Chem ch25

Worked solutions to student book questions Chapter 25 Energy from chemical reactions

Heinemann Chemistry 2 4th edition Enhanced Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. Sodium nitrate dissolves readily in water according to the equation

NaNO3(s) → NaNO3(aq); ∆H = +21.0 kJ mol–1 a Determine the energy change when 1.0 g of solid NaNO3 is dissolved in water. b Would you expect the temperature to rise or fall when sodium nitrate dissolves in

water? Explain your answer.

A1.

a n(NaNO3) = 851 =0.0117 mole

1 mole NaNO3 absorbs 21 kJ energy 0.0117 mole absorbs 0.0117 × 21 kJ = 0.25 kJ

b The temperature will fall. Since the reaction is endothermic the enthaply of reactants is higher than the enthalpy of products, i.e. the reaction absorbs energy.

Q2.

Calculate the energy released when the following quantities of ethane gas burn according to the equation:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l); ∆H = –3120 kJ mol–1 a 3.00 mol b 100 g c 10.0 L at SLC

A2. a From the equation, use stoichiometry to find energy released. 2 mol C2H6 releases 3120 kJ of energy

3.00 mol C2H6 releases 2

3120 × 3.00 kJ

So energy = 4680 kJ = 4.68 × 103 kJ

b Step 1 Calculate the amount of ethane, using n = Mm .

n(C2H6) = 1mol g 30.068g100

−

= 3.326 mol Step 2 Use stoichiometry to find energy released. 2 mol C2H6 releases 3120 kJ of energy

3.326 mol C2H6 releases 3120 × 2326.3 kJ

So energy released = 5189 kJ = 5.19 × 103 kJ



c Step 1 Calculate the amount of ethane, using n =mV

V at SLC.

n(C2H6) = 1mol L 24.5L10.0

−

= 0.4082 mol Step 2 Use stoichiometry to find energy released. 2 mol C2H6 releases 3120 kJ of energy

0.4082 mol C2H6 releases 3120 × 2

4082.0 kJ

So energy released = 636.7 kJ = 637 kJ

Q3. What volume of ethane, measured at STP, must be burnt according to the equation in Question 2 in order to yield 100 kJ of heat energy?

A3. Step 1 Use stoichiometry to find the amount of C2H6. 3120 kJ is released by 2 mol

100 kJ is released by 2 × 3120100 mol

So n(C2H6) = 2 × 3120100

= 0.641 mol

Step 2 Calculate the volume of C2H6 at STP, using n = mV

V .

V(C2H6) = n × Vm

= 0.641 mol × 22.4 L mol–1 = 1.44 L

E1.

Suggest why liquid hydrogen was chosen to fuel the Saturn rockets used in the American space program.

AE1. Hydrogen has a higher energy density than other fuels.

E2. What are the limitations of using hydrogen as fuel for cars?

AE2.

Hydrogen is more dangerous to handle. It is not manufactured on a sufficiently large enough scale. There is no existing infrastructure to store large quantities of hydrogen and to distribute it. New engines would need to be developed that can use hydrogen fuel. However, most major car manufacturers are experimenting with cars that are fuelled directly by hydrogen or by fuel cells that use hydrogen.



E3. Why is petrol preferred to coal as a fuel in cars?

AE3. Liquid fuels are easier to handle than solid fuels.

E4. Biodiesel and ethanol have been promoted as being renewable energy resources and as being carbon dioxide neutral. Which would drive your car further, one kilogram of ethanol or one kilogram of biodiesel?

AE4. The energy density of biodiesel (41.6 MJ kg–1) is higher than that of ethanol (29.7 MJ kg–1). 1 kg. of biodiesiel will drive the car further.

E5. Calculate the energy density of octane, C8H18, in MJ kg–1 if the heat of combustion of octane is –5450 kJ mol–1.

AE5.

Energy density = 5450114.2

× 1000

= 47 723 kJ kg–1 = 47.72 MJ kg–1

Q4. Calculate the energy needed to heat: a 100 mL of water from 20.0°C to 80.0°C b 250 mL of water from 25.0°C to 100.0°C c 1.5 kg of water from 20.0°C to 30.0°C d 300 g of water from 18.0°C to 100.0°C e 300 g of cooking oil from 18.0°C to 100.0°C

A4.

These questions use the formula relating specific heat capacity and energy. It is important to use the correct units. As the density of water is 1 g mL–1, 1 mL of water weighs 1 g. Remember to convert kg to g. As the temperature is only a difference, the unit is not important. The formula is: energy = specific heat capacity × mass of water × temperature rise, which can be rewritten for water as: E = 4.184 × m × ∆T a E = 4.184 × 100 g × 60.0°C = 25 104 J = 25.1 kJ b E = 4.184 × 250 g × 75.0°C = 78 450 J = 78.5 kJ



c E = 4.184 × 1500 g × 10.0°C = 62 760 J = 62.8 kJ d E = 4.184 × 300 g × 82.0°C = 102 926 J = 103 kJ e E = 2.2 × 300 g × 82.0°C = 54 120 J = 54.1 kJ

Q5.

