# Enthalpy in chemical reactions

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.50.

We can find how much heat is emitted or absorbed by a chemical reaction by working with the enthalpy of the reactants and products. For example, consider the combustion of methane with oxygen at a constant temperature of ${298\mbox{ K}}$:

$\displaystyle \mbox{CH}_{4}+2\mbox{O}_{2}\rightarrow\mbox{CO}_{2}+2\mbox{H}_{2}\mbox{O} \ \ \ \ \ (1)$

Using the table at the back of Schroeder’s book, ${\Delta H}$ for the formation of methane from elemental carbon (solid) and hydrogen (gas) is

$\displaystyle \Delta H_{CH_{4}}=-74.81\mbox{ kJ mol}^{-1} \ \ \ \ \ (2)$

${\Delta H}$ for the dissociation of methane into its elements is therefore the negative of this.

Similarly

 $\displaystyle \Delta H_{CO_{2}}$ $\displaystyle =$ $\displaystyle -393.51\mbox{ kJ mol}^{-1}\ \ \ \ \ (3)$ $\displaystyle \Delta H_{H_{2}O}\left(\mbox{gas}\right)$ $\displaystyle =$ $\displaystyle -241.82\mbox{ kJ mol}^{-1} \ \ \ \ \ (4)$

The total enthalpy change for the creation of the products in reaction 1 from their elements is

$\displaystyle \Delta H_{prod}=-393.51-2\times241.82=-877.15\mbox{ kJ mol}^{-1} \ \ \ \ \ (5)$

${\Delta H}$ for the reaction as a whole is

$\displaystyle \Delta H=\Delta H_{prod}-\Delta H_{react}=-802.34\mbox{ kJ mol}^{-1} \ \ \ \ \ (6)$

We can see this as follows. Suppose the absolute enthalpy of the elemental components (carbon, hydrogen and oxygen) is ${H_{elem}}$ and of the reactants and products ${H_{reac}}$ and ${H_{prod}}$. Then

 $\displaystyle \Delta H_{reac}$ $\displaystyle =$ $\displaystyle H_{reac}-H_{elem}\ \ \ \ \ (7)$ $\displaystyle \Delta H_{prod}$ $\displaystyle =$ $\displaystyle H_{prod}-H_{elem}\ \ \ \ \ (8)$ $\displaystyle \Delta H$ $\displaystyle =$ $\displaystyle H_{prod}-H_{reac}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Delta H_{prod}-\Delta H_{reac} \ \ \ \ \ (10)$

In 1, if all 4 compounds are gases and the temperature is the same on both sides, there is no volume change, as there are 3 moles of gas both before and after the reaction. Therefore the entire enthalpy change is due to change in internal energy ${U}$ and, assuming no ‘other’ work is done, all this energy is emitted as heat. That is, for one mole of methane

$\displaystyle \Delta U=\Delta H=Q=-802.34\mbox{ kJ } \ \ \ \ \ (11)$

If the water is produced as liquid instead of vapour, then (from the table in Schroeder), ${\Delta H_{H_{2}O}=-285.83\mbox{ kJ mol}^{-1}}$ and

$\displaystyle \Delta H=-890.36\mbox{ kJ mol}^{-1} \ \ \ \ \ (12)$

This time, the final volume is ${\frac{1}{3}}$ of the initial volume, since the 2 moles of water has condensed out as liquid with negligible volume compared to the gases. Thus the atmosphere does work

 $\displaystyle -P\Delta V$ $\displaystyle =$ $\displaystyle -RT\Delta n\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(8.31\mbox{ J K}^{-1}\right)\left(298\mbox{ K}\right)\left(-2\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4.953\mbox{ kJ} \ \ \ \ \ (15)$

The change in internal energy is therefore found from

 $\displaystyle \Delta H$ $\displaystyle =$ $\displaystyle \Delta U+P\Delta V\ \ \ \ \ (16)$ $\displaystyle \Delta U$ $\displaystyle =$ $\displaystyle -890.36+4.953\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -885.41\mbox{ kJ} \ \ \ \ \ (18)$

The difference between this value and 6 should be the latent heat of vaporization at 298 K. An approximation for this is

$\displaystyle L\approx2500.8-2.36T_{C} \ \ \ \ \ (19)$

where ${T_{C}}$ is the temperature in centigrade, so ${T_{C}=25}$ here. This gives a value of ${L=2.442\mbox{ kJ g}^{-1}}$ so for 2 moles (around 36 g) of water, we have ${L=88\mbox{ kJ}}$. The difference from the calculations here is only about 83 kJ, but I’m not sure how accurate the various values and formulas are. At least it’s close.

As a final example, suppose the Sun with a mass of around ${2\times10^{33}\mbox{ g}}$ and luminosity of ${3.839\times10^{26}\mbox{ watts}}$ used the combustion of methane and oxygen as its energy source. The molar weights of methane and molecular oxygen around 16 g and 32 g so if the Sun were composed of one part methane to two parts oxygen (by molecular number), then the mass ratio is around methane:oxygen = 1 : 4. The number of moles of methane in the Sun is therefore

$\displaystyle n_{CH_{4}}=\frac{2\times10^{33}}{5\times16}=2.5\times10^{31}\mbox{ moles} \ \ \ \ \ (20)$

Assuming the water is produced as vapour, the Sun could produce a total energy of

$\displaystyle E=2.5\times10^{31}\times802.34=2\times10^{34}\mbox{ kJ} \ \ \ \ \ (21)$

so it would burn out after a time interval of

$\displaystyle t=\frac{2\times10^{34}\mbox{ kJ}}{3.839\times10^{23}\mbox{ kJ s}^{-1}}=5.21\times10^{10}\mbox{ s}=1651\mbox{ years} \ \ \ \ \ (22)$

We can be pretty sure the Sun’s source of power isn’t chemical reactions!