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  4. Ammonium carbonate decomposes as mathrmNH2 mathrmCOONH4( mathrm~s) leftharpoons 2 mathrmNH3( mathrm~g)+ mathrmCO2( mathrm~g) For the reaction, textK textp=2.9× 10- 5 texta textt textm3 . If we start with 1 mole of the compound, the total pressure at equilibrium would be

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Q. 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

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

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

At equilibrium if partial pressure of $\text{C}\text{O}_{2} \, \, \, \text{}\text{is }\text{p}$ then that of $\text{N}\text{H}_{3}$ is $2\text{p}$

So, $\text{K}_{\text{p}} = \text{p}_{\textit{NH}_{3}}^{2} \times \text{p}_{\textit{CO}_{2}}$

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

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

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

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

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