[kgor93]

COBr2(g) <--> CO(g) + Br2(g)

Kc is 0.190 at 73 degrees C. Suppose you place 0.5 mol of COBr2 in a 2L flask and heat it to 73 degrees C. After equilibrium has been achieved, you add an additional 2 mol of CO.

a) When equilibrium is reestablished, what are the new equilibrium concentrations of COBr2, CO, and Br2?
b) How has the addition of CO affected the percentage of COBr2 that decomposed?

[sjb]

COBr2(g) <--> CO(g) + Br2(g)

Kc is 0.190 at 73 degrees C. Suppose you place 0.5 mol of COBr2 in a 2L flask and heat it to 73 degrees C. After equilibrium has been achieved, you add an additional 2 mol of CO.

a) When equilibrium is reestablished, what are the new equilibrium concentrations of COBr2, CO, and Br2?
b) How has the addition of CO affected the percentage of COBr2 that decomposed?

Take this in parts, so what is the composition of the equibrium initially, before adding the extra 2 moles of carbon monoxide?

[kgor93]

Get it in terms of 1L, so 0.5mol/2 = 0.25 mol.
COBr2 CO BR2
I 0.25 0 0
C 0.25-x +x +x
E 0.23 0.02 0.02

([CO][BR2])/[COBr2]=0.19
(2x)/(0.25-x)=0.19
x=0.02

Doesn't seem right to me, but I don't know what I'm doing wrong. Please correct me if I'm wrong because this is new to me.

[sjb]

Get it in terms of 1L, so 0.5mol/2 = 0.25 mol.
COBr2 CO BR2
I 0.25 0 0
C 0.25-x +x +x
E 0.23 0.02 0.02

([CO][BR2])/[COBr2]=0.19
(2x)/(0.25-x)=0.19
x=0.02

Doesn't seem right to me, but I don't know what I'm doing wrong. Please correct me if I'm wrong because this is new to me.

You need x multiplied by x on top, not x plus x

[kgor93]

You need x multiplied by x on top, not x plus x

Why is that? I thought since it was 1 mole each, it's 2x.

Edit: Oh, I see it now. Is it correct otherwise?

COBr2 CO Br2
I 0.25 0 0
C 0.25-x +x +x
E 0.107 0.14275 0.14275

([CO][BR2])/[COBr2]=0.19
(x^2)/(0.25-x)=0.19
x=0.14275

Then to add two more moles, would it be...
COBr2 CO Br2
I 0.107 1.14275 0.14275
C 0.107+x 1.14275-x 0.14275-x
E 0.23 2.02 0.02

([CO][BR2])/[COBr2]=0.19
((1.14275-x)(0.14275-x))/(0.107+x)=0.19
x=0.1 or 1.37

0.2 M COBr2, 1.04 M CO, 0.04 M Br2
x2
0.4 M COBr2, 2.08 M CO, 0.08 M Br2

[sjb]

I think in your second ICE table

C 0.107+x 1.14275-x 0.14275-x

should be

C 0.107+x 2.14275-x 0.14275-x,

but otherwise ok

[kgor93]

I think in your second ICE table

C 0.107+x 1.14275-x 0.14275-x

should be

C 0.107+x 2.14275-x 0.14275-x,

but otherwise ok

Wouldn't it be 1.14 because we had to divide the initial Mol count by 2 to have it in terms of 1L? That way if 2 mols are added, that would be 1 mol per liter.

[sjb]

Sorry, my mistake. Yes, you're right.

But, just a second gripe, be careful with units in your last line. You do not have e.g. 0.4 M of COBr₂, rather 0.4 moles...