Unit 3, Topic 1: Chemical Equilibrium Systems

Question 1

Open systems

Answer 1

exchange matter and energy with surroundings

Question 2

Closed systems

Answer 2

only exchange energy with surroundings – equilibrium can only occur in a closed system

Question 3

Reversible Reactions

Answer 3

• Involved in equilibrium
• A reaction is reversable if the products formed can react together and re-form the reactants
• Reactions are reversible if the products can collide with enough energy (reverse activation energy) to re-from reactants

Question 4

Static equilibrium

Answer 4

when a position of balance has been achieved and the reaction has stopped

Question 5

Dynamic Equilibrium

Answer 5

• When macroscopic properties (mass, concentration) remain the same but microscopic processes continue.

Question 6

Extent of reaction

Answer 6

• The extent of a reaction refers to how much product is formed when equilibrium is reached.

Question 7

Rate of reaction

Answer 7

• The rate of reaction is the change in concentration of products and reactants with time.
• Rate of reaction determines how quickly a reaction reaches equilibrium.

Question 8

Equilibrium position

Answer 8

refers to the relative amounts of reactants and products at equilibrium

Question 9

Le Chatelier’s Principle

Answer 9

• If a stress is applied to a system at equilibrium, the system will act to oppose this stress and reestablish equilibrium.
• This results in a shift in the position of equilibrium (favoring products or reactants)

Question 10

Temperature

Answer 10

• The only factor that has a permanent influence on the position and value of Kc (Alters rate of reaction as well)
• If the forward reaction is exothermic, increasing temperature shifts equilibrium left (decrease in Kc)
• If the forward reaction is endothermic, increasing temperature shifts equilibrium right (increase in
  Kc)
• Consider heat as a reactant/product – increasing one will invoke Le Châtelier’s Principle
• At higher temperate, the rates of both revers and forward reactions increase – collision theory
• Activation energy is greater for endothermic reactions
• Increasing temperature would generally increase the rate of endothermic reactions more

Question 11

What is the only way to change Kc

Answer 11

change in temperature

Question 12

Pressure

Answer 12

• Changing pressure only doesn’t change Kc
• If the number of molecules on both sides of the equation is equal, pressure has no effect
• If the volume of the container is doubled, pressure is halved etc.
• If pressure is increased, the system acts to oppose the stress – moves to the side with the lease molecules
• If pressure is decreased, equilibrium moves to the side with more moles
• pressure increases, the rate of reaction increases.
• The rate of reaction involving the side with more molecules becomes greater then the side with lesser molecules
• As more products form, the rate of the revers reaction increases, and the rate of the forward reaction decreases until equilibrium is reached

Question 13

Addition of an Inert Gas

Answer 13

• Total pressure of a system can be increased by adding an inert gas
• Does nothing to Kc – doesn’t change rate of reactions

Question 14

Addition or Removal of Reactants or Products

Answer 14

• If a product is added, Le Châtelier’s principle will shift to decrease the concentration of that product – results in a net forward reaction as more products will need to be reacted
• adding extra product leads to a greater rate of reaction for the forward reaction
• The forward reaction becomes greater than the rate of the reverse reaction
• the rate of the forward reaction then decreases, and equilibrium is reached
• this does not mean that now the concentration is the same as before because a net forward reaction would have occurred
• no effect on Kc

Question 15

Dilution

Answer 15

• Reducing the concentration of all particles resulting in a shift of equilibrium towards the side that produces the greatest amount of dissolved particles
• this occurs due to Le Chatelier’s principle - more particles need to be produced to counteract the dilution
• no effect on Kc

Question 16

Catalysts

Answer 16

• adding catalysts does not shift equilibrium or evoke Le Chatelier’s principle
• equilibrium is reached quicker

Question 17

Homogenous system

Answer 17

reactants and products are in the same phase (e.g. all gasses)

Question 18

Heterogeneous system

Answer 18

reactants and products are in different phases

Question 19

Kc

Answer 19

Kc describes the extent of a reaction – how much products vs how much reactants.

• Solids and liquids which are pure do not appear in a Kc expression unless all species in the system are liquids such as an esterification reaction.
• Water can be included in a reaction if it is a reaction or a product, not as a solvent.

Question 20

lower Kc

Answer 20

indicates a lower proportion of products

Question 21

higher Kc

Answer 21

indicates a greater proportion of products – the higher the Kc, the closer the reaction is to completion.

Question 22

Kc Formula

Answer 22

= [products]/[reactants]

Question 23

Reaction Quotient (Qc)

Answer 23

The reaction quotient is a measure of the reactants and products of a reaction.
Measure of if a reaction is at equilibrium * If 𝑄𝑐 = 𝐾𝑐, then the reaction is at equilibrium * Qc is defined as:
If 𝑎𝐴 + 𝑏𝐵 → 𝑐𝐶 + 𝑑𝐷
[𝐶]𝑐[𝐷]𝑑
Then 𝑄𝑐 = ([C]^c [D]^d)/([A]^a [B]^b )

Question 24

Heterogeneous Equilibria

Answer 24

If a system is heterogeneous, the concentration of a pure solid or liquid is 1

Question 25

Meaning of Kc

Answer 25

Kc indicates the extent of the reaction while the equilibrium yield describes the amount of product present at equilibrium

Question 26

Effect of Temperature on Kc

Answer 26

Temperature is the only factor that has a permanent effect on the Kc value.

Question 27

Brønsted-Lowry Model of Acids and Bases

Answer 27

Stipulates that an acid is a proton (H+) donor while a base is a proton acceptor.
Protons in an aqueous solution attach themselves to water to form hydronium (H3O+) but H+ is generally used for simplicity

Question 28

Conjugate Acid-Base Pairs

Answer 28

When a Bronsted-Lowry acid reacts by losing a H+ ion, the species it forms has the potential to act as a base

Question 29

When an acid donates its proton

Answer 29

a conjugate base is formed

Question 30

When a base accepts a proton

Answer 30

a conjugate acid if formed

Question 31

Amphiprotic Substances

Answer 31

able to act as an acid or base/able to accept and donate protons

Question 32

Monoprotic, Diprotic and Polyprotic Acids

Answer 32

• Acids which can donate one proton are monoprotic, those who can donate two are diprotic, those which can donate three are triprotic etc.
• The ionisation of polyprotic acids is stepwise as only one proton can be donated at a time
• The extent of dissociation for the first step is always greater than for each subsequent step
• Each acid in the series is progressively weaker

Question 33

Strong Acids

Answer 33

• Acids which completely ionise in water
• Acids which readily donate a proton
• Their ionisation is denoted by a singular arrow
• Strong acids are good conductors of electricity – free ions
• The stronger the acid, the weaker the conjugate base
• Strong acids are good conductors of electricity – free ions
• The stronger the acid, the weaker the conjugate base

Question 34

HCl as a strong acid

Answer 34

• Thus, HCl is a stronger acid (proton donator) then H3O+ and H2O is a stronger base (proton acceptor) then Cl- so the reaction is favoured in the forward direction

Question 35

weak acids

Answer 35

• Most acids (organic acids) are weak and only partially ionise
• Denoted by a double arrow
• The weaker the acid, the stronger the conjugate base

Question 36

strong bases

Answer 36

• Strong bases completely dissociate in water to produce hydroxide ions
• Denoted by a single arrow
• All group 1 and 2 metal hydroxides are strong bases
• Group 2 metal hydroxides have limited solubility
• The stronger the base, the weaker the conjugate acid

Question 37

Weak bases

Answer 37

• Only partially dissolves in water
• Denoted by a double arrow
• NH4+ is a stronger acid then H2O and OH- is a stronger base then NH3. Thus, the forward reaction only occurs to a small extent
• The weaker the base, the stronger the conjugate acid

Question 38

concentration

Answer 38

• Concentration refers to the relative amount of solute in a given volume of solution

Question 39

strength

Answer 39

• Strength refers to the degree of dissociation of an acid or base

Question 40

acidity of water

Answer 40

• Water undergoes self-ionisation – weak electrolyte (an electrolyte is a substance that creates an electrically conductive solution when dissolved in water)
• It dissociates equally and is thus neutral

Question 41

Water explanation

Answer 41

• 2𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂(+𝑎𝑞) + 𝑂𝐻(−𝑎𝑞)
• Thus, 𝐾𝑤 = [𝐻+][𝑂𝐻−]
• At 25oC, 𝐾𝑤 = 1 × 10−14 mol2dm-6 where
  [𝐻+] 𝑎𝑛𝑑 [𝑂𝐻−] = 1 × 10−7
• In acidic solutions:
  [𝐻+] > 1 × 10−7𝑚𝑜𝑙𝐿−1 [𝑂𝐻−] < 1 × 10−7𝑚𝑜𝑙𝐿−1
• In basic solutions:
  [𝐻+] < 1 × 10−7𝑚𝑜𝑙𝐿−1
  [𝑂𝐻−] > 1 × 10−7𝑚𝑜𝑙𝐿−1
• In neutral solutions:
  [𝐻+] = [𝑂𝐻−] = 1 × 10−7𝑚𝑜𝑙𝐿−1 * At higher temperatures, the value of Kw decreases
• Thus in water:
• 𝑝𝐾𝑤 = − log(𝑝𝐾𝑤) = 14
• 𝑝𝐾𝑤 = − log(𝑝𝐾𝑤) = (− log([𝐻+]) + (− log([𝑂𝐻−]) = 𝑝𝐻 + 𝑝𝑂𝐻
• In general
• 𝑝𝐻 + 𝑝𝑂𝐻 = 14 where, 𝑝𝐻 = −log ([𝐻3𝑂+]) and 𝑝𝑂𝐻 = −log ([𝑂𝐻−])
• Where an acid has a 𝑝𝐻 < 7 and a base has a 𝑝𝐻 > 7

Question 42

Buffers

Answer 42

• A buffer solution resists changes in pH when small amounts of acid or base is added
• A buffer can be acidic or basic and is useful in both natural and industrial applications such as in blood or in waterways.

Question 43

Acidic buffer

Answer 43

• An acid buffer solution can be made by either:
  o Adding a weak acid added to half the amount of base
  o Adding equal amounts of weak acid and its conjugate base anion (salt containing the acids conjugate base)

Question 44

A general equation for an acidic buffer in an equilibrium system is:

Answer 44

𝐻𝐴(𝑎𝑞) ⇌ H(+𝑎𝑞) + A−(𝑎𝑞)

Question 45

Thus, for an ethanoic acid buffer, the equation is:

Answer 45

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) ⇌ 𝐻3𝑂(+𝑎𝑞) + 𝐶𝐻3𝐶𝑂𝑂(−𝑎𝑞)

Question 46

Operation of Acidic Buffers

Answer 46

• The effect of a buffer can be predicted using Le Châteliers Principle

Question 47

(acidic buffer) When a small amount of acid is added to the system….

Answer 47

the equilibrium shifts left as the added H+ ions reacts with the abundant ethanoate ions to from more undissociated acid
* Therefore, the concentration of H+ ions stays relatively consistent so pH doesn’t change much

Question 48

(acidic buffer) Similarly, if base is added…

Answer 48

more weak acid dissociates to replace the H+ ions neutralised by the added base
* Thus, concentration of H+ ions stays relatively constant

Question 49

Basic Buffer

Answer 49

• Similar to acidic buffers
• Made by adding half a strong acid with a weak base or by mixing a weak base with its conjugate acid
• E.g. an ammonia buffer is:
  𝑁𝐻3(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝑁𝐻4+(𝑎𝑞) + 𝑂𝐻(−𝑎𝑞)

Question 50

Operation of Basic Buffer

Answer 50

• If acid is added, the H+ ions reacts with OH- ions to form water and equilibrium shifts right to form more OH- ions so pH stays relatively the same

Question 51

(basic buffer) If a base is added

Answer 51

the extra OH- ions react with NH4+ to form more reactants so the equilibrium shifts to the left – pH stays relatively the same

Question 52

Buffer note

Answer 52

Any buffer can be exhausted if too much acid or base is added and there is not enough of the weak acid or its conjugate base left to react with the additional ions

Question 53

Ka

Answer 53

• Ka refers to the dissociation constant of an acid
• Represents the extent of which the acid dissociates in a given temperature
• The lower the Ka, the lesser the dissociation
• Weak acids typically undergo less than 1% dissociation (𝐾𝑎 = 1 × 10−2 or lower)
• Strong acids like HCl effectively undergo 100% dissociation (𝐾𝑎 = 1 × 103 or higher)

