### Hess' Law of Constant Heat SummationUsing two equations and their enthalpiesProblems 11 - 25

Problem #11: Use the standard reaction enthalpies given below to determine ΔH° for the following reaction:

2S(s) + 3O2(g) ---> 2SO3(g)

Given:

 SO2(g) ---> S(s) + O2(g) ΔH° = +296.8 kJ 2SO2(g) + O2(g) ---> 2SO3(g) ΔH° = −197.8 kJ

Solution:

1) The two data equations are modified:

 S(s) + 2O2(g) ---> 2SO2(g) ΔH° = −593.6 kJ <--- flipped and mult. by 2 2SO2(g) + O2(g) ---> 2SO3(g) ΔH° = −197.8 kJ <--- no change

2) The 2SO2 will cancel out when the equations are added. Add the enthalpies for the answer:

−593.6 + (−197.8) = −791.4 kJ

Problem #12: Calculate ΔH° for:

2C2H4(g) + H2O(ℓ) ---> C4H9OH(ℓ)

Using:

 2CO2 + 2H2O(ℓ) ---> C2H4(g) + 3O2(g) ΔH° = +1411.1 kJ C4H9OH(ℓ) + 6O2(g) ---> 4CO2 + 5H2O(ℓ) ΔH° = −1534.7 kJ

Solution:

1) Both data equations need to be flipped, but let's not flip them yet:

 4CO2 + 4H2O(ℓ) ---> 2C2H4(g) + 6O2(g) ΔH° = +2822.2 kJ C4H9OH(ℓ) + 6O2(g) ---> 4CO2 + 5H2O(ℓ) ΔH° = −1534.7 kJ

I multiplied the first equation by 2.

2) Add the two data equations and their enthalpies:

 C4H9OH(ℓ) ---> 2C2H4(g) + H2O(ℓ) ΔH° = +1287.5 kJ

The reaction just above can now be flipped to give us our target equation and the enthalpy of the flipped reaction is −1287.5 kJ.

Problem #13: For the following reaction:

2CO(g) + 2NO(g) ---> 2CO2(g) + N2(g)

Use reactions (a) and (b) to determine ΔH

 (a) 2CO(g) + O2(g) ---> 2CO2(g) ΔH = −566.0 kJ (b) N2(g) + O2(g) ---> 2NO(g) ΔH = 180.6 kJ

Solution:

1) Flip (b) and leave (a) alone:

 (a) 2CO(g) + O2(g) ---> 2CO2(g) ΔH = −566.0 kJ (b) 2NO(g) ---> N2(g) + O2(g) ΔH = −180.6 kJ

2) When you add the two reactions, the O2(g) cancels. Add the two enthalpies for the final answer:

−746.6 kJ

Problem #14: Determine the enthalpy of reaction for the combustion of methane to carbon monoxide:

2CH4(g) + 3O2(g) ---> 2CO(g) + 4H2O(ℓ)

Use the following:

 CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ) ΔH° = −890.0 kJ 2CO(g) + O2(g) ---> 2CO2(g) ΔH° = −566.0 kJ

Solution:

1) Here are the modified equations:

 2CH4(g) + 4O2(g) ---> 2CO2(g) + 4H2O(ℓ) ΔH° = −1780.0 kJ 2CO2(g) ---> 2CO(g) + O2(g) ΔH° = +566.0 kJ

Multiplied the first by 2 and flipped the second.

−1214 kJ

Problem #15: Elemental sulfur occurs in several forms, with rhombic sulfur the most stable under normal conditions and monoclinic sulfur somewhat less stable. The standard enthalpies of combustion of the two forms to sulfur dioxide are −296.83 and −297.16 kJ/mol, respectively. Calculate the change in enthalpy for the rhombic to monoclinic transition.

Solution:

1) Using the information offer, write two combustion equations:

 S(s, rhom) + O2(g) ---> SO2(g) ΔH = −296.83 kJ S(s, mono) + O2(g) ---> SO2(g) ΔH = −297.16 kJ

By the way, the first equation is a formation reaction since sulfur's standard state is rhombic. The second equation, therefore, is not a formation reaction.

