Using three equations and their enthalpies

Problems 1 - 10

**Problem #1:** The heats of combustion of C, H_{2} and CH_{4} at 298 K and 1 atm are respectively -393 kJ/mol, -286 kJ/mol and -892 kJ/mol. What is the enthalpy of formation for CH_{4}?

**Solution:**

1) The three combustion reactions are:

C + O _{2}---> CO_{2}ΔH = -393 kJ H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = -286 kJ CH _{4}+ 2O_{2}---> CO_{2}+ 2H_{2}OΔH = -892 kJ

2) The reaction we're looking for is:

C + 2H_{2}---> CH_{4}

3) This is what the answerer on Yahoo Answers wrote:

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) + 2(-286) + (-1)(-892). I'll let you finish it, the critical thing is understanding where the 1, 2, and -1 came from.

**Problem #2:** The standard heat of combustion of benzene is -3271 kJ/mol, for CO_{2} it is -394 kJ/mol, and for H_{2}O, 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 (CO_{2} produced when C combusts and H_{2}O produced when H_{2} 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) + 3H_{2}(g) ---> C_{6}H_{6}(ℓ)

2) Set up the three combustion equations

C _{6}H_{6}+^{15}⁄_{2}O_{2}---> 6CO_{2}+ 3H_{2}OΔH = -3271 kJ C + O _{2}---> CO_{2}ΔH = -394 kJ H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = -286 kJ

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

6CO _{2}+ 3H_{2}O ---> C_{6}H_{6}+^{15}⁄_{2}O_{2}ΔH = +3271 kJ 6C + 6O _{2}---> 6CO_{2}ΔH = -2364 kJ 3H _{2}+^{3}⁄_{2}O_{2}---> 3H_{2}OΔH = -858 kJ

4) Add the three equations:

3271 + (-2364) + (-858) = +49 kJ/mol

You may check the answer here.

**Solution #2:**

1) Write the equation for the combustion of benzene:

C _{6}H_{6}+^{15}⁄_{2}O_{2}---> 6CO_{2}+ 3H_{2}OΔH = -3271 kJ

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

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

3) Inserting values and solving, we have:

-3271 = [(6) (-394) + (3) (-286)] - [(1) (x) + (15/2) (0)]The solution to the above depends on the fact that the combustion reactions for C and Hx = 49 kJ/mol

From Yahoo Answers, here is a problem like the above, one that uses ethanol. Take a look at the answer by HPV. His answer is correct and you can check it here.

**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:

2CH_{4}(g) ---> C_{2}H_{6}(g) + H_{2}(g)

**Solution:**

1) The data given are these three reactions:

H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = -285.8 kJ CH _{4}+ 2O_{2}---> CO_{2}+ 2H_{2}OΔH = -890.4 kJ C _{2}H_{6}+^{7}⁄_{2}O_{2}---> 2CO_{2}+ 3H_{2}OΔH = -1559.9 kJ

2) Manipulate the data equations as follows:

a) flip (puts hydrogen as a product)

b) multiply by 2 (gives us 2CH_{4}on the reactant side)

c) flip (puts C_{2}H_{6}as a product)

3) You then obtain these data equations:

H _{2}O ---> H_{2}+^{1}⁄_{2}O_{2}ΔH = +285.8 kJ 2CH _{4}+ 4O_{2}---> 2CO_{2}+ 4H_{2}OΔH = -1780.8 kJ 2CO _{2}+ 3H_{2}O ---> C_{2}H_{6}+^{7}⁄_{2}O_{2}ΔH = +1559.9 kJ

4) When you add the three equations above, the 4O_{2} will cancel as will the 2CO_{2} and the 4H_{2}O. 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) + CO_{2}(g)

given the following information:

3Fe _{2}O_{3}(s) + CO(g) ---> 2Fe_{3}O_{4}(s) + CO_{2}(g)ΔH = -48.26 kJ Fe _{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g)ΔH = -23.44 kJ Fe _{3}O_{4}(s) + CO(g) ---> 3FeO(s) + CO_{2}(g)ΔH = +21.79 kJ

**Solution:**

1) Changes to be made to the data equations:

a) reverse equation 1 (this puts Fe_{3}O_{4}on opposite side to compensate for switching equation 3)

b) multiply equation 2 by 3 (this will give 3Fe_{2}O_{3}, allowing it to cancel)

c) reverse equation 3 and multiply it by two (this puts FeO on the reactant side and gives us 2Fe_{3}O_{4}to cancel)Please note that no attention was paid to CO and CO

_{2}. If everything else is done correctly, they should fall into line.

