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Question

# At 87∘C, the following equilibrium is established H2(g)+S(s)⇌H2S(g) Kp=7×10−2. If 0.50 mole of hydrogen and 1.0 mole of sulfur are heated to 87∘C in 1.0 L vessel, what will be the partial pressure of H2S at equilibrium?

A
0.966 atm
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B
1.38 atm
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C
0.0327 atm
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D
9.66 atm
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Solution

## The correct option is B 0.966 atmThe equilibrium reaction is H2(g)+S(s)⇌H2S(g)Initially, 0.50 moles of hydrogen are present.At equilibrium, 0.50−x moles of hydrogen and x moles of hydrogen sulfide are present.The expression for the equilibrium constant is KP=PH2SPH2.PH2S=XH2S×P=x0.5PPH2=XH2×P=0.5−x0.5PSubstitute values in the above expression.7×10−2=x0.5−x.The number of moles of hydrogen sulphide =x=0.0327.Total pressure can be calculated from Ideal gas equation-P=nRTV=0.5×0.0821×3601=14.77atm.Partial pressure of H2S, PH2S=XH2S×P=x0.5P=0.03270.5×14.77=0.966atmHence, the correct option is A.

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