`K_(p) and K_(p)^(**)` are the equilibrium constants of the two reactions, given below
`(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)hArr NH_(3)(g)`
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
Therefore, `K_(p) and K_(p)^(**)` are related by

Options

A. `K_(p)=(K_(p)^(**))^(2)`
B. `K_(p)=sqrt((K_(p)^(**)))`
C. `K_(p)=2K_(p)^(**)`
D. `K_(p)=K_(p)^(**)`

Answer

To find the relationship between the equilibrium constants \( K_p \) and \( K_p^{**} \) for the given reactions, we will analyze each reaction and derive the expressions for their equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- Reaction 1: \(\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g)\)
- Reaction 2: \(N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)\)

2. **Write the Expression for \( K_p \) for Reaction 1:**
- The equilibrium constant \( K_p \) for Reaction 1 can be expressed as:
\[ K_p = \frac{P_{NH_3}}{(P_{N_2})^{1/2} (P_{H_2})^{3/2}} \]

3. **Write the Expression for \( K_p^{**} \) for Reaction 2:**
- The equilibrium constant \( K_p^{**} \) for Reaction 2 can be expressed as:
\[ K_p^{**} = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

4. **Relate \( K_p^{**} \) to \( K_p \):**
- Notice that \( K_p^{**} \) can be rewritten in terms of \( K_p \):
\[ K_p^{**} = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]
- If we take the square root of \( K_p^{**} \):
\[ \sqrt{K_p^{**}} = \sqrt{\frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}} = \frac{P_{NH_3}}{(P_{N_2})^{1/2} (P_{H_2})^{3/2}} \]
- This expression matches the expression for \( K_p \) derived from Reaction 1.

5. **Conclusion:**
- Therefore, we can conclude that:
\[ K_p = \sqrt{K_p^{**}} \]

### Final Relationship:
\[ K_p = \sqrt{K_p^{**}} \]