A calorimeter containing 100 mL of water is calibrated by passing a 3.00 A current through the instrument for 36.0 s at a potential difference of 3.50 V. The temperature rises by 0.82°C. Potassium hydroxide weighing 0.654 g is added to the calorimeter and dissolved rapidly by stirring. The temperature rises from 20.82°C to 22.23°C. a Calculate the calibration factor for the calorimeter and water. b Determine ∆H for the equation:

KOH(s) O(l)H 2

→ KOH(aq)

A5. Questions such as this have three parts: 1 Determination of the calibration factor. 2 Calculation of the energy change during the reaction. 3 Calculation of ∆H for the equation. a Step 1 Calculate the energy input, using E = VIt. E = 3.50 J × 3.00 A × 36.0 s = 378 J

Step 2 Calculate the calibration factor, using CF = rise etemperatur

energy .

CF = C82J 378

°

= 461 J °C–1 b Step 1 Calculate the energy released by the reaction. E = CF × ∆T = 461 J °C–1 × (22.23 – 20.82)°C = (461 × 1.41) J = 650 J

Step 2 Calculate the amount of KOH, using n = Mm .

n(KOH) = 1mol g 56.108g 654.0

−

= 0.01166 mol



Step 3 Calculate ∆H for the equation by using stoichiometry. 0.01166 mol released 650 kJ

1 mol will release kJ0.01166

650

So ∆H = –5575 kJ mol–1 = –55.8 MJ mol–1

Q6. A temperature rise of 1.78°C was observed when 1.00 × 10–3 mol of propane gas was burnt in a calorimeter. The calibration factor of the calorimeter was previously determined to be 1250 J °C–1. a Write an equation for the combustion of propane. b Calculate the energy change during the reaction. c Calculate the heat of combustion for propane in kJ mol–1. d Write a thermochemical equation for the combustion of propane.

A6. Questions such as this have three steps: Step 1: Determination of the calibration factor. Step 2: Calculation of the energy change during the reaction. Step 3: Calculation of ∆H for the equation. Sometimes one or more steps are given to you. In this question Step 1 is given. a C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) b Energy change = calibration factor × temperature change E = 1250 J °C–1 × 1.78°C = 2225 J = 2.23 kJ released by 1.00 × 10–3 mol c 1.00 × 10–3 mol released 2.23 kJ

1 mol will release 31000.123.2

−×

So ∆H = –2230 kJ mol–1 = –2.23 MJ mol–1 d C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g); ∆H = –2.23 MJ mol–1

Q7.

In a steelworks, carbon monoxide present in the exhaust gases of the blast furnace can be used as a fuel elsewhere in the plant. It reacts according to the equation:

CO(g) + 21 O2(g) → CO2(g); ∆H = –283 kJ mol–1

a Which has the greater heat content: 1 mol of CO(g) and 0.5 mol of O2(g), or 1 mol of CO2(g)?

b Write the value of ∆H for the reactions: i 2CO(g) + O2(g) → 2CO2(g) ii 2CO2(g) → 2CO(g) + O2(g)



A7. a 1 mol of CO(g) and 0.5 mol of O2(g) b i As the coefficients are twice those in the given equation, ∆H will also be

doubled. ∆H = 2 × (–283) kJ mol–1 = –566 kJ mol–1

ii As the coefficients are twice those in the given equation and the equation is reversed, ∆H will also be double, with the opposite sign.

∆H = +566 kJ mol–1 Q8. Calculate the energy released when the following quantities of hydrogen gas burn according to the equation:

2H2(g) + O2(g) →2H2O(g); ∆H = –572 kJ mol–1 a 1.00 mol b 1.00 kg c 100 L at SLC A8. a Use stoichiometry to find energy released. 2 mol H2 releases 572 kJ of energy

1.00 mol H2 releases 572 × 200.1 kJ

So energy = 286 kJ

b Step 1 Calculate the amount of H2, using n = Mm .

n(H2) = 1

3

mol g 2.01g 101.00−

×

= 497.5 mol Step 2 Use stoichiometry to find energy released. 2 mol H2 releases 572 kJ of energy

497.5 mol H2 releases 572 × 2

5.497 kJ

So energy = 142 285 kJ = 143 MJ

c Step 1 Calculate the amount of H2, using n = mV

V at SLC.

n(H2) = 1mol L 24.5g 100

−

= 4.08 mol Step 2 Use stoichiometry to find energy released. 2 mol H2 releases 572 kJ of energy

4.08 mol H2 releases 572 × 208.4 kJ

So energy = 1167.3 kJ = 1.17 MJ



Q9. What volume of hydrogen, measured at STP, must be burnt according to the equation in Question 8 in order to yield 100 kJ of heat energy?

