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Some automobiles and buses have been equipped to bum propane (C 3 H 8 ). Compare the amounts of energy that can be obtained per gram of C 3 H 8 ( g ) and per gram of gasoline, assuming that gasoline is pure octane, C 8 H 18 ( l ). (See Example 7-11.) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 89E
Textbook Problem
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Some automobiles and buses have been equipped to bum propane (C3H8). Compare the amounts of energy that can be obtained per gram of C3H8(g) and per gram of gasoline, assuming that gasoline is pure octane, C8H18(l). (See Example 7-11.) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Interpretation Introduction

Interpretation: The enthalpy change ΔH of combustion of propane to be calculated and the amount of energy per gram of C3H8 and per gram of gasoline should be compared and the disadvantages of fuel using propane instead of gasoline should be explained.

Concept Introduction:

Hess's Law:

Standard enthalpy of formation:

  • The change in enthalpy that accompanies the formation of one mole of a product from its pure elements, with all substances in their standard states is called as a standard enthalpy of formation.
  • Formula:

                 ΔHfnpΔH°f(products)-nrΔH°f(reactants)......(1)

  • In equation the enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.
  • The standard enthalpy of formation of homo atomic molecules is zero.
  • The constant pressure enthalpy change is equal to heat energy change.

Explanation of Solution

Explanation

Given data:

  • C3H8(g)+5O2(g)3CO2(g)+4H2O(l)
  • Gasoline is pure octane
  • Molecular weight of propane is 44.09 g .

To calculate: per gram of C3H8(g).

                  C3H8(g)+5O2(g)3CO2(g)+4H2O(l)

             ΔH°={3(-393.5KJ)+4(-286KJ)}-(-104KJ)=-2223KJ/mole propane

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Chapter 7 Solutions

Chemistry: An Atoms First Approach
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