Consider the following equilibria at 300 K
$\begin{array}{l} 2 A (g) + H_{2} O (g) ⇌ C (g) + 3 D (g) \\ K_{P} = 3 \times 10^{22} atm \end{array}$
The equilibrium was attained in a container by taking A, C and D at partial pressures of 2 atm each along with large amount of liquid water. If vapour pressure of water at 300 K is 38 mm of Hg then calculate partial pressure of A when equilibrium gets established. [Express your answer in terms of $10^{- 10} atm]$

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answer is 5.

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Detailed Solution

When the compound A reacts with water molecule, then reaction takes place as,

$\begin{array}{ccccc} 2 A (g) + H_{2} O (g) & ⇌ & C (g) + 3 D (g) \\ 2 & 0.05 & 2 & 2 \\ ~ 0 & 0.05 & 3 & 5 \end{array}$

$3 \times 10^{22} = \frac{3 \times 5^{3}}{5 \times 10^{- 2} \times P_{A}^{2}} \Rightarrow P_{A} = 5 \times 10^{- 10}$

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