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Raoult's Law

At certain te...

Question

At certain temperature(T) for the gas phase reaction: $2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)$ , $K_{p} = 12 \times 10^{8}$ atm. If $C l_{2},$ HCI & $O_{2}$ are mixed in such a manner that the partial pressure of each is 2 atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of $C l_{2}$ when equilibrium is attained at temperature(T)?

A

3.6 \times 10^{- 5}

atm

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B

10^{- 4}

atm

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C

3.6 \times 10^{- 3}

atm

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D

0.01

atm

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Solution

The correct option is C $3.6 \times 10^{- 3}$ atm
Assume the vapour pressure of water as $2.99 \times 10^{- 2} a t m$ .

$2 H_{2} O 0.0299 0.0299 (g) + 2 C l_{2} 2 2 - 2 x (g) ⇌ 4 H C l (g) 2 2 + 4 x + O_{2} (g) 2 2 + x$ ,

The expression for the equilibrium constant is $K_{P} = \frac{P_{O_{2}} P_{H C l}^{4}}{P_{H_{2} O}^{2} P_{C l_{2}}^{2}}$ .

Substitute values in the above expression

$K_{P} = \frac{(2 + x) (2 + 4 x)^{4}}{(0.0299)^{2} (2 - 2 x)^{2}} = 12 \times 10^{8}$

Hence $x = 0.998$ and $P_{C l_{2}} = 2 - 2 x = 3.6 \times 10^{- 3}$ .

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