Dinitrogen tetroxide equilibrium under compression
At 350 K, 0.200 mol of dinitrogen tetroxide is sealed in a 2.00 dm³ flask and allowed to reach equilibrium: N2O4(g) ⇌ 2NO2(g). Analysis shows that 0.060 mol of N2O4 remains. A second experiment at 400 K, starting with the same amount in the same volume, gives Kc = 2.4. A student then compresses the 350 K equilibrium mixture to 1.00 dm³ at constant temperature and claims that 'the value of Kc falls because the pressure increases, so the mixture becomes paler.' The laboratory has brown NO2 visible in the flask. The flask is fitted with a movable piston and held in a water bath. Assume ideal gas behaviour and that the flask volume is exactly the nominal value.
1. Calculate Kc at 350 K, with units, showing an ICE table.
Read the worked answer
Let x mol dm⁻³ be the amount of N2O4 that reacts.
ICE table (amounts in mol): N2O4 ⇌ 2NO2 Initial 0.200 0 Change –x +2x Equilibrium 0.060 0.280
From N2O4: 0.200 – x = 0.060, so x = 0.140 mol. Therefore NO2 formed = 2x = 0.280 mol.
Concentrations in 2.00 dm³: [N2O4] = 0.060 / 2.00 = 0.030 mol dm⁻³ [NO2] = 0.280 / 2.00 = 0.140 mol dm⁻³
Kc = [NO2]² / [N2O4] = (0.140)² / 0.030 = 0.0196 / 0.030 = 0.653
The units are mol dm⁻³ because Δn(gas) = 2 – 1 = 1, so the concentration units do not cancel.
2. Use the 350 K and 400 K values to deduce the sign of ΔH for the forward reaction. Then evaluate the student's compression claim: will the mixture become paler, and does Kc fall?
Read the worked answer
Kc rises from 0.653 at 350 K to 2.4 at 400 K. For an equilibrium, increasing temperature increases K only if the forward reaction is endothermic. Therefore the forward reaction is endothermic and ΔH is positive.
For the compression: Kc is unchanged because temperature is constant; K depends on temperature, not pressure or volume. The student's reason is therefore wrong.
Before any shift, compressing to 1.00 dm³ doubles both concentrations: [N2O4] = 0.060 mol dm⁻³, [NO2] = 0.280 mol dm⁻³. Q = (0.280)² / 0.060 = 0.0784 / 0.060 ≈ 1.31.
Q > Kc (0.653), so the net change is reverse: 2NO2 → N2O4. The amount of brown NO2 decreases, so the mixture becomes paler. This is also consistent with Le Chatelier's principle: higher pressure favours the side with fewer gas molecules, here N2O4.
The student's prediction of a paler mixture is correct, but the explanation that Kc falls is not.
3. At 400 K, a 1.00 dm³ flask contains 0.100 mol N2O4 and 0.200 mol NO2. Calculate Q and predict the direction of net change. Explain why the 400 K Kc cannot be used to calculate equilibrium amounts in the 350 K experiment.
Read the worked answer
In a 1.00 dm³ flask, the concentrations are numerically equal to the amounts: [N2O4] = 0.100 mol dm⁻³, [NO2] = 0.200 mol dm⁻³. Q = [NO2]² / [N2O4] = (0.200)² / 0.100 = 0.0400 / 0.100 = 0.400.
At 400 K, Kc = 2.4. Since Q < Kc, the reaction proceeds in the forward direction until Q = Kc. N2O4 is converted to NO2 and the mixture becomes darker brown.
The 400 K Kc cannot be used for the 350 K experiment because Kc is temperature-dependent. At 350 K the equilibrium constant is 0.653, not 2.4. Using the wrong Kc would predict the wrong equilibrium position even if the same starting amounts and volume were used. The calculation must use the Kc value corresponding to the temperature of that experiment.