Equilibrium?

The reaction below has an equilibrium constant, K, of 170 at 25°C.
2 NO2(g) N2O4(g)

If 2.2 10-3 mol of NO2 is present in a 22 L flask along with 2.1 10-3 mol of N2O4, is the system at equilibrium? And give answer to Q.

Answer:
If K = [N2O4(g)]/[NO2]^2 = 170

and [N2O4] = 2.1*10-3, [NO2]= 2.2*10-3

then [N2O4]/[NO2]^2 = (2.1*10-3)/(2.2*10-3)^2 = 433.88

so this is NOT at equilibrium

K = (N2O4) divided by (NO2) squared

Q = the same expression as K, except you plug in what you have as current conditions, and then compare the value to K

If the system is at equilibrium, then Q=K

If Q>K the system would shift to the left.

If Q<K the system would shift to the right.

I will assume that you want the Kc, using molarity values plugged into the K expression, instead of just moles, as the previous answer had done.

Concentration of NO2 = (2.2E-3 mol)/22L = 1E-4 mol/L

Concentration of N2O4 = (2.1E-3 mol)/22 L = 9.5E-5 mol/L

Q = (9.5E-5) divided by (1E-4) squared = 9.5E3

Since Q is larger than K, the reaction is not at equilibrium and the system would shift toward the left.

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Equilibrium?

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