A9. Step 1 Write a thermochemical equation. 2H2(g) + O2(g) → 2H2O(g); ∆H = –572 kJ mol–1 Step 2 Use stoichiometry to find the amount of H2. 572 kJ is released by 2 mol

100 kJ is released by 2 × 572100 mol

So n(H2) = 2 × 572100

= 0.3497 mol

Step 3 Calculate the volume of H2 at STP, using n = mV

V .

V(H2) = n × Vm

= 0.3497 mol × 22.4 L mol–1 = 7.83 L

Q10. Butane is used as the fuel in pocket cigarette lighters. It is a liquid when stored under pressure in the lighter, but vaporises when the valve is opened. Combustion of butane is represented by the equation:

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l); ∆H = –5772 kJ mol–1 a How much energy is evolved when 10.0 g of butane burns completely? b How much energy is evolved when 0.100 L of butane, measured at SLC, burns

completely? c Calculate the volume of butane, measured at 15°C and 108 kPa, that must be

burnt to yield 1.00 kJ of energy.

A10.

a Step 1 Calculate the amount of butane, using n = Mm .

n(C4H10) = 1mol g 58.09g 10.0

−

= 0.1721 mol Step 2 Use stoichiometry to find energy evolved. 2 mol C4H10 produces 5772 kJ

0.1721 mol C4H10 produces 5772 × 2

1721.0 kJ

So energy = 496.8 kJ = 497 kJ



b Step 1 Calculate the amount of C4H10, using n = mV

V at SLC.

n(C4H10) = 1molL 24.5L 0.100

−

= 0.004082 mol Step 2 Use stoichiometry to find energy evolved. 2 mol C4H10 produces 5772 kJ

0.004082 mol C4H10 produces 5772 × 2

004082.0 kJ

So energy = 11.78 kJ = 11.8 kJ c Step 1 Use stoichiometry to find amount of C4H10. 2 mol C4H10 produces 5772 kJ

5772

2 mol C4H10 produces 1.00 kJ

So n(C4H10) = 0.0003465 mol Step 2 Calculate the volume of C4H10, using PV = nRT.

V = P

nRT

= 108

28831.80003465.0 ××

= 0.007678 L = 7.68 mL

Q11. A hand-warmer purchased by campers and skiers consists of a sealed outer plastic bag and an inner perforated plastic bag containing powdered iron, water, salt and sawdust. When the inner bag is taken out and shaken it becomes hot because the following reaction occurs:

4Fe(s) + 3O2(g) → 2Fe2O3(s); ∆H = –823 kJ mol–1 The bag contains 15.0 g of iron that reacts completely. a Draw and label a diagram to show the energy changes that occur during the

course of the reaction (an energy profile diagram). b Calculate the energy released by the hand-warmer. c Suggest why salt is included in the contents of the bag.

A11.

a



b Step 1 Calculate the amount of Fe, using n = Mm .

n(Fe) = 1mol g 55.85g 15.0

−

= 0.2686 mol Step 2 Use stoichiometry to find energy released. 4 mol Fe produces 823 kJ

0.2686 mol Fe produces 823 × 4

2686.0 kJ

So energy = 55.26 kJ = 55.3 kJ c The solution of salt acts as an electrolyte to conduct charge during the redox

reaction between iron and oxygen. It increases the rate of reaction and hence the rate of evolution of heat. (See Heinemann Chemistry 1, Chapter 24, for a more detailed description of the corrosion process.)

Q12. An oxy-acetylene welding torch uses ethyne (acetylene) at the rate of 500 mL per minute, measured at STP. The reaction is described by:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l); ∆H = –2599 kJ mol–1 a What is the heat of combustion of ethyne in kJ mol–1 and kJ g–1? b Calculate the rate of energy production by the torch in kJ per minute.

A12. a Step 1 Calculate the heat of combustion of C2H2 by using stoichiometry in

kJ mol–1. 2 mol of C2H2 releases 2599 kJ 1 mol releases 1299.5 kJ So ∆H = –1300 kJ mol–1 (four significant figures) Step 2 Calculate the heat of combustion in kJ g–1.

∆H = 1

1

mol g 26.017mol kJ 1300

−

−−

= –49.97 kJ g –1

b Step 1 Calculate the amount of C2H2 consumed in 1 minute by using n = mV

V .

n(C2H2) = 1mol L 22.4L 500.0

−

= 0.02232 mol Step 2 Calculate the energy released per minute. Energy released per minute = energy/mol × mol/minute = 1300 kJ mol–1 × 0.02232 mol min–1 = 29.0 kJ min–1



Q13. Calculate the heat evolved when 25.00 mL of 0.100 M sulfuric acid reacts with excess sodium hydroxide solution.

