For the decomposition reaction NH2COONH4 (s) → 2NH3(g) + CO2(g) The KP = 2.9 ×10–5 atm3. The total pressure of gases at  equilibrium when 1 mole of NH2COONH4 (s) was taken to start with would be

Q: For the decomposition reaction

NH2COONH4 (s) $ \rightleftharpoons $ 2NH3(g) + CO2(g)

The KP = 2.9 ×10–5 atm3. The total pressure of gases at  equilibrium when 1 mole of NH2COONH4(s) was taken to start with would be

(A) 0.0194 atm

(B) 0.0388 atm

(C) 0.0582 atm

(D) 0.0766 atm

Solution:  NH2COONH4(s) $ \rightleftharpoons $ 2NH3(g) + CO2(g)

1                                  2                1

KP = 2.9 × 10–5 atm3

If the P is the total pressure at equilibrium

$\large K_P = (\frac{2P}{3})^2 (\frac{P}{3}) $

$\large 2.9 \times 10^{-7} = (\frac{2P}{3})^2 (\frac{P}{3}) $

$\large P^3 = 1.9575 $

P = 0.0582

Ans: (C)

Ans: