The equilibrium constant of the following are reactions
`N_2+3H_2hArr2NH_3 K_1`
`N_2+O_2hArr2NO K_2`
`H_2+1/2O_2rarrH_2O K_3`
The equilibrium constant (K) of the reaction `NH_3+5/2O_2overset(K)harr2NO+3H_2O` , will be

Options

A. `K_2K_3^3//K_1`
B. `K_2K_3//K_1`
C. `K_2^3K_3//K_1`
D. `K_1K_3^3//K_2`

Answer

To solve the problem, we need to find the equilibrium constant \( K \) for the reaction:

\[ \text{NH}_3 + \frac{5}{2} \text{O}_2 \rightleftharpoons 2 \text{NO} + 3 \text{H}_2\text{O} \]

using the provided equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) for the following reactions:

1. \( \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \) with equilibrium constant \( K_1 \)
2. \( \text{N}_2 + \text{O}_2 \rightleftharpoons 2 \text{NO} \) with equilibrium constant \( K_2 \)
3. \( \text{H}_2 + \frac{1}{2} \text{O}_2 \rightleftharpoons \text{H}_2\text{O} \) with equilibrium constant \( K_3 \)

### Step 1: Reverse the first reaction

The first reaction can be reversed to express \( \text{NH}_3 \) in terms of \( \text{N}_2 \) and \( \text{H}_2 \):

\[ 2 \text{NH}_3 \rightleftharpoons \text{N}_2 + 3 \text{H}_2 \]

The equilibrium constant for the reversed reaction will be:

\[ K' = \frac{1}{K_1} \]

### Step 2: Scale the reversed reaction

Since we need one mole of \( \text{NH}_3 \), we will divide the entire equation by 2:

\[ \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \]

The equilibrium constant for this reaction will be:

\[ K'' = \sqrt{K'} = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} \]

### Step 3: Use the second reaction

Now we will use the second reaction:

\[ \text{N}_2 + \text{O}_2 \rightleftharpoons 2 \text{NO} \]

This reaction has an equilibrium constant \( K_2 \).

### Step 4: Combine the reactions

Now we can combine the modified first reaction and the second reaction:

\[ \text{NH}_3 + \frac{5}{2} \text{O}_2 \rightleftharpoons \left( \text{N}_2 + \frac{3}{2} \text{H}_2 \right) + \text{O}_2 \rightleftharpoons 2 \text{NO} + 3 \text{H}_2\text{O} \]

### Step 5: Use the third reaction

We also need to consider the third reaction:

\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightleftharpoons \text{H}_2\text{O} \]

Since we need 3 moles of \( \text{H}_2\text{O} \), we will multiply the entire reaction by 3:

\[ 3 \text{H}_2 + \frac{3}{2} \text{O}_2 \rightleftharpoons 3 \text{H}_2\text{O} \]

The equilibrium constant for this scaled reaction will be:

\[ K''' = K_3^3 \]

### Step 6: Combine all equilibrium constants

Now we can combine all the equilibrium constants:

\[ K = K''' \cdot K_2 \cdot K'' \]

Substituting the values we derived:

\[ K = K_3^3 \cdot K_2 \cdot \frac{1}{\sqrt{K_1}} \]

### Final Result

Thus, the equilibrium constant \( K \) for the reaction \( \text{NH}_3 + \frac{5}{2} \text{O}_2 \rightleftharpoons 2 \text{NO} + 3 \text{H}_2\text{O} \) is:

\[ K = \frac{K_3^3 \cdot K_2}{\sqrt{K_1}} \]