December 03, 2024, 12:24:41 PM
Forum Rules: Read This Before Posting


Topic: Lower heating value calculation  (Read 4397 times)

0 Members and 1 Guest are viewing this topic.

Offline Heater

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Lower heating value calculation
« on: September 24, 2024, 01:39:04 PM »
I have an assigenment where i need to calculate the lover heating value of  the combustion of 100 mol of a gas mixture (MJ/Nm3)

The gas mixture is made of in mol%:
91.1% CH4
4.7% C2H6
1.7% C3H8
1.4% C4H10
0.6% N2
0.5% CO2

the enthalpy of the combustion reaction of methane, ethane, propane and butane, are given:
Methane combustion : -804,2 kJ/mol
Ethane combustion : -1428,9 kJ/mol
Propane combustion : -2045,5 kJ/mol
butane combustion : -2663.6 kJ/mol

we are also told the gasmixture behaves like an idealgas.

The solution in the book is given to be 38,8 MJ/m3 but i can't seem to get the right answer no matter what i do.

my main theory to calculating it was to calculate the ΔH of the combustion reaction using the enthalpy of the 4 gas combustions given by adding them together  and multiplying with their respective mol% and then dividing it with the normal volume of an idealgas 22,42 m3/kmol which gives me a solution of 38,93 MJ/M3 which is close to the solution but not quite it.

Does anyone know how what i am missing?

Online Hunter2

  • Sr. Member
  • *****
  • Posts: 2306
  • Mole Snacks: +190/-50
  • Gender: Male
  • Vena Lausa moris pax drux bis totis
Re: Lower heating value calculation
« Reply #1 on: September 24, 2024, 02:29:54 PM »
Calculate how many m^3 is 100 mol

Then calculate each ΔH value . Get summary and division through the volume.


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27869
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Lower heating value calculation
« Reply #2 on: September 24, 2024, 03:12:48 PM »
the enthalpy of the 4 gas combustions given by adding them together  and multiplying with their respective mol% and then dividing it with the normal volume of an idealgas 22,42 m3/kmol which gives me a solution of 38,93 MJ/M3 which is close to the solution but not quite it.

I did the same and got 38.897 MJ/m3 (so more like 38.90).

Approach is correct, I would assume this is just a typo (or lousy rounding) in the answer key.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links