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Reaction Quotient

Two equilibri...

Question

Two equilibria exist simultaneously in a vessel at $25^{\circ} C$
$\begin{matrix} N O (g) & + & N O_{2} (g) & K_{P 1} = 0.4 a t m^{- 1} ⇌ & N_{2} O_{3} (g) \end{matrix}$
$\begin{matrix} 2 N O_{2} (g) & K_{P 2} = 8 a t m^{- 1} ⇌ & N_{2} O_{4} (g) \end{matrix}$
If initially only $N O$ and $N O_{2}$ are present in a 3:5 mole ratio and the final pressure at equilibrium of $N O_{2}$ at 0.5 atm, then calculate total pressure (in atm) at equilibrium.

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Solution

If initially only $N O$ and $N O_{2}$ are present in a 3 : 5 mole ratio, then initial pressure ratio will also be 3 : 5, so let initial pressures be 3p and 5p.
$\begin{matrix} N O (g) & + & N O_{2} (g) & K_{P 1} = 0.4 a t m^{- 1} ⇌ & N_{2} O_{3} (g) 3 p & 5 p & 0 & i n i t i a l p r e s s u r e 3 p - x & 5 p - x - 2 y & x & a t e q u i l i b r i u m \end{matrix}$
And
$\begin{matrix} 2 N O_{2} (g) & K_{P 2} = 8 a t m^{- 1} ⇌ & N_{2} O_{4} (g) 5 p - x - 2 y & y \end{matrix}$

As pressure of $N O_{2} = 0.5 a t m$ as given, hence
$5 p - x - 2 y = 0.5$
So, from 2nd equilibrium $K_{P 2} = 8 = \frac{P_{N_{2} O_{4}}}{P_{N O_{2}}^{2}} \Rightarrow \frac{P_{N_{2} O_{4}}}{(0.5)^{2}} = 8 \Rightarrow P_{N_{2} O_{4}} = 2 a t m = y$
So, $5 p - x = 4.5 . . . (1)$
Again,
$K_{P 1} = \frac{P_{N_{2} O_{3}}}{P_{N O} \times P_{N O_{2}}} = \frac{x}{(3 p - x) \times 0.5} = 0.4$
$\Rightarrow p = 2 x . . . (2)$
From (1) & (2),
$x = 0.5 a t m$ and $p = 1 a t m$
So, total pressure at equilibrium
$(3 p - x) + (5 p - x - 2 y) + x + y = 8 p - x - y = 8 - 0.5 - 2 = 5.5 a t m$

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