Question

Solid ammonium carbamate dissociates into $N{H}_3$ and $C{O}_2$ as $N{H}_{2}COON{H}_4\rightleftharpoons N{H}_3\left(g\right)+C{O}_2\left(g\right).$ The total pressure at equilibrium is 5 atm. Calculate the equilibrium constnat $K_p.$

• A. $37.04at{m}^2$
• B. $22.52at{m}^2$
• C. $18.52at{m}^2$
• D. $45.04at{m}^2$

Answer 1

The correct option is C $18.52at{m}^2$

Given reaction is:
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
Given that total pressure of the gases in equilibrium, $p_T=5atm$
$K_p=p_{N{H}_3}^2.p_{C{O}_2}$
using Dalton's law of partial pressure we have
$p_i={\chi }_i\cdot p_T$
thus, $p_{N{H}_3}={\chi }_{N{H}_3}\cdot p_T$
${\chi }_{N{H}_3}=\frac{n_{N{H}_3}}{n_{N{H}_3}+n_{C{O}_2}}=\frac{2}{2+1}$
${\chi }_{N{H}_3}=\frac{2}{3}$
Similarly, ${\chi }_{C{O}_2}=\frac{1}{2+1}=\frac{1}{3}$
upon substitution we get:
$K_p={(\frac{2}{3}\times 5)}^2\times (\frac{1}{3}\times 5)=\frac{500}{27}$
$K_p=18.52at{m}^2$