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Homework #1

Chapter 9 Energy, Enthalpy, and Thermochemistry

1. When systems are in thermal contact (touching) they will eventually come to thermal

equilibrium (be the same temperature). This happens because heat (energy) flows from the

system with greater thermal energy (hot system) to the system with lower thermal energy (cold

system). When you (~36.5°C) touch a room temperature object (~25°C), you and the object are

not at thermal equilibrium. As soon as you touch, both you and the object start moving toward

equilibrium. Different materials can exchange heat at different rates. Metals feel colder than

plastic because metals have a lower specific heat capacity than plastic. This means that the

metal can transfer heat (energy) faster than plastic.

10. By convention, work is negative when the system does work, or energy flows from the system to

the surroundings. If work is positive, then the surrounding does work, or work flows from the

surroundings to the system. For the system to do work the volume must increase; therefore, Vf-

Vi =+number. If the negative sign was not in the equation then it would be opposite of

convention.

11. CP is the heat capacity at constant pressure and CV is the heat capacity at constant volume.

To understand why CP is greater than CV let’s assume that we have a monatomic ideal gas. Since

there is no potential energy (potential energy is stored in bonds and there are no bonds), ΔE will

just equal the ΔKE.

∆𝐸 = ∆(𝐾𝐸)𝑎𝑣𝑒 = 3

2𝑛𝑅∆𝑇

If we are at constant volume no work is done by or on the system, therefore, w=0. Since work is

0, all of the energy that is lost/gained goes into heat

𝑞 = 𝑛𝐶𝑉∆𝑇

Therefore, 3

2𝑛𝑅∆𝑇 = 𝑛𝐶𝑉∆𝑇

𝐶𝑉 =3

2𝑅

If we are at constant pressure, then the kinetic energy must be distributed between heat and

work. 3

2𝑛𝑅∆𝑇 = 𝑞 + 𝑤 = 𝑛𝐶𝑃∆𝑇 + −𝑃∆𝑉

For an ideal gas

𝑃∆𝑉 = 𝑛𝑅∆𝑇

Therefore, 3

2𝑛𝑅∆𝑇 = 𝑛𝐶𝑃∆𝑇 + −𝑛𝑅∆𝑇

𝐶𝑉 =5

2𝑅

This causes CP to be larger than CV.

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16. Plot (a) represents an exothermic reaction.

The lower the energy the more energetically stable the system is and the stronger the bonds in

the system.

In plot (a) the products are at a lower energy, therefore, have stronger bonds.

In plot (b) (endothermic reaction) in order for reactants to proceed to products, heat (energy)

must be introduced into the system.

20. 𝑤 = −𝑃𝑒𝑥∆𝑉

Path (12)

Step 1

P= 2.00 atm

𝑤1 = −(1.00 𝑎𝑡𝑚)(20.0 𝐿 − 10.0 𝐿) = −40.0 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −4,050 𝐽

Step 2

The volume is constant therefore w2 = 0 J

𝑤𝑡𝑜𝑡 = 𝑤1 + 𝑤2 = −4,050 𝐽 + 0 𝐽 = −4,050 𝐽

Path (34) Step 3

The volume is constant therefore w3 = 0 J

Step 4

P= 1.00 atm

𝑤4 = −(1.00 𝑎𝑡𝑚)(30.0 𝐿 − 10.0 𝐿) = −20.0 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −2,030 𝐽

𝑤𝑡𝑜𝑡 = 𝑤3 + 𝑤4 = −2,030 𝐽 + 0 𝐽 = −2,030 𝐽

The two answers are different; therefore, work is not a state function.

24. Internal energy is ΔE and is calculated by E q w

When heat is released the sign of q is - and when heat is absorbed the sign of q is +. When a gas is compressed, work is done on the gas, and the sign of work is + and when a gas expands, work is done by the gas, and the sign of work is -.

a) ∆𝐸 = 𝑞 + 𝑤 = −23 𝐽 + 100. 𝐽 = 77 𝐽

b) 𝑤 = −𝑃𝑒𝑥∆𝑉 = −(1.90 𝑎𝑡𝑚)(2.80 𝐿 − 8.30 𝐿) = 10.45 𝐿 ∙ 𝑎𝑡𝑚(101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 1,060 𝐽

∆𝐸 = 𝑞 + 𝑤 = 351 𝐽 + 1060 𝐽 = 1410 𝐽

c) 𝑤 = −𝑃𝑒𝑥∆𝑉 = −(1.00 𝑎𝑡𝑚)(29.1 𝐿 − 11.2 𝐿) = −17.9 𝐿 ∙ 𝑎𝑡𝑚(101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −1,810 𝐽

∆𝐸 = 𝑞 + 𝑤 = 1,037 𝐽 + −1,810 𝐽 = −773 𝐽 25zzz. This expansion occurs at constant pressure.

Calculate q 𝑞𝑃 = 𝑛𝐶𝑃∆𝑇

𝐶𝑃 = 20.80 𝐽

𝑚𝑜𝑙∙𝐾

𝑇𝑖 = 0.0℃ = 0.0 + 273.15 = 273.15 𝐾

𝑇𝑓 = 38.0℃ = 38.0 + 273.15 = 311.2 𝐾

∆𝑇 = 311.2 𝐾 − 273.15 𝐾 = 38.0 𝐾

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = (39.1 𝑚𝑜𝑙)(20.80 𝐽

𝑚𝑜𝑙∙𝐾)(38.0 𝐾) = 30,900 𝐽

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Calculate w

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(1.00 𝑎𝑡𝑚)(998 𝐿 − 876 𝐿) = −122 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚)

= −12,400 𝐽 Calculate ΔE

∆𝐸 = 𝑞 + 𝑤 = 30,900 𝐽 + −12,400 𝐽 = 18,500 𝐽

32. a) When gas is burned the car the engine heats up, therefore, the combustion of gasoline is exothermic.

b) The equation for water condensing on a pipe is H2O(g) H2O(l) + heat. Therefore the reaction is exothermic.

c) In order for dry ice (CO2(s)) to turn to a gas, heat must be added to the system (CO2(s) + heat CO2(g)), therefore, the sublimation of CO2 is endothermic.

d) In order for the reaction F2(g) 2F(g) to occur, bonds need to be broken which requires energy in the form of heat, therefore, this is an endothermic process.

