Answer to Question #125491 in Organic Chemistry for Francis Lopez

Question

Calculate for the energy density and carbon footprint of a reformate manufactured via continuous
catalytic regenerative reformer (Benzene, C6H6 – 17%; Toluene, C6H5CH3 – 39%; Xylene,
(CH3)2C6H4 – 44%
ΔHf C6H6(l) = 49.03 kJ/mol
ΔHf CO2(g) = -393.5 kJ/mol
ΔHf C6H5CH3(l) = 12.0 kJ/mol
ΔHf H2O(l) = -285.8 kJ/mol
ΔHf (CH3)2C6H4(l) = -24.4 kJ/mol

Answer 1

catalytic regenerative reformer (Benzene, C6H6 – 17%; Toluene, C6H5CH3 – 39%; Xylene,

(CH3)2C6H4 – 44%

ΔHf C6H6(l) = 49.03 kJ/mol

ΔHf CO2(g) = -393.5 kJ/mol

ΔHf C6H5CH3(l) = 12.0 kJ/mol

ΔHf H2O(l) = -285.8 kJ/mol

ΔHf (CH3)2C6H4(l) = -24.4 kJ/mol

Benzene C₆H₆ = 17 %

= 17 g

Mole of Benzene = 17/78

= .22

Toluene = 39%

= 39 g

Mole of toluene= 39/92

=.42

Xylene = 44 %

= 44 g

Mole of xylene = 44/106

= .42

Catalytic energy:

Mole of benzene × 49.03 + Mole of toluene × 12 + Mole of xylene × (-24.4) in KJ/ mole

= .22×49.03 +.42×12 - .42×24.4

=5.58 KJ / mole

Energy density :

Total weight given = 100 g

5.58/1000×100 = .56 KJ/kg