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CHAPTER 5 IB CHEMISTRY WARM UPS Mrs. Hilliard

CHAPTER 5 VOCABULARY1. Heat

2. Entropy

3. Enthalpy

4. System

5. Surroundings

6. Endothermic

7. Exothermic

8. Thermochemistry

9. Intensive property

10.Specific heat capacity

11.Standard enthalpy change of formation

12.Standard enthalpy change of combustion

13.Thermochemical equation

14.Bond enthalpy

CHAPTER 5 VOCABULARY1. Heat- q, a form of energy that is transferred from a warmer

body to a cooler body as a result of the temperature gradient.

2. Entropy- a measure of the distribution of total available energy between particles.

3. Enthalpy- the heat content of a system at a constant pressure.

4. System- the part of the universe that contains the reaction or process being studied.

5. Surroundings- everything in the universe except the system.

6. Endothermic- bond breaking, energy required to break the bonds.

7. Exothermic- bond making, energy is released when new chemical bonds are made.

8. Thermochemistry- the study of heat changes that occur during chemical reactions.

9. Intensive property- a physical property that remains the same no matter how much of the substance is present.

10. Specific heat capacity- the amount of heat required to raise 1 g of a substance by 1°C.

11. Standard enthalpy change of formation- the energy change upon the formation of 1 mol of a substance from its constituent elements in their standard state.

12. Standard enthalpy change of combustion- the heat evolved upon the complete combustion (burning) of 1 mol of substance.

13. Thermochemical equation- a balanced chemical equation that includes the physical states of all the reactants and the energy change, usually expressed as the change in enthalpy.

14. Bond enthalpy- the energy required to break 1 mol of bonds in gaseous covalent molecules under standard conditions.

TYPES OF REACTIONS

1. Which statement about bonding is correct? 1. Bond making is endothermic and releases energy

2. Bond making is exothermic and requires energy

3. Bond breaking is endothermic and requires energy

4. Bond breaking is endothermic and releases energy

2. Some water is heated using the heat produced by the combustion of copper metal. Which values are need to calculate the enthalpy change of reaction?

1. The mass of copper

2. The change in temperature of the water

3. The mass of the water

HEAT3. The specific heat of lead is 0.129 J g-1 K-1. What is the energy, in J, needed to

increase the temperature of 100.0 g of lead by 10.0 K?

4. A pure calcium block with a mass of 20 g is heated so that its temperature increases from 10°C to 100°C. The specific heat capacity of calcium is 6.74 x 10-1

J g-1 K-1. Which expression gives the heat energy change in kJ?

A. 20 x 6.74 x 10-1 J g-1 K-1 x 90

1000

B. 20 x 6.74 x 10-1 J g-1 K-1 x 90

C. 20 x 6.74 x 10-1 J g-1 K-1 x 363

D. 20 x 6.74 x 10-1 J g-1 K-1 x 363

1000

5. Which types of reactions are always exothermic?

1. Decomposition

2. Combustion

3. Neutralization

HEAT3. The specific heat of lead is 0.129 J g-1 K-1. What is the energy, in J, needed to

increase the temperature of 100.0 g of lead by 10.0 K? Q=mc∆T Q= (100g)(0.129 J g-1 K-1)(10K)=129J

4. A pure calcium block with a mass of 20 g is heated so that its temperature increases from 10°C to 100°C. The specific heat capacity of calcium is 6.74 x 10-1

J g-1 K-1. Which expression gives the heat energy change in kJ?

A. 20 x 6.74 x 10-1 J g-1 K-1 x 90

1000

B. 20 x 6.74 x 10-1 J g-1 K-1 x 90

C. 20 x 6.74 x 10-1 J g-1 K-1 x 363

D. 20 x 6.74 x 10-1 J g-1 K-1 x 363

1000

5. Which types of reactions are always exothermic?

1. Decomposition

2. Combustion

3. Neutralization

ENDOTHERMIC AND EXOTHERMIC REACTIONS6. Which is correct about energy changes during bond breaking

and bond formation?

7. Which statement is correct given the enthalpy level diagram below?

A. The reaction is exothermic and the reactants are more thermodynamically stable than the products.

B. The reaction is endothermic and the products are more thermodynamically stable than the reactants.

C. The reaction is endothermic and the reactants are more thermodynamically stable than the products.

D. The reaction is exothermic and the products are more thermodynamically stable than the reactants.

Bond breaking Bond formation

Endothermic and ∆H positive Exothermic and ∆H negative

Endothermic and ∆H negative Exothermic and ∆H postive

Exothermic and ∆H positive Endothermic and ∆H negative

Exothermic and ∆H negative Endothermic and ∆H positive

ENDOTHERMIC AND EXOTHERMIC REACTIONS8. Which processes are exothermic?

1. Neutralization

2. Combustion

3. Ice melting

9. Which statement is correct given the enthalpy level diagram below?

A. The reaction is exothermic and the reactants are more thermodynamically stable than the products.

B. The reaction is endothermic and the products are more thermodynamically stable than the reactants.

C. The reaction is endothermic and the reactants are more thermodynamically stable than the products.

D. The reaction is exothermic and the products are more thermodynamically stable than the reactants.

ENTHALPY10. Which is true for a chemical reaction in which the products have a higher

enthalpy than the reactants?

11. When some aqueous sodium hydroxide and aqueous hydrochloric acid were reacted together, the temperature of the surroundings was observed to increase from 25°C to 30°C. What can be deduced from this observation?

A. The reaction is endothermic and ∆H is negative.

B. The reaction is endothermic and ∆H is positive.

C. The reaction is exothermic and ∆H is negative.

D. The reaction is exothermic and ∆H is positive.

12. When some solid ammonium thiosulfate and solid barium hydroxide were reacted together, the temperature of the surroundings was observed to decrease from 35°C to 21°C. What can be deduced from this observation?

