Answer
(a) $$NH_3(g) + \frac 74 O_2(g) \leftrightharpoons \frac 32 H_2O(g) + NO_2(g)$$
(b) $$K_p'' = 1.91 \times 2.11 \times 10^{19} = 4.03 \times 10^{19}$$
Work Step by Step
1. Invert the second reaction:
$$NO(g) + \frac 12 O_2(g) \leftrightharpoons NO_2(g)$$
$$K_p' = \frac{1}{0.524} = 1.91$$
2. Add the reactions:
$$NH_3(g) + \frac 54 O_2(g) + \frac 12 O_2(g) + NO(g) \leftrightharpoons NO(g) + \frac 32 H_2O(g) + NO_2(g)$$
$$NH_3(g) + \frac 74 O_2(g) \leftrightharpoons \frac 32 H_2O(g) + NO_2(g)$$
$$K_p'' = 1.91 \times 2.11 \times 10^{19} = 4.03 \times 10^{19}$$
