### Hess' Law of Constant Heat SummationUsing standard enthalpies of formation

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.

There is another way to use Hess' Law. It does not use the full chemical equations and it is usually presented like this:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f}^{o}$ (products) − Σ $\text{ΔH}{\text{}}_{f}^{o}$ (reactants)

Here's another to write this form of Hess' Law, one that slightly varies from the above manner:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f, products}^{o}$ − Σ $\text{ΔH}{\text{}}_{f, reactants}^{o}$

The "rxn" above is a common way to abbreviate "reaction." All it means is that we are discussing the enthalpy of a generic reaction, not any specific one. I'll explain the above equation using an example problem.

Example #1: Calculate the standard enthalpy of combustion for the following reaction:

C2H5OH(ℓ) + 72O2(g) ---> 2CO2(g) + 3H2O(ℓ)

Before launching into the solution, notice I used "standard enthalpy of combustion." This is a very common chemical reaction, to take something and combust (burn) it in oxygen. It is so common that the phrase "standard enthalpy of combustion" is used alot and is given this symbol: ΔH°comb.

The key to solving this problem is to have a table of standard enthalpies of formation handy. In case you missed it, look at the equation up near the top and see the subscripted f. What we are going to do is sum up all the product enthalpies of formation and then subtract the summed up reactant enthalpies of formation. Also, we need to have the equation balanced, so be sure to remember to check for that. Fractional coefficients are OK.

Like this:

$\text{ΔH}{\text{}}_{comb}^{o}$ = [2 (−393.5) + 3 (−286)] − [(−278) + (72) (0)]

The boldfaced values are the coefficients and the other ones are the standard enthalpy of formation for the four substances involved. Since oxygen is an element in its standard state, its enthalpy of formation is zero.

Doing the math gives us $\text{ΔH}{\text{}}_{comb}^{o}$ = −1367 kJ/mol of ethyl alcohol.

The $\text{ΔH}{\text{}}_{f}^{o}$ values (−393.5, −286, −278 and zero) were looked up in a reference source. Textbooks which teach this topic will have an appendix of the values. Make sure you find it and figure out how to use it.

If you are not too clear on what the term "standard enthalpy of formation" means, please look here.

Example #2: Calculate the standard enthalpy of combustion for the following reaction:

C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(ℓ)

To solve this problem, we must know the following $\text{ΔH}{\text{}}_{f}^{o}$ values:

 C6H12O6(s) −1275.0 O2(g) zero CO2(g) −393.5 H2O(ℓ) −285.8

All the above values have units of kJ/mol because these are standard values. All standard enthalpies have the unit kJ/mol.

As a brief reminder, here is the chemical reaction for the standard enthalpy of glucose:

6C(s, graphite) + 6H2(g) + 3O2(g) ---> C6H12O6(s)

Each standard enthalpy value is associated with a chemical reaction. The reaction will always form one mole of the target substance (glucose in the example) in its standard state. The target substance is always formed from elements in their respective standard states. Note how the standard state for carbon is graphite, not diamond or buckerministerfullerene.

Remember also that all elements in their standard state have an enthalpy of formation equal to zero.

The solution

$\text{ΔH}{\text{}}_{comb}^{o}$ = [6 (−393.5) + 6 (−285.8)] − [(−1275) + (6) (0)]

The boldfaced values are the coefficients and the other ones are the standard enthalpy of formation for the four substances involved. Since oxygen is an element in its standard state, its enthalpy of formation is zero.

Doing the math gives us $\text{ΔH}{\text{}}_{comb}^{o}$ = −2801 kJ/mol of glucose.

Example #3: Calculate the standard enthalpy of formation for glucose, given the following values:

$\text{ΔH}{\text{}}_{f, comb, glucose}^{o}$ = −2800.8 kJ/mol
$\text{ΔH}{\text{}}_{f, CO2}^{o}$ = −393.5 kJ/mol
$\text{ΔH}{\text{}}_{f, H2O}^{o}$ = −285.8 kJ/mol

Solution:

−2800.8 = [ 6 (−393.5) + 6 (−285.8) ] − [ ($\text{ΔH}{\text{}}_{f, glucose}^{o}$ ) + (6) (0) ]

Did you see what I did? All the enthalpies of formation are on the right-hand side and the $\text{ΔH}{\text{}}_{comb}^{o}$ goes on the left-hand side.

By the way, this is a common test question. Be prepared.

