# How can I calculate enthalpy of formation?

Jan 7, 2016

This is not something that you normally need to calculate (unless you are given some enthalpies of reaction and a few enthalpies of formation for other compounds already); truth be told, computational chemists have already done this calculation and tabulated it, so we can just use it.

DISCLAIMER: The following example is not of enthalpy of formation, but a decent alternative---the standard molar entropy. It IS long, so it is optional reading.

CALCULATING STANDARD MOLAR ENTROPY

Unfortunately my book doesn't show how to do it for enthalpy, but it does show how to determine the standard molar entropy of nitrogen at $\text{298.15 K}$ using physical chemistry calculations, and that is actually tabulated alongside the standard enthalpy of formation in thermodynamic tables.

Just as enthalpy can be defined as:

$\Delta H = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}$

(where ${C}_{P}$ is the constant-pressure heat capacity)

...in general, entropy can be defined as:

$\textcolor{g r e e n}{\Delta \overline{S} = {\int}_{{T}_{1}}^{{T}_{2}} \frac{{\overline{C}}_{P}}{T} \mathrm{dT}}$

where $\Delta \overline{S}$ is the molar entropy and ${\overline{C}}_{P}$ is the molar constant-pressure heat capacity.

What do we need to know? Well, here is the full equation to calculate the standard molar entropy of ${\text{N}}_{2} \left(g\right)$, which is $\text{191.6 J/K}$ at $\text{298.15 K}$ and $\text{1 bar}$:

$\setminus m a t h b f \left(\Delta {S}_{{N}_{2}}^{\circ}\right)$

= \mathbf(int_("0 K")^(T_"low") (barC_P^"low")/TdT + int_(T_"low")^(T_"normal") barC_P/TdT + DeltabarS_"trs")

\mathbf(+ int_(T_"normal")^(T_"mp") barC_P/TdT + DeltabarS_"melt")

\mathbf(+ int_(T_"mp")^(T_"bp") barC_P/TdT + DeltabarS_"vap")

$\setminus m a t h b f \left(+ {\int}_{{T}_{\text{bp")^("298.15 K}}} {\overline{C}}_{P} / T \mathrm{dT}\right)$

$\setminus m a t h b f \left(+ \text{ Correction for nonideality}\right)$

It requires that we trace the entropic changes over a temperature range, from its absolute zero state all the way to its gaseous state at $\text{298.15 K}$.

Yes, it's long! But we'll go through it.

The book consulted for the content below is Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 21-4.

EVALUATING THE GENERAL INTEGRAL

Evaluating the integral is not as complicated as it seems. We have, in general:

$\Delta \overline{S} = {\int}_{{T}_{1}}^{{T}_{2}} \frac{{\overline{C}}_{P}}{T} \mathrm{dT} = \textcolor{g r e e n}{{\overline{C}}_{P} \ln \setminus \frac{{T}_{2}}{{T}_{1}}}$

And you should see this multiple times in the same main equation.

VERY LOW TEMPERATURES

At very low temperatures, obviously nitrogen is a solid, but it is so close to absolute zero; the structure needs to be treated in a special manner at such low temperatures.

Here, the Debye theory states that "the low-temperature molar [constant-pressure] heat capacity of nonmetallic solids goes as":

${\overline{C}}_{P}^{\text{low}} \left(T\right) = \frac{12 {\pi}^{4}}{5} R {\left(\frac{T}{{\Theta}_{D}}\right)}^{3}$; $0 < T \le {T}_{\text{low}}$

where ${T}_{\text{low}}$ tends to be between $10$ and $\text{20 K}$, and the Debye temperature "${\Theta}_{D}$ is a constant characteristic of the solid".

This integral evaluates as:

$\textcolor{b l u e}{\Delta {\overline{S}}_{\text{0 K"^(T_"low")) = (12pi^4)/5 R/Theta_D^3 int_("0 K")^(T_"low") T^3/TdT = |[(12pi^4 R)/(5Theta_D^3) T^3/3]|_("0 K")^(T_"low}}}$

$= \frac{{\overline{C}}_{P}^{\text{low}}}{3}$

$= \textcolor{b l u e}{\text{2.05 J/mol"*"K}}$

where ${\overline{C}}_{P}^{\text{low" = "6.15 J/mol"*"K}}$.

Next, we have the temperature change of solid nitrogen from ${T}_{\text{low}}$ to ${T}_{\text{normal}}$, where ${T}_{\text{low" = "10 K}}$ and ${T}_{\text{normal" = "35.61 K}}$. For this, the entropy is:

color(blue)(DeltabarS_(T_"low")^(T_"normal")) = int_(T_"low")^(T_"normal") barC_P/TdT = color(blue)("25.79 J/mol"*"K")

Here, solid crystalline nitrogen undergoes a phase change/"transition" within its crystalline structure, and $\textcolor{b l u e}{\Delta {\overline{S}}_{\text{trs" = "6.43 J/mol"*"K}}}$.

At this point, it has reached normal temperatures, so from here, we don't need to make any special accommodations for the entropy.

GOING THROUGH THE MELTING PHASE CHANGE

Now, we have a temperature change of solid nitrogen towards its melting point, and then the melting itself. Now, nitrogen is a liquid.

color(blue)(DeltabarS_(T_"normal")^(T_"mp")) = int_(T_"normal")^(T_"mp") barC_P/TdT = color(blue)("23.41 J/mol"*"K")

${T}_{\text{normal" = "35.61 K}}$, ${T}_{\text{mp" = "63.15 K}}$, and $\textcolor{b l u e}{\Delta {\overline{S}}_{\text{melt" = "11.2 J/mol"*"K}}}$ .

GOING THROUGH THE VAPORIZATION PHASE CHANGE

Now, we have a temperature change of liquid nitrogen from its melting point all the way towards its boiling point, and then the vaporization itself. Now, nitrogen is a gas! Almost there.

color(blue)(DeltabarS_(T_"mp")^(T_"bp")) = int_(T_"mp")^(T_"bp") barC_P/TdT = color(blue)("11.46 J/mol"*"K")

${T}_{\text{mp" = "63.15 K}}$, ${T}_{\text{bp" = "77.36 K}}$, and $\textcolor{b l u e}{\Delta {\overline{S}}_{\text{vap" = "72.0 J/mol"*"K}}}$.

HOME STRETCH: GAS AT BOILING POINT TOWARDS STANDARD TEMPERATURE

Now, we have a temperature change of gaseous nitrogen from its boiling point all the way to the standard temperature that is tabulated, $\text{298.15 K}$!

color(blue)(DeltabarS_(T_"bp")^("298.15 K")) = int_(T_"bp")^("298.15 K") barC_P/TdT = color(blue)("39.25 J/mol"*"K")

${T}_{\text{bp" = "77.36 K}}$. Finally, the correction for nonideality is just a technical convention, and it is $\textcolor{b l u e}{\text{0.02 J/mol"*"K}}$. It's tiny, but worth mentioning.

TAKE-HOME MESSAGE

Overall, you can tell that you do not have to do this on any test, ever. But it can be interesting to know where it came from.

The final result is:

$\textcolor{b l u e}{\Delta {S}_{{N}_{2}}^{\circ}} = 2.05 + 25.79 + 6.43 + 23.41 + 11.2 + 11.46 + 72.0 + 39.25 + 0.02 = \setminus m a t h b f \left(\textcolor{b l u e}{\text{191.6 J/mol"*"K}}\right)$