1
Q
Why do energy changes occur during chemical reactions?
A
Energy changes occur because chemical bonds in reactants must be broken and new bonds formed in products. Breaking bonds requires energy (endothermic), while forming bonds releases energy (exothermic). The overall enthalpy change is the difference between these two energy contributions.
2
Q
Why does bond-breaking require energy?
A
Covalent bonds arise from electrostatic attraction between nuclei and shared electrons. Energy must be supplied to overcome this attraction and separate the atoms, so bond-breaking is an endothermic process.
3
Q
Why does bond formation release energy?
A
When atoms form a bond, they move to a lower potential energy state due to electrostatic attraction between nuclei and shared electrons. The decrease in potential energy is released as energy, so bond formation is exothermic.
4
Q
How does bond energy determine whether a reaction is exothermic or endothermic?
A
Exothermic reaction: energy released during bond formation > energy required for bond breaking → ΔH negative. Endothermic reaction: energy required for bond breaking > energy released during bond formation → ΔH positive.
5
Q
State the Law of Conservation of Energy.
A
Energy cannot be created or destroyed, only transferred or converted from one form to another. Therefore, the total energy of the universe remains constant during a chemical reaction.
6
Q
How does the law of conservation of energy apply to reaction pathways?
A
Because energy is conserved, the overall enthalpy change depends only on the initial and final states, not the reaction pathway taken.
7
Q
State Hess’s Law.
A
Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway taken, provided the initial and final states are the same.
8
Q
How are energy cycles used to calculate enthalpy changes?
A
In an energy cycle, the sum of enthalpy changes around the cycle equals zero. Unknown enthalpy changes can therefore be calculated from known values.
9
Q
Define bond enthalpy.
A
Bond enthalpy is the enthalpy change required to break one mole of a specified covalent bond in gaseous molecules under standard conditions.
10
Q
Why must bond enthalpy values refer to gaseous molecules?
A
Bond enthalpy measures only energy required to break covalent bonds, so all species must be in the gas phase to eliminate additional energy changes from intermolecular forces.
11
Q
Why are bond enthalpy values average values?
A
Bond energies vary depending on the chemical environment of the bond. Therefore, tabulated bond enthalpies are averages calculated from many molecules containing that bond.
12
Q
Give an example showing why bond enthalpies are averages.
A
The two O–H bonds in water require different energies to break (502 kJ mol⁻¹ and 427 kJ mol⁻¹). Their average value is used as the average O–H bond enthalpy.
13
Q
State the relationship between bond order and bond enthalpy.
A
Bond strength generally increases with bond order: single bond < double bond < triple bond. Stronger bonds have higher bond enthalpy.
14
Q
What is the relationship between bond enthalpy and bond length?
A
Stronger bonds have higher bond enthalpies and shorter bond lengths, because atoms are held more tightly together.
15
Q
How does bond polarity affect bond enthalpy?
A
Polar X–Y bonds are often stronger than the average of the corresponding X–X and Y–Y bonds because additional electrostatic attraction arises from partial charges created by electronegativity differences.
16
Q
Why can bond enthalpy calculations differ from experimental enthalpy values?
A
Because: bond enthalpies are average values; calculations assume gaseous molecules; experimental reactions may involve liquids or solids; intermolecular forces and phase changes are not included.
17
Q
How is enthalpy change calculated using bond enthalpies?
A
ΔH = ∑E(bonds broken) − ∑E(bonds formed). Energy absorbed to break bonds minus energy released when bonds form.
18
Q
Outline the steps for calculating ΔH using bond enthalpies.
A
Write the balanced equation and structural formulas. Identify bonds broken in reactants. Identify bonds formed in products. Calculate total bond enthalpy for bonds broken and bonds formed. Apply: ΔH = ∑E(broken) − ∑E(formed).
19
Q
Why is the sign of ΔH important when calculating reaction enthalpy?
A
If ΔH is negative, the reaction is exothermic. If ΔH is positive, the reaction is endothermic.
20
Q
Why must some reactions overcome activation energy even if they are exothermic?
A
Reactant molecules must first break existing bonds before new bonds can form. This requires an initial input of energy known as activation energy.
21
Q
How can bond enthalpy explain differences in reactivity?
A
Reactions that require breaking weaker bonds occur more easily. For example, the C–I bond is weaker than other carbon-halogen bonds, so iodoalkanes react more readily in nucleophilic substitution reactions.
- Reactions that require breaking weaker bonds have lower activation energies, so they occur more readily. For example, the C–I bond is weaker than other C–X bonds, so iodoalkanes undergo nucleophilic substitution reactions more easily.
22
Q
Why are bond enthalpy values always positive?
A
Bond enthalpy values are always positive because energy must be supplied to break a chemical bond, overcoming the electrostatic attraction between nuclei and shared electrons.
23
Q
Why are bond enthalpy calculations approximate?
A
Bond enthalpy calculations are approximate because: bond enthalpies are average values; the chemical environment of bonds varies; calculations assume all substances are gaseous; intermolecular forces and phase changes are not included.
24
Q
Why are multiple bonds stronger than single bonds?
A
Multiple bonds involve more bonding electron pairs between the nuclei, increasing electrostatic attraction and resulting in higher bond enthalpy and shorter bond length.
25
How does bond enthalpy relate to reaction rates in halogenoalkanes?
The rate of nucleophilic substitution reactions depends on the strength of the C–X bond. Weaker bonds require less energy to break, so the reaction occurs more easily. Since the C–I bond is weakest, iodoalkanes are the most reactive.
26
Why do experimental enthalpy values differ from bond enthalpy calculations for methane combustion?
Bond enthalpy calculations assume all species are gaseous, but experimentally water forms as a liquid, releasing additional energy due to intermolecular forces. This causes the experimental value to be more exothermic.
27
What restriction does the law of conservation of energy place on energy cycles?
The sum of enthalpy changes around a closed energy cycle must equal zero, otherwise energy would be produced without input, which would violate conservation of energy.
28
Why do the two O–H bonds in water have different bond enthalpies?
The bond enthalpy depends on the molecular environment. After the first O–H bond in water is broken, the structure and electron distribution of the remaining molecule changes, so the second O–H bond requires a different amount of energy to break.
29
Why are intermolecular forces not included in bond enthalpy values?
Bond enthalpy refers only to breaking covalent bonds in gaseous molecules. Because the molecules are in the gas phase, intermolecular forces are absent, so their energy contributions are not included.
30
Why does the H–Cl bond have a higher bond enthalpy than the average of H–H and Cl–Cl bonds?
The H–Cl bond is polar, due to the difference in electronegativity between hydrogen and chlorine. This polarity increases electrostatic attraction between the atoms, making the bond stronger than the average of the non-polar H–H and Cl–Cl bonds.
31
Why do endothermic processes involve the separation of particles?
Endothermic processes require energy input to overcome forces of attraction between particles, such as the electrostatic attraction in chemical bonds.
32
Why do exothermic processes involve the bringing together of particles?
Exothermic processes occur when particles come together and form attractive interactions, releasing energy as the system moves to a lower potential energy state.
33
Why are bond enthalpy values measured for one mole of bonds?
Thermochemical data are expressed per mole so that energy changes can be compared consistently between different reactions and substances.
34
Why are bond enthalpy values always measured under standard conditions?
Standard conditions ensure consistent temperature, pressure, and reference states, allowing bond enthalpy values to be reliably compared.
35
What are the limitations of using average bond enthalpies to calculate reaction enthalpy?
Bond enthalpy calculations are approximate because:
bond enthalpies are average values taken from many molecules
bond strength varies depending on the chemical environment
calculations assume all substances are in the gas phase
intermolecular forces and phase changes are not included
Therefore calculated values may differ from experimental enthalpy changes.
eg. standard of water is liquid, bond enthalpy calculations assume ut us gaseous
36
How can bond enthalpy explain differences in reactivity?
The strength of the bond that must be broken affects the activation energy of the reaction. Reactions involving weaker bonds require less energy to break them, resulting in lower activation energy and faster reaction rates. For example, the C–I bond is weaker than other carbon–halogen bonds, so iodoalkanes are the most reactive in nucleophilic substitution reactions.
37
what is the relationship between the energy absorbed in bond breaking and the energy released in bond formation, and why can reactions still be exothermic or endothermic?
