Question

For the decomposition of the compound, represented as
\(\mathrm{NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g)}\)
the \(K_p = 2.9\times10^{-5}\ \mathrm{atm^3}\).
If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :

• A. \(1.94\times10^{-2}\ \mathrm{atm}\)
• B. \(5.82\times10^{-2}\ \mathrm{atm}\)
• C. \(7.66\times10^{-2}\ \mathrm{atm}\)
• D. \(38.8\times10^{-2}\ \mathrm{atm}\)

Hint:

Gases are formed in the ratio NH3 : CO2 = 2 : 1, so write partial pressures as 2/3 P and 1/3 P.

Answer 1

The Correct Option is B

Let the total pressure be P. NH₃ and CO₂ are formed in 2 : 1 ratio, so \(p_{NH_3}=\tfrac{2}{3}P\) and \(p_{CO_2}=\tfrac{1}{3}P\).
\(K_p=(p_{NH_3})^2\,p_{CO_2}=\tfrac{4}{9}\cdot\tfrac{1}{3}P^3=\tfrac{4}{27}P^3\).
\(P^3=\dfrac{27\times2.9\times10^{-5}}{4}=1.9575\times10^{-4}\).
\(P=(1.9575\times10^{-4})^{1/3}\approx5.8\times10^{-2}\) atm, i.e. \(5.82\times10^{-2}\) atm.
Answer: (2) \(5.82\times10^{-2}\ \mathrm{atm}\)