The products are usually not obtained under standard conditions but at increased temperatures. Assum — Physical Chemistry — Thermodynamics Chemistry Question
Combustion Energy
The products are usually not obtained under standard conditions but at increased temperatures. Assume for the following that the products are produced at a temperature of 100 °C and at standard pressure, while the reactants react at standard conditions.
Thermochemical data:
Propane (g): ΔfHo = –103.8 kJ mol–1, Cp = 73.6 J mol–1 K–1
Butane (g): ΔfHo = –125.7 kJ mol–1, Cp = 140.6 J mol–1 K–1
CO2 (g): ΔfHo = – 393.5 kJ mol–1, Cp = 37.1 J mol–1 K–1
H2O (l): ΔfHo = – 285.8 kJ mol–1, Cp = 75.3 J mol–1 K–1
H2O (g): ΔfHo = – 241.8 kJ mol–1, Cp = 33.6 J mol–1 K–1
O2 (g): ΔfHo = 0 kJ mol–1, Cp = 29.4 J mol–1 K–1
N2 (g): ΔfHo = 0 kJ mol–1, Cp = 29.1 J mol–1 K–1
Write down the chemical equations for the total burning of propane and butane gas in air. Indicate whether the substances are liquid (l), gaseous (g), or solid (s) under standard conditions.
Model Answer
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
Calculate the combustion energies for the burning of 1 mol of propane and butane. It can be assumed that all reactants and products are obtained under standard conditions.
Model Answer
Combustion energy (reaction enthalpy): ΔcHo = ΣpΔfHo(p) – ΣrΔfHo(r)
ΔcHo(propane) = 3(–393.5 kJ mol–1) + 4(–285.8 kJ mol–1) – (–103.8 kJ mol–1)
ΔcHo(propane) = – 2220 kJ mol–1
ΔcHo(butane) = 4(–393.5 kJ mol–1) + 5(–285.8 kJ mol–1) – (–125.7 kJ mol–1)
ΔcHo(butane) = – 2877 kJ mol–1
How much air (volume composition: 21 % of oxygen and 79 % of nitrogen) is used up in this process?
Assume that oxygen and nitrogen behave like ideal gases.
Model Answer
On the assumption that oxygen and nitrogen behave like ideal gases, the volume is proportional to the amount of substance:
n(N2) = 3.76 × n(O2)
5 mol of O2 and 18.8 mol of N2 are needed for the burning of 1 mol of propane.
6.5 mol of O2 and 24.4 mol of N2 are needed for the burning of 1 mol of butane.
When V = n R T p–1, the volumes of air are:
propane: Vair = (5 + 18.8) mol × 8.314 J K–1 mol–1 × 298.15 K × (1.013×105 Pa)–1
Vair = 0.582 m3
butane: Vair = (6.5 + 24.4) mol × 8.314 J (K mol)–1 × 298.15 K × (1.013×105 Pa)–1
Vair = 0.756 m3
Calculate the combustion energies for the burning of 1 mol of propane and butane gas in air under these conditions.
Model Answer
Under these circumstances, water is no longer liquid but gaseous. The combustion energies change due to the enthalpy of vaporization of water and higher temperature of the products.
Energy of vaporization of water at 25 °C:
ΔvHo(H2O) = ΔfHo(H2O(l)) – ΔfHo(H2O(g)) = –285.8 kJ mol–1 – (–241.8 kJ mol–1)
ΔvHo(H2O) = 44 kJ mol–1
The energy needed to increase the temperature of the products up to 100 °C is:
ΔH = Σni Cpi (T - T0)
The energy E released by burning of 1 mol of gas is:
E(propane,T) = (–2220 + 4×44) kJ + (T–T0) (3×37.1 + 4×33.6 + 18.8 mol×29.1) JK–1
E(propane, T) = –2044 kJ + (T–T0) × 792.8 J K–1 (1)
E(propane, 373.15 K) = –1984.5 kJ mol–1.
E(butane,T) = (–2877 + 5∙44) kJ + (T–T0) (4×37.1 + 5×33.6 + 24.4 mol × 29.1) JK–1
E(butane, T) = –2657 kJ + (T–T0) × 1026.4 JK–1 (2)
E(butane, 373.15 K) = –2580.0 kJ mol–1.
What is the efficiency in % of the process in 1.4 compared to 1.2 and how is the energy difference stored?
Model Answer
Efficiency of propane:
η = 1984.5 / 2220 = 89.4 %.
Efficiency of butane:
η = 2580.0 / 2877 = 89.7 %.
The energy is stored in the thermal energies of the products.
Calculate the efficiency of the combustion process as a function of the temperature of the products between 25 °C and 300 °C. Assume that the water does not condense. Plot the efficiency as a function of the temperature (reactants still react at standard conditions).
Model Answer
The combustion energies have been calculated in 1.4, equation (1), (2):
E(propane, T) = –2044 kJ + (T–T0) × 792.8 J K–1
E(butane, T) = –2657 kJ + (T–T0) × 1026.4 J K–1
The efficiencies are given by:
Propane: ηpropane(T) = 1 – 3.879×10–4 × (T–T0)
Butane: ηbutane(T) = 1 – 3.863×10–4 × (T–T0)
The plot shows that there is really no difference between the efficiencies of burning propane and butane.
Compare the combustion energy stored in a 1 liter bottle of propane and butane. Assume that the product temperature is 100 °C.
The density of liquid propane is 0.493 g cm –3, while the density of liquid butane is 0.573 g cm –3.
Model Answer
n = V ρ / M
npropane = 0.493 g cm–3 × 1000 cm3 × (44.1 g mol–1)–1 = 11.18 mol
nbutane = 0.573 g cm–3 × 1000 cm3 × (58.1 g mol–1)–1 = 9.86 mol
Ei = ni ∙ E(propane/butane, 373.15K)
E(propane) = 11.18 mol × (–1984.5 kJ mol–1) = –22.19 MJ
E(butane) = 9.86 mol × (–2580.0 kJ mol–1) = – 25.44 MJ
Despite the fact that there is less butane per volume, the energy stored in 1 dm3 of butane is higher than the energy stored in 1 dm3 of propane.