For the following gases equilibrium, `N_(2)O_(4) (g)hArr2NO_(2) (g)` , `K_(p)` is found to be equal to `K_(c)`. This is attained when:

Options

A. `0^(@)C`
B. 273 K
C. 1 K
D. 12.19 K

Answer

To solve the problem, we need to analyze the equilibrium reaction:

\[ N_2O_4 (g) \rightleftharpoons 2NO_2 (g) \]

We are given that \( K_p = K_c \) and need to determine the conditions under which this equality holds.

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constants**:
- The equilibrium constant in terms of partial pressures is given by \( K_p \).
- The equilibrium constant in terms of concentrations is given by \( K_c \).
- The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c (RT)^{\Delta N_g} \]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta N_g \) is the change in the number of moles of gas during the reaction.

2. **Calculating \( \Delta N_g \)**:
- For the reaction \( N_2O_4 \rightleftharpoons 2NO_2 \):
- The number of moles of gaseous products = 2 (from \( 2NO_2 \))
- The number of moles of gaseous reactants = 1 (from \( N_2O_4 \))
- Therefore, \( \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \).

3. **Setting Up the Equation**:
- Substituting \( \Delta N_g \) into the equation for \( K_p \):
\[ K_p = K_c (RT)^{1} \]
- This simplifies to:
\[ K_p = K_c \cdot RT \]

4. **Condition for \( K_p = K_c \)**:
- For \( K_p \) to equal \( K_c \), we set:
\[ K_c \cdot RT = K_c \]
- Dividing both sides by \( K_c \) (assuming \( K_c \neq 0 \)):
\[ RT = 1 \]

5. **Solving for Temperature \( T \)**:
- Rearranging gives:
\[ T = \frac{1}{R} \]
- Using the value of the gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \):
\[ T = \frac{1}{0.0821} \approx 12.19 \, \text{K} \]

### Final Answer:
The temperature at which \( K_p = K_c \) is approximately \( 12.19 \, \text{K} \).