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Equilibrium
Degree of Dissociation from Equilibrium Concentrations
1 question from 1 paper, each with its answer.
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- of Equilibrium
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Concepts tested
- Relating initial concentration and degree of dissociation to equilibrium concentrations
All Questions
2024 1 question
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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 $NO = 2.8 \times 10^{-3}$ M. $2NO_{(g)} \rightleftharpoons N_{2(g)} + O_{2(g)}$ If 0.1 mol L$^{-1}$ of $NO_{(g)}$ is taken in a closed vessel, what will be degree of dissociation ($\alpha$) of $NO_{(g)}$ at equilibrium?For $2NO(g) \rightleftharpoons N_2(g) + O_2(g)$: $K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{(3 \times 10^{-3})(4.2 \times 10^{-3})}{(2.8 \times 10^{-3})^2} = 1.6$ Starting with 0.1 mol $L^{-1}$ NO, let y mol $L^{-1}$ of NO dissociate: At equilibrium: [NO] = 0.1 $-$ y, $[N_2] = [O_2] = \frac{y}{2}$ $\frac{(y/2)^2}{(0.1 - y)^2} = 1.6 \Rightarrow \frac{y/2}{0.1 - y} = \sqrt{1.6} \approx 1.265$ $y \approx 0.0717$ mol $L^{-1}$ Degree of dissociation $\alpha = \frac{y}{0.1} = 0.717$