The following equilibria are given by `:`
`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_(2)O, K_(3)`
The equilibrium constant of the reaction `2NH_(3)+(5)/(2)O_(2)hArr 2NO +3H_(2)O` in terms of `K_(1) , K_(2)` and `K_(3)` is

Options

A. `K_(2)K_(3)^(3)//K_(1)`
B. `K_(1)K_(2)K_(3)`
C. `K_(1)K_(2)//K_(3)`
D. `K_(1)K_(3)^(2)//K_(2)`

Explanation

To find the equilibrium constant for the reaction:

\[ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \]

in terms of \( K_1, K_2, \) and \( K_3 \), we will manipulate the given reactions step by step.

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

1. \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) with equilibrium constant \( K_1 \)
2. \( N_2 + O_2 \rightleftharpoons 2NO \) with equilibrium constant \( K_2 \)
3. \( H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O \) with equilibrium constant \( K_3 \)

### Step 2: Manipulate the first reaction.

We need \( 2NH_3 \) on the left side. To achieve this, we can reverse the first reaction:

\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \]

The equilibrium constant for the reverse reaction is:

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

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

The second reaction is already in the desired form:

\[ N_2 + O_2 \rightleftharpoons 2NO \]

This gives us:

\[ K'' = K_2 \]

### Step 4: Manipulate the third reaction.

We need \( 3H_2O \) on the right side. To achieve this, we multiply the third reaction by 3:

\[ 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \]

The equilibrium constant for this reaction is:

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

### Step 5: Combine the reactions.

Now, we can add the manipulated reactions together:

1. \( 2NH_3 \rightleftharpoons N_2 + 3H_2 \) (from Step 2, \( K' = \frac{1}{K_1} \))
2. \( N_2 + O_2 \rightleftharpoons 2NO \) (from Step 3, \( K'' = K_2 \))
3. \( 3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O \) (from Step 4, \( K''' = K_3^3 \))

When we add these reactions, we get:

\[ 2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O \]

### Step 6: Calculate the overall equilibrium constant.

The overall equilibrium constant \( K \) for the reaction is the product of the individual equilibrium constants:

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

Substituting the values we found:

\[ K = \left( \frac{1}{K_1} \right) \cdot K_2 \cdot K_3^3 \]

Thus, we can express it as:

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

### Final Answer:

The equilibrium constant for the reaction \( 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \) is:

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

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