6th Edition 11.45

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Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

6th Edition 11.45

Postby Tam To 1B » Wed Jan 09, 2019 9:47 pm

Hello!

I'm stuck on question 11.45 part b.
(a) A sample of 2.0 mmol Cl2 was sealed into a 2.0 L reaction vessel and heated to 1000. K to study its dissociation into Cl atoms. Use the info in Table 11.2 to calculate the equilibrium composition of the mixture.
(b) If 2.0 mmol F2 was placed into the reaction vessel instead of the chlorine, what would be its equilibrium composition at 1000. K?

Would you solve part (b) the same way as part (a) just with a different K value given by the table? I'm not sure why, but I keep getting a different answer from the textbook.

Thanks in advance!

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

Re: 6th Edition 11.45

Postby Andrea Zheng 1H » Wed Jan 09, 2019 10:14 pm

yes, you should use the same process as (a) but use the value of F2 <-> 2F of 1.2*10^-4. All the other calculations should remain constant. The ICE table would be 0.001-x for F2 and 2x for 2F. This would make the Kc value (2x)^2/(0.0010-x)=1.2*10^-4. Solving for x using the quadratic formula, you would get x=1.6*10^-4, discounting the other value (negative value) you get. You would get [F2] by subtracting the given x value from 0.001 (0.0010-1.6*10^-4) and get the value of 8*10^-4. Multiplying the x value obtained (x=1.6*10^-4) by the coefficient of 2 should give you [F].


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