29. This reaction occurs at constant pressure, therefore:

𝑞𝑃 = ∆𝐻 ΔH is negative for this reaction; therefore, this reaction is exothermic and releases heat.

a) 4.00 𝑚𝑜𝑙 𝐹𝑒(−1,652 𝑘𝐽

4 𝑚𝑜𝑙 𝐹𝑒) = −1,650 𝑘𝐽

or 1,652 kJ of energy released

b) 1.00 𝑚𝑜𝑙 𝐹𝑒2𝑂3 ( −1,652 𝑘𝐽

2 𝑚𝑜𝑙 𝐹𝑒2𝑂3) = −826 𝑘𝐽

or 826 kJ of energy released

c) 1.00 𝑔 𝐹𝑒 ( 1 𝑚𝑜𝑙 𝐹𝑒

55.85 𝑔 𝐹𝑒) (−1,652 𝑘𝐽

4 𝑚𝑜𝑙 𝐹𝑒) = −7.39 𝑘𝐽

or 7.39 kJ of energy released d) This is a limiting reagent problem

Calculate the moles Fe

10.0 𝑔 𝐹𝑒 (1 𝑚𝑜𝑙 𝐹𝑒

55.85 𝑔 𝐹𝑒) = 0.179 𝑚𝑜𝑙 𝐹𝑒

Calculate the moles of O2

2.00 𝑔 𝑂2 (1 𝑚𝑜𝑙 𝑂2

31.998 𝑔 𝑂2) = 0.0625 𝑚𝑜𝑙 𝑂2

Calculate the moles of O2 that are need to fully react with 0.179 mol of Fe

0.179 𝑚𝑜𝑙 𝐹𝑒 (3 𝑚𝑜𝑙 𝑂2

4 𝑚𝑜𝑙 𝐹𝑒) = 0.134 𝑚𝑜𝑙 𝑂2

Since we only have 0.0625 moles of O2, O2 is the limiting reagent.

Fe O2 (L.R) Fe2O3

I(mol) 0.179 0.0625 0

C (mol) 0.179-4x 0.0625-3x=0 +2x

F(mol) 0.0958 0 0.0416

0.0625 − 3𝑥 = 0 𝑥 = 0.0208 Calculate the heat released

0.0625 𝑚𝑜𝑙 𝑂2 (−1,652 𝑘𝐽

3 𝑚𝑜𝑙 𝑂2) = −34.4 𝑘𝐽

or 34.4 kJ of energy released

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36. Calculate the heat that needs to be produced. Example 9.1 states 1.3×108 J of heat is consumed in order to inflate the balloon. If this heat is coming from the combustion of propane, then the propane needs to produce -1.3×108 J. However, the heat transfer process is only 60.% efficient, therefore, we will need

% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =ℎ𝑒𝑎𝑡 𝑟𝑒𝑐𝑒𝑖𝑣𝑒𝑑

ℎ𝑒𝑎𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑(100%)

60. % =−1.3 × 108 𝐽

ℎ𝑒𝑎𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑(100%)

ℎ𝑒𝑎𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 = −2.2 × 108 𝐽 = −2.2 × 105 𝑘𝐽 Calculate the mass of propane

Assume the reaction takes place in an open container, therefore, ΔH=q.

−2.2 × 105 𝑘𝐽 (1 𝑚𝑜𝑙 𝐶3𝐻8

−2,221 𝑘𝐽) (

44.11 𝑔 𝐶3𝐻8

1 𝑚𝑜𝑙 𝐶3𝐻8) = 4.4 × 103 𝑔 𝐶3𝐻8

37. Given

∆𝐻 − ∆𝐸 = 2.5 𝑘𝐽

𝑚𝑜𝑙

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) This reaction occurs at constant pressure therefore, the expression can simplify to

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) ∆𝐻 − ∆𝐸 = 𝑃∆𝑉 = −𝑤

𝑤 = −𝑃∆𝑉 = −2.5 𝑘𝐽

𝑚𝑜𝑙

Therefore, the extra 2.5 𝑘𝐽

𝑚𝑜𝑙 is a result of the work done as a result of pushing back the

atmosphere.

38. ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) All the reactions occur at constant pressure, therefore, the expression simplifies to

∆𝐻 = ∆𝐸 + P∆V If ΔV is + (more moles of gas on the products) then ΔH > ΔE If ΔV is 0 (the same number of moles of gas on the products and the reactants) then ΔH = ΔE If ΔV is - (more moles of gas on the reactants) then ΔH < ΔE

a) ΔH = ΔE b) ΔH < ΔE c) ΔH > ΔE

39. Path 1 V= Constant Calculate q

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇

𝑛 = 1.00 𝑘𝑔 𝐶2𝐻6 (1000 𝑔

1 𝑘𝑔) (

1 𝑚𝑜𝑙 𝐶2𝐻6

30.08 𝑔 𝐶2𝐻6) = 33.2 𝑚𝑜𝑙 𝐶2𝐻6

𝑇𝑖 = 25.0℃ = 25.0 + 273.15 = 298.15 𝐾 𝑇𝑓 = 75.0℃ = 75.0 + 273.15 = 348.2 𝐾

∆𝑇 = 348.2 𝐾 − 298.15 𝐾 = 50.0 𝐾

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇 = (33.2 𝑚𝑜𝑙)(44.60 𝐽

𝑚𝑜𝑙∙𝐾)(50.0 𝐾) = 7.40 × 104 𝐽 = 74.0 𝑘𝐽

Calculate w 𝑤 = −𝑃𝑒𝑥∆𝑉 Since the volume is constant the w = 0 J

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = 0 + 74.0 𝑘𝐽 = 74.0 𝑘𝐽

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Calculate ΔH ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

Since the volume is constant 𝐻 = ∆𝐸 + (∆𝑃)𝑉 For gases

𝑃𝑉 = 𝑛𝑅𝑇 Since the number of moles and volume is constant

(∆𝑃)𝑉 = 𝑛𝑅(∆𝑇) Therefore

∆𝐻 = ∆𝐸 + (∆𝑃)𝑉 = ∆𝐸 + 𝑛𝑅(∆𝑇)

∆𝐻 = 7.40 × 104 𝐽 + (33.2 𝑚𝑜𝑙) (8.3145 𝐽

𝑚𝑜𝑙∙𝐾) (50.0 𝐾)

∆𝐻 = 87,800 J = 87.8 kJ Path 2 Constant P

Calculate q

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = (33.2 𝑚𝑜𝑙)(52.92 𝐽

𝑚𝑜𝑙∙𝐾)(50.0 𝐾) = 87,800 𝐽 = 87.8 𝑘𝐽

Calculate w 𝑤 = −𝑃𝑒𝑥∆𝑉 𝑃𝑉 = 𝑛𝑅𝑇 𝑃(∆𝑉) = 𝑛𝑅(∆𝑇)

𝑤 = −𝑃𝑒𝑥∆𝑉 = − 𝑛𝑅(∆𝑇) = −(33.2 𝑚𝑜𝑙) (8.3145 𝐽

𝑚𝑜𝑙∙𝐾) (50.0 𝐾)

= −13,800 𝐽 = −13.8 𝑘𝐽 Calculate ΔE

∆𝐸 = 𝑞 + 𝑤 = 87.8 𝑘𝐽 + −13.8 𝑘𝐽 = 74.0 𝑘𝐽 Calculate ΔH

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) At constant pressure ∆𝐻 = 𝑞𝑃 = 87.8 kJ

42. Step 1 (Constant P) Calculate w

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(10.0 𝑎𝑡𝑚)(5.0 𝐿 − 10.0 𝐿) = 50. 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 5,100 𝐽

= −3.0 × 103 𝐽 Calculate q

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 Because we have a monoatomic ideal gas

𝐶𝑃 =5

2𝑅

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = 𝑛5

2𝑅∆𝑇

𝑃𝑉 = 𝑛𝑅𝑇

𝑛𝑅(∆𝑇) = 𝑃∆(𝑉)

𝑞𝑃 = 𝑛5

2𝑅∆𝑇 = −

5

2𝑃∆𝑉 = −

5

2𝑤 = −

5

2(5,100 𝐽) = −13,000 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = −13,000 𝐽 + 5,100 𝐽 = −8,000 𝐽

