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Q.

At $87^{\circ} C$ following equilibrium is established $H_{2} (g) + S (s) ⇌ H_{2} S (g); K_{p} = 7 \times 10^{- 2}$ If 0.50 mole of hydrogen and 1.0 mole of sulphur are heated to 87^oC in 1.0 L vessel, what will be the partial pressure of H₂S at equilibrium?

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a

$0 .966 atm$

b

1.38 atm

c

0.0327 atm

d

1 atm

answer is D.

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Detailed Solution

$H_{2} (g) + S (s) ⇌ H_{2} S (g) conc . at eq .; 0.5 - x - x K_{c} = \frac{[H_{2} S]}{[H_{2}]} (∵ K_{p} = K_{c} for the given reaction) \Rightarrow 7 \times 10^{- 2} = \frac{x}{0.5 - x} x = 0.0327 P_{H_{2} S} = (\frac{n_{H_{2} S}}{V}) RT = 0.0327 x 0.0821 \times 360 = 0.966 atm$

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