Question

Ammonium carbonate decomposes as

${\mathrm{NH}}_2{\mathrm{COONH}}_4(\mathrm{s})\rightleftharpoons 2{\mathrm{NH}}_3(\mathrm{g})+{\mathrm{CO}}_2(\mathrm{g})$

For the reaction, ${\text{K}}_{\text{p}}=2.9\times 10^{-5}\text{a}\text{t}{\text{m}}^3$ . If we start with 1 mole of the compound, the total pressure at equilibrium would be

• A. 0.766 atm
• B. 0.0582 atm
• C. 0.0388 atm
• D. 0.0194 atm

Answer 1

Answer: (B)

${\mathrm{NH}}_2{\mathrm{COONH}}_4(\mathrm{s})\rightleftharpoons 2{\mathrm{NH}}_3(\mathrm{g})+{\mathrm{CO}}_2(\mathrm{g})$

At equilibrium if partial pressure of $\text{C}{\text{O}}_2\text{is }\text{p}$ then that of $\text{N}{\text{H}}_3$ is $2\text{p}$

So, ${\text{K}}_{\text{p}}={\text{p}}_{{\text{NH}}_3}^2\times {\text{p}}_{{\text{CO}}_2}$

$K_p=(2p{)}^2\times p=4p^3$

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

$\text{p}=\text{1.935}\times {\text{10}}^{-2}$

hence, total pressure $\text{p}=3\text{p}$

$=5.81\times 10^{-2}=0.0581$ atm.