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Given that at 800k the concentrations are as follows: N2=3.0 × 10^–3M, O2 = 4.2 × 10^–3M and NO = 2.8 × 10^–3M, what is the equilibrium constant for the reaction N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g)?(a) 0.622(b) 0.6(c) 0.63(d) 0.94I got this question during an internship interview.Asked question is from Law of Chemical Equilibrium and Equilibrium Constant in section Equilibrium of Chemistry – Class 11

Answer»

The correct answer is (a) 0.622

For explanation: For the reaction, N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g), the EQUILIBRIUM constant is given by [NO]^2/[N2][O2]. So the equilibrium constant K is given by (2.8 × 10^–3 M)(2.8 × 10^–3 M)/(3.0 × 10^–3 M)(4.2 × 10^–3 M) = 0.622, there are no units as Δng = 0.


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