For the reaction: NH_4HS_{(s)} rightleftharpo | vedclass

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(C) The given reaction is $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$.Let the partial pressure of $NH_{3(g)}$ be $p$ and $H_2S_{(g)}$ be

Vedclass · Accepted Answer

$0.49$

Vedclass · Answer

(C) The given reaction is $NH_4HS_{(s)} ightleftharpoons NH_{3(g)} + H_2S_{(g)}$.Let the partial pressure of $NH_{3(g)}$ be $p$ and $H_2S_{(g)}$ be $p$ at equilibrium.Since the stoichiometry of the gaseous products is $1:1$,the total pressure $P_{total} = p_{NH_3} + p_{H_2S} = p + p = 2p$.Given $P_{total} = 1.4 \ atm$,so $2p = 1.4 \ atm$,which gives $p = 0.7 \ atm$.The equilibrium constant $K_p$ is defined as $K_p = p_{NH_3} 	imes p_{H_2S} = p 	imes p = p^2$.Substituting the value of $p$,$K_p = (0.7)^2 = 0.49 \ atm^2$.

For the reaction: $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$,the observed pressure of reaction mixture at equilibrium is $1.4 \ atm$ at $110 \ ^oC$. Value of $K_p$ for the reaction is ..... $atm^2$

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