18

The value of ΔH for a reaction in one direction is equal in magnitude, but opposite in sign, to ΔH for the reaction in the opposite direction, and ΔH is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, Δ H f ° , is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A number of approaches to the computation of free energy changes are possible.

By the end of this section, you will be able to:

- Define standard enthalpy of formation
- Explain Hess’s law and use it to compute reaction enthalpies

A standard enthalpy of formation $\text{\Delta}{H}_{\text{f}}^{\xb0}$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=\text{\Delta}H\text{\xb0}=\mathrm{-393.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO_{2}, also at 1 atm and 25 °C. For nitrogen dioxide, NO_{2}(*g*), $\text{\Delta}{H}_{\text{f}}^{\xb0}$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:

$$\frac{1}{2}{\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=\text{\Delta}H\text{\xb0}=\text{+33.2 kJ}$$

A reaction equation with $\frac{1}{2}$ mole of N_{2} and 1 mole of O_{2} is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO_{2}(*g*).

You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P_{4}O_{10}) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C_{2}H_{2}). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

$$3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{O}}_{3}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\text{+286 kJ}$$

$$\frac{3}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}(g)$$

For the formation of 2 mol of O_{3}(*g*), $\text{\Delta}H\text{\xb0}=\text{+286 kJ.}$ This ratio, $\left(\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\right),$ can be used as a conversion factor to find the heat produced when 1 mole of O_{3}(*g*) is formed, which is the enthalpy of formation for O_{3}(*g*):

$$\text{\Delta}\text{H}\text{\xb0 for}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mole of}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}(g)=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}}\phantom{\rule{0.2em}{0ex}}=143\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Therefore, $\text{\Delta}{H}_{\text{f}}^{\xb0}[{\text{O}}_{3}(g)]=\text{+143 kJ/mol}.$

For the reaction ${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-184.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(a) C_{2}H_{5}OH(*l*)

(b) Ca_{3}(PO_{4})_{2}(*s*)

(a) $2\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+3{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}(l)$

(b) $3\text{Ca}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{P}}_{4}(s)+4{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_{3}({\text{PO}}_{4}{)}_{2}(s)$

Note: The standard state of carbon is graphite, and phosphorus exists as P_{4}.

(a) C_{2}H_{5}OC_{2}H_{5}(*l*)

(b) Na_{2}CO_{3}(*s*)

(a) $4\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+5{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}(l);$ (b) $2\text{Na}(s)+\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{CO}}_{3}(s)$

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

This type of calculation usually involves the use of Hess’s law, which states: *If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps*. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-394}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

In the two-step process, first carbon monoxide is formed:

$$\text{C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-111}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Then, carbon monoxide reacts further to form carbon dioxide:

$$\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-283}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

The equation describing the overall reaction is the sum of these two chemical changes:

$$\begin{array}{}\\ \text{Step 1: C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\\ \underset{\xaf}{\text{Step 2: CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)}\\ \text{Sum: C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)+\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)+{\text{CO}}_{2}(g)\end{array}$$

Because the CO produced in Step 1 is consumed in Step 2, the net change is:

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)$$

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.

$$\begin{array}{ll}\text{C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\hfill & \text{\Delta}H\text{\xb0}=\mathrm{-111}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)}{\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{1em}{0ex}}}\hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-283}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-394}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

The result is shown in Figure 18.1. We see that Δ*H* of the overall reaction is the same whether it occurs in one step or two. This finding (overall Δ*H* for the reaction = sum of Δ*H* values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

Before we further practice using Hess’s law, let us recall two important features of Δ*H*.

Δ

*H*is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO_{2}(*g*) is +33.2 kJ:$$\frac{1}{2}{\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+33.2 kJ}$$When 2 moles of NO

_{2}(twice as much) are formed, the Δ*H*will be twice as large:$${\text{N}}_{2}(g)+2{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+66.4 kJ}$$In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.

Δ

*H*for a reaction in one direction is equal in magnitude and opposite in sign to Δ*H*for the reaction in the reverse direction. For example, given that:$${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-184.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$Then, for the “reverse” reaction, the enthalpy change is also “reversed”:

$$2\text{HCl}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+184.6 kJ}$$

$$\text{Fe}(s)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\mathrm{-341.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$${\text{FeCl}}_{2}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\mathrm{-57.7}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{Fe}(s)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=?$$

Looking at the reactions, we see that the reaction for which we want to find Δ*H*° is the sum of the two reactions with known Δ*H* values, so we must sum their Δ*H*s:

$$\begin{array}{lll}\text{Fe}(s)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}(s)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\mathrm{-341.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{{\text{FeCl}}_{2}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)}{\text{Fe}(s)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{1em}{0ex}}}\hfill & \hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-57.7}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-399.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

The enthalpy of formation, $\text{\Delta}{H}_{\text{f}}^{\xb0},$ of FeCl_{3}(*s*) is −399.5 kJ/mol.

$${\text{N}}_{2}(g)+2{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}(g)$$

from the following information:

$${\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=180.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{NO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-57.06}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

66.4 kJ

Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of Δ*H*) if they are difficult to determine experimentally.

