Using two equations and their enthalpies

Germain Henri Hess, in 1840, discovered a very useful principle which is named for him:

The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps.

Another way to state Hess' Law is:

If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.

A brief discussion about how Hess' Law is used, followed by some examples.

What is the enthalpy of the following reaction?

C (s, graphite) ---> C (s, diamond) ΔH° = ??? kJ

By the way, notice the presence of the degree sign, °, on the enthalpy. This indicates that the reaction is happening under standard conditions. All the examples in this tutorial will be carried out under standard conditions.

In the common chemistry laboratory, this reaction cannot be examined directly. This is because, regardless of the low enthalpy, the reaction requires a very, very high activation energy to get the reaction started and, in this case, it means both high temperature and high pressure. The consequence is that the enthalpy value cannot be determined directly in almost all labs and, in the ones that can, the process is very, very difficult.

However, Hess' Law offers a way out. If we had two (or more) reactions that could be added together, then we can add the respective enthalpies of the reactions to get what we want. Here are the two reactions we need:

C (s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH° = −393.5 kJ C (s, diamond) + O _{2}(g) ---> CO_{2}(g)ΔH° = −395.4 kJ

What I am going to do is reverse the bottom equation. This will put the C (s, diamond) on the product side, where we need it. Also, when I add the two equations together, the oxygen and carbon dioxide will cancel out. This is, of course, what we want since those two substances are not in the final, desired equation. Here are the two equations again, with the second one reversed:

C (s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH° = −393.5 kJ CO _{2}(g) ---> C (s, diamond) + O_{2}(g)ΔH° = +395.4 kJ

I want you to notice the other change I made. Look at the enthalpy for the second equation, the one I reversed. Notice how the sign has changed also. This is an absolute requirement of using Hess' Law: reversing an equation means reversing the sign on the enthalpy value.

The reason? The first, unreversed equation is exothermic. We know this from the negative in front of the 395.4. That means that the opposite, reverse equation is endothermic. Putting in enthalpy (endothermic) is the reverse, the opposite of exothermic (giving off enthalpy). Hence, we change the sign EVERY time we reverse an equation.

Now I'm ready to add the equations together. When I do this I also add the enthalpies together. Here is the added equation without anything taken out:

CO _{2}(g) + C (s, graphite) + O_{2}(g) ---> CO_{2}(g) + C (s, diamond) + O_{2}(g)ΔH° = (−393.5 kJ) + (+395.4 kJ)

Notice the items which are the same on both sides and remove them:

C (s, graphite) ---> C (s, diamond) ΔH° = +1.9 kJ

We now have the answer we desire by using the indirect means of Hess' Law and two relatively easy experiments. Thus we avoid performing a tricky, expensive, possibly dangerous experiment. However, due to Hess' discovery, we know that our indirectly obtained answer is just as valid as if we had done the experiment directly.

Here is an essay question that uses the graphite to diamond conversion:

The interconversion of graphite and diamond does not occur at room temperature. Explain how you could determine the ΔH for this transition in the laboratory.

**Example #1:** Use the following data to determine the enthalpy (ΔH°) of reaction for:

NO _{2}(g) +^{7}⁄_{2}H_{2}(g) ---> 2H_{2}O(ℓ) + NH_{3}(g)ΔH° = ??? kJ

Using the following two equations:

2NH _{3}(g) ---> N_{2}(g) + 3H_{2}(g)ΔH° = +92 kJ ^{1}⁄_{2}N_{2}(g) + 2H_{2}O(ℓ) ---> NO_{2}(g) + 2H_{2}(g)ΔH° = +170 kJ

**Solution:**

1) Notice how there is only one NH_{3} in the target equation and it's on the right-hand side. That means we have to flip our first equation and divide it by two. Like this:

^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}H_{2}(g) ---> NH_{3}(g)ΔH° = −46 kJ Notice that the sign changed on the ΔH and its numerical value was cut in half.

2) The target equation has one NO_{2} and it's on the left-hand side, so we need to flip the second equation. Like this:

NO _{2}(g) + 2H_{2}(g) --->^{1}⁄_{2}N_{2}(g) + 2H_{2}O(ℓ)ΔH° = −170 kJ Notice that the sign changed on the ΔH.

