OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीAt \( 60^\circ\text{C} \) and \( 1\text{ atm} \), \( \text{N}_2\text{O}_4(g) \) is \( 50\% \) dissociated into \( \text{NO}_2(g) \). What is the equilibrium constant \( K_p \)?
Correct answer · Option 3
\( 1.33\text{ atm} \)
Explanation
\( p_{\text{NO}_2} = \frac{2\alpha}{1+\alpha} P = \frac{1}{1.5} = \frac{2}{3}\text{ atm} \); \( p_{\text{N}_2\text{O}_4} = \frac{1-\alpha}{1+\alpha} P = \frac{0.5}{1.5} = \frac{1}{3}\text{ atm} \). \( K_p = \frac{(2/3)^2}{1/3} = \frac{4}{3} = 1.33\text{ atm} \).
