If 0.2mol of H2(g) and 2.0mol of S(s) are mixed in a 1.0 litre vessel at 90°C, the partial pressure of H2S(g) formed according to the reaction H2(g)+S(s)⇌H2S(g); Kp=6.8×10−2; would be

# If $0.2\mathrm{mol}$ of ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$ and $2.0\mathrm{mol}$ of $\mathrm{S}\left(\mathrm{s}\right)$ are mixed in a 1.0 litre vessel at 90°C, the partial pressure of ${\mathrm{H}}_{2}\mathrm{S}\left(\mathrm{g}\right)$ formed according to the reaction ; would be

1. A

0.38 atm

2. B

0.19 atm

3. C

0.6 atm

4. D

$6.8×{10}^{-2}\mathrm{atm}/\left(0.2×2\right)$

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### Solution:

${\mathrm{H}}_{2}\left(\mathrm{g}\right)+\mathrm{S}\left(\mathrm{s}\right)⇌{\mathrm{H}}_{2}\mathrm{S}\left(\mathrm{g}\right)$

Since  Thus

Solving for x, we get

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