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If $0.2 mol$ of $H_{2} (g)$ and $2.0 mol$ of $S (s)$ are mixed in a 1.0 litre vessel at 90°C, the partial pressure of $H_{2} S (g)$ formed according to the reaction $H_{2} (g) + S (s) ⇌ H_{2} S (g); K_{p} = 6.8 \times 10^{- 2}$ ; would be

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0.38 atm

0.19 atm

0.6 atm

$6.8 \times 10^{- 2} atm / (0.2 \times 2)$

answer is A.

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Detailed Solution

$H_{2} (g) + S (s) ⇌ H_{2} S (g)$

$0.2 mol - x x$

Since $Δ ν_{g} = 0, K_{p} = K_{n} = \frac{n_{H_{2} S}}{n_{H_{2}}} .$ Thus $\frac{x}{0.2 mol - x} = 0.068$

Solving for x, we get $x = 0.0127 mol; p = (\frac{0.0127 \times 0.082 \times 363}{1}) atm = 0.38 atm$

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