3) Add the two data equations together to get the target equation. Add the two enthalpies to obtain the ΔH for the target equation:

−46 + −170 = −216 kJ

**Example #2:** Calculate the enthapy for the following reaction:

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH° = ??? kJ

Using the following two equations:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH° = +180 kJ 2NO _{2}(g) ---> 2NO(g) + O_{2}(g)ΔH° = +112 kJ

**Solution:**

1) In order to solve this, we must reverse at least one equation and it turns out that the second one will require reversal. Here are both data equations with the reversal to the second:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH° = +180 kJ 2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH° = −112 kJ Notice that I have also changed the sign on the enthalpy from positive to negative.

2) Next, we add the two equations together and eliminate identical items. We also add the two enthalpies together to obtain the answer.

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH° = +68 kJ

3) Problems 10a−d and the bonus problem in this file are more examples using combinations of N_{2}/O_{2}/NO/NO_{2}.

**Example #3:** Given the following data:

2NO(g) ---> N _{2}(g) + O_{2}(g)ΔH = −180.6 kJ N _{2}(g) + O_{2}(g) + Cl_{2}(g) ---> 2NOCl(g)ΔH = +103.4 kJ

Find the ΔH of the following reaction:

2NOCl(g) ---> 2NO(g) + Cl_{2}(g)

**Solution:**

1) Flip first reaction, flip second reaction:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.6 kJ 2NOCl(g) ---> N _{2}(g) + O_{2}(g) + Cl_{2}(g)ΔH = −103.4 kJ

2) Add the equations and the ΔH values:

+180.6 + (−103.4) = +77.2

2NOCl(g) ---> 2NO(g) + Cl _{2}(g)ΔH = +77.2 kJ

Comment: you could also just add up the two reactions without flipping them, then flip the answer (remembering to change the sign on the ΔH when you do so).

**Example #4:** The compound carbon suboxide, C_{3}O_{2}, is a gas at room temperature. Use the data supplied to calculate the heat of formation of carbon suboxide.

2CO(g) + C(s) ---> C _{3}O_{2}(g)ΔH° = +127.3 kJ CO(g) $\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −110.5 kJ

**Solution:**

1) The target equation is this:

3C(s) + O_{2}(g) ---> C_{3}O_{2}(g)

2) Write the formation reaction for CO:

C(s) + ^{1}⁄_{2}O_{2}(g) ---> CO(g)$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −110.5 kJ then multiply it by two (in order to cancel 2CO when we add the equations):

2C(s) + O _{2}(g) ---> 2CO(g)ΔH = −221.0 kJ

3) Write the two equations to be added:

2CO(g) + C(s) ---> C _{3}O_{2}(g)ΔH° = +127.3 kJ 2C(s) + O _{2}(g) ---> 2CO(g)ΔH = −221.0 kJ

4) The two equations above need only be added to obtain the desired answer:

+127.3 kJ + (−221.0 kJ) = −93.7 kJ

3C(s) + O _{2}(g) ---> C_{3}O_{2}(g)$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −93.7 kJ

Comment: notice that no chemical equation was given in the problem. That is because of the subscripted 'f' on the enthalpy. That indicates the enthalpy is a formation enthalpy and, as such, already has a chemical equation in the definition of the enthalpy of formation.

**Example #5:** During discharge of a lead-acid storage battery, the following chemical reaction takes place:

Pb + PbO_{2}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2H_{2}O

Using the following two reactions:

(1) Pb + PbO _{2}+ 2SO_{3}---> 2PbSO_{4}ΔH° = −775 kJ (2) SO _{3}+ H_{2}O ---> H_{2}SO_{4}ΔH° = −113 kJ

Determine the enthalpy of reaction for the discharge reaction above.

**Solution:**

1) Multiply chemical equation (2) by 2:

2SO_{3}+ 2H_{2}O ---> 2H_{2}SO_{4}ΔH = −226 kJ

2) Switch the reactants and products in chemical reaction (2). Because of that, the sign of the change in enthalpy becomes positive. Let's number the following chemical equation as (3):

(3) 2H_{2}SO_{4}---> 2SO_{3}+ 2H_{2}O ΔH° = 226 kJ

3) Add chemical equations (1) and (3):

Pb + PbO_{2}+ 2SO_{3}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2SO_{3}+ 2H_{2}O

4) Then, add the enthalpy changes of equations (1) and (3):

Pb + PbO_{2}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2H_{2}O ΔH° = −549 kJ

Remember to cancel out 2SO_{3} because it appears on both the reactant and product side, leaving you with the desired chemical reaction.

