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The following thermodynamics data were obtained for dissolving KNO3 in water: The slope of the straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol) a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly. b) What does the entropy change indicate?

The following thermodynamics data were obtained for dissolving KNO3 in water: The slope of the straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol) a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly. b) What does the entropy change indicate?

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter17: Chemcial Thermodynamics
Section: Chapter Questions
Problem 17.16QE
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The following thermodynamics data were obtained for dissolving KNO3 in water: The slope of the straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol) a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly. b) What does the entropy change indicate?
Transcribed Image Text:The following thermodynamics data were obtained for dissolving KNO3 in water: The slope of the straight line defined by graph of In Ksp VS. 1/T is -4.0 x 103 K Average AS = +1.4 x 10² J/(K- mol) a) What is the AH? Does the process of dissolving KNO3 require heat or not? Explain briefly. b) What does the entropy change indicate?
Expert Solution
Step 1

The equation that relates equilibrium constant and Gibb's free energy is;

∆Go=-RT ln Keq∆Go=∆Ho-T∆So-RT ln Keq=∆Ho-T∆So ln Keq=-∆HoRT+∆SoRon comparing with straight line equation,y=mx+cA plot of lnKeq versus 1T gives a slope of -∆HoR and intercept of ∆SoR 

 

Step 2

a)

Given, 

slope=-∆HRslope=-4.0×103 KR=8.314 J/K mol∆H=-(Slope×R)=-(-4.0×103)×8.314∆H=3.32×104 J/ K mol

 

The process of dissolving KNO3 requires heat as the value of ∆H is positive.

It means the process is endothermic in nature.

 

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