For enthalpy calculations why is it that in some cases the "heat of formation method" is used? I know that $ΔH_f$ represents the heat required to form the compound at standard temperature and pressure ($25^oC$, $1$ atm) but why is it used?

For example, in $Ex.1$

NO(g, $300^oC$): $H$ = $\int (Cp)_{NO}dT$
where $T_1 = 25^oC$ and $T_2 = 300^oC$

The enthalpy of nitrogen monoxide is simply the integral from the initial temperature to the final temperature.

However, in $Ex.2$

$CH_4$($150^oC$): $H$ = ($ΔH_f)_{CH_4}$ +$\int (Cp)_{CH4}dT$
where $T_1 = 25^oC$ and $T_2 = 150^oC$

The enthalpy of methane is the heat of formation + the sensible heat from the initial temperature to the final temperature.

What tells me when I should add the heat of formation?

  • 1
    $\begingroup$ Unless ΔHf is zero, you should use the second equation. ΔHf would be zero for elements, but not for NO ΔHf. $\endgroup$
    – Karsten
    Jan 7, 2019 at 14:20
  • $\begingroup$ @KarstenTheis If I should always use the second equation unless the heat of formation is zero, then why didn't the book include it for NO? $\endgroup$
    – Johan
    Jan 8, 2019 at 11:21

1 Answer 1


You're confusing different enthalpies (don't worry, it seems to be rather common). Let's stick with NO as an example. The standard enthalpy of formation, $\Delta_fH^\circ$, is defined as the enthalpy of the reaction $\ce{0.5N2 + 0.5O2 = NO}$ at any given temperature. That is, if you need $\Delta_fH^\circ(298\textrm{ K})$, or $\Delta H^\circ_1$, your reaction is $$\ce{0.5N2(gas, 298) + 0.5O2(gas, 298) = NO(gas, 298)}, (1)$$ and for $\Delta_fH^\circ (T)$, or $\Delta H^\circ_2$, provided that the gases in question do not liquify at $T$, the formation reaction is $$\ce{0.5N2(gas, T) + 0.5O2(gas, T) = NO(gas, T)}. (2)$$

Remember that $\Delta_fH^\circ$ for the elements in their stable states is zero at any temperature.

Now let's say that we have to calculate $\Delta_fH^\circ(T)$ using the known value of $\Delta_fH^\circ(298\textrm{ K})$. For that we need to know the heats of three other processes: $$\ce{N2(gas, 298) = N2(gas, T)}, (3)$$ $$\ce{O2(gas, 298) = O2(gas, T)}, (4)$$ $$\ce{NO(gas, 298) = NO(gas, T)}. (5)$$

The enthalpies of $(3)$, $(4)$ and $(5)$ are basically the enthalpies of heating from $298$ to $T$. They are often called "enthalpy increments" and are defined as $\Delta_{298}^TH^\circ=\int_{298}^T C_p\,dT$ . Note that the enthalpy increments can be written between any two temperatures, not just from 298 K.

It is clear that when we add $(1)$ and $(5)$, and subtract half the $(3)$ and $(4)$, we'll get $(2)$. The same with enthalpies: $\Delta_fH^\circ(T) = \Delta_fH^\circ(298\textrm{ K}) + \int_{298}^T \Delta C_p\,dT$, where $\Delta C_p$ is the difference between $C_p$ for products and reactants of reaction $(2)$ with stoichiometric corresponding coefficients (1.0 and 0.5 in our case) taken into account.

Returning to your examples: in the first one, you are asked to find the enthalpy increment for $\ce{NO}$, which is just the amount of heat required to, well, heat it up. In the second example, you are asked to find the enthalpy of formation of $\ce{CH4}$, and I hope that my example above shows you the way of doing it. Also, if you copied the formulae from the textbook right, there's $\Delta$ missing from under the integral (should be $\int \Delta C_p(\ce{CH4})\,dT$).

A rule of thumb: when in doubt about the enthalpies of any process, write down the process thoroughly, with temperatures and phases of all reactants and products. This way it becomes much clearer. It is also useful to remember that in the real world there is no such thing as "The Enthalpy": what one has to deal with is always some change, some $\Delta$.


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