Question

Given a reaction that was exothermic as written, what enthalpy change would apply to the REVERSE reaction?

Answer 1

You mean the reverse of say a combustion reaction?

Explanation:

`CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g) + 890*kJ`

This is the standard combustion reaction. Now, ordinarily, `DeltaH` would be listed separately as a NEGATIVE quantity, to show that the reaction is exothermic, which indeed it is because we are making strong `C=O` and `O-H` bonds.

The reverse is simply:

`CO_2(g) + 2H_2O(g) + DeltaH rarr CH_4(g) + 2O_2`

`DeltaH` would be listed here as a POSITIVE quantity, to show that the reaction as written is strongly ENDOTHERMIC.

Alternatively:

`CO_2(g) + 2H_2O(g) rarr CH_4(g) + 2O_2` `DeltaH=890.4*kJ*mol^-1`

So the take home message: if the reaction as written is exothermic, in the REVERSE direction it is necessarily endothermic. Conservation of energy laws assure this outcome. Come back for another serve if you are not completely sure of what I am saying.