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Chemistry > Module 5: Physical Chemistry and Transition Elements > Flashcards

Module 5: Physical Chemistry and Transition Elements Flashcards

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1
Q

Give the two equations for rate of reaction, including units

A

rate = (change in concentration) / time

rate = (quantity reacted / produced) / time

Rate is measured in moldm^-3s^-1

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2
Q

What does [A] mean?

A

The concentration of A

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3
Q

What is the relationship between rate and the concentration of a reactant? How does it link to reaction orders?

A

Rate is proportional to the concentration of a reactant raised to a power.

rate ∝ [A]^n

For each reactant, the power is its order of reaction

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4
Q

What are the three orders of reaction?

A
  • Zero order
  • First order
  • Second order
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5
Q

What does it mean for a reactant to have zero order?

A

Concentration of the reactant has no effect on rate.

rate ∝ [A]^0

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6
Q

What does it mean for a reactant to have first order?

A

Rate is proportional to the change in concentration of the reactant. As [A] doubles, rate doubles.

rate ∝ [A] (^1)

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7
Q

What does it mean for a reactant to have second order?

A

Rate is proportional to the change in concentration squared. As [A] doubles, rate increases by a factor of 4

rate ∝ [A]^2

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8
Q

What does the rate equation tell you?

Give the rate equation for reactants A and B

A

Gives the mathematical relationship between concentrations of reactants and rate.

rate = k [A]^m [B]^n

k = rate constant
m = order of reaction for A
n = order of reaction for B

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9
Q

What does the overall order of reaction tell you and how is it found?

A

Overall order gives the overall effect of the concentrations of all reactants on the rate.

Found by adding all the orders of reactants together (powers in the rate equation)

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10
Q

How would you find the units for rate constant from a rate equation?

A

Can be found by substituting the units for all species in the rate equation in and simplifying

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11
Q

What is important to consider when determining orders of reaction from experimental results?

A

It is important to use the rate of reaction from the same point in each reaction, ideally initial rate.

You would then determine orders by the effect of changing concentrations on rate

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12
Q

What kind of experiment would you need to produce a concentration-time graph?

A

Must gain values via continuous monitoring, e.g colour change using a colorimeter (measuring absorbance)

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13
Q

How would you find the rate of reaction from a concentration-time graph?

What orders can be deduced from a concentration-time graph?

A

The gradient is the rate of reaction.

Orders 0 and 1 can be deduced from the graph if all other concentrations remain fairly constant

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14
Q

What does the concentration-time graph for a zero order reactant look like?

A

A straight line with a negative gradient.

Reaction rate doesn’t change, so gradient is equal to the rate constant k

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15
Q

What does the concentration-time graph for a first order reactant look like?

A

A curve with decreasing (negative) gradient over time as the reaction slows. Shows a pattern called exponential decay.

Half life (time for concentration to halve) is constant.

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16
Q

How would you find the rate constant from a first order concentration-time graph?

A

Draw a tangent at a particular concentration and calculate its gradient. Rate (gradient) and the concentration is subbed into the rate equation to find k.

OR

k = ln2 / half life

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17
Q

What does the concentration-time graph for a second order reactant look like?

A

Also a downward curve like first order graphs, but steeper at the start and tailing off more slowly.

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18
Q

Describe the rate-concentration graph for a zero order reactant

A

A horizontal straight line.

Rate doesn’t change with concentration.

rate = k[A]^0 so rate = k, given by the y intercept of the graph

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19
Q

Describe the rate-concentration graph of a first order reactant

A

A straight line with positive gradient through the origin.

rate = k[A], so rate is directly proportional to concentration.

rate constant k = gradient

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20
Q

Describe the rate-concentration graph of a second order reactant?

How would you find the rate constant?

A

Upward curve starting at the origin with increasing gradient.

rate = k[A]^2 so k can’t be directly obtained.

Plotting a graph of rate against concentration squared gives a straight line through the origin, where gradient = k

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21
Q

What is initial rate?

How is it found on a concentration-time graph?

A

Initial rate is the instantaneous rate at the start of a reaction when t = 0.

Found by measuring the gradient of a tangent drawn at t = 0 on the concentration-time graph

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22
Q

What type of reaction gives an easier way to find initial rate and how ?

A

Clock reactions - measures the time taken to observe a visual change.

Can be assumed average rate of reaction = initial rate provided no significant rate changes.

