1 Aqueous copper(II) sulphate contains [Cu(H2O)6]2+ ions. Aqueous ammonia is added drop by drop to a small volume of aqueous copper(II) sulphate. Two reactions take place, one after the other, as shown in the equations.

[Cu(H2O)6]2+(aq) + 2OH–(aq) Cu(OH)2(s) + 6H2O(l)

Cu(OH)2(s) + 2H2O(l) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2OH–(aq)

(a) Describe the observations that would be made as ammonia is added drop by drop until it is in an excess.

When a few drops of NH3(aq) is added, a pale blue ppt of Cu(OH)2 is formed. Upon addition of excess NH3(aq), Cu(OH)2 dissolves to give a deep blue solution of Cu(NH3)4(H2O)2

2+ since IP falls below Ksp due to a fall in [Cu2+]..................................................................................................................................................

(b) Draw the shape for the [Cu(H2O)6]2+ ion. Include the bond angles in your diagram.

Octahedral, 90o

Show dative bond in diagram

(c) Water is a simple molecule. The H—O—H bond angle in an isolated water molecule is104.5°.The diagram shows part of the [Cu(H2O)6]2+ ion and the H—O—H bond angle in the water ligand.

Explain why the H—O—H bond angle in the water ligand is 107° rather than 104.5°.

Water molecule 2 lone pairs and 2 bond pairs.Water ligand 1 lone pair and 3 bond pairs / lone pair is now a bond pair / water has one less lone pair when it is a ligand Lone pairs repel more than bond pairs .................................................................................................................................................

L_A_Level Chemistry 2815 2005_01 Q3

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Class Reg NumberAnswer Scheme

Candidate Name .......................................................................

Chemistry H2 9746Tutor TuteeRevision Exercise 22: Integrated Questions

2 One cause of low-level smog is the reaction of ozone, O3, with ethene. The smog containsmethanal, CH2O(g), and the equation for its production is shown below.

O3(g) + C2H4(g) 2CH2O(g) + ½O2(g) equation 2.1

(a) The rate of the reaction doubles when the initial concentration of either O3(g) or C2H4(g) is doubled.

(i) State the order of reaction with respect to

O3 ............1...........

C2H4 ............1...........

(ii) Write the rate equation for this reaction.

Rate = k[O3][C2H4].................................................................................................................................................

(b) For an initial concentration of ozone of 0.50 × 10–7 mol dm–3 and one of ethene of 1.0 × 10-8

mol dm–3, the initial rate of methanal formation was 1.0 × 10–12 mol dm–3 s–1.

(i) How could the initial rate of methanal formation be measured from a concentration/time graph?

Draw a tangent at t=0 on concentration-time graph. Note assumption that initial part of curve is almost linear. .................................................................................................................................................

(ii) Calculate the value of the rate constant and state the units.

(iii) The initial rate of methanal formation is different from that of oxygen formation in equation 2.1. Explain why.

2 mol CH2O forms for every 0.5 mol O2 / stoichiometry of CH2O : O2 is not 1:1.................................................................................................................................................

(iv) The experiment was repeated but at a higher temperature. What would be the effect of this change on the rate and the rate constant of the reaction?

Rate increases, k increases. Reason: There is an increase in proportion of reactant molecules having energy greater than or equal to activation energy and hence an increase in number of effective collisions per unit time. From Arrhenius’ equation of k = Ae-Ea/RT, when temperature increase, k increases, so does rate of reaction..................................................................................................................................................

2

(c) In the stratosphere, ozone forms when oxygen free radicals react with oxygen molecules.

O2 + O O3

The oxygen free radicals are initially formed as diradicals when oxygen gas, O2, is dissociated by strong ultraviolet radiation,

O2(g) 2O(g)

(i) Suggest why oxygen free radicals, O, are often called diradicals.

Each oxygen atom has two unpaired electrons..................................................................................................................................................

(ii) Draw a ‘dot-and-cross’ diagram of an ozone molecule. Show outer electrons only.

2 oxygen atoms bonded by double bondthird oxygen bonded by a covalent bond and outer shells

(iii) Chlorine free radicals formed from CFCs deplete the ozone layer in a chain reaction. Typically, 1 g of chlorine free radicals destroys 150 kg of ozone during the atmospheric lifetime of the chlorine free radical (one to two years).

