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(A) The equilibrium expression is $K_p = p_{NH_3} 	imes p_{H_2S}$.Let $p$ be the partial pressure of $H_2S$ produced at equilibrium.Then,$p_{H_2S} =

Vedclass · Accepted Answer

$p_{NH_3} = 0.665 \ atm, p_{H_2S} = 0.165 \ atm$

Vedclass · Answer

(A) The equilibrium expression is $K_p = p_{NH_3} 	imes p_{H_2S}$.Let $p$ be the partial pressure of $H_2S$ produced at equilibrium.Then,$p_{H_2S} = p$ and $p_{NH_3} = 0.5 + p$.Substituting into the $K_p$ expression: $0.11 = (0.5 + p)(p)$.This simplifies to the quadratic equation: $p^2 + 0.5p - 0.11 = 0$.Using the quadratic formula $p = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{-0.5 + \sqrt{0.5^2 - 4(1)(-0.11)}}{2(1)} = \frac{-0.5 + \sqrt{0.25 + 0.44}}{2} = \frac{-0.5 + \sqrt{0.69}}{2} \approx \frac{-0.5 + 0.83}{2} = 0.165 \ atm$.Thus,$p_{H_2S} = 0.165 \ atm$ and $p_{NH_3} = 0.5 + 0.165 = 0.665 \ atm$.

Some solid $NH_4HS$ is placed in a flask containing $0.5 \ atm$ of $NH_3$. What would be the partial pressures of $NH_3$ and $H_2S$ when equilibrium is reached for the reaction $NH_4HS_{(s)} \rightleftharpoons NH_{3_{(g)}} + H_2S_{(g)}$,given $K_p = 0.11$?

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