CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Application of Heat of Reaction: Hydrogen vs. Gasoline
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Material and Energy Balances (First chemical engineering course)
Text Reference: Felder and Rousseau, section 4.6
Concepts: Gas law, heat of combustion
Problem Motivation: Fuel cells are a promising alternative energy technology. One type
of fuel cell, a proton exchange membrane fuel cell reacts hydrogen and oxygen together
to produce electricity. Fundamental to the design of fuel cells is an understanding of heat
of reaction of different fuels. In this problem we will use the heat of reaction to determine
the energy contained in a hydrogen cylinder, and determine the equivalent number of
gallons of gasoline.
Consider the schematic of a compressed hydrogen tank feeding a proton exchange
membrane fuel cell, as seen in the figure below. The electricity generated by the fuel cell
is used here to power a laptop computer. We are interested in determining the maximum
amount of time the laptop can be operated from the compressed hydrogen tank.
Computer
(Electric Load)
H2 feed line
Air in
Anode
Gas
Chamber
Cathode
Gas
Chamber
Air / H2O out
H2 out
H2 tank
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Fuel Cell
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Problem Information
Example Problem Statement: Before carrying out the fuel cell calculation to learn the
operating time for the laptop, we will review the energy balance to calculate heats of
reaction; in this example calculation we will determine the heat of reaction of gasoline.
We will assume that liquid gasoline is 13% n-heptane (C7H16) and 87% 2,2,4 trimethyl
pentane (isooctane, C8H18), and that it is combusted to make liquid water product.
Determine the volume in gallons of 1 mol of gasoline and the energy generated for the
combustion of a gallon of gasoline. The home problem will compare this energy with that
produced by the combustion of a compressed cylinder of hydrogen gas.
Additional Information:
Hf H2O = -285.84 kJ/mol (liquid water)
Hf H2O = -241.83 kJ/mol (vapor water)
Hf C7H16 = -224.4 kJ/mol
Hf C8H18 = -259.3 kJ/mol
Hf CO2 = -393.5 kJ/mol.
Specific gravity of n-heptane = 0.684
Specific gravity of isooctane = 0.692.
Example Problem Solution:
We will assume that gasoline is composed of 87 mol% isooctane and 13% n-heptane.
Then we can calculate the heat of reaction for the individual hydrocarbons and compute a
weighted average for the gasoline.
Step 1) We first need the heat of reaction for the individual hydrocarbons. The
stoichiometry for the combustion of one mole of n-heptane is given as:
C7H16 + 11 O2 → 7 CO2 + 8 H2O
The heat of reaction is given as
Hr =  Hf,products –  Hf,reactants
Hr,C7H16 = 7 HCO2 + 8 HH2O - HC7H16 – 11 HO2
H r,C7H16 = 7 (-393.5 kJ/mol) + 8 (-285.54 kJ/mol) – (-224.4 kJ/mol) – 11(0 kJ/mol)
H r,C7H16 = -4816 kJ/mol
Step 2) The stoichiometry for the combustion of one mole of isooctane is given as:
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C8H18 + 25/2 O2 → 8 CO2 + 9 H2O
The heat of reaction is given as
Hr =  Hf,products –  Hf,reactants
Hr,C8H18 = 8 HCO2 + 9 HH2O - HC8H18 – 25/2 HO2
Hr,C8H18 = 8 (-393.5 kJ/mol) + 9 (-285.54 kJ/mol) – (-259.3 kJ/mol) – 25/2(0 kJ/mol)
Hr,C8H18 = -5461 kJ/mol
Step 3) The weighted average heat of reaction is given as:
Hr,gasoline = 0.87 Hr,C8H18 + 0.13 Hr,C7H16
Hr,gasoline = 0.87 (-5461 kJ/mol) + 0.13 (-4816 kJ/mol)
Hr,gasoline = -5370 kJ/mol
Step 4) We now need to determine the volume in gallons of 1 mol of gasoline.
First we divide up the 1 mol of gasoline into separate parts, n-heptane and isooctane.
nn-heptane = 0.13 ngasoline = 0.13 (1 mol) = 0.13 mol
and nisooctane = 0.87 ngasoline = 0.87 (1 mol) = 0.87 mol
The mass of each fuel can be determined from the molecular weights, which for nheptane is 100 g/mol and for isooctane is 114 g/mol.
Thus the mass of each fuel is
mn-heptane = 100 g/mol (0.13 mol) = 13.0 g
misooctane = 114 g/mol (0.87 mol) = 99.2 g
Step 5) The volume of each fuel can be determined from the specific gravity. Thus,
Vn-heptane = 13.0 g / (0.684 g/cm3) = 19.0 cm3
Vn-heptane = 99.2 g / (0.692 g/cm3) = 143.4 cm3
The total volume of fuel is 162.4 cm3. Using the unit conversion factors for one cubic
foot from the inside cover of Felder and Rousseau, and converting to gallons we have
162.4 cm3 (7.4805 gal / 28317 cm3) = 0.043 gal
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Step 6) Finally, we can obtain the energy per gallon is:
  5370 kJ  1 mol 
  125,000 kJ/gal
Q

