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))`

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

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

in terms of the given equilibrium constants \( K_1, K_2, \) and \( K_3 \).

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

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

### Step 2: Reverse the first reaction.

Since we need \( NH_3 \) on the reactant side, we reverse the first reaction:

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

The equilibrium constant for the reversed reaction is:

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

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

The second reaction remains unchanged:

\[ N_2 + O_2 \rightleftharpoons 2NO \]

The equilibrium constant is:

\[ K_2 \]

### Step 4: Multiply the third reaction by 3.

The third reaction needs to be multiplied by 3 to match the \( 3H_2O \) on the product side:

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

The equilibrium constant for this modified reaction becomes:

\[ K'' = K_3^3 \]

### Step 5: Combine the modified reactions.

Now, we can add the modified reactions:

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

When we add these reactions, the \( N_2 \) cancels out:

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

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

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

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

### Step 7: Simplify the expression.

Thus, we can express \( K \) 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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