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Q. At equilibrium, the concentrations of $N_2 = 3.0 x 10^{- 3} M, O_2 = 4.2 \times 10^{- 3} M $ and $NO = 2.8 \times 10^{- 3} M $ in a sealed vessel at $800 K. $ What will be Kc for the reaction, $ N_2{(g)} + O_2{(g)} \rightleftharpoons 2NO_{(g)} $ ?

Equilibrium

Solution:

For the reaction, equilibrium constant $K_c $ can be written as,
$k_{c} =\frac{\left[NO\right]^{2}}{\left[N_{2}\right]\left[O_{2}\right]} $
$=\frac{\left(2.8\times10^{-3} M^{2}\right)}{\left(3.0 \times10^{-3} M\right)\left(4.2 \times10^{-3} M\right)} =0.622$