### Hess' Law of Constant Heat SummationUsing four or more equations and their enthalpiesProblems 11 - 20

Problem #11: Calculate the enthalpy of reaction for the combustion of ethane:

2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)

Given these reactions:

 C2H4(g) + 3O2(g) ---> 2CO2(g) + 2H2O(ℓ) -1411.20 kJ C2H4(g) + H2(g) ---> C2H6(g) -137 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) -285.8 kJ H2O(ℓ) ---> H2O(g) +44.0 kJ

Solution:

1) Manipulate the four data equations as follows:

1) multiply by 2 (need to cancel 2C2H4)
2) flip, multiply by 2 (puts 2C2H6 on the reactant side)
3) multiply by 2 (need to cancel 2H2)
4) multiply by 6 (need to replace 6H2O(ℓ) with 6H2O(g))

2) The result is this:

 2C2H4(g) + 6O2(g) ---> 4CO2(g) + 4H2O(ℓ) -2822.40 kJ 2C2H6(g) ---> 2C2H4(g) + 2H2(g) +274 kJ 2H2(g) + O2(g) ---> 2H2O(ℓ) -571.6 kJ 6H2O(ℓ) ---> 6H2O(g) +264.0 kJ

 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g) -2856 kJ

Comment: the target equation is not the standard combustion equation, which has water as a liquid, not as a gas. The standard enthalpy for the combustion of hexane is -1560 kJ/mol. Multiplying -1560 by 2 and then adding +264 will yield -2856.

Problem #12: The standard enthalpy change for the reaction of SO3(g) with H2O(ℓ) to yield H2SO4(aq) is -227.8 kJ. In addition:

 S(s) + O2(g) ---> SO2(g) ΔH = -296.8 kJ SO2(g) + 1⁄2O2(g) ---> SO3(g) ΔH = -98.9 kJ

Solution:

1) The target equation is this:

H2(g) + S(s) + 2O2(g) ---> H2SO4(aq)

2) The information provided is this:

 SO3(g) + H2O(ℓ) ---> H2SO4(aq) ΔH = -227.8 kJ S(s) + O2(g) ---> SO2(g) ΔH = -296.8 kJ SO2(g) + 1⁄2O2(g) ---> SO3(g) ΔH = -98.9 kJ

3) However, a fourth equation is needed. This:

H2(g) + 12O2(g) ---> H2O(ℓ) ΔH = -285.8 kJ

This is needed to (a) introduce H2 into the mix and (b) allow H2O to be eliminated when all the data equations are added together.

4) Here is what to do to the 4 data equations:

1) leave untouched
2) leave untouched
3) leave untouched
4) leave untouched

SO3 cancels out between 1 and 3
H2O cancels out between 1 and 4
SO2 cancels out between 2 and 3

You get 2O2 from 2, 3 and 4

5) To get your ΔHf for the target reaction just add up the 4 enthalpies from the 4 data equations.

Problem #13: Determine the enthalpy of reaction for:

2CH4(g) ---> C2H4(g) + 2H2(g)

using:

 2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O(ℓ) ΔH = -3120.8 kJ CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ) ΔH = -890.3 kJ C2H4(g) + H2(g) ---> C2H6(g) ΔH = -136.3 kJ H2 + 1⁄2O2 ---> H2O ΔH = -285.8 kJ

Solution:

1) Here are the four equations with the appropriate modifications applied:

 4CO2(g) + 6H2O(ℓ) ---> 2C2H6(g) + 7O2(g) ΔH = +3120.8 kJ 4CH4(g) + 8O2(g) ---> 4CO2(g) + 8H2O(ℓ) ΔH = -3561.2 kJ 2C2H6(g) ---> 2C2H4(g) + 2H2(g) ΔH = +272.6 kJ 2H2O ---> 2H2 + O2 ΔH = +571.6 kJ

first equation ---> flipped
second equation ---> mutiplied by 4
third equation ---> flip, mutiply by 2
fourth equation ---> flip, multiply by 2

3) The rationale for this:

first equation ---> because equation 3 was flipped, to cancel the C2H6
second equation ---> need 4CO2 to canel with first equation
third equation ---> flip to put C2H4 on product side, mutiply by 2
fourth equation ---> need total of 8H2O on reactant side

Notice how O2 cancels and 2H2 is left when the four equations are added together.

