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(A) The chemical equation is $NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.Let the partial pressure of $CO_{2(g)}$ be $p$ and the parti

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$15.94$

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(A) The chemical equation is $NH_4COONH_{2(s)} ightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.Let the partial pressure of $CO_{2(g)}$ be $p$ and the partial pressure of $NH_{3(g)}$ be $2p$ due to the stoichiometry of the reaction.The equilibrium constant $K_p$ is given by $K_p = (P_{NH_3})^2 	imes (P_{CO_2})$.Substituting the values: $600 = (2p)^2 	imes (p) = 4p^3$.$p^3 = 600 / 4 = 150$.$p = (150)^{1/3} \approx 5.313 \ bar$.The total pressure at equilibrium is $P_{total} = P_{NH_3} + P_{CO_2} = 2p + p = 3p$.$P_{total} = 3 	imes 5.313 = 15.939 \ bar \approx 15.94 \ bar$.

The equilibrium constant of $NH_4COONH_2$ in a closed vessel at $400 \ K$ temperature is $600 \ bar^3$. What will be the total pressure at equilibrium (in $bar$)?
$NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$

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The equilibrium constant of $NH_4COONH_2$ in a closed vessel at $400 \ K$ temperature is $600 \ bar^3$. What will be the total pressure at equilibrium (in $bar$)?$NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$