1. Grade 11 IB CHEMISTRY
TOPIC 5: ENERGETICS

2. IB Core Objective
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho) Define: Give the precise meaning of a word, phrase or physical quantity.

3. Thermochemistry = study of energy changes in chemical reactions
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
Thermochemistry = study of energy changes in chemical reactions
Most chemical reactions absorb or evolve energy (usually as heat, sometimes as light and mechanical energy)
Energy is measured in Joules (J)

4. Enthalpy (H, heat content) = the total energy of a system
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
Enthalpy (H, heat content) = the total energy of a system
Some is stored as chemical potential energy in the chemical bonds
Energy is absorbed to BREAK bonds
Energy is released when bonds are MADE
The potential energy of the bonds changes in chemical reactions  this is ENTHALPY CHANGE

5. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

6. 1st Law of Thermodynamics:
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
1st Law of Thermodynamics:
Energy can not be created nor destroyed, but it can change form
Energy lost = Energy gained
System + Surroundings = Universe (Which is constant)
Reactions
Endothermic: Energy is used by the system
Exothermic: Energy is produced by the system
The System represents the chemical reaction

7. The surroundings are everything else (here, the cylinder and piston).
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder and piston).

8. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants

9. ΔH = (Negative) More energy was created than used. EXOTHERMIC
5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)
The change in enthalpy ΔH = the net energy after bonds are broken and remade
ΔH = (Negative) More energy was created than used. EXOTHERMIC
ΔH = (Positive) More energy was used than created. ENDOTHERMIC

10. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = - 882 kJ mol -1
IB Core Objective
State that combustion and neutralization are exothermic processes.
Combustion:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = kJ mol -1
Neutralization:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H= kJ mol-1

11. IB Core Objective
Apply the relationship between temperature change, enthalpy change and the classification of a reaction a endothermic or exothermic.
What happens to the temperature when the reaction is exothermic?
A: Temperature goes up.
What happens to the enthalpy change?
A: It is a negative value.
If the same moles of substance were reacted in a larger container of water, would the temperature and enthalpy values stay the same?
A: Enthalpy value would stay the same, the temperature however would not go up as much.

12. IB Core Objective
Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.

13. Exothermic Endothermic
Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.
Exothermic
Endothermic
Products are more stable
Heat energy is RELEASED into the SURROUNDINGS
Enthalpy
ΔH is Negative
Reactants are more stable
Heat energy is ABSORBED into the SYSTEM
Enthalpy
ΔH is Positive

14. Different from STP (Temperature is not 0 oC)
Standard State
Enthalpy is affected by multiple factors (concentration, pressure, state of reactants, temperature))
ΔHθ Under SATP (Standard Ambient Temperature and Pressure)
P = 101.3kPa
T = 25oC or 298 K
C = 1mol/dm3 (Concentration for aqueous solutions aq)
Standard state (physical state of element under these conditions)
θ (Theta) Means under SATP
Different from STP (Temperature is not 0 oC)

15. IB Core Objective
Calculate the heat energy change when the temperature of a pure substance is changed.

16. 4.18 Jg-1 K-1 Specific Heat Capacity
Calculate the heat energy change when the temperature of a pure substance is changed.
Specific Heat Capacity
The energy required to heat 1g of substance 1 degree Celsius
Specific Heat Capacity of water
1 g of H2O to heat 1 degree C requires 4.18 J
Therefore the specific heat capacity of water is 4180 J•kg-1•K-1
4.18 Jg-1 K-1
Typically we use this for labs due to the small quantities used.

17. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
Calorimetry
To measure the energy of a system is hard
To measure the energy of surroundings is easy
q = mcΔT
q = Energy change in the surroundings
m = mass of surroundings (g)
c = Specific Heat Capacity of surrounding substance
(Sometimes s is used instead of c)
ΔT = Temperature change
Energy must have come from the system
For us, this will be the mass of water used

18. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
Questions How much heat is needed to warm 250 g of water from 22°C to near its boiling point 98°C? A: 7.9 x 104 J or 79kJ. Calculate the quantity of heat used when you mix 50cm3 of 1.0M HCl and 50cm3 of NaOH in a coffee-cup calorimeter, and the temperature increases from 21.0°C to 27.5°C. Assume the density is 1.00 g/cm3. A: 2.7 kJ Calculate the enthalpy change for the above reaction in kJ/mol. A: -54 kJ/mol

19. Questions
Excess solid Magnesium is added to a 100g of a 2M solution of Copper(II) Sulphate. The temperature increased from 20.0oC to 65.0oC. (Since the solution is largely water we will assume specific heat capacity as 4.18 Jg-1K-1.
What is the enthalpy change for the reaction?
Don’t forget it must be calculated as a ratio of a full mol.
Push for answer
ΔH = kJ/mol.

20. IB Core Objective
5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.
5.2.4 Evaluate the results of experiments to determine enthalpy changes.

21. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions Evaluate the results of experiments to determine enthalpy changes.
For liquids, a calorimeter should be well insulated and the heat capacity should be low.
Calorimetry: Technique used to measure the enthalpy associated with a particular change.
Calorimetry depends on the assumption that no heat is gained from or lost to the surroundings.
Even with well insulated calorimeters, heat exchange with surroundings is a major source of error.

22. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions Evaluate the results of experiments to determine enthalpy changes.
In combustion experiments, where burning gas is used to heat liquid in a calorimeter, the error is quite large.
Temperature rises are much less than expected and thus ∆H values are less than literature values.

23. IB Core Objective
5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.

24. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.
Question: cm3 of 2 mol dm-3 aqueous sodium hydroxide is added to 30.0 cm3 of hydrochloric acid of the same concentration, the temperature increases by 12.0 C. For dilute aqueous solutions, we can assume the density is 1.00 g cm-3.
What is ∆H?
A: kJ mol-1

25. IB Core Objective
5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Students should be able to use simple enthalpy cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law.

26. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Hess’s Law
Heat of the whole = the sum of the parts
Reaction that take a direct route or multiple step route make no difference with Enthalpy.
ANALOGY
I just bought a $1200 TV, I could either pay 12 equal instalments of $100, pay 2 instalments of $600 or pay it all $1200 at once. How I do it doesn’t matter, in the end I still pay $1200.
In enthalpy terms: ∆H1 = ∆H2 + ∆H3

27. C2H2(g) 2C (s) + H2(g) 2CO2(g) + H2O(l) Calculate ΔHf for C2H2(g)
5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Calculate ΔHf for C2H2(g)
ΔHf =
ΔHf =
CO kJ mol-1 ΔH
ΔHf =
H2O -287 kJ mol-1
ΔHc =
ΔHc =
C2H2(g)
C2H kJ mol-1
2C (s) + H2(g)
A: 224 kJ mol-1
2½O2(g)
2½O2(g)
2CO2(g) + H2O(l)

28. IB Core Objective 5.4.1 Define the term average bond enthalpy.
Enthalpies are a measure of the strength of a covalent bond: the stronger the bond, the more tightly the atoms are joined together.
Breaking of a chemical bond requires energy, and is thus and endothermic process.
Average bond enthalpy: The bond enthalpy for a compound will be affected by surrounding bonds, therefore we use the average bond enthalpy.

