Using four or more equations and their enthalpies

Problems 11 - 20

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

2C_{2}H_{6}(g) + 7O_{2}(g) --> 4CO_{2}(g) + 6H_{2}O(g)

Given these reactions:

C _{2}H_{4}(g) + 3O_{2}(g) ---> 2CO_{2}(g) + 2H_{2}O(ℓ)-1411.20 kJ C _{2}H_{4}(g) + H_{2}(g) ---> C_{2}H_{6}(g)-137 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)-285.8 kJ H _{2}O(ℓ) ---> H_{2}O(g)+44.0 kJ

**Solution:**

1) Manipulate the four data equations as follows:

1) multiply by 2 (need to cancel 2C_{2}H_{4})

2) flip, multiply by 2 (puts 2C_{2}H_{6}on the reactant side)

3) multiply by 2 (need to cancel 2H_{2})

4) multiply by 6 (need to replace 6H_{2}O(ℓ) with 6H_{2}O(g))

2) The result is this:

2C _{2}H_{4}(g) + 6O_{2}(g) ---> 4CO_{2}(g) + 4H_{2}O(ℓ)-2822.40 kJ 2C _{2}H_{6}(g) ---> 2C_{2}H_{4}(g) + 2H_{2}(g)+274 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)-571.6 kJ 6H _{2}O(ℓ) ---> 6H_{2}O(g)+264.0 kJ

Add the four data equations and their enthalpies yields the answer:

2C _{2}H_{6}(g) + 7O_{2}(g) --> 4CO_{2}(g) + 6H_{2}O(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 SO_{3}(g) with H_{2}O(ℓ) to yield H_{2}SO_{4}(aq) is -227.8 kJ.
In addition:

S(s) + O _{2}(g) ---> SO_{2}(g)ΔH = -296.8 kJ SO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> SO_{3}(g)ΔH = -98.9 kJ

Please calculate ΔH_{f} for H_{2}SO_{4}(aq).

**Solution:**

1) The target equation is this:

H_{2}(g) + S(s) + 2O_{2}(g) ---> H_{2}SO_{4}(aq)

2) The information provided is this:

SO _{3}(g) + H_{2}O(ℓ) ---> H_{2}SO_{4}(aq)ΔH = -227.8 kJ S(s) + O _{2}(g) ---> SO_{2}(g)ΔH = -296.8 kJ SO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> SO_{3}(g)ΔH = -98.9 kJ

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

H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ) ΔH = -285.8 kJ

This is needed to (a) introduce H_{2} into the mix and (b) allow H_{2}O 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 untouchedSO

_{3}cancels out between 1 and 3

H_{2}O cancels out between 1 and 4

SO_{2}cancels out between 2 and 3You get 2O

_{2}from 2, 3 and 4

5) To get your ΔH_{f} for the target reaction just add up the 4 enthalpies from the 4 data equations.

**Problem #13:** Determine the enthalpy of reaction for:

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

using:

2C _{2}H_{6}(g) + 7O_{2}(g) ---> 4CO_{2}(g) + 6H_{2}O(ℓ)ΔH = -3120.8 kJ CH _{4}(g) + 2O_{2}(g) ---> CO_{2}(g) + 2H_{2}O(ℓ)ΔH = -890.3 kJ C _{2}H_{4}(g) + H_{2}(g) ---> C_{2}H_{6}(g)ΔH = -136.3 kJ H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = -285.8 kJ

**Solution:**

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

4CO _{2}(g) + 6H_{2}O(ℓ) ---> 2C_{2}H_{6}(g) + 7O_{2}(g)ΔH = +3120.8 kJ 4CH _{4}(g) + 8O_{2}(g) ---> 4CO_{2}(g) + 8H_{2}O(ℓ)ΔH = -3561.2 kJ 2C _{2}H_{6}(g) ---> 2C_{2}H_{4}(g) + 2H_{2}(g)ΔH = +272.6 kJ 2H _{2}O ---> 2H_{2}+ O_{2}ΔH = +571.6 kJ

2) What happened above was this:

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 C_{2}H_{6}

second equation ---> need 4CO_{2}to canel with first equation

third equation ---> flip to put C_{2}H_{4}on product side, mutiply by 2

fourth equation ---> need total of 8H_{2}O on reactant side

Notice how O_{2} cancels and 2H_{2} is left when the four equations are added together.

4) Adding the four equations yields:

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

2CH _{4}(g) ---> C_{2}H_{4}(g) + H_{2}(g)ΔH = +201.9 kJ

B _{2}O_{3}(s) + 3H_{2}O(g) ---> 3O_{2}(g) + B_{2}H_{6}(g)ΔH = +2035 kJ/mol H _{2}O(ℓ) ---> H_{2}O(g)ΔH = +44 kJ/mol H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH = -286 kJ/mol 2B (s) + 3H _{2}(g) ---> B_{2}H_{6}(g)ΔH = +36 kJ/mol

Find the ΔH_{f} of:

2B(s) +^{3}⁄_{2}O_{2}(g) ---> B_{2}O_{3}(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:

3O _{2}(g) + B_{2}H_{6}(g) ---> B_{2}O_{3}(s) + 3H_{2}O(g)ΔH = -2035 kJ/mol 3H _{2}O(g) ---> 3H_{2}O(ℓ)ΔH = -132 kJ/mol 3H _{2}O(ℓ) ---> 3H_{2}(g) +^{3}⁄_{2}O_{2}(g)ΔH = +858 kJ/mol 2B (s) + 3H _{2}(g) ---> B_{2}H_{6}(g)ΔH = +36 kJ/mol

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

2B(s) + ^{3}⁄_{2}O_{2}(g) ---> B_{2}O_{3}(s)ΔH _{f}= -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:

H_{2}SO_{4}(aq) + 2NaOH(aq) ---> Na_{2}SO_{4}(aq) + 2H_{2}O(ℓ)

Given the following:

H _{2}(g) + S(s) + 2O_{2}(g) ---> H_{2}SO_{4}(aq)ΔH = -907.5 kJ Na(s) + ^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> NaOH(aq)ΔH = -469.6 kJ 2Na(s) + S(s) + 2O _{2}(g) ---> Na_{2}SO_{4}(aq)ΔH = -1387.1 kJ H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(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 H_{2}SO_{4}as a reactant

eq 2 ---> put NaOH as a reactant, get two NaOH

eq 3 ---> keeps Na_{2}SO_{4}as a product, allows Na and S to be cancelled

eq 4 ---> put 2H_{2}O(ℓ) as a product, allows all H_{2}and O_{2}to be cancelled

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

H _{2}SO_{4}(aq) ---> H_{2}(g) + S(s) + 2O_{2}(g)ΔH = +907.5 kJ 2NaOH(aq) ---> 2Na(s) + H _{2}(g) + O_{2}(g)ΔH = +469.6 kJ 2Na(s) + S(s) + 2O _{2}(g) ---> Na_{2}SO_{4}(aq)ΔH = -1387.1 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(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 + NaNO_{2}--> HNO_{2}+ NaCl

Given the following:

2NaCl(s) + H _{2}O(ℓ) ---> 2HCl(g) + Na_{2}O(s)ΔH = -507 kJ NO(g) + NO _{2}(g) + Na_{2}O(s) ---> 2NaNO_{2}(s)ΔH = -427 kJ NO(g) + NO _{2}(g) ---> N_{2}O(g) + O_{2}(g)ΔH = -43 kJ 2HNO _{2}(ℓ) ---> N_{2}O(g) + O_{2}(g) + H_{2}O(ℓ)ΔH = 34 kJ

**Solution:**

1) The modifications applied to the four data equations:

HCl(g) + ^{1}⁄_{2}Na_{2}O(s) ---> NaCl(s) +^{1}⁄_{2}H_{2}O(ℓ)ΔH = +253.5 kJ (halved and reversed) NaNO _{2}(s) --->^{1}⁄_{2}NO(g) +^{1}⁄_{2}NO_{2}(g) +^{1}⁄_{2}Na_{2}O(s)ΔH = +213.5 kJ (halved and reversed) ^{1}⁄_{2}NO(g) +^{1}⁄_{2}NO_{2}(g) --->^{1}⁄_{2}N_{2}O(g) +^{1}⁄_{2}O_{2}(g)ΔH = -21.5 kJ (halved) ^{1}⁄_{2}N_{2}O(g) +^{1}⁄_{2}O_{2}(g) +^{1}⁄_{2}H_{2}O(ℓ) ---> HNO_{2}(ℓ)ΔH = -17 kJ (halved and reversed) 2) Add up:

HCl + NaNO _{2}---> HNO_{2}+ NaClΔH = +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

^{1}⁄_{2}NO had appeared in the final answer, we would multiply through by 2 to get rid of it.^{1}⁄_{2}NO 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:

CHUse the series of reactions that follow:_{4}(g) + 2O_{2}(g) ---> CO_{2}(g) + 2H_{2}O(ℓ)

