Energetics II (Topic 13) Flashcards
1
Q
What is lattice enthalpy of formation (Delta_latticeH)
A
Enthalpy change when 1 mole of a solid ionic compound is formed from it’s gasesous ions under standard conditions.
Ca^2+(g) + 2Cl^-(g) -> CaCl_2
Ions in the gaseous state forming a solid ionic compound
2
Q
What is Enthalpy change of atomisation (symbol)
A
The enthalpy change when 1 mole of gaseous atoms is made from an element in its standard state. This is ALWAYS an endothermic process. You need energy in to do this
1/2 F_2 (g) –> F (g)
Turning a standard version of an element into a single gas atom
3
Q
What is 1st electron affinity (symbol)
A
Enthalpy change when 1 mole of gaseous 1- ions are made from 1 mole of gaseous atoms
O(g) –> O^- (g)
Adding an electron to something that is neutrally charged, adding the first electron
4
Q
What is 2nd electron affinity (symbol)
A
Enthalpy change when 1 mole of gaseous 2- ions are made from 1 mole of gaseous 1- ions
O^-(g) –> O^2-(g)
Adding a second electron, adding an electron to something that is already negatively charged
5
Q
What is ionic bonding
A
Ionic bonding is formed from oppositely charged ions, there is an electrostatic attraction between the two ions. This electrostatic attraction is what formed the ionic bond
6
Q
What factors influence the strength of the ionic bond
A
The size of the charge on the ion.
The size of the ion/ionic radii
7
Q
How does size of the charge of the ion influence the strength of the ionic bond
A
The bigger the charge on an ion, the stronger the electrostatic attraction between ions. An ionic bond between a 2+ and a 2- ion is stronger than a bond between a 1+ and 1- ion.
K+ and Cl- has an mp of 770 C, Ca^2+ and O^2- has a melting point of 2572 C
8
Q
What effect does a stronger ionic bond have on melting point and boiling point
A
The stronger the ionic bond, the more energy is required to overcome electrostatic forces of attraction, so they have higher melting and boiling points.
9
Q
What do you need to break ionic bonds apart
A
Quite a lot of heat energy
10
Q
How does the size of the ion/ionic radius influence the strength of the ionic bond
A
The smaller the ion/ionic radius, the stronger the electrostatic attraction between ions.
Na+ and Cl - has an mp of 801 C, K+ and Cl- (K is bigger ion than Na) has a mp of 770 C.
11
Q
WHY does size of ion/ionic radius affect the strength of an ionic bond
A
Smaller ions can pack together more closely because of their size, so the + and - charges are much closer together. So more energy is required to overcome these stronger forces. So mp and bp increase as a result
12
Q
What is charge density
A
Higher charge in a smaller ion means the ion has a high charge density. A high charge is concentrated in a small area.
13
Q
What effect does higher charge density have on the strength of an ionic bond
A
Generally, the smaller the ion, and the higher the charge, the stronger the electrostatic attraction and therefore the higher the mp. Higher charge density means stronger ionic bond.
14
Q
What does exothermic mean
A
It gives out heat energy, meaning the system loses heat energy/enthalpy
15
Q
What does endothermic mean
A
It takes in energy, meaning the system gains heat energy, so you need energy in to make the process happen.
16
Q
What is enthalpy of formation (symbol)
A
Enthalpy from the formation of 1 mole of a solid ionic compound from elements in their standard states. This is an exothermic process
17
Q
What is enthalpy change of first ionisation
A
Energy required to remove the first outermost electron from 1 mole of gaseous atoms. Enthalpy change when 1 mole of gaseous atoms is turned into 1 mole of gaseous 1+ ions and 1 electron.
18
Q
What is the Born-Haber cycle used for and why
A
To calculate lattice enthalpies because you can’t calculate this value through experiments
19
Q
How do you start a Born-Haber cycle
A
You always start by drawing the bottom line with your solid compound which you want to form through lattice formation on top of the bottom line. Remember, always include state symbols.
______LiCl(s)_____
20
Q
What do you do after the drawing bottom line and writing the solid you want to form of a Born-Haber cycle
A
You write the first route to your product, which is through standard enthalpy change of formation above the product.
These are the two elements required to make LiCl(s) in their standard states.
