### Hess' Law of Constant Heat SummationUsing three equations and their enthalpiesProblems 1 - 10

Problem #1: The heats of combustion of C(s, graphite), H2(g) and CH4(g) at 298 K and 1 atm are respectively −393.50 kJ/mol, −285.83 kJ/mol and −890.36 kJ/mol. What is the enthalpy of formation for CH4?

Solution:

1) The three combustion reactions are:

 C(s, gr) + O2(g) ---> CO2(g) ΔH = −393.50 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = −285.83 kJ CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ) ΔH = −890.36 kJ

2) The phrase 'enthalpy of formation' tells us this reaction is called for:

C(s, gr) + 2H2(g) ---> CH4(g)

3) I saw this answer on an "answers" type website and decided to include it:

This is a Hess's Law problem. If you multiply the first reaction by 1, the second by 2, and the third by negative 1 (write it backwards) they add together to give the reaction you're looking for. So, the enthalpy of the reaction you're solving for is equal to 1(−393.50) + 2(−285.83) + (−1)(−890.36). I'll let you finish it, the critical thing is understanding where the 1, 2, and −1 came from.

4) Here is how the ChemTeam usually solves these problems:

 C(s, gr) + O2(g) ---> CO2(g) ΔH = −393.50 kJ <--- equation left untouched 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH = −571.66 kJ <--- note the doubling CO2(g) + 2H2O(ℓ) ---> CH4(g) + 2O2(g) ΔH = +890.36 kJ <--- note the sign change

5) When the three data equations are added together, duplicates on each side of the arrow will be removed. Here's before any removal:

C(s, gr) + O2(g) + 2H2(g) + O2(g) + CO2(g) + 2H2O(ℓ) ---> CO2(g) + 2H2O(ℓ) + CH4(g) + 2O2(g)

6) After 2O2(g), CO2(g), and 2H2O(ℓ) are removed:

C(s, gr) + 2H2(g) ---> CH4(g)

7) The three revised entalpies are added to arrive at the final answer:

C(s, gr) + 2H2(g) ---> CH4(g)     $\text{ΔH}{\text{}}_{f}^{o}$ = −74.80 kJ

Problem #2: The standard heat of combustion of benzene is −3271 kJ/mol, for CO2 it is −394 kJ/mol, and for H2O, it is −286 kJ/mol. Calculate the standard heat of formation of benzene.

(Note that the first bit of data is associated with the reactant (benzene) while the last two are associated with the products (CO2 produced when C combusts and H2O produced when H2 is combusted. The reason for this will be seen in solution #2.)

Solution #1:

1) The term 'standard heat of formation' tells us that this equation is the desired target:

6C(s) + 3H2(g) ---> C6H6(ℓ)

2) Set up the three combustion equations

 C6H6 + 15⁄2O2 ---> 6CO2 + 3H2O ΔH = −3271 kJ C + O2 ---> CO2 ΔH = −394 kJ H2 + 1⁄2O2 ---> H2O ΔH = −286 kJ

3) Flip equation 1, multiply equations 2 and 3:

 6CO2 + 3H2O ---> C6H6 + 15⁄2O2 ΔH = +3271 kJ 6C + 6O2 ---> 6CO2 ΔH = −2364 kJ 3H2 + 3⁄2O2 ---> 3H2O ΔH = −858 kJ

3271 + (−2364) + (−858) = +49 kJ/mol

You may check the answer here.

Solution #2:

1) Write the equation for the combustion of benzene:

 C6H6 + 15⁄2O2 ---> 6CO2 + 3H2O ΔH = −3271 kJ

2) The enthalpy of formation of beneze can be calculated thusly:

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

Where 'comb' stands for combustion.

3) Inserting values and solving, we have:

−3271 = [(6) (−394) + (3) (−286)] − [(1) (x) + (152) (0)]

x = 49 kJ/mol

The solution to the above depends on the fact that the combustion reactions for C and H2 are also the formation reactions for CO2 and H2O, with −394 kJ/mol the enthalpy of formation for CO2 and −286 kJ/mol the enthalpy of formation for H2O.

