At `25^(@)`C the equilibrium constant K,and `K_(2)` of two reaction are
`2NH_(3)hArrN_(2)+3H_(2) ,K_(1)`
`(1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3),K_(2)`
The relation `b//w` two equilibrium constant is :-

Options

A. `K_(1)=K_(2)`
B. `K_(2)=(1)/(K_(1)^(2)`
C. `K_(1)=(1)/(K_(2)^(2)`
D. `K_(1)=(1)/(K_(2)`

Answer

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we can follow these steps:

### Step 1: Write the given reactions and their equilibrium constants
1. The first reaction is:
\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \quad (K_1) \]
2. The second reaction is:
\[ \frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3 \quad (K_2) \]

### Step 2: Relate the second reaction to the first
To relate \( K_2 \) to \( K_1 \), we can manipulate the second reaction. The second reaction can be multiplied by 2 to match the stoichiometry of the first reaction:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
When we multiply the entire reaction by 2, the equilibrium constant for the reaction changes as follows:
\[ K' = K_2^2 \]
This means:
\[ K' = K_2^2 \]

### Step 3: Reverse the reaction
Now, we need to reverse this reaction to compare it with the first reaction:
\[ 2NH_3 \rightleftharpoons N_2 + 3H_2 \]
When we reverse a reaction, the equilibrium constant becomes the reciprocal:
\[ K_1 = \frac{1}{K'} \]
Substituting \( K' \) with \( K_2^2 \):
\[ K_1 = \frac{1}{K_2^2} \]

### Step 4: Final relationship between \( K_1 \) and \( K_2 \)
From the above steps, we can conclude the relationship between the two equilibrium constants:
\[ K_1 = \frac{1}{K_2^2} \]

### Summary of the solution
The relationship between the equilibrium constants \( K_1 \) and \( K_2 \) is:
\[ K_1 = \frac{1}{K_2^2} \]