For the reaction a2 + b2 → 2ab, ea(fwd) = 105 kj/mol and ea(rev) = 77 kj/mol. assuming the reaction occurs in one step, calculate δh°rxn

Answer is: 28 kJ.
Chemical reaction: A₂ + B₂ ⇄ 2AB.
Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.
ΔH(reaction) = ?
The enthalpy change of reaction is the change in the energy of the reactants to the products.
ΔH(reaction) = Ea(forward) - Ea(reverse).
ΔH(reaction) = 105 kJ/mol - 77 kJ/mol.
ΔH(reaction) = 28 kJ/mol; this is endothermic reaction (ΔH > 0).

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lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-02-24T11:57:24+01:00

In this exercise we have to use the reaction knowledge to calculate the value in joules of this reaction, like this:

The reaction have 28 kJ.

Recalling how a chemical reaction works, we have the following information given:

Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.

Then the formula for the enthalpy of a reaction is given by:

[tex]\Delta H(reaction) = Ea(forward) - Ea(reverse)\\\Delta H(reaction) = 105 kJ/mol - 77 kJ/mol\\\Delta H(reaction) = 28 kJ/mol[/tex]

See more about enthalpy at brainly.com/question/7827769