For the elementary process below, the activation energy is 31 kJ mol−1 and the enthalpy of reaction is 12 kJ mol−1. What is Ea for the reverse process, in kJ mol−1 ?
A + B ⇌ AB
I answered this question for Layla earlier today. Here is the link.
https://www.jiskha.com/questions/1817043/for-the-elementary-process-below-the-activation-energy-is-31-kj-mol-1-and-the-enthalpy
I didn't give an equation for that but
Eareverse = Eafwd - dHfwd.
Eareverse = 31 - (12) = ?
In order to calculate the activation energy (Ea) for the reverse process, we need to use the relationship between the forward and reverse reactions.
For a reversible reaction like the one given (A + B ⇌ AB), the activation energy for the reverse process (Ea reverse) can be determined using the equation:
Ea reverse = Ea forward + ΔH
Where:
- Ea reverse is the activation energy for the reverse process
- Ea forward is the activation energy for the forward process
- ΔH is the enthalpy of reaction
Given values:
- Ea forward = 31 kJ mol−1
- ΔH = 12 kJ mol−1
To find Ea reverse, we substitute the given values into the equation:
Ea reverse = 31 kJ mol−1 + 12 kJ mol−1
Ea reverse = 43 kJ mol−1
Therefore, the activation energy for the reverse process (Ea reverse) is 43 kJ mol−1.