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Question

(B) The decomposition reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.The equilibrium constant expression is $K_p = (P_{NH_3}

Vedclass · Accepted Answer

$5.82$

Vedclass · Answer

(B) The decomposition reaction is $NH_2COONH_{4(s)} ightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.The equilibrium constant expression is $K_p = (P_{NH_3})^2 	imes (P_{CO_2})$.From the stoichiometry,$NH_3$ and $CO_2$ are produced in a $2:1$ molar ratio. If $P$ is the total pressure at equilibrium,then $P_{NH_3} = \frac{2P}{3}$ and $P_{CO_2} = \frac{P}{3}$.Substituting these into the $K_p$ expression: $K_p = (\frac{2P}{3})^2 	imes (\frac{P}{3}) = \frac{4P^3}{27}$.Given $K_p = 2.9 	imes 10^{-5}$,we have $2.9 	imes 10^{-5} = \frac{4P^3}{27}$.Solving for $P$: $P^3 = \frac{2.9 	imes 10^{-5} 	imes 27}{4} = 19.575 	imes 10^{-5} = 195.75 	imes 10^{-6}$.$P = (195.75)^{1/3} 	imes 10^{-2} \ atm \approx 5.82 	imes 10^{-2} \ atm$.

For the decomposition of the compound,represented as $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the $K_p = 2.9 \times 10^{-5} \ atm^3$. If the reaction is started with $1 \ mol$ of the compound,the total pressure at equilibrium would be............ $\times 10^{-2} \ atm$.

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