A mixture of $N X_{2}$ and $H X_{2}$ in the ratio 1:3 by moles reaches equilibrium $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ at 700 K and a total pressure of 50 atm, where the mole fraction of $NH X_{3}$ is 0.30. Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , calculate $K_{c}$ in $L^{2} mo l^{- 2}$ , correct to one decimal place.

Numerical answer type, answer in L^2 mol^-2, tolerance 0.1.

Answer: 4.7 L^2 mol^-2

The remaining 0.70 of the mole fraction stays in the ratio 1:3, so

x_{N X_{2}} = 0.175

and

x_{H X_{2}} = 0.525

K_{p} = 0.3 0^{2} / (0.175 \times 0.52 5^{3} \times 5 0^{2}) = 1.422 \times 1 0^{- 3} at m^{- 2}

Δ n = - 2

, so

K_{c} = K_{p} (RT)^{2}

with

RT = 0.0821 \times 700 = 57.47

K_{c} = 4.695