Hess' Law of Constant Heat Summation
Using two equations and their enthalpies - Problems 11 - 20

Hess' Law - using two equations and their enthalpies - Problem 1 - 10      Hess' Law - using three equations and their enthalpies      Hess' Law - using standard enthalpies of formation
Hess' Law - using two equations and their enthalpies      Hess' Law - using four or more equations and their enthalpies      Hess' Law - using bond enthalpies
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Problem #11: Determine the enthalpy for the following reaction:

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

Using the following data:

2S(s) + 3O2(g) ---> 2SO3(g)ΔH = -790.4kJ
2SO2(g) + O2(g) ---> 2SO3(g)ΔH = -198.2kJ

Solution:

1) Divide first equation by two to get:

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

2) Flip second equation and divide by 2:

SO3(g) ---> SO2(g) + (1/2)O2(g)

3) For the ΔH values, remember to change signs and divide, where necessary. The enthalpy of the target equation is -296.1 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.

2) Add the enthalpies:

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

2) The equation we want is this:

S(s, romb) ---> 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 3/2
b) multiply second reaction by 1/2 and flip it.

2) The result:

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

3) Add the two reactions (the (3/2)2 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 + (1/2)O2

given:

2Fe + (3/2)O2 ---> Fe2O3ΔH = -824.2 kJ
3Fe + 2O2 ---> Fe3O4ΔH = -1118.4 kJ

Solution:

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

Here's the result:

3Fe2O3 ---> 6Fe + (9/2)O2ΔH = +2472.6 kJ
6Fe + 4O2 ---> 2Fe3O4ΔH = -2236.8 kJ

Add the two reactions together. Note that 6Fe cancels. Also note that 4O2 is (8/2)O2, so only (1/2)O2 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 + 6H2OΔH° = -1170 kJ
2N2 + 6H2O ---> 4NH3 + 3O2ΔH° = +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 ---> 4NOΔH° = +360 kJ

2) Divide through by 2 for the final answer:

N2 + O2 ---> 2NOΔH° = +180 kJ

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


Problem #19:

Determine the enthalpy for this reaction:

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

given the following reactions with known ΔH values:

(1/2)N2(g) + (1/2)O2(g) ---> NO(g)ΔH = +90.3 kJ
NO(g) + (1/2)Cl2(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, to get:

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

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

-103.4 kJ

Problem #20: Use Hess' Law to calculate the enthalpy of vaporization for ethanol, C2H5OH:

C2H5OH(ℓ) ---> C2H5OH(g)

Solution:

enthalpy of formation, gas ---> -234 kJ/mol
enthalpy of formation, liquid ---> -276 kJ/mol

ΔHvap = products - reactants

ΔHvap = -234 - (-276) = 42 kJ/mol

The value given here is 42.3 ± 0.4 kJ/mol


Hess' Law - using two equations and their enthalpies - Problem 1 - 10      Hess' Law - using three equations and their enthalpies      Hess' Law - using standard enthalpies of formation
Hess' Law - using two equations and their enthalpies      Hess' Law - using four or more equations and their enthalpies      Hess' Law - using bond enthalpies
       Thermochemistry menu