A. The reaction is endothermic and ∆H is negative.

B. The reaction is endothermic and ∆H is positive.

C. The reaction is exothermic and ∆H is negative.

D. The reaction is exothermic and ∆H is positive.

Reaction ∆H

A. Exothermic Positive

B. Exothermic Negative

C. Endothermic Positive

D. Endothermic Negative

ENTHALPY

13. Identical pieces of aluminum are added to two beakers, A and B, containing hydrochloric acid. Both acids have the same initial temperature but their volumes and concentrations differ. Which statement is correct?

A. The temperature in A and B will increase at the same rate.

B. It is not possible to predict whether A or B will have the higher maximum temperature.

C. The maximum temperature in A will be higher than in B.

D. The maximum temperature in A and B will be equal.

14. What is the energy, in kJ, released when 4.00 mol of carbon monoxide is burned according to the following equation? 2 CO (g) + O2 (g) → 2CO2 (g)

∆Hø = -564kJ

15. How much energy, in joules, is required to increase the temperature of 4.0 g of iron from 30-38°C? (Specific heat of iron is 0.450 J g-1 K-1).

0.5 g Al0.5 g Al

ENTHALPY

13. Identical pieces of aluminum are added to two beakers, A and B, containing hydrochloric acid. Both acids have the same initial temperature but their volumes and concentrations differ. Which statement is correct?

A. The temperature in A and B will increase at the same rate.

B. It is not possible to predict whether A or B will have the higher maximum temperature.

C. The maximum temperature in A will be higher than in B.

D. The maximum temperature in A and B will be equal.

14. What is the energy, in kJ, released when 4.00 mol of carbon monoxide is burned according to the following equation? 2 CO (g) + O2 (g) → 2CO2 (g) ∆Hø = -564kJ 2 x 564= 1128

15. How much energy, in joules, is required to increase the temperature of 4.0 g of iron from 30-38°C? (Specific heat of iron is 0.450 J g-1 K-1). Q=mc∆T Q= (4.0 g)(0.450 J g-1 K-

1)(8°C)=14.4J

0.5 g Al0.5 g Al

HESS’S LAW16.Consider the following reactions.

Cu2O (s) + ½ O2 (g) → 2CuO (s) ∆Hø= -163 kJ Cu2O (s) → Cu (s) + CuO (s) ∆Hø= +21 kJ What is the value of ∆Hø, in kJ, for this reaction? Cu (s) + ½ O2 (g) → CuO (s)

17.Consider the following reactions. 2Fe (s) + O2 (g) → 2 FeO (s) ∆Hø= -521 kJ 4Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) ∆Hø= -1561 kJ What is the enthalpy change, in kJ, for the reaction below?

4FeO (s) + O2 (g) → 2 Fe2O3 (s)

18. Which processes have a negative enthalpy change?

A. H2O (g) → H2O (l)

B. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

C. 2CH3OH (l) + 3 O2 (g) →2CO2 (g) + 4 H2O (l)

HESS’S LAW16.Consider the following reactions.

Cu2O (s) + ½ O2 (g) → 2CuO (s) ∆Hø= -163 kJ Cu2O (s) → Cu (s) + CuO (s) ∆Hø= +21 kJ What is the value of ∆Hø, in kJ, for this reaction? Cu (s) + ½ O2 (g) → CuO (s) ∆Hø= -163-21= -184

17.Consider the following reactions. 2Fe (s) + O2 (g) → 2 FeO (s) ∆Hø= -521 kJ 4Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) ∆Hø= -1561 kJ What is the enthalpy change, in kJ, for the reaction below? 4FeO (s) + O2 (g) → 2 Fe2O3 (s) ∆Hø= -1561 – 2(-521)

18.Which processes have a negative enthalpy change? (exotherm)

A. H2O (g) → H2O (l) Energy given off (condensation) React. higher than products for energy

B. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) (neutralization)

C. 2CH3OH (l) + 3 O2 (g) →2CO2 (g) + 4 H2O (l) (combustion reaction)

ENTHALPY CHANGES19.When 200 cm3 of 2.0 mol dm-3 HCl is mixed with 200 cm3 of 2.0 mol dm-3

KOH, the temperature of the resulting solution increases by 10°C. What will be the temperature change in °C, when 100 cm3 of these two solutions are mixed?

20.Consider the following reactions. N2 (g) + O2 (g) → 2 NO (g) ∆Hø=+150 kJ 2NO2 (g) → 2 NO (g) + O2 (g) ∆Hø= +120 kJ What is the ∆Hø value, in kJ, for the following reaction? N2 (g) + 2O2 (g) → 2 NO2 (g)

21. The standard enthalpy changes for the combustion of carbon and carbon monoxide are shown below.

C (s) + O2 (g) → CO2 (g) ∆Hø=-350 kJ CO (g) + ½ O2 (g) → CO2 (g) ∆Hø=-250 kJ What is the standard enthalpy change, in kJ, for the following reaction?

C (s) + ½ O2 (g) → CO (g)

ENTHALPY CHANGES19. When 200 cm3 of 2.0 mol dm-3 HCl is mixed with 200 cm3 of 2.0 mol dm-3 KOH, the

temperature of the resulting solution increases by 10°C. What will be the temperature change in °C, when 100 cm3 of these two solutions are mixed? 10°C, ratios are the same, same reactants and products so energy doesn’t change

20. Consider the following reactions. N2 (g) + O2 (g) → 2 NO (g) ∆Hø=+150 kJ 2NO2 (g) → 2 NO (g) + O2 (g) ∆Hø= +120 kJ What is the ∆Hø value, in kJ, for the following reaction? N2 (g) + 2O2 (g) → 2 NO2 (g) ∆Hø= 1 x (+150) + -1(+120)= 30kJ

