Chapter 11, Problem 65SCQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Chemists sometimes carry out reactions in liquid ammonia as a solvent. With adequate safety protection these reactions can be done at temperatures above ammonia’s boiling point in a sealed, thick-walled glass tube. If the reaction is being carried out at 20 °C, what is the pressure of ammonia inside the tube? (Use data from the previous question to answer this question.)

Interpretation Introduction

Interpretation:

The pressure of ammonia if the temperature is 20°C has to be determined using the given information.

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=-ΔvapH0R[1T2-1T1]

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 760mmHg we can call it as normal boiling point.

Explanation

Given in the previous question

Temperatureâ€‰(Â°C)Â VaporÂ Pressureâ€‰(mmâ€‰Hg)âˆ’68.40.132âˆ’45.40.526âˆ’33.61.000âˆ’18.72.004.75.0025.710.0050.120.00

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

â€‚Â lnâ€‰P2p1â€‰â€‰=â€‰â€‰-Î”vapH0Râ€‰[1T2-1T1]

P1â€‰â€‰=â€‰0.132â€‰mmâ€‰Hgâ€‰â€‰â€‰â€‰â€‰,â€‰â€‰P2â€‰â€‰â€‰=â€‰â€‰0.526â€‰mmâ€‰HgT1â€‰â€‰=â€‰âˆ’68.4Â°Câ€‰=â€‰204.6â€‰Kâ€‰,â€‰T2â€‰â€‰â€‰=â€‰âˆ’45.4Â°Câ€‰=â€‰227.6â€‰K

Substituting the values

â€‚Â lnâ€‰[0.5260.132]â€‰â€‰=â€‰â€‰âˆ’Î”vapH00.008314kJ/K.molâ€‰[1227

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