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

Apr 3, 2016

You mean the reverse of say a combustion reaction?

#### Explanation:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right) + 890 \cdot k J$

This is the standard combustion reaction. Now, ordinarily, $\Delta H$ 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:

$C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right) + \Delta H \rightarrow C {H}_{4} \left(g\right) + 2 {O}_{2}$

$\Delta H$ would be listed here as a POSITIVE quantity, to show that the reaction as written is strongly ENDOTHERMIC.

Alternatively:

$C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right) \rightarrow C {H}_{4} \left(g\right) + 2 {O}_{2}$ $\Delta H = 890.4 \cdot k J \cdot m o {l}^{-} 1$

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.