In the chemical reaction where N2O4(g) is converted to 2NO2(g), if 0.5 M N2O4 and 0.15 M NO2 are present in the vessel, is the reaction at equilibrium? If not, which direction would the reaction proceed? (Kc = 4.4 × 10-3) N2O4(g) 2NO2(g)

Question

a. No, Q = 4.5 ×10-2 so the reaction should proceed in the reverse direction.
b. Yes, the reaction is at equilibrium.
c. No, Q = 0.18 so the reaction should proceed in the forward direction.
d. No, Q = 9.0 × 10-2 so the reaction should proceed in the reverse direction.

I believe the answer is B.
N2O4 --->2 NO2
I 0.5 0.15
C -x 2x
E 0.5 - x 0.15 - 2x

4.4 x 10^-3 = (0.15-2x)^2/0.5 - x
x = 0.053 or 0.096
0.5 - 0.053 = 4.5 x 10^-2
0.15 - 2(0.053) = 4.4 x 10^-2

Answer 1

B is not right. You have worked the problem as if it said N2O4 was 0.5M and NO2 was 0.15, the WHAT ARE THE CONCENTRATIONS AT EQUILIBRIUM. That is the problem. They want to know if the system is at equilibrium when those concentrations are present and not what will they be at equilibrium. The way to work the problem is to clculate Qc and compare that with Kc.

Answer 2

I believe it is A.
Qc = (0.15)^2/0.5 = 0.045
Kc = 0.0044
Qc > Kc so it is not in equilibrium and will go in the reverse direction I believe?

Answer 3

If you mean it goes to the left that is right.