2) The three data equations with the changes applied:

2Fe _{3}O_{4}(s) + CO_{2}(g) ---> 3Fe_{2}O_{3}(s) + CO(g)ΔH = +48.26 kJ 3Fe _{2}O_{3}(s) + 9CO(g) ---> 6Fe(s) + 9CO_{2}(g)ΔH = -70.32 kJ 6FeO(s) + 2CO _{2}(g) ---> 2Fe_{3}O_{4}(s) + 2CO(g)ΔH = -43.58 kJ

3) Adding the three equations together gives:

6FeO(s) + 6CO(g) ---> 6Fe(s) + 6CO_{2}(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) + CO_{2}(g) ΔH = -10.94 kJ

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

3Fe_{2}O_{3}(s) + CO(g) ---> 2Fe_{3}O_{4}(s) + CO_{2}(g)

given the folowing data:

Fe _{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g)ΔH = -23.44 kJ Fe _{3}O_{4}+ CO(g) ---> 3FeO(s) + CO_{2}(g)ΔH = +21.79 kJ Fe(s) + CO _{2}(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 3Fe_{2}O_{3})

b) flip second equation and multiply by 2 (to put 2Fe_{3}O_{4}on the product side)

c) multiply third equation by 6 (to cancel Fe and FeO)Note that I have ignored the CO and CO

_{2}. If everything works out, the right amounts will be there.

2) The result:

3Fe _{2}O_{3}(s) + 9CO(g) ---> 6Fe(s) + 9CO_{2}(g)ΔH = -70.32 kJ 6FeO(s) + 2CO _{2}(g) ---> 2Fe_{3}O_{4}+ 2CO(g)ΔH = -43.58 kJ 6Fe(s) + 6CO _{2}(g) ---> 6FeO(s) + 6CO(g)ΔH = -65.64 kJ

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

3Fe_{2}O_{3}(s) + CO(g) ---> 2Fe_{3}O_{4}(s) + CO_{2}(g) ΔH = -179.54 kJ

Comment: I saw this problem on Yahoo Answers, but it had enthalpy values which were not the correct values (which are the values I used). 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) 3Fe _{2}O_{3}(s) + CO(g) ---> 2Fe_{3}O_{4}(s) + CO_{2}(g)ΔH = -48.26 kJ (b) Fe(s) + CO _{2}(g) ---> FeO(s) + CO(g)ΔH = +10.94 kJ (c) Fe _{3}O_{4}(s) + CO(g) ---> 3FeO(s) + CO_{2}(g)ΔH = +21.79 kJ

**Solution:**

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

Fe_{2}O_{3}+ 3CO ---> 2Fe + 3CO_{2}

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) 3Fe _{2}O_{3}(s) + CO(g) ---> 2Fe_{3}O_{4}(s) + CO_{2}(g)ΔH = -48.26 kJ (b) 6FeO(s) + 6CO(g) ---> 6Fe(s) + 6CO _{2}(g)ΔH = -65.64 kJ (c) 2Fe _{3}O_{4}(s) + 2CO(g) ---> 6FeO(s) + 2CO_{2}(g)ΔH = +43.58 kJ

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

3Fe_{2}O_{3}(s) + 9CO(g) ---> 6Fe(s) + 9CO_{2}(g)and the ΔH is

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

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

Fe_{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g) ΔH = -23.44 kJ

**Problem #7:** From the following data:

N_{2}+^{3}⁄_{2}O_{2}---> N_{2}O_{3}ΔH = 83.7 kJN

_{2}+ O_{2}---> 2NO ΔH = 180.4 kJ

^{1}⁄_{2}N_{2}+ O_{2}---> NO_{2}ΔH = 33.2 kJ

Calculate the enthalpy change for the reaction:

N_{2}O_{3}---> NO + NO_{2}

**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 N_{2}O_{3}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 NO

_{2}on the product side and it's the right coefficient.

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

N_{2}O_{3}---> N_{2}+^{3}⁄_{2}O_{2}ΔH = -83.7 kJ <--- note sign change on ΔH

^{1}⁄_{2}N_{2}+^{1}⁄_{2}O_{2}---> NO ΔH = 90.2 kJ <--- note ΔH divided by 2

^{1}⁄_{2}N_{2}+ O_{2}---> NO_{2}ΔH = 33.2 kJ

3) When you add the three above equations, you will recover the target equation. The N_{2} will cancel as well as the ^{3}⁄_{2}O_{2}. Add the three enthalpies for the answer.

