Using two equations and their enthalpies

Problems 1 - 10

**Problem #1:** Consider the following reaction:

N _{2}H_{4}(ℓ) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(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:

ΔH_{f}for H_{2}O(g) = −285.83 kJ/mol

ΔH_{f}for H_{2}O(ℓ) = −241.83 kJ/mol

**Solution:**

1) We need to find ΔH for this reaction:

H_{2}O(ℓ) ---> H_{2}O(g)ΔH = −285.83 − (−241.83) = −44 kJ

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

N _{2}H_{4}(ℓ) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(ℓ)ΔH = −622.2 kJ H _{2}O(ℓ) ---> H_{2}O(g)ΔH = −44 kJ

3) Multiply second equation by 2:

N _{2}H_{4}(ℓ) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(ℓ)ΔH = −622.2 kJ 2H _{2}O(ℓ) ---> 2H_{2}O(g)ΔH = −88 kJ

4) Add the two reactions (and eliminate 2H_{2}O(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 SO_{2}, 297.0 kilojoules are released. When one mole of sulfur burns to form SO_{3}, 395.8 kilojoules are released. What is the ΔH when one mole of SO_{2} is burned to form SO_{3}?

**Solution:**

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

S + O _{2}---> SO_{2}ΔH = −297.0 kJ S + ^{3}⁄_{2}O_{2}---> SO_{3}ΔH = −395.8 kJ

2) The target equation we want is this:

SO_{2}+^{1}⁄_{2}O_{2}---> SO_{3}

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:

SO _{2}---> S + O_{2}ΔH = +297.0 kJ S + ^{3}⁄_{2}O_{2}---> SO_{3}ΔH = −395.8 kJ

4) When those two equations are added, the S and an O_{2} will cancel out:

SO_{2}+^{1}⁄_{2}O_{2}---> SO_{3}

5) We add the enthalpies:

+297.0 + (−395.8) = −98.8 kJ

**Problem #3:** Calculate the enthalpy of formation for sulfur dioxide, SO_{2}. Use the following information:

S(s) + ^{3}⁄_{2}O_{2}(g) ----> SO_{3}(g)ΔH = −395.8 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH = −198.2 kJ

**Solution:**

1) This is the formation reaction for SO_{2}:

S(s) + O_{2}(g) ---> SO_{2}(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 SO_{2}on the product side, which is where we want it)

3) The result:

S(s) + ^{3}⁄_{2}O_{2}(g) ---> SO_{3}(g)ΔH = −395.8 kJ SO _{3}(g) ---> SO_{2}(g) +^{1}⁄_{2}O_{2}(g)ΔH = +99.1 kJ

4) The answer:

When the two equations are added, the SO_{3}and^{1}⁄_{2}O_{2}will cancel out, leaving the desired equation. Add the two enthalpies for the answer:

S(s) + O _{2}(g) ---> SO_{2}(g)ΔH = −296.7 kJ

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

2SO _{3}(g) ---> 2S(s) + 3O_{2}(g)ΔH = +791.6 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH = −198.2 kJ

Your assignment is to get to the answer of −296.7 kJ.

**Problem #4:** Given the following:

I _{2}(g) + 3Cl_{2}(g) ---> 2ICl_{3}(s)ΔH = −214 kJ I _{2}(s) ----> I_{2}(g)ΔH = +38 kJ

Calculate the enthalpy of formation for ICl_{3}.

**Solution #1:**

1) The target equation is this:

^{1}⁄_{2}I_{2}(s) +^{3}⁄_{2}Cl_{2}(g) ---> ICl_{3}(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}⁄_{2}I_{2}(g) +^{3}⁄_{2}Cl_{2}(g) ---> ICl_{3}(s)ΔH = −107 kJ ^{1}⁄_{2}I_{2}(s) ---->^{1}⁄_{2}I_{2}(g)ΔH = +19 kJ When you add the two equations just above, the

^{1}⁄_{2}I_{2}(g) will cancel out, leaving only the desired target equation.The enthalpy of the target equation is this:

−107 + 19 = −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 = −176Since this is for 2 moles of ICl

_{3}, diving by 2 yields the answer of −88 kJ.

