1
Q
aqueous solutions of sodium iodide become yellow in the presence of oxygen due to the slow production of iodine.
one suggested reason for this is that a low concentration of hydrogen ions in the solution produces iodine according to the equation below:
4H+ (aq) + 4I- (aq) + O2 (aq) ⇌ 2I2 (aq) + 2H2O (l)
use Le Chatelier’s principle to suggest a reagent that you could add, apart from water, to decrease the amount of yellow iodine present. explain your choice [2]
A
- add a small ‘amount’ of an alkali / sodium hydroxide / NaOH / OH- ions
- this would remove / react with hydrogen ions giving water, shifting the position of equilibrium to the left (removing iodine)
2
Q
calcium chloride, CaCl2, is soluble in water. use the values below to explain why this is the case.
equation 1: Ca2+ (g) + 2Cl- (g) —> CaCl2 (s) ∆H = -2237 kJmol^-1
equations 2: Ca2+ (g) + 2Cl- (g) + aq —> CaCl2 (aq) ∆H = -2378 kJmol^-1
A
- it is soluble because the enthalpy change of hydration is more exothermic (or more negative) than the enthalpy change of lattice making OR calculation of enthalpy of solution as -141kJmol^-1
- the enthalpy of solution would be negative / exothermic
3
Q
place the following substances in order of increasing entropy under standard conditions:
- nitrogen gas
- copper metal
- water
- air
[1]
A
copper metal, water, nitrogen gas, air
4
Q
a mixture of ammonia and ammonium chloride in aqueous solution can be used as a basic buffer solution. explain what is meant by a buffer solution and how this mixture can act as a buffer solution [3]
A
- a buffer maintains a constant pH upon addition of small amounts of acid or base
- NH4 + ⇌ NH3 + H+
- addition of H+ shifts equilibrium to left / addition of OH- reacts with H+ which shifts equilibrium to right to replace the H+
- salt dissociates completely releasing ammonium ions and ammonia sets up a reversible reaction
5
Q
some elements in the p-block form compounds where the p-block atom does not have eight electrons in its outer shell.
explain why nitrogen can only form a chloride with eight outer shell electrons, but phosphorus can form a chloride with a different number of outer shell electrons [2]
A
- nitrogen forms NCl3; phosphorus forms (PCl3 and) PCl5
- phosphorus is able to expand its octet but nitrogen is not/ phosphorus has AVAILABLE d-orbitals whilst there are none in the outer shell of nitrogen (so phosphorus can have more than 8 electrons in its outer shell and form PCl5)
6
Q
explain why aluminium forms compounds that are electron deficient. show how one of these compounds can act to gain a full outer shell [2]
A
- aluminium has three outer shell electrons so it is able to form three covalent bonds giving six outer shell electrons or an incomplete outer shell or leaving it electron deficient
- coordinate bond must be clear in dot and cross diagram - don’t accept ionic compounds
7
Q
- below is some info about the oxides of two group 4 elements
CARBON:
- there are two common oxides of carbon
- carbon dioxide is an acidic oxide and carbon monoxide can be used as a reducing agent
- both of these are gases with very low boiling temperatures
LEAD:
- there are two common oxides of lead
- lead (II) oxide is an amphoteric oxide and lead (IV) oxide can be used as an oxidising agent
- both of these are solids that exhibit a large degree of ionic character
explain the differences between the oxides of carbon and lead, giving chemical equations to illustrate their acid/base and redox properties [6QER]
A
PHYSICAL STATE:
- carbon dioxide and carbon monoxide have weak forces between separate molecules (so it is a gas)
- lead has a relatively low ionisation energy so can form positive ions; its oxides are ionic with strong attraction between ions (so they are solids)
ACID/BASE:
- carbon dioxide is a non-metal/covalent oxide so it is an acidic oxide
- lead (II) oxide reacts with both acids and bases as an amphoteric oxide
REDOX:
- stability of +2 oxidation state increases down the group / +4 most stable for top of group and +2 most stable at bottom of group / inert pair effect increases down the group
- carbon monoxide contains carbon in oxidation state +2 whilst +4 is stable, so carbon will be easily oxidised / so acts as a reducing agent
- lead (IV) oxide contains lead in oxidation state +4 whilst +2 is stable, so lead will be easily reduced / so it acts as an oxidising agent
RELEVANT EQUATIONS:
- CO as a reducing agent e.g CO + CuO —> Cu + CO2
- PbO2 as an oxidising agent e.g PbO2 + 4HCl —> PbCl2 + Cl2 + 2H2O
- CO2 reacting with a base: CO2 + 2NaOH —> Na2CO3 + H2O or CO2 + H2O —> H+ + HCO3 -
- PbO reacting with a base: PbO + 2NaOH —> Na2PbO2 + H2O
- PbO reacting with an acid: PbO + 2HNO3 —> Pb(NO3)2 + H2O
8
Q
N2O4 ⇌ 2NO2
- the equilibrium constants for the N2O4 / NO2 equilibrium measured at room temperature in some different solvents are listed below
- solvents - equilibrium constant, Kc:
• CS2 - 17.8
• CCl4 - 8.05
• CHCl3 - 5.53
• C2H5Br - 4.79
• C6H6 - 2.03
• C6H5CH3 - 1.69
- samples of 0.400mol of N2O4 are dissolved separately in 1dm^3 of each of these solvents at room temperature and the reactions allowed to reach equilibrium
- in one solution the concentration of N2O4 present at equilibrium is 5.81×10-2 moldm^-3
- find the value of the equilibrium constant in this solution and hence identify the solvent [3]
A
- concentration of NO2 = 2 x (0.400-0.0581) = 0.6838
- value of Kc = 8.05
- solvent is CCl4
9
Q
N2O4 ⇌ 2NO2
- the equilibrium constants for the N2O4 / NO2 equilibrium measured at room temperature in some different solvents are listed below
- solvents - equilibrium constant, Kc:
• CS2 - 17.8
• CCl4 - 8.05
• CHCl3 - 5.53
• C2H5Br - 4.79
• C6H6 - 2.03
• C6H5CH3 - 1.69
- the Gibbs free energy change ∆G, of this reaction is different in different solvents
- explain how the data shows this and state, with a reason, which solvent would have the most negative ∆G value for this reaction [2]
A
- CS2 given with any valid reason
- explanation in terms of position of equilibrium or amount of product formed
10
Q
4NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6 H2O (g)
∆H = -900kJmol^-1
explain why the use of a catalyst is essential in industrial processes involving exothermic equilibria [2]
A
- catalyst allows a lower temperature to be used
- this leads to an increased yield
11
Q
state why the enthalpy change of formation of Cl2 (g) is zero [1]
A
enthalpy of formation of elements in their standard state is zero
12
Q
- vanadium exhibits the properties of a typical transition element, such as having variable oxidation states and forming coloured complexes
- explain why transition elements have variable oxidation states [1]
A
- successive ionisation energies increase gradually / the energies of d-orbitals are similar
13
Q
- VO2+ ions may also be oxidised to VO2 + ions using coloured Ce4+, which is itself reduced to colourless Ce3+
- a titration using this reaction does not usually include an indictor and colorimetry is sometimes used
- state why an indicator may not be useful in this reaction and explain how colorimetry can be used to find the end point of the titration [3]
A
- indicator colour may be difficult to see due to coloured compounds / the vanadium compounds change colour during titration
- colour change of transition metals can be used in place of an indicator
- for colourimetry find a wavelength OR frequency of light absorbed ONLY BY THE VO 2+ / Ce 4+
- endpoint is when there is no absorbance at VO 2+ frequency / Ce 4+ remains in solution
14
Q
give the observation(s) expected when water is added to SiCl4 [1]
A
(white ppt and) steamy fumes
15
Q
the enthalpy of solution of sodium chloride is +4kJmol^-1. explain why this compound is soluble in water despite this value being positive [1]
A
- (Gibbs free energy is negative) as entropy increase of sodium overcomes endothermic enthalpy change
16
Q
when a solution containing the weak base ammonia is neutralised using the strong acid sulfuric acid, a solution of ammonium sulfate is formed. suggest a pH for this solution, giving a reason for your answer [2]
A
- pH between 2 and 6.5
- ammonium ion dissociates to release H+ ions / the ammonium ion exist in equilibrium with NH3 and H+ so it increases [H+] in solution
17
Q
- sodium chloride and sodium bromide both react with concentrated sulfuric acid
- describe the observations in both reactions and explain why they are different [3]
A
- any two of the following observations for (1) each:
• both release misty or steamy fumes
• only sodium bromide gives orange fumes / orange solution
• only sodium bromide gives pungent smell (or sulfur dioxide) - bromide is more easily oxidised than chloride / sulfuric acid is a strong enough oxidising agent to oxidise bromide but not strong enough to oxidise chloride
18
Q
- the concentration of a solution of hydrofluoric acid can be found by titrating against sodium hydroxide
- not all acid-base indicators would be suitable for this titration
- explain what features would make an indicator suitable for use in a weak acid-strong base titration [2]
A
- vertical part of titration curve is between 6.5 (allow +/- 0.5) and 11 (allow +/- 1) / vertical part lies in alkaline region
- select an indicator whose colour change is complete in the vertical region
19
Q
- addition of 12.5cm^3 of 0.100moldm^-3 sodium hydroxide to 25.0cm^3 of hydrofluoric acid (HF) of concentration 0.100moldm^-3 forms a buffer solution
- explain how this buffer solution works [3]
A
HF ⇌ H+ + F-
- addition of a small amount of acid reacts with fluoride ions / shifts equilibrium to left
- addition of small amount of base removes hydrogen ions and these are replaced by HF dissociating / shifting equilibrium to the right / hydroxide ions react with HF molecules
20
Q
- a student was given a pink-coloured solution containing [Co(H2O)6]2+ ions
- upon addition of hydrochloric acid the solution turned blue as [CoCl4]2- ions formed according to the equilibrium shown below:
[Co(H2O)6]2+ + 4Cl- ⇌ [CoCl4]2- + 6H2O - aqueous silver nitrate was added to the solution containing [CoCl4]2-
- state and explain the observation(s) expected [4]
A
- white precipitate
- silver chloride forms / silver ions cause a precipitate with chloride
- solution turns pink
- removal of chloride causes equilibrium to shift to left
21
Q
- H2O2 + 2H+ + 2e- ⇌ 2H2O - E^⦵=+1.77 V
- Cr2O7 2- + 14H+ + 6e- ⇌ 2Cr3+ + 7H2O - E^⦵ = +1.33V
- the concentration of both Cr2O7 2- and Cr3+ ions are 1 moldm^-3
- state and explain how the value shown on the high resistance voltmeter would change if the concentration of the Cr3+ ions were increased whilst the concentration of the Cr2O7 2- was left unchanged [2]
A
- this makes the reading less positive / smaller
- increase in concentration of chromium (III) ions will shift equilibrium to the left
22
Q
- another method of finding whether reaction is feasible (other than electrochemical cells) is to use the Gibbs free energy calculated from standard enthalpy of formation and standard entropy values
- state, giving a reason, whether Gibbs free energies or electrochemical methods are more appropriate for finding whether this reaction is feasible [2]
A
- electrochemical methods are better
- because the reaction is in solution but standard enthalpies of formation use standard states
23
Q
in born haber cycles, enthalpy changes are quotes per atom
- REMEMBER TO X2 IF ITS HAPPENING TWICE
A
ex:
atomisation of iodine = +107 kJmol^-1
Fe(s) + I2(s) —> Fe(s) + 2I(g)
+107 x2 = 214 kJmol^-1
24
Q
a mixture of ammonia and ammonium chloride in aqueous solution can be used as a basic buffer solution. explain:
A
NH3 + H2O ⇌ NH4 + + OH-
- a weak base has:
high conc - high ⇌ low - low - NH4Cl (NH4+)
- so now:
high - high ⇌ high - low - BUT equilibrium doesnt shift LHS because not a lot of OH- to react with
soo as a basic buffer:
high - high ⇌ high - low
- if addition of H+: reacts with OH-, [OH-] decreases, so equilibrium shift to RHS, so [OH] replenished
- if addition of OH-: can react with NH4 + / more OH-, so shift LHS, [OH-] remains constant, so [H+] remains constant as Kw=[H+][OH-]
25
how can you find Ka from a pH curve?
