The reaction between `N_(2) and H_(2)` to from ammonia has `K_(c)6xx10^(-2)` at the temperaturee `500^(@)C.` The numerical value of `K_(p)` for this reaction is

Options

A. `1.5xx10^(-5)`
B. `1.5xx10^(5)`
C. `1.5xx10^(-6)`
D. `1.5xx10^(6)`

Explanation

To find the numerical value of \( K_p \) for the reaction between \( N_2 \) and \( H_2 \) to form ammonia, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of ammonia is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Determine the change in moles of gas (\( \Delta n_g \))
To calculate \( \Delta n_g \), we need to find the difference between the moles of gaseous products and the moles of gaseous reactants.

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

Thus,
\[ \Delta n_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 4 = -2 \]

### Step 3: Use the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c (RT)^{\Delta n_g} \]
Where:
- \( K_c = 6 \times 10^{-2} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 500 \, ^\circ C = 500 + 273 = 773 \, K \)

### Step 4: Substitute the values into the equation
Now we can substitute the values into the equation:
\[ K_p = (6 \times 10^{-2}) \times (0.0821 \times 773)^{-2} \]

### Step 5: Calculate \( RT \)
First, calculate \( RT \):
\[ RT = 0.0821 \times 773 \approx 63.5 \]

### Step 6: Calculate \( (RT)^{\Delta n_g} \)
Now, calculate \( (RT)^{-2} \):
\[ (RT)^{-2} = (63.5)^{-2} \approx 0.000248 \]

### Step 7: Calculate \( K_p \)
Finally, calculate \( K_p \):
\[ K_p = (6 \times 10^{-2}) \times 0.000248 \approx 1.49 \times 10^{-5} \]

### Final Answer
Thus, the numerical value of \( K_p \) for this reaction is approximately:
\[ K_p \approx 1.5 \times 10^{-5} \]

---