Here's what happened:

1) First of all, this is the reaction we want an answer for:

6C(s, graphite) + 6H2(g) + 3O2(g) ---> C6H12O6(s)

We know this because the problem asks for the standard enthalpy of formation for glucose. The above chemical reaction IS the standard formation reaction for glucose. We want the enthalpy for it.

2) Here are the reactions to be added, in the manner of Hess' Law:

C6H12O6(s) + 6O2(g) ---> 6CO2(g) + 6H2O(ℓ)
C(s, gr.) + O2(g) ---> CO2(g)
H2(g) + 12O2(g) ---> H2O(ℓ)

3) Flip the first reaction and multiply the other two by six. Then add the three reactions together. If you do it right, you should recover the reaction mentioned just above in (1).

Example #4: Complete combustion of 1.00 mol of acetone (C3H6O) liberates 1790 kJ:

 C3H6O(ℓ) + 4O2(g) ---> 3CO2(g) + 3H2O(ℓ) $\text{ΔH}{\text{}}_{comb}^{o}$ = −1790 kJ

Using this information together with the data below (values in kJ/mol), calculate the enthalpy of formation of acetone.

$\text{ΔH}{\text{}}_{f, O2}^{o}$ = 0
$\text{ΔH}{\text{}}_{f, CO2}^{o}$ = −393.5
$\text{ΔH}{\text{}}_{f, H2O}^{o}$ = −285.83

Solution:

1) Hess' Law:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f, products}^{o}$ − Σ $\text{ΔH}{\text{}}_{f, reactants}^{o}$

2) Sustitute values into equation:

−1790 = [ 3 (−393.5) + 3 (−285.83) ] − [ ($\text{ΔH}{\text{}}_{f, acetone}^{o}$ ) + (4) (0) ]

−1790 = −2037.99 − ΔH°f, acetone

247.99 = −$\text{ΔH}{\text{}}_{f, acetone}^{o}$

$\text{ΔH}{\text{}}_{f, acetone}^{o}$ = −247.99 kJ/mol

To three sig figs, the value is −248 kJ/mol.

Example #5: The standard enthalpy of formation of hexane can be determined indirectly. Calculate the standard enthalpy of formation of hexane using the enthalpies of combustion (in kJ/mol) given just below.

 C6H14(ℓ) −4163.0 C(s, gr) −393.5 H2(g) −285.8

Solution using enthalpy of combustions:

1) The enthalpy of combustion for hexane, carbon and hydrogen are these chemical equations:

C6H14(ℓ) + 192O2(g) ---> 6CO2(g) + 7H2O(ℓ)
C(s, gr) + O2(g) ---> CO2(g)
H2(g) + 12O2(g) ---> H2O(ℓ)

2) To obtain the target reaction (the enthalpy of formation for hexane), we must do the following:

a) reverse the first equation
b) multiply the second equation by 6
c) multiply the third equation by 7

3) After doing that, we obtain:

 C6H14(ℓ) + 19⁄2O2(g) ---> 6CO2(g) + 7H2O(ℓ) $\text{ΔH}{\text{}}_{}^{o}$ = +4163.0 kJ 6C(s, gr) + 6O2(g) ---> 6CO2(g) $\text{ΔH}{\text{}}_{}^{o}$ = −2361.0 kJ 7H2(g) + 7⁄2O2(g) ---> 7H2O(ℓ) $\text{ΔH}{\text{}}_{}^{o}$ = −2000.6 kJ

By the way, the second equation (presented as the enthalpy of combustion of carbon) is also the equation for the formation of carbon dioxide. The third equation (presented as the combustion of hydrogen gas) is also the formation equation for water in its standard state (liquid). The moral of the story? Sometimes terms overlap. The −393.5 value is the enthalpy for the combustion of carbon. It is also the formation enthalpy for carbon dioxide.

4) Adding the above three equations gives us the equation for the formation of hexane. Add the enthalpies to obtain the enthalpy of formation for hexane::

6C(s) + 7H2(g) ---> C6H14(ℓ)    $\text{ΔH}{\text{}}_{}^{o}$ = −198.6 kJ

Solution using enthalpies of formation:

1) Write the combustion reaction for hexane:

C6H14(ℓ) + 192O2 ---> 6CO2(g) + 7H2O(ℓ)

2) State Hess' Law using standard enthalpies of formation:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f, products}^{o}$ − Σ $\text{ΔH}{\text{}}_{f, reactants}^{o}$

3) We note that the enthalpies of combustion for CO2(g) and H2O(ℓ) are also their enthalpies of formation. We insert data into Hess' Law:

−4163.0 = [(6) (−393.5) + (7) (−285.8)] − [(1) ($\text{ΔH}{\text{}}_{f, C6H14}^{o}$) + (192) (0)]

Note the enthalpy of formation for O2(g). Note how the enthalpy of formation for hexane (the desired result) is our only unknown.

