For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{c} = 0.50 mol L^{- 1}$ at 400 K. Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and calculate $K_{p}$ in atm, to two decimal places.

Numerical answer type, answer in atm, tolerance 0.005.

Answer: 16.42 atm

Δ n_{g} = 2 - 1 = 1

, so

K_{p} = K_{c} RT

RT = 0.0821 \times 400 = 32.84

, hence

K_{p} = 0.50 \times 32.84 = 16.42 atm