At 298 K
$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), K_1 = 4 \times 10^5$$
$$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g), K_2 = 1.6 \times 10^{12}$$
$$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{H}_2\text{O}(g), K_3 = 1.0 \times 10^{-13}$$
Based on above equilibria, the equilibrium constant of the reaction,
$$2\text{NH}_3(g) + \frac{5}{2}\text{O}_2(g) \rightleftharpoons 2\text{NO}(g) + 3\text{H}_2\text{O}(g)$$
is ______ $$\times 10^{-33}$$ (Nearest integer)

We need to find the equilibrium constant for:

$$2\text{NH}_3(g) + \frac{5}{2}\text{O}_2(g) \rightleftharpoons 2\text{NO}(g) + 3\text{H}_2\text{O}(g)$$

Given equilibria:

(1) $$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$$, $$K_1 = 4 \times 10^5$$

(2) $$\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}$$, $$K_2 = 1.6 \times 10^{12}$$

(3) $$\text{H}_2 + \frac{1}{2}\text{O}_2 \rightleftharpoons \text{H}_2\text{O}$$, $$K_3 = 1.0 \times 10^{-13}$$

The target reaction can be obtained by combining:

Reverse of (1) + (2) + 3 times (3)

Verification:

Reverse (1): $$2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$$

(2): $$\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}$$

3 × (3): $$3\text{H}_2 + \frac{3}{2}\text{O}_2 \rightarrow 3\text{H}_2\text{O}$$

Adding: $$2\text{NH}_3 + \text{O}_2 + \frac{3}{2}\text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O}$$

$$2\text{NH}_3 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O}$$ ✓

The equilibrium constant is:

$$ K = \frac{1}{K_1} \times K_2 \times K_3^3 $$

$$ K = \frac{1}{4 \times 10^5} \times 1.6 \times 10^{12} \times (1.0 \times 10^{-13})^3 $$

$$ K = \frac{1.6 \times 10^{12}}{4 \times 10^5} \times 10^{-39} $$

$$ K = 0.4 \times 10^{7} \times 10^{-39} = 4 \times 10^{6} \times 10^{-39} = 4 \times 10^{-33} $$

So the answer is $$4 \times 10^{-33}$$, meaning the coefficient is $$\boxed{4}$$.

This matches the recorded answer of 4.