Problem #3: The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K. Calculate (at the same temperature) the heat of reaction for the following reaction:

2CH4(g) ---> C2H6(g) + H2(g)

Solution:

1) The data given are these three reactions:

 H2 + 1⁄2O2 ---> H2O ΔH = −285.8 kJ CH4 + 2O2 ---> CO2 + 2H2O ΔH = −890.4 kJ C2H6 + 7⁄2O2 ---> 2CO2 + 3H2O ΔH = −1559.9 kJ

2) Manipulate the data equations as follows:

a) flip (puts hydrogen as a product)
b) multiply by 2 (gives us 2CH4 on the reactant side)
c) flip (puts C2H6 as a product)

3) You then obtain these data equations:

 H2O ---> H2 + 1⁄2O2 ΔH = +285.8 kJ 2CH4 + 4O2 ---> 2CO2 + 4H2O ΔH = −1780.8 kJ 2CO2 + 3H2O ---> C2H6 + 7⁄2O2 ΔH = +1559.9 kJ

4) When you add the three equations above, the 4O2 will cancel as will the 2CO2 and the 4H2O. Adding the three enthalpies yields the answer to the problem, +64.9 kJ.

Problem #4: What is the standard enthalpy of reaction for the reduction of iron(II) oxide by carbon monoxide?

FeO(s) + CO(g) ---> Fe(s) + CO2(g)

given the following information:

 3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g) ΔH = −48.26 kJ Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) ΔH = −23.44 kJ Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g) ΔH = +21.79 kJ

Solution:

1) Changes to be made to the data equations:

a) reverse equation 1 (this puts Fe3O4 on opposite side to compensate for switching equation 3)
b) multiply equation 2 by 3 (this will give 3Fe2O3, allowing it to cancel)
c) reverse equation 3 and multiply it by two (this puts FeO on the reactant side and gives us 2Fe3O4 to cancel)

Please note that no attention was paid to CO and CO2. If everything else is done correctly, they should fall into line.

2) The three data equations with the changes applied:

 2Fe3O4(s) + CO2(g) ---> 3Fe2O3(s) + CO(g) ΔH = +48.26 kJ 3Fe2O3(s) + 9CO(g) ---> 6Fe(s) + 9CO2(g) ΔH = −70.32 kJ 6FeO(s) + 2CO2(g) ---> 2Fe3O4(s) + 2CO(g) ΔH = −43.58 kJ

3) Adding the three equations together gives:

6FeO(s) + 6CO(g) ---> 6Fe(s) + 6CO2(g)

and the enthalpy for the above reaction:

+48.26 + (−70.32) + (−43.58) = −65.64 kJ

4) Dividing through by six gives the final answer:

FeO(s) + CO(g) ---> Fe(s) + CO2(g) ΔH = −10.94 kJ

Problem #5: Determine the enthalpy of the following reaction:

3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g)

given the folowing data:

 Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) ΔH = −23.44 kJ Fe3O4 + CO(g) ---> 3FeO(s) + CO2(g) ΔH = +21.79 kJ Fe(s) + CO2(g) ---> FeO(s) + CO(g) ΔH = −10.94 kJ

Solution:

1) Apply the following changes to the data equations:

a) multiply first equation by 3 (to give us 3Fe2O3)
b) flip second equation and multiply by 2 (to put 2Fe3O4 on the product side)
c) multiply third equation by 6 (to cancel Fe and FeO)

Note that I have ignored the CO and CO2. If everything works out, the right amounts will be there.

2) The result:

 3Fe2O3(s) + 9CO(g) ---> 6Fe(s) + 9CO2(g) ΔH = −70.32 kJ 6FeO(s) + 2CO2(g) ---> 2Fe3O4 + 2CO(g) ΔH = −43.58 kJ 6Fe(s) + 6CO2(g) ---> 6FeO(s) + 6CO(g) ΔH = −65.64 kJ

3) Add the three equations and their enthalpies to obtain:

3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g) ΔH = −179.54 kJ

Comment: I saw this problem on an "answers" website, but it had enthalpy values which were not the correct values (I used the correct values). Be aware of this practice (one with which I disagree). It is done to guard against someone finding the solved problem on the Internet with the correct values and just copying out the answer.

