At a certain temperature, the K, for the decomposition of H,S is 0.734 atm. H,S(g) = H,(g) + S(g) Initially, only H,S is present at a pressure of 0.246 atm in a closed container. What is the total pressure in the container at equilibrium?

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Q: atm

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Q: At a certain temperature, the K for the decompositon of H2S is 0.891. H,S(g) = H2(g) + S(g)…

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Q: At a certain temperature, the K, for the decomposition of H,S is 0.880. H, S(g) = H, (g)+S(g)…

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Q: At a certain temperature, the K, for the decomposition of H, S is 0.771. H, S(g) = H,(g)+S(g)…

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Q: At a certain temperature, the K, for the decomposition of H,S is 0.819 atm. H, S(g) = H, (g) + S(g)…

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Q: At a certain temperature, the K, for the decomposition of H,S is 0.800. H, S(g) = H,(g) + S(g)…

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Q: At a certain temperature, the K, for the decomposition of H,S is 0.840. H,S(g) =H,(g) + S(g)…

A: The reaction given is Given : initial pressure of H2S = 0.117 atm  And Kp = 0.840

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Q: When 1.50 mol CO2 and 1.50 mol H2 are placed in a 3.00-L container at 395 °C, the following reaction…

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Q: At a certain temperature, the K, for the decomposition of H, S is 0.846. H,S(g) = H,(g)+S(g)…

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Q: At a certain temperature, the K, for the decomposition of H, S is 0.725. H,S(g) = H,(g)+S(g)…

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Q: At a certain temperature, the K, for the decomposition of H,S is 0.844. H,S(g) = H,(g) +S(g)…

A: Assuming y atm of H2S is dissociated to reach equilibrium  Hence final pressure of H2S = 0.142 - y…

Q: At a certain temperature, the ?p for the decomposition of H2S is 0.885.   H2S(g)↽−−⇀H2(g)+S(g)…

A: The given data is: kp=0.885 Initial pressure of H2S=0.244 atm.