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Topic: Calculating equilibrium constant (Kc)  (Read 3513 times)

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Offline Abder-Rahman

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Calculating equilibrium constant (Kc)
« on: March 22, 2012, 04:19:07 PM »
In my lecture notes, I have a question that says the following: The reaction H2+I2 <---> 2HI has an equilibrium constant (Kc) of 50 at 700k. If a sealed flask of volume 1 dm3 containing 0.005 mol of HI and 0.1 mol of HI is allowed to reach equilibrium, calculate the equilibrium concentration of hydrogen, iodine, and hydrogen iodide.

In the table used to solve this problem (contains amount of moles at start, amount of moles at equilibrium, ...etc), the instructor used the following values for the "amount of moles at equilibrium":

H2 ---> 0.005 + 1/2x
I2 ---> 1/2x
HI ---> 0.1 - x

Why didn't we say?

H2 ---> 0.005 - x
I2 ---> -x
HI ---> 0.1 + 2x

How did the instructor come up with his values?

Thanks a lot.

Offline fledarmus

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Re: Calculating equilibrium constant (Kc)
« Reply #1 on: March 22, 2012, 04:27:57 PM »
If you solve the problem both ways, what difference do you see in the final concentrations of the compounds?

Offline Abder-Rahman

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Re: Calculating equilibrium constant (Kc)
« Reply #2 on: March 22, 2012, 05:58:48 PM »
If you solve the problem both ways, what difference do you see in the final concentrations of the compounds?

Thanks for your reply.

The sensible answer I got was: x=0.0234

And, the instructor's sensible answer was x= 0.018569

What can you conclude about this? Both ways are true?

Thanks.

Offline Borek

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Re: Calculating equilibrium constant (Kc)
« Reply #3 on: March 22, 2012, 06:48:41 PM »
Try to calculate final concentration of iodine using both approaches. Especially your approach will give you a rather surprising value.
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Offline NathanielZhu

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Re: Calculating equilibrium constant (Kc)
« Reply #4 on: March 23, 2012, 02:13:19 AM »
Correct me if I'm wrong but I remember something I read in my textbook about if we multiply the coeficcent of a balanced equation by a certain number, that number squares Kc.
Like if:
A + B = C gives kc of C/AB
But if I multiply the equation by 2,
2A+2B=2C, then it gives me (C/AB)^2

Does that have anything to do with this problem?


Offline Borek

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Re: Calculating equilibrium constant (Kc)
« Reply #5 on: March 23, 2012, 05:39:11 AM »
Does that have anything to do with this problem?

No, it is much simpler.
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