   Chapter 5, Problem 85GQ

Chapter
Section
Textbook Problem

Methanol, CH3OH, a compound that can be made relatively inexpensively from coal, is a promising substitute for gasoline. The alcohol has a smaller energy content than gasoline, but. with its higher octane rating, it burns more efficiently than gasoline in combustion engines. (It has the added advantage of contributing to a lesser degree to some air pollutants.) Compare the enthalpy of combustion per gram of CH3OH and C8H18 (isooctane), the latter being representative of the compounds in gasoline. (ΔfH° = –259.3 kJ/mol for isooctane.)

Interpretation Introduction

Interpretation:

Enthalpy change of combustion per gram of methanol and isooctane is to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy changenumber of moles

Explanation

For isooctane the balanced reaction is

C8H18+ 25/2 O28 CO2+ 9 H2O

ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

ΔrH0=[(8mol×-393.5kJ/mol)+(9mol×-285.83kJ/mol)] -[(1mol×-259.2kJ/mol+0)]

ΔrH0=-5461.3kJ

The change in enthalpy is -5461.3kJ

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy changenumber of moles

ΔrH0=-5461.3kJ/mol×1mol114.2g=-47.82J/gK

So, the enthalpy of isooctane is -47

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