**Problem #5:** Given the follow two reactions:

X _{2}+ 5Y_{2}---> 2XY_{5}ΔH _{1}3X _{2}+ Z_{2}---> 2X_{3}ZΔH _{2}

Calculate ΔH_{rxn} for the following reaction:

15Y_{2}+ 2X_{3}Z ---> 6XY_{5}+ Z_{2}

**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 15Y_{2}

eq 2 ---> flip in order to get X_{3}Z onto the reactant side

2) The result:

3X _{2}+ 15Y_{2}---> 6XY_{5}3ΔH _{1}2X _{3}Z ---> 3X_{2}+ Z_{2}−ΔH _{2}

3) Solve for the enthalpy of the desired reaction:

When the two equations above are added, the 3X_{2}will cancel out and the enthalpy is this:ΔH

_{rxn}= 3ΔH_{1}+ (−ΔH_{2})

**Problem #6:** The following two reactions are known:

Fe _{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g)ΔH = −23.44 kJ FeO(s) + CO(g) ---> Fe(s) + CO _{2}(g)ΔH = −10.94 kJ

Determine the ΔH value for the reaction below:

Fe_{2}O_{3}(s) + CO(g) ---> 2FeO(s) + CO_{2}(g)

**Solution:**

1) Manipulate the two data equations as follows:

a) leave eq 1 untouched (keeps Fe_{2}O_{3}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 CO

_{2}will follow automatically.

2) The result of the changes in step one:

Fe _{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g)ΔH = −23.44 kJ 2Fe(s) + 2CO _{2}(g) ---> 2FeO(s) + 2CO(g)ΔH = +21.88 kJ

3) Add the two equations together and . . .

. . . two of the CO, two of the CO_{2}and the two Fe will all cancel out.Add the enthalpies for the final answer of −1.56 kJ

**Problem #7:** Determine the enthalpy for this reaction:

Fe_{2}O_{3}+ 3CO ---> 2Fe + 3CO_{2}

using this information:

2Fe(s) + ^{3}⁄_{2}O_{2}(g) ---> Fe_{2}O_{3}(s)ΔH = −824.2 kJ CO(g) + ^{1}⁄_{2}O_{2}(g) ---> CO_{2}(g)ΔH = −282.7 kJ

**Solution:**

1) Do this:

a) reverse eq 1 (to put Fe_{2}O_{3}on the reactant side)

b) multiply eq 2 by three (this gets us 3CO and 3CO_{2})

2) The results of the changes:

Fe _{2}O_{3}(s) ---> 2Fe(s) +^{3}⁄_{2}O_{2}(g)ΔH= +824.2 kJ 3CO(g) + ^{3}⁄_{2}O_{2}(g) ---> 3CO_{2}(g)ΔH= −848.1 kJ

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

. . . the^{3}⁄_{2}O2 will cancel, leaving the target equation.Add the two enthalpies together for the final answer of −23.9 kJ

**Problem #8:** Calculate ΔH for the following reaction:

CH_{4}(g) + 2Cl_{2}(g) ---> CH_{2}Cl_{2}(g) + 2HCl(g)

Use the following information:

CH _{4}(g) + Cl_{2}(g) ---> CH_{3}Cl(g) + HCl(g)ΔH = −99.6 kJ CH _{3}Cl(g) + Cl_{2}(g) ---> CH_{2}Cl_{2}(g) + HCl(g)ΔH = −105.8 kJ

**Solution:**

Nothing needs to be done to either data equation. When added together, the CH_{3}Cl will cancel out, leaving the desired target equation.Add the two enthalpies for the final answer: −205.4 kJ

**Problem #9:** Find the enthalpy change for the formation of phosphorus pentachloride from its elements:

P(s) +^{5}⁄_{2}Cl_{2}(g) ---> PCl_{5}(s)

Use the following thermochemical equations.