For a specific bond, the energy absorbed when the bond is broken is equal in magnitude to the energy released when the same bond is formed, but the sign is opposite.
Example:
H₂(g) → 2H(g) ΔH = +436 kJ mol⁻¹ (bond breaking)
2H(g) → H₂(g) ΔH = −436 kJ mol⁻¹ (bond formation)
- However, chemical reactions usually involve breaking some bonds and forming different bonds. Because the types and numbers of bonds are different, the total energies do not cancel.
The overall enthalpy change is therefore:
ΔH = ΣE(bonds broken) − ΣE(bonds formed)
- If more energy is released forming bonds than absorbed breaking bonds → exothermic (ΔH < 0)
- If more energy is absorbed breaking bonds than released forming bonds → endothermic (ΔH > 0)
38
Hess’s Law
Hess’s law states that the enthalpy change for a chemical reaction is independent of the pathway taken between the initial and final states, provided the reactants, products, and conditions are the same.
39
Why is Hess’s law valid?
Hess’s law is a consequence of the law of conservation of energy. Because energy cannot be created or destroyed, the total enthalpy change must depend only on the initial and final states of a system.
40
What does Hess’s law imply about reaction pathways?
A reaction occurring in one step or several intermediate steps will have the same overall enthalpy change.
41
Why is the enthalpy change of a complete energy cycle equal to zero?
Because the starting and ending states are identical, there is no overall change in the system’s energy, so: ΣΔH=0
42
How can Hess’s law be represented using an energy cycle?
For a cycle: ΔH1 + ΔH2 = ΔH3 because the sum of enthalpy changes around a complete cycle equals zero.
43
What does the equation 0=ΔH1 + ΔH2 − ΔH3 represent?
It represents conservation of energy in a reaction cycle, showing that the enthalpy change around a closed cycle must equal zero.
44
Why would violating Hess’s law violate conservation of energy?
If the enthalpy change depended on the pathway, it would be possible to construct cycles that continuously produced energy, which would violate the law of conservation of energy.
45
Why are energy cycles useful in chemistry?
Energy cycles allow chemists to calculate enthalpy changes for reactions that cannot be measured directly in the laboratory.
46
What requirement must reactions in an enthalpy cycle satisfy?
Each reaction in the cycle must be balanced chemically, even if the reaction step itself is hypothetical.
47
What is meant by a hypothetical step in an enthalpy cycle?
A hypothetical step is a reaction that may not occur directly in reality, but it is included in the cycle because it helps calculate the overall enthalpy change.
48
How does reversing a chemical equation affect the enthalpy change?
Reversing the direction of a reaction reverses the sign of ΔH. Example: A→B ΔH=−200 B→A ΔH=+200
49
How does multiplying a chemical equation affect ΔH?
If the equation is multiplied by a factor, the enthalpy change must also be multiplied by the same factor. Example: A→B ΔH=−100 2A→2B ΔH=−200
50
What are the three main steps when solving Hess’s law problems?
Write the target reaction with the unknown ΔH. Manipulate known equations so they combine to give the target reaction. Add the enthalpy changes of the manipulated equations.
51
Why must the starting and final conditions be identical when applying Hess’s law?
Because enthalpy depends on the state of the substances, the initial and final conditions must be identical for Hess’s law to apply.
52
Why can enthalpies of formation be calculated using Hess’s law?
Because formation reactions may not occur directly, Hess’s law allows them to be calculated indirectly using enthalpies of combustion or other known enthalpy changes.
53
How can enthalpies of combustion be used in Hess’s law calculations?
By constructing an enthalpy cycle involving combustion reactions, the unknown reaction enthalpy can be calculated from the known enthalpies of combustion of the substances involved.
54
Why are combustion reactions commonly used in Hess’s law calculations?
Because enthalpies of combustion are easily measured experimentally and widely tabulated, making them useful reference reactions.
55
Why do combustion cycles often involve CO₂ and H₂O?
Because combustion of most organic substances produces carbon dioxide and water, which allows reactions to be connected in an enthalpy cycle.
56
What does an energy level diagram show in Hess’s law problems?
An energy level diagram shows: relative enthalpy levels of reactants and products, the enthalpy change between states, how different reaction pathways connect energetically.
