Question

A certain reaction has an enthalpy of Δ=44 kJ and an activation energy of a=61 kJ. What is the activation energy of the reverse reaction?

Answer 1

Answer:

E = 17 kJ

Explanation:

The enthalpy of the reaction is:

$\Delta H = E_{p} - E_{r}$  

Where:

Ep: is the energy of the products

Er: is the energy of the reactants

Similarly, the enthalpy of the reaction is related to the activation energy forward ($E_{F}$) and to the activation energy reverse ($E_{R}$) as follows:

$\Delta H = E_{F} - E_{R}$  

Having that ΔH = 44 kJ and $E_{F}$ = 61 kJ, the activation energy of the reverse reaction is:

$E_{R} = E_{F} - \Delta H = 61 kJ - 44 kJ = 17 kJ$  

Therefore, the activation energy of the reverse reaction is 17 kJ.

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