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ΔG° = -RTln(K)    =>-3.9 × 105 J/Mol.  = - 8.314 J/K .Mol × 298K ln (K)   By Solving K = 9.77× 10^68 ist this right  ln(K) = -(-3.9 × 10^5 J/mol) / (8.314 J/K.mol × 298 K)  ln(K) = 157.41 Taking the exponential of both sides gives: K = e^157.41  K =  2.3 X 10^68

ΔG° = -RTln(K)    =>-3.9 × 105 J/Mol.  = - 8.314 J/K .Mol × 298K ln (K)   By Solving K = 9.77× 10^68 ist this right  ln(K) = -(-3.9 × 10^5 J/mol) / (8.314 J/K.mol × 298 K)  ln(K) = 157.41 Taking the exponential of both sides gives: K = e^157.41  K =  2.3 X 10^68

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter18: Principles Of Chemical Reactivity: Entropy And Free Energy
Section18.5: Entropy Changes And Spontaneity
Problem 2RC
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ΔG° = -RTln(K) 

 

=>-3.9 × 105 J/Mol.  = - 8.314 J/K .Mol × 298K ln (K)

 

By Solving K = 9.77× 10^68

ist this right 

ln(K) = -(-3.9 × 10^5 J/mol) / (8.314 J/K.mol × 298 K) 

ln(K) = 157.41

Taking the exponential of both sides gives:

K = e^157.41 

K =  2.3 X 10^68

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