Please help with the following!! Remarkably, a catalytic system that uses light
ID: 492730 • Letter: P
Question
Please help with the following!!
Remarkably, a catalytic system that uses light energy to hydrolyze water into hydrogen H2(g) and O(g) has been devised (Science 334. 2011: 645-648) to capture and efficiently store solar energy. Hydrogen is "clean burning" in that it yields only H2O. In this problem you will contrast the combustion of 1 kg H2(g) to that of 1 kg of n-octane. a. Calculate the enthalpy, entropy, and free-energy change on burning 1 kg of H2(g) to H2O(l) at 25 degree C and 1 bar. b. Calculate the enthalpy, entropy, and free-energy change on burning 1 kg of n-octane(g) to H2O(I) and CO2(g) at 25 degree C and 1 bar.Explanation / Answer
We need number of moles of hydrogen: n = m/MW = 1000 g / 2.016 g/mol = 496 mol.
One mole of Hydrogen produces one mole of Water
Therefore number of moles of water produced = 496 mol
The standard enthalpy (heat) of formation for liquid water at 298K (25o) is -286 kJ mol-1.
Hrxn = n X H0
Hrxn = (496 mol) X (-285.83 kJ/mol) = -142 X 10-6 J
The standard free energy change for liquid water at 298K (25o) is -237.129 kJ mol-1.
Grxn = n G0
Grxn = (496 mol)(-237.129 kJ/mol) = -118 MJ
S rxn = nisio - njsjo
i=prod j=prod
= (496 mol)(69.91 J/K·mol) – {(496 mol)(130.684 J/K·mol) + (248 mol)(205.138 J/K·mol)}
= -81.0 kJ/K.
This is negative because 1.5 mole of gas is forming 1 mole of liquid
.....................................................................................................................................................................
We need number of moles of octane is n = (1000 g)/(114.3 g/mol) = 8.75 mol.
The reaction is
C8H18(g) + 25/2 O2(g) 8CO2(g) + 9H2O(l)
The Hf and Gf for each reactant and product can be obtained from the appendix in the book, and from those we can get H0 and G0 for the reactions:
H0 = H0product - H0rectant
H0 = 8(-393.509 kJ/mol) + 9(-285.83 kJ/mol) – 1(-208.45 kJ/mol) – 12.5(0 kJ/mol) = -5512 kJ/mol.
G0 = G0product - G0rectant
G0 = 8(-394.359 kJ/mol) + 9(-237.129 kJ/mol) – 1(16.4 kJ/mol) – 12.5(0 kJ/mol) = -5305 kJ/mol.
The values for the 8.75 mol of reactants are
Hrxn = n X H0 = (8.75 mol)(-5512 kJ/mol) = -48.2 MJ
Grxn = n X G0 = (8.75 mol)(-5238 kJ/mol) = -46.4 MJ
We can calculate S just as we did in part a):
S rxn = nisio - = njsjo
= (8)(8.75 mol)(213.74 J/K·mol) +(9)(8.75 mol)(69.91 J/K·mol) – (8.75 mol)(466.73 J/K·mol) –
(12.5)(8.75 mol)(205.138 J/K·mol) = -6.05 J/K·mol.
This is negative because the entropy of the liquid water is much less than that of the gases.