Means initial rate is proportional to 1/t

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23
Q

How is the assumption that average rate = initial rate for a clock reaction made more accurate?

A

The shorter the time taken for a clock reaction, the more accurate the measured average rate is to initial rate.

It is an approximation but fairly accurate provided less than 15% of the reaction has occurred

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24
Q

Give a common type of clock reaction

A

An iodine clock - as iodine forms, the solution turns brown.

Starch is usually added so the solution turns black instead

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25
Q

Why do reactions often occur in multiple steps?

A

Reactions can only occur when particles collide with the sufficient activation energy, so for reactions with more than two reactants, each reactant would have to collide with each other at the same time with enough energy to complete in one step.

This is very unlikely

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26
Q

What is a reaction mechanism?

A

The series of steps that make up an overall reaction

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27
Q

What is the rate-determining step?

A

The slowest step in a multi-step reaction, as each stage occurs at a different rate.

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28
Q

Describe how you would start to predict a reaction mechanism from the rate equation

A
  • Rate equation only contains species involved in the rate-determining step.
  • Orders in the rate equation match the moles of species involved in the rate determining step
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29
Q

How does rate constant k change with temperature? Why?

A

As temperature increases, k increases (because rate increases).
For many reactions, an increases of 10 degrees C doubles rate and k.

This mostly happens because the number of molecules exceeding the activation energy increases (from the shift in Boltzmann distribution), not just particles moving faster.

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30
Q

What is the Arrhenius equation?

A

k = Ae^(-Ea/RT)

k = rate constant
Ea = activation energy
R = gas constant = 8.314
T = temperature in kelvin

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31
Q

What is the exponential factor in the Arrhenius equation?

A

e^(-Ea/RT)

Represents the proportion of molecules exceeding the activation energy

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32
Q

What is the pre-exponential factor of the Arrhenius equation?

A

A

Takes into account the frequency of collisions with the correct orientation. Is a constant because the frequency of collisions increases very very little with temperature

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33
Q

How would you determine Ea and A graphically using the Arrhenius equation?

A

Must use the logarithmic form of the Arrhenius equation.

lnK = -(Ea/RT) + lnA

So, a plot of lnK against 1/T gives a downward straight line where the gradient is (-Ea/R) and lnA is the y intercept

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34
Q

Give the Kc expression for the equation:

N2 + 3H2 <–> 2NH3

A

Kc = ([NH3]^2) / ([N2][H2]^3)

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35
Q

What does the value of the equilibrium constant Kc indicate?

A

The larger the Kc value, the further the position of equilibrium towards the products

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36
Q

How would you find the units for Kc?

A

By substituting the concentration units for all species into the Kc expression, then simplifying

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37
Q

What is a homogenous equilibrium?

A

All equilibrium species are in the same state

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38
Q

What is a heterogenous equilibrium?

How would you write the Kc expression?

A

The equilibrium species have different states.

Any solid / liquid species are omitted from Kc expressions as their concentrations are essentially constant

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39
Q

How would you calculate Kc from initial quantities of reactants?

A
  • Set up a RICE table (Ratio, initial moles, change in moles, equilibrium moles).
  • Calculate the change in moles for all species (one equilibrium quantity will be given, then the rest are found from the reaction ratio)
  • Calculate the equilibrium moles for all species using initial and change
  • Divide equilibrium moles by the total volume to find concentrations
  • Substitute the concentrations into the Kc expression to find Kc
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40
Q

What is the relationship between the equilibrium constants Kc and Kp?

A

Kc is in terms of concentration, while Kp is the equilibrium constant in terms of partial pressures (so usually used for equilibria involving gases)

Kp has a direct relationship to Kc as concentration and pressure are proportional

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41
Q

What is mole fraction and how is it written?

A

Mole fraction is a gas’ proportion by volume to the total volume of gases in a mixture. The sum of all mole fractions in a gas mixture must equal 1.

Mole fraction of A is written as x(A)

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42
Q

Give the equation for mole fraction

A

x(A) = (number of moles of A) / (total number of moles in mixture)

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43
Q

What is partial pressure?

How is it written?

A

The contribution that a gas makes to the total pressure.
The sum of all partial pressures equals the total pressure.

The partial pressure of A is written as p(A)

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44
Q

Give the equation for partial pressure

A

partial pressure = mole fraction x total pressure

p(A) = x(A) x P

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45
Q

How would you write the Kp expression for the equation:

H2 + I2 <–> 2HI

A

Kp = p(HI)^2 / (p(H2) x p(I2))

Kp expressions are written the same as Kc expressions but using partial pressures.

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46
Q

How would you find the units for Kp?

A

Suitable units for partial pressures are kPa, Pa or atm as long as each gas has the same unit

Kp units are found by substituting all partial pressure units into the Kp equation and simplifying, with any non-gas species being ignored

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47
Q

What does the value of the equilibrium constant K mean?

A

The magnitude of K indicates the extent of the equilibrium.

  • K = 1 means an equilibrium halfway between reactants and products.
  • K = 100 means an equilibrium well in favour of the products
  • K = 0.01 means an equilibrium well in favour of the reactants
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48
Q

What can change equilibrium constant K?

A

The only factor that can change K is temperature.

At a set temperature, K does not change despite any change in concentration / pressure / catalyst.

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49
Q

For an exothermic reaction, describe and explain how Kc/Kp changes with increasing temperature

A
  • K decreases with increasing temperature, decreasing equilibrium yield of products.
  • Generally, K = products/reactants
  • If temperature increases, K decreases, so the ratio is greater than K and the system is no longer in equilibrium
  • To correct the ratio, the amount of product must decrease and reactants must increase, so P.O.E shifts towards reactants until a new equilibrium is reached
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50
Q

For an endothermic reaction, describe how Kc/Kp changes with increasing temperature

A
  • K increases with increasing temperature, increasing equilibrium yield of products.
  • Generally, K = products / reactants
  • If temperature increases, Kp/Kc increases, so the ratio is less than K and the system is no longer in equilibrium
  • To correct the ratio, amount of products must increase and reactants must decrease, so P.O.E shifts towards products until a new equilibrium is reached
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51
Q

What happens to the value of Kc/Kp in concentration / pressure changes?

How does this relate to position of equilibrium?

A

K is unaffected by concentration and pressure changes.

Because of this, the equilibrium position shift from le Chatelier’s principle occurs to fix the equilibrium

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52
Q

For the equation:
Kc = [NO2]^2 / [N2O4]

Explain what happens to the equilibrium when [N2O4] increases

A
  • If [N2O4] increases, the ratio is now less than Kc, so the system is no longer in equilibrium
  • To fix the ratio, [NO2] must increase and [N2O4] must decrease until the ratio is restored.
  • Overall to do this, P.O.E shifts to the right to produce more NO2.

The same explanation is used for concentration decreases but in reverse.
Both explanations are the same for partial pressures instead of concentrations.

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53
Q

For the equation:
Kp = p(NO2)^2 / p(N2O4)

Explain what happens to the equilibrium when pressure is increased

A
  • Increasing pressure increases p(NO2) and p(N2O4) by the same amount.
  • As p(NO2) is squared, the top of the expression increases more than the bottom.
  • The ratio is now greater than Kp, so the system is no longer in equilibrium.
  • p(NO2) must decrease and p(N2O4) must increase until the ratio is restored.
  • Overall, P.O.E shifts to the left to produce more N2O4
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54
Q

How does the presence of a catalyst affect equilibrium constants?

A

Catalysts have no effect on k, as they affect the rate of the forward and reverse reactions equally.

Position of equilibrium does not change, equilibrium is just reached more quickly

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55
Q

What is a Bronsted-Lowry acid?

A

A proton donor

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56
Q

What is a Bronsted-Lowry base?

A

A proton acceptor

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57
Q

What are conjugate acid-base pairs?

A

In solution, the dissociation of ions forms conjugate acid-base pairs.

The pair contains 2 species that can be interconverted by the transfer of a proton

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58
Q

In the equation:
HCl + OH- <–> H2O + Cl-

Label the conjugate acid-base pairs

A

HCl: acid 1
OH-: base 2
H2O: acid 2
Cl-: base 1

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59
Q

Explain why dissociation of acid/base ions doesn’t occur without water

A

In aqueous solution, dissociation requires a proton to be transferred to a base.

Water acts as this base, accepting a H+ to form oxonium (H3O+) ions

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60
Q

Give the two ionic equations for neutralisation

A

H+ + OH- –> H2O
H3O+ + OH- –> 2H2O

Both valid as H3O+ and H+ are interchangeable.

