Q.For the reaction N2O4

The key idea is that the equilibrium constant $K$ changes with temperature according to Le Chatelier’s principle and the van’t Hoff equation.

Step 1 – Sign of $\Delta H$

$K$ increases from 50 at 400 K to 1700 at 500 K. Since $K$ rises with temperature, the forward reaction absorbs heat — it is endothermic. So option (i) is correct, (ii) is wrong.

Step 2 – Direction of reaction at 400 K

For the given partial pressures, the reaction quotient is

$Q_p = \frac{(P_{\mathrm{NO_2}})^2}{P_{\mathrm{N_2O_4}}} = \frac{(20)^2}{2} = 200.$ …

The key to this problem is understanding what the equilibrium constant tells us about the reaction's nature and direction. When K increases with temperature, the forward reaction absorbs heat — that's the definition of endothermic. Let's walk through each option carefully.

1. Temperature dependence of K and reaction enthalpy

   For the reaction N₂O₄(g) ⇌ 2NO₂(g), K = 50 at 400 K and K = 1700 at 500 K. A huge jump in K as temperature rises means the equilibrium shifts strongly toward products at higher temperature. According to Le Chatelier's principle, if increasing temperature favors the forward reaction, that forward reaction must absorb heat. So the reaction is endothermic. Option (i) is correct; (ii) is wrong.

2. Predicting the direction of reaction using Q vs K

   At 400 K, we are given partial pressures: P(NO₂) = 20 bar, P(N₂O₄) = 2 bar. The reaction quotient Q is:

Since K = 50 at 400 K, we have Q > K. When Q exceeds K, the system has too much product relative to equilibrium. To re-establish equilibrium, the reaction proceeds backward — NO₂ combines to form more N₂O₄. So more N₂O₄ will indeed be formed. Option (iii) is correct. …