21. The standard enthalpy changes for the combustion of carbon and carbon monoxide are shown below.

C (s) + O2 (g) → CO2 (g) ∆Hø=-350 kJ CO (g) + ½ O2 (g) → CO2 (g) ∆Hø=-250 kJ What is the standard enthalpy change, in kJ, for the following reaction? C (s) + ½ O2 (g) → CO (g) ∆Hø=-350+ 250 = -100kJ mol -1

BOND ENTHALPY22.Using the equation below:

C (s) + O2 (g) → CO2 (g) ∆Hø=-400 kJ mol -1

Mn (s) + O2 (g) → MnO2 (s) ∆Hø= -520 kJ mol -1

What is ∆H, in kJ, for the following reaction? MnO2 (s) + C (s) → Mn (s) + CO2 (g)

23.2.0 g of sodium hydroxide, NaOH, was added to 98.0 g of water. The temperature of the solution increased from 12.0 °C to 17.5°C. The specific heat capacity of the solution is 4.18 J g-1K-1. Give the expression for the heat evolved in kJ mol-1.

24.Which equation represent the bond enthalpy for the H- Cl bond in hydrogen chloride?

A. HCl (g)→ H (g) + Cl (l)

B. HCl (g)→ H (g) + Cl (g)

C. HCl (g)→ H (g) + ½ Cl2 (g)

D. HCl (g)→ H (g) + ½ Cl2 (l)

BOND ENTHALPY22.Using the equation below:

C (s) + O2 (g) → CO2 (g) ∆Hø=-400 kJ mol -1

Mn (s) + O2 (g) → MnO2 (s) ∆Hø= -520 kJ mol -1

What is ∆H, in kJ, for the following reaction? MnO2 (s) + C (s) → Mn (s) + CO2 (g) ∆Hø= -400 + 520= 120 kJ

23.2.0 g of sodium hydroxide, NaOH, was added to 98.0 g of water. The temperature of the solution increased from 12.0 °C to 17.5°C. The specific heat capacity of the solution is 4.18 J g-1K-1. Give the expression for the heat evolved in kJ mol-1. 5.5°C x 100.0 g x 4.18 J g-1 °C-1 x 40 g mol-1 / 1000 (conversion from J to kJ)

24. Which equation represent the bond enthalpy for the H- Cl bond in hydrogen chloride?

A. HCl (g)→ H (g) + Cl (l)

B. HCl (g)→ H (g) + Cl (g)

C. HCl (g)→ H (g) + ½ Cl2 (g)

D. HCl (g)→ H (g) + ½ Cl2 (l)

RXN RATE25.Consider the reaction between zinc and hydrochloric acid. Which factors

will affect the reaction rate?

A. The number of reactant particles that collide with the appropriate geometry

B. The collision frequency of the reactant particles

C. The number of reactant particles with E > Ea

26. In a reaction that occurs in 100 g of aqueous solution, the temperature of the reaction mixture increases by 15°C. If 0.25 mol of the limiting reagent is consumed, what is the enthalpy change (in kJ mol -1) for the reaction? Assume the specific heat capacity of the solution = 3.7 kJ kg -1 K -1.

27.Which equation best represents the bond enthalpy of HCl?

A. 2HCl (g) → H2 (g) + Cl2 (g)

B. HCl (g) → ½ H2 (g) + ½ Cl2 (g)

C. HCl (g) → H+ (g) + Cl- (g)

D. HCl (g) → H (g) + Cl (g)

RXN RATE25.Consider the reaction between zinc and hydrochloric acid. Which factors

will affect the reaction rate?

A. The number of reactant particles that collide with the appropriate geometry

B. The collision frequency of the reactant particles

C. The number of reactant particles with E > Ea

26. In a reaction that occurs in 100 g of aqueous solution, the temperature of the reaction mixture increases by 15°C. If 0.25 mol of the limiting reagent is consumed, what is the enthalpy change (in kJ mol -1) for the reaction? Assume the specific heat capacity of the solution = 3.7 kJ kg -1 K -1.

-0.100 kg x 3.7 kJ kg -1 K -1x 15°C/ 0.25 mol

27.Which equation best represents the bond enthalpy of HCl?

A. 2HCl (g) → H2 (g) + Cl2 (g)

B. HCl (g) → ½ H2 (g) + ½ Cl2 (g)

C. HCl (g) → H+ (g) + Cl- (g)

D. HCl (g) → H (g) + Cl (g)

RXNS28.Which combination is correct for a

chemical reaction that absorbs heat from the surroundings?

29.Which combination is correct for a chemical reaction that releases heat into the surroundings?

30.Which process represents the C- Cl bond enthalpy in tetrachloromethane?

A. CCl4 (l) → C (g) + 4Cl (g)

B. CCl4 (l) → C (g) + 2Cl2 (g)

C. CCl4 (g) → CCl3 (g) + Cl (g)

D. CCl4 (g) → C (g) + 4Cl (g)

Type of reaction ∆H at constant pressure

A. Endothermic Positive

B. Endothermic Negative

C. Exothermic Positive

D. Exothermic Negative

Type of reaction ∆H at constant pressure

A. Endothermic Positive

B. Endothermic Negative

C. Exothermic Positive

D. Exothermic Negative

RXNS31.Use the average bond enthalpies below to calculate the enthalpy

change, in kJ, for the following reaction. H2 (g) + I2 (g) → 2HI (g)

32.Use the average bond enthalpies below to calculate the enthalpy change, in kJ, for the following reaction.

H2 (g) + Br2 (g) → 2HBr (g)

33.Which of the following reactions are endothermic?

A. CH4 + 2O2 → CO2 + 2H2O

B. NaOH + HCl → NaCl + H2O

C. Cl2 → 2Cl

Bond Bond energy/ kJ mol-1

I-I 151

H-H 436

H-I 298


H-H 436

Br- Br 193

H- Br 366

RXNS31.Use the average bond enthalpies below to calculate the enthalpy

change, in kJ, for the following reaction. H2 (g) + I2 (g) → 2HI (g)∑(BE bonds broken) - ∑(BE bonds formed)

(436 + 151) – 2 (298) = -9

32.Use the average bond enthalpies below to calculate the enthalpy change, in kJ, for the following reaction.

