CHEMISTRY (LL) W/CNCT   >BI<
CHEMISTRY (LL) W/CNCT >BI<
13th Edition
ISBN: 9781260572384
Author: Chang
Publisher: MCG
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Chapter 6, Problem 6.143QP

From the following data, calculate the heat of solution for KI:

Chapter 6, Problem 6.143QP, From the following data, calculate the heat of solution for KI:

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The heat of solution for KI has to be calculated.

Concept Introduction:

Heat of hydration can be defined as the change in enthalpy associated with the process of hydration.  Applying Hess’s law, it is possible to estimate ΔHsoln is the sum of two related quantities, lattice energy and the heat of hydration.  The heat of hydration is given by the equation,

ΔHsoln = U+ΔHhydration

Answer to Problem 6.143QP

The heat of solution for KI is 43kJ/mole.

Explanation of Solution

The ΔHhydr of KI is calculated below,

(1)Na+(g)+Cl-(g)Na+(aq)+Cl-(aq) ΔHhydr = (4.0-788)kJ/mole = -784.0kJ/mole(2)Na+(g)+I-(g)Na+(aq)+I-(aq) ΔHhydr = (-5.1-686)kJ/mole = -691.1kJ/mole(3)K+(g)+Cl-(g)K+(aq)+Cl-(aq) ΔHhydr = (17.2-699)kJ/mole = -681.8kJ/mole

The equation (2)and(3) are added together and then subtracted from the first equation.  The heat of hydration of KI is given as,

(2)Na+(g)+I-(g)Na+(aq)+I-(aq) ΔH=-691.1kJ/mole(3)K+(g)+Cl-(g)K+(aq)+Cl-(aq) ΔH=-681.8kJ/mole(1)Na+(aq)+Cl-(aq)Na+(g)+Cl-(g) ΔH=+784.0kJ/mole_K+(g)+I-(g)K+(aq)+I-(aq) ΔH=-588.9kJ/mole

The heat of solution is calculated as,

ΔHsoln = U+ΔHhydrΔHsoln = (632kJ/mole-588.9kJ/mole)ΔHsoln = 43kJ/mole

The heat of solution for KI is 43kJ/mole.

Conclusion

The heat of solution for KI was calculated as 43kJ/mole.

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Chapter 6 Solutions

CHEMISTRY (LL) W/CNCT >BI<

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