Question

Solid ammonium carbamate dissociated according to the given reaction
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
Total pressure of the gases in equilibrium is $5$ atm. Hence $K_p$.

• A. $18.5$
• B. $25$
• C. $1/5$
• D. $12.5$

Answer 1

The correct option is A $18.5$

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, $p_T=5$ atm

$K_p=p_{N{H}_3}^2.p_{C{O}_2}$

using Dalton's law of partial pressure we have

$p_i={\chi }_i.p_T$

thus, $p_{N{H}_3}={\chi }_{N{H}_3}.p_T$

${\chi }_{N{H}_3}=\frac{n_{N{H}_3}}{n_{N{H}_3}+n_{C{O}_2}}=\frac{2}{2+1}$

${\chi }_{N{H}_3}=\frac{2}{3}$

Similarly, ${\chi }_{C{O}_2}=\frac{1}{2+1}=\frac{1}{3}$

upon substitution we get:

$K_p=(\frac{2}{3}\times 5{)}^2\times (\frac{1}{3}\times 5)=\frac{500}{27}$

$K_p=18.52$