Question

In: Physics

Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy...

Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy of formation at RTP) is zero. Every kilogram of aluminum has (at RTP) an entropy of 1.05 kJ/K, whereas every kilogram of oxygen has an entropy of 6.41 kJ/K. Aluminum burns fiercely, forming an oxide (Al2O3) and releasing energy. The standard enthalpy of formation of the oxide is –1.67 GJ/kmol. The entropy of the oxide is 51.0 kJ/K per kilomole. According to the second law of thermodynamics, the entropy of a closed system suffering any transformation cannot diminish. It can, at best, remain unchanged as the result of the transformation, or else it must increase. If you add up the entropies of the aluminum and of the oxygen, you will discover that the sum exceeds the entropy of the oxide formed. This means that when aluminum combines with oxygen, only part of the chemical energy can be transformed into electricity, while the rest must appear as the heat associated with the missing entropy. That part that can (ideally) be converted into electricity is called the available energy.

Calculate the available energy of the aluminum/oxygen reaction (MJ/kmol).

Calculate the percentage of available to total energy for this reaction (%).

Note: The molecular weights of Al and O2 are 27 and 32 kg/kmol, respectively.

Solutions

Expert Solution


Related Solutions

standard enthalpy change of formation
a. Define standard enthalpy change of formation.b. Calculate the standard enthalpy change of formation of methane from the following standard enthalpy changes of combustion:carbon = –394 kJ mol–1hydrogen = –286 kJ mol–1methane = –891 kJ mol–1c. Calculate the standard enthalpy change of combustion of methane using the following bond energies:E(C — H) = +412 kJ mol–1E(O — O) = +496 kJ mol–1E(C — O) = +805 kJ mol–1E(O — H) = +463 kJ mol–1
A.) the standard enthalpy of formation of the reaction C2H4(g)+H2O(l)=C2H5OH(l) B.) Use standard enthalpies of formation...
A.) the standard enthalpy of formation of the reaction C2H4(g)+H2O(l)=C2H5OH(l) B.) Use standard enthalpies of formation to calculate the standard enthalphy change of: the reaction of methane gas, CH4, with chlorine rine to form liquid chloroform, CHCl3. Gaseuous hydrogen chloridee is the other product. C.) Use standard enthalpies of formation to calculate the standard enthalphy change of PCl3(g)+HCl(g)=PCl5(g)+H2(g)
Use bond-energy data to calculate the enthalpy of formation for eah of the following compounds at...
Use bond-energy data to calculate the enthalpy of formation for eah of the following compounds at 25 C a) n-Octane, C8H18(g) b) Napthalene, C10H8(g) c) Formaldehyde, H2CO(g) d) Formic acid, HCOOH(g) Give the most likely reason for the largest discrepancies between your calculated values and the ones in the tables.
calculate (a) the standard enthalpy, ( b) the standard Gibbs energy, of the reaction 4NO2(g) +...
calculate (a) the standard enthalpy, ( b) the standard Gibbs energy, of the reaction 4NO2(g) + O2(g) → 2N2O5(g)
at 25c the standard enthalpy of formation of HF(aq) is -320.1 kj/mol. of OH(aq) it is...
at 25c the standard enthalpy of formation of HF(aq) is -320.1 kj/mol. of OH(aq) it is -229.6Kj/mol. of F (aq) it is -329.2 Kj/mol.of H2O(l) it is -285.9Kj/mol. a. calculate the standard enthalpy of neutralization of HF. HF (aq) + OH- (aq) -->F-(aq) +H2O (l) b.using the value of -56.2kj as the standard enthalpy change for the reaction H+(aq)+ OH(aq)->H2O(l) c. calculate the standard enthalpy of change for the reaction HF->H+F
Calculate the standard enthalpy of formation of solid AlCl3 from the following data: 2 Al(s) +...
Calculate the standard enthalpy of formation of solid AlCl3 from the following data: 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)                  ΔHo = -1048 kJ HCl(aq) → HCl(g)                  ΔHo = 75 kJ Cl2(g) + H2(g) → 2 HCl(g)                      ΔHo = -186 kJ AlCl3(s) → AlCl3(aq)                                                       ΔHo = -323 kJ
Write a chemical equation for the formation reaction and then calculate the standard free energy of...
Write a chemical equation for the formation reaction and then calculate the standard free energy of formation of each of the following compounds from the enthalpies of formation and the standard molar entropies, using ∆G◦ r = ∆H◦ r - T∆S◦ r: (a) NH3(g) (b) H2O(g) (c) CO(g) (d) NO2(g)
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change...
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process? C2 H6(g) ⟶ H2(g) + C2 H4(g)
Part A Calculate the standard enthalpy change for the reaction 2A+B⇌2C+2D where the heats of formation...
Part A Calculate the standard enthalpy change for the reaction 2A+B⇌2C+2D where the heats of formation are given in the following table: Substance ΔH∘f (kJ/mol) A -241 B -391 C 197 D -497 Express your answer in kilojoules. Hints ΔH∘rxn = 273 kJ SubmitMy AnswersGive Up Correct Part B For the reaction given in Part A, how much heat is absorbed when 2.80 mol of A reacts? Express your answer numerically in kilojoules. Hints kJ SubmitMy AnswersGive Up Part C...
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2H2O(l) A. Write the standard enthalpy of formation equations for...
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2H2O(l) A. Write the standard enthalpy of formation equations for those that apply: B. Manipulate these equations and calculate the enthalpy of reaction: C. Use the enthalpy of reaction equation and the enthalpy of formations to check your work for (B)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT