Physical Chemistry
Physical Chemistry
3rd Edition
ISBN: 9780321812001
Author: ENGEL, Thomas/ Reid
Publisher: Pearson College Div
Question
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Chapter 5, Problem 5.1CP
Interpretation Introduction

Interpretation:The condition for which ΔS <0 for a spontaneous process needs to be explained.

Concept Introduction:The Gibb’s equation of thermodynamic purposed a relation between ΔS , ΔH and ΔG with temperature. With the help of this equation,one can predict the change in ΔS , ΔH and ΔG . The equation is represented as follows:

  ΔrGo=ΔrHoTΔrSo

For any reaction, the ΔH and ΔS can be calculated with the help of:

  ΔrS°= ΣΔrS°product - ΔrS°reactantΔrH°= ΣΔrH°product - ΔrH°reactant

Expert Solution & Answer
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Answer to Problem 5.1CP

For a spontaneous reaction, ΔS+ΔSsurr>0 .Hence, the process will be spontaneous only if the ΔS<0 and ΔSsurr > 0 so that ΔSsurr > ΔS 

Explanation of Solution

A system is the substance that is used to study in thermodynamic. The remaining part of universe is said to be surrounding. Overall one can say that universe is made from system and surrounding which are separated by an imaginary boundary. The entropy change for surrounding can be written as ΔSsurr and the entropy change for system is ΔSsys Hence ΔSuni represents the change in entropy of the universe and equal to the sum of ΔSsurr and ΔSsys .

  ΔSuni = ΔSsurr + ΔSsys

According to the second law of thermodynamics, a spontaneous process causes the entropy of the universe to increase. Overall the net change in entropy for the system and the surrounding must be positive.

Hence, for a spontaneous reaction, ΔS+ΔSsurr>0 . Hence, the process will only spontaneous if the ΔS<0 and ΔSsurr > 0 and ΔSsurr > ΔS 

Conclusion

For a spontaneous reaction, ΔS+ΔSsurr>0 .Hence, the process will only spontaneous if the ΔS<0 and ΔSsurr > 0 and ΔSsurr > ΔS  .

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