Calculate ΔH ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) At constant pressure

∆𝐻 = 𝑞 = −13,000 𝐽

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Step 2 (Constant V) Calculate w

𝑤 = −𝑃𝑒𝑥∆𝑉 = 0 Since the volume is constant the w = 0 kJ Calculate q

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇 Because we have a monoatomic ideal gas

𝐶𝑉 =3

2𝑅

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇 = 𝑛3

2𝑅∆𝑇 =

3

2(5.0 𝐿)(20.0 𝑎𝑡𝑚 − 10.0 𝑎𝑡𝑚)

= 75 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 7,600 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = 0 𝐽 + 7,600 𝐽 = 7,600 𝐽

Calculate ΔH ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

Since the volume is constant ∆𝐻 = ∆𝐸 + (∆𝑃)𝑉 = 75 𝐿 ∙ 𝑎𝑡𝑚 + (20.0 𝑎𝑡𝑚 − 10.0 𝑎𝑡𝑚)(5.0 𝐿)

= 125 𝐿 ∙ 𝑎𝑡𝑚 ( 101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 12,700 𝐽

Step 3 (Constant P) Calculate w

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(20.0 𝑎𝑡𝑚)(25.0 𝐿 − 5.0 𝐿) = −400. 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚)

= −40,500 𝐽 Calculate q

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = 𝑛5

2𝑅∆𝑇 =

5

2𝑃∆𝑉 = −

5

2𝑤 = −

5

2(−40,500 𝐽) = 101,000 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = 101,000 𝐽 + −40,500 𝐽 = 61,000 𝐽

Calculate ΔH ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) At constant pressure

∆𝐻 = 𝑞 = 101,000 𝐽 Calculate totals

𝑤𝑡𝑜𝑡 = 5,100 𝐽 + 0 𝐽 + −40,500 𝐽 = −35,400 𝐽 𝑞𝑡𝑜𝑡 = −13,000 + 7,600 𝐽 + 101,000 𝐽 = 96,000 𝐽 ∆𝐸𝑡𝑜𝑡 = −8,000 + 7,600 𝐽 + 61,000 𝐽 = 61,000 𝐽 ∆𝐻𝑡𝑜𝑡 = −13,000 + 12,700 𝐽 + 101,000 𝐽 = 101,000 𝐽 43. There are two types of calorimetry: constant volume and constant pressure.

Constant Pressure (Coffee Cup) The constant pressure calorimeters are usually open to the atmosphere (coffee cup calorimeter). In a constant pressure calorimeter you measure the temperature of the solution directly to get the value of q (q=nCPΔT). If the solution (system) heats up (exothermic reaction) then q is negative. If the solution (system) cools down (endothermic reaction) then q is positive. When pressure is constant ΔH=q. Therefore, by knowing the temperature change of the calorimeter, the moles of substance, and the heat capacity of the substance, ΔH can be calculated.

Constant Volume (Bomb) The constant volume calorimeters have an air tight container called a bomb that reactants can be placed in and then it is charged (filled with) with O2 gas. The bomb is

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then placed in a water bath and the initial temperature of the water bath is taken. An electrical current is passed through the bomb initiating a combustion reaction in the bomb. Combustion reaction are exothermic, therefore, the water around the bomb (system) heats up. Since volume is constant ΔE=q and q=nCΔT.

44. You are trying to find ΔEwater=- ΔEAg (ΔE=q+w). Since neither the silver or the water change volume, ∆𝐸 = 𝑞, qwater=- qAg

a) Calculate q 𝑞𝐴𝑔 = 𝑚𝐶𝑃∆𝑇

∆𝑇 = 297 𝐾 − 273 𝐾 = 24𝐾 = 24℃

𝑞𝐴𝑔 = 𝑚𝐶𝑃∆𝑇 = (150.0 𝑔) (0.24 𝐽

𝑔∙℃) (24℃) = 9.0 × 102 𝐽

∆𝐸𝑤𝑎𝑡𝑒𝑟 = −𝑞𝐴𝑔 = −9.0 × 102 𝐽

The negative sign shows that energy is going from the water to the silver. Or 9.0×102 J of energy is required.

b) Calculate q 𝑞𝐴𝑔 = 𝑚𝐶𝑃∆𝑇

𝑚 = 1.0 𝑚𝑜𝑙 𝐴𝑔 (107.87 𝑔 𝐴𝑔

1 𝑚𝑜𝑙 𝐴𝑔) = 110 𝑔 𝐴𝑔

𝑞𝐴𝑔 = 𝑚𝐶𝑃∆𝑇 = (110 𝑔) (0.24 𝐽

𝑔∙℃) (1.0℃) = 26 𝐽

or 𝐶 = 26 𝐽

𝑚𝑜𝑙∙℃

c) You are asked to find m ∆𝐸𝑤𝑎𝑡𝑒𝑟 = −𝑞𝐴𝑔 = −𝑚𝐶𝑃∆𝑇

𝑚 =𝐶𝑃∆𝑇

−∆𝐸𝑤𝑎𝑡𝑒𝑟

∆𝐸𝑤𝑎𝑡𝑒𝑟 = −1.25 𝑘𝐽 (1000 𝐽

1 𝑘𝐽) = −1.25 × 103 𝐽

The negative sign shows that the energy is leaving the water. ∆𝑇 = 15.2 𝐾 − 12.0 𝐾 = 3.2 𝐾 = 3.2℃

𝑚 =−(−1.25 × 103 𝐽)

(0.24 𝐽𝑔∙℃

) (3.2℃)= 1,600 𝑔

46. The question wants you to calculate: 𝐶𝑃𝑚𝑒𝑡𝑎𝑙

.

Since the system is a closed system the energy that the metal loses must go into the water. −∆𝐸𝐻2𝑂 = ∆𝐸𝑚𝑒𝑡𝑎𝑙

The water and the metal will not change volume making the work 0, therefore, ∆𝐸 = 𝑞 and all the heat that leaves the metal will enter the water. 𝐸 = 𝑚𝐶𝑃∆𝑇

−∆𝐸𝐻2𝑂 = ∆𝐸𝑚𝑒𝑡𝑎𝑙

−𝑚𝐻2𝑂𝐶𝑃𝐻2𝑂∆𝑇𝐻2𝑂 = 𝑚𝑚𝑒𝑡𝑎𝑙𝐶𝑃𝑚𝑒𝑡𝑎𝑙

∆𝑇𝑚𝑒𝑡𝑎𝑙

∆𝑇𝐻2𝑂 = 18.3℃ − 15.0℃ = 3.3℃

∆𝑇𝑚𝑒𝑡𝑎𝑙 = 18.3℃ − 75.0℃ = −56.7℃

−(150.0 𝑔) (4.18 𝐽

𝑔∙℃) (3.3℃) = (150.0 𝑔)𝐶𝑃𝑚𝑒𝑡𝑎𝑙

(−56.7℃)

𝐶𝑃𝑚𝑒𝑡𝑎𝑙= 0.24

𝐽

𝑔∙℃

8

48. The question wants you to calculate: 𝑇𝑓.

The system is at constant pressure, therefore, ∆𝐻 = 𝑞 and all the heat that leaves the metals will enter the water.