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61
Q

Give the three main types of acids

A
  • Monobasic - can donate one H+ per molecule
  • Dibasic - can donate two H+ per molecule
  • Tribasic - can donate three H+ per molecule
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62
Q

What is the pH scale used for and why?

A

A logarithmic scale used to measure H+ concentration.

H+ are found in concentrations of 10^-1 to 10^-14, which is difficult to manage, so pH converts it to a number between 1 and 14

63
Q

What is the equation for pH?

A

pH = -log10([H+])

64
Q

What is the equation for [H+]?

A

[H+] = 10^-pH

65
Q

What does a decrease of one on the pH scale mean?

A

An increase in [H+] by a factor of 10

66
Q

How would you calculate the pH of a strong acid?

A

For strong acids, [HA] = [H+] as the acid completely dissociates.

This means pH = -log10[HA]

However, you must first multiply the [HA] by however many H+ ions the acid can dissociate per molecule.

67
Q

What is the acid dissociation constant Ka?

A

A type of equilibrium constant (works exactly the same as Kc and Kp).

Used for the dissociation of weak acids, as they partially dissociate, forming a reversible reaction.

68
Q

What does the value of Ka indicate?

A

The strength of the acid.

A larger Ka means a stronger acid (as there is a higher concentration of products, as the position of equilibrium is further to the right)

69
Q

Give the generic Ka expression for any weak acid.

A

For a weak acid HA,
HA <–> H+ + A-

So:
Ka = [H+][A-]/[HA]

70
Q

What is pKa and why is it used?

A

Ka values have the same problems as [H+], where there is a large range of values with negative indices, which are difficult to work with.

pKa is used to turn Ka values into a manageable scale.

71
Q

What does the value of pKa indicate?

A

The acid strength.

A smaller pKa value means a stronger acid.

72
Q

Give the equation for pKa

A

pKa = -log10(Ka)

73
Q

Give the equation for Ka, from pKa

A

Ka = 10^-pKa

74
Q

Give and explain the Ka equation for a weak acid at equilibrium

A

Ka = ([H+]eqm x [A-]eqm) / ([HA]start - [H+]eqm)

When HA molecules dissociate, H+ and A- ions are made in equal concentrations.
The equilibrium concentration of acid would therefore be [HA]start - [H+]eqm

75
Q

Give the simplified Ka equation for a weak acid at equilibrium.

What assumptions have been made to simplify it?

A

Ka = ([H+]^2) / [HA]

  • The amount of H+ from water is extremely small, so can be neglected. We therefore assume [H+]eqm = [A-] eqm, so [H+][A-] becomes [H+]^2
  • As dissociation of weak acids is small, we assume [HA]eqm = [HA]start
76
Q

How would you determine Ka of a weak acid experimentally?

A

Prepare a standard solution of weak acid with a known concentration, and measure pH with a pH meter.
Calculate [H+] from pH and substitute values into the simplified Ka equation

77
Q

Give some situations where the weak acid approximations break down

A
  • [H+] = [A-] is not valid for very weak acids (pH > 6) or very dilute solutions, as the dissociation of water is significant compared to the dissociation of acid
  • [HA]eqm = [HA]start is not valid for stronger acids (Ka > 0.001) and very dilute solutions, as [H+] becomes significant and changes [HA]
78
Q

What happens in solutions of pure water? Give the equation and label the conjugate pairs

A

Water ionises slightly, acting as both an acid and a base.

H2O + H2O <–> H3O+ + OH-

H2O = acid 1
H2O = base 2
H3O+ = acid 2
OH- = base 1

79
Q

Give the Ka equation for pure water.

How does this relate to Kw?

A

Ka = [H+][OH-] / [H2O]

[H2O] is a constant as all pure water has the same concentration.

Kw is the ionic product of water, and is another way of saying Ka of water.

Kw = [H+][OH-]

80
Q

How does Kw change with temperature and what is its value at 25 degrees C?

A

Like all equilibrium constants, Kw changes with temperature.

At 25 degrees, Kw = 1x10^-14 mol^2dm^-6

81
Q

Why is Kw important?

A

It sets up the neutral point on the pH scale and controls concentrations of H+ and OH- in aqueous solutions

82
Q

When does a solution become alkaline?

A

When [OH-] > [H+]

83
Q

How would you calculate the pH of a strong base?

