Explanation

(b)
Given, $\left\lbrack H_{2} \right\rbrack = 8.0$ mol/L
$\left\lbrack I_{2} \right\rbrack = 3.0$ mol/L
$\lbrack HI\rbrack = 28$ mol/L
$K = ?$
$H_{2} + I_{2} \rightleftharpoons 2HI$
$\therefore$ $K = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\left\lbrack I_{2} \right\rbrack} = \frac{(28)^{2}}{(8) \times (3)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{28 \times 28}{24}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 32.66$