4) Adding the four equations yields:

4CH4(g) ---> 2C2H4(g) + 2H2(g)

5) Adding the four modified enthalpies yields:

(+3120.8 kJ) + (-3561.2 kJ) + (+272.6 kJ) + (+571.6 kJ) = +403.8 kJ

6) The chemical equation must be divided by two (in order to obtain the target equation) and so the enthalpy must also be divided by two:

 2CH4(g) ---> C2H4(g) + H2(g) ΔH = +201.9 kJ

Problem #14: Given:
 B2O3(s) + 3H2O(g) ---> 3O2(g) + B2H6(g) ΔH = +2035 kJ/mol H2O(ℓ) ---> H2O(g) ΔH = +44 kJ/mol H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = -286 kJ/mol 2B (s) + 3H2(g) ---> B2H6(g) ΔH = +36 kJ/mol

Find the ΔHf of:

2B(s) + 32O2(g) ---> B2O3(s)

Solution:

1) After various reversing of the equations and multiplications (or not!), the result (don't forget to apply changes to the enthalpy changes too) is:

 3O2(g) + B2H6(g) ---> B2O3(s) + 3H2O(g) ΔH = -2035 kJ/mol 3H2O(g) ---> 3H2O(ℓ) ΔH = -132 kJ/mol 3H2O(ℓ) ---> 3H2(g) + 3⁄2O2(g) ΔH = +858 kJ/mol 2B (s) + 3H2(g) ---> B2H6(g) ΔH = +36 kJ/mol

2) Adding these equations and canceling out the common terms on both sides, we get:

 2B(s) + 3⁄2O2(g) ---> B2O3(s) ΔHf = -1273 kJ/mol

3) The various reversings and multiplications were these:

eq 1 ---> reversed
eq 2 ---> reversed, multiplied by 3
eq 3 ---> reversed, multiplied by 3
eq 4 ---> unchanged

Problem #15: Determine the enthalpy of reaction for this reaction:

H2SO4(aq) + 2NaOH(aq) ---> Na2SO4(aq) + 2H2O(ℓ)

Given the following:

 H2(g) + S(s) + 2O2(g) ---> H2SO4(aq) ΔH = -907.5 kJ Na(s) + 1⁄2H2(g) + 1⁄2O2(g) ---> NaOH(aq) ΔH = -469.6 kJ 2Na(s) + S(s) + 2O2(g) ---> Na2SO4(aq) ΔH = -1387.1 kJ H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH = -285.8 kJ

Solution:

1) These are the changes to the data equations:

eq 1 ---> reversed
eq 2 ---> reversed, multiplied by 2
eq 3 ---> unchanged
eq 4 ---> multiplied by 2

2) These are the rationales for the changes:

eq 1 ---> put H2SO4 as a reactant
eq 2 ---> put NaOH as a reactant, get two NaOH
eq 3 ---> keeps Na2SO4 as a product, allows Na and S to be cancelled
eq 4 ---> put 2H2O(ℓ) as a product, allows all H2 and O2 to be cancelled

3) Here are the changes to the data equations (note the changes to the enthalpies):

 H2SO4(aq) ---> H2(g) + S(s) + 2O2(g) ΔH = +907.5 kJ 2NaOH(aq) ---> 2Na(s) + H2(g) + O2(g) ΔH = +469.6 kJ 2Na(s) + S(s) + 2O2(g) ---> Na2SO4(aq) ΔH = -1387.1 kJ 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH = -571.6 kJ

4) Sum up the enthalpies for the final answer:

ΔH = 907.5 + 469.6 + (-1387.1) + ( -571.6) = -581.6 kJ

Problem #16: Find the enthalpy of reaction for the equation:

HCl + NaNO2 --> HNO2 + NaCl

Given the following:

 2NaCl(s) + H2O(ℓ) ---> 2HCl(g) + Na2O(s) ΔH = -507 kJ NO(g) + NO2(g) + Na2O(s) ---> 2NaNO2(s) ΔH = -427 kJ NO(g) + NO2(g) ---> N2O(g) + O2(g) ΔH = -43 kJ 2HNO2(ℓ) ---> N2O(g) + O2(g) + H 2O(ℓ) ΔH = 34 kJ

Solution:

1) The modifications applied to the four data equations:

 HCl(g) + 1⁄2Na2O(s) ---> NaCl(s) + 1⁄2H2O(ℓ) ΔH = +253.5 kJ (halved and reversed) NaNO2(s) ---> 1⁄2NO(g) + 1⁄2NO2(g) + 1⁄2Na2O(s) ΔH = +213.5 kJ (halved and reversed) 1⁄2NO(g) + 1⁄2NO2(g) ---> 1⁄2N2O(g) + 1⁄2O2(g) ΔH = -21.5 kJ (halved) 1⁄2N2O(g) + 1⁄2O2(g) + 1⁄2H2O(ℓ) ---> HNO2(ℓ) ΔH = -17 kJ (halved and reversed)

 HCl + NaNO2 ---> HNO2 + NaCl ΔH = +428.5 kJ

Comment: notice how the solution (which the ChemTeam did not write) blithely ignores the idea that one-half of, say, an NO molecule cannot exist. However, one-half of a mole of NO molecules can exist and, in any event, all the one-half coefficients cancel out on the way to a final answer. Now, if a 12NO had appeared in the final answer, we would multiply through by 2 to get rid of it. 12NO is fine on the way to the final answer, but not in the final answer.

Problem #17: Calculate the reaction enthalpy, ΔH, for the following reaction:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(ℓ)
Use the series of reactions that follow:
 C(s) + 2H2(g) ---> CH4(g) ΔH = −74.8 kJ C(s) + O2(g) ---> CO2(g) ΔH = −393.5 kJ 2H2(g) + O2(g) ---> 2H2O(g) ΔH = −484.0 kJ H2O(ℓ) ---> H2O(g) ΔH = 44.0 kJ

Solution:

1) Reverse the first reaction

CH4(g) ---> C(s) + 2H2(g) , ΔH = 74.8 kJ

2) Add the second reaction to the line above

CH4(g) + O2(g) ---> 2H2(g) + CO2(g) , ΔH = -318.7 kJ

3) Note that the C(s) canceled out. Add the third reaction to the line above

CH4(g) + 2O2(g) ---> 2H2O(g) + CO2(g), ΔH = -802.7 kJ

4) Note that the 2H2(g) canceled out. Reverse the fourth reaction and add two of it.

CH4(g) + 2O2(g) ---> 2H2O(ℓ) + CO2(g), ΔH = -890.7 kJ

Note that the 2H2O(g) canceled out.

Problem #18: Determine the heat of formation of calcium carbonate from the thermochemical equations given below.

 Ca(OH)2(s) ---> CaO(s) + H2O(ℓ) ΔH = 65.2 kJ Ca(OH)2(s) + CO2(g) ---> CaCO3(s) + H2O(ℓ) ΔH = -113.2 kJ C(s) + O2(g) ---> CO2(g) ΔH = -393.5 kJ 2Ca(s) + O2(g) ---> 2CaO(s) ΔH = -1270.2 kJ

No solution will be provided. The correct answer is -1207.0 kJ

Problem #19: Calculate the standard enthalpy of formation of potassium chloride given the following:

 K(s) ---> K(g) +89 kJ K(g) ---> K+(g) + e¯ +418 kJ 1⁄2Cl2(g) ---> Cl(g) +122 kJ Cl(g) + e¯ ---> Cl¯(g) -349 kJ KCl(s) ---> K+(g) + Cl¯(g) +717 kJ

Solution:

1) The standard enthalpy of formation for KCl is associated with this chemical reaction:

K(s) + 12Cl2(g) ---> KCl(s)

2) What must be done to the data equations?

eq 1 ---> K(s) is already on the reactant side. This is what we want.
eq 2 ---> K(g) is on the reactant side, ready to cancel K(g) in eq 1. This is what we want.
eq 3 ---> 12Cl2(g) is on the reactant side. This is what we want.
eq 4 ---> Cl(g) is on the reactant side, ready to cancel Cl(g) in eq 3. This is what we want.
eq 5 ---> this equation must be flipped in order to have KCl(s) on the product side.

3) The result:

 K(s) ---> K(g) +89 kJ K(g) ---> K+(g) + e¯ +418 kJ 1⁄2Cl2(g) ---> Cl(g) +122 kJ Cl(g) + e¯ ---> Cl¯(g) -349 kJ K+(g) + Cl¯(g) ---> KCl(s) -717 kJ

4) When the five equations just above are added, the KCl(s) formation equation results. Add the five enthalpies:

89 + 418 + 122 + (-349) + (-717) = -437 kJ

This is the standard enthalpy of formation for KCl(s)

Problem #20: Make the following equation:

CS2(ℓ) + 3Cl2(g) ---> CCl4(ℓ) + S2Cl2(ℓ)

from:

CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g)
S(s) + 12Cl2(g) ---> 12S2Cl2(ℓ)
C(s) + 2Cl2(g) ---> CCl4(ℓ)
S(s) + O2(g) ---> SO2(g)
C(s) + O2(g) ---> CO2(g)

Solution:

Notice that reaction enthalpies aren't included. It's just a bit more practice in manipulating data equations to get the desired equation. Here's the rearranged set:

CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g)
2S(s) + Cl2(g) ---> S2Cl2(ℓ) (mult by 2)
C(s) + 2Cl2(g) ---> CCl4(ℓ)
2SO2(g) ---> 2S(s) + 2O2(g) (flip, mult by 2)
CO2(g) ---> C(s) + O2(g) (flip)

Bonus Problem: Given:

 Br2(ℓ) + 5F2(g) ---> 2BrF5(ℓ) ΔH° = -918.0 kJ BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g) ΔH° = 125.2 kJ 2NaBr(s) + F2(g) ---> 2NaF(s) + Br2(ℓ) ΔH° = -316.0 kJ NaBr(s) + F2(g) ---> NaF(s) + BrF(g) ΔH° = -216.6 kJ

calculate ΔH° for the reaction:

BrF3(ℓ) + F2(g) ---> BrF5(ℓ)

Solution:

1) Manipulate the four data equations:

 1⁄2Br2(ℓ) + 5⁄2F2(g) ---> BrF5(ℓ) ΔH° = -459.0 kJ (divide by 2) BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g) ΔH° = 125.2 kJ 3NaBr(s) + 3⁄2F2(g) ---> 3NaF(s) + 3⁄2Br2(ℓ) ΔH° = -474.0 kJ (mult by 3⁄2) 3NaF(s) + 3BrF(g) ---> 3NaBr(s) + 3F2(g) ΔH° = 649.8 kJ (flip, mult by 3)

2) The reasons:

(a) get BrF5 with a coefficient of 1
(b) leave unchanged, BrF3 is on correct side and with coefficient of 1
(c) make 3NaF to cancel with 4th equation
(d) cancel the 3BrF in the second data equation

Notice that there are now 4F2 on the left (from 52 + 32). When the equations are added, that will cancel with the 3F2 on the right, giving us one F2 on the left, which is what we want.

Notice also that there will be 32Br2 on each side.

3) Add the four changed enthalpies for the final answer of -158 kJ.