 mol  0.043 gal 
We will compare this value with the energy generated in the combustion of a compressed
tank of hydrogen gas.
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Home Problem Statement:
A gas cylinder is full of hydrogen gas at room temperature and 2000 psig pressure. This
cylinder has been proposed as a fuel tank for a fuel cell powered vehicle. Note that the
volume of the cylinder is 49.9 L.
a. Determine the energy content in kJ within the cylinder when the hydrogen is
combusted to make liquid water. Use the ideal gas law, PV = nRT.
b. How much gasoline in gallons is this energy equivalent to? Assume liquid
gasoline has a heat of reaction of Hr,gasoline -5370 kJ/mol, an average molecular
weight of 112 g/mol, and a density of 0.69 g/cm3.
c. If the hydrogen in the tank is combusted with oxygen to make water, how much
liquid water is produced?
d. Suppose you have a 100 W fuel cell (which can power a laptop computer). What
is the maximum amount of time you can run the fuel cell (hint: use the total
energy from part a).
Additional information:
Hf H2O = -285.84 kJ/mol (liquid water)
Hf H2O = -241.83 kJ/mol (vapor water)
Home Problem Solution:
Part a. To begin we need to determine the heat of reaction when hydrogen and oxygen
are mixed to form liquid water. The overall reaction stoichiometry is given as:
H2 + ½ O2 → H2O
The heat of reaction is determined from the sum of the heats of formation of the products
minus the sum of the heats of formation of the reactants. That is,
Hr =  Hf,products –  Hf,reactants
For pure hydrogen and oxygen the heat of formation is zero, while for liquid water the
heat of formation is -285.84 kJ/mol of water. Thus, H = -285.84 kJ/mol of hydrogen
fuel.
The number of moles is given from the ideal gas law as n = PV/RT. Thus, we have
 2000psi  1atm  49.9L 


mol K



  278 mol

n  

 14.696psi 
 0.08206 L atm  298 K 
The number of moles of hydrogen in the cylinder is n = 278 mol.
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Thus the energy release is Q = -n H = 278 mol (-285.84 kJ/mol) = 79000 kJ.
Part b. Given the heat of reaction of gasoline, the number of moles of gasoline necessary
to produce the same energy is obtained by: ngasoline = Qhydrogen/Hgasoline. Thus, ngasoline =
79000 kJ / (5370 kJ/mol) = 14.7 mol gasoline.
To determine the number of gallons of gasoline, we need to convert from the number of
moles to volume. This can be done by multiplying by molecular weight (to give mass)
and then dividing by the density (to give volume). Then, unit conversions on volume can
give us gallons.
The mass of gasoline is 14.7 mol (112 g/mol) = 1646 g
The volume of gasoline is 1646 g / (0.69 g/cm3) = 2386 cm3
Converting to gallons we have
2386 cm3 (7.4805 gal / 28317 cm3) = 0.63 gal. This is the energy equivalent of a
compressed cylinder of hydrogen initially filled at 2000 psi.
Part c. The volume of water can be calculated from the fact that for each mole of
hydrogen that reacts one mole of water is formed. Thus, since there were n = 278 moles
of hydrogen gas, there are 278 moles of water produced.
 278 mol  18 g  cm 3  7.4805 gal 
  1.32 gal


Volume  

3 

 mol  1g  28317 cm 
Part d. We note that energy is power multiplied by time. At a power level of 100 W, the
79000 kJ of energy from part a can be used for
 79000 kJ  1000 J 
 1 W  1 h 



  219 h
Time  


 kJ  100 W  1 J/s  3600 s 
This time is the maximum time the laptop computer can be operated from a hydrogen
cylinder. This answer assumes that all of the energy from the compressed tank is
converted into electricity. In future problems we will learn about fuel cell efficiency,
which will reduce the time by about a factor of 2-3.
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