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 \( [N_2] \) and \( [O_2] = 0.5 \, \text{mol} \)

### Step-by-Step Solution:

1. **Identify the Reaction and Constants**:
The reaction is \( N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \). We know \( K_c \) and need to find \( K_p \).

2. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot R^T \cdot (Δn) \]
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

3. **Calculate \( Δn \)**:
For the reaction:
- Moles of products = 2 (from \( 2NO \))
- Moles of reactants = 1 (from \( N_2 \)) + 1 (from \( O_2 \)) = 2
Therefore,
\[ Δn = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

4. **Convert Temperature to Kelvin**:
The temperature is given as \( 800^\circ C \). To convert this to Kelvin:
\[ T(K) = 800 + 273 = 1073 \, K \]

5. **Substitute Values into the Equation**:
Since \( Δn = 0 \), the equation simplifies to:
\[ K_p = K_c \cdot R^T \cdot (0) \]
This means:
\[ K_p = K_c \]

6. **Final Calculation**:
Since \( K_c = 0.1 \):
\[ K_p = 0.1 \]

### Conclusion:
The value of \( K_p \) at \( 800^\circ C \) is \( 0.1 \).

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