2) The equation we want is this:

S(s, rhom) ---> S(s, mono)

3) Reversing equation 2 will get us what we want. Change the sign on the second enthalpy and add:

−296.83 kJ + 297.16 kJ = +0.33 kJ

Problem #16: Given the following information:

 2H2(g) + O2(g) ---> 2H2O(g) ΔH = −483.6 kJ 3O2(g) ---> 2O3(g) ΔH = +285.4 kJ

Determine the ΔH of this reaction:

3H2(g) + O3(g) ---> 3H2O(g)

Solution:

1) Do the following:

a) Multiply first reaction by 32
b) Multiply second reaction by 12 and flip it.

2) The result:

 3H2(g) + 3⁄2O2(g) ---> 3H2O(g) ΔH = −725.4 kJ O3(g) ---> 3⁄2O2(g) ΔH = −142.7 kJ

3) Add the two reactions (the 32O2 cancels out). Add the two enthalpies for the final answer:

−868.1 kJ

Problem #17: Calculate the value of heat of reaction for the equation:

3Fe2O3 ---> 2Fe3O4 + 12O2

given:

 2Fe + 3⁄2O2 ---> Fe2O3 ΔH = −824.2 kJ 3Fe + 2O2 ---> Fe3O4 ΔH = −1118.4 kJ

Solution:

flip first reaction and multiply by three
multiply second reaction by two

Here's the result:

 3Fe2O3 ---> 6Fe + 9⁄2O2 ΔH = +2472.6 kJ 6Fe + 4O2 ---> 2Fe3O4 ΔH = −2236.8 kJ

Add the two reactions together. Note that 6Fe cancels. Also note that 4O2 is 82O2, so only 12O2 results. Add +2472.6 and −2236.8 to get the kJ for the reaction.

Problem #18: Calculate the enthalpy of the following reaction:

N2 + O2 ---> 2NO

Given:

 4NH3 + 5O2 ---> 4NO + 6H2O ΔH° = −1170 kJ 2N2 + 6H2O ---> 4NH3 + 3O2 ΔH° = +1530 kJ

Solution:

1) The reactants and the product are in the correct relative places. Add the two data equations together to obtain this:

 2N2 + 2O2 ---> 4NO ΔH° = +360 kJ

2) Divide through by 2 for the final answer:

 N2 + O2 ---> 2NO ΔH° = +180 kJ

Note: you could have divided both data equations by 2 and then added. The result would have been the same.

Problem #19:

Determine the enthalpy for this reaction:

2NOCl(g) ---> N2(g) + O2(g) + Cl2(g)

given the following reactions with known ΔH values:

 1⁄2N2(g) + 1⁄2O2(g) ---> NO(g) ΔH = +90.3 kJ NO(g) + 1⁄2Cl2(g) ---> NOCl(g) ΔH = −38.6kJ

Solution:

1) Both equations need to be reversed. Both equations need to be multiplied by 2. We can actually do those actions after adding the equations together. When we add the equations, we get:

 1⁄2N2(g) + 1⁄2O2(g) + 1⁄2Cl2(g) ---> NOCl(g) ΔH = +51.7 kJ

2) Reversing the equation and multiplying by 2 yields this enthalpy:

-103.4 kJ

Problem #20: Determine ΔH for this reaction:

PCl5(s) ---> PCl3(g) + Cl2(g)

given the following reactions and their enthapies of reaction:

 P4(s) + 6Cl2(g) ---> 4PCl3(g) ΔH = −2439 kJ 4PCl5(s) ---> P4(s) + 10Cl2(g) ΔH = +3438 kJ

Solution:

1) The two reactions can be added together without any modifications. The result is:

4PCl5(s) ---> 4PCl3(g) + 4Cl2(g)   ΔH = 999 kJ

2) Divide through by 4 to obtain the final answer:

PCl5(s) ---> PCl3(g) + Cl2(g)   ΔH = 249.8 kJ

Problem #21: The following two reactions are known:

 N2(g) + O2(g) ---> 2NO(g) ΔH = +180.50 kJ 2NO2(g) ---> N2(g) + 2O2(g) ΔH = −66.36 kJ

Determine the ΔH value for the reaction below:

2NO(g) + O2(g) ---> 2NO2(g)

Solution:

1) Make these modifications:

a) first equation ---> flip it
b) second equation ---> flip it

2) The result:

 2NO(g) ---> N2(g) + O2(g) ΔH = −180.50 kJ N2(g) + 2O2(g) ---> 2NO2(g) ΔH = +66.36 kJ

3) Add the two modified equations and . . .

