studyingpeach
Badges: 11
Rep:
?
#1
Report Thread starter 2 weeks ago
#1
Hey guys, any help would be appreciated:

If the initial [H2] = 0.200 mol dm-3, [I2] = 0.200 mol dm-3 and Kc = 55.6 at 250C calculate the equilibrium concentrations of all molecules.

I keep on getting stuck because the volume hasn't been given? Any tips?
0
reply
studyingpeach
Badges: 11
Rep:
?
#2
Report Thread starter 2 weeks ago
#2
(Original post by laloon)
Seems that to answer the question we will need to solve a quadratic?In essence, the volume doesn't need to be known as we are given concentrations...
We know that Kc = [HI]^2 / [H2][I2] … from the equilibrium...
If final [HI] is x, then the final [H2] and [I2] is 0.2 - (x/2) <=== look at the mole ratio of the rxn...
then its just a matter a substitution and solving a quadratic I guess unless you sqrt during the working out
Thanks for the help, I worked it out in the end
0
reply
Pigster
Badges: 19
Rep:
?
#3
Report 2 weeks ago
#3
(Original post by studyingpeach)
Hey guys, any help would be appreciated:

If the initial [H2] = 0.200 mol dm-3, [I2] = 0.200 mol dm-3 and Kc = 55.6 at 250C calculate the equilibrium concentrations of all molecules.

I keep on getting stuck because the volume hasn't been given? Any tips?
Assuming you're talking about H2 + I2 <-> 2HI (as if you're not my answer doesn't apply)...

Kc = (n(HI)/V)2 / (n(H2)/V) x (n(I2)/V)

Hopefully you notice the V terms cancel out. But as has been said, you're using concs anyway, but if the volume was 1 dm3, then the values for the concs are the same as for the amounts, so could be interchanged.

No quadratics are needed.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How would you want Freshers' Fairs to work this year?

Completely online (9)
14.52%
A mix of in-person and online options (9)
14.52%
As many in-person events as is safe to do (38)
61.29%
I don't plan on attending a Freshers' Fair (6)
9.68%

Watched Threads

View All