Explanation

(a)
$2NH_{3} \rightleftharpoons N_{2} + 3H_{2}$
Initial moles $\ \ \ \ \ a\ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ 0$
At equilibrium$(a - 2x)\ \ x\ \ \ 3x$
Initial pressure of $NH_{3}$ of ‘$a$’$mole\ = \ 15\ atm\ at\ 27℃$. The pressure of ’$a$’ mole of $NH_{3} = p\ atm\ at\ 347℃$
$\therefore\frac{15}{300} = \frac{p}{620}$
$\therefore p = 31\ atm$
At constant volume and at $347℃$,
mole $\propto$ pressure
$a \propto 31$ (before equilibrium)
$\therefore(a - 2x) \propto 50$ (after equilibrium)
$\therefore\ \frac{(a - 2x)}{a} = \frac{50}{31}$
$\therefore x = \frac{19}{62}a$
$\therefore\%\ of\ NH_{3}$ decomposed $= \frac{2x}{a} \times 100$
$= \frac{2 \times 19a}{62 \times a} \times 100$
$= 61.33\%$