**Example #6:** Calculate ΔH for the reaction:

CO(g) + 2H_{2}(g) ---> CH_{3}OH(g)

Given the following information:

2C(s) + O _{2}(g) ---> 2CO(g)ΔH = −221.0 kJ 2C(s) + O _{2}(g) + 4H_{2}(g) ---> 2CH_{3}OH(g)ΔH = −402.4 kJ

**Solution #1:**

1) Subtract the first equation from the second equation:

2C(s) + O_{2}(g) + 4H_{2}(g) − [2C(s) + O_{2}(g)] ---> 2CH_{3}OH(g) − 2CO(g) ΔH = −402.4 kJ − (−221.0 kJ

2) Simplify:

4H_{2}(g) ---> 2CH_{3}OH(g) − 2CO(g) ΔH = −181.4 kJ

3) Rearrange:

2CO(g) + 4H_{2}(g) ---> 2CH_{3}OH(g) ΔH = −181.4 kJ

4) Divide through by 2:

CO(g) + 2H_{2}(g) ---> CH_{3}OH(g) ΔH = −90.7 kJ

**Solution #2:**

This is the usual way the ChemTeam uses to solve Hess' Law problems.

1) Do these modifications to the data equations:

Reverse first data equation and divide by 2 (this gets one CO on the left).

Divide second data equation by 2 (this gives us 2H_{2}and one CH_{3}OH).

2) The results of step 1 are:

CO(g) ---> C(s) + ^{1}⁄_{2}O_{2}(g)ΔH = +110.5 kJ C(s )+ ^{1}⁄_{2}O_{2}(g) + 2H_{2}(g) ---> CH_{3}OH(g)ΔH = −201.2 kJ

3) Add the two equations together. The C and the ^{1}⁄_{2}O_{2} will cancel and the target equation is what is left. Add the two enthalpies for the final answer.

CO(g) + 2H_{2}(g) ---> CH_{3}OH(g) ΔH = −90.7 kJ

**Example #7:** The standard enthalpy changes of combustion of glucose and ethanol are given as −2820 and −1368 kJ mol¯^{1} respectively. Glucose, C_{6}H_{12}O_{6}, can be converted into ethanol.

C_{6}H_{12}O_{6}(s) ---> 2C_{2}H_{5}OH(ℓ) + 2CO_{2}(g)

What is the standard enthalpy change for the above reaction?

**Solution:**

1) Write the combustion equations for glucose and ethanol:

C _{6}H_{12}O_{6}(s) + 6O_{2}(g) ---> 6CO_{2}(g) + 6H_{2}O(ℓ)ΔH = −2820 kJ C _{2}H_{5}OH(ℓ) + 3O_{2}(g) ---> 2CO_{2}(g) + 3H_{2}O(ℓ)ΔH = −1368 kJ Note that I do not include mol¯

^{1}when the enthalpy is associated with a chemical equation.

2) Adjust our two data equations so, when added together, they result in the target equation:

C _{6}H_{12}O_{6}(s) + 6O_{2}(g) ---> 6CO_{2}(g) + 6H_{2}O(ℓ)ΔH = −2820 kJ 4CO _{2}(g) + 6H_{2}O(ℓ) ---> 2C_{2}H_{5}OH(ℓ) + 6O_{2}(g)ΔH = +2736 kJ The second data equation has been reversed and multiplied by 2. Note that the enthalpy has been doubled and the sign changed.

3) Adding the resulting data equations will give us our target equation. Add the enthalpies for the final answer:

C _{6}H_{12}O_{6}(s) ---> 2C_{2}H_{5}OH(ℓ) + 2CO_{2}(g)ΔH = −84 kJ Note that all the O

_{2}and H_{2}O cancel out as well as 4 of the CO_{2}, when the two data equations are added together.

**Example #8:** Given the following thermochemical equations:

4NH _{3}(g) + 7O_{2}---> 4NO_{2}(g) + 6H_{2}O(g)ΔH = −1132 kJ 6NO _{2}(g) + 8NH_{3}(g) ---> 7N_{2}(g) + 12H_{2}O(g)ΔH = −2740 kJ

Determine the ΔH for this reaction:

4NH_{3}(g) + 3O_{2}---> 2N_{2}(g) + 6H_{2}O(g)

**Solution:**

The key (in the ChemTeam's opinion) to solving this problem lies with the NO_{2}. We know that it does not appear in the final answer, so what will get the NO_{2} coefficients equal? The answer is to use a factor of 2 and a factor of 3.