Question 54

Ka for Polyprotic Acids

Answer 54

• After every step, the Ka value decreases and it becomes harder at each stage to lose a proton as the charge on the acid species becomes negative
• E.g. Stepwise dissociation of 𝐻3𝑃𝑂4
  Step 1: 𝐻3𝑃𝑂4(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂(+𝑎𝑞) + 𝐻2𝑃𝑂4−(𝑎𝑞) 𝐾𝑎 = 7.1 × 10−3
  Step 2: 𝐻2𝑃𝑂4−(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂(+𝑎𝑞) + 𝐻𝑃𝑂42(−𝑎𝑞) 𝐾𝑎 = 6.3 × 10−8
  Step 3: 𝐻𝑃𝑂42(−𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐻3𝑂(+𝑎𝑞) + 𝐻𝑃𝑂43(−𝑎𝑞) 𝐾𝑎 = 4.2 × 10−13

Question 55

Kb for Bases

Answer 55

• Strength of a base is indicated by Kb
• Represents the extent of which the base dissociates in a given temperature

Question 56

Indicators

Answer 56

• Indicators have a different colours in acidic and basic solutions and are used to distinguish between acids and bases
• Used to determine the pH of a solution

Question 57

pH Range of an Indicator

Answer 57

• pH range is related to the dissociation constant of the acid-base indicator
• In solution, the acid form of the indicator is in equilibrium with its conjugate base:
  𝐻𝐼𝑛(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐼𝑛(−𝑎𝑞) + 𝐻3𝑂(+𝑎𝑞) Where the dissociation constant for the indicator is:
  𝐾𝑎 = [H+][In-]/[HIn]
• The indicator changes colour when the concentration of the acid, HIn, is equal to its conjugate base,
  In-. Thus, colour changes when [𝐻+] = 𝐾𝑎 when 𝑝𝐾𝑎 = 𝑝𝐻

Question 58

Universal Indicator

Answer 58

• Mixture of a bunch of indicators
• Changes through a range of colours
• Indicator megazord

Question 59

Bromothymol Blue

Answer 59

• Yellow in acidic solutions and blue in basic solutions
• 𝐻𝐵𝐵(𝑎𝑞) + 𝐻2𝑂(𝑙) ⇌ 𝐵𝐵(−𝑎𝑞) + 𝐻3𝑂(+𝑎𝑞)
• Transition point of bromothymol blue is at pH 7

Question 60

Methyl Orange

Answer 60

• Weak acid – red in acidic solutions and yellow in basic ones
• Changes in colour between pH 3.4 and 4.4

Question 61

Phenolphthalein

Answer 61

• Synthetic indicator
• Changes colour from clear to magenta at pH
  8.3 to 10.0

Question 62

Indicator Range

Answer 62

The range of pH values over which an indicator changes colour

Question 63

Equivalence Point

Answer 63

The point in a neutralisation reaction where the stoichiometric ratio between the acid and base is in balance
Indicators are used to visualise this point though they are not the most accurate

Question 64

End Point

Answer 64

• The point during the titration when the indicator changes colour
• Often not the same as the equivalence point
• The indicator used must have an end point around the equivalence point of the reaction

Question 65

Determining Transition Point and pH Range of an Indicator

Answer 65

• The transition point is when the concentration of the acid and basic forms of the indicator are equal
• [𝐻+] = 𝐾𝑎
• Thus, pH at the transition point is:
• 𝑝𝐻 = 𝑝𝐾𝑎 = −log ([𝐻+])

Question 66

Volumetric analysis

Answer 66

a technique to determine the concentration of an acid or a base

Question 67

Standard solution

Answer 67

a solution with an accurately known concentration. Prepared from a primary standard

Question 68

Primary standard

Answer 68

a pure substance that can have the number of moles calculated by its mass

Question 69

Titration curves

Answer 69

• shows a change in pH value versus volume when an acid is titrated against a base or visa versa

Question 70

Half Equivalence Point

Answer 70

• The point at which 𝑝𝐻 = 𝑝𝐾𝑎 where the analyte is a weak acid
• The point at which 𝑝𝑂𝐻 = 𝑝𝐾𝑏 where the analyte is a weak base
• At half the volume needed to reach equivalence point
• No half equivalence point for a strong acid-strong base titration as it fully dissociates

Question 71

titration calculations

Answer 71

• When given a list of titre volumes, the average titre is used
• Average titre value is three concordant titres (titres within 0.10 mL from highest to lowest of each other or 0.05mL within each

Question 72

Uncertainty and Errors

Answer 72

• Accuracy is how close a measurement is to the true value
• Precision is how close a set of measurements are to each other
• A systematic error affects all measurements and is not eliminated by multiple trials
• A random error has equal change of being greater or lower than the true value