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A Level Chemistry > Physical Chemistry And Transition Elements > Flashcards

Physical Chemistry And Transition Elements Flashcards

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1
Q

Units for rate of reaction?
+ Calc for rate

A

Mol dm^-3 s^-1

Change in reaction/ time taken = rate

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2
Q

How do you measure rate of reaction?
How can you “stop” reaction chemically/ physically to measure for RofR

A

Measured by measuring some property of a reactant/ product at various times during reaction
E.g
CHEMICALLY
-titrating samples
- Quenching reactions by adding another chemical to remove one reactant
-Cooling sample
- diluting it massively

PHYSICALLY (measured by)
- change in volume/ pressure of gas
- changes in light absorbed/transmitted
- change in thermal conductivity
- change in mass

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3
Q

State the rate equation?

A

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

m = order of reaction with respect to A
n = order of reaction with respect to *
K= rate constant - separate for every reaction

*the oder is the power to which the concentration of a reactant is raised in the rate equation

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4
Q

If a 1st order reaction
1) write rate equation
2) units for K

A

1st order
Rate = K[A]
K is in S^-1
(See notes for more info on why)

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5
Q

If a 2nd order reaction
1) write rate equation
2) units for K

A

2nd order
Rate= K[A]^2
K is in Mol^-1 dm^3 s^-1
(See notes for more info on why)

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6
Q

To form concentration time graph what do you measure
(Continuous monitoring)?

A

If gas produced
- monitor by collecting gas produced in gas syringe
- monitor the mass loss

If colour change
- monitor using colorimeter

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7
Q

Sketch how a concentration (y) time (x) graph would look for
1) zero order reaction
2) first order reaction
3) second order reaction

A

See note for graphs
1) zero order
Produces a straight line with a negative gradient

2) first order
Downward curve with a decreasing gradient over time but with a constant half life

3) second order
Steep downward curve, steeper at the start but trailing off more slowly with not constant half life

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8
Q

Sketch how a rate (y) concentration (x) graph would look for
1) zero order reaction
2) first order reaction
3) second order reaction

A

See note for graphs
1) zero order
Rate remains constant- straight horizontal line

2) first order
Consistent linear increase

3) second order
Upwards curve

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9
Q

Def of half life?

A

Half-life (t1/2)
- the time taken for half of a reactant to be used up.

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10
Q

In a first order reaction what is the trend of half lives?
+ what is this called?

A

First order reactions have a constant half-life with the concentration halving every half life

+this pattern is called exponential decay.

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11
Q

How can you calculate the rate constant from half-life?

A

K = In 2
———
t 1/2

In 2 (natural log2 found on calculator on right of log

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12
Q

How do you calculate K from a rate (y) concentration (x) graph?

A

The rate constant is equal to the gradient of the graph

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13
Q

Things to consider when suggesting a mechanism?

A

1) the rate-determining step of reaction must fit with the rate equation
2) all of the equation in the mechanism must balance
3) the different steps of the reaction must add up to he overall equation

(Often several right answers for these questions)

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14
Q

State Arrhenius equation
(Given in exam)

A

K = Ae ^-Ea/RT

K= rate constant
Ea = activation energy (J mol-1)
T= temperature (K)
R= gas constant (8.314 J K-1 mol -1)
A= pre-exponential factor (same units as K)
*essentially a measure of the number of collisions)

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15
Q

Determination of Ea and A graphically from Arrhenius equation

A

Y = m x + c

LnK = -Ea. 1.
——. — +. LnA
R. T

If lnK is plotted against 1/t gradient is -Ea/R so we can calculate the activation energy

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16
Q

General equilibrium constant equation

A

Kc = Right
——
Left

aW (aq) + bX (aq) (reverse reaction arrow) cY(aq) = dZ(aq)

Kc = [Y]^c [Z] ^d
——————
[W]^a [X]^b

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17
Q

Position of equilibrium if more products than reactants + Kc size

A

When product concentration is high and reactant conc is low
Equilibrium will lie towards the left hand side
Kc will be small

