# How can I calculate enthalpy of formation?

##### 1 Answer

**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 *is* actually tabulated alongside the standard enthalpy of formation in thermodynamic tables.

Just as enthalpy can be defined as:

#DeltaH = int_(T_1)^(T_2) C_PdT# (where

#C_P# is the constant-pressure heat capacity)

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

#color(green)(DeltabarS = int_(T_1)^(T_2) (barC_P)/TdT)# where

#DeltabarS# is themolar entropyand#barC_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

#\mathbf(DeltaS_(N_2)^@)#

#= \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")#

#\mathbf(+ int_(T_"bp")^("298.15 K") barC_P/TdT)#

#\mathbf(+ " Correction for nonideality")#

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

*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:

#DeltabarS = int_(T_1)^(T_2) (barC_P)/TdT = color(green)(barC_Pln\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":

#barC_P^"low"(T) = (12pi^4)/5 R(T/(Theta_D))^3# ;#0 < T <= T_"low"#

where

This integral evaluates as:

#color(blue)(DeltabarS_"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")#

#= (barC_P^"low")/3#

#= color(blue)("2.05 J/mol"*"K")# where

#barC_P^"low" = "6.15 J/mol"*"K"# .

Next, we have the **temperature change** of solid nitrogen from

#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

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")#

**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")#

**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,

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

**correction for nonideality** is just a technical convention, and it is

**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:

#color(blue)(DeltaS_(N_2)^@ )= 2.05 + 25.79 + 6.43 + 23.41 + 11.2 + 11.46 + 72.0 + 39.25 + 0.02 = \mathbf(color(blue)("191.6 J/mol"*"K"))#