A

For strong alkalis, [OH-] = [base] if they are monoacidic.

[H+] = Kw / [OH-]

So, at 25 degrees C,
[H+] = (1x10^-14)/[OH-]

Then substitute [H+] into the pH equation

84
Q

What is a buffer solution?

A

A system that minimises pH changes when small amounts of an acid or base are added.

As they work, the pH does change but only by a small amount

85
Q

What are the components of a buffer solution and what are they used for?

A

Buffers contain a weak acid to remove added alkali and its conjugate base to remove added acid.

When one component has all reacted, the solution therefore loses its buffering ability to either added acid or alkali

86
Q

What are the two ways to prepare a buffer solution?

A
  • From a weak acid and its salt
  • By partial neutralisation of a weak acid
87
Q

Describe how you would prepare a buffer solution from a weak acid and its salt

A

When the acid is added to water, it partially dissociates and the amount of conjugate base ions in the solution is very small.

By adding a salt of the weak acid, its dissociation in water into ions is complete and is the source of the conjugate base in the solution

88
Q

Describe how you would prepare a buffer solution by partial neutralisation of a weak acid

A
  • Add an aqueous alkali solution to an excess of the weak acid.
  • The acid is partially neutralised, leaving some acid left over unreacted.
  • Forms a solution made from unreacted weak acid and its salt
89
Q

Give the generic equation of a buffer solution

A

HA <–> H+ + A-

90
Q

Describe the action of a buffer solution when acid is added

A

For the equation:
HA <–> H+ + A-

  • [H+] increases
  • H+ reacts with the conjugate base A-
  • Equilibrium position shifts to the left, removing most of the added H+ ions
  • [H+] stays relatively constant so so does pH
91
Q

Describe the action of a buffer solution when alkali is added

A

For the equation:
HA <–> H+ + A-

  • [OH-] increases
  • The small concentration of H+ in the solution reacts with the added OH- ions, forming water
  • HA dissociates, shifting the position of equilibrium to the right, restoring most of the removed H+ ions
    -[H+] stays relatively constant so so does pH
92
Q

What would you consider when choosing the weak acid for a buffer solution?

A

A buffer is most effective at removing either added acid or alkali when [HA] = [A-].

  • The pH of the buffer solution is the same as the pKa of HA
  • The operating pH is usually over 2 pH units, centred at the pKa
  • The ratio of [HA] and [A-] can then be adjusted to fine-tune the pH
93
Q

What is the equation for calculating [H+] of a buffer solution, and why is it not just the Ka equation?

A

For buffer solutions, the assumption that [H+] = [A-] is no longer true, as A- is an added component.

So, the Ka equation is rearranged to give:

[H+] = Ka x ([HA]/[A-])

94
Q

How would you calculate the pH of a buffer solution made from a weak acid and its salt?

A
  • Calculate the moles of acid and added base
  • Calculate the concentrations of acid and base using the new total volume
  • Use the Ka equation to calculate [H+]
  • Calculate pH using the log equation
95
Q

How would you calculate the pH of a buffer solution made by partial neutralisation of a weak acid?

A
  • Calculate the moles of conjugate base ions in the solution using the moles of alkali added and the neutralisation equation
  • Calculate the moles of acid left in the solution, using moles of acid added - moles of alkali added
  • Use the Ka equation to find [H+]
  • Calculate pH using the log equation
96
Q

Why are buffer solutions needed in the body?

A

They keep pH relatively constant, as different parts of the body need specific pHs to function, e.g for enzymes

97
Q

Give an example of a buffer solution used in the body

A

Blood plasma must be between pH 7.35 - 7.45.

The carbonic acid - hydrogencarbonate buffer system maintains this

H2CO3 <–> H+ + HCO3-

98
Q

Describe how the carbonic acid - hydrogencarbonate buffer system reacts upon the addition of acid

A
  • [H+] increases and reacts with HCO3-.
  • Equilibrium position shifts to the left, removing most of the added H+ ions, keeping pH relatively constant
99
Q

Describe how the carbonic acid - hydrogencarbonate buffer system reacts upon the addition of alkali

A
  • [OH-] increases and reacts with the small concentration of H+ to form water
  • H2CO3 dissociates, so the position of equilibrium shifts to the right to restore most of the used H+ ions, keeping pH relatively constant`
100
Q

How does the carbonic acid - hydrogencarbonate buffer system lead to CO2 being exhaled by the lungs?

