For `N_(2)+3H_(2)hArr2NH_(3)` equlibrium constant is k then equlibrium constant for `2N_(2)+6H_(2)hArr4NH_(3)` is

Options

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

Answer

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

\[ 2N_2 + 6H_2 \rightleftharpoons 4NH_3 \]

given that the equilibrium constant for the reaction:

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

is \( K \).

### Step-by-Step Solution:

1. **Identify the Original Reaction and Its Equilibrium Constant**:
The original reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
The equilibrium constant for this reaction is given as \( K \).

2. **Relate the New Reaction to the Original Reaction**:
The new reaction can be derived from the original reaction by multiplying the entire original reaction by 2:
\[ 2(N_2 + 3H_2 \rightleftharpoons 2NH_3) \]
This gives us:
\[ 2N_2 + 6H_2 \rightleftharpoons 4NH_3 \]

3. **Determine the New Equilibrium Constant**:
When a balanced chemical equation is multiplied by a factor, the equilibrium constant for the new reaction is raised to the power of that factor. Since we multiplied the original reaction by 2, the new equilibrium constant \( K' \) is given by:
\[ K' = K^2 \]

4. **Conclusion**:
Therefore, the equilibrium constant for the reaction \( 2N_2 + 6H_2 \rightleftharpoons 4NH_3 \) is:
\[ K' = K^2 \]

### Final Answer:
The equilibrium constant for the reaction \( 2N_2 + 6H_2 \rightleftharpoons 4NH_3 \) is \( K^2 \).

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