Question

If $a$ is the fraction of $N_2{O}_4\rightleftharpoons 2{NO}_2$ molecules dissociated in the reaction ($N_2{O}_4\rightleftharpoons 2{NO}_2$), then at equilibrium, the total number of moles of $N_2{O}_4$ and ${NO}_2$ present will be

• A. $a$
• B. $1+a$
• C. $2a$
• D. $1-a$

Answer 1

The correct option is A $a$

$N_2{O}_4\rightleftharpoons 2N{O}_2$

Let $x$ be the initial molecules present.

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$x$ $-$ Total $=x$

$x-a$ $2a$ Total $=x+a$

Excess$x+a-x=a$

Hence, the number of moles is $a.$