Using three 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.

**Example #1:** Calculate the enthalpy for this reaction:

2C(s) + H _{2}(g) ---> C_{2}H_{2}(g)ΔH° = ??? kJ

Given the following thermochemical equations:

C _{2}H_{2}(g) +^{5}⁄_{2}O_{2}(g) ---> 2CO_{2}(g) + H_{2}O(ℓ)ΔH° = −1299.5 kJ C(s) + O _{2}(g) ---> CO_{2}(g)ΔH° = −393.5 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −285.8 kJ

**Solution:**

1) Determine what we must do to the three given equations to get our target equation:

a) first eq: flip it so as to put C_{2}H_{2}on the product side

b) second eq: multiply it by two to get 2C

c) third eq: do nothing. We need one H_{2}on the reactant side and that's what we have.

2) Rewrite all three equations with changes applied:

2CO _{2}(g) + H_{2}O(ℓ) ---> C_{2}H_{2}(g) +^{5}⁄_{2}O_{2}(g)ΔH° = +1299.5 kJ 2C(s) + 2O _{2}(g) ---> 2CO_{2}(g)ΔH° = −787 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −285.8 kJ Notice that the ΔH values changed as well.

3) Examine what cancels:

2CO_{2}⇒ first & second equation

H_{2}O ⇒ first & third equation^{5}⁄_{2}O_{2}⇒ first & sum of second and third equation

4) Add up ΔH values for our answer:

+1299.5 kJ + (−787 kJ) + (−285.8 kJ) = +226.7 kJ

**Example #2:** Calculate the enthalpy of the following chemical reaction:

CS_{2}(ℓ) + 3O_{2}(g) ---> CO_{2}(g) + 2SO_{2}(g)

Given:

C(s) + O _{2}(g) ---> CO_{2}(g)ΔH = −393.5 kJ/mol S(s) + O _{2}(g) ---> SO_{2}(g)ΔH = −296.8 kJ/mol C(s) + 2S(s) ---> CS _{2}(ℓ)ΔH = +87.9 kJ/mol

**Solution:**

1) What to do to the data equations:

leave eq 1 untouched (want CO_{2}as a product)

multiply second eq by 2 (want to cancel 2S, also want 2SO_{2}on product side)

flip 3rd equation (want CS_{2}as a reactant)

2) The result:

C(s) + O _{2}(g) ---> CO_{2}(g)ΔH = −393.5 kJ/mol 2S(s) + 2O _{2}(g) ---> 2SO_{2}(g)ΔH = −593.6 kJ/mol <--- note multiply by 2 on the ΔH CS _{2}(ℓ) ---> C(s) + 2S(s)ΔH = −87.9 kJ/mol <--- note sign change on the ΔH

3) Add the three revised equations. C and 2S will cancel.

4) Add the three enthalpies for the final answer.

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

SrO(s) + CO _{2}(g) ---> SrCO_{3}(s)ΔH = −234 kJ 2SrO(s) ---> 2Sr(s) + O _{2}(g)ΔH = +1184 kJ 2SrCO _{3}(s) ---> 2Sr(s) + 2C(s, gr) + 3O_{2}(g)ΔH = +2440 kJ

Find the ΔH of the following reaction:

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

**Solution:**

1) Analyze what must happen to each equation:

a) first eq ⇒ flip it (this put the CO_{2}on the right-hand side, where we want it)b) second eq ⇒ do not flip it, divide through by two (no flip because we need to cancel the SrO, divide by two because we only need to cancel one SrO)

c) third equation ⇒ flip it (to put the SrCO

_{3}on the other side so we can cancel it), divide by two (since we need to cancel only one SrCO_{3})Notice that what we did to the third equation also sets up the Sr to be cancelled. Why not also multiply first equation by two (to get 2SrO for canceling)? Because we only want one CO

_{2}in the final answer, not two. Notice also that I ignored the oxygen. If everything is right, the oxygen will take care of itself.

2) Apply all the above changes (notice what happens to the ΔH values):

SrCO _{3}(s) ---> SrO(s) + CO_{2}(g)ΔH = +234 kJ SrO(s) ---> Sr(s) + ^{1}⁄_{2}O_{2}(g)ΔH = +592 kJ Sr(s) + C(s, gr) + ^{3}⁄_{2}O_{2}(g) ---> SrCO_{3}(s)ΔH = −1220 kJ

3) Here is a list of what gets eliminated when everything is added:

SrCO_{3}, SrO, Sr,^{1}⁄_{2}O_{2}The last one comes from

^{3}⁄_{2}O_{2}on the left in the third equation and^{1}⁄_{2}O_{2}on the right in the second equation.