Calculate how many ozone molecules are destroyed in this chain reaction by a single chlorine free radical before the free radical is destroyed.

No of moles of O3 in 150 kg = 150 x 103/48 = 3.13 x 103 mol No of moles of Cl radicals in 1 g = 1 /35.5 = 2.82 x 10−2 mol

1 mol Cl destroys 3.13 x 103/2.82 x 10−2 = 1.11 x 105 mol O3

1 Cl radical destroys 1.11 x 105 O3 molecules

L_A_Level_Chemistry 2816_01 Jan 05 Q23 This question is about haemoglobin and the photosynthetic pigment chlorophyll-a. The

structure of chlorophyll-a is shown below.

3

(a)(i) State one way in which the structure of chlorophyll-a is similar to that of haemoglobin.

Metal atom/ion bound in porphyrin/haem-like ring structure/complex.................................................................................................................................................

(ii) Suggest one way in which the structure of chlorophyll-a differs from that of haemoglobin.

Mg (instead of Fe)/ not protein/other difference.................................................................................................................................................

(b) Suggest the formula of one alcohol that would be released by hydrolysis of an ester group in chlorophyll-a.

CH3OH or C20H39OH.................................................................................................................................................

(c) Describe the quaternary structure of haemoglobin.

Four protein/polypeptide chains (2 alpha and 2 beta), each with a haem group, group together to form full protein.................................................................................................................................................

L_A_Level_Chemistry 2815_02 Q34 Unwanted metals can find their way into food, either from the soil or from machinery during

harvesting and processing. Ions of metals such as copper, iron and nickel catalyse the breakdown of fats in food.

(a) Copper, iron and nickel are transition metals. What property of transition metals allows them to behave as catalysts?

Ability to exist in different OS / Availability of 3d and 4s orbitals .................................................................................................................................................

(b) A class of food additives known as sequestrants can be used to remove these metal ions.

Typical sequestrants are the calcium and sodium salts of H4 edta.

The structure of the edta4– ion is shown below.

The edta4– ion acts as a hexadentate ligand. On the structure above, label with an asterisk (*) the six sites that form bonds to the metal ion.

(c) Underline the term below that describes the shape of the complex formed between a nickel(II) ion and an edta4– ion.

linear octahedral square planar tetrahedral

(d) The reaction of edta4– ions with nickel(II) ions can be represented by the following equation.

[Ni(H2O)6]2+(aq) + edta4– (aq) ⇌ [Ni(edta)]2– (aq) + 6H2O(l)

4

(i) Write an expression for the equilibrium constant, K, for the reaction.

K = [Ni(edta)2-] / [Ni(H2O)62+][edta4-]

(ii) Equilibrium constants for ligand exchange reactions are usually called stability constants, Kstab. The value of Kstab for the above reaction is 2.00 × 1019 mol–1 dm3 at 25 °C.

What does this tell you about the position of equilibrium at 25 °C? Explain your reasoning.

.................................................................................................................................................

.................................................................................................................................................

.................................................................................................................................................

(iii) The forward reaction in the above equilibrium is slightly exothermic. At temperatures higher than 25°C, edta4– ions sequester nickel(II) ions less effectively. Explain why this is so.

.................................................................................................................................................

.................................................................................................................................................

.................................................................................................................................................

(e) The amount of nickel(II) ion in a solution can be found by titrating with a solution of edta4– of known concentration using a suitable indicator.

25.0 cm3 of a solution containing nickel(II) ions reacted with exactly 22.0 cm3 of a 0.100 mol dm–3 solution of edta4– ions.

Calculate the concentration of nickel(II) ions in the solution.

L_A_Level Chemistry 2849 Jan 05 Q5 5 In your answers to the questions that follow, show all of your working.

At room temperature and pressure, r.t.p., 1 mol of gas molecules has a volume of 24 dm3. Whilst digging his garden, a chemistry student found what appeared to be a piece of bronze, possibly from the Bronze Age. The student knew that bronze was an alloy of copper with other metals including tin. He carried out three experiments on samples of the bronze.

5

(a) Experiment 1

He dissolved a small piece of the bronze, weighing 0.28 g, in concentrated (16 mol dm–3) nitric acid, HNO3. 5 cm3 of a blue solution C containing Cu2+ ions was formed together with a brown gas with the molecular formula NO2.