Problem #6: Iron metal can be produced in a blast furnace through a complex series of reactions involving reduction of iron(III) oxide with carbon monoxide. The overall reacton is this:

iron(III)oxide + carbon monoxide ---> iron + carbon dioxide

Use the equations below to calculate ΔH for the overall equation.

 (a) 3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g) ΔH = −48.26 kJ (b) Fe(s) + CO2(g) ---> FeO(s) + CO(g) ΔH = +10.94 kJ (c) Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g) ΔH = +21.79 kJ

Solution:

1) Let's get a balanced equation for our target equation:

Fe2O3 + 3CO ---> 2Fe + 3CO2

2) Rearrange the three data equations so that, when added, they give the target equation:

a) leave untouched
b) flip, multiply by 6
c) multiply by 2

3) This results in:

 (a) 3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g) ΔH = −48.26 kJ (b) 6FeO(s) + 6CO(g) ---> 6Fe(s) + 6CO2(g) ΔH = −65.64 kJ (c) 2Fe3O4(s) + 2CO(g) ---> 6FeO(s) + 2CO2(g) ΔH = +43.58 kJ

4) When the three equations are added together, this results in:

3Fe2O3(s) + 9CO(g) ---> 6Fe(s) + 9CO2(g)

and the ΔH is

−48.26 + (−65.64) + 43.58 = −70.32 kJ

5) To get the final answer, divide everything by 3:

Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g)  ΔH = −23.44 kJ

Problem #7: From the following data:

 N2 + 3⁄2O2 ---> N2O3 ΔH = 83.7 kJ N2 + O2 ---> 2NO ΔH = 180.4 kJ 1⁄2N2 + O2 ---> NO2 ΔH = 33.2 kJ

Calculate the enthalpy change for the reaction:

N2O3 ---> NO + NO2

Solution:

1) We modify the data equations based on the idea of reproducing the target equation when the data equations are added together.

a) We need to flip the first data equation. We do this to put N2O3 on the reactant side.

b) We will divide the second data equation by 2. We do this to get NO (which is already a product, it is where we want it) rather than 2NO.

c) The third data equation will be left alone. We have NO2 on the product side and it's the right coefficient.

2) When I apply the above changes, I get this:

 N2O3 ---> N2 + 3⁄2O2 ΔH = −83.7 kJ <--- note sign change on ΔH 1⁄2N2 + 1⁄2O2 ---> NO ΔH = 90.2 kJ <--- note ΔH divided by 2 1⁄2N2 + O2 ---> NO2 ΔH = 33.2 kJ

3) When you add the three above equations, you will recover the target equation. The N2 will cancel as well as the 32O2. Add the three enthalpies for the answer.

Problem #8: Determine the standard enthalpy of formation for NO:

12N2(g) + 12O2(g) ---> NO(g)
using the following three data equations:
 N2(g) + 3H2(g) ---> 2NH3(g) ΔH = −91.8 kJ 4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g) ΔH = −906.2 kJ H2(g) + 1⁄2O2(g) ---> H2O(g) ΔH = −241.8 kJ

Solution:

In the second equation, the 4NO is going to drive what I do to solve this problem. At the very end, I'm going to divide by 4.

First equation: multiply it by 2. That is going to get me 4NH3 which will cancel with the 4NH3 in the second equation.

Second equation: leave it alone

Third equation: flip it and multiply by 6. That gets H2O on the left-hand side and the 6 will cancel the 6H2O in the second equation.

Notice that I'm not paying any attention to N2 or O2. That's because, if I do everything else right, they will come out correctly as well. Here are the equations with the changes:

 2N2(g) + 6H2(g) ---> 4NH3(g) ΔH = −183.6 kJ 4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g) ΔH = −906.2 kJ 6H2O(g) ---> 6H2(g) + 3O2(g) ΔH = +1450.8kJ

When you add the three equations together, the 6H2 cancels as well as the 4NH3 and 3O2 (of the 5O2). Here is what results:

 2N2(g) + 2O2(g) ---> 4NO(g) ΔH = +361 kJ

Last step: divide the equation by 4 to get the target equation. Divide the enthalpy by 4 to obtain +90.25 kJ

Is this the correct answer? Of course it is!

