The Use of Heats of Formation

Heats of formation reactions for each of the compounds involved in a chemical change may be rearranged, using the law of Hess to calculate the overall enthalpy change for the process. Using the heats of formation for methane, carbon dioxide, and liquid water from the Warm Up, and the previous example, we can determine the overall enthalpy change for the combustion of methane as follows:

CH4(g) ➝ C(s) + 2 H2(g) ∆Ho = +75 kJ/molrxn

C(s) + O2(g) ➝ CO2(g) ∆Ho f = –394 kJ/molrxn

2 H2(g) + O2(g) ➝ 2 H2O(l) ∆Ho f = –572 kJ/molrxn

CH4(g) + 2 O2(g) ➝ CO2(g) + 2 H2O(l) ∆Ho comb = –891 kJ/molrxn

Close examination of this process reveals a very useful algorithm that may be used to calculate the overall enthalpy change for any reaction:

∆H o (overall) = ∑nH o f(products) – ∑ nH o f (reactants)

The symbol Σ means “sum of” and n is the coefficient of each species in a balanced equation. It is critical to keep in mind that:

The H o f for a pure element in its standard state is always equal to zero

Observe the state of each species in a balanced equation carefully when using tabular values to complete these calculations, as the state of a substance affects its enthalpy of formation value. The formula above can be used to calculate the overall enthalpy change for the combustion of methane as follows:

∆H o comb = [1 x Hf CO2(g) + 2 x HfH2O(l)] – [1 x HfCH4(g) + 2 x HfO2(g)] thus: ∆Ho comb = [1 mol CO2 (–394 kJ/ ) + 2 mol H2O (–286 kJ/ )] –mol rxn mol CO2 mol rxn mol H2O [1 mol CH4 ( –75 kJ ) + 2 mol O2 ( 0 kJ )] = –891 kJ/molrxn mol rxn mol CH4 mol rxn mol O2

Notice the answer is the same as that we would have produced using the law of Hess. Subtracting the heat of formation of the reactant, methane, has the same effect as reversing the equation. The heats of formation of the products are summed in both methods. Multiplication of heats of formation by the coefficient occurs in both methods as well.

Sample Problem — Heats of Formation

Given the following heats of formation, ΔHo f of SO2(g) = –296.8 kJ/molf and ΔH o f of SO3(g) = –396.0 kJ/molf , what is the enthalpy change for the following reaction: 2 SO2(g) + O2(g) ➝ 2 SO3(g) ?

What To Think about

1. Determine the ΔH o f (heats of formation) values for each of the species involved in the reaction (with the exception of those in elemental form as those = 0).

Note: Generally these values will be found in a table of standard enthalpies of formation

2. Multiply products and reactants by their coefficients and subtract the sum of the reactants from the sum of the products.

How To Do It

ΔH o f of SO2(g) = –296.8 kJ/mol and

ΔH o f of SO3(g) = –396.0 kJ/mol and

ΔH o f of O2 = 0 kJ/mol (pure element) 2 mol SO mol SO2 (–296.8 kJ)+ 1 mol O2 ( O kJ )] mol rxn mol SO2 mol rxn mol O2 = –198.4 kJ/mol

Practice Problems— Heats of Formation

Use the values in Table 4.5.1 to solve the practice problems below (values are listed alphabetically by name in this table):

1. Determine the enthalpy change for each of the following reactions:

(a)C3H8(g) + H2(g) ➝ C2H6(g) + CH4(g)

(b)2 H2S(g) + 3 O2(g) ➝ 2 SO2(g) + 2 H2O(l)

2. Use the following information along with the tabular data to calculate the molar enthalpy of formation of ZnS(s).

2 ZnS(s) + 3 O2(g) ➝ 2 ZnO(s) + 2 SO2(g) ΔH o f = – 878.2 kJ/molrxn

The Use of Bond Energy

A third way to calculate enthalpy changes is to use bond energy values. Bond energy is the energy required to break a bond. Bond energy values are frequently presented in tables. Most Chemistry texts list bond energies as positive values in units of kJ/mole. Such values represent the amount of energy that must be input into the system to break the bond. To represent the value of the energy released when product bonds form, a negative sign must precede the value given in Table 4.5.2.

Bond energy is the enthalpy change required to break a given bond in one mole of gaseous molecules. Bond energy is always positive and is expressed in units of kJ/mole.

The bond energy of a particular bond between two atoms, for example C-C, will vary depending on the “chemical environment” the bond is in, what other bonds are adjacent to it, and what lone pairs and bonding pairs of electrons are nearby. The bond energy for H-Cl, determined from hydrogen chloride gas, HCl, will differ from that determined within the PVC precursor, C2H5Cl, even when it is gasified. This means the bond energies typically reported in tables are really average bond energies gathered for each bond in a large variety of different compounds.

