A 10 litre box contains $O_{3}$ and $O_{2}$ at equilibrium at $2000 K .$ $K_{p} = 4 \times 10^{14}$ atm for $2 O_{3} (g) ⇌ 3 O_{2} (g)$ Assume that $P_{O_{2}} ≫ P_{O_{3}}$ and if total pressure is 8 atm, then partial pressure of $O_{3}$ will be:

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$9.71 \times 10^{- 6} atm$

$8 \times 10^{- 5} atm$

$9.71 x 10^{- 2} atm$

$11.3 \times 10^{- 7} atm$

answer is B.

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Detailed Solution

Here, the complete pressure;

$\begin{array}{l} P_{O_{2}} + P_{O_{3}} = 8 \\ P_{O_{2}} ≃ 8 \end{array}$

$\begin{array}{l} K_{P} = \frac{{(P_{O_{2}})}^{3}}{{(P_{O_{3}})}^{2}} \Rightarrow 4 \times 10^{14} = \frac{8^{3}}{{(P_{O_{3}})}^{2}} \\ \Rightarrow P_{O_{3}} = 11.3 \times 10^{- 7} \end{array}$

Now, Equilibrium constant;

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