H+(aq) + OH–(aq) → H2O(l); ∆H = –57.2 kJ mol–1

A13.

Step 1 Calculate the amount of H2SO4, using n = c × V. n(H2SO4) = 0.100 M × 0.02500 L = 0.002500 mol Step 2 When H2SO4 reacts with a strong base it is fully ionised. For every mole of

H2SO4 that reacts, 2 mol H+ ions are produced. Use the thermochemical equation to calculate the heat released.

1 mol H2SO4 releases (2 × 57.2) kJ 0.002500 mol H2SO4 releases (2 × 57.2) × 0.002500 kJ So energy = 0.286 kJ = 286 J

Q14. Glucose is produced by photosynthesis according to the following equation:

6CO2(g) + 6H2O(l)→ C6H12O6(aq) + 6O2(g); ∆H = +2803 kJ mol–1 a Write a thermochemical equation for the reaction of glucose with oxygen. b How much energy is released when 5.00 g of glucose is oxidised?

A14.

a C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l); ∆H = –2803 kJ mol–1

b Step 1 Calculate the amount of C6H12O6, using n = Mm .

n(C6H12O6) = 1mol g 180g 00.5

−

= 0.0278 mol Step 2 Using stoichiometry, calculate the energy released. 1 mol C6H12O6 releases 2803 kJ 0.0278 C6H12O6 releases 2803 × 0.0278 kJ So energy = 77.9234 kJ = 77.9 kJ

Q15. Recycled ammonia has been proposed as a new source of hydrogen for electricity generation. In a closed system, solar energy is used to dissociate ammonia into hydrogen and nitrogen gas. The reverse reaction releases heat for power generation. Under standard conditions, the reaction is:

2NH3(l)) → N2(g) + 3H2(g); ∆H = 92.4 kJ mol–1 a Is ammonia produced or consumed when power is being generated? b What mass of hydrogen gas is required to power a 100 W light bulb for an hour?

(Hint: 1 W = 1 J s–1) c How would the temperature and pressure affect the equilibrium position of the

reaction?



A15. a Ammonia is produced. b energy required = 100 × 1 × 60 × 60

= 360 kJ From the equation 3 mole H2 produces 92.4 kJ x mole H2 produces 360 kJ

x = 4.92

3360× = 11.69 mol

mass H2 = 11.69 × 2 g = 23.4 g c Increased temperature shifts the reaction forward to produce more hydrogen and

nitrogen gases. Increased pressure shifts the reaction backward (reverse) to produce more ammonia.

Q16.

Water evaporating from the skin after swimming causes us to feel cool. Account for the fact that during this process the heat content of the water is increasing.

A16. The change of state from liquid to gaseous water absorbs energy from the skin and leaves us feeling cool. The absorbed energy present in the evaporated water causes the heat content of the water to increase.

Q17.

On a hot day a swimmer leaves the water, walks across the sand and, when entering a car, sits on the metal buckle of the seat belt. Explain how the data in Table 24.1 are related to the swimmer’s experiences.

A17.

Metal and sand have lower specific heats than water; this means that metal and sand reach higher temperatures than water when they absorb energy from the sun. The high thermal conductivity of metal allows the rapid conduction of heat from the hot metal buckle of the seat belt to the skin of the hapless swimmer.

Q18.

The moist brown coal used in coal-fired power stations in Victoria is not as good a fuel as the black coal used in power stations in New South Wales. Only about one-third of the substances present in brown coal are combustible. What other factor reduces the heat energy available from this fuel?

A18. Absorption of energy by the water present in brown coal reduces the heat energy available from this fuel.

Q19. Sand and water are both used to smother fires. Which material would absorb the most heat from a fire? Why?



A19. Water, because it has a higher specific heat (4.184 J °C–1 g–1) than sand (0.48 J °C–1 g–1).

Q20.

By referring to Table 25.1, calculate the energy required to raise the temperature of: a 1.00 kg of iron from 25.0 to 100.0°C b 500 g of water from 25.0 to 100.0°C c 0.500 kg of ethanol from 50.0 to 100.0°C d 500 g of aluminium from 50.0 to 100.0°C

A20. These questions use the formula relating specific heat capacity and energy. It is important to use the correct units. The formula is: energy = specific heat capacity × mass of water × temperature rise, which can be rewritten as: E = SHC × m × ∆T. a E = 0.473 × (1.00 × 103) × 75.0°C = 35 475 J = 35.5 kJ b E = 4.184 × 500 × 75.0°C = 156 900 J = 157 kJ c E = 2.40 × (0.500 × 103) × 50.0°C = 60 000 J = 60.0 kJ d E = 1.97 × 500 × 50.0°C = 49 250 J = 49.3 kJ

Q21.

A 500 mL volume of water in an electric kettle was heated using 136 kJ of electrical energy. If the water was initially at 15°C, calculate its final temperature.