H2 (g) + Br2 (g) → 2HBr (g)∑(BE bonds broken) - ∑(BE bonds formed)(436 + 193) – 2 (366) = -103

33.Which of the following reactions are endothermic?

A. CH4 + 2O2 → CO2 + 2H2O

B. NaOH + HCl → NaCl + H2O

C. Cl2 → 2Cl


I-I 151

H-H 436

H-I 298


H-H 436

Br- Br 193

H- Br 366

WRITTEN TEST34. The data to the right is from an experiment

used to measure the enthalpy change for the combustion of 1 mole of glucose, C6H12O6 (s). The time-temperature data was taken from a data- logging software program. Mass of sample of glucose, m= 0.3251 g

Heat capacity of the system, Csystem= 2.713 kJ K-1. Calculate ∆T, for the water, surrounding the chamber in the calorimeter.

35. Determine the amount, in moles, of glucose.

36. Calculate the enthalpy change for the combustion of 1 mole of glucose.

37. Using table 13 of the data booklet, calculate the percentage experimental error based on the data used in this experiment.

23.0

23.5

24.0

24.5

WRITTEN TEST34. The data to the right is from an experiment used to

measure the enthalpy change for the combustion of 1 mole of glucose, C6H12O6 (s). The time-temperature data was taken from a data- logging software program. Mass of sample of glucose, m= 0.3251 g

Heat capacity of the system, Csystem= 2.713 kJ K-

1. Calculate ∆T, for the water, surrounding the chamber in the calorimeter. Final – initial = 24.8°C- 23.0°C= 1.8°C

35. Determine the amount, in moles, of glucose. 0.3251g x 1mol/ 180.18g= 0.001804 mol or 1.804 x 10-3 mol

36. Calculate the enthalpy change for the combustion of 1 mole of glucose. ∆Hc= c∆T/mol

∆Hc= 2.713 kJ K-1)(1.8°C)/ 0.001804 mol∆Hc= -2706.98kJ/mol or -2.7 x 10-3 kJ/mol

37. Using table 13 of the data booklet, calculate the percentage experimental error based on the data used in this experiment.

% error= (theoretical)- (experimental) / theoretical x 100

-2803- (-2706.93)/ -2803 x 100= 3.4%

23.0

23.5

24.0

24.5

WRITTEN TEST38. A hypothesis is suggested that TNT, 2-

methyl-1,3,5- trinitrobenzene, is a powerful explosive because it has:

A large enthalpy of combustion

A large volume of gas generated upon combustion

A high reaction rate

Use your answer in 36 and the following data to evaluate this hypothesis. 23.0

23.5

24.0

24.5

Equation for combustion Relative rate of

combustion

Enthalpy of combustion/

kJ mol-1

Glucose C6H12O6 (s) + 6O2 (g) → 6CO2

(g) + 6 H2O (g)

Low

TNT 2C7H5N3O6 (s) → 7 C(s) + 7CO

(g) + 3N2 (g) + 5 H2O (g)

High 3406

WRITTEN TEST38. A hypothesis is suggested that TNT, 2-methyl-1,3,5- trinitrobenzene, is a powerful explosive

because it has:

A large enthalpy of combustion

A large volume of gas generated upon combustion

A high reaction rate

Use your answer in 36 and the following data to evaluate this hypothesis.

1. Enthalpy of combustion is higher in glucose than TNT so enthalpy is not an important part of the explosive power.

2. The amount of gas (CO2 + H2O) in glucose is higher in mol than the amount of gas (CO2 + H2O + N2) in TNT

3. High relative rate of combustion for TNT is important and higher than glucose.


combustion

Enthalpy of combustion/

kJ mol-1

Glucose C6H12O6 (s) + 6O2 (g) → 6CO2

(g) + 6 H2O (g)

Low

TNT 2C7H5N3O6 (s) → 7 C(s) + 7CO

(g) + 3N2 (g) + 5 H2O (g)

High 3406

ENTHALPY OF METHANOL39. In an experiment to measure the enthalpy change of combustion of

methanol, a student heated a copper calorimeter containing 200 cm3 of water with a spirit lamp and collected the following data.

Initial temperature of water: 12°C

Final temperature of water: 22°C

Mass of methanol burned: 1.32 g

Density of water: 1.00 g cm-3

Use the data to calculate the heat evolved when the methanol was combusted.

40.CalcuIate the enthalpy change of combustion per mole of methanol.

41.Suggest two reasons why the result is not the same as the value in the Data Booklet.

ENTHALPY OF METHANOL39. In an experiment to measure the enthalpy change of combustion of methanol, a student

heated a copper calorimeter containing 200 cm3 of water with a spirit lamp and collected the following data.

Initial temperature of water: 12°C

Final temperature of water: 22°C

Mass of methanol burned: 1.32 g

Density of water: 1.00 g cm-3

Use the data to calculate the heat evolved when the methanol was combusted. Q=mc∆T Q= (200 g)(4.18 J g-1 K-1)(10°C)= 8360J

40. CalcuIate the enthalpy change of combustion per mole of methanol. 1.32 g methanol/ 32.05 g/mol CH3OH= 0.0412 mol CH3OH ∆H= Q/mol ∆H= 8360J/0.0412 mol = 202912.62 J/mol or 202.91 kJ/mol

41. Suggest two reasons why the result is not the same as the value in the Data Booklet. 1. Heat loss 2. Incomplete combustion 3. Heat absorbed by calorimeter not

included

DIAZENE AND ETHANOL42. The equation for the reaction between diazene and oxygen is given below.

2N2H2 (g) + O2 (g) → 2N2 (g) + 2 H2O (g) Use the bond enthalpy values from Table 11 of the Data Booklet to

determine the enthalpy change for this reaction.

