3.1.8 - Thermodynamics Flashcards
1
Q
Define ‘enthalpy change’
A
Heat energy change at a constant pressure.
2
Q
Define ‘standard enthalpy change’
A
Enthalpy change measured under standard conditions.
3
Q
Define ‘enthalpy of atomisation’
A
Enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state.
4
Q
Define ‘1st ionisation enthalpy’
A
The enthalpy change when 1 mole of gaseous 1+ ions is formed from its gaseous atoms.
5
Q
Define ‘2nd ionisation enthalpy’
A
The enthalpy change when 1 mole of gaseous 2+ ions are formed from gaseous 1+ ions.
6
Q
Define ‘1st electron affinity’
A
The enthalpy change when 1 mole of gaseous 1- ions are formed from gaseous atoms.
7
Q
Define ‘2nd electron affinity’
A
The enthalpy change when 1 mole of gaseous 2- ions is formed from gaseous 1- ions.
8
Q
Define ‘bond dissociation enthalpy’
A
The enthalpy change when 1 mole of covalent bonds is completely separated to form gaseous atoms.
9
Q
Define ‘lattice enthalpy of formation’
A
The enthalpy change when 1 mole of solid ionic compound is formed from gaseous ions.
10
Q
Define ‘lattice enthalpy of dissociation’
A
The enthalpy change when 1 mole of solid ionic compound dissociates into gaseous ions.
11
Q
State 2 factors that form the perfect ionic model.
A
- Ions are perfect spheres. 2. With evenly distributed charge.
12
Q
In what scenario is covalent character more likely for an ionic compound?
A
- Cation has high charge density. 2. Anion has low charge density.
13
Q
How does covalent character change the lattice enthalpy?
A
Larger than expected as it requires more energy to break/form.
14
Q
State the 2 factors that lattice enthalpy depend on and how.
A
- Size of ions (bigger size, charges are further apart, weaker attraction). 2. Charge on the ion (bigger charge, stronger attraction).
15
Q
What are the 3 main factors that impact entropy?
A
- Physical State. 2. Dissolving. 3. Number of particles.
16
Q
What is the formula for entropy?
A
ΔS = Σ S_prod - Σ S_react.
17
Q
If ΔS is +ve, is a reaction likely? What if it’s endothermic?
A
It means a reaction is likely even if it’s endothermic.
18
Q
Using ΔG, how can you tell if a reaction is feasible?
A
- If ΔG is -ve (or 0) the reaction is feasible. 2. If ΔG is +ve the reaction is not feasible.
19
Q
Recall briefly how we use calorimetry to measure energy stores in sample.
A
- Dry sample weighed + burnt in pure oxygen. 2. In a sealed container. 3. Temperature increase of the fixed volume of water used to calculate energy released.
20
Q
Define Enthalpy of formation
A
The enthalpy change when one mole of a compound is formed from its elements in its standard states under standard conditions.
21
Q
Define Lattice enthalpy of formation
A
The enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions under standard conditions.
22
Q
In the Born-Haber cycle, what are up arrows?
A
Endothermic.
23
Q
Endothermic is a
A
Positive enthalpy change.
24
Q
Why would the lattice enthalpy be different from the theoretical value?
A
Doesn’t follow a perfectly ionic model and has some covalent characteristics in between.
25
Define electron affinity for chlorine.
The enthalpy change / heat energy change / ΔH for the formation of one mole of chloride ions from chlorine atoms.
26
What is covalent character caused by?
A larger distortion in the negative ion.
27
Why are experimental values usually higher than theoretical values?
1. Theoretically, the greater the polarization the bigger the charge, and the larger the ion the larger the distortion. 2. The larger the difference in enthalpy, the greater the polarization and covalency.
28
What are the 2 factors required for a substance to dissolve?
1. For something to dissolve the substance bonds must break. 2. New bonds form between the solvent and the substance.
29
Explain what happens in the enthalpy of solution.
