Question

For the reaction equilibrium, $ {{N}_{2}}{{O}_{4}}(g)2NO(g) $ the concentrations of $ {{N}_{2}}{{O}_{4}} $ and $ N{{O}_{2}} $ at equilibrium are $ 4.8\times {{10}^{-2}} $ and $ 1.2\times {{10}^{-2}}mol\,\,{{L}^{-1}} $ respectively. The value of $ {{K}_{c}} $ for the reaction is

• A. $ 3.3\times {{10}^{2}}mol\,\,{{L}^{-1}} $
• B. $ 3\times {{10}^{-1}}mol\,\,{{L}^{-1}} $
• C. $ 3\times {{10}^{-3}}mol\,\,{{L}^{-1}} $
• D. $ 3\times {{10}^{3}}mol\,\,{{L}^{-1}} $

Answer 1

Answer: (C)

$ {{K}_{c}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{(1.2\times {{10}^{-2}})}^{2}}}{4.8\times {{10}^{-2}}} $ $ =3\times {{10}^{-3}}mol\,\,{{L}^{-1}} $