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18
Q

Position of equilibrium if more reactants than products + Kc size

A

When reactant concentration is high and product conc is low
Equilibrium will lie towards the right hand side
Kc will be large

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19
Q

Why does equilibrium shift if change occurs
(Explain Le Chatelier’s using Kc)

A

Kc remains constant for a given temperature
This is why the reaction will oppose the change

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20
Q

How does Kc value change in different temps 1) for endothermic reaction
2) for exothermic reaction

A

For endothermic reaction
- Kc increases at higher temperatures
For exothermic reaction
-Kc decreases at higher temperatures

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21
Q

Change of equilibrium equation with heterogenous reaction

A

When different sates are involved then equilibrium = heterogenous
But products in liquid or solid states are omitted (as they concentrations are effectively unchanged)

This means only gas or aqueous reactants are included

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22
Q

Annotate an equilibrium graph

A

See notes

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23
Q

How do you work out equilibria of reaction

A

See notes

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24
Q

Describe the effect of condition changes on equilibrium
General
1) temp change
2) change pressure

A

1) temp change
- cooling temperature will favour the reaction that produces … to counteract the change because the …. Reaction is exothermic therefore the equilibrium shifted from left to right to reastablish the Kc value leading to more … being produce
(Opposite for higher temps but it depends on the reaction)

2) pressure change
- higher pressure will decrease production of … as to oppose the increase and re-establish the Kc value the reaction will shift equilibrium from the right to left favour reaction that produces less molecules.

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25
Q

Observation of ammonium chloride heating
(Reversible reaction)

A

White solid app reads to sublime reforming as a white gas on the cooler area of the tube towards to top

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26
Q

Observations of NaOH hen H2SO4
(Reversible reaction)

A

Brown solution turns colourless on addition of NaOH
Brown solution reappears on addition of H2SO4

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27
Q

Observations of addition of c. HCl then H20
(Reversible reaction)

A

Pink solution turns blue on addition of HCl
Pink solution reappraisal on dilution with distilled water

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28
Q

Obervations on addition of H2SO4 then NaOH
(Reversible reaction)

A

Yellow solution turns orange on addition of H2SO4
Yellow solution reappears on addition of NaOH

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29
Q

When do you use Kc or Kp

A

Kc - reversible reactions in solid/ liquid

Kp - for gaseous reactions based on the volume of its container

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30
Q

Def of partial pressure of gas

A

The pressure that a gas would exert if it alone was present in the mixture

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31
Q

Process to Calc Kp value

A

(Possibly need to calculate moles of each gas)

1) Calc mole fractions of each compound
Mole / total moles

2) find particular pressure of each compound
Mole fraction x total pressure

3) calculate Kp value like Kc
But use partial pressures instead of concentrations

Units = whatever pressure is measured in (KPa or atm)

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32
Q

Methods of continuous monitoring of reaction
(2 ways)

A

1) monitoring by gas collection

2) monitoring by mass loss

These can be used to plot a concentration- time graph

33
Q

How does colorimeter work?

A

The Wavelength of light passes through coloured solution is controlled using a fixture. The amount of light absorbed by a solution is measured.

34
Q

Which step in a multi-step reaction is slowest? how would you know this step from the rate equation?

A

The rate-determining step is the slowest.

The rebate equation only includes reacting species involved in the rate-determining step.

35
Q

How to predict a reaction mechanism

A
  • the rate equation only includes reacting species involved in the rate-determining step
  • the orders in the rate equation match the number of species involved in the rate-determining step
36
Q

Two types of equilibria

A

Homogeneous
Heterogeneous

37
Q

Def of homogenous equilibrium?

