Consider the following reaction in a sealed vessel at equilibrium with concentrations of M, M and M. If 0.1 of is taken in a closed vessel, what will be the degree of dissociation of at equilibrium? [2024]

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Solution

Q.
Consider the following reaction in a sealed vessel at equilibrium with concentrations of

$N_{2} = 3.0 \times 10^{- 3}$ M, $O_{2} = 4.2 \times 10^{- 3}$ M and $N O = 2.8 \times 10^{- 3}$ M.

$2 N O_{(g)} ⇌ N_{2 (g)} + O_{2 (g)}$

If 0.1 $mol$ $L^{- 1}$ of ${NO}_{(g)}$ is taken in a closed vessel,

what will be the degree of dissociation $(α)$ of ${NO}_{(g)}$ at equilibrium [2024]

1 0.00889

2 0.0889

3 0.8889

4 0.717

Ans.
(4)

$\begin{matrix} 2 {NO}_{(g)} & ⇌ & N_{2 (g)} & + \begin{matrix} O_{2 (g)} \end{matrix} \\ 0.1 & 0 & 0 \\ 0.1 - x & \frac{x}{2} & \frac{x}{2} \end{matrix}$

$K_{c} = \frac{\frac{x}{2} \times \frac{x}{2}}{{(0.1 - x)}^{2}}$

$Again, K_{c} = \frac{[N_{2}] [O_{2}]}{[NO]^{2}} = \frac{3 \times 10^{- 3} \times 4.2 \times 10^{- 3}}{{(2.8 \times 10^{- 3})}^{2}} = 1.607$

$\Rightarrow \frac{x^{2}}{4 {(0.1 - x)}^{2}} = 1.607 \Rightarrow \frac{x}{2 (0.1 - x)} = \sqrt{1.607}$

$\Rightarrow \frac{x}{0.2 - 2 x} = 1.26 \Rightarrow x = 0.252 - 2.52 x$

$\Rightarrow 3.52 x = 0.252 \Rightarrow x = \frac{0.252}{3.52} = 0.0716$

$Degree of dissociation, α = \frac{0.0716}{0.1} = 0.716$

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