`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 can follow these steps:

### Step 1: Write the two reactions and their equilibrium constants

1. The first reaction is:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightleftharpoons NH_3(g) \]
The equilibrium constant for this reaction is denoted as \( K_p \).

2. The second reaction is:
\[ N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g) \]
The equilibrium constant for this reaction is denoted as \( K_p^{**} \).

### Step 2: Write the expressions for the equilibrium constants

For the first reaction, the expression for \( K_p \) is:
\[ K_p = \frac{P_{NH_3}}{(P_{N_2})^{1/2} (P_{H_2})^{3/2}} \]

For the second reaction, the expression for \( K_p^{**} \) is:
\[ K_p^{**} = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

### Step 3: Relate the two reactions

Notice that the second reaction can be derived from the first reaction by multiplying the entire first reaction by 2. When you multiply a balanced chemical equation by a factor, the equilibrium constant is raised to the power of that factor.

Thus, if we multiply the first reaction by 2, we get:
\[ N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g) \]

This means:
\[ K_p^{**} = (K_p)^2 \]

### Step 4: Solve for \( K_p \)

From the relationship \( K_p^{**} = (K_p)^2 \), we can express \( K_p \) in terms of \( K_p^{**} \):
\[ K_p = \sqrt{K_p^{**}} \]

### Final Answer

Thus, the relationship between \( K_p \) and \( K_p^{**} \) is:
\[ K_p = \sqrt{K_p^{**}} \]