57
Why is the enthalpy change for hydrogen reacting with oxygen the same whether the reaction occurs in one step or multiple steps?
Because the initial and final states are the same, the total energy change must also be the same according to Hess’s law and conservation of energy.
58
What does Hess’s law allow chemists to calculate that may not be measurable experimentally?
It allows calculation of enthalpy changes of reactions that cannot be directly measured, such as formation reactions or unstable intermediate reactions.
59
How can Hess’s law be applied to calculate the enthalpy of formation of propane?
By constructing an enthalpy cycle involving the combustion of propane, carbon, and hydrogen, and combining their enthalpies of combustion.
60
Why can the enthalpy change of formation of CS₂ be calculated using Hess’s law?
Because the formation reaction cannot easily be measured directly, but it can be calculated using known combustion enthalpies of carbon, sulfur, and CS₂.
61
Why is the enthalpy of hydration of ethene calculated using combustion enthalpies?
Because the hydration reaction cannot be easily measured directly, but the enthalpies of combustion of ethene and ethanol are known, allowing the reaction enthalpy to be determined indirectly.
62
What does Hess’s law show about the relationship between mechanism and enthalpy change?
The mechanism (route) of the reaction does not affect the overall enthalpy change, only the initial and final states matter.
63
What conditions must be the same for Hess’s law to apply?
The initial and final states, reactants, products, and conditions must be the same.
64
What is the key idea behind solving Hess’s law problems?
Rearrange known thermochemical equations so that, when added together, they produce the target equation. Then add the corresponding enthalpy changes.
65
Why can Hess’s law be used for reactions that cannot be measured directly?
Because enthalpy is a state function, so the enthalpy change depends only on the initial and final states, not on whether the reaction occurs directly or through indirect steps.
66
Why is Hess’s law possible in terms of thermodynamic properties?
Hess’s law works because enthalpy is a state function, meaning the enthalpy change depends only on the initial and final states of a system, not on the pathway taken.
67
Why must the initial and final states be the same when applying Hess’s law?
Because Hess’s law applies only when the reactants, products, and conditions are identical, ensuring the same initial and final enthalpy states.
68
Why can thermochemical equations be added or manipulated in Hess’s law calculations?
Because enthalpy is an extensive property, so enthalpy changes scale with the stoichiometry of the reaction and can be added algebraically when reactions are combined.
69
What thermodynamic data are used in Reactivity 1.2.3 calculations?
Standard enthalpy changes of combustion (ΔH°c) and formation (ΔH°f) are used in thermodynamic calculations.
70
Where can values of enthalpy of combustion and formation be found?
They are provided in the IB Chemistry Data Booklet (Tables 8, 13 and 14).
71
Why have scientists developed databases of standard enthalpy of formation values?
To allow the calculation of enthalpy effects of chemical reactions using thermodynamic data.
72
Define the standard enthalpy change of formation (ΔH°f).
The standard enthalpy change of formation is the enthalpy change when one mole of a substance is formed from its elements in their standard states at 298 K (25°C) and 100 kPa (1.00 × 10⁵ Pa).
73
What conditions define standard thermodynamic measurements?
Temperature = 298 K (25°C) Pressure = 100 kPa (1.00 × 10⁵ Pa)
74
What is the standard state of an element?
The standard state of an element is its most stable form under standard conditions (298 K and 100 kPa).
75
Why is the standard enthalpy of formation important?
It: gives a measure of the stability of a substance relative to its elements can be used to calculate enthalpy changes of all reactions, whether real or hypothetical.
76
What is the standard enthalpy of formation of an element in its most stable form?
ΔHf° = 0
77
Why is the standard enthalpy of formation of an element in its standard state equal to zero?
Because no chemical reaction occurs when an element is formed from itself, so no enthalpy change occurs.
78
What is the standard enthalpy of formation of O₂(g)?
ΔHf°(O₂(g)) = 0 because it is the standard state of oxygen.
79
Why is the standard enthalpy of formation of O₃(g) not zero?
Because ozone is not the most stable form of oxygen, so energy is required to form it.
