Question

For the decomposition of the compound, represented as
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
the $K_p=2.9\times 10^{-5}at{m}^3.$
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)

$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$

$K_p=\frac{{\left(P_{N{H}_3}\right)}^2\times \left(P_{C{O}_2}\right)}{P_{N{H}_2}COON{H}_4\left(s\right)}$

$={\left(P_{N{H}_3}\right)}^2\times \left(P_{c{o}_2}\right)$

As evident by the reaction, $N{H}_{3}andC{O}_2$ are formed in molar ratio of 2:1. Thus if P is the total pressure of the system at equilibrium, then

$P_{N{H}_3}=\frac{2\times p}{3}{P}_{c{o}_2}=\frac{1\times P}{3}$

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

Given $K_p=2.9\times 10^{-5}$

$\therefore 2.9\times 10^{-5}=\frac{4p^3}{27}$

$p^3=\frac{2.9\times 10^{-5}\times 27}{4}$

$P={\left(\frac{2.9\times 10^{-5}\times 27}{4}\right)}^{1/3}=5.82\times 10^{-2}atm$