   # Vapor pressure data are given here for octane, C 8 H 18 . Use the Clausius–Clapeyron equation to calculate the molar enthalpy of vaporization of octane and its normal boiling point. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 11, Problem 22PS
Textbook Problem
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## Vapor pressure data are given here for octane, C8H18. Use the Clausius–Clapeyron equation to calculate the molar enthalpy of vaporization of octane and its normal boiling point.

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization and boiling point of octane has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

lnP=ΔvapH0RT+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

lnP2p1=-ΔvapH0R1T2-1T1

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHgwe can call it as normal boiling point.

### Explanation of Solution

Given:

TemperatureoCVapor PressuremmHg2513.65045.375127.2100310.8

Temperatures and corresponding pressures are given. We can calculate the molar enthalpy of vaporization by using the Clausius-Clapeyron equation

lnP2p1=-ΔvapH0R1T2-1T1

• The molar enthalpy of vaporization is calculated using given data.

P1=13.6mmHg,P2=45.3mmHgT1=25°C=298K,T2=50°C=323K

Substituting the values

ln45.313.6=ΔvapH00.008314kJ/K.mol1323K1298Kln(1.2032)=ΔvapH00.008314kJ/K.mol1323K1298KΔvapH0=38

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