43. In some countries, ethanol is mixed with gasoline to produce a fuel for cars called gasohol. Define the term average bond enthalpy.

44.Use the information in Table 11 of the Data Booklet to determine the standard enthalpy change for the complete combustion of ethanol.

CH3CH2OH (g) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

45. The standard enthalpy change for the complete combustion of propane, C3H8, is -2219 kJ mol-1. Calculate the amount of energy produced in kJ when 1 g of ethanol and 1 g of propane is burned completely in air.

DIAZENE AND ETHANOL42. The equation for the reaction between diazene and oxygen is given below.

2N2H2 (g) + O2 (g) → 2N2 (g) + 2 H2O (g) Use the bond enthalpy values from Table 11 of the Data Booklet to determine the enthalpy change

for this reaction. 4 N-H = 4 x 391= 1564, 2 N=N = 2x 470= 940, 1 O=O = 498. Total bonds broken 3002. Bonds formed 2 N=N = 2x 470= 940, 4 O-H = 4 x 463 = 1852. Total bonds formed 2792. 3002-2792= +210 kJ/mol

43. In some countries, ethanol is mixed with gasoline to produce a fuel for cars called gasohol. Define the term average bond enthalpy. The energy required to break 1 mol of bonds in gaseous covalent molecules. It is averaged for the same bond in a number of similar compounds.

44. Use the information in Table 11 of the Data Booklet to determine the standard enthalpy change for the complete combustion of ethanol. CH3CH2OH (g) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

5 C-H = 5 x 414= 2070, 1 O-H = 463, 1 C-C = 346, 1 C- O = 358, 3 O=O = 3 x 498= 1494. Total bonds broken 4731. Bonds formed 4 C=O = 4 x 804= 3216, 6 O-H = 6 x 463 = 2778. Total bonds formed 5994. 4731-5994= -1263 kJ/mol

45. The standard enthalpy change for the complete combustion of propane, C3H8, is -2219 kJ mol-1. Calculate the amount of energy produced in kJ when 1 g of ethanol and 1 g of propane is burned completely in air. -2219 kJ mol-1/ 44.11g mol-1 = -50.31 kJ (since both are 1 g) -1367 kJ mol-1/ 46.08 g mol-1 = -29.67 kJ Both answers must be given

CUSO446. The data below are from an experiment to measure the

enthalpy change for the reaction of aqueous copper (II) sulfate, CuSO4 (aq) and zinc, Zn (s).

Cu2+ (aq) + Zn(s) Cu(s) + Zn2+ (aq)100cm3 of 1.00 mol dm-3 copper (II) sulfate solution

was placed in a polystyrene cup and zinc powder was added after 120 seconds. The temperature- time data was taken from a data logging software program. The initial 20 readings were at 19.5°C. A straight line has been drawn through some of the data points. The equation for this line is given by the data logging software as T= -0.100t + 26.5

where T is the Temperature at time t.The heat produced by the reaction can be calculated from the temperature change, ∆T, using the expression below.

Heat change= Volume of CuSO4 (aq) x Specific heat capacity of H2O x ∆T

Describe two assumptions made in using this expression to calculate heat changes.

Linear fit for selected

data. T= -0.100t + 26.5

T Temperature

t time

CUSO446. The data below are from an experiment to measure the enthalpy change for the reaction of aqueous copper (II) sulfate, CuSO4 (aq) and zinc, Zn (s).

Cu2+ (aq) + Zn(s) Cu(s) + Zn2+ (aq)100cm3 of 1.00 mol dm-3 copper (II) sulfate solution was

placed in a polystyrene cup and zinc powder was added after 120 seconds. The temperature- time data was taken from a data logging software program. The initial 20 readings were at 19.5°C. A straight line has been drawn through some of the data points. The equation for this line is given by the data logging software as T= -0.100t + 26.5

where T is the Temperature at time t.The heat produced by the reaction can be calculated

from the temperature change, ∆T, using the expression below. Heat change= Volume of CuSO4 (aq) x Specific heat capacity of H2O x ∆T

Describe two assumptions made in using this expression to calculate heat changes.

1. No heat is lost 2. The specific heat capacity of zinc is zero or negligible 3. No heat is absorbed by zinc

4. The density of the solution is 1.0 g/ cm3


data. T= -0.100t + 26.5

T Temperature

t time

CUSO447. Use the data presented by the data logging

software to deduce the temperature change, ∆T, which would have occurred if the reaction had taken place instantaneously with no heat loss.

48. State the assumption made in #47.

49. Calculate the heat, in kJ, produced during the reaction using the expression given in #46.

50. The color of the solution changed from blue to colorless. Deduce the amount, in moles, of zinc which reacted in the polystyrene cup.

51. Calculate the enthalpy change, in kJ mol-1, for this reaction.


data. T= -0.100t + 26.5

T Temperature

t time

CUSO447. Use the data presented by the data logging software to

deduce the temperature change, ∆T, which would have occurred if the reaction had taken place instantaneously with no heat loss. (final temp – initial temp) 26.5- 19.5= 7°C

48. State the assumption made in #47. Temperature decreases at a uniform rate (when above room temperature)

49. Calculate the heat, in kJ, produced during the reaction using the expression given in #46. 100g x 4.18J g-1K-1 x 7°C= 2926 J= 2.9 kJ

50. The color of the solution changed from blue to colorless. Deduce the amount, in moles, of zinc which reacted in the polystyrene cup. n=M x dm3 n= 1.00 mol dm-3 x 0.100 dm3 = 0.100 mol

51. Calculate the enthalpy change, in kJ mol-1, for this reaction. 2.9 kJ/ 0.100 mol = 29 kJ/mol


data. T= -0.100t + 26.5

T Temperature

t time

METHANOL52. Methanol is made in large quantities as it is used in the production of polymers and in

fuels. The enthalpy of combustion of methanol can be determined theoretically or experimentally.