When you examine a number of different average bond energies, such as those represented in this table, several patterns are noticeable. Consider the bond between hydrogen and members of the halogen family, for example. Because the highest energy bond is more difficult to break, HF behaves as a weak acid, while HCl through HI release their hydrogen ions more easily and so are strong acids.

A second pattern that is obvious is that bond energies depend not only on the elements involved in the bond, but also on whether the bond is single, double, or triple. Triple bonds are the most energetic and hence the strongest, while single bonds are the weakest and least energetic.

Bond energy and strength also relate to bond length. Triple bonds involve more electron energy-level overlap and hence are shorter.

Because bond energies are averages, the answer to an enthalpy problem determined using bond energies might not be exactly the same as one calculated using heats of formation, but it will closely approximate it. Since bond energy is a state function, there is a simple algorithm available for calculating enthalpy changes using bond energies. It is similar to the one described for the calculation of enthalpy changes.

ΔHreaction = ∑ nB.E. (bonds broken) – ∑ nB.E. (bonds formed)

Because the reactants and products appear in the opposite order in this equation compared to calculation of enthalpy from heats of formation students frequently make errors in the calculation of enthalpy changes using bond energies. The most common incorrect result has the correct magnitude, but the wrong sign. The easiest way to ensure you fully understand how to determine the overall bond energy change is to always remember that:

Bond breaking requires energy while bond forming releases energy!

Once the structural formulas of all species in a balanced equation are written down, the bonds that are broken and the bonds that are formed can be determined. The broken bonds are assigned positive bond energy values and each newly formed bond is assigned a negative one. Then the sum of the bond energies is calculated. For example, in the sample problem from the law of Hess,

Four C-H bonds and two O=O bonds are broken while two C=O bonds and four O-H bonds are formed. Using tabular bond energy values, we would calculate:

While these two values are not exactly the same, they are both exothermic and are within 2% of each other. For reactions occurring in the gas state, bond energies give very close approximations to the enthalpy changes calculated with standard ΔHo f values.

Sample Problem — Enthalpy Changes from Average Bond Energy Values

Use Table 4.5.2 to determine the enthalpy change for the Haber process for the production of ammonia:

What to Think About

1. Consider the bonding arrangement of the reactants and products. Structural formulas are very useful.

How To Do It

Sample Problem — Continued

What to Think About

2. Multiply the bond energies by the appropriate coefficients and apply the correct signage (+) for bonds breaking and (–) for bonds forming. Count the number of bonds carefully.

Note: Should a bond remain intact, there is no need to break and reform it. Simply ignore it in the calculation.

3. Do the math!

4. I t’s always a good idea to consider the feasibility of your answer.

How To Do It

1 mol N=N ( 945 kJ ) + 3 mol H-H (432 kJ ) mol rxn molN=N mol rxn molH-H –6 mol N-H ( 391 kJ ) mol rxn molN-H

= –105 kJ/molrxn

As we are forming ammonia, an exothermic answer makes sense. The enthalpies of formation table give a ΔHf as –45.9 kJ/mol of NH3. The reaction forms two moles of ammonia, so our answer is close enough to be reasonable.

Practice Problems — Enthalpy Changes from Average Bond Energy Values

Use Table 4.5.2 to produce a full thermochemical equation for each of the molecular representations below.

It is comforting to know that there are multiple ways to use tabular data to solve a problem in thermochemistry. But for most chemists, the real joy lies in working in the laboratory to do things such as creating the tabular data to begin with. This is the realm of calorimetry, and it will be the topic of our next section.

4.5 Review Questions

1. Complete the following table about symbols and sign conventions in the equation, ΔE = q + w.

System gains thermal energy Work

ΔE +

2. Indicate whether each of the following is a state function or is pathway dependent:

(a) A plane’s total mileage meter

(b) A plane’s altimeter (indicates altitude)

(c) Work

(d) ΔH

(e) Your monthly cell phone bill

3. Calculate the change in internal energy for a gas that absorbs 28.5 J of heat and then performs 16.4 J of work.

5. Though we are unaware of it at rest, breathing requires work. A 70 kg man likely has an “empty” lung volume of 2210 mL.

(a) If 50.7 J of work is required to take a single breath against a constant external atmospheric pressure of 1.00 atm, what is the man’s “full” lung volume?

4. A gas expands from a volume of 1.4 L to 6.7 L against an external pressure of 1.00 atm. During this process, 565 J of heat is transferred from the surroundings to the gas

(a) How much work has been done?

(b) During exercise, the elastic lungs expand to a “full” volume of 5250 mL. How much work is required to take a single breath during exercise?

(b) What is the change in internal energy of the system?

6. The following is a molecular representation of a state change occurring inside a piston-cylinder assembly against a constant external pressure of 1.05 atm.

