If the value of equilibrium constant `K_(c)` for the reaction, `N_(2)+3H_(2)hArr2NH_(3)` is 7. The equilibrium constant for the reaction `2N_(2)+6H_(2)hArr4NH_(3)` will be

Options

A. (a) 49
B. (b) 7
C. (c) 14
D. (d) 28

Explanation

To find the equilibrium constant for the reaction \(2N_2 + 6H_2 \rightleftharpoons 4NH_3\), we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \(K_c\) for the first reaction.
The first reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
The equilibrium constant \(K_c\) for this reaction is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Given that \(K_c = 7\).

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

### Step 3: Relate \(K_c'\) to \(K_c\).
To relate \(K_c'\) to \(K_c\), we can express \(K_c'\) in terms of \(K_c\):
\[ K_c' = \frac{[NH_3]^4}{[N_2]^2[H_2]^6} = \frac{([NH_3]^2)^2}{([N_2]^1)^2([H_2]^3)^2} \]
This shows that \(K_c'\) can be expressed as:
\[ K_c' = \left(\frac{[NH_3]^2}{[N_2][H_2]^3}\right)^2 = (K_c)^2 \]

### Step 4: Substitute the value of \(K_c\) into the equation for \(K_c'\).
Since we know \(K_c = 7\), we can substitute this value:
\[ K_c' = (7)^2 = 49 \]

### Conclusion:
The equilibrium constant for the reaction \(2N_2 + 6H_2 \rightleftharpoons 4NH_3\) is:
\[ K_c' = 49 \]