Chapter 13, Problem 73GQ

### 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

# Calculate the enthalpies of solution for Li2SO4 and K2SO4. Are the solution processes exothermic or endothermic? Compare them with LiCl and KCl. What similarities or differences do you find?

Interpretation Introduction

Interpretation: The enthalpy of solution of Li2SO4 and K2SO4 is to be identified and to find out whether the solution process is endothermic or exothermic. The enthalpy of solution of Li2SO4 and is to be compared with the enthalpy of solution of LiCl and KCl

Concept introduction:

Enthalpy of solution: In the process, the change in enthalpy while dissolving the solute in the solvent.

The enthalpy of solution is calculated by the formula

ΔsolnH0=Σ(nΔfH0(product))Σ(nΔfH0(reactant))

Explanation

Given,

EnthalpyÂ ofÂ formationÂ ofÂ Li2SO4â€‰inÂ theÂ solidÂ stateÂ =â€‰âˆ’1436.4Â kJ/molEnthalpyÂ ofÂ formationÂ ofÂ Li2SO4inÂ theÂ solutionÂ Â Â Â =Â âˆ’1464.4Â kJ/molEnthalpyÂ ofÂ formationÂ ofÂ K2SO4â€‰inÂ theÂ solidÂ stateÂ =â€‰âˆ’1437.7Â kJ/molEnthalpyÂ ofÂ formationÂ ofÂ K2SO4inÂ theÂ solutionÂ Â Â Â =Â âˆ’1413.0Â kJ/mol

The enthalpy of solution is calculated by the formula

Î”solnH0â€‰=â€‰Î£(nÎ”fH0(product))â€‰âˆ’â€‰Î£(nÎ”fH0(reactant))

The enthalpy of solution of Li2SO4 is,

â€‚Â Î”solnH0â€‰=Î”fHo[Li2SO4(aq)]â€‰-â€‰Î”fHo[Li2SO4(s)]â€‰(whereâ€‰n=1forâ€‰both)â€‰=â€‰(âˆ’1464.4Â kJ/mol)â€‰âˆ’â€‰(âˆ’1436.4Â kJ/mol)=â€‰âˆ’28â€‰Â kJ/mol

The enthalpy of solution of K2SO4 is 24.7â€‰Â kJ/mol

The value of Î”solnH0 is negative and hence it is an exothermic process

The enthalpy of solution of K2SO4 is,

â€‚Â Î”solnH0â€‰=Î”fHo[K2SO4(aq)]â€‰-â€‰Î”fHo[K2SO4(s)]â€‰(whereâ€‰n=1forâ€‰both)â€‰=â€‰(âˆ’1413

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