1. Substances react with water and the bonds are broken, creating moving ions. 2. Water then reacts with these ions, hydrating them, forming bonds between them; where the negative ion is pulled towards the delta + hydrogen, whereas the positive ion is pulled towards the delta - oxygen. 3. The structure breaks down.
30
What is the condition for dissolving to occur?
The new bonds must be the same strength or greater than those broken. If not, dissolving will not occur.
31
What entropy do solids have?
Solids have a low entropy, as there is less disorder between the molecules. Particles are arranged neatly in rows.
32
How do you know which molecule has the highest covalent character?
The greater the experimental value, the greater the covalent character.
33
What does it mean when there is an increase in entropy?
1. Increase in disorder within the system, where there must be a greater number of moles in the products than reactants. 2. Entropy increase is positive.
34
What does it mean when there is a decrease in entropy?
Loss of moles, and goes from gas to solids. Decrease is negative and unfavorable.
35
What is entropically unfavorable?
Entropy decrease (-) is unfavorable.
36
Does CO2 have a high entropy?
Yes, less structured and a gas; so much higher entropy.
37
Will exothermic reactions with a positive entropy be feasible?
Exothermic reactions with a positive entropy will always be feasible whatever the temperature.
38
Formula to find the boiling point of a substance.
ΔH = TΔS.
39
Provide the formula to find temperature from the Gibbs-free energy equation.
T = ΔH / ΔS.
40
State why a value for the enthalpy change of solution of magnesium oxide is not found in any data books.
Magnesium oxide reacts with water.
41
Explain in terms of molecules, why the entropy is zero when the temperature is zero Kelvin.
1. Molecules do not have enough energy to collide. 2. At 0K the particles are stationary and there is no disorder.
42
Equation to calculate entropy from free Gibbs.
ΔS = (ΔH – ΔG) / T.
43
Formula to calculate delta H from free Gibbs.
ΔH = TΔS.
44
Write an equation to show how nitrogen monoxide is formed in an internal combustion engine.
NO + O2 ----> 2NO.
45
Suggest one reason why a sample of magnesium appears to be stable in air at room temperature.
Because MgO has a protective layer around it.
46
Suggest the condition under which carbon would have an entropy of 0.
0K.
47
Is the reaction feasible when Delta H is negative and Delta S is positive?
Feasible at any temperature.
48
Is the reaction feasible when Delta H is negative and Delta S is negative?
Feasible below a certain temperature.
49
Is the reaction feasible when Delta H is positive and Delta S is negative?
Never feasible.
50
Define the enthalpy of atomisation.
Enthalpy change when one mole of gaseous atoms is formed from its elements in standard states.
51
Define the enthalpy of hydration.
The enthalpy change when one mole of gaseous ions become hydrated (fully dissolved in water).
52
Define bond dissociation enthalpy.
Enthalpy change when one mole of a covalent bond is broken down into its gaseous states.
53
Define the lattice enthalpy of dissociation.
The enthalpy change when one mole of a solid ionic compound is broken down into its constituent ions in the gas phase.
54
Define the enthalpy of vaporisation.
Enthalpy change when one mole of a liquid is turned into a gas.
55
Define enthalpy of fusion.
The enthalpy change when one mole of a solid is turned into a liquid.
56
The more polarization the more =
Covalent character.
57
Explain the term perfectly ionic.
1. Ions are perfect spheres. 2. No covalent interactions.
58
Explain in terms of electrons why the complexes are different colours.
ΔE = hv = hc/plancks constant. Different wavelengths of light are different colours. ΔE is different for different ligands; so the colour of the complex is different.
59
A calculation based on the perfect ionic model gives a smaller numerical value than the value calculated. Explain the difference.
1. Silver iodide contains covalent character. 2. The forces holding the lattice together are stronger.
60
Define the enthalpy of atomisation and give an example.
The enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state. For example, the atomisation of sodium from solid to gaseous atoms (Na(s)→Na(g)) has an enthalpy change of +148 kJ/mol.
61
What is bond dissociation enthalpy?
Bond dissociation enthalpy is the energy required to break one mole of a covalent bond in a molecule, producing gaseous atoms or radicals. For example, breaking one mole of chlorine Cl2(g)→2Cl(g) requires +242 kJ/mol.
62
Differentiate between the first and second ionisation enthalpies.
The first ionisation enthalpy is the energy needed to remove one electron from each atom in one mole of gaseous atoms. The second ionisation enthalpy is the energy required to remove one electron from each ion in one mole of gaseous 1+ ions, forming 2+ ions.
63
What are standard conditions?
298K/25°C, 100kPa/1 atm, 1 moldm-3.
64
Define the enthalpy change of formation.
The enthalpy change when 1 mole of a compound is formed from its constituent elements under standard conditions, with all reactants and products in their standard states.
65
Define the enthalpy change of combustion.
The enthalpy change when 1 mole of a substance is burned completely in oxygen.
66
Define the enthalpy change of atomisation.
The enthalpy change that accompanies the formation of 1 mole of gaseous atoms from the element in its standard state.
67
Define the first ionisation enthalpy.
The enthalpy change when 1 mole of gaseous atoms loses 1 electron per atom to form 1 mole of gaseous 1+ ions.
68
Define the second ionisation enthalpy.
The enthalpy change when 1 mole of gaseous 1+ ions loses 1 electron per atom to form 1 mole of gaseous 2+ ions.
69
Define the first electron affinity.
The enthalpy change when 1 mole of gaseous atoms gains 1 electron per atom to form 1 mole of gaseous 1- ions.
70
Define the second electron affinity.
The enthalpy change when 1 mole of gaseous 1- ions gains 1 electron per atom to form 1 mole of gaseous 2- ions.
71
Define the mean bond enthalpy.
The enthalpy change when 1 mole of gaseous molecules each breaks a covalent bond to form 2 gaseous ions or free radicals.
72
Define the lattice enthalpy of formation.
The enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions.
73
Define the lattice enthalpy of dissociation.
The enthalpy change when 1 mole of a solid ionic compound dissociates into its gaseous ions.
74
Define the enthalpy of hydration.
The enthalpy change when 1 mole of gaseous ions dissolves completely in water.
75
Define the enthalpy of solution.
The enthalpy change when 1 mole of an ionic solid dissolves in enough solvent to ensure that the dissolved ions are well separated and do not interact with one another.
76
Define the enthalpy of vaporisation.
The enthalpy change when 1 mole of a liquid evaporates to form 1 mole of a gas.
77
Define the enthalpy of sublimation.
The enthalpy change when 1 mole of a solid sublimes to form 1 mole of a gas.
78
Why are ionisation enthalpies always positive (endothermic)?
Energy needs to be put in to pull an electron away from the nucleus to overcome the electrostatic attraction.
79
Why are first electron affinities generally exothermic?
1. First electron affinities are typically exothermic because energy is released when an electron is added to a neutral atom. 2. This occurs because the electron is attracted to the nucleus.
80
What happens when a mole of a solid sublimes?
A mole of a solid sublimes to form 1 mole of a gas.
81
Why are ionisation enthalpies always positive?
Energy needs to be put in to pull an electron away from the nucleus to overcome the electrostatic attraction.
82
Why are first electron affinities generally exothermic?
First electron affinities are typically exothermic because energy is released when an electron is added to a neutral atom. This occurs because the electron is attracted to the positively charged nucleus, resulting in a release of energy as the electron is captured into the atom's electron cloud.
83
Why are second electron affinities and subsequent ones generally endothermic?
Second and subsequent electron affinities are endothermic because energy must be added to overcome the electrostatic repulsion between an incoming electron and an already negatively charged ion. This repulsion requires external energy to force the electron into the electron cloud of the negatively charged ion.