- use the 1/2 neutralisation point
- Ka = [H+] at this point
- so if neutralisation point = 20cm^3 HCl added
- 1/2 neutralisation point = 10cm^3
- pH at 10cm^3 = 9.6
- [H+] = 10^-9.6
- Ka = 10^-9.6
26
Ka = [H+] at the 1/2 neutralisation point
27
what does it mean if there are 2 vertical regions on a pH curve?
- 2 points of neutralisation
- so 2 H+
28
- a graph shows the change in pH during a reaction between 0.10moldm^-3 sodium hydroxide and 0.10 moldm^-3 ethanoic acid at room temperature
- volume of NaOH(aq) added on x axis of graph, pH on y axis
- using the details included in the description, describe the experiment that was carried out in order to plot the graph [4]
- add ethanoic acid to conical flask
- add NaOH to burette
- read the initial volume of NaOH in the burette
- and take the initial pH reading - using a pH probe
- add 5cm^3 of NaOH to the conical flask
- and measure the pH again
- repeat
29
if given the Ka, how do you work out the [H+] at the 1/2 neutralisation point
Ka = [H+] at half neutralisation point
30
- draw a titration curve obtained when 50.0cm^3 of a 0.10 moldm^-3 solution of a strong base is added gradually to 25.0cm^3 of a 0.10 moldm^-3 solution of a weak acid
- the weak acid has a Ka value of 1.80x10^-5 moldm^-3 at 298K
- give the pH values at key points during the titration and explain their significance [6QER]
- vertical region must be above 7 (weak acid, strong base)
- Ka = 1.80 x10^-5 moldm^-3
- as Ka = [H+] at half neutralisation point
- pH at half neutralisation point = -log[H+] = pH 4.7
- pH4.7 = 12.5cm^3
- full neutralisation point = 25cm^3
- volume of base added (cm^3) on x axis, pH on y axis
- vertical region, above pH7, at 25cm^3
- goes through pH4.7 at 12.5cm^3
- start at pH of weak acid
- end at pH of strong base
31
write the expression for the ionic products of water, Kw [1]
Kw = [H+][OH-]
32
when CCl4 is added to water two separate layers are formed and there is no reaction, whilst SiCl4 react very violently with water. give the reason for this difference [1]
- SiCl4 has available d-orbitals whilst carbon does not (all water can bond to Si atom)
33
the entropy change of the reaction shown below is positive. give a reason for this:
Na2CO3 (aq) + 2HCl —> 2NaCl (aq) + CO2 (g) + H2O (l)
[1]
forms a gas and gases have a greater entropy than liquids (and solids)
34
- in a born haber cycle
- Ca2+(g) + Cl2(g) + 2e- —> Ca2+(g) + 2Cl(g) + 2e-
- — 242kJmol^-1 —>
- give the value of the standard enthalpy change of atomisation of of chlorine
121 kJ mol^-1
35
- in born haber cycle
- Ca(g) + Cl2(g) —> Ca2+(g) + Cl2(g) + 2e-
—1735kJmol^-1—>
- a student incorrectly states that the second ionisation energy of calcium is 1735kJmol^-1 as this is the energy needed to form Ca2+(g) in the born-haber cycle
- suggest a possible value for the second ionisation energy of calcium, giving a reason for your answer [2]
- any value in range 900-1600kJmol^-1 (1)
- (1) for either:
• ionisation energies increase for successive ionisation energies (so value must be more than half the ionisation of the first two electrons)
• second ionisation energy is greater than first ionisation energy
36
- the reaction occurring when calcium chloride is used as a drying agent is shown below:
CaCl2 + xH2O —> CaCl2.xH2O
- a laboratory manual suggests a MINIMUM temperature of 200°C to remove all the water from the hydrated calcium chloride. at this temperature the reaction occurring is as follows:
CaCl2.xH2O —> CaCl2 + xH2O
- a student calculates the enthalpy change when calcium chloride becomes dehydrates as 78.0 kJmol^-1
- assuming that this data is correct, calculate a value for the entropy change when calcium chloride becomes dehydrated at this temperature [3]
- at minimum temp: ∆G=0 so equation gives ∆S = ∆H / T
- change units giving T = 473K and 78000 Jmol^-1
- ∆S = 165 JK^-1mol^-1
37
- calcium halides react with concentrated sulfuric acid in a similar manner to sodium halides
- state the observation(s) when concentrated sulfuric acid is added to calcium chloride and calcium bromide. explain why the observations are different [3]
CaCl2
- misty fumes (and white solid) (1)
CaBr2
- will produce orange fumes / solution as well as misty fumes (and white solid)
- bromide is easier to oxidise than chloride (so can be oxidised by the sulfuric acid)
38
- one method of performing this experiment involves sampling and quenching
- explain what is meant by the term ‘quenching’ and why it needs to be carried out [2]
- quenching is the sudden stopping / significant slowing of a chemical reaction in a sample
- ensure sample composition does not change between taking sample and analysis OR during analysis
39
- one method of measuring the concentration of thiosulfate ions in solution is by titration using a solution containing iodine
- suggest an indicator for this titration [1]
starch
40
explain why transition metal complexes such as [Ir(CO)2I2]- can act as homogenous catalysts [2]
award (1) each for any two of following:
- transition metals have empty orbitals so can form bonds to reactant molecules / form complex
- transition metals have variable oxidation states (so they can oxidise/reduce the reactants)
- products are released and the metal returns to the original oxidation states
41
- the half equation for oxalic acid is:
(COOH)2 —> 2CO2 + 2H+ + 2e-
- oxalic acid solutions may be analysed using acid-base titrations
- the simplified pH curve for the titration of a solution of oxalic acid against sodium hydroxide is given below
.
