91Ó°ÊÓ

One mole of H₂

1/2 mole of O₂

The enthalpy change for reaction where one mole of hydrogen molecules combines with half a mole of oxygen molecules to produce water is $\mathrm{Δ}H=-2.86×{10}^5\mathrm{J}$, assuming that reactant gases and the resulting water are both at $25°\mathrm{C}$and $1\mathrm{atm}$pressure. Because the water will be in the form of vapour at first, it will need to release heat in order to condense into a liquid and then cool to room temperature. This results in a decrease in the thermal energy U of the system because U depends on the temperature difference, $T_f<T_i$. As well, the atmosphere will fill in the volume originally occupied by the reactant gases, doing work $PV$on the system. The enthalpy change is the total heat emitted by the system as a result of these two mechanisms.

The energy resulting from the $PV$ work is (assuming that the volume of the liquid water is negligible compared to the initial volume) is:

the final volume volume of the water and we neglect it $V_f=0$, since it is very small, the reactant gases and the resulting water are both at $25°\mathrm{C}$and $1\mathrm{atm}$pressure, so the work (from ideal gas law) is therefore:

where $n=0.5\mathrm{mol}\left(H\right)+1\mathrm{mol}\left(O\right)=1.5\mathrm{mol},R=8.31\mathrm{J}·{\mathrm{K}}^{-1}·{mol}^{-1}$ and $T=25°\mathrm{C}=298°\mathrm{K}$, so:

The change in enthalpy in the reaction is given by:

Substitute, so:

The work, $P\mathrm{Δ}V$, contribution is:

The contribution of amount of heat come from the work done by the collapsing atmosphere is $1.31\%$, where the rest for $98.69\%$ comes from the increase of internal energy of the system.