80
Example:
ΔHf°(O₃(g)) = +142 kJ mol⁻¹
81
standard enthalpy of formation of carbon in its standard state
ΔHf∘(C(graphite))=0
82
Why does diamond have a non-zero enthalpy of formation?
Because diamond is not the most stable allotrope of carbon.
83
ΔHf∘(C(diamond))
=+1.90 kJ mol−1
84
Would allotropes of an element such as graphite and diamond have different ΔH values?
Yes. Different bonding structures and energies cause different enthalpy values and stability.
85
Why is graphite more stable than diamond?
Graphite has lower internal energy due to its bonding structure, making it thermodynamically more stable than diamond.
86
What rule must be followed when writing a standard enthalpy of formation equation?
The reaction must form exactly one mole of the compound from its elements.
87
What elements are used to form ethanol in its formation reaction?
Carbon → C(graphite) Hydrogen → H₂(g) Oxygen → O₂(g)
88
Write the thermochemical equation for the standard enthalpy of formation of ethanol.
2C(graphite)+3H2(g)+O2(g)→C2H5OH(l) ΔHf∘=−278 kJ mol−1
89
Why can fractional coefficients appear in formation equations?
Because the equation must produce exactly one mole of product, which sometimes requires fractional amounts of elements.
90
Define the standard enthalpy change of combustion (ΔH°c).
The standard enthalpy change of combustion is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions (298 K and 100 kPa).
91
How can enthalpy of combustion be determined experimentally?
By burning a substance and measuring the temperature increase of a known mass of water heated by the combustion.
92
What are two uses of enthalpy of combustion values?
Comparing energy output of fuels; Calculating enthalpy changes of other reactions.
93
What formula is used to calculate enthalpy change using formation enthalpies?
ΔHreaction∘=ΣΔHf∘(products)−ΣΔHf∘(reactants)
94
What principle should always be remembered when calculating reaction enthalpies?
ΔH=H(products)−H(reactants)
95
Write the combustion reaction of benzene.
2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l)
96
How is the enthalpy change of the benzene combustion reaction calculated using formation enthalpies?
ΔHrxn=[12ΔHf∘(CO2)+6ΔHf∘(H2O)]−[2ΔHf∘(C6H6)]
97
Calculate ΔH° for the combustion of benzene per mole.
Given: ΔHf∘(C6H6)=49, ΔHf∘(CO2)=−393.5, ΔHf∘(H2O)=−285.8. Calculation: ΔH=−6534.8 kJ for 2 mol benzene =−3267.4 kJ mol−1.
98
Why can formation and combustion enthalpies be used to calculate other reaction enthalpies?
Because enthalpy is a state function, so the enthalpy change depends only on the initial and final states, not on the pathway.
99
Why must the combustion equation represent one mole of the substance?
Because the standard enthalpy of combustion is defined for the combustion of one mole of the substance.
100
Why may fractional coefficients appear in formation equations?
Because the standard enthalpy of formation refers to the formation of one mole of the compound, which may require fractional coefficients.
101
What does the standard enthalpy of formation indicate about a compound?
It indicates the stability of the compound relative to its elements in their standard states.
102
What is the standard state of an element?
The standard state of an element is its most stable form at 298 K and 1.00 × 10⁵ Pa.
103
What is the standard enthalpy of formation of an element in its standard state?
ΔHf∘=0 because no reaction occurs when an element forms itself.
104
Which elements exist as diatomic molecules in their standard state?
H2, N2, O2, F2, Cl2, Br2, I2
105
Why do many non-metals exist as diatomic molecules in their standard state?
Because forming X–X covalent bonds lowers the energy, making the diatomic molecule more stable than individual atoms.
106
What is the standard state of most metals?
Most metals exist as solid metallic lattices (M(s)) in their standard state.
107
What is an important exception to the “solid metal” rule for standard states?
Mercury: Hg(l)
108
What are allotropes?
Allotropes are different structural forms of the same element in the same physical state.
109
Which allotrope is the standard state of carbon?
C(graphite)
110
Why do allotropes have different enthalpies of formation?
Because they have different bonding structures and energies, giving them different thermodynamic stability.
111
Why does diamond have ΔH°f ≠ 0 but graphite has ΔH°f = 0?
Graphite is the most stable allotrope → ΔH°f = 0. Diamond is less stable → ΔH°f > 0.
