The standard free energy of activation of a reaction A is 88.6 kJ mol–1 (21.2 kc
ID: 887669 • Letter: T
Question
The standard free energy of activation of a reaction A is 88.6 kJ mol–1 (21.2 kcal mol–1) at 298 K. Reaction B is one million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ mol–1 (2.39 kcal mol–1) more stable than the reactants. (a) What is the standard free energy of activation of reaction B?(b) What is the standard free energy of activation of the reverse of reaction A? (c) What is the standard free energy of activation of the reverse of reaction B?
Explanation / Answer
standard free energy of activation of reaction B
log (rate A/rate B) = (GB - GA )/ 2.3RT
log (1/1000000) = (GB - 88.6 kJ/mol) / 2.3(0.008314 kJ/K mol)(298K)
-6 = (GB - 88.6)/5.6984156
GB = -34.19 + 88.6
= 54.41 kJ/mol
standard free energy of activation of the reverse of reaction A
88.6kJ/mol + 10kJ/mol = 98.6kJ / mol
standard free energy of activation of the reverse of reaction B
54.41 kJ/mol + 10 kJ/mol = 64.41kJ/mol
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.