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At 87∘C 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 sulphur are heated to 87oC in 1.0 L vessel, what will be the partial pressure of H2S at equilibrium?

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a

0.966atm

b

1.38 atm

c

0.0327 atm

d

1 atm

answer is D.

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Detailed Solution

H2(g) + S(s) ⇌ H2S(g) conc.at eq.; 0.5-x - x Kc=H2 SH2 ∵Kp=Kc for the given reaction ⇒7×10-2=x0.5-x x =0.0327 PH2 S=nH2 SVRT =0.0327 x 0.0821×360 = 0.966atm

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