112
What determines the standard state of gaseous elements?
The standard state is the most stable molecular form at 298 K and 1.00 × 10⁵ Pa.
113
What are the standard states of the halogens?
F2(g), Cl2(g), Br2(l), I2(s)
114
What rule should you always apply when determining ΔH°f of an element?
Ask: Is the element in its most stable form at 298 K and 1.00 × 10⁵ Pa? If yes: ΔHf∘=0 If not → ΔH°f ≠ 0.
115
How can the standard enthalpy change of combustion be determined experimentally?
By burning a known mass of a substance and measuring the temperature increase of a known mass of water heated by the combustion. The heat released is calculated using: q=mcΔT.
116
Why can enthalpy changes of combustion be used to compare fuels?
Because they measure the energy released when one mole of fuel burns completely, allowing comparison of the energy output of different fuels.
117
Why are bond enthalpy calculations most accurate for gas-phase reactions?
Because bond enthalpy values are measured for gaseous molecules, so calculations are most accurate when all reactants and products are also gases. Phase changes introduce additional energy changes that bond enthalpy values do not include.
118
What formula is used to calculate the enthalpy change of a reaction from standard enthalpies of combustion?
ΔH∘=ΣΔHc∘(reactants)−ΣΔHc∘(products)
119
What formula is used to calculate the enthalpy change of a reaction from standard enthalpies of formation?
ΔH∘=ΣΔHf∘(products)−ΣΔHf∘(reactants)
120
Why is the combustion formula written as reactants minus products?
Because in the combustion cycle, both reactants and products are converted to the same combustion products, and the energy cycle gives: ΔH∘+ΣΔHc∘(products)=ΣΔHc∘(reactants) So: ΔH∘=ΣΔHc∘(reactants)−ΣΔHc∘(products)
121
Why is the formation formula written as products minus reactants?
Because in the formation cycle, both reactants and products are formed from the same elements, and the energy cycle gives: ΣΔHf∘(reactants)+ΔH∘=ΣΔHf∘(products) So: ΔH∘=ΣΔHf∘(products)−ΣΔHf∘(reactants)
122
What must always be done before using enthalpy data in these formulas?
Write the balanced chemical equation first, because the enthalpy values are given per mole and must be multiplied by the relevant stoichiometric coefficients.
123
Why must standard enthalpy values be multiplied by stoichiometric coefficients?
Because standard enthalpy values are given per mole, so in a balanced equation they must be multiplied by the number of moles of each substance present.
124
What is the enthalpy of combustion of oxygen in these calculations?
ΔHc∘(O2)=0 because oxygen does not combust further and is already included in the combustion of the other substances.
125
Why can oxygen be given an enthalpy of combustion of zero?
Because the energy released when oxygen reacts is already included in the enthalpies of combustion of the other reactants and products.
126
Why do the formulas for calculating ΔH using combustion data and formation data look reversed?
Because the energy cycles are constructed differently.
In a combustion cycle, reactants and products are both converted to the same combustion products, so
ΔH∘=ΣΔHc∘ (reactants)−ΣΔHc∘ (products)
In a formation cycle, reactants and products are formed from the same elements, so
ΔH∘=ΣΔHf∘(products)−ΣΔHf∘(reactants)
127
ow is the enthalpy of formation of glucose calculated from combustion data?
128
How is the enthalpy change of combustion of propane calculated from formation data?
129
Why do the formulas for calculating ΔH using combustion data and formation data look reversed?
Because the energy cycles are constructed differently.
Combustion cycle: both reactants and products go to the same combustion products
Formation cycle: both reactants and products come from the same elements
130
What is the difference between reaction products and combustion products in enthalpy cycles?
- Reaction products are the substances formed in the reaction being studied.
- Combustion products are the substances formed when reactants and products are completely burned in oxygen, usually CO₂ and H₂O.
In a combustion cycle, both reactants and products are converted to the same combustion products.
131
What should always be remembered when using standard enthalpy data in calculations?
- write the balanced equation
- use the correct formula
- multiply values by stoichiometric coefficients
- include ΔH° = 0 for elements in their standard states, such as
O2 (g)
132
133
Why is the second electron affinity usually positive?
