Le Chatelier Reasoning in a Sealed NO2/N2O4 System
A rigid 2.00 L flask at 298 K is filled with a mixture of NO2 (brown) and N2O4 (colorless) that has reached equilibrium: N2O4(g) ⇌ 2NO2(g), Kc = 4.6 × 10⁻³ at 298 K. The equilibrium mixture is visibly pale brown. A technician performs three separate experiments on identical flasks: (A) compresses the mixture to 1.00 L at constant temperature; (B) adds 0.010 mol of N2O4 at constant volume; (C) raises the temperature to 350 K, at which the mixture becomes noticeably darker brown. The technician expects compression to darken the gas because 'more molecules per liter means more color.'
1. Predict and justify the color change on compression (Experiment A), using both a Le Chatelier argument and the reaction quotient Qc.
Read the worked answer
Le Chatelier: Halving the volume doubles all concentrations. The side with more gas moles (2 mol NO2 vs 1 mol N2O4) is disfavored by increased pressure, so the system shifts toward N2O4 - fewer colored NO2 molecules.
Qc check: Qc = [NO2]²/[N2O4]. Doubling both concentrations gives Qc' = (2[NO2])²/(2[N2O4]) = 2 × ([NO2]²/[N2O4]) = 2Kc. Since Qc > Kc, the reaction shifts left (toward N2O4) to restore equilibrium.
Net color: Although concentration of everything doubles instantaneously (which would briefly deepen color), the equilibrium shift consumes NO2. The fraction of NO2 falls enough that the equilibrium brown color is lighter than before compression. The technician's intuition is wrong: pressure favors the dimer, so the gas pales.
2. For Experiment C, the mixture darkens at higher temperature. Determine the sign of ΔH for the forward reaction and explain how this connects to the temperature dependence of Kc.
Read the worked answer
Darker brown means more NO2 at equilibrium, so heating shifts the equilibrium to the right (toward products). By Le Chatelier, adding heat favors the endothermic direction, so the forward reaction (N2O4 → 2NO2) is endothermic, ΔH > 0.
This is consistent with the van't Hoff relationship: for an endothermic reaction, ln K increases with increasing 1/T being unfavorable - concretely, Kc rises as T rises. Since Kc = [NO2]²/[N2O4] grows, the equilibrium ratio of NO2 to N2O4 increases, producing the darker color.
Note the contrast with Experiment A: compression changes color via a pressure/volume shift at fixed K, whereas heating changes the value of K itself. Both must be reasoned separately.
3. In Experiment B, 0.010 mol N2O4 is injected at constant volume. Does the equilibrium concentration of NO2 rise, fall, or stay the same? Justify with Qc.
Read the worked answer
Adding N2O4 raises the denominator of Qc = [NO2]²/[N2O4], making Qc < Kc momentarily, so the system shifts right and produces additional NO2. Thus [NO2] at the new equilibrium is higher than before the injection.
Important subtlety: the fraction of N2O4 that dissociates may decrease, but the absolute [NO2] increases because the added reactant pushes the reaction forward. This is the standard trap - students often say 'adding reactant dilutes the effect' and predict [NO2] falls. The correct prediction from Qc is that [NO2] increases (though not by the full amount a naive 1:2 stoichiometry would suggest, since the shift is partial).