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Reaction Quotient

At a certain ...

Question

At a certain temperature, equilibrium constant $K_{c} = 4 \times 10^{- 2}$ for the reaction $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$
If we take $1.5 m o l$ of NO and $2 m o l$ each of $N_{2} a n d O_{2}$ in $5 L$ vessel, what would be the equilibrium concentration of $N O$ (in $m o l / L$ )?

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Solution

$Q_{c} = \frac{(1.5 / 5)^{2}}{(2 / 5)^{2}} = 0.5625 ∵ Q_{c} > K_{c}$
Therefore, the reaction will proceed in backward direction $\begin{matrix} N_{2} (g) & + & O_{2} (g) & ⇌ & 2 N O (g) I n i t i a l m o l e s & 2 & 2 & 1.5 C o n c . a t e q^{m} & \frac{2 + x}{5} & \frac{2 + x}{5} & \frac{1.5 - 2 x}{5} \end{matrix}$
$K_{c} = \frac{[N O]^{2}}{[N_{2}] [O_{2}]} = \frac{{(\frac{1.5 - 2 x}{5})}^{2}}{{(\frac{2 + x}{5})}^{2}}$
$\frac{1.5 - 2 x}{2 + x} = \sqrt{0.04} = 0.2$
$1.5 - 2 x = 0.4 + 0.2 x$
$x = 0.5$
$∴ Equilibrium concentration of N O = \frac{1.5 - 2 x}{5} = 0.1 m o l / L$

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Reaction Quotient

Standard XIII Chemistry

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