Equation 5.1 represents the equation for the reaction between copper and concentrated nitric acid.

Cu + 4H+ + 2NO3– Cu2+ + 2NO2 + 2H2O equation 4.1

The student analysed the blue colour from the Cu2+ ions in solution C using a colorimeter. He found out that the concentration of Cu2+ ions in solution C was 0.68 mol dm–3.

The student concentrated the solution and obtained some blue crystals of a compound A with a percentage composition by mass of Cu, 26.29%; N, 11.60%; O, 59.63%; H, 2.48%. This composition included 3 waters of crystallisation.

(i) Calculate the percentage of copper in the bronze relic.

moles of Cu = 0.68 x 5/1000 = 0.0034

mass of Cu = 0.0034 x 63.5 = 0.216 g

% Cu = 0.216/0.28 = 77.1 %

(ii) Calculate the empirical formula of A.

ratios:Cu = 26.29/63.5 = 0.41 1N= 11.6/14 = 0.83 OR 2O = 59.63/16 = 3.73 9H= 2.48/1 = 2.48 6

empirical formula = CuN2O9H6

Cu(NO3)2.3H2O

(b) Experiment 2

The student dissolved another small piece of the bronze relic in dilute (8 mol dm–3) nitric acid. A blue solution containing Cu2+ ions was again formed but this time a colourless gas was produced with the molecular formula NO. Equation 5.2 represents the unbalanced equation for this second reaction.

Cu + H+ + NO3– → Cu2+ + NO + H2O equation 5.2

By considering oxidation numbers, balance equation 5.2.

3Cu + 8H+ + 2NO3– 3Cu2+ + 2NO + 4H2O

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(c) Experiment 3

The student heated a third small piece of the bronze relic with concentrated sulphuric acid. The copper in the bronze relic reacted to produce a blue solution and 90 cm3 of a gas B, measured at r.t.p.. The mass of the gas B collected was 0.24 g.

(i) Suggest a possible identity of gas B.

moles of A = 90/24000 = 3.75 x 10–3

Mr of A = 0.24/ 3.75 x 10–3 = 64

Gas is SO2

(ii) Suggest a likely balanced equation for this reaction.

Cu + 2H2SO4 CuSO4 + SO2 + 2H2O /Cu + 4H+ + SO4

2– Cu2+ + SO2 + 2H2O /Cu + 3H+ + HSO4

– Cu2+ + SO2 + 2H2O.................................................................................................................................................

L_A_Level Chemistry 2816_01 Jun 056 The scheme below shows the final stages in the synthesis of cypermethrin, which is widely

used today as an insecticide.

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(a)

(i) Complete the structure of the organic compound C in the box above.

Draw in acyl chloride group

(ii) Name the functional group that is formed in stage 3.

Ester.................................................................................................................................................

(iii) Identify suitable reagents to carry out stage 2.

HCN, trace NaCN.................................................................................................................................................

(iv) State the type of chemical reaction occurring in stage 2.

Nucleophilic addition.................................................................................................................................................

(b) One company manufacturing the insecticide uses an enzyme to catalyse the reaction in stage 2. Cypermethrin made by this method can be used in smaller doses than cypermethrin made by the normal laboratory conditions.

8

Suggest an explanation.

Since the synthesis had created a chiral centre, the laboratory method would give a 50/50 mixture of optical isomers, compared to the stereospecific enzyme-catalysed reaction. Hence, only one of the optical isomers would be active as the insecticide..................................................................................................................................................

(c) Cypermethrin does not remain on crops because it is broken down in the environment by hydrolysis. It can also be rapidly hydrolysed in the laboratory using an acid catalyst.

(i) State how you would carry out acid hydrolysis of cypermethrin in the laboratory.

Dilute HCl, reflux.................................................................................................................................................

(ii) Suggest the structures of two organic compounds that could be produced by acid hydrolysis of cypermethrin.

Note acid hydrolysis of nitrile group too

A_GCE_Chemistry 2814_01 Jun 07

7 Compounds A and B, which are isomers with molecular formula C6H12O4, react readily with potassium carbonate under room conditions to give a colourless gas. Reaction of A and B with hot acidified potassium manganate (VII) yields products C and D respectively.

