Chemistry Question from JEE Main 2023 · 29 January, Shift 2

At $298 \, \text{K}$: $N_2 + 3H_2 \rightleftharpoons 2NH_3, \, K_1 = 4 \times 10^5$ $N_2 + O_2 \rightleftharpoons 2NO, \, K_2 = 1.6 \times 10^{12}$ $H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O, \, K_3 = 1.0 \times 10^{-13}$ Based on the above equilibria, the equilibrium constant of the reaction: $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ is $\_\_\_\_\_ \times 10^{-33}$ (nearest integer).