Common rocket engines that power spacecraft used for the transportation of space probes to Earth’s o — Physical Chemistry — Thermodynamics Chemistry Question
Thermochemistry of rocket fuels
Common rocket engines that power spacecraft used for the transportation of space probes to Earth’s orbit or to leave its gravitational field rely on simple nitrogen-based fuels. Methylhydrazine and 1,1-dimethylhydrazine (also known as unsymmetrical dimethylhydrazine) are commonly used in combination with nitrogen dioxide (or fuming nitric acid) for this purpose. Despite the relatively high toxicity of the given hydrazine derivatives, these compounds possess several beneficial characteristics which make them most suitable for powering rocket engines in the outer space. First, all hydrazine derivatives form hypergolic (spontaneously ignitable) mixtures with nitrogen dioxide, making it possible for the engine to work without any additional ignition system. Next, the melting temperatures of both methylhydrazine and 1,1-dimethylhydrazine are sufficiently low, so these species remain liquid even at extreme conditions.
To investigate the thermochemical properties of selected derivatives of hydrazine, the following calorimetric experiments were performed. Samples of liquid hydrazine, methylhydrazine, and 1,1-dimethylhydrazine, each weighing 1 gram, were combusted in an adiabatic bomb calorimeter operating at a constant volume, in a stoichiometric amount of oxygen. The temperature in the calorimeter was initially 298.15 K, and it increased by 8.25 K, 12.55 K and 14.76 K during the experiments with the respective individual compounds. By calibration, the heat capacity of the calorimeter was determined to be 2.04 kJ K−1.
Assuming that all three hydrazine derivatives react with oxygen to yield molecular nitrogen, water vapour and, if relevant, carbon dioxide at 298.15 K and 101 325 Pa, calculate the enthalpies of combustion for the reactants at the given conditions. Consider all gaseous species participating in the reaction to behave as ideal gas, and neglect any differences between the enthalpy and internal energy of all condensed phases.
Model Answer
Calculation of the number of moles corresponding to 1 g of the samples:
ni = mi / Mi M0M = 32.05 g mol−1; M1M = 46.07 g mol−1; M2M = 60.10 g mol−1.
n0M = 31.20 mmol; n1M = 21.71 mmol; n2M = 16.64 mmol.
Calculation of combustion heat: qi = Ccal × ΔTi
q0M = 16.83 kJ; q1M = 25.60 kJ; q2M = 30.11 kJ.
Calculation of the molar internal energies of combustion: ΔcUi = −qi / ni
ΔcombU0M = −539.40 kJ mol−1; ΔcombU1M = −1 179.48 kJ mol−1;
ΔcombU2M = −1 809.64 kJ mol−1.
Bomb calorimeter combustion reactions with the stoichiometric coefficients added:
Hydrazine N2H4(l) + O2(g) → N2(g) + 2 H2O(g)
Methylhydrazine N2H3CH3(l) + 2.5 O2(g) → N2(g) + CO2(g) + 3 H2O(g)
1,1-Dimethylhydrazine N2H2(CH3)2(l) + 4 O2(g) → N2 (g) + 2 CO2 (g) + 4 H2O(g)
Calculation of the molar enthalpies of combustion: ΔcHi = ΔcUi + Δc n(gas)RTstd
ΔcombH0M = −534.44 kJ mol−1;
ΔcombH1M = −1 173.29 kJ mol−1;
ΔcombH2M = −1 802.20 kJ mol−1.
Calculate the reaction enthalpies for combustion reactions of the three selected fuels with dinitrogen tetroxide at 298.15 K and 101 325 Pa (again, water vapour is produced). Consider all of the reacting hydrazine species to be liquid, mimicking the chemical processes occurring in rocket engines. Dinitrogen tetroxide enters the reaction in gaseous state. Use the standard enthalpies of formation of gaseous water (−241.83 kJ mol−1), carbon dioxide (−393.52 kJ mol−1) and dinitrogen tetroxide (9.08 kJ mol−1).
Model Answer
Calculation of the molar enthalpies of formation:
ΔformH0M = 2 ΔformHH2O,g − ΔcombH0M = +50.78 kJ mol−1
ΔformH1M = 3 ΔformHH2O,g + ΔformHCO2 − ΔcombH1M = +54.28 kJ mol−1
ΔformH2M = 4 ΔformHH2O,g + 2 ΔformHCO2 − ΔcombH2M = +47.84 kJ mol−1
Rocket engines combustion reactions:
Hydrazine N2H4(l) + 1/2 N2O4(l) → 2 H2O(g) + 3/2 N2(g)
Methylhydrazine N2H3CH3 (l) + 5/4 N2O4 (l) → CO2 (g) + 3 H2O(g) + 9/4 N2(g)
1,1-Dimethylhydrazine N2H2(CH3)2 (l) + 2 N2O4 (l) → 3 N2 (g) + 4 H2O (g) + 2 CO2 (g)
Calculation of molar reaction enthalpies, related to one mole of hydrazine derivatives:
ΔreH0M = (2 ΔformHH2O,g − 1/2 ΔformHN2O4 − ΔformH0M) = −538.98 kJ mol−1
ΔreH1M = (ΔformHCO2 + 3 ΔformHH2O,g − 5/4 ΔformHN2O4 − ΔformH1M) = −1 184.64 kJ mol−1
ΔreH2M = (2 ΔformHCO2 + 4 ΔformHH2O,g − 2 ΔformHN2O4 − 1 ΔformH2M) = −1 820.36 kJ mol−1
Extensive calorimetric experiments were performed for the relevant low-temperature phases of all chemical compounds present in the given systems at temperatures ranging from the vicinity of absolute zero up to ambient temperature. From these measurements, the absolute values of the standard molar entropies (at 298.15 K and 101 325 Pa) were evaluated according to the third law of thermodynamics:
Calculate standard reaction Gibbs energies for the three combustion reactions with dinitrogen tetroxide, estimate the corresponding equilibrium constants and predict qualitatively the extent of reactions occurring at 101 325 Pa and 298.15 K. Assume that the reactions start from stoichiometric amounts of reactants, and water is produced in its standard state – liquid. Use the given standard entropy values and the molar vaporization enthalpy of water at 298.15 K, which amounts to 40.65 kJ mol−1.
