N2O4 Dissociation: Incomplete Equilibration and Temperature Effects
A sealed, rigid 1.00 L flask is charged with dinitrogen tetroxide to an initial pressure of 0.100 atm at 298 K. Thirty seconds later, when the gas has turned pale brown, the gauge reads 0.132 atm. Using the usual ICE treatment, a student obtains Kp = 0.060 for N2O4(g) ⇌ 2 NO2(g), but the accepted value at 298 K is 0.14. The arithmetic is checked twice with no error found. The flask had been rinsed with acetone and used without drying, and the pressure was read while the flask was still being handled. In a second trial the same amount of N2O4 is sealed in the flask, immersed in a 320 K bath, and left for an hour; the color deepens noticeably and the final pressure is 0.165 atm. A classmate concludes that "heating speeds up the forward reaction, so the equilibrium shifts right."
1. Reproduce the Kp calculation and use it to diagnose the discrepancy with the accepted value.
Read the worked answer
Let x be the pressure of N2O4 that dissociates. The ICE table gives P(N2O4) = 0.100 - x and P(NO2) = 2x, so the total is 0.100 + x = 0.132 atm, hence x = 0.032 atm.
P(N2O4) = 0.068 atm; P(NO2) = 0.064 atm.
Kp = P(NO2)^2 / P(N2O4) = (0.064)^2 / 0.068 = 0.004096 / 0.068 = 0.060.
The arithmetic is right, so the model, not the math, must be examined. Running the accepted Kp = 0.14 backwards: 4x^2/(0.100 - x) = 0.14 gives 4x^2 + 0.14x - 0.014 = 0, and x = 0.044 atm, for a total pressure of 0.144 atm at true equilibrium.
The measured 0.132 atm lies between the starting pressure of 0.100 atm and the equilibrium value of 0.144 atm. That is the signature of a system still approaching equilibrium, not a system at a wrong equilibrium. Thirty seconds is short for a dissociation that is slow without a catalyst. The acetone residue would have raised the total pressure, which is the wrong direction to explain the deficit, so it is not the leading explanation - though it does corrupt the measurement.
2. Analyze the 320 K trial. Calculate Kp at that temperature and determine what it shows about the enthalpy of dissociation, including whether the total pressure alone proves a shift.
Read the worked answer
The amount of gas and the volume are unchanged, so the starting pressure at 320 K scales with temperature: 0.100 atm × (320/298) = 0.107 atm. The total is 0.107 + x = 0.165 atm, so x = 0.058 atm.
P(N2O4) = 0.049 atm; P(NO2) = 0.116 atm.
Kp(320 K) = (0.116)^2 / 0.049 = 0.01346 / 0.049 = 0.27.
Kp rises from 0.14 at 298 K to about 0.27 at 320 K, so dissociation is endothermic. A van't Hoff estimate with ΔH° ≈ +57 kJ/mol predicts a factor near five over 22 K, more than the roughly twofold observed, which is another sign that an hour at 320 K may still be short of equilibrium.
Crucially, the pressure rise by itself proves nothing. Even with no shift at all, a fixed amount of gas in a fixed volume gains pressure in proportion to temperature, and 0.100 atm would become 0.107 atm. The evidence for a shift is the deepening brown color, which reports an increased NO2 concentration, combined with the fact that the observed pressure exceeds the no-shift prediction.
3. Critique the classmate's statement that heating speeds the forward reaction and therefore shifts the equilibrium right.
Read the worked answer
The statement gets the direction right and the cause wrong, and the distinction is exactly what an equilibrium question is testing.
Heating increases the rate constants of both the forward and the reverse step; there is no such thing as heating only one direction of an elementary step. If a system were at equilibrium, both rates would rise and the two new rates would be equal again - the system would stay at equilibrium at the new temperature. A shift occurs only because K itself is temperature-dependent, through ΔG° = -RT ln K and the van't Hoff relation d(ln K)/dT = ΔH°/RT^2. For an endothermic reaction, K grows with temperature, so the equilibrium composition moves toward products. The shift is a thermodynamic consequence, not a kinetic one.
There is a kinetic shadow of the same fact: for an endothermic reaction the forward step has the larger activation energy, so its rate constant grows faster with temperature. But that argument only tells you which way the transient goes while the system is out of equilibrium; it does not set the new position.
The correct phrasing is that raising the temperature increases K for an endothermic reaction, so the equilibrium mixture contains a larger fraction of NO2 - which is what the darker color shows.