Argon, Catalysts and Colour: Judging Shifts in the N2O4/NO2 Equilibrium
A 2.00 L sealed glass flask held at 298 K contains 0.0800 mol of colourless N2O4(g). The flask has a piston that can be locked in place or left free to move, and a septum through which gases can be injected. At 298 K the equilibrium N2O4(g) ⇌ 2 NO2(g) has Kc = 4.6×10⁻³ and ΔH° = +57 kJ mol⁻¹; NO2 is brown and N2O4 is colourless.
After the mixture reaches equilibrium a student makes two predictions. First, that injecting 0.500 mol of argon with the piston locked will turn the mixture browner, because 'the total pressure rises, so the system shifts to the side with fewer gas moles'. Second, that adding a platinum catalyst and warming the flask to 325 K will darken the gas only because the gas expands. A second student disagrees with both predictions. The argon is assumed not to react with either oxide and to be insoluble in the mixture.
1. Calculate the equilibrium concentration of N2O4 and of NO2 in the original flask, and the percentage of the N2O4 that has dissociated. State any approximation you make.
Read the worked answer
Initial concentration: [N2O4]₀ = 0.0800 mol / 2.00 L = 0.0400 mol L⁻¹.
ICE table (concentrations in mol L⁻¹):
| N2O4 | 2 NO2 | |
|---|---|---|
| initial | 0.0400 | 0 |
| change | −x | +2x |
| equilibrium | 0.0400 − x | 2x |
Kc = (2x)² / (0.0400 − x) = 4.6×10⁻³
4x² = 4.6×10⁻³(0.0400 − x) → 4x² + 4.6×10⁻³x − 1.84×10⁻⁴ = 0
x = [−4.6×10⁻³ + √(2.12×10⁻⁵ + 2.944×10⁻³)] / 8 = (−4.6×10⁻³ + 5.445×10⁻²) / 8 = 6.2×10⁻³
So [NO2] = 2x = 1.25×10⁻² mol L⁻¹ and [N2O4] = 0.0400 − 0.0062 = 3.38×10⁻² mol L⁻¹.
Check: (1.25×10⁻²)² / 3.38×10⁻² = 4.6×10⁻³ ✓
Degree of dissociation: 6.2×10⁻³ / 0.0400 = 15.5%, i.e. about 16% of the N2O4 has dissociated. This also gives 0.0250 mol of NO2 in the flask.
The approximation [N2O4]eq ≈ [N2O4]₀ would give x = 6.8×10⁻³ and is not acceptable here because Kc is not small enough relative to 0.0400 mol L⁻¹; the quadratic must be solved.
2. Evaluate the student's first prediction. Does adding 0.500 mol of argon change the position of the equilibrium (a) with the piston locked, and (b) with the piston free and the total pressure held at 1.00 atm? Use the reaction quotient.
Read the worked answer
The position of equilibrium is set by the partial pressures (or concentrations) of the species in the balanced equation, never by the total pressure of an inert gas.
(a) Constant volume (piston locked). Adding argon leaves n(N2O4), n(NO2) and V unchanged, so p(N2O4) = nRT/V and p(NO2) = nRT/V are unchanged, and therefore
Q = p(NO2)² / p(N2O4) is still equal to Kp.
No shift occurs and the brown colour does not deepen. The student has confused total pressure with the partial pressures that enter Q; argon raises the total pressure but appears in no term of Q.
(b) Constant total pressure (piston free). Argon added at constant total pressure makes the volume grow. The mole fraction of each oxide falls while the total pressure stays the same, so each partial pressure, and hence the concentrations, fall. Because Q = p(NO2)²/p(N2O4) ∝ (n/V)² / (n/V) = 1/V, Q decreases below Kp and the system shifts towards more gas moles, i.e. to the right.
However, the prediction is only accidentally right: the shift is caused by dilution (falling partial pressures), not by the rise in total pressure, and the extra NO2 is partly cancelled by the larger volume, so the visible darkening is weak.
3. Deal separately with the catalyst and with the warming to 325 K. Which of these actually changes the colour of the mixture, and why? What is wrong with the 'expansion' explanation?
Read the worked answer
Catalyst. A platinum catalyst provides a lower-activation-energy path in both directions. It therefore makes the system reach equilibrium faster, but the forward and reverse rates are lowered by the same factor at equilibrium, so the ratio of forward to reverse rate constants, and hence K, is unchanged. The equilibrium concentrations, and therefore the colour, are unchanged. A catalyst raises the rate constant, not the equilibrium constant.
Warming to 325 K. The reaction is endothermic (ΔH° = +57 kJ mol⁻¹), so the equilibrium constant itself changes with temperature: K increases as T rises (van 't Hoff behaviour). The new equilibrium mixture has a larger NO2/N2O4 ratio, so the gas really does become browner.
Note that Le Chatelier's principle alone is not enough here: the honest statement is that K is a function of T, so a new equilibrium with a higher NO2 fraction is established. It is worth adding that if one reasoned only from 'higher T means higher total pressure, shift to fewer moles', one would predict a shift to the left, i.e. the opposite of what is observed. The temperature dependence of K dominates.
The expansion claim is wrong on two counts. First, the flask is sealed and rigid at this stage, so there is no expansion and no dilution. Second, if the gas did expand at constant amount, the mixture would become paler, not darker.