29. IB Core Objective
5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.

30. Energy Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.
463 kJ/mol
463 kJ/mol
436 kJ/mol
It takes energy to break bonds
Energy is produced upon bond formation
ΔH = Energy IN (Bonds broken) – Energy OUT(Bonds formed)
ΔH = 1(H-H 436) + ½ (O=O 496) – 2(O-H 463)
ΔH = 684 – 926
ΔH = (Energy is left over, Exothermic) H H
496 kJ/mol O O

31. ΔH = {Bonds broken – Bonds formed}
5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.
Bond
Bond Enthalpy kJ/mol
H-H(g)
436
Cl-Cl(g)
242
F-F(g)
158
H-Cl(g)
431
H-F(g)
562
ΔH = {Bonds broken – Bonds formed}
Sometimes Bond energy is effected by surrounding bonds so an average must be used.
Calculate the Enthalpy of change for the following:
CH4(g) + O2(g)  CO2(g) + H2O(g)
Compare to enthalpy of combustion value (Data book)
Bond
Avg. Bond Enthalpy kJ/mol
C-C(g)
348
C=C(g)
612
C=C(g) (in benzene)
518
C-H(g)
412
C=O(g)
743
O-H(g)
463
N-H(g)
388
O=O(g)
496
Difference calculating enthalpy using heat of formation/ combustion and calculating it using bond enthalpies is due to the average bond enthalpy used and the fact that heat of formation/combustion requires things in there standard state...ie H2O is liquid not gas...it takes more energy to vaporize this.
Why the difference?

32. Given that the enthalpy change for the reaction
5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.
Given that the enthalpy change for the reaction
N2(g) + 3Cl2(g) → 2NCl3(g) is +688kJ/mol
Calculate the enthalpy of the N-Cl bond, given that the bond enthalpies in the nitrogen molecule and the chlorine molecule are 944kJ/mol and 242kJ/mol.
A: 164 kJ/mol

33. IB HL Objective
Define and apply the terms standard state, standard enthalpy change of formation (∆Hfѳ) and standard enthalpy change of combustion (∆Hcѳ)
You have already learned about standard state. What are the conditions?
A: Form normally found at 298K, kPa, 1 mol dm-3

34. Hθc = Standard enthalpy change of combustion
Define and apply the terms standard state, standard enthalpy change of formation (∆Hfѳ) and standard enthalpy change of combustion (∆Hcѳ)
Hθc = Standard enthalpy change of combustion
Enthalpy change when one mole of compound is burned in excess oxygen under standard conditions. (Always exothermic).
Hθf = Standard enthalpy change of formation
Amount of energy released or absorbed in the formation of one mol of compound from elements in their normal states.
By using the definition of standard state, the enthalpy of formation of an element in the standard state is zero.

35. ∆Hѳ = ∑∆Hfѳ(products) - ∑∆Hfѳ(reactants)
IB HL Objective
Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.
∆Hѳ = ∑∆Hfѳ(products) - ∑∆Hfѳ(reactants)

36. Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.
For the following reaction, find the enthalpy of formation: 2 C(graphite) + 3H2(g) + ½O2(g) → C2H5OH(l) A: ∆Hfѳ = -277 kJ mol-1 (All the reactants are in their elemental, standard state, so they would be zero!) Find the enthalpy of formation for the following reaction

37. Find the enthalpy of formation for the following reaction:
Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.
Find the enthalpy of formation for the following reaction:
NH4NO3 (s) → N2O(g) +2H2O(l)
The enthalpies for formation for the above compounds are: NH4NO3 (s) :-366 kJ mol-1, N2O(g) +82 kJ mol-1, and H2O(l) -285 kJ mol -1
A: -122 kJ mol-1
When we have learned more about Hess’s Law, we will return to this later…

38. Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.
Find the formation of combustion of benzene using enthalpies of formation values. First, balance the equation: C6H6(l) + O2(g) → CO2(g) + H2O(g) A: C6H6(l) + O2(g) → 6CO2(g) + 3H2O(l) The enthalpies of formation are: CO2(g): kJ mol-1, H2O(g): kJ, benzene (find in your data booklet), O2??? A: kJ mol-1 (Compare this answer with the data booklet).

39. IB HL Objective
Define and apply the terms lattice enthalpy and electron affinity.
Lattice enthalpy: Enthalpy change to convert one mole of a solid ionic compound into gaseous ions or vice versa.
Electron affinity: Enthalpy change when one mole of gaseous atoms or anions gain electrons to form a mole of negatively charged gaseous ions.