C is optically inactive and gives an orange precipitate with 2,4-dinitrophenylhydrazine. With alkaline aqueous iodine followed by acidification, C gives compound E, C5H6O5, which is a symmetrical molecule.

D, another optically inactive molecule, has molecular formula C6H10O5. D reacts with excess phosphorus pentachloride to form F with the evolution of vigorous white fumes. F reacts with ethane-1,2-diol to give a compound G, a diester.

Deduce the structures of compounds A to F, explaining your reasoning.

8 Hydrocarbon U (Mr = 120) was isolated. Upon irradiation of U with uv light in the presence of bromine gas, a product V is isolated. Upon heating with aqueous potassium hydroxide, V yields a compound W, which is sparingly soluble in water.

When W reacts with hot acidified potassium dichromate, compound X was formed, which

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gives an orange precipitate with 2,4-dinitrophenylhydrazine. X produces a silver mirror upon reaction with Tollen’s reagent, but does not give any precipitate when heated with Fehling’s solution. X also gives a yellow precipitate with a characteristic smell when heated with alkaline aqueous iodine.

However, when W reacts with hot acidified potassium manganate (VII), compound Y is produced. Y readily dissolves in sodium hydroxide and is isomeric with benzene-1,4-dicarboxylic acid. Strong heating of Y under certain conditions causes each molecule of Y to lose one water molecule to form the structure below.

Deduce the structures of compounds U to Y, explaining your reasoning.

9 Various hormones are involved in the regulation of the developmental transition stages of insects. One such hormone, Precor, has the molecular formula C19H34O3. The following partial structure of Precor is known:

In refluxing Precor with dilute H2SO4, two compounds, S, C16H28O3, and T, C3H8O, are formed. T gives a yellow precipitate when warmed with alkaline aqueous iodine. On heating with acidified KMnO4, one mole of S forms one mole of U, C11H22O3, one mole of V, C3H4O, and 2 moles of colourless gas X, which is non-polar but capable of permanent dipole-dipole interaction when its molecules are in the correct orientation and in the solid state. X undergoes hydrolysis to form a weak acid. U reacts with T when refluxed under acidic conditions to form W, C14H28O3.

(a) Draw the structures of Precor, S, T, U, V, W and X.

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(b) Write all relevant balanced equations and explain all chemical reactions involved.

(c) Draw the mechanism for the reaction of Precor to produce S and T.

(d) Draw the mechanism for the hydrolysis of X.

11 Ammonia is manufactured by direct synthesis in the Haber Process:

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = –92 kJ mol–1

(a) Write an expression for the equilibrium constant, Kc, for this reaction and give its units.

Kc = [NH3]2 / [N2][H2]3 mol-2 dm6

(b) When 3 mol of hydrogen and 1 mol of nitrogen were allowed to reach equilibrium in a vessel of 1 dm3 capacity at 500°C and 1000 atm pressure, the equilibrium mixture contained 0.27 mol of N2, 0.81 mol of H2 and 1.46 mol of NH3.

Calculate Kc at this temperature.

Kc = 1.462 / (0.27)(0.813) = 14.9 mol-2 dm6 (to 3sf)

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(c) Predict and explain the effect of an increase in temperature on:

(i) the value of Kc;

Kc will decrease since by LCP, an increase in temperature causes equilibrium position to shift left to favour the reverse endothermic reaction to absorb the excess heat, causing Kc to decrease. .................................................................................................................................................

(ii) the rate of the forward reaction.

Rate of forward reaction will increase since an increase in temperature leads to an increase in average kinetic energy of the reactant molecules, resulting in more effective collisions per unit time, and that a larger proportion of molecules have energy greater than or equal to Ea for the reaction to occur. .................................................................................................................................................

(d) When ammonium salts are dissolved in water, the following reaction occurs.

NH4+ (aq) + H2O(l) ⇌ NH3(aq) + H3O+ (aq)

(i) Write an expression for the dissociation constant, Ka, for NH4+ (aq).

Ka = [NH3][H3O+] / [NH4+]

(ii) Calculate the pH of a solution of ammonium chloride of concentration 0.100 mol dm–3 at 298 K, the Ka value for NH4

+ being 5.62 × 10–10 mol dm–3 at this temperature.

