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:

$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ for H_{2}O(g) = −285.83 kJ/mol

$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ 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 These are actually formation reactions. I'll ignore that since it's not germane to the problem.

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:

$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −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}O_{2}will cancel, leaving the target equation.Add the two enthalpies together for the final answer of −23.9 kJ

**Problem #8:** Calculate the standard enthalpy of reaction for the following reaction:

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

Given:

4Fe(s) + 3O _{2}(g) ---> 2Fe_{2}O_{3}(s)ΔH = −1648 kJ 2CO _{2}(g) ---> 2CO(g) + O_{2}(g)ΔH = +565.4 kJ

**Solution:**

1) What needs to be done:

(a) flip first reaction and divide by 2. This gets one Fe_{2}O_{3}on the reactant side and two Fe on the product side.(b) flip second reaction and multiply by 1.5. This gets the proper amount and placement for CO and CO

_{2}.

2) Here are the results:

3) Add the two reactions. Note that the

Fe _{2}O_{3}(s) ---> 2Fe(s) +^{3}⁄_{2}O_{2}(g)ΔH = +824 kJ 3CO(g) + ^{3}⁄_{2}O_{2}(g) ---> 3CO_{2}(g)ΔH = −848.1 kJ

Fe _{2}O_{3}(s) + 3CO(g) ---> 2Fe(s) + 3CO_{2}(g)ΔH° = −24.1 kJ

**Problem #9:** 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 #10:** 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

**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 $\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$
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