A $3.0 L$ flask holds $0.30$ mol $NO X_{2}$ and a $2.0 L$ flask holds $0.10$ mol $N X_{2} O X_{4}$ , at the same temperature. For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{c} = 0.0065 mol L^{- 1}$ at this temperature. When the stopcock between the flasks is opened, the gases mix instantly. Calculate $Q_{c} / K_{c}$ at that instant (one decimal).

Numerical answer type, tolerance 0.1.

Answer: 27.7

Total volume is

5.0 L

[NO X_{2}] = 0.30/5.0 = 0.060

and

[N X_{2} O X_{4}] = 0.10/5.0 = 0.020 mol L^{- 1}

Q_{c} = (0.060)^{2} /0.020 = 0.180

Q_{c} / K_{c} = 0.180/0.0065 = 27.692

, i.e. 27.7, so the reaction shifts strongly backward.