Hess’s Law: One Reaction

Question

The standard enthalpy change for the following reaction is 3.35×103 kJ at 298 K.

2 Al2O3(s) 4 Al(s) + 3 O2(g) ΔH° = 3.35×103 kJ

What is the standard enthalpy change for this reaction at 298 K?

2 Al(s) + 3/2 O2(g) Al2O3(s)

Answer 1

2Al₂O₃(s) ===> 4Al(s) + 3O₂(g) ... ∆Hº = 3.35x10³ kJ

2Al(s) + 3/2O₂(g) ===> Al₂O₃(s) ... ∆Hº = ?

The second equation for which we are asked to find ∆Hº is the reverse of the given equation and it has been multiplied by 1/2. Therefore, to find the new value of ∆Hº we need to take the square root of the original value and then change the sign

∆Hº = sqrt 3.35x101³

∆Hº = -57.9 kJ