Consider the following reaction in a sealed vessel at equilibrium with concentrations of $N_2=3.0 imes10^{-3}$ M, $O_2=4.2 imes10^{-3}$ M and $NO=2.8 imes10^{-3}$ M. $2NO_{(g)} ightleftharpoons N_{2(g)}+O_{2(g)}$. 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?

Question

Consider the following reaction in a sealed vessel at equilibrium with concentrations of $N_2=3.0	imes10^{-3}$ M, $O_2=4.2	imes10^{-3}$ M and $NO=2.8	imes10^{-3}$ M. $2NO_{(g)}ightleftharpoons N_{2(g)}+O_{2(g)}$. 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?

Dr. Abubacker · Accepted Answer

[Solution by Claude - not from source] $K_c=\frac{3.0	imes4.2}{2.8^2}=1.607$. With 0.1 M NO: $K_c=\left[\frac{\alpha}{2(1-\alpha)}ight]^2\Rightarrow\frac{\alpha}{2(1-\alpha)}=1.268\Rightarrow\alpha\approx0.717$. Answer (2).

Explanation

Reviewed by Dr. Abubacker

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