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

using the following three data equations:^{1}⁄_{2}N_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> NO(g)

N _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH = -91.8 kJ 4NH _{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(g)ΔH = -906.2 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(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 4NH

_{3}which will cancel with the 4NH_{3}in the second equation.Second equation: leave it alone

Third equation: flip it and multiply by 6. That gets H

_{2}O on the left-hand side and the 6 will cancel the 6H_{2}O in the second equation.Notice that I'm not paying any attention to N

_{2}or O_{2}. That's because, if I do everything else right, they will come out correctly as well. Here are the equations with the changes:

2N _{2}(g) + 6H_{2}(g) ---> 4NH_{3}(g)ΔH = -183.6 kJ 4NH _{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(g)ΔH = -906.2 kJ 6H _{2}O(g) ---> 6H_{2}(g) + 3O_{2}(g)ΔH = +1450.8kJ When you add the three equations together, the 6H

_{2}cancels as well as the 4NH_{3}and 3O_{2}(of the 5O_{2}). Here is what results:

2N _{2}(g) + 2O_{2}(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!

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

NO + O ---> NOgiven the following reactions_{2}

O _{2}---> 2OΔH = 495 kJ 2O _{3}---> 3O_{2}ΔH = -427 kJ NO + O _{3}---> NO_{2}+ O_{2}ΔH = -199 kJ

**Solution:**

O ---> ^{1}⁄_{2}O_{2}ΔH = -247.5 kJ (flipped, gets O as a reactant, divide by 2, gets one O) ^{3}⁄_{2}O_{2}---> O_{3}ΔH = +213.5 kJ (flipped, divide by 2, sets up for O _{3}to cancel)NO + O _{3}---> NO_{2}+ O_{2}ΔH = -199 kJ

Note that ^{3}⁄_{2}O_{2} also cancels when the above three equations are added. Add the three enthalpies for the answer.

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

NGiven the following equations_{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)

N _{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)ΔH = -115 kJ 2NH _{3}(g) + 4H_{2}O(ℓ) ---> 2NO_{2}(g) + 7H_{2}(g)ΔH = -142.5 kJ H _{2}O(ℓ) ---> H_{2}(g) +^{1}⁄_{2}O_{2}(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 H_{2}will cancel as will the NH_{3}and H_{2}O. The final answer is -82.7 kJ.

**Bonus Problem:** The standard molar enthalpy of formation, ΔH_{f}° , 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 (B_{2}H_{6}) using the following thermochemical information:

a) 4B(s) + 3O _{2}(g) ---> 2 B_{2}O_{3}(s)ΔH° = -2509.1 kJ b) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)ΔH° = -571.7 kJ c) B _{2}H_{6}(g) + 3O_{2}(g) ----> B_{2}O_{3}(s) + 3H_{2}O(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 + 3H_{2}---> B_{2}H_{6}

Remember that formation means forming one mole of our target substance. This means that a one MUST be in front of the B_{2}H_{6}

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^{3}⁄_{2}

equation (c) - flip

Why?

equation (a) - this gives us 2B (from 4B) for our final equation

equation (b) - this gives us 3H_{2}for our final equation

equation (c) - this puts B_{2}H_{6}on the right-hand side of the final equation

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

a) 2B(s) + ^{3}⁄_{2}O_{2}(g) ---> B_{2}O_{3}(s)ΔH° = -1254.55 kJ b) 3H _{2}(g) +^{3}⁄_{2}O_{2}(g) ---> 3H_{2}O(ℓ)ΔH° = -857.55 kJ c) B _{2}O_{3}(s) + 3H_{2}O(ℓ) ---> B_{2}H_{6}(g) + 3O_{2}(g)ΔH° = +2147.5 kJ

Add the three equations and the ΔH° values:

2B + 3H _{2}---> B_{2}H_{6}ΔH° = +35.4 kJ

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

a) 4B(s) + 3O _{2}(g) ---> 2B_{2}O_{3}(s)ΔH° = -2543.8 kJ b) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(g)ΔH° = -484.0 kJ c) B _{2}H_{6}(g) + 3O_{2}(g) ----> B_{2}O_{3}(s) + 3H_{2}O(ℓ)ΔH° = -2032.9 kJ

Calculate the standard enthalpy of formation of gaseous diborane (B_{2}H_{6}).

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.