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

Problem #1: Consider the following reaction:

 N2H4(ℓ) + O2(g) ---> N2(g) + 2H2O(ℓ) ΔH = −622.2 kJ

Given the following data, calculate the heat of reaction for the same reaction where water is a gaseous product instead of a liquid:

$\text{ΔH}{\text{}}_{f}^{o}$ for H2O(g) = −285.83 kJ/mol
$\text{ΔH}{\text{}}_{f}^{o}$ for H2O(ℓ) = −241.83 kJ/mol

Solution:

1) We need to find ΔH for this reaction:

H2O(ℓ) ---> H2O(g)

ΔH = −285.83 − (−241.83) = −44 kJ

2) Now, we use the two reactions we have:

 N2H4(ℓ) + O2(g) ---> N2(g) + 2H2O(ℓ) ΔH = −622.2 kJ H2O(ℓ) ---> H2O(g) ΔH = −44 kJ

3) Multiply second equation by 2:

 N2H4(ℓ) + O2(g) ---> N2(g) + 2H2O(ℓ) ΔH = −622.2 kJ 2H2O(ℓ) ---> 2H2O(g) ΔH = −88 kJ

4) Add the two reactions (and eliminate 2H2O(ℓ)) and add the enthalpies to get the final answer:

ΔH = −622.2 kJ + −88 kJ = −710.2 kJ

Problem #2: When one mole of sulfur burns to form SO2, 297.0 kilojoules are released. When one mole of sulfur burns to form SO3, 395.8 kilojoules are released. What is the ΔH when one mole of SO2 is burned to form SO3?

Solution:

1) Let's write out our information in a chemical way:

 S + O2 ---> SO2 ΔH = −297.0 kJ S + 3⁄2O2 ---> SO3 ΔH = −395.8 kJ

These are actually formation reactions. I'll ignore that since it's not germane to the problem.

2) The target equation we want is this:

SO2 + 12O2 ---> SO3

3) To obtain the target equation, what I will do is flip the first equation, remembering to also change the sign on the enthalpy. Here are the equations with the first one flipped:

 SO2 ---> S + O2 ΔH = +297.0 kJ S + 3⁄2O2 ---> SO3 ΔH = −395.8 kJ

4) When those two equations are added, the S and an O2 will cancel out:

SO2 + 12O2 ---> SO3

+297.0 + (−395.8) = −98.8 kJ

Problem #3: Calculate the enthalpy of formation for sulfur dioxide, SO2. Use the following information:

 S(s) + 3⁄2O2(g) ----> SO3(g) ΔH = −395.8 kJ 2SO2(g) + O2(g) ---> 2SO3(g) ΔH = −198.2 kJ

Solution:

1) This is the formation reaction for SO2:

S(s) + O2(g) ---> SO2(g)

2) What should be done to the two data equations:

eq 1 ---> nothing
eq 2 ---> reverse it and divide by 2 (this puts one SO2 on the product side, which is where we want it)

3) The result:

 S(s) + 3⁄2O2(g) ---> SO3(g) ΔH = −395.8 kJ SO3(g) ---> SO2(g) + 1⁄2O2(g) ΔH = +99.1 kJ

When the two equations are added, the SO3 and 12O2 will cancel out, leaving the desired equation. Add the two enthalpies for the answer:

 S(s) + O2(g) ---> SO2(g) ΔH = −296.7 kJ

5) Comment: try the exact same problem as above, but with slightly different data equations:

 2SO3(g) ---> 2S(s) + 3O2(g) ΔH = +791.6 kJ 2SO2(g) + O2(g) ---> 2SO3(g) ΔH = −198.2 kJ

Problem #4: Given the following:

 I2(g) + 3Cl2(g) ---> 2ICl3(s) ΔH = −214 kJ I2(s) ----> I2(g) ΔH = +38 kJ

Calculate the enthalpy of formation for ICl3.

Solution #1:

1) The target equation is this:

12I2(s) + 32Cl2(g) ---> ICl3(s)

2) What needs to be done to the two data equations

eq 1 ---> divide by 2
eq 2 ---> divide by 2

3) The result:

 1⁄2I2(g) + 3⁄2Cl2(g) ---> ICl3(s) ΔH = −107 kJ 1⁄2I2(s) ----> 1⁄2I2(g) ΔH = +19 kJ

When you add the two equations just above, the 12I2(g) will cancel out, leaving only the desired target equation.

The enthalpy of the target equation is this:

$\text{ΔH}{\text{}}_{f}^{o}$ = −88 kJ

Solution #2:

Keeping mind that the enthalpy of formation is always for 1 mole of the product and in standard states, can simply add both the data equation's enthalpies:
−214 + 38 = −176

Since this is for 2 moles of ICl3, diving by 2 yields the answer of −88 kJ.

Problem #5: Given the follow two reactions:

 X2 + 5Y2 ---> 2XY5 ΔH1 3X2 + Z2 ---> 2X3Z ΔH2

Calculate ΔHrxn for the following reaction:

15Y2 + 2X3Z ---> 6XY5 + Z2

Solution:

Comment: A completely made-up problem? Doesn't matter. The technique for solving is the same.

