${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$
$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and
$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is
$\therefore \mathrm{k}_{\mathrm{f}}=2.1 \times 10^{-3}$
$\mathrm{k}_{\mathrm{b}}=4.2 \times 10^{-4}$
$\mathrm{K}_{\mathrm{c}}=\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=5.0$
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[Given : Molal elevation constant of water $K _{ b }=0.5\, \,K\, kg\, mol ^{-1}$ boiling point of pure water $\left.=100^{\circ} C \right]$
