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 K
B. 273 K
C. 1 K
D. 12.18 K

Answer

To solve the problem, we need to find the temperature at which the equilibrium constants \( K_p \) and \( K_c \) for the reaction

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

are equal.

### Step-by-Step Solution:

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 ideal 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.

2. **Calculate \( \Delta n \)**:
For the reaction:
- Reactants: 1 mole of \( N_2O_4 \)
- Products: 2 moles of \( NO_2 \)

Therefore, the change in moles \( \Delta n \) is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \]

3. **Set Up the Equation**:
Since we are given that \( K_p = K_c \), we can substitute this into the equation:
\[ K_p = K_c \implies K_c \cdot R^T \cdot \Delta n = K_c \]
Dividing both sides by \( K_c \) (assuming \( K_c \neq 0 \)):
\[ R^T \cdot \Delta n = 1 \]

4. **Substitute \( \Delta n \)**:
Substitute \( \Delta n = 1 \) into the equation:
\[ R^T = 1 \]

5. **Solve for Temperature \( T \)**:
Rearranging gives:
\[ T = \frac{1}{R} \]
Now substituting the value of \( R \):
\[ R = 0.0821 \, \text{L·atm/(K·mol)} \]
Therefore:
\[ T = \frac{1}{0.0821} \approx 12.18 \, \text{K} \]

6. **Conclusion**:
The temperature at which \( K_p \) equals \( K_c \) is approximately \( 12.18 \, \text{K} \).

### Final Answer:
The correct option is **12.18 Kelvin**.

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