Solid ammonium carbamate decomposes as $NH X_{2} COONH X_{4} (s) 2 NH X_{3} (g) + CO X_{2} (g)$ . In an evacuated vessel at 300 K the total pressure at equilibrium is 0.60 atm. Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , $K_{c}$ (in $mo l^{3} L^{- 3}$ ) is approximately

Answer: A

The gases form in the ratio 2:1, so

p_{NH X_{3}} = \frac{2}{3} (0.60) = 0.40

atm and

p_{CO X_{2}} = 0.20

atm.

K_{p} = (0.40)^{2} (0.20) = 0.032 at m^{3}

Δ n = 3

, so

K_{c} = K_{p} / (RT)^{3}

with

RT = 24.63

K_{c} = 2.14 \times 1 0^{- 6} mo l^{3} L^{- 3}