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
21
Q
How is the Born-Haber cycle organized
A
The higher up the step, the more energy the step has. This is why we put Li(s) + 1/2 Cl_2 (g) just above LiCl(s), because that step has more energy.
22
Q
What does the direction of the arrows in the Born-Haber cycle mean
A
The direction the arrow is pointing signals the energy change. For example in the step below, the arrow going from standard enthalpy of formation to the product is pointing down. This means that the process is exothermic, so it gives out heat energy.
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
If the arrow was pointing up, that means the process would be endothermic, so it takes in heat energy. Each type of enthalpy change has the quality of either being exothermic or endothermic.
23
Q
How do you know whether an enthalpy change process is exothermic or endothermic
A
Forming bonds is exothermic, breaking bonds is endothermic. Each type of enthalpy change process has its own exothermic vs endothermic process, But memorize the quality for each process.
24
Q
When drawing arrows in a Born-Haber cycle, what should you make sure to do
A
Make sure that the tip and the tail of the arrow touches the level lines. Otherwise you may lose marks
25
When drawing arrows in a Born-Haber cycle, what should you make sure to do
Make sure that the tip and the tail of the arrow touches the level lines. Otherwise you may lose marks
26
How does the Born-Cycle work in analogy form?
Born-Haber is a cycle of different events. If you wanted to form LiCl(s), you just go down ROUTE 2 through standard enthalpy of formation, or you could go up and around ROUTE 2 and form it through standard LATTICE enthalpy of formation. In the same way, you can get to school by going from the springs/spinneys route, or you could go around the Springs Souk route. You end up at the same location, but one route may be longer than the other.
27
What is the goal when drawing a Born-Haber cycle?
You start with your solid compound at the bottom, and draw route 1 above it, standard enthalpy of formation of that compound. Then, you must go from that step to the phase where you have a positive gaseous ion and a negative gaseous ion to allow for LATTICE enthalpy of formation. This is the end goal step of a Born-Haber cycle. Each step brings you closer to this finish line.
28
After drawing the bottom line and route 1 of the Born-Haber cycle, what must you do next
After that, you have to work towards getting your elements from their standard states, to their gaseous ion forms. In this example, we must do the first enthalpy of atomisation of chlorine in order to get it from 1/2 Cl_2(g) to just Cl (g). All atomisation reactions are endothermic. You need heat to convert, so we need energy in. This is why the arrow is pointing up. We are going from a diatomic molecule to a single atom because we only need 1 chlorine for LiCl(s)
_A___Li(s) + Cl (g)_______
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
29
After converting one of your elements from it's diatomic form to a single atom through enthalpy change of atomisation, what is your next step in your Born-Haber cycle
The goal is to convert all elements from their standard states into their charged gaseous forms. In our example, we already have Cl(g) in its single atom gaseous form, so we must now convert Li(s) into its gaseous form.
This is also enthalpy change of atomisation (endothermic) since we are turning lithium solid, into lithium gas
_A___Li(g) + Cl (g)______
_A___Li(s) + Cl (g)______
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
30
What determines the order of the steps in a Born-Haber cycle when it is in the same section i.e in converting the elements from their standard states to their gaseous forms
The order of these steps is determined by their energy levels, so which ever one has the lowest energy increase will be put first. If turning 1/2 Cl_2 (g) to Cl (g) is a lower energy increase than turning Li(s) to Li(g), then the Cl(g) step will go below, and the Li(g) step will go above
31
What are the sections of steps in a Born-Haber cycle
1. Desired Compound on bottom line
2. Standard Enthalpy of formation Showing route 1
3. Going from standard enthalpy of formation to elements in their gaseous forms through standard enthalpy of atomisation
4. Providing a charge to atoms in their gaseous forms through 1st ionisation energy THEN 1st electron affinity.
5. Turning charged gaseous ions into the lattice through standard enthalpy of LATTICE formation
32
After both elements are in their gaseous form, what is the next step in the Born Haber cycle
You have to provide charge for the elements to turn them into ions. They must be gaseous ions in order for standard LATTICE formation enthalpy to work. We remove an electron from lithium through enthalpy change of 1st ionisation, giving it a positive charge, resulting in a free electron. Ionisation is always endothermic. This free electron is used in the next step.
