For $2 NH X_{3} (g) N X_{2} (g) + 3 H X_{2} (g)$ at 700 K, $K_{c} = 2.3 \times 1 0^{- 3} mo l^{2} L^{- 2}$ . Take $R = 0.0831 L bar mo l^{- 1} K^{- 1}$ and calculate $K_{p}$ in $ba r^{2}$ , correct to one decimal place.

Numerical answer type, answer in bar^2, tolerance 0.1.

Answer: 7.8 bar^2

Δ n = 4 - 2 = + 2

, so

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

RT = 0.0831 \times 700 = 58.17

L bar/mol, and

K_{p} = 2.3 \times 1 0^{- 3} \times (58.17)^{2} = 7.783 ba r^{2}