−∆𝐻𝐻2𝑂 = ∆𝐻𝐴𝑙 + ∆𝐻𝐹𝑒

∆𝐻 = 𝑞 = 𝑚𝐶𝑃∆𝑇 Therefore,

−𝑚𝐻2𝑂𝐶𝑃𝐻2𝑂∆𝑇𝐻2𝑂 = 𝑚𝐴𝑙𝐶𝑃𝐴𝑙

∆𝑇𝐴𝑙 + 𝑚𝐹𝑒𝐶𝑃𝐹𝑒∆𝑇𝐹𝑒

−(97.3 𝑔) (4.18 𝐽

𝑔∙℃) (𝑇𝑓 − 22.0℃)

= (5.0 𝑔) (0.89 𝐽

𝑔∙℃) (𝑇𝑓 − 100.0℃)

+ (10.0 𝑔) (0.45 𝐽

𝑔∙℃) (𝑇𝑓 − 100.0℃)

𝑇𝑓 = 23.7℃

51. The question wants you to calculate ΔHrxn.

Reaction of interest Ag+(aq) + Cl-(aq) AgCl (s)

They want you to calculate ΔHrxn Constant pressure calorimeter ∆𝐻𝑟𝑥𝑛 = 𝑞𝑟𝑥𝑛

𝑞𝑟𝑥𝑛 = −(𝑞𝑠𝑜𝑙 + 𝑞𝑐𝑎𝑙) 𝑞 𝑐𝑎𝑙 = 0 ∆𝐻𝑟𝑥𝑛 = −𝑞 𝑠𝑜𝑙 = −𝑚𝐶∆𝑇

∆𝐻𝑟𝑥𝑛 = −𝑚𝐶∆𝑇

∆𝐻𝑟𝑥𝑛 = −(100. 𝑔) (4.18 𝐽

𝑔∙℃) (23.40℃ − 22.60℃) = −334 𝐽 = −0.334 𝑘𝐽

Calculate ΔHrxn per mol (limiting reagent problem) Calculate moles of Ag+

50.0 𝑚𝐿 𝐴𝑔𝑁𝑂3 (1 𝐿

1000 𝑚𝐿) (

0.100 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3

1 𝐿 𝐴𝑔𝑁𝑂3 ) (

1 𝑚𝑜𝑙 𝐴𝑔+

1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3) = 0.00500 𝑚𝑜𝑙 𝐴𝑔+

Calculate moles of Cl-

50.0 𝑚𝐿 𝐻𝐶𝑙 (1 𝐿

1000 𝑚𝐿) (

0.100 𝑚𝑜𝑙 𝐻𝐶𝑙

1 𝐿 𝐻𝐶𝑙 ) (

1 𝑚𝑜𝑙 𝐶𝑙_

1 𝑚𝑜𝑙 𝐻𝐶𝑙) = 0.00500 𝑚𝑜𝑙 𝐶𝑙+

No limiting reagent both reactants are used up. Since all of the reactants and products have a coefficient of one divided by the moles of any one of the reactants or products

∆𝐻𝑟𝑥𝑛 =−0.334 𝑘𝐽

0.00500 𝑚𝑜𝑙= −66.8

𝑘𝐽

𝑚𝑜𝑙

54. The question wants you to calculate ΔHrxn.

Reaction of interest NH4NO3(s) NH4

+(aq) + NO3-(aq)

You need to calculate ΔHrxn. Since it is a coffee cup calorimeter, pressure is constant and ΔHrxn=qrxn. The question also tells you that qcal=0. The heat from the reaction will go into the solution therefore qrxn=-qsol and ∆𝐻𝑟𝑥𝑛 = −𝑞𝑠𝑜𝑙.

Ag+ Cl- AgCl

I(mol) 0.00500 0.00500 0

F(mol) 0 0 0.00500

9

∆𝐻𝑟𝑥𝑛 = −𝑚𝑠𝑜𝑙𝐶𝑠𝑜𝑙∆𝑇𝑠𝑜𝑙 𝑚𝑠𝑜𝑙 = 1.60 𝑔 + 75.0 𝑔 = 76.5 𝑔 ∆𝑇𝑠𝑜𝑙 = 23.34℃ − 25.00℃ = −1.66℃

∆𝐻𝑟𝑥𝑛 = −(76.6 𝑔) (4.18 𝐽

℃∙𝑔) (−1.66℃) = 532 𝐽

Calculate ΔH per mol

1.60 𝑔 𝑁𝐻4𝑁𝑂3 (1 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3

80.06 𝑔 𝑁𝐻4𝑁𝑂3) = 0.0200 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3

Since there is only one mole of NH4NO3 in the equation, divide by 0.0200 mol.

∆𝐻𝑟𝑥𝑛 =532 𝐽

0.0200 𝑚𝑜𝑙= 2.66 × 104

𝐽

𝑚𝑜𝑙= 26.6

𝑘𝐽

𝑚𝑜𝑙

55. The question wants you to calculate Tf of solution. The reaction of interest is: CaCl2(s) Ca2+(aq) + 2Cl-(aq) ΔH=-81.5 kJ

Since there the system is at constant pressure ΔHrxn=qrxn. The heat from the reaction will go into the solution therefore qrxn=-qsol and ∆𝐻𝑟𝑥𝑛 = −𝑞𝑠𝑜𝑙

∆𝐻𝑟𝑥𝑛 = 11.0 𝑔 𝐶𝑎𝐶𝑙2 (1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2

110.98 𝑔 𝐶𝑎𝐶𝑙2) (

−81.5 𝑘𝐽

1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2) = −8.08 𝑘𝐽 = −8,080 𝐽

∆𝐻𝑟𝑥𝑛 = −𝑞𝑠𝑜𝑙 = −𝑚𝑠𝑜𝑙𝐶𝑠𝑜𝑙∆𝑇𝑠𝑜𝑙 𝑚𝑠𝑜𝑙 = 11.0 𝑔 + 125 𝑔 = 136 𝑔

−8,080 J = −(136 𝑔) (4.18 𝐽

℃∙𝑔) ∆𝑇

∆𝑇 = 14.2℃ = 𝑇𝑓 − 25.0℃

𝑇𝑓 = 39.2℃

56. The question wants you to calculate Tf of solution.

The reaction of interest is: 2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2H2O(l) ΔH=-118 kJ

Constant pressure calorimeter ∆𝐻𝑟𝑥𝑛 = 𝑞𝑟𝑥𝑛 𝑞𝑟𝑥𝑛 = −(𝑞𝑠𝑜𝑙 + 𝑞𝑐𝑎𝑙) 𝑞 𝑐𝑎𝑙 = 0 ∆𝐻𝑟𝑥𝑛 = −𝑞 𝑠𝑜𝑙 = −𝑚𝐶∆𝑇

Calculate ΔHrxn in J Find the limiting reagent

Therefore, use HCl to calculate heat.