4) Add the equations and the ΔH values:

+234 + (+592) + (−1220) = −394

C(s, gr) + O _{2}(g) ---> CO_{2}(g)ΔH° _{f}= −394 kJ

Notice the subscripted f. This is the formation reaction for CO_{2} and its value can be looked up, either in your textbook or online.

**Example #4:** Given the following information:

2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH = −116 kJ 2N _{2}(g) + 5O_{2}(g) + 2H_{2}O(ℓ) ---> 4HNO_{3}(aq)ΔH = −256 kJ N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +183 kJ

Calculate the enthalpy change for the reaction below:

3NO _{2}(g) + H_{2}O(ℓ) ---> 2HNO_{3}(aq) + NO(g)ΔH = ???

**Solution:**

1) Analyze what must happen to each equation:

a) first eq ⇒ flip; multiply by^{3}⁄_{2}(this gives 3NO_{2}as well as the 3NO which will be necessary to get one NO in the final answer)b) second eq ⇒ divide by 2 (gives two nitric acid in the final answer)

c) third eq ⇒ flip (cancels 2NO as well as nitrogen)

2) Comment on the oxygens:

a) step 1a above puts^{3}⁄_{2}O_{2}on the right

b) step 1b puts^{5}⁄_{2}O_{2}on the left

c) step 1c puts^{2}⁄_{2}O_{2}on the rightIn addition, a and c give

^{5}⁄_{2}O_{2}on the right to cancel out the^{5}⁄_{2}O_{2}on the left.

3) Apply all the changes listed above:

3NO _{2}(g) ---> 3NO(g) +^{3}⁄_{2}O_{2}(g)ΔH = +174 kJ N _{2}(g) +^{5}⁄_{2}O_{2}(g) + H_{2}O(ℓ) ---> 2HNO_{3}(aq)ΔH = −128 kJ 2NO(g) ---> N _{2}(g) + O_{2}(g)ΔH = −183 kJ

4) Add the equations and the ΔH values:

+174 + (−128) + (−183) = −137 kJ

3NO _{2}(g) + H_{2}O(ℓ) ---> 2HNO_{3}(aq) + NO(g)ΔH = −137 kJ

**Example #5:** Calculate ΔH for this reaction: CH_{4}(g) + NH_{3}(g) ---> HCN(g) + 3H_{2}(g)

given:

N _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH = −91.8 kJ C(s) + 2H _{2}(g) ---> CH_{4}(g)ΔH = −74.9 kJ H _{2}(g) + 2C(s) + N_{2}(g) ---> 2HCN(g)ΔH = +270.3 kJ

**Solution:**

1) Analyze what must happen to each equation:

a) first eq ⇒ flip and divide by 2 (puts one NH_{3}on the reactant side)

b) second eq ⇒ flip (puts one CH_{4}on the reactant side)

c) third eq ⇒ divide by 2 (puts one HCN on the product side)

2) Rewite all equations with the changes:

NH _{3}(g) --->^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}H_{2}(g)ΔH = +45.9 kJ <--- note sign change & divide by 2 CH _{4}(g) ---> C(s) + 2 H_{2}(g)ΔH = +74.9 kJ <--- note sign change ^{1}⁄_{2}H_{2}(g) + C(s) +^{1}⁄_{2}N_{2}(g) ---> HCN(g)ΔH = +135.15 kJ <--- note divided by 2

3) What cancels when you add the equations:

^{1}⁄_{2}N_{2}(g) ⇒ first and third equations

C(s) ⇒ second and third equations^{1}⁄_{2}H_{2}(g) on the left side of the third equation cancels out^{1}⁄_{2}H_{2}(g) on the right, leaving a total of 3H_{2}(g) on the right (which is what we want)

4) Calculate the ΔH for our reaction:

(+45.9 kJ) + (+74.9 kJ) + (+135.15) = +255.95 kJRounded off to three sig figs gives +260. kJ (note use of explicit decimal point)

**Example #6:** Determine the heat of reaction for the oxidation of iron:

2Fe(s) +^{3}⁄_{2}O_{2}(g) ---> Fe_{2}O_{3}(s)

given the thermochemical equations:

2Fe(s) + 6H _{2}O(ℓ) ---> 2Fe(OH)_{3}(s) + 3H_{2}(g)ΔH = +322 kJ Fe _{2}O_{3}(s) + 3H_{2}O(ℓ) ---> 2Fe(OH)_{3}(s)ΔH = +289 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)ΔH = –572 kJ

**Solution:**

1) Here's what needs to be done:

2Fe(s) + 6H _{2}O(ℓ) ---> 2Fe(OH)_{3}(s) + 3H_{2}(g)ΔH = +322 kJ <--- nothing was done 2Fe(OH) _{3}(s) ---> Fe_{2}O_{3}(s) + 3H_{2}O(ℓ)ΔH = −289 kJ <--- reversed equation 3H _{2}(g) +^{3}⁄_{2}O_{2}(g) ---> 3H_{2}O(ℓ)ΔH = –858 kJ <--- multiplied through by ^{3}⁄_{2}

2) Adding up the equations gives the target equation. Adding the enthalpies gives us our answer:

2Fe(s) +^{3}⁄_{2}O_{2}(g) ---> Fe_{2}O_{3}(s) ΔH = −825 kJ

Note how the multiplying factor doesn't have to be an integer value.