A21.

Step 1 Write the expression relating energy consumed to the temperature rise of water.

Energy (J) = 4.184 × mass (g) × ∆T(°C) Step 2 Calculate the ∆T. Remember that 1 mL of water has a mass of 1 g.

∆T = mass4.184

energy×

= 5004.184

000 136×

= 65.01°C Step 3 Calculate the final temperature. Final temperature = (15 + 65.01)°C = 80.01°C = 80°C



Q22. An ‘instant’ cold pack sold for the treatment of sporting injuries contains solid ammonium nitrate and a weak inner plastic bag full of water. Punching the pack breaks the inner bag and causes the ammonium nitrate to dissolve.

NH4NO3(s) → NH4+(aq) + NO3

–(aq); ∆H = +25 kJ mol–1 A bag containing 25.0 g of ammonium nitrate was punched and dropped into a 2.00 L container of water at a temperature of 19.00°C. a Calculate the lowest temperature that the water could reach. b Why would this temperature not be achieved in practice?

A22. a Step 1 Calculate the amount of NH4NO3.

n(NH4NO3) = 1mol g 80.036g 0.25

−

= 0.3126 mol Step 2 Calculate the energy absorbed by using stoichiometry. 1 mol absorbs 25.0 kJ 0.3126 mol absorbs (0.3126 × 25.0) kJ So energy = 7.815 kJ = 7815 J Step 3 Write the expression relating energy consumed to the temperature rise of

water. Energy (J) = 4.184 × mass (g) × ∆T(°C) Step 4 Calculate ∆T, assuming the density of the solution is 1 g mL–1.

∆T = 20004.184

7815×

= 0.9339°C Step 5 Calculate the lowest temperature of the solution. Tlowest = (19.00 – 0.9339)°C = 18.066°C = 18.07°C b The water would not reach the calculated minimum temperature because the

imperfect insulation of the container would allow heat from the surrounding environment to enter.

Q23. Surf on, play on! Super-cool ‘hotsuits’ are a new Australian invention that use heat-releasing liquid gel packs in a wet suit. The packs are placed in pockets near parts of the body where there is substantial flow of blood under the skin. Surfers can get over an hour more playtime in cold conditions. The packs contain supercooled sodium ethanoate in water. Sodium ethanoate solidifies at 54°C but it can be supercooled and stay as a stable liquid gel at much lower temperatures until disturbed by vigorous motion. The gel packs can be re-used by boiling them and allowing them to cool. a Would you expect the reaction to be exothermic or endothermic? Explain. b The gel packs crystallise as they release heat. Write a feasible reaction for this

process.



c Footballers sometimes wear ice vests to cool them down in warm weather. Suggest a thermochemical alternative to ice.

A23. a Exothermic b Na+(aq) + CH3COO–(aq)) → CH3COONa (s); ∆H negative c An endothermic reaction, similar to that in a cold pack (e.g.

NH4NO3(s) → NH4+(aq) + NO3

–(aq); ∆H = 25 kJ mol–1)

Q24.

The equation for dissolving crystals of ammonium chloride in water at SLC is: NH4Cl(s) + H2O → NH4

+(aq) + Cl–(aq); ∆H = 15.15 kJ mol–1 a Is this reaction endothermic or exothermic? b Will the temperature of the solution increase or decrease as the reaction

proceeds? Explain. c Which is greater in this reaction, the energy required to break bonds in the

reactants or the energy released when new bonds are formed? Explain. d Calculate the energy absorbed when 2.00 g of NH4Cl is dissolved in:

i 100 mL water ii 1000 mL water

e Predict the expected temperaure change for each case in part d.

A24.

a Endothermic as ∆H is positive. b Decrease, endothermic reactions absorb energy from their surroundings c Energy is absorbed overall (endothermic) therefore the energy required to break

bonds is greater than the energy released when new bonds made.

d n(NH4Cl) = 5.530.2

energy absorbed = 5.530.2 × 15.15 kJ

= 0.566 kJ The amount of energy absorbed does not depend on the volume of the solution. 0.566 kJ will be absorbed in both cases. i 0.566 kJ ii 0.566 kJ

e i The specific heat of water is 4.18 J oC–1 4.18 J raises temp. of 1 g water by 1oC 418 J raises temp. of 100 g water by 1oC 566 J raises temp. of 100 g water by

418566 = 1.35°C

ii 566 J raises temp. of 100 g of water by 1.35°C. The same amount of heat energy will raise the temperature of 1000 g of water by 0.135°C.



Q25. What is the calibration factor of a calorimeter if a current of 1.20 A at a potential difference of 5.67 V raises its temperature by 1.25°C in 60.0 seconds?

