The synthesis of ammonia is given as:
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-92.6 kJ mol^(-1)` given `K_(c)=1.2` and temperature `(T)=375^(@)C`
The relationship between `K_(p)` and `K_(c)` for this reaction is

Options

A. `K_(c)=K_(p)(RT)^(-2)`
B. `K_(p)=K_(c)(RT)^(-1)`
C. `K_(p)=K_(c)(RT)^(2)`
D. `K_(p)=K_(c)(RT)^(4)`

Explanation

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

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

we will follow these steps:

### Step 1: Understand the Definitions
- \( K_c \) is the equilibrium constant based on concentrations.
- \( K_p \) is the equilibrium constant based on partial pressures.

### Step 2: Write the Expression for \( K_c \)
For the reaction, the expression for \( K_c \) is given by:

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

### Step 3: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

\[ K_p = K_c \cdot R^T \cdot (Δn) \]

where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( Δn \) is the change in the number of moles of gas, calculated as \( n_{products} - n_{reactants} \).

### Step 4: Calculate \( Δn \)
For the given reaction:
- Moles of products = 2 (from \( 2NH_3 \))
- Moles of reactants = 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4

Thus,

\[ Δn = n_{products} - n_{reactants} = 2 - 4 = -2 \]

### Step 5: Substitute Values into the Equation
Now we can substitute \( Δn \) into the relationship:

\[ K_p = K_c \cdot R^T \cdot (-2) \]

### Step 6: Rearranging the Equation
To express \( K_c \) in terms of \( K_p \):

\[ K_c = \frac{K_p}{R^T \cdot (-2)} \]

### Conclusion
Thus, the relationship between \( K_p \) and \( K_c \) for this reaction is:

\[ K_c = K_p \cdot R^T \cdot 2 \]