A

The body produces more acidic materials than alkaline, which HCO3- converts to H2CO3.

To prevent H2CO3 build-up, the body converts it to carbon dioxide, which is then exhaled by the lungs

101
Q

Describe and explain the shape of a pH titration curve, when a base is being added to an acid

A
  • When base is first added, the acid is in great excess and pH increases only slightly (first section has a low pH with a low gradient)
  • As the vertical section approaches, pH increases more quickly as acid is used up more quickly
  • Eventually, pH increases rapidly for a small volume of base, making the vertical section
    -After this, the pH rises very slightly as the base is now in excess
102
Q

What is the equivalence point and where is it found on a pH titration curve?

A

The volume of one solution that exactly reacts with the volume of the other solution in the titration.

It is found in the centre of the vertical section of the pH titration curve

103
Q

What is an acid-base indicator?

A

A weak acid that has a distinctively different colour from its conjugate base.

At the end point of a titration, [HA] = [A-], so the colour of the solution will be in between the two colours

104
Q

Describe and explain how methyl orange works as an acid-base indicator

A

Methyl orange is red in acidic conditions and yellow in alkali.

When a strong base is added to a strong acid, methyl orange is initially red (H+ forces equilibrium position to the left towards the acid).

When OH- ions are added, they react with H+, and HA dissociates, moving position of equilibrium to the right, turning the solution less red.

At the end point, methyl orange turns orange, and then yellow as more base is added.

105
Q

What is the link between different indicators and the pH at which they change colour?

A

Different indicators have different Ka values, and so change colour over different pH ranges.

At the end point, [HA] = [A-], so Ka = [H+] and pKa = pH

106
Q

How would you choose the indicator for a pH titration?

A

To choose the indicator for a titration, the colour change must coincide with the vertical section of the pH titration curve.

No indicator is suitable for a weak acid - weak base titration, as there is no vertical section

107
Q

What is lattice enthalpy?

Is it exothermic or endothermic?

A

The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions.

A measure of the strength of ionic bonds in a giant ionic lattice.

Is an exothermic change

108
Q

What are Born-Haber cycles made from?

A
  • The formation of gaseous atoms
  • The formation of gaseous ions
  • The formation of the compound, both from gaseous ions (lattice enthalpy) or its elements in their standard states (enthalpy of formation)
109
Q

What is a Born-Haber cycle most commonly used for?

A

To calculate lattice enthalpy using other known energy changes, as lattice enthalpy can’t be measured directly.

Can also be used to find other missing values

110
Q

What is the enthalpy change of atomisation?

Is it exothermic or endothermic?

A

The enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state under standard conditions.

It is endothermic as energy is used to break bonds

111
Q

What is the first electron affinity?

Is it exothermic or endothermic?

A

The enthalpy change when one electron is added to each atom in 1 mole of gaseous atoms to form 1 mole of 1- ions.

Always exothermic because the electron being added is attracted to the nucleus

112
Q

How do successive electron affinities work?

Are they endothermic or exothermic?

A

They are defined in the same way as first electron affinity, just with a 1- ion turning into 2-, etc.

Are endothermic because the second (etc) electron is being gained by a negative ion, so energy is needed to overcome the repulsion

113
Q

How would you use a Born-Haber cycle to calculate a missing value?

A
  • Draw a line from the tail of the arrow representing the missing enthalpy to its tip.
  • For each enthalpy value, if both arrows are pointing the same way, keep the sign, else flip it.
  • Then add all the values together (taking into account any steps where the enthalpy is being applied multiple times)
114
Q

What is the enthalpy change of solution?

Give an equation representing it

A

The enthalpy change when 1 mole of solute dissolves in a solvent.

Na+Cl-(s) + aq –> Na+(aq) + Cl-(aq)

115
Q

How does enthalpy of solution work?

Is it exothermic or endothermic?

A

The oppositely charged ions attract each other in the giant ionic lattice, but get surrounded by water and separated when dissolved. The partially negative oxygen atom in water is attracted to the cation, and the partially positive hydrogen atom in water is attracted to the anion.

Can be exothermic or endothermic

116
Q

How would you determine enthalpy of solution from experimental results?

A

Use the equations:
- q = mc(change in T)
- enthalpy = q/n

Where m is the mass that is changing temperature, so the whole solution, not just the water

117
Q

What enthalpies are involved when a solid ionic compound dissolves?