A25. Step 1 Calculate the energy input, using E = VIt. E = 5.67 V × 1.20 A × 60.0 s = 408.24 J


energy

CF = C1.25

J 408.24°

= 326.59 J °C–1 = 327 J °C–1

Q26. When 1.01 g of solid potassium nitrate was dissolved in a calorimeter, the temperature dropped by 0.672°C. When the calorimeter was calibrated, a current of 1.50 A applied at a potential difference of 5.90 V for 60.0 seconds caused a temperature rise of 0.456°C. a Write an equation for the dissolution of potassium nitrate. b Determine the calibration factor of the calorimeter. c Calculate the energy change during the reaction. d Write the thermochemical equation for the reaction.

A26.

a KNO3(s) ⎯⎯⎯ →⎯ O(l)H2 K+(aq) + NO3–(aq)

Questions such as this have three parts: 1 Determination of the calibration factor. 2 Calculation of the energy change during the reaction. 3 Calculation of ∆H for the equation. b Step 1 Calculate the energy input, using E = VIt. E = 5.90 J × 1.50 A × 60.0 s = 531 J


energy .

CF = C0.456J 531°

= 11 645 J °C–1 = 1.16 kJ °C–1 c E = CF × ∆T = 1.1645 kJ °C–1 × 0.672°C = 0.7825 kJ = 783 J



d Step 1 Calculate the amount of KNO3, using n = Mm .

n(KNO3) = 1mol g 101g 01.1

−

= 0.01 mol Step 2 Calculate ∆H for the equation, using stoichiometry. 0.01 mol absorbed 783 J

1 mol will absorb 0.01783 J

So ∆H = +78 300 J mol–1 = +78.3 kJ mol–1 Step 3 Write the thermochemical equation. KNO3(s) → K+(aq) + NO3

–(aq); ∆H = +78.3 kJ mol–1

Q27.

When a current of 1.50 A passed through the electric heater of a calorimeter for a time of 25.1 s at a potential difference of 8.05 V, the temperature of the calorimeter rose by 0.399°C. 7.62 × 10–4 mol of ethene gas was then burnt in the calorimeter. The temperature rose from 20.12°C to 21.54°C. a Write an equation for the combustion of ethene. b Calculate ∆H for the equation in part a.

A27.

a C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) b Step 1 Calculate the energy input, using E = VIt. E = 8.05 J × 1.50 A × 25.1 s = 303.1 J Step 2 Calculate the calorimeter constant.

CF = C0.399J 303.1°

= 759.6 J °C–1

Step 3 Calculate the energy released by the reaction. E = CF × ∆T = 759.6 J °C–1 × 1.42°C = 1079 J Step 4 Calculate ∆H for the equation in part a. 7.62 × 10–4 mol ethene releases 1.079 kJ

1 mol releases 4107.621.079

−× kJ

So ∆H = –1416 kJ mol–1



Q28. A calorimeter is used to measure energy changes that occur during chemical reactions. a Why should a calorimeter be well insulated? b Why is it necessary to calibrate a calorimeter? c Try to devise a method for calibrating a calorimeter that does not involve the use

of an electric heater. d The combustion of a 3.00 g piece of wood in a bomb calorimeter results in an

observed temperature rise of 1.6°C. The calibration factor of the calorimeter was 30.0 kJ °C–1. Calculate the heat of combustion of the wood.

A28. a A calorimeter should be well insulated to ensure that no thermal energy enters or

leaves the apparatus. Temperature changes in the calorimeter must reflect only energy changes occurring as a result of a chemical reaction in the calorimeter, and must not be affected by energy exchanges with the external environment.

b Energy changes that take place in a calorimeter are determined by measuring the change in temperature of its contents, since no instrument is commonly available to measure energy changes directly. A calorimeter is calibrated to determine the amount of energy required to change the temperature of the contents by 1°C. This calibration factor depends on the construction of the particular calorimeter and must be determined by experiment.

c By performing a reaction with a known enthalpy change in a calorimeter and measuring the temperature change that takes place, the calibration factor of the calorimeter can be calculated.

d Step 1 Calculate the heat released by the wood. Heat released = calibration factor × ∆T = 30.0 kJ °C–1 × 1.6°C = 48.0 kJ Step 2 Calculate the heat of combustion of the wood.

Heat of combustion = g 3.00kJ 48.0

= 16 kJ g–1

Q29. Like a calorimeter, your classroom requires a certain amount of energy to raise its temperature by one degree. a Explain how you could determine the ‘calibration factor’ of your classroom. b Could a classroom be used as a calorimeter to measure the energy released by

students during a lesson? Why, or why not?



A29. a One approach would be to heat one of two otherwise identical classrooms with

electric radiators of known power (e.g. 5000 W) for an hour and compare the temperature change. The difference in temperature between the heated and unheated rooms could be used to calculate a calibration factor. The unheated classroom could be used as a control to account for varying external temperatures.

b In theory, a classroom could be used as a rather imperfect calorimeter. Human-size calorimeters have been used to determine a person’s energy outputs. In practice this would be difficult because of the uncontrolled variables, such as fluctuating external temperatures and the poor insulation of a classroom. Heat losses and/or gains via radiation and convection through ventilation systems would affect any measurements.

