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 need to manipulate the given reactions and their equilibrium constants. Here’s the step-by-step solution:

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

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

2. **Formation of Nitric Oxide:**
\[ N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \]

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

### Step 2: Reverse the first reaction to express NH₃ in terms of N₂ and H₂.

Reversing the first reaction gives:
\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \]
The equilibrium constant for the reversed reaction is:
\[ K' = \frac{1}{K_1} \]

### Step 3: Write the second reaction as it is.

The second reaction remains unchanged:
\[ N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \]

### Step 4: Modify the third reaction to match the stoichiometry needed.

We need to use the formation of water. To match the stoichiometry, we multiply the third reaction by 3:
\[ 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \]
The equilibrium constant for this modified reaction is:
\[ K'' = (K_3)^3 \]

### Step 5: Combine the reactions.

Now, we will combine the modified reactions:

1. From the reversed first reaction:
\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \quad \left(\frac{1}{K_1}\right) \]

2. From the second reaction:
\[ N_2 + O_2 \rightleftharpoons 2NO \quad (K_2) \]

3. From the modified third reaction:
\[ 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \quad (K_3)^3 \]

Adding these reactions together, we can cancel out \(N_2\) and \(3H_2\):
- The \(N_2\) from the second reaction cancels with the \(N_2\) from the reversed first reaction.
- The \(3H_2\) from the reversed first reaction cancels with \(3H_2\) from the modified third reaction.

The overall reaction becomes:
\[ 2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O \]

### Step 6: Write the equilibrium constant for the overall reaction.

The equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions:
\[ K = K_2 \cdot (K_3)^3 \cdot \frac{1}{K_1} \]

### Final Answer:
The equilibrium constant for the oxidation of ammonia by oxygen to give nitrogen monoxide is:
\[ K = \frac{K_2 \cdot (K_3)^3}{K_1} \]

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