CH3OH (l) + 1½ O2 (g) → CO2 (g) + 2 H2O (g) Using the information from Table 11 of the Data Booklet, determine the theoretical enthalpy of combustion of methanol.

53. The enthalpy of combustion of methanol can also be determined experimentally in a school laboratory. A burner containing methanol was weighed and used to heat water in a test tube as illustrated below. The following data were collected.

Calculate the amount, in mol, of methanol burned.

Initial mass of burner and methanol/ g 95.371

Final mass of burner and methanol/ g 90.538

Mass of water in test tube/g 20.000

Initial temperature of water/ °C 23.4°C

Final temperature of water/ °C 41.8°C

METHANOL52. Methanol is made in large quantities as it is used in the production of polymers and in

fuels. The enthalpy of combustion of methanol can be determined theoretically or experimentally.

CH3OH (l) + 1½ O2 (g) → CO2 (g) + 2 H2O (g) Using the information from Table 11 of the Data Booklet, determine the theoretical enthalpy of combustion of methanol. 3 C-H = 3 x 414= 1242, 1 C-O = 358, 1 O-H = 463, 1.5 O=O = 1.5 x 498= 747. Total bonds broken 2810. Bonds formed 2 C=O = 2x 804= 1608, 4 O-H = 4 x 463 = 1852. Total bonds formed 3460. 2810-3460 = -650 kJ/mol

53. The enthalpy of combustion of methanol can also be determined experimentally in a school laboratory. A burner containing methanol was weighed and used to heat water in a test tube as illustrated below. The following data were collected.

Calculate the amount, in mol, of methanol burned.4.833g methanol x 1 mol/

32.05g methanol= 0.151 mol burned






METHANOL54.Calculate the heat absorbed, in kJ, by the water.

55.Determine the enthalpy change, in kJ mol-1, for the combustion of 1 mole of methanol.

56. The Data Booklet value for the enthalpy of combustion of methanol is -726 kJ mol-1. Suggest why this value differs from the values calculated in #52 and #55.






METHANOL54.Calculate the heat absorbed, in kJ, by the water.

Q= mc∆T 20g x 4.18J g-1K-1 x 18.4°C= 1538.24J= 1.54kJ

55.Determine the enthalpy change, in kJ mol-1, for the combustion of 1 mole of methanol. -1.54kJ/0.151 mol= -10.2 kJ/mol (neg. due to exotherm rxn)

56. The Data Booklet value for the enthalpy of combustion of methanol is -726 kJ mol-1. Suggest why this value differs from the values calculated in #52 and #55. #52- Bond enthalpies are average values and differ slightly from one compound to another depending on neighboring atoms. #55-Heat is lost to the surroundings or not all heat transferred to the water.






PROPENE57. Two students were asked to use information from the Data Booklet to

calculate a value for the enthalpy of hydrogenation of propene to form propane. C3H6 (g) + H2 (g) → C3H8 (g)Karen used the average bond enthalpies from Table 11. Jessica used the values of enthalpies of combustion from Table 13. Calculate the value for the enthalpy of hydrogenation of propene obtained using the average bond enthalpies given in Table 11.

58. Jessica arranged the values she found in Table 13 into an energy cycle. Calculate the value for the enthalpy of hydrogenation of propene from the energy cycle. ∆Hø (hydrogenation)

C3H6 (g) + H2 (g) C3H8 (g)

2CO2 (g) + H2O(l)

3O

2

-13

79

kJ m

ol-1

2O

2

-11

31

kJ m

ol-1

PROPENE57. Two students were asked to use information from the Data Booklet to calculate a value

for the enthalpy of hydrogenation of propene to form propane.C3H6 (g) + H2 (g) → C3H8 (g)

Karen used the average bond enthalpies from Table 11. Jessica used the values of enthalpies of combustion from Table 13. Calculate the value for the enthalpy of hydrogenation of propene obtained using the average bond enthalpies given in Table 11. 1 C-C= 346, 1 C=C = 614, 6 C-H = 6 x 414= 2484, 1 H-H = 436. Total bonds broken 3880. 2 C-C = 2 x 346= 692, 8 C-H = 8 x 414 = 3312. Total bonds formed 4004. 3880-4004= -124 kJ/mol

58. Jessica arranged the values she found in Table 13 into an energy cycle. Calculate the value for the enthalpy of hydrogenation of propene from the energy cycle.

∆Hø (hydrogenation)

C3H6 (g) + H2 (g) C3H8 (g)

3CO2 (g) + 4H2O(l)

(-1379+ -1131) – (-2219)= -291 kJ/ mol

3O

2

-13

79

kJ m

ol-1

2O

2

-11

31

kJ m

ol-1

PROPANE59.Suggest one reason why Karen’s answer is slightly less accurate than

Jessica’s answer.

60.Karen then decided to determine the enthalpy of hydrogenation of cyclohexene to produce cyclohexane. Use the average bond enthalpies to deduce a value for the enthalpy of hydrogenation of cyclohexene.

C6H10 (l) + H2 (g) → C6H12 (l)

61. The percentage difference between these two methods (average bond enthalpies and enthalpies of combustion) is greater for cyclohexene than it was for propene. Karen’s hypothesis was that it would be the same. Determine why the use of average bond enthalpies is less accurate for the cyclohexene equation shown above, than it was for propene. Deduce what extra information is needed to provide a more accurate answer.

PROPANE59. Suggest one reason why Karen’s answer is slightly less accurate than Jessica’s answer.

The actual values for the specific bonds may be different from the average values.