40. Na(s) → Na(g) ∆Hѳ +103kJ mol-1 Exothermic or endothermic?
Define and apply the terms lattice enthalpy and electron affinity.
Standard enthalpy change of atomization (also known as standard enthalpy of vaporization). This is the enthalpy required to change one mole of atoms from the standard state to the gaseous state.
Na(s) → Na(g) ∆Hѳ +103kJ mol-1
Exothermic or endothermic?

41. First ionization energy (remember this?)
Define and apply the terms lattice enthalpy and electron affinity.
First ionization energy (remember this?)
Na(g) → Na+(g) + e- ∆Hѳ = +494 kJ mol-1
Enthalpy atomization of Cl
½Cl2(g) → Cl(g) ∆Hѳ = +121 kJ mol-1
First electron affinity of Cl
Cl(g) + e- → Cl-(g) ∆Hѳ = -364 kJ mol-1

42. IB HL Objective
Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.

43. Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.
Na+(g) + Cl-(g)
Na+ (g)
Cl-(g)
LATTICE ENTHALPY
Na(g) Cl (g)
ΔHof =
Na(s) + ½Cl2(g)
NaCl(S)

44. Formation of NaCl(s) from its gaseous elements.
Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.
Affinity
Formation of NaCl(s) from its gaseous elements.
1) Na(s)  Na(g) ΔHoat = 103 kJ/mol
2) ½Cl2(g)  Cl(g) ΔHoat = ½(242) kJ/mol
3) Cl(g) + e-  Cl(g)- ΔHo = -364 kJ/mol
4) Na(g)  Na+(g) ΔHoat = 500 kJ/mol
Find the enthalpy of formation. Literature value for lattice enthalpy can be found in the data booklet.
A: kJ/mol

45. The second ionization energy of magnesium is +1450 kJ/mol.
Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.
Draw a Born-Haber cycle for the formation of magnesium oxide, and calculate the enthalpy of formation.
The enthalpy of atomization of magnesium is +150 kJ/mol, and for oxygen it is +249 kJ/mol.
The second ionization energy of magnesium is kJ/mol.
Use the data booklet to find other relevant information.
A: -547 kJ/mol

46. IB HL Objective
Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.
The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased attractive forces.

47. Lattice enthalpy of MgO > NaCl. Why? A: Increased ionic charge.
Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.
The greater the charge on the ions, the greater the electrostatic attraction and thus greater the lattice enthalpy, and vice versa.
The larger the ions, then the greater the separation of the charges and the lower the lattice enthalpy, and vice versa.
Lattice enthalpy of MgO > NaCl. Why?
A: Increased ionic charge.
Lattice enthalpy of KBr<NaCl. Why?
A: KBr has larger ions.

48. IB HL Objective
Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character.
A significant difference between the two values indicates covalent character.

49. Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character.
Energy cycles such as Hess’s Law and the Born-Haber cycle are used to find unknown values, provided that the other values are known.
The Born-Haber cycle assumes that the salt formed has 100% ionic character.
Look at Table 13 in your data booklet. There are two tables showing different enthalpy values. What do they label them as?
Lattice enthalpies can be calculated experimentally by using the Born-Haber cycle.
They can also be calculated theoretically from the size, charge, and packing of the ions.

50. Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character.
When the electronegativity difference is lower, will the compound exhibit greater or less ionic character?
A: Less.
When the electronegativity difference is less, there will be a greater amount of covalent bonding, and this will result in a higher experimental value for lattice enthalpy than theoretical.
Look at electronegativity differences between NaCl and AgI. Which would you guess has a greater difference in experimental vs. theoretical lattice enthalpy values?
A: AgI

51. IB HL Objective
State and explain the factors that increase the entropy in a system.
Systems naturally tend towards disorder.
Entropy: A measure of the degree of disorder or randomness in a system.