*(i)* $\text{ClF}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=?$

Use the reactions here to determine the Δ*H*° for reaction *(i)*:

*(ii)* $2{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}(g)+2{\text{F}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(ii)}^{\xb0}=\mathrm{-49.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iii)* $2\text{ClF}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\text{O}(g)+{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iii)}^{\xb0}=\text{+214.0 kJ}$

*(iv)* ${\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iv)}^{\xb0}=\text{+236.2 kJ}$

$$\text{ClF}(g)+\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}(214.0)=\text{+107.0 kJ}$$

Next, we see that F_{2} is also needed as a reactant. To get this, reverse and halve reaction *(ii)*, which means that the Δ*H*° changes sign and is halved:

$$\frac{1}{2}{\text{O}}_{2}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\text{+24.7 kJ}$$

To get ClF_{3} as a product, reverse *(iv)*, changing the sign of Δ*H*°:

$$\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\text{\u2212236.2 kJ}$$

Now check to make sure that these reactions add up to the reaction we want:

$$\begin{array}{lll}\text{ClF}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}(g)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\text{+107.0 kJ}\hfill \\ \frac{1}{2}{\text{O}}_{2}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}(g)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\text{+24.7 kJ}\hfill \\ \frac{\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)}{\text{ClF}(g)+{\text{F}}_{2}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)\phantom{\rule{8em}{0ex}}}\hfill & \hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-236.2}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-104.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

Reactants $\frac{1}{2}{\text{O}}_{2}$ and $\frac{1}{2}{\text{O}}_{2}$ cancel out product O_{2}; product $\frac{1}{2}{\text{Cl}}_{2}\text{O}$ cancels reactant $\frac{1}{2}{\text{Cl}}_{2}\text{O;}$ and reactant $\frac{3}{2}{\text{OF}}_{2}$ is cancelled by products $\frac{1}{2}{\text{OF}}_{2}$ and OF_{2}. This leaves only reactants ClF(*g*) and F_{2}(*g*) and product ClF_{3}(*g*), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified Δ*H*° values will give the desired Δ*H*°:

$$\text{\Delta}H\text{\xb0}=(\mathrm{+107.0}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(24.7\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\mathrm{-236.2}\phantom{\rule{0.2em}{0ex}}\text{kJ})=\mathrm{-104.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

*(i)* $2\text{Al}(s)+3{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=?$

Use the reactions here to determine the Δ*H*° for reaction *(i)*:

*(ii)* $\text{HCl}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{HCl}(aq)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(ii)}^{\xb0}=\mathrm{-74.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iii)* ${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iii)}^{\xb0}=\mathrm{-185}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iv)* ${\text{AlCl}}_{3}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{AlCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iv)}^{\xb0}=\mathrm{+323}\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$

*(v)* $\text{2Al}(s)+6\text{HCl}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}(aq)+3{\text{H}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(v)}^{\xb0}=\mathrm{-1049}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

−1407 kJ

We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and *n* standing for the stoichiometric coefficients:

$$\text{\Delta}{H}_{\text{reaction}}^{\xb0}={\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{products})-{\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{reactants})$$

The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.

$$3{\text{NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=?$$

$$\text{\Delta}{H}_{\text{reaction}}^{\xb0}={\displaystyle \sum n\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}\text{(products)}}-{\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{reactants})$$

$$\begin{array}{l}\\ \\ \\ =\left[2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}(aq)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-207.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}(aq)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol NO}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+90.2 kJ}}{\overline{)\text{mol NO}(g)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+33.2 kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-285.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}}\right]\\ =[2\times \left(\mathrm{-206.64}\right)+90.25]-[3\times 33.2+-\left(\mathrm{-285.83}\right)]\\ =\mathrm{\u2013323.03}+186.23\\ =\mathrm{-136.80}\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$$

$$3{\text{NO}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}(g)+{\text{3O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{1}^{\xb0}=\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$${\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{2}^{\xb0}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\mathrm{[-1}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{H}}_{2}\text{O})]$$

$${\text{H}}_{2}(g)+{\text{N}}_{2}(g)+3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{3}^{\xb0}=\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}[2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{HNO}}_{3})]$$

$$\frac{1}{2}{\text{N}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{4}^{\xb0}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}[1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\text{NO})]$$

Summing these reaction equations gives the reaction we are interested in:

$${\text{3NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)$$

Summing their enthalpy changes gives the value we want to determine:

$$\begin{array}{cc}\hfill \text{\Delta}{H}_{\text{rxn}}^{\xb0}& =\text{\Delta}{H}_{1}^{\xb0}+\text{\Delta}{H}_{2}^{\xb0}+\text{\Delta}{H}_{3}^{\xb0}+\text{\Delta}{H}_{4}^{\xb0}=(\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+285.8 kJ})+(\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+90.2 kJ})\hfill \\ & =\mathrm{-138.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}$$

So the standard enthalpy change for this reaction is Δ*H*° = −138.4 kJ.

