Question

For the decomposition of the compound, represented as
$NH_{2}COONH_{4}\left(s\right) {\rightleftharpoons} 2NH_{3}\left(g\right) + CO_{2}\left(g\right)$
the $K_{p} = 2.9\times10^{-5} atm^{3}.$
If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be :

• A. $1.94 \times 10^{-2}\, atm$
• B. $5.82 \times 10^{-2} \,atm$
• C. $7.66 \times 10^{-2} \,atm$
• D. $38.8 \times 10^{-2} \,atm$

Answer 1

Answer: (B)

$NH_{2} COONH_{4} \left(s\right) {\rightleftharpoons} 2NH_{3} \left(g\right)+CO_{2} \left(g\right)$

$K_{p} =\frac{\left(P_{NH_3}\right) ^{2} \times\left(P_{CO_2}\right)}{P_{NH_2}COONH_{4} \left(s\right)}$

$=\left(P_{NH_3}\right)^{2} \times\left(P_{co_2}\right)$

As evident by the reaction, $NH_{3} and CO_{2}$ are formed in molar ratio of 2:1. Thus if P is the total pressure of the system at equilibrium, then

$P_{NH_3}=\frac{2\times p}{3} P_{co_2} =\frac{1\times P}{3}$

$K_{p} =\left(\frac{2p}{3}\right)^{2} \times\frac{p}{3}=\frac{4p^{3}}{27}$

Given $K_{p} =2.9\times10^{-5}$

$\therefore \quad2.9\times10^{-5}=\frac{4 p^{3}}{27}$

$p^{3} =\frac{2.9\times10^{-5}\times27}{4}$

$P=\left(\frac{2.9\times10^{-5}\times27}{4}\right)^{ 1/3} =5.82\times10^{-2} atm$