March 28, 2024, 08:10:19 AM
Forum Rules: Read This Before Posting


Topic: Calculating Energy Density and Carbon Footprint given Heats of Formation  (Read 641 times)

0 Members and 1 Guest are viewing this topic.

Offline CodingSource

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
This is the problem that I'm attempting to solve:

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


What I did first is to compute for the equivalent moles from the percentages that are given assuming that they all constitute to 100 g.

Benzene = 78.11 g/mol, Toluene = 92.14 g/mol, Xylene = 106.16 g/mol

For benzene, I multiplied 17 g C6H6 to 1/78.11 g. This gave me 1700/7811 mol. Doing the same thing for the others resulted to 1950/4607 mol for toluene and 550/1327 mol for xylene.

Now, I looked for the heat of reaction by obtaining the summation of the energy produced by these chemicals. From the given, I disregarded the CO2 and H2O, since they are not involved in catalytic reformation anyway, and proceeded with my calculation:

(1700/7881 mol)(49.03 kJ/1 mol) + (1950/4607 mol)(12.0 kJ/1 mol) + (550/1327 mol)(-24.4 kJ/1 mol) = 5.637167165 kJ

To arrive at the energy density, I divided the newly-attained value for energy to 100 g. Then, I converted the value to MJ/kg. My answer is 0.56 MJ/kg. However, I feel that I did something wrong along the way.

I am clueless about the carbon footprint. Upon searching, I did not see a standard formula for it.

Heeeeelp.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
I imagine the "energy density" as the heat produced when burning the fuel in air to make CO2 and H2O. A subtlety is that H2O can be gaseous or liquid, but your data mentions ΔHf for the liquid. And I'd take all reactants and products at conventional temperature.

The carbon footprint is more debatable, since CO2 can be emitted at every step of the production of the fuel, prior to its consumption. It's the old story of hydrogen fuel: consuming it releases no CO2, but its production presently does (it wouldn't have to) during the partial burning of CH4.

Here you have apparently no data about the previous steps, so "carbon footprint" could be understood as "during the combustion". So give some ratio between the emitted CO2 and the combustion heat.

Sponsored Links