## Hess’ Law and Thermochemical Calculations

Read this short section which supplements the video you just watched on Hess's Law.

Even before the science of thermodynamics developed in the late nineteenth century, it was observed that the heats associated with chemical reactions can be combined in the same way to yield the heat of another reaction. For example, the standard enthalpy changes for the oxidation of graphite and diamond can be combined to obtain $\Delta H^{o}$ for the transformation between these two forms of solid carbon, a reaction that cannot be studied experimentally.

C(graphite) + O2(g)→ CO2(g)     $\Delta H^{o}$ = –393.51 kJ mol–1

C(diamond) + O2(g)→ CO2(g)     $\Delta H^{o}$ = –395.40 kJ mol–1

Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields

C(graphite) → C(diamond)     $\Delta H^{o}$ = 1.89 kJ mol–1

This principle, known as Hess’ law of independent heat summation is a direct consequence of the enthalpy being a state function. Hess’ law is one of the most powerful tools of chemistry, for it allows the change in the enthalpy (and in other thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small base of experimental data.

Germain Henri Hess (1802-1850) was a Swiss-born professor of chemistry at St. Petersburg, Russia. He formulated his famous law, which he discovered empirically, in 1840. We know little about his other work in chemistry.

#### Standard Enthalpies of Combustion

Because most substances cannot be prepared directly from their elements, heats of formation of compounds are seldom determined by direct measurement. Instead, Hess’ law is employed to calculate enthalpies of formation from more accessible data. The most important of these are the standard enthalpies of combustion. Most elements and compounds combine with oxygen, and many of these oxidations are highly exothermic, making the measurement of their heats relatively easy.

For example, by combining the heats of combustion of carbon, hydrogen, and methane, we can obtain the standard enthalpy of formation of methane, which as we noted above, cannot be determined directly.

#### Problem Example 1

Use the following heat of formation/combustion information to estimate the standard heat of formation of methane CH4.

C(graphite) + O2(g) → CO2(g)     $\Delta H^{o}$ = –393 kJ mol–1(P1-1)

H2(g) + ½O2(g) → H2O(g)     $\Delta H^{o}$ = –286 kJ mol–1. (P1-2)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)     $\Delta H^{o}$ = –890 kJ mol–1.(P1-3)

Solution: The standard heat of formation of methane is defined by the reaction

C(graphite) + 2H2(g) → CH4(g)     $\Delta H^{o}$ = ??? (P1-4)

Our task is thus to combine the top three equations in such a way that they add up to (4).

1. Begin by noting that (3), the combustion of methane, is the only equation that contains the CH4 term, so we need to write it in reverse (not forgetting to reverse the sign of $\Delta H^{o}$!) so that CH4 appears as the product.

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)     $\Delta H^{o}$ = +890 kJ mol–1 (P1-3 Rev)

2. Since H2O does not appear in the net reaction (4), add two times (2) to cancel these out. Notice that this also cancels one of the oxygens in (3 Rev):

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)     $\Delta H^{o}$ = +890 kJ mol–1(P1-3 Rev)

2 H2(g) + O2(g)→ 2H2O(g)     $\Delta H^{o}$ = –484 kJ mol–1 (P1-2)

3. Finally, get rid of the remaining O2 and CO2 by adding (1); this also adds a needed C:

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)     $\Delta H^{o}$ = +890 kJ mol–1 (P1-3 Rev)

2 H2(g) + O2(g)→ 2H2O(g)     $\Delta H^{o}$ = –572 kJ mol–1 (P1-2)

C(graphite) + O2(g)→ CO2(g)     $\Delta H^{o}$ = –393 kJ mol–1 (P1-1)

4. So our creative cancelling has eliminated all except the substances that appear in (4). Just add up the enthalpy changes and we are done:

C(graphite) + 2H2(g) → CH4(g)     $\Delta H^{o}$ = ??? (P1-4)

(The tabulated value is –74.6 kJ mol–1)

Source: Stephen Lower, http://www.chem1.com/acad/webtext/energetics/CE-4.html