For the reaction `N_(2(g)) + O_(2(g))hArr2NO_((g))`, the value of `K_(c)` at `800^(@)C` is `0.1`. When the equilibrium concentrations of both the reactants is `0.5` mol, what is the value of `K_(p)` at the same temperature

Options

A. 0.5
B. 0.1
C. 0.01
D. 0.025

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:
- \( K_c = 0.1 \) at \( 800^\circ C \)
- Equilibrium concentrations of both reactants \( [N_2] = [O_2] = 0.5 \) mol

### Step 1: Understand the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ K_p = K_c \cdot R^T \cdot \Delta n \]

Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas

### Step 2: Calculate \( \Delta n \)

For the reaction:

- Moles of products: \( 2 \) (from \( 2NO \))
- Moles of reactants: \( 1 + 1 = 2 \) (from \( N_2 + O_2 \))

Thus,

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

### Step 3: Convert temperature to Kelvin

Given temperature is \( 800^\circ C \). To convert to Kelvin:

\[ T = 800 + 273 = 1073 \, K \]

### Step 4: Substitute values into the equation for \( K_p \)

Now substituting the values into the equation:

\[ K_p = K_c \cdot R^T \cdot \Delta n \]

Substituting the known values:

\[ K_p = 0.1 \cdot (0.0821 \, \text{L·atm/(K·mol)})^{1073} \cdot 0 \]

Since \( \Delta n = 0 \):

\[ K_p = 0.1 \cdot 1 = 0.1 \]

### Conclusion

Thus, the value of \( K_p \) at the same temperature is:

\[ \boxed{0.1} \]