_A_____________________________Li+ (g) + Cl(g) + e-
_A___Li(g) + Cl (g)______
_A___Li(s) + Cl (g)______
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
33
After one of your elements has been ionised with a positive charge, what is the next step in the Born Haber cycle
Now, the free electron that was ionised from the element which has gained a positive charge. This free electron is given to the other element to form a negative ion. This is done through enthalpy change of 1st electron affinity. This is an EXOTHERMIC PROCESS, SINCE WE ARE FORMING BONDS. Breaking bonds is endothermic. But the electron is joining the second element to make a negative ion, which means it is forming a bond. Since the previous step was the highest energy step, that is the top level. This electron affinity step is going back down in energy as it is exothermic.
Cl(g) gains an electron, turning it into a Cl- (g) ion.
_A_____________________________Li+ (g) + Cl(g) + e-______________________
_A___Li(g) + Cl (g)______ __Li+ (g) + Cl- (g)_____v_
_A___Li(s) + Cl (g)______
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)____
34
When is something considered 1st electron affinity or 1st ionisation energy
1st electron affinity is when you are adding an electron to a neutral atom, as it is the first electron added before becoming negative. 1st ionisation energy when you are removing an electron from a neutral atom, as this is the first electron removed before becoming positive.
35
What is the FINAL step of the Born-Haber cycle
Going from gaseous ions to forming your solid compound with standard enthalpy of LATTICE FORMATION. This is the finish line we were aiming towards. This is EXOTHERMIC, because we are FORMING BONDS, even if they are ionic bonds, we are still forming bonds. This step actually takes us all the way to the bottom line, as we have just formed our desired solid compound.
_A_____________________________Li+ (g) + Cl(g) + e-______________________
_A___Li(g) + Cl (g)______ __Li+ (g) + Cl- (g)_____v_
_A___Li(s) + Cl (g)______
__Li(s) + 1/2 Cl_2 (g)___
_v_______________________________LiCl(s)______________________________v_
36
What is the difference between standard enthalpy change of formation and standard enthalpy change of LATTICE FORMATION
In standard enthalpy change of formation, two species combine covalently from their standard states. In standard enthalpy change of LATTICE FORMATION, two gaseous IONS combine electrostatically.
37
What is the difference in enthalpy between route 1 and route 2 of a Born-Haber cycle
There is no difference in total enthalpy. It works like the Hess's cycle. As long as reactants and products are the same, the total enthalpy change is the same regardless of the route.
38
How do you calculate the value of Lattice enthalpy of formation from a Born-Haber cycle with given values
Put all the values next to each step, if you are going along the arrow, keep the sign of the value given the same. If you are going against the direction of the arrow, then flip the sign.
39
How do you make sure the given values match up to the processes in your Born-Haber cycle. For example, you are given enthalpy of atomisation of Cl_2 (g)= 242, but on your cycle, the step needs the atomisation of 1/2 Cl_2 (g)
Simply alter the value given to the desired result which is needed for the step. DO NOT FORGET TO DO THIS. From the example, you simply do 242/2 = 121, which is the value you use. Remember to check if you are going against or along the arrow's direction to see which sign to put.
40
What is the analogy to help you solve Born-Haber cycle problems
Imagine your 2 routes as Sheikh Zayed and Al Khail Road. Sheikh zayed is faster and more direct, but today Sheikh Zayed is blocked off because of the 10km run for the 30x30 challenge. So you have to take Al Khail road to get to your destination. If you are asked to solve the energy for one step, you have to go all the way around and use algebra to find the value which is asked for.
41
How do you test if you got the right answer when solving Born-Haber cycle problems
Think about which step they are asking for and whether it is exothermic or endothermic. For example, they ask for standard enthalpy of LATTICE formation, which you know is exothermic. If your answer is on the right lines, you should get a negative value for your answer.
42
Once you have worked out the final value on a Born-Haber cycle problem, how do you test it a final time to see if it is correct.
Follow the Born-Haber cycle from one side to the other, you can choose ANY step as your starting point, using the arrow rules. If the answer to your calculation = 0, then the value you worked out is correct!
43
What is the difference between a regular Born-Haber cycle and a Born-Haber cycle that is between ions of 2+ and 2- charge.