. . . N2(g) will cancel as well as one O2(g) when the two equations are added.

ΔH = −180.50 + (+66.36) = −114.14 kJ

Problem #22: The following two reactions are known:

 N2(g) + O2(g) ---> 2NO(g) ΔH = +180.50 kJ N2(g) + 2O2(g) ---> 2NO2(g) ΔH = +66.36 kJ

Calculate for enthalpy change for:

NO2(g) ---> NO(g) + 12O2(g)

Solution:

1) Make these modifications:

Divide first equation by 2
Flip second equation and divide by 2

2) The result:

 1⁄2N2(g) + 1⁄2O2(g) ---> NO(g) ΔH = +90.25 kJ NO2(g) ---> 1⁄2N2(g) + O2(g) ΔH = −33.18 kJ

3) Add the two modified equations and . . .

. . . 12N2(g) will cancel as well as 12O2(g) when the two equations are added.

ΔH = −180.50 + (+66.36) = +57.07 kJ

Problem #23: Use Hess's law to calculate ΔH for the following reaction:

N2(g) + 2O2(g) ---> 2NO2(g)

using the following three data equations:

 N2(g) + O2(g) ---> 2NO(g) ΔH = +180.7 kJ 2NO(g) + O2(g) ---> 2NO2(g) ΔH = −113.1 kJ 2N2O(g) ---> 2N2(g) + O2(g) ΔH = −163.2 kJ

Solution:

1) Add the following two equations:

 N2(g) + O2(g) ---> 2NO(g) ΔH = +180.7 kJ 2NO(g) + O2(g) ---> 2 NO2(g) ΔH = −113.1 kJ

2) And get this result:

N2(g) + 2O2(g) ---> 2NO2(g) ΔH = +67.6 kJ

3) You do not need the third equation because N2O is not involved in the target equation and cannot be cancelled out with either of the first two equations, It is only present as a distractor and to make you worry because you're not using it in getting the answer.

4) Notice also that there are slightly different ΔH values used for the same data equation. This is because many enthalpy values have been determined more than once by experiment and the exact value used in the problem depends on which compilation of values the question writer (usually not the ChemTeam!) used.

Problem #24: Given the following reactions:

 N2(g) + 2O2(g) ---> 2NO2(g) ΔH = +66.36 kJ 2NO(g) + O2(g) ---> 2NO2(g) ΔH = −114.2 kJ

Determine the enthalpy of the reaction of nitrogen and oxygen to produce nitric oxide

N2(g) + O2(g) ---> 2NO(g)

Solution:

1) Manipulate the data equation as follows:

 N2(g) + 2O2(g) ---> 2NO2(g) ΔH = +66.36 kJ <--- leave unchanged 2NO2(g) ---> 2NO(g) + O2(g) ΔH = +114.2 kJ <--- flipped equation, changed sign on ΔH

2) After you flip the second equation, you will add the equations together. Notice that the 2NO2 cancels out as well as one of the O2. What remains is your target equation.

3) Add the two enthalpies for the answer. Make sure to use the changed enthalpy, the one associated with the flipped second equation. The answer is +180.6 kJ.

Problem #25: Calculate ΔHrxn for the following reaction:

CaO(s) + CO2(g) ---> CaCO3(s)
Use the following reactions and given ΔH values:
 Ca(s) + CO2(g) + 1⁄2O2(g) ---> CaCO3(s) ΔH = −812.8 kJ 2Ca(s) + O2(g) ---> 2CaO(s) ΔH = −1269.8 kJ

Solution:

1) Look at the first data equation. It has one CaCO3 as a product. That's what we want, so leave the equation alone.

2) Look at the second data equation. It has CaO as a product and we want it as a reactant, so flip the equation. Also, it has 2Ca and we want one, not two. Divide the equation by 2.

3) The result:

 Ca(s) + CO2(g) + 1⁄2O2(g) ---> CaCO3(s) ΔH = −812.8 kJ CaO(s) ---> Ca(s) + 1⁄2O2(g) ΔH = +634.9 kJ

Notice how the sign on the ΔH changed and it was divided by 2.

4) When you add the two equations, the Ca and the 32O2 will cancel, leaving the desired target equation. Add the two enthalpies for the answer.