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Effective Equilibrium Constant

When NO and N...

Question

When $N O$ and $N O_{2}$ are mixed, the following equilibria are readily obtained;
$2 N O_{2} ⇌ N_{2} O_{4}, K_{P} = 6.8 a t m^{- 1} a n d N O + N O_{2} ⇌ N_{2} O_{3}$
In an experiment when $N O$ and $N O_{2}$ are mixed in the ratio of 1 : 2, the find total pressure was 5.05 atm and the partial pressure of $N_{2} O_{4}$ was 1.7 atm. Calculate the equilibrium partial pressure of NO in atmosphere.

A

3 atm

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B

2 atm

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C

4 atm

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D

1 atm

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Solution

The correct option is D 1 atm
For I equilibrium $2 N O_{2} ⇌ N_{2} O_{4}$
$K_{P} = \frac{P_{N_{2} O_{4}}}{(P_{N O_{2}})^{2}} = 6.8$ ....(i)
$P_{N_{2} O_{4}} = 1.7 a t m ∴ By eq. (i) P_{N O_{2}}^{'} = 0.5 a t m$
The equilibria are maintained using $N O$ and $N O_{2}$ in the ratio 1 : 2
$\begin{matrix} For II equilibrium & N O & + & N O_{2} ⇌ N_{2} O_{3} Initial pressure & P & 2 P & 0 Pressure at equilibrium & (p - x) & (2 P - x - 3.4) & x \end{matrix}$
3.4 atm of $N O_{2}$ are used for I equilibrium to have $P_{N_{2} O_{4}} = 1.7 a t m$
At equilibrium (P - x) 0.5 x

( $∵ P_{N O_{2}}$ is same for both the equilibria since both reactions are at equilibrium at a time)
Total pressure at equilibrium (Given 5.05 atm)
$= P_{N O} + P_{N O_{2}} + P_{N_{2} O_{3}} + P_{N_{2} O_{4}}$
$= P - x + 0.5 + x + 1.7$
$5.05 = P + 2.20$
$5.05 = P + 2.20$
$P = 2.85 a t m$
$2 \times 2.85 - x - 3.4 = 0.5$
$x = 1.80 a t m$
$P_{N O} = P - x$
$P_{N O} = 2.85 - 1.80 = 1.05 a t m \approx 1 a t m$

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Effective Equilibrium Constant

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