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Topic: Rates of reactions, help  (Read 2900 times)

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Offline matrix_the01

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Rates of reactions, help
« on: September 12, 2009, 10:35:08 AM »
Hi I've had some trouble answering the question below, could I have a few suggestions on how to answer it,
as nobody seems to have a set way of answering it.

The following reaction has reached equilibrium in a closed container:
C(s) + H2O(g)  ::equil:: CO(g) + H2(g)  :delta:H>0
The pressure of the system is decreased by increasing the volume of the container.
How will the [H2](concentration of H2) and the value of Kc be affected when the
new equilibrium is established?
(Assume that the temperature of the system remains unchanged).

Any help would be appreciated! ;D

Offline orgoclear

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Re: Rates of reactions, help
« Reply #1 on: September 12, 2009, 11:57:02 AM »
What is the equilibrium constant of the expression?
Kc = [CO][H2]/[H2O]
if you increase the volume of the container by x times (say) then the pressure will become 1/x times for each component.
i.e. concentration (assuming that the no. of moles is unchanged) is decreased by x times
So, Qc = ([CO]/x)([H2]/x)/([H2O]/x) = [CO][H2]/[H2O]x
So Qc<Kc.
So the reaction will proceed in the forward direction.

Kc of course will remain unaffected. Kc depends only on temperature

[H2] will be more that [H2](at prev. equilibrium)/x but you cant say its relation between the initial concentration unless you know the quantitative values of Kc, how many times vol. was increased, and the initial moles taken or the equilibrium concentrations

Offline matrix_the01

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Re: Rates of reactions, help
« Reply #2 on: September 12, 2009, 11:56:22 PM »
Hey, orgoclear, thanks for the reply  :)

I'm still a bit confused...
You say the [H2] increases.
But does the the change in concentration (H2) affect anything?
c=n/v
If V increases, and n (H2) decreases (since the reverse reaction is favored), shouldn't the concentration decrease. 8-o
c (H2) = n (decrease)/V (increase)

The only reason I'm asking is in some questions like this, they ignore the affect of volume on the concentration calculation, so
I'm just seeking clarity.

Thanks for your help again !

Offline orgoclear

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Re: Rates of reactions, help
« Reply #3 on: September 13, 2009, 12:11:03 AM »
see carefully what I have written.
Let the concentration of H2 before volume was increased be C
If the volume increases x times then,
the concentration of H2 after this equilibrium is reached will be more than C/x

It may or may not be more than C itself

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