Question

2 SO2(g) + O2(g) <==> 2SO3(g)

Initially, 1.6mol SO3 is placed in a 3.0L container. At equilibrium, [O2] = 0.15M. What is the value of Keq?

A. 0.26
B. 1.2
C. 4.0
D. 43

My work:
...2SO2...+...O2...<==>...2SO3
I 0.53M.....-----.........-----
C +0.15.....+0.15........-0.15
E 0.53+0.15..0.15........-0.15

Keq = (-0.15)^2/(0.53+0.15)^2(0.15)= .321M

what am I doing wrong?

Answer 1

oh I realized, I have my 0.53M on 2SO2 and not 2SO3.

Keq = (0.53-0.30)^2/(0.30)^2(0.15) = 0.26

Is this correct? :)

Answer 2

Looks like the second set up is correct but you punched in the wrong numbers on your calculator. I get about 4 or close to that.

Answer 3

I get 3.918... so does that mean I round up to 4.0? so the answer would be C?

Answer 4

Yes and no.
3.9 is the correct answer and I would not round up to 4 since you are allowed two s.f. in the answer. But you choose 4.0 as the answer since that's the closest of the choices.