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

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}, 1300 calories are released. When one mole of sulfur burns to form SO_{3}, 3600 calories 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 = -1.3 kcal S + ^{3}⁄_{2}O_{2}---> SO_{3}ΔH = -3.6 kcal

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 = +1.3 kcal S + ^{3}⁄_{2}O_{2}---> SO_{3}ΔH = -3.6 kcal

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:

+1.3 plus -3.6 = -2.3 kcal or -2300 cal

Comment: note that calories were used rather than Joules. This in no way affects the solution technique.

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

Calculate ΔH_{rxn} for the following reaction:

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

3 X _{2}+ 15 Y_{2}---> 6 XY_{5}3ΔH _{1}2 X _{3}Z ---> 3 X_{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 #8a:** 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 #8b:** 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 #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 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