Q. Consider the reaction, $NO_{2} \rightleftharpoons \frac{1}{2}N_{2} + O_{2},\ K_{1};N_{2}O_{4} \rightleftharpoons 2NO_{2},K_{2}$ Give the equilibrium constant for the formation of $N_{2}O_{4}\ from\ N_{2}\ and\ O_{2}$.

A) $\frac{1}{K_{1}^{2}} + \frac{1}{K_{2}}$
B) $\frac{1}{K_{1}K_{2}}$
C) $\sqrt{\frac{1}{K_{1}K_{2}}}$
D) $\frac{K_{2}}{K_{1}}$
Explanation

(a)
Given, $NO_{2} \rightleftharpoons \frac{1}{2}N_{2} + O_{2},\ \ \ \ \ \ \ \ \ \ \ \ \ K_{1}$ …(i)
$N_{2}O_{4} \rightleftharpoons 2NO_{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{2}$ …(ii)
Required reaction
$N_{2} + 2O_{2} \rightleftharpoons N_{2}O_{4},\ \ \ \ \ \ \ \ \ \ \ \ \ \ K$ …(iii)
By squaring Eq. (i) we get,
$2NO_{2} \rightleftharpoons N_{2} + {2O}_{2},\ \ \ \ \ \ \ \ \ \ \ \ \ K_{1}^{2}$ …(iv)
Now, Eq (iii) is obtained by inverting Eq. (ii) and Eq. (iv) and by adding them as.
$N_{2} + 2O_{2} \rightleftharpoons 2{NO}_{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{K_{1}^{2}}$
$2NO_{2} \rightleftharpoons N_{2}O_{4},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{K_{2}}$
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$N_{2} + 2O_{2} \rightleftharpoons N_{2}O_{4},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{K_{1}^{2}} + \frac{1}{K_{2}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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