The equilibrium constant for the reaction
`N_2+3H_2 hArr 2NH_3` is K , then the equilibrium constant for the equilibrium `2NH_3hArr N_2+3H_2` is

Options

A. `sqrtk`
B. `sqrt(1/k)`
C. `1/k`
D. `1/(K^2)`

Answer

To solve the problem, we need to understand the relationship between the equilibrium constants of a reaction and its reverse reaction. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the given reaction and its equilibrium constant**:
The reaction given is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
The equilibrium constant for this reaction is denoted as \( K \).

2. **Write the expression for the equilibrium constant \( K \)**:
The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
where \([NH_3]\), \([N_2]\), and \([H_2]\) are the molar concentrations of ammonia, nitrogen, and hydrogen, respectively.

3. **Consider the reverse reaction**:
The reverse reaction of the given reaction is:
\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \]
We need to find the equilibrium constant for this reverse reaction, which we will denote as \( K' \).

4. **Write the expression for the equilibrium constant \( K' \)**:
For the reverse reaction, the equilibrium constant \( K' \) can be expressed as:
\[ K' = \frac{[N_2][H_2]^3}{[NH_3]^2} \]

5. **Relate \( K' \) to \( K \)**:
Since \( K' \) is the equilibrium constant for the reverse reaction, it is the reciprocal of \( K \):
\[ K' = \frac{1}{K} \]

### Final Answer:
Thus, the equilibrium constant for the reaction \( 2NH_3 \rightleftharpoons N_2 + 3H_2 \) is:
\[ K' = \frac{1}{K} \]