The equilibrium constant for the reversible reaction `N_(2)+3H_(2) hArr 2NH_(3)` is K and for the reaction `(1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3)` , the equilibrium constant is `K',.K` and `K'` will be related as

Options

A. `K=K'`
B. `K'=sqrt(K)`
C. `K=sqrt(K')`
D. `KxxK'=1`

Explanation

To determine the relationship between the equilibrium constants \( K \) and \( K' \) for the given reactions, we can follow these steps:

### Step 1: Write the Equilibrium Expressions
For the first reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
The equilibrium constant \( K \) is given by:
\[ K = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

For the second reaction:
\[ \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \]
The equilibrium constant \( K' \) is given by:
\[ K' = \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}} \]

### Step 2: Relate the Two Equilibrium Constants
To relate \( K' \) to \( K \), we can manipulate the expression for \( K' \). Notice that \( K' \) can be rewritten in terms of \( K \):
\[ K' = \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}} = \frac{[NH_3]}{(N_2)^{1/2}(H_2)^{3/2}} = \frac{[NH_3]}{(N_2)^{1/2}(H_2)^{3/2}} = \frac{[NH_3]}{(N_2)^{1/2}(H_2)^{3/2}} = \frac{1}{\sqrt{K}} \]

### Step 3: Simplify the Expression
From the expression for \( K' \), we can see that:
\[ K' = K^{1/2} \]
This means that \( K' \) is the square root of \( K \).

### Conclusion
Thus, the relationship between \( K \) and \( K' \) is:
\[ K' = \sqrt{K} \]