Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 19, Problem 19.87QA
Interpretation Introduction

To find:

a) The fuel values of diethyl ether and butanol based on the thermochemical data in Appendix 4.

b) Which has the higher fuel values?

Expert Solution & Answer
Check Mark

Answer to Problem 19.87QA

Solution:

a) Fuel value of diethyl ether is  36.74kJg, and fuel value of butanol is 36.10  kJ/g.

b) Diethyl ether has a higher fuel value than butanol.

Explanation of Solution

1) Concept:

To calculate the fuel value of the compound, we need to use the heat of combustion of fuel and molar mass. First we will write the combustion reaction of diethyl ether and butanol. Then we will calculate the Hcomb0 from the thermodynamic value, Hf0, of compounds present in the reaction. By using Hcomb0 and the molar mass of the fuel, we will calculate the 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) Butanol, Hf0=-327.3 kJ/mol 

ii) Butanol, molar mass=74.122 g/mol 

iii) Diethyl ether, Hf0=-279.6 kJ/mol 

iv) Diethyl ether, molar mass=74.122 g/mol

v) O2, Hf0 = 0.0 kJ/mol 

vi) CO2 (g), Hf0=-393.5 kJ/mol 

vii) H2O (l), Hf0=-285.8 kJ/mol 

4) Calculations:

Write the balanced combustion reaction of diethyl ether.

CH3CH2-O-CH2CH3 l+6 O2g 4 CO2g+5 H2O l

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

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

Hcomb0 = 4 × -393.5kJmol+5 × -285.8kJmol-1 × -279.6kJmol+6 × 0.0kJmol

Hcomb0 =-3003.0 kJ--279.6 kJ= -2723.4 kJ

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

Fuel Value= Hcomb0molar mass ×number of moles of fuel

Fuel Value= 2723.4 kJ74.122gmol ×1 mol= 36.74 kJ /g

The fuel value of diethyl ether is 36.74 kJ/g.

Write the balanced combustion reaction of butanol.

CH3CH2CH2CH2-OH l+6 O2g 4 CO2g+5 H2O l

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

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

Hcomb0 = 4 × -393.5kJmol+5 × -285.8kJmol-1 × -327.3kJmol+6 × 0.0kJmol

Hcomb0 =-3003.0 kJ--327.3 kJ= -2675.7 kJ

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

Fuel Value= Hcomb0molar mass ×number of moles of fuel

Fuel Value= 2675.7 kJ74.122gmol ×1 mol= 36.10 kJ /g

The fuel value of butanol is 36.10 kJ/g.

The fuel value of diethyl ether is higher than butanol.

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

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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