Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 11 Intermolecular Forces and Liquids - Study Questions - Page 435a: 21


See the answer below.

Work Step by Step

a) $80.1°C$, the temperature in which the vapor pressure is equal to atmosphere's pressure, which is 760 mmHg. b) Performing a regression, we obtain: $log_{10} P(mmHg)=-1747.4/T(K)+7.835$ Or reversing it, it follows: $T(°C)=-1747.4/(log_{10} P(mmHg)-7.835)-273$ For 250mmHg: $48.4°C$ For 600mmHg: $72.5°C$ c) From the regression: $\Delta H_v=-1747.4\ K×0.008314\ kJ/mol.K/log_{10} e=33.45\ kJ/mol$ The $log_{10} e$ factor is necessary to convert the base 10 logarithmic to a natural log one.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.