Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 19, Problem 19.90QA
Interpretation Introduction

To find:

Which has the higher fuel value, propanol or isopropanol?

Expert Solution & Answer
Check Mark

Answer to Problem 19.90QA

Solution:

Propanol has the higher fuel value than isopropanol.

Explanation of Solution

1) Concept:

To calculate the fuel value of a compound, we need to use heat of combustion of fuel and molar mass. First we will write the combustion reaction of propanol and isopropanol. Then we will calculate the Hcomb0 from the thermodynamic value, Hf0, of compounds present in the reaction. By using Hcomb0 and molar mass of the fuel, we will calculate the fuel value. From the fuel values, we will check which has the higher fuel value.

2) Formula:

i) Hcomb0 = m ×Hf, product0-n ×Hf, reactant0

ii) Fuel Value= Hcomb0molar mass ×number of moles of fuel

3) Given:

i) Propanol, Hf0=-302.6 kJ/mol 

ii) Propanol, molar mass=60.095 g/mol 

iii) Isopropanol, Hf0=-318.1 kJ/mol 

iv) Isopropanol, molar mass=60.095 g/mol 

v) O2, Hf0 = 0.0 kJ/mol 

vi) CO2 (g)Hf0=-393.5 kJ/mol    Chemistry: An Atoms-Focused Approach (Second Edition), Chapter 19, Problem 19.90QA

vii) H2O (l), Hf0=-285.8 kJ/mol 

4) Calculations:

Write the balanced combustion reaction of propanol.

2 CH3CH2CH2-OH l+9 O2g 6 CO2g+8 H2O l

Calculate the  Hcomb0 of propanol by using the thermodynamic value given from Appendix 4.

Hcomb0 = m ×Hf, product0-n ×Hf, reactant0

Hcomb0 = 6 × -393.5kJmol+8 × -285.8kJmol-2 × -302.6kJmol+9 × 0.0kJmol

Hcomb0 =-4647.4 kJ--605.2 kJ= -4042.2 kJ

If we divide the absolute value of Hcomb0 by twice the molar mass of propanol, we will get the fuel value.

Fuel Value= Hcomb0molar mass ×number of moles of fuel

Fuel Value= 4042.2 kJ60.095 gmol ×2 mol= 33.63 kJ /g

The fuel value of propanol is 33.63 kJ/g.

Write the balanced combustion reaction of isopropanol.

2 CH3CHOHCH3 l+9 O2g 6 CO2g+8 H2O l

Calculate the  Hcomb0 of propanol by using the thermodynamic value given from Appendix 4.

Hcomb0 = m ×Hf, product0-n ×Hf, reactant0

Hcomb0 = 6 × -393.5kJmol+8 × -285.8kJmol-2 × -318.1kJmol+9 × 0.0kJmol

Hcomb0 =-4647.4 kJ--636.2 kJ= -4011.2 kJ

If we divide the absolute value of Hcomb0 by twice the molar mass of ispropanol, we will get the fuel value.

Fuel Value= Hcomb0molar mass ×number of moles of fuel

Fuel Value= 4011.2 kJ60.095 gmol ×2 mol= 33.37 kJ /g

The fuel value of isopropanol is 33.37 kJ/g.

The fuel value of propanol is higher than isopropanol.

Conclusion:

The fuel value is calculated from the thermodynamic data, combustion reaction, and molar mass.

