The value of `K_(c)` at 900 K temp for following reaction is 5
`N_(2(g))+O_(2(g)) hArr 2NO_((g))` if equilibrium `conc^(n)` of `N_(2)` and `O_(2)` is 0.5M then value of `K_(p)` is :-

Options

A. 0.02
B. 0.2
C. 5
D. `(5)/(RT)`

Explanation

To solve the problem, we need to find the value of \( K_p \) for the reaction:

\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

given that \( K_c = 5 \) at 900 K and the equilibrium concentrations of \( N_2 \) and \( O_2 \) are both 0.5 M.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Values:**
- The reaction is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]
- Given \( K_c = 5 \) at 900 K.

2. **Calculate \( \Delta N_g \):**
- \( \Delta N_g \) is calculated as:
\[ \Delta N_g = N_p - N_r \]
- Where \( N_p \) is the number of moles of products and \( N_r \) is the number of moles of reactants.
- In this reaction:
- Products (NO): 2 moles
- Reactants (N2 + O2): 1 + 1 = 2 moles
- Therefore:
\[ \Delta N_g = 2 - 2 = 0 \]

3. **Use the Relationship Between \( K_p \) and \( K_c \):**
- The relationship is given by:
\[ K_p = K_c \cdot R^{\Delta N_g} \]
- Where \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)).

4. **Substitute \( \Delta N_g \) into the Equation:**
- Since \( \Delta N_g = 0 \):
\[ K_p = K_c \cdot R^0 \]
- And \( R^0 = 1 \):
\[ K_p = K_c \cdot 1 \]
\[ K_p = K_c \]

5. **Final Calculation:**
- Since \( K_c = 5 \):
\[ K_p = 5 \]

### Conclusion:
The value of \( K_p \) is **5**.