0.0500 𝑚𝑜𝑙 𝐻𝐶𝑙 (−118 𝑘𝐽

2 𝑚𝑜𝑙 𝐻𝐶𝑙) = −2.95 𝑘𝐽 = −2,950 𝐽

∆𝐻𝑟𝑥𝑛 = −𝑚𝐶∆𝑇

−2,950 𝐽 = −(400. 𝑔) (4.18 𝐽

𝑔∙℃) (𝑇𝑓 − 25.0℃)

𝑇𝑓 = 26.8℃

Since the reaction is exothermic the final temperature should be greater than the initial temperature.

HCl (L.R.) Ba(OH)2 BaCl2 H20

I(mol) (.1000 L)(0.500 M)= 0.0500

(0.3000 L)(0.100 M)= 0.0300

0

F(mol) 0 0.0050 0.0250

10

57. a) This question should have read: What is the heat capacity of the calorimeter?

The energy from the reaction goes into the calorimeter.

∆𝐸𝑟𝑥𝑛 = −∆𝐸𝑐𝑎𝑙

Since this is a bomb calorimeter volume is constant and ∆𝐸 = 𝑞.

𝑞𝑐𝑜𝑚 = −𝑞𝑐𝑎𝑙 = −𝐶𝑐𝑎𝑙∆𝑇 𝑞𝑐𝑜𝑚 = 𝐸𝑐𝑜𝑚 = −802 𝑘𝐽

𝑚𝑜𝑙 (Constant Volume)

Calculate ΔEcom in kJ (CH4(g) + 2O2(g) CO2(g) + 2H2O(l))

6.79 𝑔 𝐶𝐻4 ( 1 𝑚𝑜𝑙 𝐶𝐻416.50 𝑔 𝐶𝐻4

) ( −802 𝑘𝐽

1 𝑚𝑜𝑙 𝐶𝐻4) = −339 𝑘𝐽

𝑞𝑐𝑜𝑚 = −𝐶𝑐𝑎𝑙∆𝑇 −339 𝑘𝐽 = −𝐶𝑐𝑎𝑙(10.8℃)

𝐶𝑐𝑎𝑙 = 31.4 𝑘𝐽

℃

a) The question wants you to calculate ΔEcom(C2H2) 𝑞𝑐𝑜𝑚 = 𝐸𝑐𝑜𝑚 = −𝑞𝑐𝑎𝑙 = −𝐶𝑐𝑎𝑙∆𝑇

𝐸𝑐𝑜𝑚 = − (31.4 𝑘𝐽

℃) (16.9℃) = −531 𝑘𝐽

Convert to 𝑘𝐽

𝑚𝑜𝑙

Determine the moles of C2H2

6.79 𝑔 𝐶2𝐻2 (1 𝑚𝑜𝑙 𝐶2𝐻2

26.04 𝑔 𝐶2𝐻2) = 0.484 𝑚𝑜𝑙 𝐶2𝐻2

−531 𝑘𝐽

0.484 𝑚𝑜𝑙= −1.10 × 103

𝑘𝐽

𝑚𝑜𝑙

58. a) ∆𝐸𝑟𝑥𝑛 = −∆𝐸𝑐𝑎𝑙

Since this is a bomb calorimeter volume is constant and ∆𝐸 = 𝑞.

𝑞𝑐𝑜𝑚 = −𝑞𝑐𝑎𝑙 = −𝐶𝑐𝑎𝑙∆𝑇

𝑞𝑐𝑜𝑚 = 0.1584 𝑔 𝑏𝑒𝑛𝑧𝑜𝑖𝑐 𝑎𝑐𝑖𝑑 (−26.42 𝑘𝐽

1 𝑔 𝑏𝑒𝑛𝑧𝑜𝑖𝑐 𝑎𝑐𝑖𝑑) = −4.18 𝑘𝐽

−4.18 𝑘𝐽 = −𝐶𝑐𝑎𝑙(2.54℃)

𝐶𝑐𝑎𝑙 = 1.64 𝑘𝐽

℃

Calculate ΔEcom for vanillin

𝐸𝑐𝑜𝑚 = −𝐶𝑐𝑎𝑙∆𝑇 = −(1.64 𝑘𝐽

℃)(3.25℃) = −5.35 𝑘𝐽

Convert to per g and per mol

𝐸𝑐𝑜𝑚 =−5.35 𝑘𝐽

0.2130 𝑔= −2.51 𝑘𝐽

𝑔

𝐸𝑐𝑜𝑚 = −2.51 𝑘𝐽

𝑔(152.148 𝑔 𝐶8𝐻8𝑂3

1 𝑚𝑜𝑙 𝐶8𝐻8𝑂3) = −3820 𝑘𝐽

𝑚𝑜𝑙

60. The question wants you to calculate w and ΔE. Calculate work.

𝑤 = −𝑃𝑒𝑥∆𝑉 Originally the substance is a liquid and then it converts to a gas. Liquids take up essential no volume when compared to their gas counterparts. Therefore, the change in volume will equal the volume of the gas. Determine the volume of the gas

𝑇 = 80. ℃ = 80. +273.15 = 353 𝐾 𝑃𝑉 = 𝑛𝑅𝑇

𝑉 =𝑛𝑅𝑇

𝑃=

(1 𝑚𝑜𝑙)(0.08206 𝐿∙𝑎𝑡𝑚𝑚𝑜𝑙∙𝐾

)(353. 𝐾)

1 𝑎𝑡𝑚= −2,940 𝐿

11

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(1 𝑚𝑜𝑙)(29.0 𝐿 − 0 𝐿) = 29.0 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −2,940 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤

Pressure is constant 𝑞 = ∆𝐻

∆𝐻 = 30.7 𝑘𝐽

𝑚𝑜𝑙

Since we only have 1 mol ∆𝐻 = 30.7 𝑘𝐽 = 30,700 𝐽 ∆𝐸 = 𝑞 + 𝑤 = −2,940 𝐽 + 30,700𝐽 = 27,800 𝐽

66. 1

3(3𝐹𝑒𝑂(𝑠) + C𝑂2(𝑔) → 𝐹𝑒3𝑂4(𝑠) + 𝐶𝑂(𝑔)) ∆𝐻𝑎 = −1

3∆𝐻3 = −1

3(18 𝑘𝐽) = −6.0 𝑘𝐽

1

6(2𝐹𝑒3𝑂4(𝑠) + C𝑂2(𝑔) → 3𝐹𝑒2𝑂3(𝑠) + 𝐶𝑂(𝑔)) ∆𝐻𝑏 = −1

6∆𝐻2 = −1

6(39 𝑘𝐽) = 6.5 𝑘𝐽

1

2(3𝐹𝑒2𝑂3(𝑠) + 3CO(𝑔) → 2Fe(𝑠) + 3𝐶𝑂2(𝑔)) ∆𝐻𝑐 =

1

2∆𝐻1 =

1

2(−23 𝑘𝐽) = −12 𝑘𝐽

𝐹𝑒𝑂(𝑠) + 𝐶𝑂(𝑔) → 𝐹𝑒(𝑠) + 𝐶𝑂2(𝑔) ∆𝐻 = ∆𝐻𝑎 + ∆𝐻𝑏 + ∆𝐻𝑐 ∆𝐻 = −6.0 𝑘𝐽 + 6.5 𝑘𝐽 − 12 𝑘𝐽 = −12 𝑘𝐽

