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Q.

The vapour pressure of mercury is 0.002 mm Hg at $27^{\circ} C . K_{c}$ for the process $Hg (l) ⇌ Hg (g)$ is

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a

$8 .12 \times 10^{- 5}$

b

$1 .068 \times 10^{- 7}$

c

$6 .48 \times 10^{- 5}$

d

0.002

answer is D.

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Detailed Solution

$K_{c} = [Hg (g)]; P = CRT C = (\frac{0.002}{760}) \times \frac{1}{(0.0821 \times 300)} = 1.068 \times 10^{- 7} M$

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