Because the second electron is added to an already negatively charged ion, so energy is needed to overcome electrostatic repulsion.
134
When using ionization energies or electron affinities in a Born–Haber cycle, can the first value simply be multiplied by the ionic charge?
No. You must use the separate ionization energies or electron affinities for each step.
135
What types of ionic compounds should you be able to interpret Born–Haber cycles for?
You should be able to interpret and determine values from Born–Haber cycles for ionic compounds containing univalent ions and divalent ions.
136
What enthalpy terms may be used for atomization in a Born–Haber cycle?
Atomization may involve: sublimation/atomization enthalpy for metals, bond enthalpy (bond dissociation) for non-metals.
137
What does the syllabus say about constructing a complete Born–Haber cycle?
The construction of a complete Born–Haber cycle will not be assessed, but you must be able to interpret cycles and determine values from them.
138
What is the enthalpy change for electron transfer from Na(g) to Cl(g), and what does it show?
Na(g)+Cl(g)→Na+(g)+Cl−(g) ΔH=+147 kJ mol−1 This shows that electron transfer alone is endothermic and energetically unfavourable.
139
Why is the formation of NaCl(s) from gaseous ions so exothermic?
Because there is strong electrostatic attraction between oppositely charged gaseous ions, so lattice formation releases a large amount of energy.
140
How is lattice enthalpy expressed in the IB data booklet?
In the IB data booklet, lattice enthalpy is expressed as the endothermic separation of one mole of an ionic solid into gaseous ions.
141
What is good practice when writing the enthalpy of an endothermic step in a Born–Haber cycle?
It is good practice to explicitly include the positive sign (+) for an endothermic enthalpy change.
142
How is the enthalpy of atomization of oxygen obtained in a Born–Haber cycle?
The enthalpy of atomization of oxygen is half the bond enthalpy of O₂: 1/2 O₂(g)→O(g) So: ΔHatom(O)=1/2 E(O=O)
143
What is the lattice enthalpy of MgO from the Born–Haber cycle shown?
ΔH_lattice (MgO)=+3800 kJ mol−1 (using the IB convention of ion separation)
144
Write the equations for the first and second electron affinities of oxygen.
First electron affinity: O(g)+e−→O−(g) This is exothermic. Second electron affinity: O−(g)+e−→O2−(g) This is endothermic because of repulsion between negatively charged species.
145
How are ionization energies used for metals forming 2+ or 3+ ions in Born–Haber cycles?
For a metal forming a 2+ or 3+ ion, you must include the first, second, and if necessary third ionization energies. Example: Mg(g)→Mg+(g)+e− Mg+(g)→Mg2+(g)+e−
146
How are electron affinities used for non-metals forming 2− ions in Born–Haber cycles?
For a non-metal forming a 2− ion, you must include the first and second electron affinities. Example: O(g)+e−→O−(g) O−(g)+e−→O2−(g)
147
What are the main Born–Haber steps for LiF?
Li(s)→Li(g) atomization of lithium Li(g)→Li+(g)+e− first ionization energy F2(g)→F(g) atomization of fluorine F(g)+e−→F−(g) first electron affinity Li+(g)+F−(g)→LiF(s) lattice formation
148
How is enthalpy of solution related to lattice enthalpy and hydration enthalpy?
For dissolving an ionic solid: ΔH_soln=lattice enthalpy+hydration enthalpy In the example given: ΔH_soln=788−784=4 kJ mol−1
149
What does it mean that enthalpy is a state function?
A state function depends only on the state of the system, not how that state was reached. So in a Born–Haber cycle, the total enthalpy change is the same regardless of the route taken.
150
What is a Born–Haber cycle?
A Born–Haber cycle is an application of Hess’s law used to show energy changes in the formation of an ionic compound. It allows lattice enthalpy to be calculated indirectly from other enthalpy changes.
151
Why are Born–Haber cycles needed?
Born–Haber cycles are needed because lattice enthalpy cannot usually be measured directly, so it must be calculated indirectly using Hess’s law and other experimentally known enthalpy changes.
152
Define the first ionization energy.
The first ionization energy is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous positive ions.
153
Define the first electron affinity.
The first electron affinity is the enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous negative ions.