A
  • The whole process is the enthalpy of solution
    -The ionic lattice is broken up to form separate gaseous ions (opposite of lattice enthalpy)
  • The gaseous ions interact with water molecules to form hydrated aqueous ions (enthalpy of hydration)

You can use these to construct an energy cycle like a Born-Haber cycle to find missing values

118
Q

What is the enthalpy change of hydration?

Give an equation representing it

A

The enthalpy change when gaseous ions dissolve in water to form 1 mole of aqueous ions.

Na+(g) + aq –> Na+(aq)

119
Q

What are some factors affecting lattice enthalpy?

A
  • As ionic radius increases, ionic attraction decreases. Lattice enthalpy becomes less negative, so melting point decreases.
  • As ionic charge increases, ionic attraction increases. Lattice enthalpy becomes more negative, so melting point increases.

Lattice enthalpy gives a good indication of the melting point of an ionic compound

120
Q

What are some factors affecting enthalpy of hydration?

A
  • As ionic radius increases, attraction between the ion and water molecule decreases. Enthalpy of hydration becomes less negative
  • As ionic charge increases, attraction between the ion and water molecule increases. Enthalpy of hydration becomes more negative
121
Q

How would you predict the solubility of an ionic compound?

A

To dissolve an ionic compound, energy equal to lattice enthalpy is taken in to overcome ionic attraction, and energy equal to enthalpy of hydration is released to surround ions with water.

If the enthalpies of hydration are greater than the lattice enthalpy, the overall energy change is exothermic and the compound is likely to dissolve. However, not always.

122
Q

What is entropy?

A

Describes the dispersal of energy within the chemicals in the system. Measured in JK^-1mol^-1.

Greater entropy means more disorder, and energy is more spread out.

123
Q

Describe the entropies of the three states

Are they positive or negative?

A

There is a natural tendency for energy to disperse, so:

Gases have the highest entropy, then liquids, then solids.

At 0K, substances have an entropy value of 0. Every other entropy value is positive.

124
Q

How would you predict the entropy change of a reaction?

A

If a system changes to become more random, change in entropy (S) is positive.

A reaction with more moles of gaseous products than reactants will have positive entropy change, and vice versa.

The same is true from a solid to liquid/aqueous

125
Q

How would you calculate the entropy change of a reaction?

A

Entropy change = (entropy of products) - (entropy of reactants)

126
Q

Define feasibility

A

Whether a reaction is able to occur, and is energetically feasible (spontaneous)

127
Q

What is free energy?

A

The overall energy change during a reaction.

Made from enthalpy change (heat transfer between chemical system and surroundings) and the dispersal of energy within the chemical system at that temperature.

128
Q

What is the Gibbs’ equation?

A

Free energy = enthalpy change - T x entropy change

  • Free energy is measured in kJmol-1
  • Enthalpy change is measured in kJmol-1
  • Temperature is measured in K
  • Entropy change is measured in Jmol-1, so must be converted to kJ first
129
Q

What must the free energy be for a reaction to be feasible?

A

For a reaction to be feasible, there must be a decrease in free energy, so change in free energy must be negative.

A reaction first becomes feasible at free energy change = 0

130
Q

What is the most important part of the Gibbs’ equation and why?

A

Enthalpy change, as it is usually much larger than entropy change (as enthalpy is measured in kJ but entropy in J)

131
Q

Why is the Gibbs’ equation not a perfect indicator of feasibility?

A

It doesn’t take into account activation energy, so some reactions with a negative free energy change don’t occur spontaneously.

It also doesn’t take into account kinetics or rate of reaction

132
Q

What is an oxidising agent?

What is a reducing agent?

A

Oxidising agents accept electrons from the species being oxidised

Reducing agents donate electrons to the species being reduced

133
Q

Describe how you would balance a redox equation using oxidation numbers

A

There must be no overall change in oxidation number, so first balance out the increases / decreases.

Then, balance the other atoms without changing the redox species

134
Q

Give two examples of redox titrations

A
  • Manganate (VII) (KMnO4) is used to analyse reducing agents, e.g Fe2+, ethanedioic acid
  • Iodine-thiosulfate (2S2O3 2- + I2 –> 2I- + S4O6 2-) uses a starch indicator. At the end point, the blue-black colour disappears as iodine is used up. Used to analyse oxidising agents, e.g ClO- and Cu2+
135
Q

What is an electrochemical cell? What is its other name?