53. Entropy values for gas>liquid>solid.
State and explain the factors that increase the entropy in a system.
Entropy values for gas>liquid>solid.
Gas pressure increases, then entropy decreases (reduces volume for gas particles to move in).
When a solid or liquid dissolves in a solvent, entropy increases.
Increase the number of moles increases the entropy.
Heating a substance increases the entropy, since this increases the movement of particles.

54. IB HL Objective
Predict whether the entropy change (∆S) for a given reaction or process is positive or negative.
ΔS = ∑S(Products) – ∑S(Reactants)
So if the products have more disorder than the reactants, would the entropy change be positive or negative?
A: Positive

55. Consider the following equation: NH4Cl(s) → NH3(g) + HCl(g)
Predict whether the entropy change (∆S) for a given reaction or process is positive or negative.
Consider the following equation:
NH4Cl(s) → NH3(g) + HCl(g)
State ∆S as positive or negative, and explain why.
A: Positive. (∆S=+285 J K-1 mol-1) More moles on product side, so more disorder. Also, solid has become a gas, so more disorder.

56. IB HL Objective
Calculate the standard entropy change for a reaction (∆Sѳ) using standard entropy values (∆Sѳ).
ΔS = ∑S(Products) – ∑S(Reactants)

57. Calculate the entropy change for the combustion of methane (CH4).
Calculate the standard entropy change for a reaction (∆Sѳ) using standard entropy values (∆Sѳ).
Calculate the entropy change for the combustion of methane (CH4).
First, write the equation.
Make a prediction if it will be positive or negative, and explain why.
Next, plug in the values and calculate. You can find the entropy value for methane in your data booklet. O2 is 205, CO2 is 214, and H2O is 70.
A: -242 J K-1 mol-1

58. IB HL Objective
Predict whether a reaction or process will be spontaneous by using the sign of ∆Gѳ.
A reaction will be spontaneous when ∆Gѳ has a negative value.
Spontaneous: Once started, a reaction will continue without any extra energy having to be added.
Remember redox? An electrochemical cell can create energy through a spontaneous reactions. So if Eѳcell is positive, what would ∆Gѳ be?
A: Negative

59. Is the combustion of coal spontaneous at room temperature?
Predict whether a reaction or process will be spontaneous by using the sign of ∆Gѳ.
Is the combustion of coal spontaneous at room temperature?
Yes. The activation energy is different than spontaneous reactions. You have to get the coal going, but once you do, it continues to react.
What about the combustion of diamond at room temperature?
Yes, although it has a very high activation energy. Once this is reached, it will be a spontaneous reaction.

60. IB HL Objective 15.4.2 Calculate ∆Gѳ for a reaction using the equation
∆Gѳ = ∆Hѳ – T∆Sѳ
and by using values of the standard free energy change of formation, ∆Gfѳ.

61. ∆Hѳ = -395.4 kJ mol-1. ∆Sѳ = +6.6 J K-1 mol-1.
Calculate ∆Gѳ for a reaction using the equation ∆Gѳ = ∆Hѳ – T∆Sѳ and by using values of the standard free energy change of formation, ∆Gfѳ.
Let’s go back to the combustion of diamond at room temperature (298K):
C (s) + O2(g) → CO2(g)
∆Hѳ = kJ mol-1.
∆Sѳ = +6.6 J K-1 mol-1.
What is ∆Gѳ?
A: kJ mol-1
Would this reaction be spontaneous at all temperatures? Why or why not? Refer to the equation!
A: If ∆Hѳ is negative and ∆Sѳ is positive, then ∆Gѳ will always be negative.

62. IB HL Objective
Predict the effect of a change in temperature on the spontaneity of a reaction, using standard entropy and enthalpy changes and the equation:
∆Gѳ = ∆Hѳ – T∆Sѳ
If both ∆Sѳ and ∆Hѳ are positive, then spontaneity depends on T. (Spontaneous at higher temperatures).
If both ∆Sѳ and ∆Hѳ are negative, then spontaneity depends on T. (Spontaneous at lower temperatures).