60. Karen then decided to determine the enthalpy of hydrogenation of cyclohexene to produce cyclohexane. Use the average bond enthalpies to deduce a value for the enthalpy of hydrogenation of cyclohexene.C6H10 (l) + H2 (g) → C6H12 (l)

10 C-H= 10 x 414= 4140, 5 C-C = 5 x 346= 1730, 1 C=C = 507, 1 H-H =436 Total bonds broken= 6813 12 C-H= 12 x 414= 4968, 6 C-C = 6 x 346= 2076= 7044 (6813-7044)= -231kJ/mol

61. The percentage difference between these two methods (average bond enthalpies and enthalpies of combustion) is greater for cyclohexene than it was for propene. Karen’s hypothesis was that it would be the same. Determine why the use of average bond enthalpies is less accurate for the cyclohexene equation shown above, than it was for propene. Deduce what extra information is needed to provide a more accurate answer.

1. Average bond enthalpies do not apply to the liquid state. 2. The extra info needed is the enthalpy of vaporization or condensation of cyclohexene and cyclohexane.

COPPER (I) SULFATE62. If anhydrous copper (I) sulfate powder is left in the atmosphere it slowly absorbs water vapor giving the

blue pentahydrated solid. Cu2SO4 (s) + 5H2O(l) → Cu2SO4 ∙ 5H2O(s)It is difficult to measure the enthalpy change for this reaction directly. However, it is possible to

measure the heat changes directly when both anhydrous and pentahydrated copper (I) sulfate are separately dissolved in water, and then use an energy cycle to determine the required enthalpy change value, ∆Hx, indirectly. To determine ∆H1 a student placed 15.0 g of water in a cup made of expanded polystyrene and used a data logger to measure the temperature. After 2 minutes she dissolved 2.37 g of anhydrous copper (I) sulfate in the water and continued to record the temperature while continuously stirring. She obtained the following results. Calculate the amount, in mol, of anhydrous copper (I) sulfate dissolved in the 15.0 g of water.

∆Hx

Cu2SO4 (s) + 5H2O(l) Cu2SO4 ∙ 5H2O(s)

Cu2SO4 (s)

∆H1 ∆H2

COPPER (I) SULFATE62. If anhydrous copper (I) sulfate powder is left in the atmosphere it slowly absorbs water vapor giving the

blue pentahydrated solid. Cu2SO4 (s) + 5H2O(l) → Cu2SO4 ∙ 5H2O(s)It is difficult to measure the enthalpy change for this reaction directly. However, it is possible to

measure the heat changes directly when both anhydrous and pentahydrated copper (I) sulfate are separately dissolved in water, and then use an energy cycle to determine the required enthalpy change value, ∆Hx, indirectly. To determine ∆H1 a student placed 15.0 g of water in a cup made of expanded polystyrene and used a data logger to measure the temperature. After 2 minutes she dissolved 2.37 g of anhydrous copper (I) sulfate in the water and continued to record the temperature while continuously stirring. She obtained the following results. Calculate the amount, in mol, of anhydrous copper (I) sulfate dissolved in the 15.0 g of water.

∆Hx

Cu2SO4 (s) + 5H2O(l) Cu2SO4 ∙ 5H2O(s)

Cu2SO4 (s)

2.37 g Cu2SO4 x 1 mol/ 223.11g Cu2SO4 = 0.0106 mol Cu2SO4

∆H1 ∆H2

COPPER (I) SULFATE63.Determine what the temperature rise would have been, in °C, if no heat

had been lost to the surroundings.

64.Calculate the heat change, in kJ, when 2.37 g of anhydrous copper (I) sulfate is dissolved in the water.

65.Determine the value of ∆H1 in kJ mol -1.

66. To determine ∆H2, 5.34 g of pentahydrated copper (I) sulfate was dissolved in 13.35 g of water. It was observed that the temperature of the solution decreased by 2.30°C. Calculate the amount, in mol, of water in 5.34 g of pentahydrated copper (I) sulfate.

COPPER (I) SULFATE63. Determine what the temperature rise would have been, in °C, if no heat had been lost to the surroundings.

Final temp- initial temp = 26.1°C – 19.1°C = 7 °C

64. Calculate the heat change, in kJ, when 2.78 g of anhydrous copper (I) sulfate is dissolved in the water.Q= mc∆T Q= 15g x 4.18J/g°C x 7°C= 438.9J = 0.44 kJ or 17.78g x 4.18J/g°C x 7°C=

520.2428 J = 0.52 kJ

65. Determine the value of ∆H1 in kJ mol -1. 0.44kJ/0.0106mol= -41.5 kJ/mol (exothermic or temp. rose) or 0.52kJ/ 0.0106 mol= -49.1 kJ/mol

66. To determine ∆H2, 5.34 g of pentahydrated copper (I) sulfate was dissolved in 13.35 g of water. It was observed that the temperature of the solution decreased by 2.30°C. Calculate the amount, in mol, of water in 5.34 g of pentahydrated copper (I) sulfate.

5.34 g Cu2SO4 ∙ 5H2O(s) /313.21 Cu2SO4 ∙ 5H2O(s)= 0.017 molCu2SO4 ∙ 5H2O

COPPER (I) SULFATE67.Determine the value of ∆H2 in kJ mol -1.

68.Using the values obtained for ∆H1 in #65 and #67, determine the value for ∆Hx in kJ mol -1.

69. The magnitude (the value without the + or – sign) found in a data book for ∆Hx is 78.0 kJ mol -1. Calculate the percentage error obtained in this experiment. (If you did not obtain an answer for the experimental value of ∆Hx then use the value 70.0 kJ mol -1, but this is not the true value.)

70. The student recorded in her qualitative data that the anhydrous copper (I) sulfate she used was pale blue rather than completely white. Suggest a reason why it might have had this pale blue color and deduce how this would have affected the value she obtained for ∆Hx.

