The standard enthalpy change for the following reaction is -1.39×103 kJ at 298 K.

Question

The standard enthalpy change for the following reaction is -1.39×10³ kJ at 298 K. 

3 Mn(s) + 2 O2(g)Mn3O4(s)    ΔH° = -1.39×103 kJ

What is the standard enthalpy change for the reaction

1. Mn₃O₄(s)3 Mn(s) + 2 O₂(g)

at 298 K?

 kJ

Answer 1

The reaction in question is simply the reverse of the reaction that is given. Thus, the ∆Hº will have the same magnitude but with the opposite sign.

∆Hº = +1.39x103 kJ