67. 𝐶4𝐻4(𝑔) + 5𝑂2(𝑔) → 4C𝑂2(𝑔) + 2𝐻2𝑂(𝑙) ∆𝐻1 = −2,341 𝑘𝐽

𝑚𝑜𝑙

𝐶4𝐻8(𝑔) + 6𝑂2(𝑔) → 4C𝑂2(𝑔) + 4𝐻2𝑂(𝑙) ∆𝐻2 = −2,755 𝑘𝐽

𝑚𝑜𝑙

𝐻2(𝑔) +1

2𝑂2(𝑔) → 𝐻2𝑂(𝑙) ∆𝐻3 = −286

𝑘𝐽

𝑚𝑜𝑙

𝐶4𝐻4(𝑔) + 5𝑂2(𝑔) → 4C𝑂2(𝑔) + 2𝐻2𝑂(𝑙) ∆𝐻𝑎 = ∆𝐻1 = −2,341 𝑘𝐽

𝑚𝑜𝑙

4C𝑂2(𝑔) + 4𝐻2𝑂(𝑙) → 𝐶4𝐻8(𝑔) + 6𝑂2(𝑔) ∆𝐻𝑏 = −∆𝐻2 = 2,755 𝑘𝐽

𝑚𝑜𝑙

2 (𝐻2(𝑔) +1

2𝑂2(𝑔) → 𝐻2𝑂(𝑙)) ∆𝐻𝑐 = 2∆𝐻3 = 2 (−286

𝑘𝐽

𝑚𝑜𝑙) = −572

𝑘𝐽

𝑚𝑜𝑙

𝐶4𝐻2(𝑔) + 2𝐻2(𝑔) → 𝐶4𝐻8(𝑔) ∆𝐻 = ∆𝐻𝑎 + ∆𝐻𝑏 + ∆𝐻𝑐

∆𝐻 = −2,341 𝑘𝐽

𝑚𝑜𝑙+ 2,755

𝑘𝐽

𝑚𝑜𝑙+ −572

𝑘𝐽

𝑚𝑜𝑙= −158

𝑘𝐽

𝑚𝑜𝑙

68. 𝑁𝑂(𝑔) + 𝑂3(𝑔) → N𝑂2(𝑔) + 𝑂2(𝑔) ∆𝐻𝑎 = ∆𝐻3 = −199 𝑘𝐽 1

2(3𝑂2(𝑔) → 2𝑂3(𝑔)) ∆𝐻𝑏 = −

1

2∆𝐻1 = −

1

2(−427 𝑘𝐽) = 214 𝑘𝐽

1

2(2𝑂(𝑔) → 𝑂3(𝑔)) ∆𝐻𝑐 = −

1

2∆𝐻2 = −

1

2(495 𝑘𝐽) = −248 𝑘𝐽

𝑁𝑂(𝑔) + 𝑂(𝑔) → 𝑁𝑂2(𝑔) ∆𝐻 = ∆𝐻𝑎 + ∆𝐻𝑏 + ∆𝐻𝑐 ∆𝐻 = −199 𝑘𝐽 + 214 𝑘𝐽 + −248 𝑘𝐽 = −233 𝑘𝐽

69. 𝐶6𝐻4(𝑂𝐻)2(𝑎𝑞) → 𝐶6𝐻4𝑂2(𝑔) + 𝐻2(𝑔) ∆𝐻𝑎 = ∆𝐻1 = 177.4 𝑘𝐽

𝐻2𝑂2(𝑎𝑞) → 𝐻2(𝑔) + 𝑂2(𝑔) ∆𝐻𝑏 = −∆𝐻2 = 191.2 𝑘𝐽

2 (𝐻2(𝑔) +1

2𝑂2(𝑔) → 𝐻2𝑂(𝑔)) ∆𝐻𝑐 = 2∆𝐻3 = 2(−241.8 𝑘𝐽) = −483.6 𝑘𝐽

2(𝐻2𝑂(𝑔) → 𝐻3𝑂(𝑙)) ∆𝐻𝑑 = 2∆𝐻4 = 2(−43.8 𝑘𝐽) = −87.6 𝑘𝐽

𝐶6𝐻4(𝑂𝐻)2(𝑎𝑞) + 𝐻2𝑂2(𝑎𝑞) → 𝐶6𝐻4𝑂2(𝑔) + 2𝐻2𝑂(𝑙) ∆𝐻 = ∆𝐻𝑎 + ∆𝐻𝑏 + ∆𝐻𝑐 + ∆𝐻𝑑 ∆𝐻 = 177.4 𝑘𝐽 + 191.2 𝑘𝐽 + −483.6 𝑘𝐽 + −87.6 𝑘𝐽 = −202.6 𝑘𝐽

12

70. 𝑃4(𝑠) + 6𝐶𝑙2(𝑔) → 4𝑃𝐶𝑙3(𝑔) ∆𝐻𝑎 = ∆𝐻1 = −1,225.6 𝑘𝐽 𝑃4𝑂10(𝑠) → 𝑃4(𝑠) + 5𝑂2(𝑔) ∆𝐻𝑏 = −∆𝐻2 = 2,967.3 𝑘𝐽

6(𝑃𝐶𝑙5(𝑔) → 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔)) ∆𝐻𝑐 = −6∆𝐻3 = 6(−84.2 𝑘𝐽) = 505.2 𝑘𝐽

10 (𝑃𝐶𝑙3(𝑔) +1

2𝑂2(𝑔) → 𝐶𝑙3𝑃𝑂(𝑔)) ∆𝐻𝑑 = 10∆𝐻4 = 10(−285.7 𝑘𝐽) = −2,857 𝑘𝐽

𝑃4𝑂10(𝑠) + 6𝑃𝐶𝑙5(𝑔) → 10𝐶𝑙3𝑃𝑂(𝑔) ∆𝐻 = ∆𝐻𝑎 + ∆𝐻𝑏 + ∆𝐻𝑐 + ∆𝐻𝑑 ∆𝐻 = −1,225.6 𝑘𝐽 + 2,967.3 𝑘𝐽 + 505.2 𝑘𝐽 + −2,857 𝑘𝐽 = −610.1 𝑘𝐽

71. Na(s) + ½Cl2(g) NaCl(s) H2(g) + ½O2(g) H2O(l) 6C(graphite,s) + 6H2(g) + 3O2(g) C6H12O6(l) Pb(s) + S(s) + 2O2(g) PbSO4(s) 76.

∆𝐻𝑟𝑥𝑛° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐)

∆𝐻𝑟𝑥𝑛° = 𝑛𝐴𝑙2𝑂3

∆𝐻𝑓°(𝐴𝑙2𝑂3) + 𝑛𝐴𝑙𝐶𝑙3

∆𝐻𝑓°(𝐴𝑙𝐶𝑙3) + 𝑛𝑁𝑂∆𝐻𝑓

°(𝑁𝑂) − 𝑛𝐴𝑙∆𝐻𝑓°(𝐴𝑙)

− 𝑛𝑁𝐻4𝐶𝑙𝑂4∆𝐻𝑓

°(𝑁𝐻4𝐶𝑙𝑂4)

∆𝐻𝑟𝑥𝑛° = (1) (−1,676

𝑘𝐽

𝑚𝑜𝑙) + (1) (−704

𝑘𝐽

𝑚𝑜𝑙) + (3) (90.

𝑘𝐽

𝑚𝑜𝑙) − (3) (0

𝑘𝐽

𝑚𝑜𝑙) − (3) (−295

𝑘𝐽

𝑚𝑜𝑙)

= −2,677 𝑘𝐽

𝑚𝑜𝑙

77.

∆𝐻𝑟𝑥𝑛° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐)

∆𝐻𝑟𝑥𝑛° = 𝑛𝐻2𝑂∆𝐻𝑓

°(𝐻2𝑂) + 𝑛𝑁2∆𝐻𝑓

°(𝑁2) + 𝑛𝐶𝑂2∆𝐻𝑓

°(𝐶𝑂2) − 𝑛𝑁2𝑂4∆𝐻𝑓

°(𝑁2𝑂4)

− 𝑛𝑁2𝐻3𝐶𝐻3∆𝐻𝑓

°(𝑁2𝐻3𝐶𝐻3)

∆𝐻𝑟𝑥𝑛° = (12) (−242

𝑘𝐽

𝑚𝑜𝑙) + (9) (0

𝑘𝐽

𝑚𝑜𝑙) + (4) (−393.5

𝑘𝐽

𝑚𝑜𝑙) − (5) (−20

𝑘𝐽

𝑚𝑜𝑙) − (4) (54

𝑘𝐽

𝑚𝑜𝑙)

= −4,594 𝑘𝐽

𝑚𝑜𝑙

Compound ΔH° ( 𝑘𝐽

𝑚𝑜𝑙)

Al(s) 0

NH4ClO4(s) -295

Al2O3(s) -1676

AlCl3(s) -704

NO(g) 90.

H2O(g) -242

Compound ΔHf°( 𝑘𝐽

𝑚𝑜𝑙)

N2O4(l) -20.

N2H3CH3(l) 54

H2O(g) -242

N2(g) 0

CO2(g) -393.5

13

81.

∆𝐻𝑟𝑥𝑛° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐)

∆𝐻𝑟𝑥𝑛° = 𝑛𝑁2

∆𝐻𝑓°(𝑁2) + 𝑛𝐻𝐹∆𝐻𝑓

°(𝐻𝐹) + 𝑛𝐶𝑙2∆𝐻𝑓

°(𝐶𝑙2) − 𝑛𝐶𝑙𝐹3∆𝐻𝑓

°(𝐶𝑙𝐹3) − 𝑛𝑁𝐻3∆𝐻𝑓

°(𝑁𝐻3)

−1,196 𝑘𝐽 = (1) (0 𝑘𝐽

𝑚𝑜𝑙) + (6) (−271

𝑘𝐽

𝑚𝑜𝑙) + (1) (0

𝑘𝐽

𝑚𝑜𝑙) − (2) (∆𝐻𝑓

°(𝐶𝑙𝐹3))

− (2) (−46 𝑘𝐽

𝑚𝑜𝑙)

∆𝐻𝑓°(𝐶𝑙𝐹3) = −169

𝑘𝐽

𝑚𝑜𝑙

82. Combustion of C2H4(g)

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) ΔHcom = -1,411.1molkJ

∆𝐻𝑐𝑜𝑚° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐)

∆𝐻𝑐𝑜𝑚° = 𝑛𝐶𝑂2

∆𝐻𝑓°(𝐶𝑂2) + 𝑛𝐻2𝑂∆𝐻𝑓

°(𝐻2𝑂) − 𝑛𝐶2𝐻4∆𝐻𝑓

°(𝐶2𝐻4) − 𝑛𝑂2∆𝐻𝑓

°(𝑂2)

−1,411.1 𝑘𝐽 = (2) (−393.5 𝑘𝐽

𝑚𝑜𝑙) + (2) (−285.8

𝑘𝐽

𝑚𝑜𝑙) − (1) (∆𝐻𝑓

°(𝐶2𝐻4)) − (3) (0 𝑘𝐽

𝑚𝑜𝑙)

∆𝐻𝑓°(𝐶2𝐻4) = 52.5

𝑘𝐽

𝑚𝑜𝑙

85. The question wants you to calculate the mass of C2H2 needed to heat 1 gallon of water 10˚C.

Calculate the heat (energy) needed to raise 1 gallon of water 10˚C.

𝑞 = 𝑚𝐶∆𝑇 Calculate the mass of water

1.00 𝑔𝑎𝑙 𝐻2𝑂 (3.7854 𝐿

1 𝑔𝑎𝑙 ) (10−3 𝑚3

1 𝐿)(100 𝑐𝑚

1 𝑚)

3(1.00 𝑔 𝐻2𝑂

1 𝑐𝑚3 ) = 3,790 𝑔

Calculate the heat

𝑞 = 𝑚𝐶∆𝑡 = (3,790 𝑔) (4.18 𝐽

𝑔∙℃) (10.0℃) = 158,000 𝐽 = 158 𝑘𝐽

Calculate the amount of heat needed since the process is only 80.0% efficient ℎ𝑒𝑎𝑡 𝑔𝑜𝑖𝑛𝑔 𝑖𝑛𝑡𝑜 𝑤𝑎𝑡𝑒𝑟

𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑100% = % 𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡

𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 =ℎ𝑒𝑎𝑡 𝑔𝑜𝑖𝑛𝑔 𝑖𝑛𝑡𝑜 𝑤𝑎𝑡𝑒𝑟

% 𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡100% =

158 𝑘𝐽

80.0100% = 198 𝑘𝐽

Calculate the amount of acetylene need to make 198 kJ of heat C2H2(g) + 5

2O2(g) 2CO2(g) + H2O(l) ΔH°com = -1,300. kJ

−198 𝑘𝐽 (1 𝑚𝑜𝑙 𝐶2𝐻2

−1,300 𝑘𝐽) (

26.04 𝑔 𝐶2𝐻2

1 𝑚𝑜𝑙 𝐶2𝐻2) = 3.97 𝑔 𝐶2𝐻2

Compound ΔHf°( 𝑘𝐽

𝑚𝑜𝑙)

NH3(g) -46

N2(g) 0

HF(g) -271

Cl2(g) 0

14

88. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) (remember enthalpies of combustion always form liquid water)

∆𝐻𝑐𝑜𝑚° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∆𝐻𝑓°(𝑟𝑒𝑎𝑐)

∆𝐻𝑐𝑜𝑚° = 𝑛𝐶𝑂2

∆𝐻𝑓°(𝐶𝑂2) + 𝑛𝐻2𝑂∆𝐻𝑓

°(𝐻2𝑂) − 𝑛𝐶2𝐻5𝑂𝐻∆𝐻𝑓°(𝐶2𝐻5𝑂𝐻) − 𝑛𝑂2

∆𝐻𝑓°(𝑂2)

