The standard free energy of activation of Reaction A is 86.5kJ mol-1(20.7kcal mo
ID: 851152 • Letter: T
Question
The standard free energy of activation of Reaction A is 86.5kJ mol-1(20.7kcal mol-1) at 298K. Reaction B is one million times faster than Reaction A at the same temperature. The products of each reaction are 10.0kJ mol-1(2.39kcal 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 the reverse of Reaction B?
Please show the equation, all steps including factoring, and units to get full rating. Please don't send me to another answered question, I've seen them and they aren't fully answered. Thank you!
Explanation / Answer
a) factor = e^(G1 - G2)/RT , where
R is the gas law constant 1.98 x 10-3 kcal/mol
&
T is your 298 Kelvin
1000000 = e^ ( 20.7 - G2)/1.98x 10-3 *298.
we get G2 = 20.7 - 8.1517 = 12.548kcal mol
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