Note that this result was obtained by (1) multiplying the $\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.

−1368 kJ/mol

By the end of this section, you will be able to:

- Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
- Explain how temperature affects the spontaneity of some proceses

As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

$$\text{\Delta}G=\text{\Delta}H-T\text{\Delta}S$$

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since *T* is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

**Both Δ**This condition describes an endothermic process that involves an increase in system entropy. In this case, Δ*H*and Δ*S*are positive.*G*will be negative if the magnitude of the*T*Δ*S*term is greater than Δ*H*. If the*T*Δ*S*term is less than Δ*H*, the free energy change will be positive. Such a process is*spontaneous at high temperatures and nonspontaneous at low temperatures.***Both Δ**This condition describes an exothermic process that involves a decrease in system entropy. In this case, Δ*H*and Δ*S*are negative.*G*will be negative if the magnitude of the*T*Δ*S*term is less than Δ*H*. If the*T*Δ*S*term’s magnitude is greater than Δ*H*, the free energy change will be positive. Such a process is*spontaneous at low temperatures and nonspontaneous at high temperatures.***Δ**This condition describes an endothermic process that involves a decrease in system entropy. In this case, Δ*H*is positive and Δ*S*is negative.*G*will be positive regardless of the temperature. Such a process is*nonspontaneous at all temperatures.***Δ**This condition describes an exothermic process that involves an increase in system entropy. In this case, Δ*H*is negative and Δ*S*is positive.*G*will be negative regardless of the temperature. Such a process is*spontaneous at all temperatures.*

These four scenarios are summarized in Figure 18.2.

$$\text{2C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{2CO}(g)$$

How does the spontaneity of this process depend upon temperature?

$$\text{4Fe}(s)+{\text{3O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{2Fe}}_{2}{\text{O}}_{3}(s)$$

How does the spontaneity of this process depend upon temperature?

Δ*H* and Δ*S* are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its Δ*G*) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which Δ*G* is plotted on the *y* axis versus *T* on the *x* axis:

$$\text{\Delta}G=\text{\Delta}H-T\text{\Delta}S$$

$$y=b+mx$$

Such a plot is shown in Figure 18.3. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative Δ*G*) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the *x*-intercept of the line, that is, the value of *T* for which Δ*G* is zero:

$$\text{\Delta}G=0=\text{\Delta}H-T\text{\Delta}S$$

$$T=\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}H}{\text{\Delta}S}$$

So, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which Δ*G* for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

$${\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(g)$$

When this process is at equilibrium, Δ*G* = 0, so the following is true:

$$0=\text{\Delta}H\text{\xb0}-T\text{\Delta}S\text{\xb0}\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}T=\phantom{\rule{0.2em}{0ex}}\frac{\text{\Delta}H\text{\xb0}}{\text{\Delta}S\text{\xb0}}$$

Using the standard thermodynamic data from Appendix G,

$$\begin{array}{ccc}\hfill \text{\Delta}H\text{\xb0}& =\hfill & \text{1 mol}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}\left({\text{H}}_{2}\text{O}(g)\right)\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}\left({\text{H}}_{2}\text{O}(l)\right)\hfill \\ & =\hfill & (\text{1 mol})-\text{241.82 kJ/mol}-(\text{1 mol})(\text{\u2212241.82 kJ/mol})=\text{44.01 kJ}\hfill \\ \hfill \text{\Delta}S\xb0& =\hfill & \text{1 mol}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{S}^{\xb0}\left({\text{H}}_{2}\text{O}(g)\right)\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{S}^{\xb0}\left({\text{H}}_{2}\text{O}(l)\right)\hfill \\ & =\hfill & (\text{1 mol})\text{188.8 J/K\xb7mol}-(\text{1 mol})\text{70.0 J/K\xb7mol}=\text{118.8 J/K}\hfill \\ \hfill T& =\hfill & \frac{\text{\Delta}H\text{\xb0}}{\text{\Delta}S\text{\xb0}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{44.01\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}}{118.8\phantom{\rule{0.2em}{0ex}}\text{J/K}}\phantom{\rule{0.2em}{0ex}}=370.5\phantom{\rule{0.2em}{0ex}}\text{K}=97.3\phantom{\rule{0.2em}{0ex}}\text{\xb0C}\hfill \end{array}$$

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

313 K (accepted value 319 K)

Previous Citation(s)

Flowers, P., et al. (2019). Chemistry: Atoms First 2e. https://openstax.org/details/books/chemistry-atoms-first-2e (12.1-12.4, 13.4)

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Access it online or download it at https://edtechbooks.org/general_college_chemistry_2/review_of_thermodyna.