Answer:
E = 17 kJ
Explanation:
The enthalpy of the reaction is:
[tex] \Delta H = E_{p} - E_{r} [/tex]
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 ([tex]E_{F}[/tex]) and to the activation energy reverse ([tex]E_{R}[/tex]) as follows:
[tex] \Delta H = E_{F} - E_{R} [/tex]
Having that ΔH = 44 kJ and [tex]E_{F}[/tex] = 61 kJ, the activation energy of the reverse reaction is:
[tex]E_{R} = E_{F} - \Delta H = 61 kJ - 44 kJ = 17 kJ[/tex]
Therefore, the activation energy of the reverse reaction is 17 kJ.
I hope it helps you!
The activation energy of the reverse reaction is 17 kJ.
The given parameters;
The activation energy of the reverse reaction is calculated by applying the following formula as shown below;
[tex]\Delta H = E_f - E_r\\\\E_r = E_f - \Delta H[/tex]
where;
[tex]E_r = 61 - 44\\\\E_r = 17 \ kJ[/tex]
Thus, the activation energy of the reverse reaction is 17 kJ.
Learn more here:https://brainly.com/question/9061190
