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.

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:

DHf H2O = -285.84 kJ/mol (liquid water)

DHf H2O = -241.83 kJ/mol (vapor water)

DHf C7H16 = -224.4 kJ/mol

DHf C8H18 = -259.3 kJ/mol

DHf 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

DHr = S DHf,products – S DHf,reactants

DHr,C7H16 = 7 DHCO2 + 8 DHH2O - DHC7H16 – 11 DHO2

DH r,C7H16 = 7 (-393.5 kJ/mol) + 8 (-285.54 kJ/mol) – (-224.4 kJ/mol) – 11(0 kJ/mol)

DH r,C7H16 = -4816 kJ/mol

Step 2) The stoichiometry for the combustion of one mole of isooctane is given as:

C8H18 + 25/2 O2 → 8 CO2 + 9 H2O

The heat of reaction is given as

DHr = S DHf,products – S DHf,reactants

DHr,C8H18 = 8 DHCO2 + 9 DHH2O - DHC8H18 – 25/2 DHO2

DHr,C8H18 = 8 (-393.5 kJ/mol) + 9 (-285.54 kJ/mol) – (-259.3 kJ/mol) – 25/2(0 kJ/mol)

DHr,C8H18 = -5461 kJ/mol

Step 3) The weighted average heat of reaction is given as:

DHr,gasoline = 0.87 DHr,C8H18 + 0.13 DHr,C7H16

DHr,gasoline = 0.87 (-5461 kJ/mol) + 0.13 (-4816 kJ/mol)

DHr,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 n-heptane 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

Step 6) Finally, we can obtain the energy per gallon is:

We will compare this value with the energy generated in the combustion of a compressed tank of hydrogen gas.

1st draft J.M. Keith June 5, 2007

Revision J. M. Keith July 25, 2007

Page 2