A

Equilibrium species that all have the same state/phase

E.g all gases or all liquids

38
Q

Def of heterogeneous equilibrium?
+ how is equilibrium constant different

A

Equilibrium species that have different states/phases

E.g carbon is solid but others gases

+equilibrium only includes gases and solutions
All other states omitted

39
Q

When is K constant (equilibrium)

A

at a set temperature K is constant and does not change despite any modifications to concentration, pressure or presence of catalyst

40
Q

The Brønsted-Lowry model for acids

A

A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry alkali is a proton acceptor

41
Q

Position of equilibrium in dissociation strong acid (why are single arrows Ortern used?)

A

A strong acid has equilibrium positioned well over to the right-hand side (products).

A single arrow used to indicate that the forward reaction effectively goes to completion

42
Q

Describe conjugate acid-base pair with HCL(aq) and CL-(aq)

A

A conjugate acid-base pair contains two species that can be interconverted by transfer of a proton

  • in the forward direction, HCL releases a proton to form its conjugate base Cl-
  • in the reverse direction, Cl- accepts a proton to forms its conjugate acid, HCl
43
Q

Suggest why …. Is unlikely to be a one step reaction
2 marks

A

1- not all the reactants involved in rate equation
2- collisions are unlikely to occur with more that 2 ions/species at once

44
Q

Outline a series of experiments that the student could have carried out using initial rates method.
How could the results be used to show that the reaction is first Oder with respect to both I- & S2O8 ^2-
4 marks

A

Vary [S2O8 ^-2] while keeping [I-] constant
Vary [I-] while keeping [S2O8 ^-2]
Rate = concentration
————————
Time

45
Q

Def of spectator ions

A

Ions that do not change in the reaction and therefore left out of the ionic equation

46
Q

Outline the ionic equation between and acid and carbonate

A

2H+.+ CO3^2- —> H2O + CO2

47
Q

Def of strong acid

A

An acid that completely dissociates into its ions when in solution

48
Q

Def of weak acid

A

An acid that only slightly dissociates into its ions when in solution

49
Q

What is Ka and when is it used

A

Ka = the acid dissociation constant

It is used for weak acids and bases that only slightly dissociate in solution to form an equilibrium mixture

50
Q

Formula for Ka

A

Ka = [H+] [A-]
——————
[HA]

H+ > the number of H+ in solution
A- > the ion concentration
HA > acid concentration

51
Q

How can Ka be found from pKa?

A

Ka = 10 ^-pKa

pKa = -log10(Ka)

A low value for pKa or equivalently a large Ka indicates a strong acid

52
Q

Formula to calculate pH

A

pH = -log10 [H+]

[H+] = 10^-pH

53
Q

The [H+] is equivalent to ….. of strong acids as it …. In solution

A

The concentration of H+ ions is equivalent to the concentration of a strong acid as it completely dissociates to ions in solution

54
Q

What dilution gives you a pH increase by 1 unit

A

If you dilute a strong acid 10 times its pH will increase by 1 unit

Weak acids don’t fully dissociate so 10 times dilution would increase the pH by less than 1 unit

55
Q

What is the ionic production of water (Kw)

A

Kw = [H+][OH-]

56
Q

-

A
57
Q

-

A
58
Q

How do you calculate the pH of a strong base from its concentration

A

pH = -log10 (1x10^-14)
————-
[OH-]

pH = pKw - pOH

pOH = -log10[OH-]

59
Q

Calculation relationships of Ka, pKa, [H+] and other elements
1) if HA in excess
2) if A- in excess
3) if HA = A-

A

HA in excess - Use [HA] and [A-] along with Ka to find [H+], then pH

A- in excess - Use Kw to find [H+], then pH

HA = A- - in this case, pKa is equal to pH, therefore find pKa

60
Q

Def of buffer solution

A

Buffer solution is a system that minimises pH changes on addition of small amounts of acid or base

61
Q

What are the limitations of Ka related calculations for ‘stronger’ weak acids

A

We assume
1- [H+] = [A-]
2) assume CH3COOH is a constant

62
Q

What is a buffer solution made from

A

It is formed from
1- a weak acid and its salt
2- excess weak acid and a strong alkali
This produces a mixture containing H+ ions and a reservoir of OH- ions which helps to resist any changes in pH.

