Using two equations and their enthalpies

Problems 11 - 20

**Problem #11:** Use the standard reaction enthalpies given below to determine ΔH° for the following reaction:

2S(s) + 3O_{2}(g) ---> 2SO_{3}(g)

Given:

SO _{2}(g) ---> S(s) + O_{2}(g)ΔH° = +296.8 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH° = −197.8 kJ

**Solution:**

1) The two data equations are modified:

S(s) + 2O _{2}(g) ---> 2SO_{2}(g)ΔH° = −593.6 kJ <--- flipped and mult. by 2 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH° = −197.8 kJ <--- no change

2) The 2SO_{2} will cancel out when the equations are added. Add the enthalpies for the answer:

−593.6 + (−197.8) = −791.4 kJ

**Problem #12:** Calculate ΔH° for:

2C_{2}H_{4}(g) + H_{2}O(ℓ) ---> C_{4}H_{9}OH(ℓ)

Using:

2CO _{2}+ 2H_{2}O(ℓ) ---> C_{2}H_{4}(g) + 3O_{2}(g)ΔH° = +1411.1 kJ C _{4}H_{9}OH(ℓ) + 6O_{2}(g) ---> 4CO_{2}+ 5H_{2}O(ℓ)ΔH° = −1534.7 kJ

**Solution:**

1) Both data equations need to be flipped, but let's not flip them yet:

4CO _{2}+ 4H_{2}O(ℓ) ---> 2C_{2}H_{4}(g) + 6O_{2}(g)ΔH° = +2822.2 kJ C _{4}H_{9}OH(ℓ) + 6O_{2}(g) ---> 4CO_{2}+ 5H_{2}O(ℓ)ΔH° = −1534.7 kJ I multiplied the first equation by 2.

2) Add the two data equations and their enthalpies:

C _{4}H_{9}OH(ℓ) ---> 2C_{2}H_{4}(g) + H_{2}O(ℓ)ΔH° = +1287.5 kJ The reaction just above can now be flipped to give us our target equation and the enthalpy of the flipped reaction is −1287.5 kJ.

**Problem #13:** For the following reaction:

2CO(g) + 2NO(g) ---> 2CO_{2}(g) + N_{2}(g)

Use reactions (a) and (b) to determine ΔH

(a) 2CO(g) + O _{2}(g) ---> 2CO_{2}(g)ΔH = −566.0 kJ (b) N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = 180.6 kJ

**Solution:**

1) Flip (b) and leave (a) alone:

(a) 2CO(g) + O _{2}(g) ---> 2CO_{2}(g)ΔH = −566.0 kJ (b) 2NO(g) ---> N _{2}(g) + O_{2}(g)ΔH = −180.6 kJ

2) When you add the two reactions, the O_{2}(g) cancels. Add the two enthalpies for the final answer:

−746.6 kJ

**Problem #14:** Determine the enthalpy of reaction for the combustion of methane to carbon monoxide:

2CH_{4}(g) + 3O_{2}(g) ---> 2CO(g) + 4H_{2}O(ℓ)

Use the following:

CH _{4}(g) + 2O_{2}(g) ---> CO_{2}(g) + 2H_{2}O(ℓ)ΔH° = −890.0 kJ 2CO(g) + O _{2}(g) ---> 2CO_{2}(g)ΔH° = −566.0 kJ

**Solution:**

1) Here are the modified equations:

2CH _{4}(g) + 4O_{2}(g) ---> 2CO_{2}(g) + 4H_{2}O(ℓ)ΔH° = −1780.0 kJ 2CO _{2}(g) ---> 2CO(g) + O_{2}(g)ΔH° = +566.0 kJ Multiplied the first by 2 and flipped the second.

2) Add the enthalpies:

−1214 kJ

**Problem #15:** Elemental sulfur occurs in several forms, with rhombic sulfur the most stable under normal conditions and monoclinic sulfur somewhat less stable. The standard enthalpies of combustion of the two forms to sulfur dioxide are −296.83 and −297.16 kJ/mol, respectively. Calculate the change in enthalpy for the rhombic to monoclinic transition.

**Solution:**

1) Using the information offer, write two combustion equations:

S(s, romb) + O _{2}(g) ---> SO_{2}(g)ΔH = −296.83 kJ S(s, mono) + O _{2}(g) ---> SO_{2}(g)ΔH = −297.16 kJ

2) The equation we want is this:

S(s, romb) ---> S(s, mono)

3) Reversing equation 2 will get us what we want. Change the sign on the second enthalpy and add:

−296.83 kJ + 297.16 kJ = +0.33 kJ

**Problem #16:** Given the following information:

2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(g)ΔH = −483.6 kJ 3O _{2}(g) ---> 2O_{3}(g)ΔH = +285.4 kJ

Determine the ΔH of this reaction:

3H_{2}(g) + O_{3}(g) ---> 3H_{2}O(g)

**Solution:**

1) Do the following:

a) Multiply first reaction by^{3}⁄_{2}

b) Multiply second reaction by^{1}⁄_{2}and flip it.