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Chapter 19 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 19 - Prob. 19.11VPCh. 19 - Prob. 19.12VPCh. 19 - Prob. 19.13QACh. 19 - Prob. 19.14QACh. 19 - Prob. 19.15QACh. 19 - Prob. 19.16QACh. 19 - Prob. 19.17QACh. 19 - Prob. 19.18QACh. 19 - Prob. 19.19QACh. 19 - Prob. 19.20QACh. 19 - Prob. 19.21QACh. 19 - Prob. 19.22QACh. 19 - Prob. 19.23QACh. 19 - Prob. 19.24QACh. 19 - Prob. 19.25QACh. 19 - Prob. 19.26QACh. 19 - Prob. 19.27QACh. 19 - Prob. 19.28QACh. 19 - Prob. 19.29QACh. 19 - Prob. 19.30QACh. 19 - Prob. 19.31QACh. 19 - Prob. 19.32QACh. 19 - Prob. 19.33QACh. 19 - Prob. 19.34QACh. 19 - Prob. 19.35QACh. 19 - Prob. 19.36QACh. 19 - Prob. 19.37QACh. 19 - Prob. 19.38QACh. 19 - Prob. 19.39QACh. 19 - Prob. 19.40QACh. 19 - Prob. 19.41QACh. 19 - Prob. 19.42QACh. 19 - Prob. 19.43QACh. 19 - Prob. 19.44QACh. 19 - Prob. 19.45QACh. 19 - Prob. 19.46QACh. 19 - Prob. 19.47QACh. 19 - Prob. 19.48QACh. 19 - Prob. 19.49QACh. 19 - Prob. 19.50QACh. 19 - Prob. 19.51QACh. 19 - Prob. 19.52QACh. 19 - Prob. 19.53QACh. 19 - Prob. 19.54QACh. 19 - Prob. 19.55QACh. 19 - Prob. 19.56QACh. 19 - Prob. 19.57QACh. 19 - Prob. 19.58QACh. 19 - Prob. 19.59QACh. 19 - Prob. 19.60QACh. 19 - Prob. 19.61QACh. 19 - Prob. 19.62QACh. 19 - Prob. 19.63QACh. 19 - Prob. 19.64QACh. 19 - Prob. 19.65QACh. 19 - Prob. 19.66QACh. 19 - Prob. 19.67QACh. 19 - Prob. 19.68QACh. 19 - Prob. 19.69QACh. 19 - Prob. 19.70QACh. 19 - Prob. 19.71QACh. 19 - Prob. 19.72QACh. 19 - Prob. 19.73QACh. 19 - Prob. 19.74QACh. 19 - Prob. 19.75QACh. 19 - Prob. 19.76QACh. 19 - Prob. 19.77QACh. 19 - Prob. 19.78QACh. 19 - Prob. 19.79QACh. 19 - Prob. 19.80QACh. 19 - Prob. 19.81QACh. 19 - Prob. 19.82QACh. 19 - Prob. 19.83QACh. 19 - Prob. 19.84QACh. 19 - Prob. 19.85QACh. 19 - Prob. 19.86QACh. 19 - Prob. 19.87QACh. 19 - Prob. 19.88QACh. 19 - Prob. 19.89QACh. 19 - Prob. 19.90QACh. 19 - Prob. 19.91QACh. 19 - Prob. 19.92QACh. 19 - Prob. 19.93QACh. 19 - Prob. 19.94QACh. 19 - Prob. 19.95QACh. 19 - Prob. 19.96QACh. 19 - Prob. 19.97QACh. 19 - Prob. 19.98QACh. 19 - Prob. 19.99QACh. 19 - Prob. 19.100QACh. 19 - Prob. 19.101QACh. 19 - Prob. 19.102QACh. 19 - Prob. 19.103QACh. 19 - Prob. 19.104QACh. 19 - Prob. 19.105QACh. 19 - Prob. 19.106QACh. 19 - Prob. 19.107QACh. 19 - Prob. 19.108QACh. 19 - Prob. 19.109QACh. 19 - Prob. 19.110QACh. 19 - Prob. 19.111QACh. 19 - Prob. 19.112QACh. 19 - Prob. 19.113QACh. 19 - Prob. 19.114QACh. 19 - Prob. 19.115QACh. 19 - Prob. 19.116QACh. 19 - Prob. 19.117QACh. 19 - Prob. 19.118QACh. 19 - Prob. 19.119QACh. 19 - Prob. 19.120QACh. 19 - Prob. 19.121QACh. 19 - Prob. 19.122QACh. 19 - Prob. 19.123QACh. 19 - Prob. 19.124QACh. 19 - Prob. 19.125QACh. 19 - Prob. 19.126QACh. 19 - Prob. 19.127QACh. 19 - Prob. 19.128QACh. 19 - Prob. 19.129QACh. 19 - Prob. 19.130QACh. 19 - Prob. 19.131QACh. 19 - Prob. 19.132QACh. 19 - Prob. 19.133QACh. 19 - Prob. 19.134QACh. 19 - Prob. 19.135QACh. 19 - Prob. 19.136QACh. 19 - Prob. 19.137QACh. 19 - Prob. 19.138QACh. 19 - Prob. 19.139QACh. 19 - Prob. 19.140QACh. 19 - Prob. 19.141QACh. 19 - Prob. 19.142QA
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