When NO and NO₂ are mixed, the following equilibria is established
$2 N O_{2 (g)} ⇌ N_{2} O_{4 (g)} K_{p} = 6.8 a t m$ and $N O_{(g)} + N O_{2 (g)} ⇌ N_{2} O_{3 (g)}$
In an experiment, when $N O$ and $N O_{2}$ are mixed in the molar ratio of $1 : 2$ , the final total pressure is $5.05 a t m$ and the partial pressure of $N_{2} O_{4}$ is $1.7 a t m$ , what is the partial pressure of NO in atm.

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Detailed Solution

At equilibrium

$\begin{array}{l} 2 N O_{2 (g)} ⇌ N_{2} O_{4 (g)} K_{p} = 6.8 a t m ........ (i) \\ 2 p - 3.4 - x 1.7 a t m \\ N O_{(g)} + N O_{2 (g)} ⇌ N O_{23 (g)} ........... (i i) \\ P - x 2 p - x - 3.4 x \end{array}$

From equation (i) $6.8 = \frac{1.7}{p_{N O_{2}}^{2}}; ∴ P_{N O_{2}} = 0.5 a t m$
From equation (i) & (ii) $2 p - x - 3.4 = 0.5 = P_{N O_{2}}$
$∴$ Total $P = 5.05 = 0.5 + 1.7 + P$
$∴$ $P = 2.85 a t m$