PCl _{5}(s) ---> PCl_{3}(g) + Cl_{2}(g)ΔH = 87.9 kJ 2P(s) + 3Cl _{2}(g) ---> 2PCl_{3}(g)ΔH = −574 kJ

**Solution:**

1) Reverse the first data equation:

PCl_{3}(g) + Cl_{2}(g) ---> PCl_{5}(s) ΔH = −87.9 kJ

2) Divide the second data equation by 2:

P(s) +^{3}⁄_{2}Cl_{2}(g) ---> PCl_{3}(g) ΔH = −287 kJ

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

PCl_{3}cancels and the enthalpy of the target reaction is −374.9 kJ

**Problem #10a:** The following two reactions are known:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.50 kJ 2NO _{2}(g) ---> N_{2}(g) + 2O_{2}(g)ΔH = −66.36 kJ

Determine the ΔH value for the reaction below:

2NO(g) + O_{2}(g) ---> 2NO_{2}(g)

**Solution:**

1) Make these modifications:

a) first equation ---> flip it

b) second equation ---> flip it

2) The result:

2NO(g) ---> N _{2}(g) + O_{2}(g)ΔH = −180.50 kJ N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH = +66.36 kJ

3) Add the two modified equations and . . .

. . . N_{2}(g) will cancel as well as one O_{2}(g) when the two equations are added.ΔH = −180.50 + (+66.36) = −114.14 kJ

**Problem #10b:** The following two reactions are known:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.50 kJ N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH = +66.36 kJ

Calculate for enthalpy change for:

NO_{2}(g) ---> NO(g) +^{1}⁄_{2}O_{2}(g)

**Solution:**

1) Make these modifications:

Divide first equation by 2

Flip second equation and divide by 2

2) The result:

^{1}⁄_{2}N_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> NO(g)ΔH = +90.25 kJ NO _{2}(g) --->^{1}⁄_{2}N_{2}(g) + O_{2}(g)ΔH = −33.18 kJ

3) Add the two modified equations and . . .

. . .^{1}⁄_{2}N_{2}(g) will cancel as well as^{1}⁄_{2}O_{2}(g) when the two equations are added.ΔH = −180.50 + (+66.36) = +57.07 kJ

**Problem #10c:** Use Hess's law to calculate ΔH for the following reaction:

N_{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)

using the following three data equations:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.7 kJ 2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH = −113.1 kJ 2N _{2}O(g) ---> 2N_{2}(g) + O_{2}(g)ΔH = −163.2 kJ

**Solution:**

1) Add the following two equations:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.7 kJ 2NO(g) + O _{2}(g) ---> 2 NO_{2}(g)ΔH = −113.1 kJ

2) And get this result:

N_{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g) ΔH = +67.6 kJ

3) You do not need the third equation because N_{2}O 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 #10d:** Given the following reactions:

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH = +66.36 kJ 2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH = −114.2 kJ

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

N_{2}(g) + O_{2}(g) ---> 2NO(g)

**Solution:**

1) Manipulate the data equation as follows:

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH = +66.36 kJ <--- leave unchanged 2NO _{2}(g) ---> 2NO(g) + O_{2}(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 2NO_{2} cancels out as well as one of the O_{2}. 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.

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

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH _{1}NO(g) + ^{1}⁄_{2}O_{2}(g) ---> NO_{2}(g)ΔH _{2}

which of the statements below correctly describes the ΔH_{f}° for NO_{2}(g)?

(A) ΔH_{1}

(B) ΔH_{2}

(C) ΔH_{2}+^{1}⁄_{2}ΔH_{1}

(D) ΔH_{2}-^{1}⁄_{2}ΔH_{1}

(E) ΔH_{2}+ ΔH_{1}

**Solution:**

1) The target equation we want is the formation reaction for NO_{2}(g):

^{1}⁄_{2}N_{2}(g) + O_{2}(g) ---> NO_{2}(g)

2) We get that by doing this:

^{1}⁄_{2}N_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> NO(g)^{1}⁄_{2}ΔH_{1}NO(g) + ^{1}⁄_{2}O_{2}(g) ---> NO_{2}(g)ΔH _{2}

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

^{1}⁄_{2}ΔH_{1}+ ΔH_{2}answer choice C