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

Options

A. T = 1 K
B. T = 12.18 K
C. T = 27.3 K
D. T = 273 K

Explanation

To solve the problem, we need to determine the conditions under which \( K_p \) is equal to \( K_c \) for the equilibrium reaction:

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

### 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 equation:
\[ K_p = K_c (RT)^{\Delta N_g} \]
where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the change in the number of moles of gas, calculated as the moles of products minus the moles of reactants.

2. **Calculate \( \Delta N_g \)**:
For the given reaction:
- Moles of gaseous products (NO2) = 2
- Moles of gaseous reactants (N2O4) = 1
Therefore,
\[ \Delta N_g = 2 - 1 = 1 \]

3. **Set Up the Condition for \( K_p = K_c \)**:
For \( K_p \) to equal \( K_c \), the equation becomes:
\[ K_p = K_c (RT)^{\Delta N_g} \implies K_p = K_c (RT)^{1} \]
This implies:
\[ K_p = K_c \cdot RT \]
For \( K_p \) to equal \( K_c \), we need:
\[ RT = 1 \]

4. **Solve for Temperature \( T \)**:
Rearranging the equation gives:
\[ T = \frac{1}{R} \]
Substituting the value of \( R \):
\[ T = \frac{1}{0.0821} \approx 12.18 \text{ K} \]

5. **Conclusion**:
Thus, \( K_p \) is equal to \( K_c \) at a temperature of approximately **12.18 K**.

### Final Answer:
The condition under which \( K_p \) is equal to \( K_c \) is attained when the temperature is approximately **12.18 K**.

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