- four students undertook the titration in different ways:
• alice used a pH probe to measure the pH after addition of each portion of sodium hydroxide solution
- brychan used the indicator phenolphthalein in his titration; phenolphthalein has a pH range of 8.2 to 10.0
- carys used the indicator methyl orange in her titration; methyl orange has a pH range of 3.2 to 4.4
- david used the indicator cresol red in his titration; cresol red has a pH range 1.8 to 2.8
- state and explain which, if any of the methods chosen will allow the students to find valid data to calculate the concentration of oxalic acid
- briefly explain why oxalic acid has a pH curve of this shape
(you are not required to carry out any calculation)
[6QER]nn
- oxalic acid has two acidic protons
- the Ka/acidity/ease of removal of each proton is different so they give two different vertical regions
- first vertical region occurs at half the volume of second as each occurs after removing same number of protons
- flat region from 5-10 and 20-25 due to formation of buffer Alice, Brychan and David methods will work
- indicators only work if they change colour within a vertical region
- Alice‘s method allows the first and second equivalence points to be found by plotting a curve
- brychan’s indicator will change colour fully during the second vertical region
- david’s indicator will change colour fully during the first vertical region
-carys’s method will not work as indicator will change colour gradually as sodium hydroxide is added
42
- give the colours of the precipitates formed when sodium hydroxide solution is added to solutions containing the following ions:
• Mg2+
• Fe2+
• Cr3+
• Pb2+
[2]
- the method given opposite is incomplete
- suggest an additional step that would allow the remaining solutions formed to be identified. give the expected observations. state the property of the metals that allows them to be distinguished in this way [3]
- Mg2+ = white ppt
- Fe2+ = (dark) green ppt
- Cr3+ = (grey) green ppt
- Pb2+ = white ppt
.
- add excess sodium hydroxide (1)
(1) for either:
- magnesium hydroxide white ppt remains but lead hydroxide ppt dissolves (giving a colourless solution)
- iron (II) hydroxide green ppt remains but chromium hydroxide ppt dissolves (giving a dark green solution)
- lead and chromium are amphoteric (iron and magnesium are not)
43
- potassium dichromate (VI) is used as a reagent in acid solution in a wide range of redox reactions
- the dichromate(VI) ions can be reduced using EXCESS metallic zinc in acid conditions
• Elfed performs the experiment in a sealed tube under an atmosphere of pure nitrogen gas
• Fatima performs the experiment in a standard test tube
- the final products are different in each case
- use the standard electrode potentials below to identify the products in both cases. give reasons for your answers:
.
Cr3+(aq) + 3e- ⇌ Cr(s) E⦵ = -0.78
Zn2+(aq) + 2e- ⇌ Zn(s) E⦵ =-0.76
Cr3+ (aq) + e- ⇌ Cr2+ (aq) E =-0.42
O2(g) + 2H2O(l) + 4e- ⇌ 4OH-(aq) E⦵ = +0.40
Cr2O7 2-(aq) + 14H+(aq) + 6e- ⇌ 2Cr3+ (aq) + 7H2O(l) E⦵ = +1.33
- Elfed’s reaction produces Cr2+
- dichromate(VI) can be reduced to Cr3+ and further to Cr2+ as E⦵ values for these two processes are more positive than that for Zn2+/Zn/ EMF values for reactions are positive (2.09V and 0.34V)
- Fatima’s reaction produces Cr3+
- oxygen oxidises Cr2+ back to Cr3+ as E⦵ value is more positive than that for Cr3+/Cr2+ / EMF value for reaction is positive (0.82V)
44
what are the units of Kw?
mol^2dm^-6
because Kw = [H+][OH-]
45
- the value for Kw at 298k is 1.0 x10^-14
- explain why the pH of pure water at this temperature has a value of 7 [2]
- in pure water [H+] = [OH-]
- [H+] = √1.0x10^-14
- pH = -log(10^-7) = 7
46
calculate the pH of the final solution if 10cm^3 of 0.10moldm^-3 hydrochloric acid is added to 990cm^3 of pure water [2]
- final vol = 1000cm^3
- so acid diluted by a factor of 100
- so final conc of acid is 0.001
- (0.1 x 10) / 1000 = 0.001
- pH = -log0.001 = 3
47
- calculate the pH of a solution which is 0.010moldm^-3 with respect to ethanoic acid and 0.020moldm^-3 with respect to sodium ethanoate at 298K
- Ka for ethanoic acid = 1.78×10^-5 moldm^-3 at 298k [3]
- 1.78 x10^-5 = ([H+] x 0.02) / 0.01
- [H+] = 8.90 x10^-6
= 5.05
48
define pH [1]
pH = -log[H+]
49
- when sodium hydroxide is added to a solution of potassium dichromate, a colour change occurs without a redox reaction occurring
- give the formula of the new chromium containing ion and the colour of the solution formed [2]
CrO4 2-
- yellow
50
2[Na2CO3.NaHCO3.2H2O] (s) —> 3Na2CO3 (s) + CO2 (g) + 5H2O (l)
- the above reaction is used commercially to obtain sodium carbonate
- state one environmental disadvantage of this reaction as indicated by the equation, and state what could be done to overcome this problem [2]
- CO2 is produced (and released into the air) and this contributes to the greenhouse effect / increases acidity of sea
- it should be trapped / a use found for it
51
- give the chemical name of a chlorine-containing compound of commercial or industrial importance
- state the use made of this compound [1]
- (sodium) chlorate (I) as bleach
- (sodium) chlorate (V) as weedkiller
etc
52
- planners have to ensure a secure supply of energy in the future
- it has been suggested that the use of fossil fuel should be reduced, the use of renewable energy increased and that energy efficiency should be greatly improved
- by considering both the benefits and the difficulties involved, discuss whether you think that these suggestions are realistic [4 QWC]
BENEFITS:
- stops fossil fuels running out
- reduces CO2 emission/greenhouse emission/global warming/effect of global warming
- reduces SO2 emission/acid rain
- there will be an investment in new technology
DIFFICULTIES
- dependent on fossil fuel/unlikely to meet current demand
- renewable energy currently more expensive
- reliability of supply from renewables
- major development in energy efficiency technologies required
.
- consideration and discussion of benefits/difficulties
53
- the forwards rate of reaction is exothermic. normally the process is carried out at a temperature of around 900°C
- suggest why this temperature is used [5 QWC]
- if temperature too low, the reaction is too slow
- if temperature too high, yield is too low
- compromise temperature - acceptable rate and yield
54
- sodium iodide is very soluble in water at room temperature
- for a compound to be very soluble in water the value of the enthalpy of _______ will be greater than the enthalpy of ________
- hydration
- lattice breaking
55
write an equation, including state symbols, that represents the enthalpy change of lattice formation of solid magnesium chloride [1]
Mg2+ (g) + 2Cl- (g) —> MgCl2 (s)
56
- when chlorine is bubble into cold sodium hydroxide solution, the following reaction occurs:
Cl2 (g) + 2NaOH (aq) —> NaCl (aq) + NaOCl (aq) + H2O (l)
- show that this is a disproportionation reaction [2]
- oxidation state of chlorine at start is 0; at end is -1 (in NaCl) and +1 (in NaOCl)
- chlorine (or same element) has been both oxidised and reduced
57
- one proposal to reduce the carbon dioxide emissions from public transport is to replace diesel trains with trains powered by hydrogen fuel cells
- give one other advantage of using hydrogen fuel cells [1]
more efficient release of energy from fuel
58
- ammonia is an example of a weak base
- state what is meant by a base and explain why the ammonia molecule is able to act as a base [2]
- a base is a proton acceptor
- lone pair on nitrogen (can form coordinate bonds with H+)
59
- ammonia can be used as part of a mixture that forms a buffer
- suggest a compound that could be added to ammonia solution to form a buffer [1]
- ammonium salt of any strong acid e.g ammonium sulfate, ammonium nitrate, ammonium chloride
60
suggest a use for a buffer [1]
- storing/using enzymes (at constant pH)
- storage of biological molecules
- fermentation
- dyeing
61
- state one structural similarity and one structural difference between hexagonal boron nitride and graphite [1]
similarity:
- layers of hexagonal sheets present in both
difference:
- layers in register in boron nitride and out of register in graphite / atoms of adjacent layers above one another in boron nitride but not in graphite
62
- state one bonding similarity and one bonding difference between hexagonal boron nitride and graphite [1]
similarity:
- each atom is bonded by covalent bonds to three others in both graphite and BN
- delocalised electrons in graphite but not in BN / bonds are non-polar in graphite but polar in BN
63
∆H = -197 kJmol^-1
∆S = -187 JK^-1mol^-1
- a student calculates the value at which ∆G = 0 and states that this is the minimum temperature needed for the reaction to occur
- find the temperature which ∆G=0 and state, giving a reason, whether this is the MINIMUM temperature needed for the reaction to occur [3]
T = ∆H / ∆S
T = 1053 K
- student is incorrect
- ∆G is negative below this temperature so the reaction is feasible at a lower temperature
64
- samples of 0.040 mol of SO2 and 0.040 mol of O2 are placed in a sealed vessel of volume 2.00dm^3
- the reaction is allowed to come to equilibrium, giving a concentration of 0.014 moldm^-3 of SO3
- calculate the value of the equilibrium constant, Kc, under these conditions giving its unit [4]
- inital conc of SO2 and O2 = 0.020moldm^-3
- equilibrium concentrations:
[SO2] = 0.006 moldm^-2
[O2] = 0.013 moldm^-3
- Kc = (0.014)^2 / (0.006)^2 x (0.013) = 419 mol^-1dm^3
65
- the standard electrode potential for the half equation below is +0.80V
- Ag+ (aq) + e- ⇌ Ag (s)
- when a piece of copper metal is placed in a solution of silver nitrate, a displacement reaction occurs
- write the ionic equation for this displacement reaction [1]
Cu + 2Ag+ —> Cu2+ + 2Ag
66
- an experiment produced 14.5mg of Fe2O3
- the total error when measuring the mass of Fe2O3 by difference was 0.2mg
- find the percentage error in this measurement [1]
0.2 / 14.5 x100 = 1.4%
67
- method one = % error = 1.4%
- method two = % error = 0.8%
.
- method one says conc of Fe2+ in test water A is 7.27x10^-3 moldm^-3
- method two says conc of Fe2+ in test water A is 7.6x10^-3
.
- two students compared their results and decided that they are in agreement
- calculate the % difference between the concentration of Fe2+ ions found in method one and two, and hence show whether the students’ decision is valid [2]
- % difference = 4.5%
- students’ decision is NOT valid:
• total of percentage errors for both methods is 2.2% (1.4% + 0.8%) and the percentage difference is greater than this
68
- the rate equation for the decomposition for N2O5 to form O2 and NO2 is first order overall
a) write the rate equation for this reaction [1]
b) suggest a rate-determining step for this reaction [1]
a) rate = k[N2O5]
b) accept any balanced equation that has one N2O5 as reactant
e.g N2O5 —> NO2 + NO + O2
69
give a reason why the entropy of Hg (l) is greater than that of Au (s) [1]
- particles have greater freedom in liquid mercury compared to solid gold (so they have less order in liquid and higher entropy)
70
describe the structure of graphite and explain why it’s soft [2]
- layers / sheets of hexagons of C atoms bonded together
- weak forces between layers (allowing layers to move and making it soft)
71
- the bond energy of Cl2 is 242kJmol^-1
- state the enthalpy of atomisation of chlorine [1]
121 kJmol^-1
72
- iron is an example of a transition element
- it has two main oxidation states in its compounds
- write the electronic structure of an Fe2+ ion and use it to show why iron is classed as a transition metal [2]
1s2 2s2 2p6 3s2 3p6 3d6
- partially filled d-orbitals
73
explain why iron has more than one common oxidation state in its compounds [1]
- the energy of the (4s and) 3d orbitals are all similar
- the ionisation energies to remove the (4s and) 3d electrons are similar
74
explain why the complex ion [Fe(H2O)]3+ and [Fe(H2O)5(OH)]2+ are not the same colour [2]
- different ligands cause different amount of d-orbital splitting
- so different frequencies/ wavelengths of light are absorbed (and different frequencies / wavelengths are transmitted/reflected)
75
- suggest how you would select an appropriate wavelength to find the concentration of [Fe(H2O)6]3+ in the equilibrium mixture [1]
find a wavelength absorbed by [Fe(H2O)6] 3+ but not by any other species in the mixture
76
- aqueous iron (II) compounds form a precipitate of a different colour with aqueous sodium hydroxide
- in a test tube, the precipitate can change colour on standing
- use the standard electrode potential values below to explain this change :
Fe3+ + e- ⇌ Fe2+ E = +0.77V
O2 + 4H+ + 4e- ⇌ 2H2O E = +1.23
[2]
- OXYGEN (from the air) can oxidise Fe2+ to Fe3+ (turning the ppt brown)
- bc oxygen has a more +ve standard electrode potential than Fe3+ so it is a stronger oxidising agent
- the EMF for the reaction between O2 and Fe2+ is positive / +0.45V and positive reactions are feasible
77
- in a blast furnace Fe2O3 is reduced by CO to form iron metal
- write an equation for this process [1]
Fe2O3 + 3CO —> 2Fe + 3CO2
78
- in a titration, dewi had access to 5 concentrations of sodium thiosulfate: 2.0 moldm^-3, 1.00 moldm^-3, 0.500 moldm^-3, 0.200 molddm^-3 and 0.0500 moldm^-3
- the teacher tells Dewi that he has selected an inappropriate concentration of sodium thiosulfate solution
- suggest which concentration of sodium thiosufate he should have chosen.
- give 2 reasojs for your answer [2]
- 0.200 moldm^-3
- use a lower concentration so titration volume is greater - smaller % error / more accurate
- cannot use too low a concentration / 0.0500 moldm^-3 as volume would be too large (for a standard burette)
79
- chemists often use an approximate rule that the rate of a reaction doubles when this temperature is increased by 10°C
- show that this rue is true if the reaction is warmed from 298K to 308K. the activation energy for the reaction is 52.8kJmol^-1 [3]
- A will be same as similar temperature
- e^(-52800 / 308R) = 1.0986 x10^-9
- e^ (-52800 / 298R) = 5.4981 x10^-10
1.0986 x10^-9 is double 5.4981 x10^-10 so the true is true under these conditions
80
- the acid from chloride ions is commonly called perchloric acid,HClO4
- it is a very strong acid and is commonly classed as a superacid as it is a stronger acid than sulfuric acid
- state how the Ka value of a stronger acid will compare with that of a weaker acid, giving a reason [1]
Ka of stronger acid would be GREATER than that for weaker acid as stronger acids have GREATER DISSOCIATION
81
- for a particular scenario - out of two methods - one method was better because their measurements were to more significant figures / more precise / have higher resolution
82
- one common salt of perchloric acid is ammonium perchlorate
- suggest a pH value for a solution of ammonium perchlorate. give a reason for your answer [2]
- pH 2-6
- ammonium ion (partially) dissociates to release H+ ion
- NH4 + ⇌ NH3 + H+
83
- one method measures the vol of CO2 by measuring the volume of the gas released using a gas syringe and the other by measuring the mass lost as Co2 is released by placing the reaction flask on digital balance
.
- someone suggests using an excess magnesium in place of calcium carbonate for the experiments
- this will produce hydrogen gas in place of carbon dioxide
- explain which of the two methods will suffer the greatest loss in accuracy [2]
- wont affect method one as the same vol of gas would be produced
- will make method two experiment less accurate as the mass released would be much smaller (as H2 has much smaller Mr than CO2)
84
HA ⇌ H+ + A-
H L L
(+ NaA)
H L L
- find mol of A- by moles of salt added
85
- assume no volume change
- [H+] = [A-] because 1:1 ratio
- assume conc at start of HA same at equilibrium bc dissociation so small
86
- HY ⇌ H+ + Y- with NaOH
- conc H+ decreases
- equilibrium will shift to RHS by what it went down by due to salt added
- find conc of HY and Y- before adding NaOH
- after adding NaOH
- put back into Ka equation
- find pH
- shouldnt have changed much
87
cH2SO4
- with Cl- :
- with Br- :
- with I- :
CL- :
- misty fumes HCl
- white solid NaHSO4
- not redox
BR- :
- misty fumes HBr
- white solid NaHSO4
- orange/brown solution
- sulfur (yellow solid)
- SO2 choking gas
I- :
- misty fumes HI
- white solid
- SO2 choking gas
- yellow solid sulfur
- I2 black solid/purple vapour
- H2S = rotten egg smell
88
I
I
I
v
Cl-
Br-
I-
more -ve E value
so better reducing agent
89
- AlCl3 can form a dimer because it doesnt have a full octet
- aluminium is electron deficient
90
- aluminium chloride forms Al2Cl6 dimers using coordinate bonding
- explain why the coordinate bonds form [2]
- Al is electron deficient / doesnt have a full outer shell
- lone pair on Cl
91
- hexagonal boron nitride is sometimes called white graphite
- give one difference between the physical properties of hexagonal boron nitride and graphite [1]
- boron nitride doesnt conduct electricity but graphite does
92
- the elements of the p block show a variety of oxidation states within the same group, whilst some elements of the d block can show an even wider range of oxidation states
• some oxides of group 4 have an oxidation state of +4 such as CO2, SiO2, and PbO2 and others have an oxidation state of +2 such as CO and PbO. there is no stable oxide with the formula SiO
• some chlorides of group 5 have a +5 oxidation state such as PCl5 and some have an oxidation state of +3 such as NCl3 and PCl3. there is no stable chloride with the formula NCl5
• manganese forms of range of oxides including MnO, Mn2O3 and MnO2 which have oxidation states of +2, +3, +4 respectively
- explain why these elements can form the oxidation states shown and why SiO and NCl5 do not form [6QER]
- oxidation states in group 4 are stable at +4 at the top of the group and +2 at the bottom
- due to the inert pair effect
- the inert pair effect increases down the group
- PbO more stable and PbO2 / PbO2 is an oxidising agent/CO2 more stable than CO/CO is a reducing agent
- SiO doesn’t form as this is a +2 oxidation state which cannot form this high in the group
- +5 oxidation state possible in third period of group 5 but not in period 2
- this is because phosphorus can expand its octet/phosphorus has available d-orbitals
- NCl5 not possible as N has no availability d orbitals so it cannot expand its octet
- one characteristic of transition metals is that they form of variety of oxidation states
- Mn can form a range of oxidation states as 3d orbitals (and 4s) have similar energies and similar ionisation energies
93
- tin (II) ions, Sn2+ (aq), can be used as reducing agents for a range of metal ions
- in one study of the reduction of Pt 4+ ions to Pt 2+ ions in an acid solution, the reaction was found to be first order with respect to both Pt 4+ and Sn 2+ with a rate constant of 895 mol^-1dm^3 at 298k
.
- a student suggests studying this reaction using sampling and quenching with samples taken every 30 seconds for 5 minutes
- explain why this would not be an appropriate sampling interval [2]
- (rate constant is large so) the reaction is very fast has a high rate
- reaction will be complete/most would have reacted in 30 seconds
94
- tin (II) ions, Sn2+ (aq), can be used as reducing agents for a range of metal ions
- in one study of the reduction of Pt 4+ ions to Pt 2+ ions in an acid solution, the reaction was found to be first order with respect to both Pt 4+ and Sn 2+ with a rate constant of 895 mol^-1dm^3 at 298k
.
- a student decides to use colourimetry to study the rate of this reaction
- explain what the student should consider when choosing an appropriate wavelength of light for their colorimetry experiment [1]
- frequency that is absorbed by one reactant only but not by its product (or vice versa)
(accept frequency absorbed by only one species)
95
- HOCl can act as an oxidising agent
2HOCl (aq) + 2H+ + 2e- ⇌ Cl2 (g) + 2H2O (l)
E = +1.63V
- give the formulae of all the metal ions in the table that can be oxidised by HOCl (aq)
- explain why you have chosen these ions [2]
Fe3+ + e- ⇌ Fe2+ E= +0.77V
Tl3+ + e- ⇌ TI+ E=+1.25
Pb4+ + 2e- ⇌ Pb2+ E= +1.69
Co3+ + e- ⇌ Co2+ E=+1.82
- Tl+ and Fe2+
- HClO is more positive than for these half cells so can oxidise the ions
OR
- positive EMF values calculated for the reactions
96
- substituted propanoic acids, CH3CHXCOOH have different acid strengths
- the table below gives some information about three substituted propanoic acids:
X=H - the pH of a 0.50 moldm^-3 aqueous solution is 2.59
X=Cl - Ka is 1.48 x10^-3 moldm^-3
X=OH - pKa = 3.86
- place these acids in order of increasing acid strength. you must show your working [3]
X=H:
[H+] = 2.57x10^-3
Ka= [H+]^2 / 0.50
Ka = 1.32x10^-5
X=OH:
pKa = -log[Ka]
10^-3.86 = 1.38 x10^-4
Ka = 1.38 x10^-4
X=H < X=OH < X=Cl
97
- a buffer is produced by mixing 100cm³ of ethanoic acid (Ka= 1.8x10^-5 moldm^-3) of concentration 0.80moldm^-3 with 50 cm³ of aqueous sodium hydroxide concentration 0.8moldm^-3 at a temperature of 298K
- calculate the initial pH of the aqueous sodium hydroxide [2]
[H+] = Kw / [OH-] = 1.25x10^-14
pH = 13.9
98
- a buffer is produced by mixing 100cm³ of ethanoic acid (Ka= 1.8x10^-5 moldm^-3) of concentration 0.80moldm^-3 with 50 cm³ of aqueous sodium hydroxide concentration 0.8moldm^-3 at a temperature of 298K
- explain how buffers such as this work [3]
CH3COOH ⇌ CH3COO- + H+
award (1) for any 2 of:
- CH3COONa —> CH3COO- + Na+
- addition of small amount of H+ - this shifts equilibrium to left to remove H+
- addition of small amount of OH- - this removes H+ and shifts equilibrium to right to produce more H+
99
circle all the acids in the following equation:
CH3COOH + NH3 ⇌ CH3COO- + NH4 +
CH2COOH and NH4 +
100
in a standard hydrogen electrode, hydrogen gas at 1atm is pumped in. what is the name of the solution in the beaker and state its concentration [1]
HCl at 1 moldm^-3
101
- carbon tetrachloride does not react with cold water, however silicon (IV) chloride reacts violently
- explain why the reactivity of these compounds is so different [1]
- silicon possess available d-orbitals whilst there are none in the outer shell of carbon (therefore carbon cannot bond)
102
- a 25cm^3 sample of sodium hydroxide solution was exactly neutralised by 24.25cm^3 of sulfuric acid of concentration 0.176moldm^-2
- an alternative method to calculate the concentration of the sodium hydroxide solution is to use its pH
- a student measures the pH of the same sodium hydroxide solution as being 13.5, which was then used to find the conc of NaOH
.
state and explain which method gives the more precise value [1]
- titration is more precise as a change of 0.1 in pH gives a large change in concentration/ the resolution of the burette or pipette is much greater than that of the pH probe
103
- electrochemical uses a solution reaction
- gibbs free energy data is for solids/liquids
104
increase in pH by one unit is a decrease in [H+] by a factor of 10
105
if see pH in order of rates, calculate H+ first
106
units of rate in rate equation:
moldm^-3s^-1
107
you cant measure ∆Hlatt experimentally - only theoretically by born-haber cycles for example
108
the change of state to work out entropy change is more important than moles
109
in an acidic buffer with a weak acid and its salt, why is the salt e.g sodium ethanoate, necessary in the buffer solution?
- because it dissociates fully when it dissolves to produce Na+ ions (which are not involved in buffer action) and CH3COO- ions
- weak acids are not very dissociated and there would not be otherwise enough CH3COO- ions to ‘mop up’ the H+ ions added
110
basic buffers:
ammonia and ammonium chloride
NH3 + H2O ⇌ NH4 + + OH-
NH4Cl added which increases conc of NH4 + present in solution
111
- a buffer solution is made that contains 0.400moldm^-3 methanoic acid, HCOOH, and 0.600moldm^-3 sodium methanoate, HCOO-Na+
- the Ka for methanoic acid is 1.6x10^-4moldm^-3
- what is the pH of this buffer
HA ⇌ H+ + A-
Ka = [H+][A-] / [H-]
1.6x10^-4 = [H+]0.6 / 0.4
[H+] = 1.06667 x10^-4
pH = 3.97
112
what is the formula of sodium ethanoate?
CH3CH2COONa
(Mr = 96.05)
113
- aqueous propanoic acid and sodium propanoate can form a buffer solution
- a student requires a buffer of pH 4.46 for an experiment
- he adds solid sodium propanoate to 500cm^3 of aqueous propanoic acid of concentration 0.210 moldm^-3
- calculate the mass of sodium propanoate needed assuming no change in volume
(- Ka for propanoic acid is 1.35x10^-5 moldm^-3)
[4]
HA ⇌ H+ + A-
c HA = 0.210
c H+ = 3.467 x10^-5
c A- = ???? c of sodium propanoate
Ka = [H+][A-] / [HA]
c A- = 0.08177
n A- = 0.08177 x 500/1000 = 0.0409
Mr = 96.05
mass = 3.928g
114
- a buffer is formed by mixing 15cm³ of 0.1moldm^-3 sodium hydroxide and 30cm³ of 0.6moldm^-3 propanoic acid (CH3CH2COOH)
- the Ka for propanoic acid is 1.35×10^-5
- calculate the pH of this buffer solution 
CH3CH2COOH ⇌ CH3COO- + H+
(before adding NaOH)
n = 0.018 ⇌ 0 0
(after adding NaOH) :
n NaOH added = 1.5x10^-3
reacts with H+, so H+ decreases by 1.5x10^-3 but equilibrium shifts right to minimise this, so moles H+ = 1.5x10^-3 and moles of CH3CH2COOH = 0.018 - 1.5x10^-3 = 0.0165
0.0165 ⇌ 1.5x10^-3. 1.5x10^-3
conc of all = (divide by 45/1000)
0.36666 ⇌ 0.033. 0.0333
Ka = [H+][A-] / [HA]
1.35x10^-5 = [H+]0.033 / 0.36666
[H+] = 1.485x10^-4
pH = 3.83
115
write the equation for the second ionisation energy of boron [1]
B+ (g) —> B 2+ (g) + e-
116
suggest a pH value for a dilute solution of ammonium chloride, giving a reason for your answer [2]
- pH in range 2.5-6.5
- ammonium ion will release H+ ions into solution making solution acidic
117
- catalysts: CuO and Fe2O3
- explain why these transition metal oxides are able to act as catalysts [2]
- gases can be adsorbed onto catalyst
- this brings the gas reagents together
- or the transition metal oxidise or reduce the gas molecules making them more reactive
——- for gas reactants and solid catalysts
118
explain why silicon can form only one chloride (SiCl4) but lead can form two chlorides (PbCl2 and PbCl4) [2]
- inert pair effect
- becomes more significant down group so lead can form +2 but silicon can’t
119
- explain how an indicator works and give reasons why two different indicators are used to identify the end points for neutralisation of sodium hydroxide and sodium carbonate
- suggest how you would select appropriate indicators for this experiment [6QER]
- indicators are weak acids or bases where protonated and deprotonated forms are different colours
- indicator changes colour of a range of pH
- different indicators change colour over different pH ranges
- the colour change range must lie with a vertical region of the titration curve for indicator to be useful
- to select appropriate indicators, calculate OR find experimentally the pH of the equivalence points and their vertical regions
120
suggest how the equivalence point for the titration of a weak acid and a weak base such as sodium carbonate can be found experimentally [2]
- use pH probe to measure pH as acid is added
- plot results to find equivalence point for/ no sharp increase in pH
121
- when measuring the standard electrode potential for the Zn2+(aq)I Zn(s) system a piece of zinc metal is placed in an aqueous solution containing Zn2+ (aq)
- explain why a similar method would not be appropriate for the Li+ (aq) I Li (s) system [1]
lithium metal would react with the solution
122
- the rate constant for the reaction in solution at 30°C is 1.752x10^-2 dm^3mol^-1s^-1
a) give the overall order of the reaction [1]
b) explain what information about the mechanism can be found from:
• the overall order of reaction
• the orders with respect to each reactant [3]
a) second order
b)
- rate data tells us about the rate determining/slowest step of the reaction mechanism
- second order shows there are two molecules/ the number of molecules involved (in the rate determining step)
- order with respect to each reactant tells how many of each reactant molecules are present in the rate determining step
123
- the first ionisation energy of caesium is 376 kJmol^-1
- this value can be found from the frequency of a line in the atomic spectrum of caesium
- calculate the frequency of this line in THz
(1 THz = 1000GHz)
[3]
- 376/6.02x10^23 = 6.25x10^-22 kJ
- 6.25x10^-19 J
- f= E/h = 942 THz
124
explain why copper (I) compounds are not coloured [2]
- d orbitals are full in Cu+
- so electrons can’t move / no d-d transitions
125
- when blue crystals of CuSO4.5H2O are heated they form a white solid
- upon addition of water they return to their original blue colour
- explain these observations [2]
- copper ions only coloured when ligands / water binds to them
- heating removes (water) ligands and adding water replaces them
126
- the thiosulfate ions act as reducing agents
2S2O3 2- —> S4O6 2- + 2e-
a) write the equation for the reduction of iodine by thiosulfate ions [1]
b) suggest a suitable indicator for this titration, giving the colour change expected at the end-point [2]
a) 2S2O3 2- + I2 —> S4O6 2- + 2I-
b)
- starch
- blue-black to colourless
127
- the entropy of concentrated acids etc is lower than dilute acids
a level chemistry
flashcards
Decks in class (21)
# Cards
-
3.1 redox and standard electrode potential74
-
3.2 redox reactions30
-
3.3 chemistry of the p-block102
-
3.4 chemistry of the d-block transition metals132
-
3.5 chemical kinetics53
-
3.6 enthalpy changes for solids and solutions53
-
3.7 entropy and feasibility of reactions41
-
3.8 equilibrium constants26
-
3.9 acid-base equilibria62
-
4.1 stereoisomerism33
-
4.2 aromaticity37
-
4.3 alcohols and phenols93
-
4.4 aldehydes and ketones72
-
4.5 carboxylic acids and their derivatives82
-
4.6 amines87
-
4.7 amino acids, peptides and proteins51
-
4.8 organic synthesis and analysis145
-
equations37
-
unit 5 practical207
-
unit 3127
-
unit 4117
