If `N_(2)+3H_(2)overset(K)(hArr)2NH_(3)` then `2N_(2)+6H_(2)overset(K')(hArr)4NH_(3);K'` is equal to

Options

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

Explanation

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

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

### Step 2: Write the second reaction and its equilibrium constant.
The second reaction is:
\[ 2N_2 + 6H_2 \rightleftharpoons 4NH_3 \]
We need to express the equilibrium constant \( K' \) for this reaction.

### Step 3: Write the expression for \( K' \).
The equilibrium constant \( K' \) for the second reaction is given by:
\[ K' = \frac{[NH_3]^4}{[N_2]^2[H_2]^6} \]

### Step 4: Relate \( K' \) to \( K \).
Notice that the second reaction is essentially the first reaction multiplied by 2. When we multiply a balanced chemical equation by a factor, the equilibrium constant is raised to the power of that factor. Therefore, since we multiplied the first reaction by 2, we can relate \( K' \) to \( K \) as follows:
\[ K' = K^2 \]

### Conclusion
Thus, the relationship between the equilibrium constants is:
\[ K' = K^2 \]