4) Do some solving of the above.

−4163.0 = (−2361) + (−2000.6) − $\text{ΔH}{\text{}}_{f, C6H14}^{o}$

198.6 = −$\text{ΔH}{\text{}}_{f, C6H14}^{o}$

$\text{ΔH}{\text{}}_{f, C6H14}^{o}$ = −198.6 kJ

Example #6: Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows:

4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(g)

Given the following standard enthalpies of formation (given in kJ/mol), calculate the enthalpy of the above reaction:

 NH3(g) −45.90 NO2(g) +33.1 H2O(ℓ) −241.83

Note that water is given as a gas. The usual problem of this type uses water as a liquid. Not in this one.

Solution:

Use Hess' Law:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = [ (4) (+33.1) + (6) (−241.83) ] − [ (4) (−45.90) + (7) (0) ]

$\text{ΔH}{\text{}}_{rxn}^{o}$ = −1134.98 kJ = −1135 kJ

Note that the units kJ/mol are NOT used.

Example #7: The standard enthalpy change, ΔH°, for the thermal decomposition of silver nitrate according to the following equation is +78.67 kJ:

AgNO3(s) ---> AgNO2(s) + 12O2(g)

The standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO2(s)

Solution:

1) Let's write what we know:

 AgNO3(s) ---> AgNO2(s) + 1⁄2O2(g) ΔH° = +78.67 kJ Ag(s) + 1⁄2N2(g) + 3⁄2O2(g) ---> AgNO3(s) $\text{ΔH}{\text{}}_{f}^{o}$ = −123.02 kJ

2) Let's write the formation equation for AgNO2(s):

Ag(s) + 12N2(g) + O2(g) ---> AgNO2(s)   $\text{ΔH}{\text{}}_{f}^{o}$ = ???

3) Determine the unknown value by adding the two equations listed in step 1:

+78.67 kJ + (−123.02 kJ) = −44.35 kJ (this is the answer)

When the two equations are added together, the AgNO3(s) cancels out as does 12O2(g) and we are left with the formation equation for AgNO2(s), the equation given in step 2.

Example #8: Using standard enthalpies of formation, calculate the heat of combustion per mole of gaseous water formed during the complete combustion of ethane gas.

The enthalpies of formation needed are:

 C2H6(g) −84.68 O2 (g) zero CO2 (g) −393.5 H2O (g) −241.8

Solution:

1) The balanced equation for the combustion of C2H6 (ethane) is:

2C2H6 + 7O2 ---> 4CO2 + 6H2O

2) The enthalpy of the reaction is:

[sum of enthalpies of formation of products] − [sum of enthalpies of formation of reactants]

[(2 moles CO2) (−393.5 kJ/mole) + (6 moles H2O) (−241.8 kJ/mole)] − [(2 moles C2H6) (−84.68 kJ/mole) + (7 moles O2) (0 kJ/mole)]

−2238 kJ − (−169 kJ) = −2069 kJ

3) However, that's the heat produced when we make 6 moles of H2O(g). Therefore,

−2069 kJ / 6 moles H2O = −345 kJ / mole H2O

Example #9: The ΔH for the following reaction equals −89 kJ:

IF7 + I2 ---> IF5 + 2IF

In addition, these two standard enthalpies of formation are known:

IF7 = −941 kJ
IF5 = −840 kJ

Determine the $\text{ΔH}{\text{}}_{f}^{o}$ for IF.

Solution #1:

1) The enthalpy of the reaction is:

[sum of enthalpies of formation of products] − [sum of enthalpies of formation of reactants]

2) Inserting values into the above, we find:

−89 = [(−840) (1) + (2x)] − [(−941) (1) + (0) (1)]

−89 = 101 + 2x

2x = −190

x = −95 kJ

Solution #2:

1) Here are all three data reactions written out in equation form:

 1⁄2I2 + 7⁄2F2 ---> IF7 $\text{ΔH}{\text{}}_{f}^{o}$ = −941 kJ 1⁄2I2 + 5⁄2F2 ---> IF5 $\text{ΔH}{\text{}}_{f}^{o}$ = −840 kJ IF7 + I2 ---> IF5 + 2IF ΔH° = −89 kJ

and here is the target equation:

12I2 + 12F2 ---> IF ΔHf = ?