**Example #7:** Using the following thermochemical equations, calculate the standard enthalpy of combustion for one mole of liquid acetone (C_{3}H_{6}O).

3C(s) + 3H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> C_{3}H_{6}O(ℓ)ΔH° = −285.0 kJ C(s) + O _{2}(g) ---> CO_{2}(g)ΔH° = −394.0 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −286.0 kJ

**Solution:**

1) The combustion of liquid acetone is the target equation. Write (and balance) it:

C_{3}H_{6}O(ℓ) + 4O_{2}(g) ---> 3CO_{2}(g) + 3H_{2}O(ℓ)

2) The first data equation needs to be reversed, so as to put acetone on the reactant side. Here are all three data equations with the first one changed:

C _{3}H_{6}O(ℓ) ---> 3C(s) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g)ΔH° = +285.0 kJ C(s) + O _{2}(g) ---> CO_{2}(g)ΔH° = −394.0 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −286.0 kJ Note the sign change in the enthalpy when the equation is reversed.

3) The second data equation needs to be changed to create a situation where the 3C(s) will cancel when the equations are added together:

C _{3}H_{6}O(ℓ) ---> 3C(s) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g)ΔH° = +285.0 kJ 3C(s) + 3O _{2}(g) ---> 3CO_{2}(g)ΔH° = −1182.0 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −286.0 kJ Note that the enthalpy was also multiplied by three.

4) The 3H_{2}(g) (present in the first data equation) also needs to be removed from the final answer. Another multiplication by 3 is used:

C _{3}H_{6}O(ℓ) ---> 3C(s) + 3H_{2}(g) +^{1}⁄_{2}O_{2}(g)ΔH° = +285.0 kJ 3C(s) + 3O _{2}(g) ---> 3CO_{2}(g)ΔH° = −1182.0 kJ 3H _{2}(g) +^{3}⁄_{2}O_{2}(g) ---> 3H_{2}O(ℓ)ΔH° = −858.0 kJ Note that the enthalpy was also multiplied by three.

5) Add the three data equations together to recover the target equation laid out in step 1. Note the following:

^{1}⁄_{2}O_{2}(g) will cancel from each side, leaving 4O_{2}(g) on the left-hand side.The enthalpies are added together to obtain the final answer of −1755 kJ.

**Example #8:** The standard enthalpy change of formation of propane is impossible to measure directly. That is because carbon and hydrogen will not directly react to make propane. However, standard enthalpy changes of combustion are relatively easy to measure.

C _{3}H_{8}(g)ΔH _{1}= −2219.9 kJC(s, gr) ΔH _{2}= −393.5 kJH _{2}(g)ΔH _{3}= −285.8 kJ

Determine the enthalpy of formation for propane.

**Solution:**

1) The chemical equation of interest is this:

3C(s, gr) + 4H_{2}(g) ---> C_{3}H_{8}(g) ΔH = ???

2) Write the chemical equations for combustion of the three chemical species given:

C _{3}H_{8}(g) + 5O_{2}(g) ---> 3CO_{2}(g) + 4H_{2}O(ℓ)ΔH _{1}C(s, gr) + O _{2}---> CO_{2}(g)ΔH _{2}H _{2}+^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH _{3}

3) Modify the three data equations so as to reproduce the target equation:

3CO _{2}(g) + 4H_{2}O(ℓ) ---> C_{3}H_{8}(g) + 5O_{2}(g)−ΔH _{1}(reversed equation)3C(s, gr) + 3O _{2}---> 3CO_{2}(g)3ΔH _{2}(multiplied by 3)4H _{2}+ 2O_{2}(g) ---> 4H_{2}O(ℓ)4ΔH _{3}(multiplied by 4)

4) Add the enthalpies for the answer:

−ΔH_{1}+ 3ΔH_{2}+ 4ΔH_{3}−(−2219.9) + (3) (−393.5) + (4) (−285.8) = −103.8 kJ

Answer may be verified here.