2) The result:

3H _{2}(g) +^{3}⁄_{2}O_{2}(g) ---> 3H_{2}O(g)ΔH = −725.4 kJ O _{3}(g) --->^{3}⁄_{2}O_{2}(g)ΔH = −142.7 kJ

3) Add the two reactions (the ^{3}⁄_{2}O_{2} cancels out). Add the two enthalpies for the final answer:

−868.1 kJ

**Problem #17:** Calculate the value of heat of reaction for the equation:

3Fe_{2}O_{3}---> 2Fe_{3}O_{4}+^{1}⁄_{2}O_{2}

given:

2Fe + ^{3}⁄_{2}O_{2}---> Fe_{2}O_{3}ΔH = −824.2 kJ 3Fe + 2O _{2}---> Fe_{3}O_{4}ΔH = −1118.4 kJ

**Solution:**

flip first reaction and multiply by three

multiply second reaction by twoHere's the result:

3Fe _{2}O_{3}---> 6Fe +^{9}⁄_{2}O_{2}ΔH = +2472.6 kJ 6Fe + 4O _{2}---> 2Fe_{3}O_{4}ΔH = −2236.8 kJ Add the two reactions together. Note that 6Fe cancels. Also note that 4O

_{2}is^{8}⁄_{2}O_{2}, so only^{1}⁄_{2}O_{2}results. Add +2472.6 and −2236.8 to get the kJ for the reaction.

**Problem #18:** Calculate the enthalpy of the following reaction:

N_{2}+ O_{2}---> 2NO

Given:

4NH _{3}+ 5O_{2}---> 4NO + 6H_{2}OΔH° = −1170 kJ 2N _{2}+ 6H_{2}O ---> 4NH_{3}+ 3O_{2}ΔH° = +1530 kJ

**Solution:**

1) The reactants and the product are in the correct relative places. Add the two data equations together to obtain this:

2N _{2}+ 2O_{2}---> 4NOΔH° = +360 kJ

2) Divide through by 2 for the final answer:

N _{2}+ O_{2}---> 2NOΔH° = +180 kJ

Note: you could have divided both data equations by 2 and then added. The result would have been the same.

**Problem #19:**

Determine the enthalpy for this reaction:

2NOCl(g) ---> N_{2}(g) + O_{2}(g) + Cl_{2}(g)

given the following reactions with known ΔH values:

^{1}⁄_{2}N_{2}(g) +^{1}⁄_{2}O_{2}(g) ---> NO(g)ΔH = +90.3 kJ NO(g) + ^{1}⁄_{2}Cl_{2}(g) ---> NOCl(g)ΔH = −38.6kJ

**Solution:**

1) Both equations need to be reversed. Both equations need to be multiplied by 2. We can actually do those actions after adding the equations together. When we add the equations, we get:

^{1}⁄_{2}N_{2}(g) +^{1}⁄_{2}O_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> NOCl(g)ΔH = +51.7 kJ

2) Reversing the equation and multiplying by 2 yields this enthalpy:

-103.4 kJ

**Problem #20:** Use Hess' Law to calculate the enthalpy of vaporization for ethanol, C_{2}H_{5}OH:

C_{2}H_{5}OH(ℓ) ---> C_{2}H_{5}OH(g)

**Solution:**

enthalpy of formation, gas ---> −234 kJ/mol

enthalpy of formation, liquid ---> −276 kJ/molΔH

_{vap}= products − reactantsΔH

_{vap}= −234 − (−276) = 42 kJ/molThe value given here is 42.3 ± 0.4 kJ/mol

**Bonus Problem:** Calculate ΔH_{rxn} for the following reaction:

CaO(s) + COUse the following reactions and given ΔH values:_{2}(g) ---> CaCO_{3}(s)

Ca(s) + CO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> CaCO_{3}(s)ΔH = −812.8 kJ 2Ca(s) + O _{2}(g) ---> 2CaO(s)ΔH = −1269.8 kJ

**Solution:**

1) Look at the first data equation. It has one CaCO_{3} as a product. That's what we want, so leave the equation alone.

2) Look at the second data equation. It has CaO as a product and we want it as a reactant, so flip the equation. Also, it has 2Ca and we want one, not two. Divide the equation by 2.

3) The result:

Ca(s) + CO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> CaCO_{3}(s)ΔH = −812.8 kJ CaO(s) ---> Ca(s) + ^{1}⁄_{2}O_{2}(g)ΔH = +634.9 kJ Notice how the sign on the ΔH changed and it was divided by 2.

4) When you add the two equations, the Ca and the ^{3}⁄_{2}O_{2} will cancel, leaving the desired target equation. Add the two enthalpies for the answer.