3. One suggestion for solving fuel-shortage problems is to use electric power (o

Question

3. One suggestion for solving fuel-shortage problems is to use electric power (obtained from solar energy) to electrolyze water to form H2(g) and o2(g). The hydrogen could be used as a pollution-free fuel. In the following, enter AH and AG in kJ and AS in J K to two significant figures. Enter the quantities in the order that they appear. (a) Calculate AH, AS, and AG in kJ for burning 1 kg of H2(g) to H20 (l) at 25 oC and 1 atm. (b) Calculate AH, AS, and AG for burning kg of n-octane (g) to H2o() and Co2 (g) at 25°C and 1 atm.

Explanation / Answer

Calculate H, S and G on burning 1 kg of H2(g), to H2O(l) at 25 ºC and 1 atm.

(b) Calculate these for the burning of 1 kg of n-octane(g) to H2O(l) and CO2(g) at 25 ºC and 1 atm.

a)

# of moles of hydrogen: n = m/MW = (1000 g)/(2.016 g/mol) = 496 mol.

The reaction produces one mole of water per mole of hydrogen consumed, so this is also the # of moles of water produced. Since this is just the formation reaction for water, we can look up the tabulated values for

H0 f and G0 f on

Hrxn = n H0 f = (496 mol)(-285.83 kJ/mol) = -142 MJ

Grxn = n G0 f = (496 mol)(-237.129 kJ/mol) = -118 MJ

S rxn = (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.

Note that this is negative, as expected, because 1.5 mole of gas is forming 1 mole of liquid. b) The # 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 = 8(-393.509 kJ/mol) + 9(-285.83 kJ/mol) – 1(-208.45 kJ/mol) – 12.5(0 kJ/mol) = -5512 kJ/mol.

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 H0 = (8.75 mol)(-5512 kJ/mol) = -48.2 MJ

Grxn = n G0 = (8.75 mol)(-5238 kJ/mol) = -46.4 MJ

s = = = (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 mostly because the entropy of the liquid water is much less than that of the gases

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