8. Determine the enthalpy change for the following reaction: CH4(g) + 4 Cl2(g) ➝ CCl4(l) + 4 HCl(g) Using the following information:

C(s) + 2 H2(g) ➝ CH4(g) ΔHo f = –74.6 kJ/molrxn

C(s) + 2 Cl2(g) ➝ CCl4(g) ΔHo f = –95.7 kJ/molrxn

Indicate whether each of the following is (+), (–), 0 or impossible to determine. Justify your answer in each case.

H2(g) + Cl2(g) ➝ 2 HCl(g) ΔHo f = –92.3 kJ/molrxn

7. In the 1970s, scientists became concerned about the depletion of the ozone layer due to the affects of aerosol propellants such as chlorofluorocarbons. Chlorofluorocarbons are now known to absorb highenergy radiation, producing free radical chlorine atoms. These chlorine atoms have a catalytic effect on the removal of ozone from our atmosphere.

Use the following two reactions and their enthalpy changes to calculate the enthalpy change for the removal of ozone by free radical oxygen. (This is the net equation involved in ozone destruction in the atmosphere.) O atoms (free radicals) are present due to O2 dissociation by highenergy radiation. All species are gaseous.

O3 + Cl ➝ O2 + ClO ΔH o = –126 kJ/molrxn

ClO + O ➝ Cl + O2 ΔH o = –268 kJ/molrxn ===================================

9. The Born-Haber cycle is a hypothetical series of steps that represent the formation of an ionic compound from its constituent elements. According to the law of Hess, the enthalpy changes for each step in the series should sum to give the overall enthalpy change for the formation of the ionic compound from its elements in their standard states. The enthalpy change for the formation of an ionic compound from its ions in the gas state is called the lattice energy of the compound. Here is the Born-Haber cycle for cesium fluoride.

ΔH3 = 79.4 kJ/mol (–1 2 Dissociation energy)

ΔH2 = 375.7 kJ/mol (Ionization energy)

ΔH1 = 76.5 kJ/mol (Enthalpy of sublimation)

ΔH4 = –328.2 kJ/mol (Electron affinity) ΔH5 = –756.9 kJ/mol (– Lattice energy) ΔHf = –553.5 kJ/mol (Enthalpy of formation)

Use the Born-Haber cycle to show how five equations may be added together using the law of Hess to give the overall enthalpy of formation for cesium fluoride from the following equation: Cs(

). Label each of the processes involved. Be sure to use five steps The first step is done for you as an example.

10. The Ostwald process for making nitric acid involves multiple steps, beginning with ammonia reacting with oxygen, forming nitrogen monoxide and water. Determine the ΔH o rxn for this first step in the Ostwald process. Begin by writing a balanced chemical reaction. Then use the standard enthalpy of formation values in Table 4.5.1.

12. Consider the following unbalanced molecular representation of an oxidation-reduction reaction. Translate this molecular representation to a balanced formula equation. Then use the table of standard enthalpy of formation values from Table 4.5.1 to determine the overall enthalpy change for the balanced molecular equation.

11. Use the standard enthalpy of formation values in Table 4.5.1 to determine the overall enthalpy change for the reaction depicted by the following molecular representation:

13. Solid baking soda (sodium bicarbonate or sodium hydrogen carbonate) neutralizes sulfuric acid solution spills, producing sodium sulfate, water, and carbon dioxide. Write a complete, balanced thermochemical equation for this process. The enthalpy values required may be found in Table 4.5.1.

14. The Law of Hess Practice Problem 2 involves steps in the production of polyvinylchloride (PVC) plastic. The addition of hydrogen chloride gas, HCl, to ethene, C2H4, forms the monomer, C2H5Cl(l) which is used to make the polymer, PVC. Given the overall enthalpy change of –65.0 kJ/molrxn and a value in Table 4.5.1, determine the enthalpy of formation for the monomer, C2H5Cl(l).

16. (a) Use a highlighter or some other method to indicate the bonds that break on the reactant side and form on the product side of this reaction. Remember the note in the sample problem about enthalpy changes from average bond energy values in Table 4.5.2 — if a bond remains intact there is no need to break and reform it. Calculate the enthalpy change using the bond energy values in Table 4.5.2. CH2—CH2

15. Chlorine gas can react in a substitution reaction with the gaseous monomer mentioned in question 14 as follows:

C2H5Cl(g) + Cl2(g) ➝ C2H4Cl2(g) + HCl(g)

(a) Calculate the ∆H˚ for the reaction above, using values from Table 4.5.2.

(b) Repeat for this reaction.

(b) Repeat the calculation done in part (a) using ΔH o f values from Table 4.5.2.

(c) Compare the answer to part (a) with par t (b). Comment on the comparison.

(c) Considering bond energy is a state function, combine steps (a) and (b) into one reaction and calculate the overall enthalpy change.

17. The following shows the structural formulas for the reaction: