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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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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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1 Answer

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  • For a first order reaction, $ k = \frac{1}{t} log_e\frac{C_0}{C} $
 
 
Answer: 5.0
 
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 $

 

answered Nov 16, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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