63
Q

Sketch curve for strong acid into strong base titration

A

Start high (10-12), go low (1,3) pH range of 6 -> 8 vertically neutrality at 7
See notes for specific curve shape

64
Q

Sketch curve for weak acid into strong base titration

A

Start high (10-12), go lowish (2,3), pH range of 3 ->5, vertical at neutrality (centre 8)
See notes for curve shape

65
Q

Sketch curve for strong acid into weak base titration

A

Start high ish (8-10), go low (1-3), pH range of 3 -> 5, vertically at neutrality (centre= 5)
See notes for graph shape

66
Q

At half neutralisation what is the relationship between pKa and pH

A

pKa = pH at high-neutralisation

67
Q

When is methyl orange a good indicator to use in titrations

A

Changes over pH 3.5-4.5 red -> yellow
Good for strong acids/ weak base

68
Q

When is phenolphthalein a good indicator to use in titrations

A

Changes over pH 8.5 - 10 colourless -> pink
Good for weak acid/strong base

69
Q

suggest why is the reaction unlikely to take one step (when rate equation given)

A

1- not all the reactants involved in rate equation
2- collision unlikely with more than 2 ions/species

70
Q

outline a series of experiments that the student could have carried out using initial rates method
(2I- + S2O3^2 —> I2 + 2SO4^2-)
how could the results be used to show that the reaction is first order with respect to both I- and S2O8^2-

A

Vary [S2O8^2-] while keeping [I-] constant
Vary [I-] while keeping [S2O8^2-]
Rate = concentration
———————-
Time

71
Q

how would the Kc value change if chemist increase pressure at constant temp

A

Kc value would remain the same because it is temperature dependent

72
Q

how would you calculate
PH of acid in solution from the pka value
Use
Calculate pH of 0.015 mol dm-3 of pyruvic acid (pka = 2.39)

A

1) calculate Ka from pka
Ka = 10 ^-pKa
Ka = 10^-2.39 = 4.0738x10^-3

2) calculate [H+] with Ka and HA
[H+] = square root of Ka X HA
[H+] = square root of (4.0738x10^-3)(0.015) = 7.817x10^-3

3) calculate pH using [H+]
PH = -log[H+]
PH= -log( 7.817x10-3) = 2.11

2.11

73
Q

ionic production of water

A

Kw = [H+][OH-]

74
Q

calculate to [OH-] of water using [H=]

A

14 = 10^-A + 10^-B
So add up powers to get concentration of the other

75
Q

calculating pH of strong bases

A

POH = -log(OH-)
14 = pH + pOH
PH = 14 - pOH

76
Q

concentration of strong bases with pH given

A

Work out the pOH by doing 14 - pH = pOH
Then [OH-] using [OH] = 10^-pOH
[OH] then used to find concentration

77
Q

pH of water with Kw value

A

Kw = [H+][OH-]
Kw = [H+]^2
square root of Kw = [H+]
PH = -log [H+]

78
Q

healthy blood at pH 7.4 has hydrogencarbonate : carbonic acid ratio of 10.5 : 1
a patient is admitted to hospital with blood pH of 7.2

Calculate the hydrogen carbonate : carbonic acid ratio of their blood

A

1) work out the [H+] of pH 7.4
[H+] = 10^-pH = 3.98x10-8
2) work out Ka value using original ratio
Ka = [H+] [HCO3-]. (Ans)(10.5)
—————- —————— = 4.18x10^-7
[H2CO3]. 1
3) work out [H+] of new ratio
=6.31x10^-8
4) then rearrange to get ratio
[HCO3-]. Ka. 4.18x10-7. 6.6
———— = ——- = ————- = ——-
[H2CO3] [H+]. 6.31x10-8. 1

6.6 : 1

A Level Chemistry (11 decks)

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