Q30. A camper uses a butane stove to heat water in a metal cup for a drink of coffee. The camper boils 150 mL of water that was initially at 20.0°C.

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l); ∆H = –5718 kJ mol–1 a Calculate the minimum volume of butane, measured at SLC, required to boil the

water. b Why would more butane be required in practice than your answer to part a?

A30. a Step 1 Write a balanced equation. 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l); ∆H = –5718 kJ mol–1 Step 2 Calculate ∆T. ∆T = (100.0 – 20.0)°C = 80°C Step 3 Calculate the energy required to boil the water. Energy = 4.18 × 150 g × 80.0°C = 50 160 J Step 4 Use stoichiometry to calculate the amount of butane. 2 mol butane produced 5718 kJ

2 × 5718

160.50 mol butane produced 50.160 kJ

So n(C5H12) = 0.01754 mol

Step 5 Calculate the volume of C5H12, using n = mV

V .

V(C5H12) = 0.01754 mol × 24.5 L mol–1 = 0.42973 L = 430 mL b When heating water in the metal cup, not all the heat energy from the burning

butane is transferred to the water. Some energy radiates out sideways and some is used to heat the cup. As a result, more butane is required in practice than calculated in part a of this question.



Q31. A mixture of 0.0100 mol hydrogen gas and an unknown amount of methane was burnt completely in a calorimeter to produce a temperature rise of 5.75°C. The calibration factor of the calorimeter was 1500 J °C–1. If the heats of combustion of hydrogen gas and methane are 286 kJ mol–1 and 890 kJ mol–1 respectively: a write the thermochemical equations for the two combustion reactions b calculate the amount of methane in the mixture.

A31.

a 2H2(g) + O2(g) → 2H2O(g); ∆H = –572 kJ mol–1 CH4(g) + 2O2(g) → CO2(g) + 2H2O(g); ∆H = –890 kJ mol–1

b Step 1 Calculate the energy released by the reaction. E = ∆T × CF = 5.75°C × 1500 J °C–1 = 8625 J = 8.625 kJ Step 2 Use stoichiometry to calculate the energy released by 0.0100 mol H2. 2 mol H2 produced 572 kJ

0.0100 mol H2 produced (572 × 2

0100.0 ) kJ

So energy released by H2 = 2.860 kJ Step 3 Calculate the energy released by CH4. Energy released by CH4 = (8.625 – 2.860) kJ = 5.765 kJ Step 4 Use stoichiometry to calculate the amount of CH4. 1 mol CH4 produced 890 kJ

890765.5 mol CH4 produced 5.765 kJ

So n(CH4) = 0.006478 mol = 6.48 × 10–3 mol

Q32. Methane and propane are used as alternative transport fuels. Given the heats of combustion of methane and propane are 890 kJ mol–1 and 2220 kJ mol–1 respectively, calculate the energy released per gram by each fuel.

A32. Step 1 Calculate the heat of combustion of CH4 in kJ g–1.

∆H = 1

1

mol g 16.001mol kJ 890

−

−−

= –55.6 kJ g –1 Step 2 Calculate the heat of combustion of C3H8 in kJ g–1.

∆H = 1

1

mol g 44.001mol kJ 2220

−

−−

= –50.5 kJ g –1



Q33. Ethanol produced from sugar is used as a substitute for petrol in Brazil. Heats of combustion of ethanol (CH3CH2OH) and octane (C8H18), a major component of petrol, are listed in Table 25.2. a Write equations to represent the combustion reactions of ethanol and octane. b Calculate the heats of combustion of each fuel in units of kJ g–1. c If the fuels were the same price per gram, which fuel would represent the better

value? d Ten per cent blends of ethanol in petrol are available from some outlets in

Australia. Would cars using a 10% blend of ethanol in petrol travel a greater or smaller distance per tank than those using unblended petrol?

e Comment on the prospects of replacing petrol with ethanol produced from sugar or cellulose and using it as fuel for cars.

A33.

a CH3CH2OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l); ∆H = –1367 kJ mol–1

C8H18(g) + 225 O2(g) → 8CO2(g) + 9H2O(l); ∆H = –5450 kJ mol–1

b Heat of combustion of CH3CH2OH in kJ g–1

∆H = 1

1

mol g 46.068mol kJ 1367

−

−−

= –29.7 kJ g –1 Heat of combustion of C8H18 in kJ g–1

∆H = 1

1

mol g 114.224mol kJ 5450

−

−−

= –47.7 kJ g –1 c octane d shorter e Replacing petrol with ethanol would be feasible if petrol prices became too high

or supplies of petrol became scarce. In such circumstances, production of ethanol could be increased to meet demand from motorists. Cars in Brazil run on fuel made this way. However, the amount of farming land required to provide supplies of ethanol could limit the usefulness of this approach.



