Explanation

(a)
$H_{2} + \ \ \ \ \ \ \ \ \ \ I_{2} \rightleftharpoons 2HI$
Initial concentration $4.5\ \ \ \ \ \ \ \ \ \ 4.5\ \ \ \ \ \ \ 0$
$(4.5 - x)(4.5 - x)\ 2x$
From equation, $2x = 3$
$\therefore x = \frac{3}{2} = 1.5$
So, concentration at equilibrium
$\left\lbrack H_{2} \right\rbrack = 4.5 - 1.5 = 3$
$\left\lbrack I_{2} \right\rbrack = 4.5 - 1.5 = 3$
$\lbrack HI\rbrack = 3$
$\therefore K_{c} = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\left\lbrack I_{2} \right\rbrack} = \frac{3 \times 3}{3 \times 3} = 1$