At 87oC, the following equilibrium is established: H2(g)+S(s)⇌H2S(g),Kc=8×10−2
If 0.3 mole of hydrogen and 2 mole of sulphur are heated to 87oC in a 2 L vessel, what will be the partial pressure of H2S at equilibrium.
A
0.32 atm
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B
1.43 atm
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C
0.62 atm
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D
4.0 atm
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Solution
The correct option is A 0.32 atm For reaction, H2(g)+S(s)⇌H2S(g) 0.3−xx KC=[H2S(g)][H2(g)]⇒8×10−2=x0.3−x 0.024−0.08x=x 0.024=1.08x x=0.022 P=nRTV PH2S=0.022×0.08×3602⇒≈0.32atm