By the way, you could also start by dividing the second equation by 4. I will leave you to contemplate what to do with the first and third data equations after dividing equation 2 by 4.

Problem #9: Use Hess's Law to determine ΔH for the reaction:

NO + O ---> NO2
given the following reactions
 O2 ---> 2O ΔH = +495 kJ 2O3 ---> 3O2 ΔH = −142.3 kJ NO + O3 ---> NO2 + O2 ΔH = −199 kJ

Solution:

 O ---> 1⁄2O2 ΔH = −247.5 kJ (flipped, gets O as a reactant, divide by 2, gets one O) 3⁄2O2 ---> O3 ΔH = +71.15 kJ (flipped, divide by 2, sets up for O3 to cancel) NO + O3 ---> NO2 + O2 ΔH = −199 kJ

Note that 32O2 cancels when the above three equations are added.

−247.5 + +71.15 + (−199) = −375.35 kJ

Rounding off to −375 kJ seems like the best choice for the final answer.

Problem #10: Using Hess' Law, determine the ΔH of the following reaction:

N2(g) + 2O2(g) ---> 2NO2(g)
Given the following equations
 N2(g) + 3H2(g) ---> 2NH3(g) ΔH = −115 kJ 2NH3(g) + 4H2O(ℓ) ---> 2NO2(g) + 7H2(g) ΔH = −142.5 kJ H2O(ℓ) ---> H2(g) + 1⁄2O2(g) ΔH = −43.7 kJ

Solution:

Use the first two data equations without modifying them. For the third data equation, reverse it and multiply it by four. The H2 will cancel as will the NH3 and H2O. The final answer is −82.7 kJ.

Bonus Problem: The standard molar enthalpy of formation, $\text{ΔH}{\text{}}_{f}^{o}$ , of diborane cannot be determined directly because the compound cannot be prepared by reaction of boron and hydrogen. However, the value can be calculated. Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical information:

 (a) 4B(s) + 3O2(g) ---> 2B2O3(s) ΔH° = −2509.1 kJ (b) 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH° = −571.7 kJ (c) B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(ℓ) ΔH° = −2147.5 kJ

Solution:

1) An important key is to know what equation we are aiming for. The answer is in the word 'formation:'

2B + 3H2 ---> B2H6

Remember that formation means forming one mole of our target substance. This means that a one MUST be in front of the B2H6

2) In order to get to our formation reaction, the following must happen to equations (a), (b) and (c):

equation (a) - divide through by two
equation (b) - multiply through by 32
equation (c) - flip

3) Why?

equation (a) - this gives us 2B (from 4B) for our final equation
equation (b) - this gives us 3H2 for our final equation
equation (c) - this puts B2H6 on the right-hand side of the final equation

4) The above manipulations have consequences for the coefficients AND the ΔH° values. Rewrite equations (a), (b) and (c):

 a) 2B(s) + 3⁄2O2(g) ---> B2O3(s) ΔH° = −1254.55 kJ b) 3H2(g) + 3⁄2O2(g) ---> 3H2O(ℓ) ΔH° = −857.55 kJ c) B2O3(s) + 3H2O(ℓ) ---> B2H6(g) + 3O2(g) ΔH° = +2147.5 kJ

5) Add the three equations and the ΔH° values to get:

 2B + 3H2 ---> B2H6 $\text{ΔH}{\text{}}_{f}^{o}$ = +35.4 kJ

Here is the same diborane question, with slightly different numbers:

 a) 4B(s) + 3O2(g) ---> 2B2O3(s) ΔH° = −2543.8 kJ b) 2H2(g) + O2(g) ---> 2H2O(g) ΔH° = −484.0 kJ c) B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(ℓ) ΔH° = −2032.9 kJ

Calculate the standard enthalpy of formation of gaseous diborane (B2H6).

Just remember:

With all Hess's Law (of heat summation) problems, the chemical reactions given must add up to the final chemical equation. The key to these problems is that whatever you do to the reaction equation, you must do to the ΔH value. So, for example, if you reverse the equation, you must reverse the sign of ΔH. If you multiply the equation by 2, then you must multiply the ΔH value by 2.