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Chemistry - all papers > 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)

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

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

37
Q

What is a homogenous equilibrium?

A

All equilibrium species are in the same state

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

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
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

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)

42
Q

Give the equation for mole fraction

A

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

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)

44
Q

Give the equation for partial pressure

A

partial pressure = mole fraction x total pressure

p(A) = x(A) x P

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.

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

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
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.

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
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
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

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.

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
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

55
Q

What is a Bronsted-Lowry acid?

A

A proton donor

56
Q

What is a Bronsted-Lowry base?

A

A proton acceptor

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

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

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

60
Q

Give the two ionic equations for neutralisation

A

H+ + OH- –> H2O
H3O+ + OH- –> 2H2O

Both valid as H3O+ and H+ are interchangeable.

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
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

Chemistry - all papers (2 decks)

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