   # For rubidium Δ H vap o = 69.0 KJ/mol at 686°C, its boiling point. Calculate ∆ S °, q, w, and ∆ E for the vaporization of 1.00 mole of rubidium at 686°C and 1.00 atm pressure. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 96CWP
Textbook Problem
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## For rubidium Δ H vap o = 69.0   KJ/mol at 686°C, its boiling point. Calculate ∆S°, q, w, and ∆E for the vaporization of 1.00 mole of rubidium at 686°C and 1.00 atm pressure.

Interpretation Introduction

Interpretation:

The values of ΔS,q,w and ΔE for the vaporization of one mole of rubidium at its boiling point is to be calculated.

### Explanation of Solution

Given

The standard enthalpy of vaporization of rubidium at its boiling point is 69.0kJ/mol.’

The boiling point of rubidium is 686°C.

The pressure of the system is 1atm.

The conversion of temperature from °C into Kelvin is done as,

T(K)=T(°C)+273.15

Hence, the conversion of temperature from 686°C into Kelvin is,

T(K)=(686+273.15)K=1059.15K

The conversion of enthalpy from kJ/mol into J/mol is done as,

1kJ/mol=103J/mol

Hence, the conversion of enthalpy from 69.0kJ/mol into J/mol is,

69.0kJ/mol=69.0×103J/mol

The entropy change of vaporization of the given liquid is calculated by the formula,

ΔSvap=ΔHvapT

Where,

• ΔSvap is the entropy of vaporization

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