1) Manipulate the data equations:

12NH _{3}(g) + 21O_{2}---> 12NO_{2}(g) + 18H_{2}O(g)ΔH = −3396 kJ <--- times by 3 12NO _{2}(g) + 16NH_{3}(g) ---> 14N_{2}(g) + 24H_{2}O(g)ΔH = −5480 kJ <--- times by 2 The two and the three help get the NO

_{2}to a coefficient of 12, which happens to be the least common multiple between the 4 and the 6 that were the original coefficients of NO_{2}.Also, note that no equation should be reversed. The NH

_{3}and the O_{2}are on the reactant side (and NOT on the product side) in both data equations. The N_{2}and the H_{2}O are both on the product side (and NOT on the reactant side) in both data equations.That means that NO

_{2}will be the only substance that will cancel out.

2) Add:

28NH_{3}(g) + 21O_{2}---> 14N_{2}(g) + 42H_{2}O(g) ΔH = −8876 kJ

3) Divide through by 7:

4NH_{3}(g) + 3O_{2}---> 2N_{2}(g) + 6H_{2}O(g) ΔH = −1268 kJ

4) One day, it occurred to me to use 1.5 instead of 3 and to use 1 instead of 2. After all, 3:2 and 1.5:1 are the same ratio. Here are the two data equations:

6NH _{3}(g) + 10.5O_{2}---> 6NO_{2}(g) + 9H_{2}O(g)ΔH = −1698 kJ <--- times by 1.5 6NO _{2}(g) + 8NH_{3}(g) ---> 7N_{2}(g) + 12H_{2}O(g)ΔH = −2740 kJ <--- times by 1

5) Add:

14NH _{3}(g) + 10.5O_{2}---> 7N_{2}(g) + 21H_{2}O(g)ΔH = −4438 kJ

6) Divide through by 3.5 to arrive at the final answer (given above in step 3). I think using the whole numbers at the start makes it easier to realize that you need to divide through by 7, as opposed to the 3.5 that is needed when using 1.5 and 1 as the factors.

**Example #9:** Use the given information to find the standard molar heat of formation of phosphorus pentachloride.

P _{4}(s) + 6Cl_{2}(g) ---> 4PCl_{3}(s)ΔH = −1270.7 kJ PCl _{3}(s) + Cl_{2}(g) ---> PCl_{5}(s)ΔH = −137.1 kJ

**Solution #1:**

1) The equation we are seeking is this:

^{1}⁄_{4}P_{4}(s) +^{5}⁄_{2}Cl_{2}(g) ---> PCl_{5}(s)

2) Multiply the first equation by ^{1}⁄_{4}:

^{1}⁄_{4}P_{4}(s) +^{3}⁄_{2}Cl_{2}(g) ---> PCl_{3}(s)ΔH = −317.675 kJ PCl _{3}(s) + Cl_{2}(g) ---> PCl_{5}(s)ΔH = −137.1 kJ

3) When the two equations just above are added, the target equation is formed. Add the two enthalpies and round off as needed for the answer of −454.8 kJ

**Solution #2:**

1) Multiply second equation by 4:

P _{4}(s) + 6Cl_{2}(g) ---> 4PCl_{3}(s)ΔH = −1270.7 kJ 4PCl _{3}(s) + 4Cl_{2}(g) ---> 4PCl_{5}(s)ΔH = −548.4 kJ

2) Add the two reactions:

P_{4}(s) + 10Cl_{2}(g) ---> 4PCl_{5}(s) ΔH = −1819.1 kJ

3) Determine the standard heat of formation for phosphorus pentachloride:

−1819.1 kJ ––––––––– = −454.775 kJ/mol = −454.8 kJ/mol (to four sig figs) 4 mol

**Example #10:** The enthalpy changes for the formation of aluminium oxide and iron(III) oxide from their elements are:

2Al + ^{3}⁄_{2}O_{2}---> Al_{2}O_{3}ΔH° = −1675.7 kJ 2Fe + ^{3}⁄_{2}O_{2}--> Fe_{2}O_{3}ΔH° = −824.2 kJ

Using Hess's Law calculate the enthalphy change for this target reaction:

Fe _{2}O_{3}+ 2Al ---> Al_{2}O_{3}+ 2FeΔH° = ???

**Solution:**

1) Reverse second equation:

2Al + ^{3}⁄_{2}O_{2}---> Al_{2}O_{3}ΔH° = −1675.7 kJ Fe _{2}O_{3}---> 2Fe +^{3}⁄_{2}O_{2}ΔH° = +824.2 kJ

2) Add the two equations together and cancel out the ^{3}⁄_{2}O_{2} that appears on both sides. Add the two enthalpies for the enthaly change for the target reaction.

Fe _{2}O_{3}+ 2Al ---> Al_{2}O_{3}+ 2FeΔH° = −851.5 kJ