C(s) + 2H _{2}(g) ---> CH_{4}(g)ΔH = −74.8 kJ C(s) + O _{2}(g) ---> CO_{2}(g)ΔH = −393.5 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(g)ΔH = −484.0 kJ H _{2}O(ℓ) ---> H_{2}O(g)ΔH = 44.0 kJ

**Solution:**

1) Reverse the first reaction

CH_{4}(g) ---> C(s) + 2H_{2}(g) , ΔH = 74.8 kJ

2) Add the second reaction to the line above

CH_{4}(g) + O_{2}(g) ---> 2H_{2}(g) + CO_{2}(g) , ΔH = -318.7 kJ

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

CH_{4}(g) + 2O_{2}(g) ---> 2H_{2}O(g) + CO_{2}(g), ΔH = -802.7 kJ

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

CH_{4}(g) + 2O_{2}(g) ---> 2H_{2}O(ℓ) + CO_{2}(g), ΔH = -890.7 kJNote that the 2H

_{2}O(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) + H_{2}O(ℓ)ΔH = 65.2 kJ Ca(OH) _{2}(s) + CO_{2}(g) ---> CaCO_{3}(s) + H_{2}O(ℓ)ΔH = -113.2 kJ C(s) + O _{2}(g) ---> CO_{2}(g)ΔH = -393.5 kJ 2Ca(s) + O _{2}(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}⁄_{2}Cl_{2}(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) +^{1}⁄_{2}Cl_{2}(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 --->^{1}⁄_{2}Cl_{2}(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}⁄_{2}Cl_{2}(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 kJThis is the standard enthalpy of formation for KCl(s)

This page show a value of -436.68 kJ

**Problem #20:** Make the following equation:

CS_{2}(ℓ) + 3Cl_{2}(g) ---> CCl_{4}(ℓ) + S_{2}Cl_{2}(ℓ)

from:

CS_{2}(ℓ) + 3O_{2}(g) ---> CO_{2}(g) + 2SO_{2}(g)

S(s) +^{1}⁄_{2}Cl_{2}(g) --->^{1}⁄_{2}S_{2}Cl_{2}(ℓ)

C(s) + 2Cl_{2}(g) ---> CCl_{4}(ℓ)

S(s) + O2(g) ---> SO_{2}(g)

C(s) + O_{2}(g) ---> CO_{2}(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:

CS_{2}(ℓ) + 3O_{2}(g) ---> CO_{2}(g) + 2SO_{2}(g)

2S(s) + Cl_{2}(g) ---> S_{2}Cl_{2}(ℓ) (mult by 2)

C(s) + 2Cl_{2}(g) ---> CCl_{4}(ℓ)

2SO_{2}(g) ---> 2S(s) + 2O_{2}(g) (flip, mult by 2)

CO_{2}(g) ---> C(s) + O_{2}(g) (flip)

**Bonus Problem:** Given:

Br _{2}(ℓ) + 5F_{2}(g) ---> 2BrF_{5}(ℓ)ΔH° = -918.0 kJ BrF _{3}(ℓ) + Br_{2}(ℓ) ---> 3BrF(g)ΔH° = 125.2 kJ 2NaBr(s) + F _{2}(g) ---> 2NaF(s) + Br_{2}(ℓ)ΔH° = -316.0 kJ NaBr(s) + F _{2}(g) ---> NaF(s) + BrF(g)ΔH° = -216.6 kJ

calculate ΔH° for the reaction:

BrF_{3}(ℓ) + F_{2}(g) ---> BrF_{5}(ℓ)

**Solution:**

1) Manipulate the four data equations:

^{1}⁄_{2}Br_{2}(ℓ) +^{5}⁄_{2}F_{2}(g) ---> BrF_{5}(ℓ)ΔH° = -459.0 kJ (divide by 2) BrF _{3}(ℓ) + Br_{2}(ℓ) ---> 3BrF(g)ΔH° = 125.2 kJ 3NaBr(s) + ^{3}⁄_{2}F_{2}(g) ---> 3NaF(s) +^{3}⁄_{2}Br_{2}(ℓ)ΔH° = -474.0 kJ (mult by ^{3}⁄_{2})3NaF(s) + 3BrF(g) ---> 3NaBr(s) + 3F _{2}(g)ΔH° = 649.8 kJ (flip, mult by 3)

2) The reasons:

(a) get BrF_{5}with a coefficient of 1

(b) leave unchanged, BrF_{3}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 equationNotice that there are now 4F

_{2}on the left (from^{5}⁄_{2}+^{3}⁄_{2}). When the equations are added, that will cancel with the 3F_{2}on the right, giving us one F_{2}on the left, which is what we want.Notice also that there will be

^{3}⁄_{2}Br_{2}on each side.

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