2) What we need to do is add the three data equations together in such a way as to recover the target equation:

a) leave equation 1 untouched
b) flip equation 2
c) leave equation 3 untouched.

3) The result of the above is this:

I2 + F2 ---> 2IF

and

ΔH = −941 + (+840) + (−89) = −190 kJ

4) However, this is not the enthalpy of formation, since that value is always for one mole of the product. This is the answer:

ΔHf = −190 / 2 = −95 kJ

Example #10: What is the enthalpy change for the following reaction?

SiCl4(ℓ) + 2H2(g) ---> Si(s) + 4HCl(g)

Use the following standard enthalpies of formation:

SiCl4(ℓ); −687 kJ mol¯1
HCl(g); −92 kJ mol¯1

Solution:

ΔH = [0 + (4) (−92)] − [−687 + (2) (0)]

The zeros are the enthalpies for H2 and Si. These are elements in their standard sate and in that case, the enthalpy of formaton is always zero.

ΔH = +319 kJ

Example #11: The combustion of ethylene glycol is shown:

(CH2OH)2(ℓ) + 52O2(g) ---> 2CO2(g) + 3H2O(ℓ);   ΔH° = −1191 kJ/mol

Determine the standard enthalpy of formation for ethylene glycol.

Solution:

1) The first thing to do is look up standard enthalpies of formation for the other three substances involved:

oxygen ---> zero (by definition)
carbon dioxide ---> −393.52 kJ/mol (source)
water ---> −285.83 kJ/mol (source)

2) Next, we write Hess' Law in the form that uses standard enthalpies of formation:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f}^{o}$ (products) − Σ $\text{ΔH}{\text{}}_{f}^{o}$ (reactants)

3) And then, we put in values and solve:

−1191 = [(2) (−393.52) + (3) (−285.83)] − [(1) (x) + (52) (0)]

−1191 = −1644.53 − x

x = −453.5 kJ/mol (to 4 sig figs)

4) We can look up the value for the standard enthalpy of formation for ethylene glycol.

Example #12: Determine the standard heat of formation for methyl bromide, CH3Br(g), given the following equation:

CH3Br(g) + H2(g) ---> CH4(g) + HBr(g)   ΔH° = −73.60 kJ

Solution:

1) The first thing to do is write the formation equation for methyl bromide:

C(s, gr) + 32H2(g) + 12Br2(ℓ) ---> CH3Br(g)

2) Since CH4(g) and HBr(g) do not appear in the final answer, we need equations that will include them. Since we are discussing formation equations, let's go look up their formation enthalpies:

C(s, gr) + 2H2(g) ---> CH4(g)    $\text{ΔH}{\text{}}_{f}^{o}$ = −74.87 kJ   (found here)

12H2(g) + 12Br2(ℓ) ---> HBr(g)  $\text{ΔH}{\text{}}_{f}^{o}$ = −36.29 kJ   (found here)

3) Use Hess' Law:

 CH4(g) + HBr(g) ---> CH3Br(g) + H2(g) ΔH° = +73.60 kJ C(s, gr) + 2H2(g) ---> CH4(g) $\text{ΔH}{\text{}}_{f}^{o}$ = −74.87 kJ 1⁄2H2(g) + 1⁄2Br2(ℓ) ---> HBr(g) $\text{ΔH}{\text{}}_{f}^{o}$ = −36.29 kJ

Note that the first equation was reversed. Nothing was done to the other two equations.

4) The above equations, when added, will produce the formation equation for methyl bromide. Add the enthalpies to obtain:

$\text{ΔH}{\text{}}_{f}^{o}$ = −37.56 kJ/mol

Data for methyl bromide may be found here.

Example #13: Use Hess' Law to calculate the enthalpy of vaporization for ethanol, C2H5OH:

C2H5OH(ℓ) ---> C2H5OH(g)

Solution:

enthalpy of formation, gas ---> −234 kJ/mol
enthalpy of formation, liquid ---> −276 kJ/mol

ΔHvap = products − reactants

ΔHvap = −234 − (−276) = 42 kJ/mol

The value given here is 42.3 ± 0.4 kJ/mol

Example #14: Use standard enthalpies of formation to calculate the enthalpy change (in kJ) for the reduction of iron(III) oxide to iron at 298 K and 1 atm. (Calculate it for the reaction as written, namely 2 moles of iron(III) oxide and 3 moles of carbon.)

2Fe2O3(s) + 3C(s) ---> 4Fe(s) + 3CO2(g)

Solution:

1) Let us assume that the carbon is in its standard state of graphite (as opposed to diamond or buckminsterfullerene). That means:

$\text{ΔH}{\text{}}_{f, carbon}^{o}$ = 0

$\text{ΔH}{\text{}}_{f, iron}^{o}$ = 0

2) We must look up the standard enthalpy of formation for the other two substances:

$\text{ΔH}{\text{}}_{f, Fe2O3}^{o}$ = −825.50 kJ/mol (source)

$\text{ΔH}{\text{}}_{f, CO2}^{o}$ = −393.51 kJ/mol (source)

3) Hess' Law:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f, products}^{o}$ − Σ $\text{ΔH}{\text{}}_{f, reactants}^{o}$

4) Sustitute values into equation:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = [(3) (−393.51 kJ) + (4) (0)] − [(2) (−825.50 kJ) + (3) (0)]

$\text{ΔH}{\text{}}_{rxn}^{o}$ = 470.47 kJ

Note: Do not write kJ/mol. The kJ produced are for the reaction as written. In other words, 470.47 kJ are produced when two moles of iron(III) oxide and three moles of carbon are reacted.

Example #15: Using the standard enthalpies of formation to determine the enthalpy of reaction for:

2H2O(ℓ) + 2SO2(g) ---> 2H2S(g) + 3O2(g)

Solution:

1) Since the example does not provide enthalpy of formation values, we must look them up. The ChemTeam's usual source is the NIST Chemistry WebBook:

$\text{ΔH}{\text{}}_{f, H2O}^{o}$ = −285.83 kJ/mol (source)

$\text{ΔH}{\text{}}_{f, SO2}^{o}$ = −296.84 kJ/mol (source)

$\text{ΔH}{\text{}}_{f, H2S}^{o}$ = −20.50 kJ/mol (source)

$\text{ΔH}{\text{}}_{f, O2}^{o}$ = zero kJ/mol (by definition)

2) State Hess' Law

$\text{ΔH}{\text{}}_{rxn}^{o}$ = Σ $\text{ΔH}{\text{}}_{f, products}^{o}$ − Σ $\text{ΔH}{\text{}}_{f, reactants}^{o}$

3) Put values in place and solve:

$\text{ΔH}{\text{}}_{rxn}^{o}$ = [(2 mol) (−20.50 kJ/mol) + (3 mol) (0 kJ/mol)] − [(2 mol) (−285.83 kJ/mol) + (2 mol) (−296.84 kJ/mol)

$\text{ΔH}{\text{}}_{rxn}^{o}$ = (−41.00 kJ) − (−1165.34 kJ)

$\text{ΔH}{\text{}}_{rxn}^{o}$ = +1124.34 kJ

4) A popular reaction for standard enthalpy questions is the reverse of the reaction just discussed. Here is a search. If you look at any of the examples, be aware that the enthalpy values are often going to be slightly different than the ones I used above.

Bonus Example: Given the following information:

 $\text{ΔH}{\text{}}_{f}^{o}$ kJ/mol $\text{ΔH}{\text{}}_{f}^{o}$ kJ/mol Li2O(s) 597.9 Li+(aq) −278.5 Na2O(s) −416 Na+(aq) −240 K2O(s) −361 K+(aq) −251 CO(g) −110.5 CO2(g) −393.5 H2O(ℓ) −286 OH¯(aq) −230 CCl4(ℓ) −135 SiO2(s) −911 LiCl(s) −409 NaCl(s) −411 KCl(s) −436 Cl¯(aq) −167

Calculate ΔH for the following reaction:

2Li(s) + 2H2O(ℓ) ---> 2LiOH(aq) + H2(g)

Solution:

1) The key is to see the meaning of 2LiOH(aq):

2LiOH(aq) ---> 2Li+(aq) + 2OH¯(aq)

2) That means that, in reality, we want the ΔH for this reaction:

2Li(s) + 2H2O(ℓ) ---> 2Li+(aq) + 2OH¯(aq) + H2(g)

3) We need the following reactions:

 Li(s) ---> Li+(aq) + e¯ ΔH = −278.5 kJ/mol e¯ + 1⁄2H2(g) + 1⁄2O2(g) ---> OH¯(aq) ΔH = −230 kJ/mol

4) Rewrite the revised target equation:

2Li(s) + 2H2O(ℓ) ---> 2Li+(aq) + 2OH¯(aq) + H2(g)

5) Use Hess' Law utilizing the revised target equation:

ΔH = [(2) (−278.5) + (2) (−230) + (0)] − [(2) (0) + (2) (−286)]

ΔH = −445 kJ