1) Changes to be applied:

eq 1 ---> multiply by 3 in order to get 15Y2
eq 2 ---> flip in order to get X3Z onto the reactant side

2) The result:

 3X2 + 15Y2 ---> 6XY5 3ΔH1 2X3Z ---> 3X2 + Z2 −ΔH2

3) Solve for the enthalpy of the desired reaction:

When the two equations above are added, the 3X2 will cancel out and the enthalpy is this:

ΔHrxn = 3ΔH1 + (−ΔH2)

Problem #6: The following two reactions are known:

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

Determine the ΔH value for the reaction below:

Fe2O3(s) + CO(g) ---> 2FeO(s) + CO2(g)

Solution:

1) Manipulate the two data equations as follows:

a) leave eq 1 untouched (keeps Fe2O3 as a reactant)
b) flip eq 2 and multiply by two (puts 2FeO on the product side)

If those are the correct steps, the placement and amounts of CO and CO2 will follow automatically.

2) The result of the changes in step one:

 Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) ΔH = −23.44 kJ 2Fe(s) + 2CO2(g) ---> 2FeO(s) + 2CO(g) ΔH = +21.88 kJ

3) Add the two equations together and . . .

. . . two of the CO, two of the CO2 and the two Fe will all cancel out.

Problem #7: Determine the enthalpy for this reaction:

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

using this information:

 2Fe(s) + 3⁄2O2(g) ---> Fe2O3(s) ΔH = −824.2 kJ CO(g) + 1⁄2O2(g) ---> CO2(g) ΔH = −282.7 kJ

Solution:

1) Do this:

a) reverse eq 1 (to put Fe2O3 on the reactant side)
b) multiply eq 2 by three (this gets us 3CO and 3CO2)

2) The results of the changes:

 Fe2O3(s) ---> 2Fe(s) + 3⁄2O2(g) ΔH = +824.2 kJ 3CO(g) + 3⁄2O2(g) ---> 3CO2(g) ΔH = −848.1 kJ

3) Add the two modified equations together and . . .

. . . the 32O2 will cancel, leaving the target equation.

Add the two enthalpies together for the final answer of −23.9 kJ

Problem #8: Calculate the standard enthalpy of reaction for the following reaction:

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

Given:

 4Fe(s) + 3O2(g) ---> 2Fe2O3(s) ΔH = −1648 kJ 2CO2(g) ---> 2CO(g) + O2(g) ΔH = +565.4 kJ

Solution:

1) What needs to be done:

(a) flip first reaction and divide by 2. This gets one Fe2O3 on the reactant side and two Fe on the product side.

(b) flip second reaction and multiply by 1.5. This gets the proper amount and placement for CO and CO2.

2) Here are the results:

 Fe2O3(s) ---> 2Fe(s) + 3⁄2O2(g) ΔH = +824 kJ 3CO(g) + 3⁄2O2(g) ---> 3CO2(g) ΔH = −848.1 kJ
3) Add the two reactions. Note that the 32O2 cancels. Add the two enthalpies:
 Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) ΔH° = −24.1 kJ

Problem #9: Calculate ΔH for the following reaction:

CH4(g) + 2Cl2(g) ---> CH2Cl2(g) + 2HCl(g)

Use the following information:

 CH4(g) + Cl2(g) ---> CH3Cl(g) + HCl(g) ΔH = −99.6 kJ CH3Cl(g) + Cl2(g) ---> CH2Cl2(g) + HCl(g) ΔH = −105.8 kJ

Solution:

Nothing needs to be done to either data equation. When added together, the CH3Cl will cancel out, leaving the desired target equation.

Problem #10: Find the enthalpy change for the formation of phosphorus pentachloride from its elements:

P(s) + 52Cl2(g) ---> PCl5(s)

Use the following thermochemical equations.

 PCl5(s) ---> PCl3(g) + Cl2(g) ΔH = 87.9 kJ 2P(s) + 3Cl2(g) ---> 2PCl3(g) ΔH = −574 kJ

Solution:

1) Reverse the first data equation:

PCl3(g) + Cl2(g) ---> PCl5(s) ΔH = −87.9 kJ

2) Divide the second data equation by 2:

P(s) + 32Cl2(g) ---> PCl3(g) ΔH = −287 kJ

3) Add the two chemical equations from steps 1 and 2:

PCl3 cancels and the enthalpy of the target reaction is −374.9 kJ

Bonus Problem: Given the following two reactions at 298 K and 1 atm pressure:

 N2(g) + O2(g) ---> 2NO(g) ΔH1 NO(g) + 1⁄2O2(g) ---> NO2(g) ΔH2

which of the statements below correctly describes the $\text{ΔH}{\text{}}_{f}^{o}$ for NO2(g)?

(A) ΔH1
(B) ΔH2
(C) ΔH2 + 12ΔH1
(D) ΔH2 - 12ΔH1
(E) ΔH2 + ΔH1

Solution:

1) The target equation we want is the formation reaction for NO2(g):

12N2(g) + O2(g) ---> NO2(g)

2) We get that by doing this:

 1⁄2N2(g) + 1⁄2O2(g) ---> NO(g) 1⁄2ΔH1 NO(g) + 1⁄2O2(g) ---> NO2(g) ΔH2

3) Adding the two above equations yields the target equation. When we add the equations together, we also add the enthalpies together:

12ΔH1 + ΔH2