# For the reversible reaction equilibrium is $N_2(g) + O_2(g) \overset{\overset{k_r}{\rightleftharpoons}}{_{k_f}} 2NO(g),$ $C_0 = Ce^{-2.1 \times 10^{-3} t}$ for the forward reaction and $C'_0 = C'e^{-4.2 \times 10^{-4}} t$ for the backward reaction, hence, if both forward and backward reactions are first order reactions then $K_c$ for the above equilibrium is:

(a) 5.0

(b) 2.0

(c) 0.5

(d) 2

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• For a first order reaction, $k = \frac{1}{t} log_e\frac{C_0}{C}$

Since the reaction $N_2(g) + O_2(g) \overset{\overset{k_f}{\rightleftharpoons}}{_{k_r}} 2NO(g),$ is a first order reaction,

$k = \frac{1}{t} log_e\frac{C_0}{C}$

Given, for the forward reaction $C_o = Ce^{-2.1 \times 10^-3 t}$
$\therefore k_f = \frac{1}{t} log_e e^{-2.1 \times 10^{-3} t} = \frac{-2.1 \times 10^{-3} t}{t} = -2.1 \times 10^{-3}$

and, for the backward reaction $C_o' = C'e^{-4.2 \times 10^-4 t}$
$\therefore k_r = \frac{1}{t} log_e e^{-4.2 \times 10^{-4} t} = -4.2 \times 10^{-4}$

$K = \frac{k_f}{k_r} = \frac{-2.1 \times 10^{-3}}{-4.2 \times 10^{-4}} = \frac{210}{42} = 5.0$

edited Nov 23, 2013