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Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

Simplified, Enthalpy is the measure of how much energy is trapped within the bonds of the molecules we’re measuring.

The enathalpy equation

\[ H=E+PV \]

However we can modify this a little bit since we only care about \(\Delta H\), buy substituting \(\Delta E=q+w\)

\[ \Delta H=q+w+\Delta(PV) \]

And we can assume (since it will almost always be the case) that pressure will remain constant

\[ \Delta H=q+w+P\Delta V \]

And we can assume that the only work done is pressure-volume related and we can substitute in \(w=-P\Delta V\)

\[ \Delta H=q+-P\Delta V+P\Delta V \]

Since \(-P\Delta V\) and \(P\Delta V\) cancel, we are left with

\[ \Delta H=q \]

Enthalpy change (\(\Delta H\)) tells us about the heat transfer (exo or endothermic nature of the system), but nothing about its rate or spontaneity.

Enthalpy is useful because it is a state function. That is, the way that the system got from state 1 to state 2 will not effect the change in enthalpy. This makes it more useful a measure than heat alone.

Standard Enthalpy of Formation

  • It is difficult to measure enthalpy, so we measure change in enthalpy instead
  • By convention, the standard state of formation is 0 for any elemental material in its most stable form
  • Using that benchmark, the enthalpy of formation of other materials can be determined
  • The energy to form 1 mol of product from its elements in their most stable form under standard state conditions is the standard enthalpy of formation \(\Delta H_f^\circ\)

Hess’ Law

States that since enthalpy is a state function, we can calculate it for a substance by building a thermochemical equation from other known substances * Because of this, we can combine multiple thermochemical equations together to find out the ΔH° of a specific reaction

E.g. Find the enthalpy of formation for the following reaction:

\[ \ce{2C_{(s)} + H2_{(g)} -> C2H2_{(g)}} \]

Using the following equations we can calculate this:

\[ \begin{gather} \ce{C_{(s)}+{O2}_{(g)} -> CO2_{(g)}\hskip{5cm}\Delta H^\circ=−393.5\:kJ\cdot mol^{-1}}\\ \ce{H2_{(g)} + 1/2O{2(g)} -> H2O_{(l)}\hskip{4cm}\Delta H^\circ=−285.8\:kJ\cdot mol^{-1}}\\ \ce{2C2H2_{(g)} + 5O2_{(g)} -> 4CO2_{(g)} + 2H2O_{(l)}\hskip{1cm}\Delta H^\circ=−2598.8\:kJ\cdot mol^{-1}}\\ \end{gather} \]

The overall equation looks like: $$ \begin{gather} \ce{2C_{(s)} + 2O2_{(g)} + H2_{(g)} + ½O2_{(g)} + 2CO2_{(g)} + H2O_{(l)} -> 2CO2_{(g)} + H2O_{(l)} + 2.5O2_{(g)} + C2H2_{(g)}} \end{gather} $$

To give: $$ \ce{2C_{(s)} + H2_{(g)} -> C2H2_{(g)}} $$

We want \((2\times +ve\:R_1)+(1\times +ve\:R_2)+(0.5\times −ve\:R_3)\): $$ \Delta H^\circ=(−787)+(−285.8)+(1299.4)=226.2:kJ\cdot mol^{-1} $$