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

a)

for 1 kg of H2, i.e. 1000 g of H2, 500 mol of H2

H2 + 1/2O2 = H2O(l)

i) dH = Hproducts - Hreactants

HRxn = (-285.8) - (0 + 1/2*0) = - 285.8 kJ/mol

HRxn total = (500)(-285.8) = -142,900 kJ

ii) dS = Hproducts - Hreactants

SRxn = (S-H2O) - (S-H2+ 1/2*S-O2)

SRxn = (70.0) - (130.7 + 1/2*205.5) = -163.45 J/K-mol

Srxn total = 500*(-163.45) = -81,725 J/K

iii) dG = dH - T*dS

dG = (-285.8) - (298)(-163.45/1000) = -237.0919 kJ/mol

dGtotal = 500*(-237.09) = -118,545 kJ

b)

for 1 kg of octane, 1000 g

mol = mass/MW = 1000/114 = 8.7719 mol of octane

C8H18 + 25/2O2 = 8CO2 + 9H2O

i) dH = Hproducts - Hreactants

HRxn = (8*-393.5 + 9*-285.8) - (208.5 + 25/2*0) = - 5928.7 kJ/mol

HRxn total = 8.7719*(-5928.7) = -52,005.96353 kJ

ii) dS = Hproducts - Hreactants

SRxn = (8*S-CO2 + 9*S-H2O) - (1*S-octane+ 25/2*S-O2)

SRxn = (8*213.8 + 9*70.0) - (361.20 + 25/2*205.5) = -589.55 J/K-mol

Srxn total = 8.7719*-589.55 = -5,171.473 J/K

iii) dG = dH - T*dS

dG = (-5928.7) - (298)(-589.55/1000) = -5,753.0141 kJ/mol

dG total = 8.7719*-5753.0141 = -50,464.8 kJ

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