**Example #9:** Determine the standard enthalpy of formation for butane, using the following data:

C _{4}H_{10}(g) +^{13}⁄_{2}O_{2}(g) ---> 4CO_{2}(g) + 5H_{2}O(g)ΔH _{1}= −2657.4 kJC(s, gr) + O _{2}(g) ---> CO_{2}(g)ΔH _{2}= −393.5 kJ2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(g)ΔH _{3}= −483.6 kJ

Comment: note that the first and third equations are not standard combustion equations. The water in each equation is as a gas. In standard combustion equations, water is a liquid (its standard state).

**Solution:**

1) The equation for the formation of butane is as follows:

4C(s, gr) + 5H_{2}(g) ---> C_{4}H_{10}(g)

2) The three data equations are modified as follows:

4CO _{2}(g) + 5H_{2}O(g) ---> C_{4}H_{10}(g) +^{13}⁄_{2}O_{2}(g)−ΔH _{1}4C(s, gr) + ^{8}⁄_{2}O_{2}(g) ---> 4CO_{2}(g)4ΔH _{2}5H _{2}(g) +^{5}⁄_{2}O_{2}(g) ---> 5H_{2}O(g)2.5ΔH _{3}

4) The enthalpies are added together:

−ΔH_{1}+ 4ΔH_{2}+ 2.5ΔH_{3}−(−2657.4) + (−1574) + (−1209) = -125.6 kJ

Is it correct? Of course it is.

**Example #10:** Calculate the enthalpy of formation for acetylene (C_{2}H_{2}), given the following data:

C(s, gr) + O _{2}(g) ---> CO_{2}(g)ΔH = −393.5 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH = −285.8 kJ 2C _{2}H_{2}(g) + 5O_{2}(g) ---> 4CO_{2}(g) + 2H_{2}O(ℓ)ΔH = −2598 kJ

**Solution:**

1) The first thing to do is state the formation equation for acetylene:

2C(s, gr) + H_{2}(g) ---> C_{2}H_{2}(g)Remember, a formation reaction has all substances in their standard states and only one mole of product is produced.

2) Manipulate the data equations:

eq 1 ---> do not flip, multiply by 2 (gets the 2C we need on the reactant side)

eq 2 ---> leave untouched (keeps H_{2}on the reactant side and in the desired amount)

eq 3 ---> flip, divide by 2 (puts C_{2}H_{2}on the product side in the desired amount)

3) The result:

2C(s, gr) + 2O _{2}(g) ---> 2CO_{2}(g)ΔH = −787.0 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH = −285.8 kJ 2CO _{2}(g) + H_{2}O(ℓ) ---> C_{2}H_{2}(g) +^{5}⁄_{2}O_{2}(g)ΔH = +1299 kJ

4) Adding the three reactions together yields the desired equation. Adding the three enthalpies yields the enthalpy of formation for acetylene:

2C(s, gr) + H _{2}(ℓ) ---> C_{2}H_{2}(g)ΔH = +226.2 kJ

**Bonus Example:** Given:

2C _{2}H_{6}+ 7O_{2}---> 4CO_{2}+ 6H_{2}OΔH = −3119.7 kJ (1) 2H _{2}+ O_{2}---> 2H_{2}OΔH = −478.84 kJ (2) 2CO + O _{2}---> 2CO_{2}ΔH = −565.98 kJ (3)

please calculate delta H for the following reaction:

C_{2}H_{6}+ O_{2}---> 3H_{2}+ 2CO

**Solution:**

Comment: this is not the usual ChemTeam manner of solving Hess' Law problems. Which is why I coped it, so as to allow you to analyze how another brain approaches these problems. Pay close attention to the reasoning going on in step 4.

1) Multiply equation (2) by 3 and designate as equation (4):

6H _{2}+ 3O_{2}---> 6H_{2}OΔH = −1436.52 kJ (4)

2) Multiply equation (3) by 2 and designate as equation (5):

4CO + 2O _{2}---> 4CO_{2}ΔH = −1131.96 kJ (5)

3) Add equation (4) and equation (5) and designate as equation (6):

6H _{2}+ 4CO + 5O_{2}---> 6H_{2}O + 4CO_{2}ΔH = −2568.48 kJ (6)

4) Subtract equation (6) from equation (1) and designate as equation (7):

2C _{2}H_{6}+ 7O_{2}− (6H_{2}+ 4CO + 5O_{2}) ---> 4CO_{2}+ 6H_{2}O − (6H_{2}O + 4CO_{2})ΔH = −551.22 kJ 2C _{2}H_{6}+ 2O_{2}− 6H_{2}− 4CO ---> 4CO_{2}− 4CO_{2}+ 6H_{2}O − 6H_{2}OΔH = −551.22 kJ 2C _{2}H_{6}+ 2O_{2}---> 6H_{2}+ 4COΔH = −551.22 kJ (7)

5) Dividing equation (7) by 2:

C_{2}H_{6}+ O_{2}---> 3H_{2}+ 2CO ΔH = −275.61 kJ