Consider the combustion of hydrogen gas, commonly used propel spacecraft to Eart
ID: 520524 • Letter: C
Question
Consider the combustion of hydrogen gas, commonly used propel spacecraft to Earth's orbit. 2 H_2 (g) + O_2 (g) rightarrow 2 H_2O (g) Ea = 295 kJ/mol Delta Ep (aka Delta H_rxn) = -483 kJ/mol Is this reaction endothermic or exothermic? What about the reverse process? Are the bonds collectively stronger in the reactants or the products? Rank the three molecules in terms of internal potential energy (i.e. sum of all bond energies) If you were not given Delta Ep for this reaction, explain how you would still be able to predict the sign of itExplanation / Answer
The given reaction is
2 H2 (g) + O2 (g) ------> 2 H2O (g); Ea = 295 kJ/mol, Ep (aka Hrxn) = -483 kJ/mol.
a) The enthalpy of the reaction (Hrxn) is negative. Therefore, the formation of H2O from H2 and O2 releases heat energy into the surroundings and thus, the surroundings will have a higher temperature after the reaction, i.e, the surroundings get warmer. The reaction is therefore exothermic.
Since the formation of H2O is exothermic, the reverse reaction, i.e, splitting of H2O into H2 and O2 will absorb heat energy from the surroundings and hence the surroundings are cooler after the reaction. Since heat is absorbed, the reaction is endothermic.
b) In order to answer this question, we will qualitatively figure out the entropy change of the system. The reactant side contains 3 moles of gases while the product side has 2 moles. The entropy of a system is directly proportional to the number of moles (and molecules therefore). The higher the number of molecules, the higher will be the randomness in the system and hence the higher will be the entropy. Therefore, the reactants have a higher entropy than the product and the entropy change for the reaction is negative.
Since both the enthalpy change and the entropy change are negative, therefore, at lower temperatures, the free energy change for the reaction will be negative. A negative free energy change indicates a spontaneous reaction. Therefore, at low temperatures, H2O is spontaneously formed from H2 and O2. A product is spontaneously or easily formed from the reactants only when the bonds in the product are stronger than the bonds in the reactants. Therefore, bonds in the reactants break up easily and re-orient to form stronger bonds in the products.
c) The bond energies can be easily obtained from tables as:
H-H: 432 kJ/mol
O=O: 494 kJ/mol
O-H: 459 kJ/mol.
We have 2 moles of H2 (2 H-H bonds) and 1 mole of O2 (1 O=O bond) on the reactant side. The sum total of the bond energies is (2*432 + 1*494) kJ = 1358 kJ.
The product side has 2 moles of H2O. Each mole of H2O has two O-H bonds; therefore 1 mole of H2O has an energy of (2*459) kJ = 918 kJ.
Therefore, the three molecules can be ranked in order of energy as
H2 < O2 < H2O