Model Answer
Calculation of the standard molar reaction enthalpies, related to one mole of hydrazine derivatives:
ΔreH°0M = ΔreH0M − 2 ΔvapHH2O = −620.28 kJ mol−1
ΔreH°1M = ΔreH1M − 3 ΔvapHH2O = −1 306.59 kJ mol−1
ΔreH°2M = ΔreH2M − 4 ΔvapHH2O = −1 982.96 kJ mol−1
Calculation of the standard molar reaction entropies, related to one mole of hydrazine derivatives:
ΔreS°0M = (2 SH2O,l + 3/2 SN2 − 1/2 SN2O4 − S0M) = 200.67 J K−1 mol−1
ΔreS°1M = (SCO2 + 3 SH2O,l + 9/4 SN2 − 5/4 SN2O4 − S1M) = 426.59 J K−1 mol−1
ΔreS°2M = (2 SCO2 + 4 SH2O,l + 3 SN2 − 2 SN2O4 − S2M) = 663.69 J K−1 mol−1
Calculation of standard molar reaction Gibbs energies:
ΔreG°0M = ΔreH°0M − T° × ΔreS°0M = −680.11 kJ mol−1
ΔreG°1M = ΔreH°1M − T° × ΔreS°1M = −1 433.77 kJ mol−1
ΔreG°2M = ΔreH°2M − T° × ΔreS°2M = −2 180.84 kJ mol−1
Estimation of the equilibrium constants for combustion reactions:
K0M = e^274.37 ≈ 1 × 10^119
K1M = e^578.41 ≈ 1 × 10^251
K2M = e^879.79 ≈ 1 × 10^382
Equilibrium constants are practically equal to infinity; the equilibrium mixture of the outlet gases contains reaction products only.
In which direction do the total pressure and temperature affect the given chemical equilibria? In other words, does an increase in pressure or in temperature lead to an increase or a decrease in the extent of the reaction?
Model Answer
All reactions increase the number of the moles of gaseous species, so increasing the pressure will suppress the extent of the reaction (though negligibly for such values of K). All reactions are strongly exothermic, so increasing the temperature will affect the equilibrium in the same direction as pressure.
Within the adiabatic approximation, calculate the flame temperature for a 1:1:1 molar mixture of the three fuels reacting with 3.75 moles of N2O4 for the case when the reactants enter a combustion chamber at 298.15 K in the liquid state. The flame temperature can be calculated, assuming that the combustion reaction formally occurs at 298.15 K and all of the heat released by the reaction (enthalpy) is then consumed to warm up the gaseous products (including water vapour) to the resulting temperature of the flame. Approximate the isobaric heat capacities of the relevant compounds with the following constants.
Model Answer
Summarizing the chemical equation representing the fuel mixture combustion:
N2H4(l) + N2H3CH3(l) + N2H2(CH3)2(l) + 3.75 N2O4(l) → 6.75 N2(g) + 9 H2O(g) + 3 CO2(g)
– (Δre0M + Δre1M + Δre2M) = (6.75 p (N2) + 9 p (H2O) + 3 p (CO2))(Tf – T 0), solve for Tf :
Tf = 4 288.65 K
Compare the calculated flame temperature obtained above for the mixture of fuels, using an analogous value corresponding to burning pure liquid 1,1-dimethylhydrazine in an oxygen atmosphere.
Model Answer
Burning of 1,1-dimethylhydrazine with oxygen can be expressed as:
N2H2(CH3)2(l) + 4 O2(g) → N2(g) + 2 CO2(g) + 4 H2O(g)
– ΔcombH2M = (Cp(N2) + 4 Cp(H2O) + 2 Cp(CO2))(Tx − T0) ,
solve for Tx: Tx = 5 248.16 K
The critical temperature of oxygen is 154.6 K and the melting temperature of 1,1-dimethylhydrazine is 216.0 K. Is there a temperature range in which the same liquid–fuel engine could be used for this alternative fuel setup?
Model Answer
There is no temperature range of coexistence of both liquid oxygen and 1,1-dimethylhydrazine, either 1,1-dimethylhydrazine is liquid and O2 is a supercritical fluid, or O2 is liquid and 1,1-dimethylhydrazine is solid.
Explain the extraordinarily high thermodynamic efficiency of rocket engines when compared to the other representatives of thermal engines (e.g. steam or Diesel engine) and support your answer with a quantitative argument.
Model Answer
Very high working temperatures maximize the temperature difference term in relation to the hypothetical efficiency of the Carnot engine. Assuming the low temperature equals T°, we get: η = (Tf − T°) / Tf = 93.0%.