The following equilibrium constants are given
`N_2+3H_2 harr 2NH_3,K_1`
`N_2+O_2 harr 2NO , K_2`
`H_2+1/2 O_2 harr H_2O , K_3`
The equlibrium constant for the oxidation of `NH_3` by oxygen to given NO is

Options

A. `(K_1K_2)/(K_3)`
B. `K_2K_3^3//K_1`
C. `(K_2K_3^2)/(K_1)`
D. `(K_2^2K_3)/(K_1)`

Answer

To find the equilibrium constant for the oxidation of ammonia (NH₃) by oxygen (O₂) to produce nitrogen monoxide (NO), we will use the given equilibrium reactions and their constants. Let's break down the steps:

### Step 1: Write down the given reactions and their equilibrium constants.

1. **Reaction 1:**
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad (K_1) \]

2. **Reaction 2:**
\[ N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \]

3. **Reaction 3:**
\[ H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O \quad (K_3) \]

### Step 2: Write the target reaction.

We need to find the equilibrium constant for the following reaction:
\[ 2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O \]

### Step 3: Manipulate the given reactions to obtain the target reaction.

To obtain the target reaction, we will manipulate the given reactions as follows:

1. **Multiply Reaction 3 by 3** (to get 3 moles of water):
\[ 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \quad (K_3^3) \]
The equilibrium constant becomes \( K_3^3 = K_3^3 \).

2. **Add Reaction 2**:
\[ N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \]

3. **Subtract Reaction 1** (to cancel out the nitrogen and hydrogen):
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad (-K_1) \]

### Step 4: Combine the manipulated reactions.

Now we combine the manipulated reactions:
- From \( 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \) (multiplied by 3)
- Plus \( N_2 + O_2 \rightleftharpoons 2NO \)
- Minus \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \)

### Step 5: Write the final combined reaction.

Combining these gives:
\[ 2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O \]

### Step 6: Write the expression for the equilibrium constant.

The equilibrium constant for the combined reaction is given by:
\[ K = \frac{K_2 \cdot K_3^3}{K_1} \]

### Final Result

Thus, the equilibrium constant for the oxidation of ammonia by oxygen to give NO is:
\[ K = \frac{K_2 \cdot K_3^3}{K_1} \]