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
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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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