For $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ , $K_{c} = 0.50 L^{2} mo l^{- 2}$ at $400 K$ . Use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ . A mixture at $400 K$ has partial pressures $p_{N X_{2}} = 2.0$ , $p_{H X_{2}} = 6.0$ and $p_{NH X_{3}} = 1.5$ atm. Which conclusion is correct?

ABackward, because $Q_{p} = 5.21 \times 1 0^{- 3} at m^{- 2}$ exceeds $K_{p} = 4.64 \times 1 0^{- 4} at m^{- 2}$
BForward, because $Q_{p} = 5.21 \times 1 0^{- 3} at m^{- 2}$ is below $K_{c} = 0.50$
CForward, because $Q_{p} = 5.21 \times 1 0^{- 3} at m^{- 2}$ exceeds $K_{p} = 4.64 \times 1 0^{- 4} at m^{- 2}$
DBackward, because $K_{p} = K_{c}$ here and $Q_{p} > K_{c}$

Answer: A

K_{p} = K_{c} (RT)^{Δ n}

with

Δ n = - 2

gives

K_{p} = 0.50/ (0.0821 \times 400)^{2} = 4.64 \times 1 0^{- 4} at m^{- 2}

Q_{p} = (1.5)^{2} / (2.0 \times 6. 0^{3}) = 5.21 \times 1 0^{- 3} at m^{- 2}

. Since

Q_{p} > K_{p}

the net reaction is backward. Comparing

Q_{p}

directly with

K_{c}

would wrongly suggest a forward reaction.