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} $)

Question

Find the solubility product of $ \mathrm{Ba(OH)_2} $?
Where $ S = 1.73 \times 10^{-14} $.

Answer 1

The relationship between $K_p$ and $K_c$ is given by:

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

For the reaction $ \text{N}_2\text{O}_4(g) \leftrightharpoons 2\text{NO}_2(g) $, the change in the number of moles of gas ($\Delta n_g$) is calculated as:

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

Rearranging the formula to solve for $K_c$ and substituting the given values:

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

Thus, $K_c$ is $2 \times 10^{-2}$.

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: \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

Correct Answer: 2

Solution and Explanation

Top Questions on Equilibrium

Questions Asked in JEE Main exam

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: \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

Correct Answer: 2

Solution and Explanation

Top Questions on Equilibrium

Questions Asked in JEE Main exam

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: \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \))