Using standard enthalpies of formation

Hess' Law: two equations and their enthalpies | Hess' Law: bond enthalpies |

Hess' Law: three equations and their enthalpies | Thermochemistry menu |

Hess' Law: four or more 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.

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

ΔH°_{rxn}= Σ ΔH°_{f}(products) − Σ ΔH°_{f}(reactants)

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

ΔH°_{rxn}= Σ ΔH°_{f, products}− Σ ΔH°_{f, reactants}

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:

C_{2}H_{5}OH(ℓ) +^{7}⁄_{2}O_{2}(g) ---> 2CO_{2}(g) + 3H_{2}O(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:

ΔH°_{comb}= [2(−393.5) +3(−286)] − [(−278) + () (0)]^{7}⁄_{2}

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 ΔH°_{comb} = −1367 kJ/mol of ethyl alcohol.

The ΔH°_{f} 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:

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

To solve this problem, we must know the following ΔH°_{f} values:

C _{6}H_{12}O_{6}(s)−1275.0 O _{2}(g)zero CO _{2}(g)−393.5 H _{2}O(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) + 6H_{2}(g) + 3O_{2}(g) ---> C_{6}H_{12}O_{6}(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

ΔH°_{comb}= [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 ΔH°_{comb} = −2801 kJ/mol of glucose.

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

ΔH°_{comb, glucose}= −2800.8 kJ/mol

ΔH°_{f, CO2}= −393.5

ΔH°_{f, H2O}= −285.8

**Solution:**

−2800.8 = [6(−393.5) +6(−285.8) ] minus [ (ΔH°_{f, glucose}) +(6)(0) ]

Did you see what I did. All the enthalpies of formation are on the right-hand side and the ΔH°_{comb} 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) + 6H_{2}(g) + 3O_{2}(g) ---> C_{6}H_{12}O_{6}(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:

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

C(s, gr.) + O_{2}(g) ---> CO_{2}(g)

H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(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 (C_{3}H_{6}O) liberates 1790 kJ:

C_{3}H_{6}O(ℓ) + 4O_{2}(g) ---> 3CO_{2}(g) + 3H_{2}O(ℓ); ΔH°_{comb, acetone}= −1790 kJ

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

ΔH°_{f, O2}= 0

ΔH°_{f, CO2}= −393.5

ΔH°_{f, H2O}= −285.83

**Solution:**

1) Hess' Law:

ΔH°_{rxn}= Σ ΔH°_{f, products}minus Σ ΔH°_{f, reactants}

2) Sustitute values into equation:

−1790 = [3(−393.5) +3(−285.83) ] − [ (ΔH°_{f, acetone}) +(4)(0) ]−1790 = −2037.99 - ΔH°

_{f, acetone}247.99 = - ΔH°

_{f, acetone}ΔH°

_{f, acetone}= −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.

C _{6}H_{14}(ℓ)−4163.0 C(s, gr.) −393.5 H _{2}(g)−285.8

Before the solution is given, a bit of discussion: the enthalpy of combustion for hexane, carbon and hydrogen are these chemical equations:

C_{6}H_{14}(ℓ) +^{19}⁄_{2}O_{2}(g) ---> 6CO_{2}(g) + 7H_{2}O(ℓ)

C(s, gr) + O_{2}(g) ---> CO_{2}(g)

H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)

To obtain the target reaction (see just below, in the solution), we must do the following:

a) reverse the first equation

b) multiply the second equation by 6

c) multiply the third equation by 7

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 equation for carbon dioxide.

Last point: notice how the enthalpy of combustion focuses on the reactant while the standard enthalpy of formation focuses on the product.

**Solution:**

1) Write the equation for the formation of hexane:

6C(s) + 7H_{2}(g) ---> C_{6}H_{14}(ℓ)

Use Hess' Law:

ΔH°_{rxn}= Σ ΔH°_{comb, products}minus Σ ΔH°_{comb, reactants}ΔH°

_{rxn}= [ (4136) ] − [(6)(−393.5) +(7)(−285.8) ]ΔH°

_{rxn}= −198.6 kJ/mol

This question can also be found on Yahoo Answer's chemistry section.

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

4NH_{3}(g) + 7O_{2}(g) ---> 4NO_{2}(g) + 6H_{2}O(g)

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

NH _{3}(g)−45.90 NO _{2}(g)+33.1 H _{2}O(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:

ΔH°_{rxn}= [(4)(+33.1) +(6)(−241.83) ] minus [(4)(−45.90) +(7)(0) ]ΔH°

_{rxn}= −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:

AgNO_{3}(s) ---> AgNO_{2}(s) +^{1}⁄_{2}O_{2}(g)

The standard enthalpy of formation of AgNO_{3}(s) is −123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO_{2}(s)

**Solution:**

1) Let's write what we know:

AgNO _{3}(s) ---> AgNO_{2}(s) +^{1}⁄_{2}O_{2}(g)ΔH° = +78.67 kJ Ag(s) + ^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}O_{2}(g) ---> AgNO_{3}(s)ΔH° _{f}= −123.02 kJ

2) Let's write the formation equation for AgNO_{2}(s):

Ag(s) +^{1}⁄_{2}N_{2}(g) + O_{2}(g) ---> AgNO_{2}(s) ΔH°_{f}= ???

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 AgNO_{3}(s) cancels out as does ^{1}⁄_{2}O_{2}(g) and we are left with the formation equation for AgNO_{2}(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:

C _{2}H_{6}(g)−84.68 O _{2}(g)zero CO _{2}(g)−393.5 H _{2}O (g)−241.8

**Solution:**

1) The balanced equation for the combustion of C_{2}H_{6} (ethane) is:

2C_{2}H_{6}+ 7O_{2}---> 4CO_{2}+ 6H_{2}O

2) The enthalpy of the reaction is:

[sum of enthalpies of formation of products] minus [sum of enthalpies of formation of reactants][(2 moles CO

_{2})(−393.5 kJ/mole) + (6 moles H_{2}O)(−241.8 kJ/mole)] − [(2 moles C_{2}H_{6})(−84.68 kJ/mole) + (7 moles O_{2})(0 kJ/mole)]−2238 kJ − (−169 kJ) = −2069 kJ

3) However, that's the heat produced when we make 6 moles of H_{2}O(g). Therefore,

−2069 kJ / 6 moles H_{2}O = −345 kJ / mole H_{2}O

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

IF_{7}+ I_{2}---> IF_{5}+ 2IF

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

IF_{7}= −941 kJ

IF_{5}= −840 kJ

Determine the ΔH_{f}° for IF.

**Solution #1:**

1) The enthalpy of the reaction is:

[sum of enthalpies of formation of products] minus [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}⁄_{2}I_{2}+^{7}⁄_{2}F_{2}---> IF_{7}ΔH _{f}= −941 kJ^{1}⁄_{2}I_{2}+^{5}⁄_{2}F_{2}---> IF_{5}ΔH _{f}= −840 kJIF _{7}+ I_{2}---> IF_{5}+ 2IFΔH = −89 kJ

and here is the target equation:

^{1}⁄_{2}I_{2}+^{1}⁄_{2}F_{2}---> IF ΔH_{f}= ?

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 eqation 2

c) leave equation 3 untouched.

3) The result of the above is this:

I_{2}+ F_{2}---> 2IFand

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:

ΔH_{f}= −190 / 2 = −95 kJ

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

SiCl_{4}(ℓ) + 2H_{2}(g) ---> Si(s) + 4HCl(g)

Use the following standard enthalpies of formation:

SiCl_{4}(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 H

_{2}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:

(CH_{2}OH)_{2}(ℓ) +^{5}⁄_{2}O_{2}(g) ---> 2CO_{2}(g) + 3H_{2}O(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:

ΔH°_{rxn}= Σ ΔH°_{f}(products) − Σ ΔH°_{f}(reactants)

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

−1191 = [(2) (−393.52) + (3) (−285.83)] − [(1) (x) + (^{5}⁄_{2}) (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.

**Bonus Problem:** Given the following information:

ΔH _{f}kJ/mol ΔH _{f}kJ/mol Li _{2}O(s)597.9 Li ^{+}(aq)−278.5 Na _{2}O(s)−416 Na ^{+}(aq)−240 K _{2}O(s)−361 K ^{+}(aq)−251 CO(g) −110.5 CO _{2}(g)−393.5 H _{2}O(ℓ)−286 OH¯(aq) −230 CCl _{4}(ℓ)−135 SiO _{2}(s)−911 LiCl(s) −409 NaCl(s) −411 KCl(s) −436 Cl¯(aq) −167

Calculate ΔH for the following reaction:

2Li(s) + 2H_{2}O(ℓ) ---> 2LiOH(aq) + H_{2}(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) + 2H_{2}O(ℓ) ---> 2Li^{+}(aq) + 2OH¯(aq) + H_{2}(g)

3) We need the following formation reactions:

Li(s) ---> Li ^{+}(aq) + e¯ΔH _{f}= −278.5 kJ/mole¯ + 1/2H _{2}(g) + 1/2O_{2}(g) ---> OH¯(aq)ΔH _{f}= −230 kJ/mol

4) Rewrite the revised target equation:

2Li(s) + 2H_{2}O(ℓ) ---> 2Li^{+}(aq) + 2OH¯(aq) + H_{2}(g)

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

ΔH = [(2) (−278.5) + (2) (−230) + (0)] − [(2) (0) + (2) (−286)]ΔH = −445kJ

Hess' Law: two equations and their enthalpies | Hess' Law: bond enthalpies |

Hess' Law: three equations and their enthalpies | Thermochemistry menu |

Hess' Law: four or more equations and their enthalpies |