The formation of amoonia from nitrogen and hydrogen gases can be written by the following two equations:
a. `1/2 N_(2)(g)+3/2H_(2)(g) lt lt NH_(3)(g)`
b. `1/3 N_(2)(g)+H_(2)(g) lt lt 2/3 NH_(3)(g)`
The two equations have equilibrium constants `K_(1)` and `K_(2)` respectively. The relationship between the equilibrium constant is

Options

A. `K_(1)=K_(2)^(2)`
B. `K_(1)^(3)=K_(2)^(2)`
C. `K_(1)^(2//3)=K_(2)`
D. `K_(1)=K_(2)^(3//2)`

Explanation

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we will analyze the two equations step by step.

### Step 1: Write down the two equilibrium reactions
The two reactions given are:
1. \( \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g) \) (Equation A)
2. \( \frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g) \) (Equation B)

### Step 2: Write the expressions for the equilibrium constants
The equilibrium constant \( K \) for a reaction is expressed in terms of the concentrations of the products and reactants.

For Equation A:
\[ K_1 = \frac{[NH_3]}{[N_2]^{1/2} [H_2]^{3/2}} \]

For Equation B:
\[ K_2 = \frac{[NH_3]^{2/3}}{[N_2]^{1/3} [H_2]} \]

### Step 3: Relate the two equations
To relate \( K_1 \) and \( K_2 \), we can manipulate Equation A to derive Equation B.

We can multiply the entire Equation A by a factor of \( \frac{2}{3} \):
\[ \frac{2}{3} \left( \frac{1}{2} N_2 + \frac{3}{2} H_2 \right) \rightleftharpoons \frac{2}{3} NH_3 \]

This gives us:
\[ \frac{1}{3} N_2 + H_2 \rightleftharpoons \frac{2}{3} NH_3 \]

### Step 4: Determine the relationship between \( K_1 \) and \( K_2 \)
When we multiply the reaction by a factor, the equilibrium constant is raised to the power of that factor. In this case, since we multiplied the entire reaction by \( \frac{2}{3} \), we have:
\[ K_2 = K_1^{\frac{2}{3}} \]

### Conclusion
Thus, the relationship between the equilibrium constants is:
\[ K_2 = K_1^{\frac{2}{3}} \]

### Final Answer
The correct option is:
**K_2 = K_1^{\frac{2}{3}}**

---