NH_4HS_{(s)} rightleftharpoons NH_{3(g)} + H_ | vedclass

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(B) For the reaction $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$,the equilibrium constant $K_p$ is given by the product of the partial pr

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$0.0625$

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(B) For the reaction $NH_4HS_{(s)} ightleftharpoons NH_{3(g)} + H_2S_{(g)}$,the equilibrium constant $K_p$ is given by the product of the partial pressures of the gaseous products.$K_p = P_{NH_3} 	imes P_{H_2S}$Since the stoichiometry of the reaction is $1:1$ for the gases,at equilibrium,$P_{NH_3} = P_{H_2S} = p$.The total equilibrium pressure $P_{total} = P_{NH_3} + P_{H_2S} = p + p = 2p$.Given $P_{total} = 0.5 \ atm$,we have $2p = 0.5 \ atm$,so $p = 0.25 \ atm$.Therefore,$K_p = p 	imes p = (0.25 \ atm) 	imes (0.25 \ atm) = 0.0625 \ atm^2$.

$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?

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$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?