The standard free energy of activation of a reaction A is 88.5 kJ mol^(-1) at 29
ID: 791115 • Letter: T
Question
The standard free energy of activation of a reaction A is 88.5 kJ mol^(-1) at 298K. Reaction B is one hundred million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ 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?
My answers are:
A) 48.6 kJ mol^(-1) [incorrect]
B) 98.5 kJ mol^(-1) [correct answer]
C) 58.6 kJ mol^(-1) [incorrect]
Explanation / Answer
use the equation
K 1=A e^(Ea1/RT) ---(1)
K2 =A e^(Ea2/RT) ------(2)
2/1 = ln 10^8 *RT = -Ea2 +Ea1
- Ea2 = 45.638 KJ -88.5 kJ
A)Ea2 =42.86 KJ <----------answer
B) 88.5 +10 =98.5 <---answer
C)52.861 KJ <--------answer