In the process `N_(2)+3H Leftrightarrow 2NH_(3)`, the initial concentration of Nitrogen and Hydrogen are one mole per litre and 3 moles per litre respectively. The equilibrium constant of the reaction is x. Then `K_(C)" for "2NH_(3) Leftrightarrow N_(2)+3H_(2)` is

Options

A. x
B. 2x
C. 1/x
D. 3x

Answer

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

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

given that the equilibrium constant for the reaction:

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

is \( x \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the first reaction:**
The equilibrium constant \( K_c \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
According to the problem, this is equal to \( x \):
\[ x = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

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

3. **Relate the two equilibrium constants:**
The relationship between the equilibrium constants for a reaction and its reverse is:
\[ K_c' = \frac{1}{K_c} \]
Therefore, substituting \( K_c = x \):
\[ K_c' = \frac{1}{x} \]

4. **Final answer:**
Thus, the equilibrium constant \( K_c \) for the reaction \( 2NH_3 \rightleftharpoons N_2 + 3H_2 \) is:
\[ K_c' = \frac{1}{x} \]