Question

For the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), \( K_p = 0.492 \, \text{atm} \) at 300 K. \( K_c \) for the reaction at the same temperature is _____ \( \times \, 10^{-2} \).
(Given: \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

Answer 1

Correct Answer: 2

To convert \(K_p\) to \(K_c\):

\[ K_p = K_c \cdot (RT)^{\Delta n_g} \]

For the reaction \( \text{N}_2\text{O}_4(g) \leftrightharpoons 2\text{NO}_2(g) \),

\[ \Delta n_g = 2 - 1 = 1 \]

Therefore:

\[ K_c = \frac{K_p}{RT} = \frac{0.492}{0.082 \times 300} = 2 \times 10^{-2} \]

So, the correct answer is: $2 \times 10^{-2}$

Answer 2

Step 1: Relation between \( K_p \) and \( K_c \).

\[ K_p = K_c (RT)^{\Delta n} \] where \(\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}\)

Step 2: Calculate Δn.

\[ \Delta n = 2 - 1 = 1 \]

Step 3: Substitute values.

\[ K_p = K_c (RT)^{1} \] \[ K_c = \frac{K_p}{RT} \]

Step 4: Substitute numerical values.

\[ K_c = \frac{0.492}{0.082 \times 300} \] \[ K_c = \frac{0.492}{24.6} = 0.02 = 2 \times 10^{-2} \]

Final Answer:

\[ \boxed{K_c = 2 \times 10^{-2}} \]