A flask is initially filled with pure $N_{2} O_{g} (g)$ having pressure 2 bar and following equilibria are established.
$N_{2} O_{3} (g) ⇌ {NO}_{2} (g) + NO (g)$ $K_{P_{1}} = 2.5 bar$ ?
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$ $K_{P_{y}} =$ ?
If at equilibrium partial pressure of NO(g) was found to be 1.5 bar, then:

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Equilibrium partial pressure of $N_{2} O_{3} (g)$ is 0.5 bar

Equilibrium partial pressure of ${NO}_{2} (g)$ is 0.83 bar

Equilibrium partial pressure of $N_{2} O_{4}$ is 0.33 bar

Value of $K_{P_{2}}$ is 0.48 bar

answer is A, B, D, C.

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Detailed Solution

Here, we have,

$N_{2} O_{3} ⟶ {NO}_{2} + NO$

$\begin{array}{ll} 2 - x x - y x \\ 2 {NO}_{2} ⟶ N_{2} O_{4} \\ x - y y / 2 \end{array}$

$x = 1.5$ (given)

$\begin{array}{l} K_{P_{1}} = \frac{(3 - y) \times 1.5}{0.5} \\ P_{N_{2} O_{5}} = 2 - 1.5 = 0.5 bar \\ P_{{NO}_{2}} = 0.83 bar \\ y = 1.5 - 0.83 = 0.67 \\ P_{N_{2} O_{4}} = 0.3 bar \\ K_{P_{2}} = \frac{0.33}{(0.82)^{2}} \\ = 0.48 \end{array}$

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