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At $87^{\circ} C$ the following equilibrium is established. $H_{2} (g) + S (s) ⇌ H_{2} S (g); K_{c} = 0 .08$
If 0.3 mole hydrogen and 2 mole sulphur are heated to 87^oC in a 2 L vessel, what will be the concentration of H₂S at equilibrium?

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$0 .011 M$

0.022 M

0.044 M

0.08 M

answer is B.

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Detailed Solution

$K_{c} = \frac{[H_{2} S (g)]}{[H_{2} (g)]} \Rightarrow 8 \times 10^{- 2} = \frac{x}{0.3 - x} \Rightarrow x = 0.022 [H_{2} S (g)] = \frac{0.022}{2} = 0.011 M$

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