5.1 g NH_4SH is introduced in a 3.0 L evacu | vedclass

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Question

(D) The chemical equation for the decomposition is: $NH_4SH_{(s)} \leftrightarrow NH_{3(g)} + H_2S_{(g)}$Initial moles of $NH_4SH = \frac{5.1 \ g}{51

Vedclass · Accepted Answer

$0.242 \ atm^2$

Vedclass · Answer

(D) The chemical equation for the decomposition is: $NH_4SH_{(s)} \leftrightarrow NH_{3(g)} + H_2S_{(g)}$Initial moles of $NH_4SH = \frac{5.1 \ g}{51 \ g/mol} = 0.1 \ mol$.Moles decomposed $= 30\% 	imes 0.1 \ mol = 0.03 \ mol$.According to the stoichiometry,moles of $NH_3$ formed $= 0.03 \ mol$ and moles of $H_2S$ formed $= 0.03 \ mol$.Temperature $T = 327 + 273 = 600 \ K$.Partial pressure of $NH_3$ $(P_{NH_3})$ $= \frac{nRT}{V} = \frac{0.03 	imes 0.082 	imes 600}{3} = 0.492 \ atm$.Partial pressure of $H_2S$ $(P_{H_2S})$ $= \frac{nRT}{V} = \frac{0.03 	imes 0.082 	imes 600}{3} = 0.492 \ atm$.$K_p = P_{NH_3} 	imes P_{H_2S} = 0.492 	imes 0.492 \approx 0.242 \ atm^2$.

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