Question

Why is enthalpy a state function?

Answer 1

Because it is frequently written as a change in enthalpy, not a singular value. You can start anywhere and end anywhere, and as long as the difference is the same, you have the same change. You can read more about this here. (There's a more rigorous calculus proof but you probably don't want to see it.)

For the sake of simplicity, let's take a look at `DeltaH_"rxn"` and consider two different reaction mechanisms that lead to the same product:

UNCATALYZED REACTION

`2"H"_2"O"_2 stackrel("slow")(->) 2"H"_2"O" + "O"_2`

`color(green)(DeltaH_"rxn" = -"196.0 kJ/mol")`, given for this reaction as-written

CATALYZED REACTION:

`cancel(MnO_2) + H_2O_2 + cancel(2H^(+)) => cancel(Mn^(2+)) + 2H_2O + O_2`
`cancel(Mn^(2+)) + 2H_2O_2 rightleftharpoons cancel(Mn(OH)_2) + cancel(2H^(+)) + color(red)(O_2)`
`cancel(Mn(OH)_2) + H_2O_2 => cancel(MnO_2) + 2H_2O`
`stackrel("---------------------------------------------------------------------------------------------")(2H_2O_2 stackrel("fast")(->) 2H_2O + O_2)`

(note though that the source had the incorrect mechanism, so I had to correct the imbalanced second step.)

You can see that they both turn out to begin at the same reactants and end at the same products. The only difference is the way to get there.

The important thing is that the catalyst and intermediates cancel out. This is due to the fact that their formation and their destruction follow the first law of thermodynamics for conservation of energy.

As a result, the reaction that remains after canceling these reaction components out is the same before and after state. The only difference is the reaction mechanism and the reaction rate.

If then we consider enthalpy from a bond-breaking and bond-making standpoint, the same bonds are broken and the same ones are made. The rest cancel each other out numerically.