∆𝐻𝑐𝑜𝑚° = (2) (−393.5

𝑘𝐽

𝑚𝑜𝑙) + (3) (−286

𝑘𝐽

𝑚𝑜𝑙) − (1) (−278

𝑘𝐽

𝑚𝑜𝑙) − (3) (0

𝑘𝐽

𝑚𝑜𝑙) = −1,367

𝑘𝐽

𝑚𝑜𝑙

(−1,367 𝑘𝐽

1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 ) (

1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻

40.08 𝑔 𝐶2𝐻5𝑂𝐻) = −29.67

𝑘𝐽

𝑔

103. Work is done by the system when the volume expands (more moles of gas of the products than reactants). Work is done by the surroundings when the volume decreases (more moles of gas of the reactants than products). No work is done when the volume is constant (equal moles of gas of the reactants and products). a) Surroundings does work on system b) No work c) System does work on surroundings d) Surroundings does work on system e) System does work on surroundings

107. For formation you need to form 1 mole of product For combustion you need to combust 1 mole of compound For neutralization you need to form 1 mole of H2O For solution you need to dissolve 1 mole of substance

a) 2Al(s) + 3

2O2(g) Al2O3(s)

b) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) c) 1

2Ba(OH)2(aq) + HCl(aq) 1

2BaCl2(aq) + H2O(l)

d) 2C(graphite,s) + 3

2H2(g) +1

2Cl2(g) C2H3Cl(g)

e) C6H6(l) + 15

2O2(g) 6CO2(g) + 3H2O(l)

f) NH4Br(s) NH4+(aq) + Br-(aq)

Compound ΔHf°( 𝑘𝐽

𝑚𝑜𝑙)

C2H5OH(l) -278

O2(g) 0

CO2(g) -393.5

H2O(l) -286

15

125. Path 1 (Step 1) Constant Pressure Calculate work (work will have a negative sign because the system expanded)

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(33.2 𝑎𝑡𝑚)(55.0 𝐿 − 15.0 𝐿) = −120. 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚)

= −12,200 𝐽 Calculate heat

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 Because we have a monoatomic ideal gas

𝐶𝑃 =5

2𝑅

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = 𝑛5

2𝑅∆𝑇

𝑃𝑉 = 𝑛𝑅𝑇 𝑛𝑅(∆𝑇) = 𝑃(∆𝑉)

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = 𝑛5

2𝑅∆𝑇 =

5

2𝑃∆𝑉 = −

5

2𝑤 = −

5

2(−12,200 𝐽) = 30,500 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = 30,500 𝐽 + −12,200 𝐽 = 18,300 𝐽

Calculate ΔH (constant pressure) ∆𝐻 = 𝑞𝑃 = 30,500 J

Path 1 (Step 2) Calculate work (work will have a positive sign because the system contracts)

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(6.00 𝑎𝑡𝑚)(20.0 𝐿 − 55.0 𝐿) = 210. 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 21,300 𝐽

Calculate ΔE (all of the energy of the gas is in kinetic energy) (𝐾𝐸)𝑎𝑣𝑒 = 3

2𝑅𝑇 (per mol)

(𝐾𝐸)𝑎𝑣𝑒 = 𝑛3

2𝑅𝑇 (per atom)

∆𝐸 = ∆(𝐾𝐸)𝑎𝑣𝑒 = 𝑛3

2𝑅∆𝑇

𝑃𝑉 = 𝑛𝑅𝑇

𝑛𝑅(∆𝑇) = ∆(𝑃𝑉)

∆𝐸 = ∆(𝐾𝐸)𝑎𝑣𝑒 =3

2∆(𝑃𝑉) =

3

2(𝑃2𝑉2 − 𝑃1𝑉1)

=3

2((6.00 𝑎𝑡𝑚)(20.0 𝐿) − (3.00 𝑎𝑡𝑚)(55.0 𝐿))

= −67.5 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −6,840 𝐽

Calculate q ∆𝐸 = 𝑞 + 𝑤 −6,840 𝐽 = 𝑞 + 21,300 𝐽 𝑞 = −28,100 𝐽

Calculate ΔH ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) = ∆𝐸 + (𝑃2𝑉2 − 𝑃1𝑉1)

= −67.5 𝐿 ∙ 𝑎𝑡𝑚 + ((6.00 𝑎𝑡𝑚)(20.0 𝐿) − (3.00 𝑎𝑡𝑚)(55.0 𝐿))

= −113 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = −11,400 𝐽

Calculate Totals 𝑤𝑡𝑜𝑡 = −12,200 𝐽 + 21,300 𝐽 = 9,100 𝐽 𝑞𝑡𝑜𝑡 = 30,2 = 500 𝐽 + −28,100 𝐽 = 2,400 𝐽 ∆𝐸𝑡𝑜𝑡 = 18,300 𝐽 + −6,840 𝐽 = 11,500 𝐽 ∆𝐻𝑡𝑜𝑡 = 30,500 𝐽 + −11,400 𝐽 = 19,100 𝐽

Path 2 (step 1) constant volume Calculate work since volume is constant work=0 𝑤 = −𝑃𝑒𝑥∆𝑉 = 0

16

Calculate heat

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇 Because we have a monoatomic ideal gas

𝐶𝑉 =3

2𝑅

𝑞𝑉 = 𝑛𝐶𝑉∆𝑇 = 𝑛3

2𝑅∆𝑇 =

3

2(∆𝑃)𝑉 =

3

2(6.00 𝑎𝑡𝑚 − 3.00 𝑎𝑡𝑚)(15.0 𝐿)

= 67.5 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 6,840 𝐽

Calculate ΔE ∆𝐸 = 𝑞 + 𝑤 = 6,840 𝐽 + 0 𝐽 = 6,840 𝐽 Calculate ΔH ∆𝐻 = ∆𝐸 + (∆𝑃)𝑉 = 67.5 𝐿 ∙ 𝑎𝑡𝑚 + (6.00 𝑎𝑡𝑚 − 3.00 𝑎𝑡𝑚)(15.0 𝐿)

= 112.5 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 11,400 𝐽

Path 2 (Step 2) Constant Pressure Calculate work (work will have a positive sign because the system contracts)

𝑤 = −𝑃𝑒𝑥∆𝑉 = −(6.00 𝑎𝑡𝑚)(20.0 𝐿 − 15.0 𝐿) = −30.0 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚)

= −3.0 × 103 𝐽 Calculate heat

𝑞𝑃 = 𝑛𝐶𝑃∆𝑇 = 𝑛5

2𝑅∆𝑇 = −

5

2𝑤 = −

5

2(−3.0 × 103 𝐽) = 7,500 𝐽

Calculate ΔE (all of the energy of the gas is in kinetic energy) ∆𝐸 = 𝑞 + 𝑤 = 7,500 𝐽 + −3.0 × 103 𝐽 = 4,500 𝐽

Calculate ΔH ∆𝐻 = 𝑞 = 7,500 𝐽

Calculate Totals 𝑤𝑡𝑜𝑡 = 0 𝐽 + −3.0 × 103 𝐽 = −3.0 × 103 𝐽 𝑞𝑡𝑜𝑡 = 6,840 𝐽 + 7,500 𝐽 = 14,300 𝐽 ∆𝐸𝑡𝑜𝑡 = 6,840 𝐽 + 4,500 𝐽 = 11,300 𝐽 ∆𝐻𝑡𝑜𝑡 = 11,400 𝐽 + 7,500 𝐽 = 18,900 𝐽

ΔE and ΔH are the same for both pathways therefore they are state variables. Note: The numbers are slightly off due to rounding errors. w and q are different depending on the pathway, therefore they are not state variables.