154
Define lattice enthalpy.
Lattice enthalpy is the enthalpy change when one mole of a solid ionic compound is separated into gaseous ions under standard conditions.
155
How do I identify a formation cycle?
A formation cycle has the elements in their standard states at the bottom. The arrows from the bottom to the compounds represent standard enthalpies of formation, ΔHf°.
156
How do I identify a combustion cycle?
A combustion cycle has the combustion products at the bottom, usually CO₂ and H₂O. The arrows to the bottom represent standard enthalpies of combustion, ΔHc°.
157
What does the top arrow in a Hess cycle represent?
The top arrow represents the enthalpy change of the reaction being studied. It may be written as ΔHr°, ΔHc°, or another specific reaction enthalpy depending on the question.
158
Do I always need to draw a Hess cycle to solve the problem?
No. If the question only asks you to calculate ΔH, the formula is usually enough. If the question asks you to construct, draw, or complete a Hess cycle, then you must draw it to gain those marks.
159
In a combustion cycle, is the top right always CO₂ + H₂O?
No. The top right box is the products of the reaction being studied. The bottom box is the combustion products, usually CO₂ and H₂O. The top right equals the bottom only if the reaction itself is a combustion reaction.
160
What determines how many times a ΔHf° or ΔHc° value is used in a Hess calculation?
The stoichiometric coefficient in the balanced equation determines how many times each enthalpy value is used. Standard enthalpy values are given per mole, so they must be multiplied by the number of moles in the equation.
161
How do I know if an element has ΔHf° = 0?
If the element is in its most stable form under standard conditions, 298 K and 100 kPa, then its standard enthalpy of formation is zero.
162
What is the correct atomization equation for fluorine in a Born–Haber cycle?
The correct atomization equation is: 1/2 F₂(g) → F(g) This represents the formation of one mole of gaseous fluorine atoms from fluorine in its standard state.
- the enthalpy change that occurs when one mole of gaseous atoms is formed from the elements inits standard state
163
What is the difference between a formation cycle and a combustion cycle?
In a formation cycle, both reactants and products are related to the same elements in their standard states, so the formula is: ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants) In a combustion cycle, both reactants and products are related to the same combustion products, usually CO₂ and H₂O, so the formula is: ΔH° = ΣΔHc°(reactants) − ΣΔHc°(products)
164
Why is the top arrow in a Hess cycle not always called ΔHc° or ΔHf°?
Because the top arrow represents the enthalpy change of the specific reaction being studied. It is only called ΔHc° if the reaction is combustion, and only called ΔHf° if the reaction is formation.
165
How do I decide whether to use formation data or combustion data in a Hess problem?
Use the type of data given in the question. If the question gives ΔHf° values, use the formation formula. If the question gives ΔHc° values, use the combustion formula. The method depends on the data provided, not just the type of reaction.
166
Why do arrows pointing up not always mean formation enthalpy?
Arrow direction alone does not determine the enthalpy type. You must look at what is at the bottom of the cycle. If the bottom is elements, upward arrows usually represent formation enthalpies. If the bottom is combustion products, the arrows are part of a combustion cycle instead.
167
What should I check first when I see a Hess cycle diagram?
First check what is at the bottom of the cycle. If the bottom is elements, it is a formation cycle. If the bottom is CO₂ and H₂O, it is a combustion cycle. This tells you which formula and which type of enthalpy data to use.
IB chemistry HL
flashcards
Decks in class (23)
# Cards
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Atomic structure21
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The periodic table: Classification of Elements145
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2.2-2.3 The Nuclear atom and electronic configurations118
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functional groups organic chemistry1
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the metallic model39
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Particulate nature of matter and separation techniques (Pre-IB)50
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Kinetic molecular theory and changes of state (pre-IB)52
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1 1.4 Counting particles by mass: mole49
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1 1.5 Ideal gases55
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Nuclear atom110
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Electronic configuration123
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the ionic model141
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metallic model43
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polyatomic ions17
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periodic trends80
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The covalent model378
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covalent model pt.2 IMF104
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covalent model - resonance structures + benzene37
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Measuring enthalpy changes18
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1.1121
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1.2167
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..8
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Reaction rates0
