To find the equilibrium constant $K$ for the reaction $2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O,$ we need to relate it to the given reactions:$\\$1. $N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1 \\$2. $N_2 + O_2 \rightleftharpoons 2NO \quad K_2 \\$3. $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3 \\$We need to manipulate these reactions to obtain the target reaction.$\\$Step 1: Reverse the first reaction to get $2NH_3 \rightleftharpoons N_2 + 3H_2. \\$The equilibrium constant for the reverse reaction is $\frac{1}{K_1}. \\$Step 2: Use the second reaction as it is: $N_2 + O_2 \rightleftharpoons 2NO. \\$The equilibrium constant remains $K_2. \\$Step 3: Multiply the third reaction by 3 to get $3H_2 + \frac{3}{2} O_2 \rightleftharpoons 3H_2O. \\$The equilibrium constant for this modified reaction is $K_3^3. \\$Now, add these reactions:$\\$• $2NH_3 \rightleftharpoons N_2 + 3H_2 \quad \left(\frac{1}{K_1}\right) \\$• $N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \\$• $3H_2 + \frac{3}{2} O_2 \rightleftharpoons 3H_2O \quad (K_3^3) \\$Adding these gives:$\\ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \\$The equilibrium constant for the overall reaction is the product of the individual constants:$\\ K = \frac{1}{K_1} \cdot K_2 \cdot K_3^3 \\$Simplifying, we get:$\\ K = \frac{K_2 K_3^3}{K_1} \\$Therefore, the correct option is Option 1: $\frac{K_2 K_3^3}{K_1}.$