Q34. 300 mL of LPG is burnt to heat 400 mL of water from 18.0°C to 34.7°C. What is the heat of combustion of the LPG in kJ L–1?

A34. Step 1 Write the expression relating energy consumed to the temperature rise of

water. Energy (J) = 4.184 × mass (g) × ∆T (°C) Step 2 Calculate the energy released by 300 mL of LP gas. E = (4.184 × 400 × 16.7) J = 2795 J Step 3 Use stoichiometry to calculate the ∆H in kJ L–1. 0.300 L releases 2795 J

1 L releases 300.0

2795 J

So ∆H = 300.0

2795 J L–1

= 931 666 J L–1 = 93.2 kJ L–1

Q35. A 0.254 g sample of black coal was burnt in a bomb calorimeter. The calorimeter contained 300 mL of water which rose from 18.25°C to 24.92°C. a Using the specific heat capacity of water, calculate the heat of combustion of the

coal. b Is your answer likely to be higher or lower than the actual value? Explain.

A35. a Step 1 Write the expression relating energy consumed to the temperature rise of

water. Energy (J) = 4.184 × mass (g) × ∆T (°C) Step 2 Calculate the energy released by the coal. E = 4.184 × 300 g × (24.92 – 18.25)°C = (4.184 × 300 × 6.67) J = 8372 J Step 3 Use stoichiometry to calculate the ∆H. 0.254 g of coal produces 8372 J of energy

1.00 g of coal produces (8.372 × 254.000.1 ) kJ

So ∆H = 32.96 kJ g–1 = 33.0 kJ g–1 b The calculated heat of combustion of coal is likely to be lower than the actual

value because it is calculated on the assumption that the water in the calorimeter absorbs all of the energy from the reaction. Metal, glass and insulation in the calorimeter vessel would also absorb some energy from the reaction. Calibration of the calorimeter would reduce this error.



Q36. A particular car travels 20 000 km in a year and operates at a 20% efficiency in transforming the chemical energy in fuel into mechanical energy (motion). The car has a fuel consumption of 10 L per 100 km and its petrol tank holds 50 L. Using data from Table 25.2, calculate: a the amount of energy supplied by a full tank of petrol b the amount of energy released as waste heat during the year c the cost of the fuel that is converted to waste heat, if petrol costs $1.40 cents per

litre

A36. a Step 1 Calculate the energy content of the tank of petrol. Energy content = 50 L × 34 200 kJ L–1 = 1 710 000 kJ = 1.7 × 106 kJ b Step 1 Calculate the volume consumed in 1 year.

Volume = 20 000 km × km100L 10

= 2000 L Step 2 Calculate the energy content of this petrol. E = 2000 L × 34 200 kJ L–1 = 68 400 000 kJ Step 3 Calculate the waste energy. Energy wasted = 80% of available energy

= 68 400 000 kJ × 10080

= 54 720 000 kJ = 5.5 × 107 kJ c Step 1 Calculate the volume of petrol producing waste energy. Waste litres = 80% of total volume

= 2000 L × 10080

= 1600 L Step 2 Calculate the cost of waste litres. Cost = 1600 × $0.70 = $1120

Q37. a Use data from Table 25.2 to compare the energy supplied by one gram of black

coal with that supplied by one gram of octane (petrol). Which is the better fuel on this basis?

b Describe one application in which coal is preferred to petrol and another in which petrol is preferred. Explain the reasons behind the choice of fuel in each case.



A37. a Step 1 Calculate the energy per gram of octane.

Energy = 1

1

mol g 114.2mol kJ 5450

−

−

= 47.7 kJ g–1 Step 2 Compare with energy per gram of black coal. Coal = 33 kJ g–1 Octane = 47.7 kJ g–1 ∴ Octane is the better fuel on this basis. b Coal is preferred to petrol as a fuel for power stations in Victoria because it is

cheaper, easily accessible, and will be available further into the future. Octane is preferred to coal for use as a fuel in vehicles because it has less mass per joule of energy supplied, is cleaner and easier to handle, and undergoes more consistent combustion.

Q38. A foam cup containing 200 mL of black coffee was heated from 20°C to 100°C in microwave oven that operates at 1200 W (1 W (watt) is equivalent to an energy consumption for 1 J s–1). a Suppose that there was no heat loss. Calculate the time required to heat the

coffee. b Calculate the calibration factor of the foam cup when it contained 200 mL of

solution. A38. a E = 4.184 × mass of solution × temperature change = 4.184 × 200 × 80 J = 66 944 J Time = energy/watts

= 669441200

= 55.8 seconds = 56 seconds

b Calibration factor = ET∆

= 6694480

= 837 J °C–1

Chem ch25

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