosbykins View Post
[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 )
p21bass
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#3
Nov29-10, 02:36 PM
P: 137
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
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Nov29-10, 03:08 PM
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URGENT: chemistry qs: write the equillibrim constant for the reverse reaction


Quote Quote by p21bass View Post
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
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#5
Nov29-10, 03:35 PM
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Quote Quote by zhermes View Post
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?
crosbykins
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#6
Nov29-10, 03:38 PM
P: 53
Quote Quote by Borek View Post
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
Borek
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#7
Nov29-10, 03:40 PM
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[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
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#8
Nov29-10, 03:48 PM
P: 53
Quote Quote by Borek View Post
[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.
so K1 * K2 =1

that means K 2 must equal 1/2*10^ 4
Borek
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#9
Nov29-10, 04:34 PM
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Wasn't that hard.
sjb-2812
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#10
Nov30-10, 02:21 PM
P: 418
Quote Quote by crosbykins View Post
so K1 * K2 =1

that means K 2 must equal 1/2*10^ 4
Not quite, try multiplying K1 and K2 to check.


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