For `N_(2)+3H_(3) hArr 2NH_(3)+"Heat"`

Options

A. `K_(p)=K_(c)`
B. `K_(p)=K_(c)RT`
C. `K_(p)=K_(c)(RT)^(-2)`
D. `K_(p)=K_(c)(RT)^(-1)`

Explanation

To find the relationship between \( K_p \) and \( K_c \) for the reaction:

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 + \text{Heat} \]

we will follow these steps:

### Step 1: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 2: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3} \]

### Step 3: Relate \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ K_p = K_c \cdot (RT)^{\Delta n} \]

where \( \Delta n \) is the change in the number of moles of gas, calculated as:

\[ \Delta n = \text{moles of products} - \text{moles of reactants} \]

### Step 4: Calculate \( \Delta n \)
For the given reaction:

- Moles of products (NH3) = 2
- Moles of reactants (N2 + 3H2) = 1 + 3 = 4

Thus,

\[ \Delta n = 2 - 4 = -2 \]

### Step 5: Substitute \( \Delta n \) into the equation
Now substituting \( \Delta n \) into the relationship:

\[ K_p = K_c \cdot (RT)^{-2} \]

This can also be written as:

\[ K_p = \frac{K_c}{(RT)^2} \]

### Conclusion
The final relationship between \( K_p \) and \( K_c \) for the reaction is:

\[ K_p = K_c \cdot (RT)^{-2} \]