84
How does the lattice enthalpy change as the ionic radius increases?
As the ionic radius increases, the lattice enthalpy becomes less negative, indicating a weaker lattice. This is because larger ions have their positive and negative charges further apart, reducing the electrostatic force of attraction between them and thus weakening the lattice enthalpy.
85
What is the symbol for entropy and in what units is it measured?
The symbol for entropy is 𝑆, and it is measured in joules per kelvin (J K−1) per mole.
86
What is the Gibb's free energy equation at 0K?
At 0K, the Gibb's free energy equation is ΔG = ΔH.
87
How does entropy change as temperature increases?
As temperature increases, entropy generally increases. This is because higher temperatures provide more energy to the particles, causing them to move more vigorously and occupy more possible states, thereby increasing disorder within the system.
88
Define the term 'enthalpy change.'
Enthalpy change refers to the heat or energy change that occurs at constant pressure during a chemical reaction.
89
What are the steps to complete a Born-Haber cycle for the formation of calcium chloride?
Starting from calcium and chlorine gas, you move to gaseous calcium ions and chloride ions, including the necessary ionisation energies, electron affinities, and finally to solid calcium chloride, reflecting lattice formation.
90
Write the equation that represents the enthalpy of solution for magnesium chloride.
MgCl2(s) → Mg2+(aq) + 2Cl–(aq)
91
Why is the hydration enthalpy of Ca2+ less exothermic compared to Mg2+?
Ca2+ has a larger ionic radius and a lower charge density compared to Mg2+, resulting in weaker electrostatic interactions with water molecules, thus a less exothermic hydration enthalpy.
92
Explain why the standard entropy value for carbon dioxide is greater than that for carbon.
Carbon dioxide has a higher standard entropy compared to carbon because it is a gas, which is more disordered than solid carbon, resulting in more ways for its particles to be arranged and distribute energy.
93
Why does the hydration enthalpy become less exothermic from lithium to potassium in Group 1 ions?
As the ionic radius increases from lithium to potassium, the charge density decreases. This reduction in charge density means weaker electrostatic interactions between the ion and water molecules, resulting in less exothermic hydration enthalpies.
94
What does it imply about a reaction if it is exothermic and has a negative entropy change?
A reaction that is exothermic and has a negative entropy change is feasible below a certain temperature. Above this temperature, the positive 𝑇Δ𝑆 term outweighs the negative Δ𝐻, making ΔG positive and the reaction non-spontaneous.
95
Explain why the hydration enthalpy is less exothermic from lithium to potassium.
This is because as you move from lithium to potassium in Group 1, the ionic radius increases. Larger ions have a lower charge density, which weakens their electrostatic attraction to the water molecules, resulting in less exothermic hydration enthalpies.
96
Why might the theoretical lattice enthalpy differ from the experimental lattice enthalpy?
Theoretical lattice enthalpies are calculated assuming perfectly ionic interactions and spherical ions. Experimental values can differ due to partial covalent character in the bonding and ion polarization, which are not considered in the perfect ionic model.
97
How does temperature affect the feasibility of a reaction that is exothermic with a negative entropy change?
If a reaction is exothermic (Δ𝐻<0) and has a negative entropy change (Δ𝑆<0), the reaction is feasible below a certain temperature. Above this temperature, the term 𝑇Δ𝑆 becomes significant enough to make Δ𝐺 positive, thus making the reaction non-spontaneous.
98
Explain how temperature affects the feasibility of a reaction with ΔS = 25 J K-1 mol-1 and ΔH = 180 kJ mol-1.
Use the equation ΔG = ΔH - TΔS to find the temperature at which ΔG = 0: T = ΔH / (ΔS / 1000) = 180 kJ mol-1 / (25 / 1000) J K-1 mol-1 = 7200 K. The reaction becomes feasible (ΔG ≤ 0) at temperatures above 7200 K.