In a 2+ and 2- charge Born Haber cycle, there are some extra steps. The first difference is that there are 2 ionisation steps for the ion which will become positive. So Mg (g) going to Mg+ (g) is step 1, and Mg+ (g) going to Mg 2+ (g) is step 2. Both are endothermic so they go up as normal. The second difference is that when the second element becomes negatively charged, the first electron affinity is going down, exothermic as normal, but the second electron affinity goes back up, it is endothermic. Then the final step of enthalpy change of LATTICE FORMATION is exothermic, arrow goes down as normal.
_A____________________________Mg^2+ (g) + 2O (g) + 2e-________________________________________
_A___Mg+ (g) + 2O(g) + e-_ __________A____Mg 2+ (g) + O^2- (g)____
_A___Mg(s) + 2O (g) __ _V_______Mg 2+ (g) + O- (g) + e-________
______Mg(s) + O_2 (g)__
_v_________________________________________MgO(s)___________________________________v______
44
Which order do you do enthalpy of ionisation and enthalpy of electron affinity in a Born-Haber cycle between ions of 2+ and 2- charge.
First, ionise the element that is going to become a 2+ charge fully. So do a step with 1st ionisation energy enthalpy, then a step with 2nd ionisation energy enthalpy. Get your electrons both free first, giving you 2e-. After THAT, you do the electron affinity for the second element which is going to become a 2- charge. Remember, going from O^1- to O^2- is ENDOTHERMIC, arrow goes up.
45
In a Born-Haber cycle between 2+ and 2- charge ions, why is 2nd electron affinity endothermic
Because energy is needed to add an electron to a negative ion. Repulsive forces try to prevent you doing this between 2 negative species, so you have to add energy in to force the electron to bond, meaning the step is endothermic.
46
Why can theoretical and experimental values of lattice enthalpies be different
This depends on how purely ionic the compound is. The more ionic the bonding, the closer the experimental value will be to the theoretical. There are instances of ionic bonding which are impure, which have some other bonding type, like covalent, mixed in.
47
What is the perfectly ionic model that we assume when calculating theoretical lattice enthalpies
1. Ions that are perfectly spherical
2. The charge is evenly distributed in this sphere (point charge)
48
What does it mean when you carry out an experiment and the theoretical lattice enthalpy is different than the value you calculated through experimental methods
The compound being experimented on DOESN'T follow the perfectly ionic model and has some covalent characteristics. Non-purely ionic compounds are more common than purely ionic ones.
49
What happens in an impurely ionic compound to make it impure
Most of the time, the positive ion distorts the charge distribution in the negative ion. We say the positive polarises the negative ion.
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50
What is the relation between more polarisation of an ionic compound and how purely ionic it is
The more polarisation we get, the more covalent character there will be, and therefore it will be less purely ionic
51
What can lattice enthalpy values tell us about the bonding of an ionic compound and how
How purely ionic a compound is. The closer the experimental lattice enthalpy value is to the theoretical, the more purely ionic a compound is.
52
What do you see when comparing theoretical and experimental values of lattice enthalpy, how are the 2 figures different?
In all examples, experimental value shows a higher lattice enthalpy than the 'purely ionic' theoretical value.
53
What does larger difference in experimental vs theoretical lattice enthalpies indicate
The larger difference indicates that more covalent character is being displayed. This is caused by larger distortions in the negative ion. The bigger the difference in lattice enthalpy, the more polarisation you have, the greater the covalent character.
54
What type of cations (+) are more polarising, tend to be polarise other anions (-) more, and why
Smaller cations are more polarising than larger ones. Smaller cations have a higher charge density as the charge is concentrated in a smaller area. The cation pulls electrons towards itself more readily
55
What types of anions (-) tend to be polarised more easily
Larger anions with a large charge are polarised much more easily than smaller, lower charged anions.
A large ionic radius with a 2- charge anion is way more polarised than a small ionic radius with 1- charge
56
Why are larger anions with larger charge more easily polarised
Large ionic radius means electrons further away from nucleus + larger charge means electrons repel each other more. So they are more easily pulled towards the cations.
57
How do you remember whether a cation or anion is positive or negative
CATion is PAWSotive. Anion sounds like onion makes you cry, so negative.
58
What is electronegativity
The ability for an atom to attract electrons towards itself in a covalent bond. The further up and right you go in the periodic table, the more electronegative the element is.
59
When can a covalent compound indicate more ionic characteristics
The bigger the difference in electronegativity, the more ionic a compound will be. A difference of 0 will be a purely covalent compound.
2.2 - 4.0 (Highly ionic characteristics) vs 3.0 - 3.0 (Purely covalent characteristics)
60
Summarize how to tell whether an ionic compound has more covalent characteristics and whether a covalent compound has more ionic characteristics.
For Ionic Compounds: The larger the difference between theoretical and experimental lattice enthalpy values, the more covalent characteristics the compound has
For covalent compounds: The larger the difference in electronegativity between the elements in the compound, the more ionic characteristics the compound has.
61
What is enthalpy change of solution (Delta_(solution)H
We are gonna add the tiniest amount of solvent to the ionic substance until it is just about dissolved and the solution is exactly saturated. This means that if you add any more solvent to dilute the solution, there will be no further enthalpy change seen.
the enthalpy change when 1 mole of an ionic substance is dissolved in the minimum amount of solvent to ensure no further enthalpy change is observed upon further solution.
62
What must happen for a substance to dissolve
1. Substance bonds must break (endothermic)
2. New bonds formed between solvent and substance (exothermic)
63
How does an ionic lattice in solid form dissolve in water
The ionic lattice is in solid form in water. Substance bonds are broken to create free moving ions. Bonds formed between ions and the respective polar sides of a water molecule. The ions are hydrated.
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64
How do solutes dissolve in water
The delta+ Hydrogen is attracted to negative ions and the delta- Oxygen is attracted to positive ions. The structure starts to break down
65
What is hydration
When the water molecules surround the ions from a solid ionic lattice
66
What is hydration
When the water molecules surround the ions from a solid ionic lattice
67
What condition must be met for an ionic substance to be able to dissolve in a solvent, and what happens if this condition is not met
For this to happen, the new bonds formed must be the same strength or greater than those broken. If not, then a substance is very unlikely to dissolve.
68
What is the enthalpy of solution usually for soluble substances
Exothermic enthalpies of solution for this reason. They need to have a lower bond strength than the bond formed between it' s ions and the solvent for it to be soluble.
69
How much ionic compounds likely to dissolve in water
Most ionic compounds dissolve in polar solvents like H2O.
70
What enthalpy values are needed to calculate enthalpy change of solution
Lattice dissociation enthalpy & Enthalpy of hydration
71
What is lattice dissociation enthalpy
Enthalpy change from the breaking up of an ionic compound
72
What is enthalpy change of hydration (Delta_(hyd)H
Enthalpy change when 1 mole of aqueous ions is made from 1 mole of gaseous ions
73
How do you calculate enthalpy of solution using enthalpy of dissociation and enthalpy of hydration
Putting it in a Hess's cycle-like structure and going along the arrow following the arrow rules. The data is given to you
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74
What are the assumptions made when finding enthalpy of solution from enthalpy of dissociation and enthalpy of hydration
Assume the order that the enthalpies occur. Assume that we do the following
1. Break the solid lattice up into it's gaseous ions first (lattice dissociation)
2. Dissolve the gaseous ions in water (enthalpy of hydration)
75
What affects the enthalpy change of hydration
The charge of the ion
The size of the ion
76
How does the charge of the ion affect enthalpy change of hydration
The larger the charge, the greater the enthalpy of hydration
Larger charge = Greater hydration enthalpy
77
Why does size of charge of ion affect the enthalpy change of hydration
Ions with higher charge attract water molecules more strongly as electrostatic attraction is stronger. More energy released when bond is made, which means they have more exothermic energy
78
How does size of ion affect enthalpy change of hydration
The smaller the ion, the greater the enthalpy of hydration
Smaller ion = Greater hydration enthalpy
79
Why does size of ion affect enthalpy change of hydration
Smaller ions have higher charge density than larger ions. They attract water molecules more strongly hence there is more exothermic enthalpy of hydration.
80
How does charge density affect enthalpy of hydration
Higher charge density = Greater enthalpy of hydration. So any factors that influence charge density like charge magnitude of ion or size of ion will have an affect on hydration enthalpy. Stronger attraction between water and the ion. This means we have more exothermic enthalpy of hydration
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81
What is entropy
The measure of disorder in a system. How chaotic something is
82
What is the explanation for entropy relating to particles
Entropy (S) is the number of ways energy can be shared out between particles. The more disorder there is, the higher the level of entropy.
83
How does entropy change when going from solid -> liquid ->
Solids have the lowest level of disorder/entropy, gases have the highest level of disorder/entropy
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84
Why do solids have the lowest level of entropy
Because the particles are arranged neatly in rows. There is a very limited number of ways in which the particles can be arranged
85
Why do liquids and gases have a higher level of entropy
Liquids and gases are more disordered because there are more ways in which they can be arranged.
86
What factors can affect entropy
Number of particles.
State of Matter or Phase of the Substance
87
How does number of particles affect entropy
If a reaction is in the same state but more moles are produced, then entropy increases. There are ore ways which energy can be distributed.
88
What is the analogy linked to entropy and bedrooms
If you have a bedroom, its easier to allow the bedroom to get messy, than it is to put it back into order. It takes more energy and time to put things back into place, but it takes very little time to messy up a room. It's a lot lower energy to allow things to go to disorder.
89
What entropy level do particles tend towards and why
Particles always try and tend to a more disordered arrangement because it doesn't require much energy to do that.
90
How do endothermic reactions feel
As they go, they get colder
91
Can a reaction still occur if it is enthalpically unfavourable
A reaction can be spontaneous (feasible) even if it is enthalpically unfavourable (endothermic, needs energy to continue)
92
How can an endothermic (enthalpically unfavourable) reaction still occur
They can still spontaneously react if changes in entropy overcome changes in enthalpy, since increasing entropy is energetically favourable
93
Enthalpically, why should the reaction between hydrated Barium Hydroxide and Ammonium Chloride not be able to occur
It is an endothermic reaction, meaning it is enthalpically unfavourable without putting energy in. (+164 kJ/mol). Usually in this type of reaction energy must be supplied in for the reaction to progress
94
Why does the reaction between Barium Hydroxide and Ammonium Chloride still occur, despite it being endothermic
It is entropically favourable for the reaction to occur.
95
How is it entropically favourable for the reaction between Barium Hydroxide and Ammonium Chloride to occur
Ba(OH)_2.8H_2O(s) + 2NH_4Cl(s) -> 2NH_3(g) + 10H_2O(l) + BaCl_2(s)
There are 3 moles on the left hand side on the reactants, and 13 moles on the right hand side on the products, so the products side has higher entropy due to more moles, so it is more entropically favourable.
The left hand side starts with 2 solids, but the right hand side has a liquid, a gas and a solid. Increased disorder so entropically favourable.
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96
What are the units of entropy
J K^-1 mol^-1
97
What is the equation for entropy change of the system (ΔS_system)
ΔS_system = S_products - S_reactants
98
Under which conditions are entropy values given in
Standard entropy (S^θ).
1 mole of substance
100kPa
298K
99
What qualities make a reaction entropically favourable
If on the products there is:
More moles than the reactants
More disordered states of products than reactants (gas is more disordered than solid)
100
What sign must ΔS have for the reaction to be possible at room temperature (standard entropy conditions)
If ΔS is positive, the reaction is entropically feasible, so it will occur at room temperature.
101
Why is entropy calculated using the entropy changes of the surroundings and system
Energy is transferred from the actual reaction vessel to the surroundings
102
What is the equation for total entropy change (ΔS_total)
ΔS_total = ΔS_system + ΔS_surroundings
103
What is the equation to calculate entropy change of surroundings (ΔS_surroundings)
ΔS_surroundings = -ΔH/T
ΔH is in Jmol^-1, so you may have to convert from kJmol^-1
T is in K (kelvin)
104
What must you remember to do when calculating ΔS_surroundings
Convert ΔH from kJmol^-1 to Jmol^-1 if you need to
105
How do you convert kJmol^-1 to Jmol^-1
Convert +25.7 kJmol^-1 to Jmol^-1
You multiply the given value in kJ mol^-1 by 10^3
25.7 = kJ mol^-1
25.7 x 10^3 = J mol^-1
106
What are standard conditions
Temperature - 298K
Pressure - 100kPa
1 mole of substance
107
What does Gibbs Free Energy (ΔG) tell you
Tells us if a reaction is feasible or not. It takes into account enthalpy and entropy
108
What is the equation and units for Gibbs Free Energy
ΔG = ΔH - TΔS_system
109
What are the units for the components of the Gibbs Free Energy equation
ΔG: Jmol^-1
ΔH: Jmol^-1
T: K
S_system: J K^-1 mol^-1
110
What are the units for change in free energy (ΔG)
J mol^-1
111
What is the basic rule for Gibbs Free Energy (ΔG)
A reaction is feasible in theory if ΔG is NEGATIVE or ZERO.
112
What does ΔG=0 tell us
The minimum free energy required for the reaction to be feasible
113
Why might you not observe a reaction even if it is calculated to be feasible
You may not observe a reaction occurring due to the activation energy being too high, or the rate of reaction being very slow
114
What is an example of a feasible reaction that is not able to be observed
Iron rusts when it comes into contact with oxygen. But the rate of this reaction is very slow, so while it is feasible, it is only seen after weeks or months.
115
What does a negative ΔG tell you about the reaction
The reaction is feasible AT the given temperature
116
What value of ΔG does an EXOTHERMIC (-ve ΔH value) reaction with a positive ΔS value have
ΔG is always going to be negative, even with temperature of 0K, which is the lowest temperature possible. This means it will always be feasible.
This is because ΔH is negative and ΔS is positive which means the second half of the equation is negative, making ΔG always negative.
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What value of ΔG does an ENDOTHERMIC (+ve ΔH value) reaction with a negative ΔS value have
ΔG is always going to be positive. This means the reaction will NEVER be feasible, at any temp
118
What value of ΔG does an ENDOTHERMIC (+ve ΔH value) reaction with a positive ΔS value have
ΔG will only be negative at HIGHER temperatures, which means the reaction will only be feasible at higher temperatures.
Decomposition of sodium hydrogen carbonate is an endothermic process (+ΔH), but we are going to a more random arrangement, solid to gas (+ΔS). This only happens above a certain temperature, because the second half, -TΔS_system, has to be large enough to not be cancelled out by positive ΔH.
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What value of ΔG does an EXOTHERMIC (-ve ΔH value) reaction with a negative ΔS value have
ΔG will only be negative at LOWER temperatures, which means the reaction will only be feasible at lower temperatures.
Freezing water is an exothermic process (-ΔH), so energy is released to freeze water. But we are going to a more ordered arrangement, so ΔS is negative as well. So water only freezes below a certain temperature, because with a low T value, ΔS_system will not be large enough to make the equations result a positive value for ΔG. (Remember, ΔS is negative, so a double negative becomes a positive)
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How do you use logic to determine the relationship between ΔH, ΔS and ΔG values
Use the Gibbs Free Energy Equation
ΔG = ΔH - TΔS_system
Treat it like balancing a seesaw. If ΔH is negative and ΔS is positive, then the reaction will always be exothermic no matter the value of T, as ΔG will always be a negative number.
121
How do you calculate the temperature which a reaction just becomes 0.
Use ΔG = ΔH - TΔS_system, and rearrange to make T the subject. When ΔG is zero, the reaction has just become feasible, so you have to rearrange
0=ΔH - TΔS_system
T = ΔH/ΔS_system
Remember, a reaction is just feasible when ΔG is 0.
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Why would calculating the minimum temperature for a reaction to be feasible, temp when, ΔG=0 be useful
We want to try to use as little energy as we can, and carry out our reactions at the minimum temperature possible.
We save the environment by using less fossil fuels, since we don't need to get to an unnecessarily high temperature. And we save money because the fuel to get temperatures cost a lot of money.
This equation is about being more efficient with your reactions.
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What is the equilibrium constant
The concentration ratio between reactants and products at equilibrium at a certain temperature.
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What is the equilibrium constant when ΔG is negative
Reaction is theoretically feasible, equilibrium constants are large, GREATER THAN 1.
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What is the equilibrium constant (k) when ΔG is positive
Reaction is NOT theoretically feasible, equilibrium constants are small, LESS THAN 1.
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What is the equation linking ΔG with equilibrium constant
ΔG = -RT lnk
k - equilibrium constant
R value is given to you (e.g 8.31)
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How do you predict what the value of ΔG will be when given the equilibrium constant
If equilibrium constant is greater than 1, then you can predict ΔG to be negative, if equilibrium constant is less than 1, you can predict ΔG to be positive.
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What is the symbol for equilibrium constant
K
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How do you calculate the equilibrium constant (K) when given the value of ΔG
Rearrange the equation
ΔG = -RT lnk
to make lnk the subject.
lnK = ΔG/-RT.
Make sure to work out the units of the equilibrium constant as they always change.