A

Voltaic cells convert chemical energy into electrical energy using redox reactions

Made from two half cells connected together, allowing electrons to flow between them

136
Q

Describe a metal/metal ion half cell

A

Metal rod is dipped into the solution of its aqueous ion.

Represented using a vertical line for the phase boundary between them, e.g Zn2+(aq) | Zn(s)

At the phase boundary (where metal is in contact with ions), an equilibrium is set up. The forward reaction shows reduction

137
Q

Describe an ion/ion half cell

A

Contains ions of the same element in different oxidation states, e.g Fe3+ (aq) + e- <=> Fe2+(aq)

There is no solid metal to transport electrons in/out the half cell, so an inert platinum electrode is used

138
Q

In an operating redox cell, which is the positive and negative electrode?

A

The electrode with the more reactive metal loses electrons and is oxidised - this is the negative electrode.

The less reactive metal gains electrons and is reduced - this is the positive electrode

139
Q

What is a standard electrode potential and what is it used for?

A

The e.m.f (voltage) of a half-cell connected to a standard hydrogen half cell under standard conditions (298K, 100kPa, all solutions at 1moldm^-3)

Measures the tendency for the half cell to be reduced and gain electrons

140
Q

Describe the standard hydrogen electrode

A

Has an E of 0V, uses an inert platinum electrode to allow electrons into or out of the half cell. Uses a glass tube to add hydrogen gas to the solution of H+ ions.

141
Q

How do you measure a standard electrode potential?

A

Half cell is connected to a standard hydrogen electrode using a wire and voltmeter. Solutions are connected with a salt bridge (an electrolyte that doesn’t react with either solution) to complete the circuit.

E is the reading off the voltmeter

142
Q

What does the sign of E (electrode potential) mean?

A

The more negative the E, the greater the tendency to lose electrons and be oxidised. Metals tend to have negative E values and lose electrons

143
Q

How do you calculate E(cell)?

A

E(cell) = E(+ electrode) - E(- electrode)

144
Q

How do you combine two half cell equations?

A

The electrons are balanced, and the system with the more negative E is flipped (since it is oxidised).

Overall, oxidising agents are reduced, are so are on the left, as the forward reaction must be reduction

145
Q

How would you predict which oxidising/reducing agents react with which?

A

Oxidising agents only react with systems with a more negative E, and reducing agents only react with system s with a more positive E

146
Q

What are some limitations of using E values to make predictions?

A
  • Electrode potentials give the thermodynamic feasibility of a reaction, but no indication of the reaction rate
  • Standard electrode potentials are measured at 1moldm^-3. At other concentrations, E is different
  • Conditions may not be standard, affecting E values
147
Q

For the half equation:

Zn2+(aq) + 2e- <=> Zn(s) E = -0.76V

What happens if the concentration of Zn2+ increases?

A

Position of equilibrium shifts to the right, removing electrons from the system and making the electrode potential less negative, and vice versa

148
Q

What are the different types of storage cells?

A
  • Primary cells
  • Secondary cells
  • Fuel cells
149
Q

Describe primary storage cells

A

Non-rechargeable and designed to be used once. Electrical energy is produced at the electrodes, but when the chemicals are used up, a voltage stops being produced

150
Q

Describe secondary storage cells

A

Rechargeable. Chemicals can be regenerated by reversing the reaction, so the cell can be used again, e.g lithium-ion batteries

151
Q

Describe how fuel cells work

A

Use the energy from the reaction of a fuel with oxygen to create a voltage. The fuel and oxygen flow into the cell and the products flow out, but the electrolyte remains. Can operate continuously provided reactants are supplied, so don’t need recharging

152
Q

Give the different redox equations involved in an alkaline hydrogen/oxygen fuel cell

A

Oxidation:
H2 + 2OH- –> 2H2O + 2e-

Reduction:
1/2O2 + H2O + 2e- –> 2OH-

Overall:
H2 + 1/2O2 –> H2O

153
Q

Give the different redox equations involved in an acidic hydrogen/oxygen fuel cell

A

Oxidation:
H2 –> 2H+ + 2e-

Reduction:
1/2O2 + 2H+ + 2e- –> H2O

Overall:
H2 + 1/2O2 –> H2O

Chemistry (4 decks)

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