equilibrium constant is The value of \( K_{ p } \) for the equilibrium \( NH _{4} H S_{(s)} \rightleftharpoons NH _{3(g)}+ H _{2} S _{(g)} \) at a certain temperature and total 2 atm pressure at equilibrium will be \( atm ^{2} \)

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equilibrium constant is The value of \( K_{ p } \) for the equilibrium \( NH _{4} H S_{(s)} \rightleftharpoons NH _{3(g)}+ H _{2} S _{(g)} \) at a certain temperature and total 2 atm pressure at equilibrium will be \( atm ^{2} \)

The value of \( K_{ p } \) for the equilibrium \( NH _{4} H S_{(s)} \rightleftharpoons NH _{3(g)}+ H _{2} S _{(g)} \) at a certain temperature and total 2 atm pressure at equilibrium will be _____\( atm ^{2} \)

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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

We have given,

Total pressure at equilibrium = 2 atm.

Constant temperature = NH₄HS_(s) \(\rightleftharpoons\)NH_3(g) + H₂S_(g)

K_p = \(\frac{P_{NH_3}\times P_{H_2S}}{P_{NH_4HS}}\)

∵ Partial pressure of solid substance is considered to be 1.

∴ K_p = \(P_{NH_3}\times P_{H_2S}\)

K_p = \((P_{NH_3})^2\)

(From the stoichiometry of reaction)

∵ Total pressure = 2 atm

∴ \(P_{NH_3}\) = \(\frac{2\,atm}{2}\) = 1 atm

\(P_{H_2S}\) = \(\frac{2\,atm}{2}\) = 1 atm

∴ K_p = (1)²

K_p = 1atm²

Hence, the equilibrium constant K_p = 1atm²

equilibrium constant is The value of \( K_{ p } \) for the equilibrium \( NH _{4} H S_{(s)} \rightleftharpoons NH _{3(g)}+ H _{2} S _{(g)} \) at a certain temperature and total 2 atm pressure at equilibrium will be \( atm ^{2} \)

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