COPPER (I) SULFATE67. Determine the value of ∆H2 in kJ mol -1. Q= mc∆T Q= 13.35g x 4.18J/g°C x 2.3°C=

128.35J = 0.13kJ 0.13/0.017= 7.6kJ/mol or Q= 16.13g x 4.18J/g°C x 2.3°C= 155.07J = 0.16kJ 0.16kJ/ 0.017mol= 9.4 kJ/mol

68. Using the values obtained for ∆H1 in #65 and #67, determine the value for ∆Hx in kJ mol -

1. ∆Hx = ∆H1- ∆H2 ∆Hx = -41.5- 7.6= -49.1

69. The magnitude (the value without the + or – sign) found in a data book for ∆Hx is 68.0 kJ mol -1. Calculate the percentage error obtained in this experiment. (If you did not obtain an answer for the experimental value of ∆Hx then use the value 60.0 kJ mol -1, but this is notthe true value.) % error = theoretical – experimental/ theoretical x 100= -68- (-49.1)/ -68 x 100 = 29.26% or -68-(-60)/ -68 x 100 = 11.76%

70. The student recorded in her qualitative data that the anhydrous copper (I) sulfate she used was pale blue rather than completely white. Suggest a reason why it might have had this pale blue color and deduce how this would have affected the value she obtained for ∆Hx. The anhydrous copper (I) sulfate absorbed water from the air. The value would be less exothermic or less negative than expected as temperature increase would be lower or less heat evolved when anhydrous copper (I) sulfate is dissolved in water.

GLUCOSE71. The data below is from an

experiment used to measure the enthalpy change for the combustion of 1 mole of glucose C6H12O6 (s). The time- temperature data was taken from a data- logging software program. Mass of sample of glucose, m= 0.3653 g Heat capacity of the system, Csystem = 2.35kJ K-1

Calculate ∆T, for the water, surrounding the chamber in the calorimeter.

GLUCOSE71. The data below is from an

experiment used to measure the enthalpy change for the combustion of 1 mole of glucose C6H12O6 (s). The time- temperature data was taken from a data- logging software program. Mass of sample of glucose, m= 0.3653 g Heat capacity of the system, Csystem = 2.35kJ K-1 Calculate ∆T, for the water, surrounding the chamber in the calorimeter. Final- initial= 23.78-22.01= 1.77°C

GLUCOSE72.Determine the amount, in moles, of glucose.

73.Calculate the enthalpy change for the combustion of 1 mole of glucose.

74.Using Table 13 of the Data Booklet, calculate the percentage experimental error based on the data used in this experiment.

75.A hypothesis is suggested that TNT, 2-methyl-1,3,5- trinitrobenzene, is a powerful explosive because it has:

A. A large enthalpy of combustion

B. A high reaction rate

C. A large volume of gas generated upon combustion

Use your answer in # 73 and the following data to evaluate this hypothesis:


combustion

Enthalpy of

combustion/ kJ mol-1

Glucose C6H12O6 (s) + 6O2 (g) → 6 CO2 (g) + 6H2O (g) Low

TNT C7H5N3O6 (s) → 7CO (g) + 7 C (s) + 5H2O (g) + 3N2 (g) High 3406

GLUCOSE72. Determine the amount, in moles, of glucose. 0.3653 g C6H12O6 (s) x 1 mol/ 180.18g C6H12O6 = 0.002027 mol

73. Calculate the enthalpy change for the combustion of 1 mole of glucose. ∆Hc= c∆T/mol2.35kJ K-1 x 1.77°C / 0.002027= -2052.05kJ/mol (combustion reaction always exothermic)

74. Using Table 13 of the Data Booklet, calculate the percentage experimental error based on the data used in this experiment. -2803-(-2052.05)/-2803 x 100= 26.79%

75. A hypothesis is suggested that TNT, 2-methyl-1,3,5- trinitrobenzene, is a powerful explosive because it has:

A. A large enthalpy of combustion

B. A high reaction rate

C. A large volume of gas generated upon combustion

Use your answer in # 73 and the following data to evaluate this hypothesis:

1. Enthalpy of combustion is higher in glucose than TNT so enthalpy is not an important part of explosive power.

2. High relative rate of combustion for TNT is important and higher than glucose.

3. The volume of gas (CO2 (g) + 6H2O) in glucose is higher in mol than volume of gas (CO2 + H2O + N2)

4. of TNT. Equation for combustion Relative rate of

combustion

Enthalpy of

combustion/ kJ mol-1

Glucose C6H12O6 (s) + 6O2 (g) → 6 CO2 (g) + 6H2O (g) Low

TNT C7H5N3O6 (s) → 7CO (g) + 7 C (s) + 5H2O (g) + 3N2 (g) High 3406

COMBUSTION REACTIONS76.The standard enthalpy change of three combustion reactions is

given below in kJ. 2C2H6 (g) + 7O2 (g) → 4 CO2 (g) + 6H2O (l) ∆Hø = -31152H2 (g) + O2 (g) → 2H2O (l) ∆Hø = -537C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ∆Hø = -1431 Based on the above information, calculate the standard change

in enthalpy, ∆Hø, for the following reaction. C2H6 (g) → C2H4 (g) + H2 (g)

COMBUSTION REACTIONS76.The standard enthalpy change of three combustion reactions is given

below in kJ. 2C2H6 (g) + 7O2 (g) → 4 CO2 (g) + 6H2O (l) ∆Hø = -3115/2C2H6 (g) + 3 ½ O2 (g) → 2 CO2 (g) + 3H2O (l) 2H2 (g) + O2 (g) → 2H2O (l) ∆Hø = +537/2H2O (l) → H2 (g) + ½ O2 (g) C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ∆Hø = +1431

2CO2 (g) + 2H2O (l) → C2H4 (g) + 3O2 (g) Based on the above information, calculate the standard change in

enthalpy, ∆Hø, for the following reaction. C2H6 (g) → C2H4 (g) + H2 (g)

∆Hø = -3115/2= -1557.5 + 268.5 + 1431= +142kJ/mol

chapter 5 ib chemistry - weebly · 2018. 10. 13. · 11. standard enthalpy change of formation- the...

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