For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{p} = 0.14$ atm at a given temperature. In a vessel $p_{N X_{2} O X_{4}} = 0.75$ atm. The reaction goes backward (dimerisation) only if $p_{NO X_{2}}$ exceeds a critical value. Find this critical $p_{NO X_{2}}$ in atm (two decimals).

Numerical answer type, answer in atm, tolerance 0.01.

Answer: 0.32 atm

Backward reaction needs

Q_{p} > K_{p}

p_{NO X_{2}}^{2} /0.75 > 0.14

, so

p_{NO X_{2}} > 0.14 \times 0.75 = 0.1050 = 0.3240

atm, i.e.

0.32

atm to two decimals.