5.62 x 10-10 = [H3O+]2 / 0.100

[H3O+] = 7.50 x 10-6 mol dm-3

pH = -lg (7.50 x 10-6) = 5.12 (to 3sf)

A2 Chemistry Equilibria Q12, 1312 This question concerns acid-base equilibria.

(a)(i) Calculate the concentration, in mol dm–3, of a solution of hydrochloric acid, HCl, which has a pH of 1.13.

[H+] = 10-1.13 = 0.0741 mol dm-3

Since HCl is a strong acid, it undergoes complete dissociation in water, hence [H+] = [HCl]

[HCl] = 0.0741 mol dm-3

(ii) Calculate the concentration, in mol dm–3, of a solution of chloric(l) acid, HOCl, which has a pH of 4.23. Chloric(l) acid is a weak acid with Ka = 3.72 × 10–8 mol dm–3

.

3.72 x 10-8 = (10-4.23)2 / [HOCl][HOCl] = 0.0932 mol dm-3

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(b) The pH of 0.100 mol dm–3 sulphuric acid is 0.98.

(i) Calculate the concentration of hydrogen ions, H+, in this solution.

pH = 0.98

[H+] = 10-0.98 = 0.105 mol dm-3

(ii) Write equations to show the two successive ionisations of sulphuric acid, H2SO4, in water.

H2SO4 HSO4- + H+

HSO4- SO4

2- + H+

.................................................................................................................................................

(iii) Suggest why the concentration of hydrogen ions is not 0.20 mol dm–3 in 0.100 mol dm–3

sulphuric acid.

Second ionisation is suppressed by first ionisation..................................................................................................................................................

(c) Many industrial organic reactions produce hydrogen chloride as an additional product. This can be oxidised to chlorine by the Deacon process:

4HCl(g) + O2(g) ⇌ 2Cl2(g) + 2H2O(g) ΔH = –115 kJ mol–1.

0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of volume 10.0 dm3 in the presence of a copper(I) chloride catalyst at 400 ºC. At equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride.

(i) Write an expression for the equilibrium constant Kc.

(ii) Calculate the value of Kc at 400 ºC.

(d) State and explain the effect, if any, on the position of equilibrium in (c) of:

(i) decreasing the temperature;

13

Equilibrium position shifts right to favour the forward exothermic reaction to increase the temperature of the system by LCP..................................................................................................................................................

(ii) decreasing the volume;

Decreasing the volume is akin to increasing pressure. Hence, equilibrium position shifts right by LCP to decrease the total pressure of the system by favouring the forward reaction which produces fewer no. of moles of gas..................................................................................................................................................

(iii) removing the catalyst.

A catalyst has no effect as it only alters the rate of the reaction not the position of equilibrium / it alters the rate of the forward and reverse reactions equally..................................................................................................................................................

A2 Chemistry13 Integrated circuits or ‘silicon chips’ are tiny electronic circuits produced within and upon a

single crystal of silicon. Most of the stages of chip manufacture involve chemical reactions many of which are ultraclean chemical processes.

(a) The starting point in chip manufacture is silicon wafers – thin, polished slices of a single crystal of silicon. Silicon chips must be coated with a protective film – usually silicon(IV) oxide (SiO2) but sometimes silicon nitride (Si3N4)

(i) Account for the difference in electrical conductivity among silicon, diamond and graphite. Explain your answer.

Diamond is an insulator; graphite is a moderate conductor and silicon is a semi-conductor .................................................................................................................................................

(ii) Suggest a reason why the name silicon(IV) oxide is preferable to silicon dioxide.

Silicon(IV) oxide is preferable to silicon dioxide as it does not imply that discrete silicon dioxide molecules (cf CO2) exist..................................................................................................................................................

(iii) Which sample will have a greater percentage by mass of silicon: 5.5 g of silicon nitride or 6.1 g of silicon (IV) oxide? Show your working clearly.

% Si in Si3N4 = (28.1 x 3) / (28.1 x 3 + 14.0 x 4) x 100% = 60.1%

Mass of Si in Si3N4 = 60.1/100 x 5.5 = 3.30 g

% Si in SiO2 = 28.1 / (28.1 + 16.0 x 2) x 100% = 46.8 %

Mass of Si in SiO2 = 46.8 / 100 x 6.1 = 2.85 g

5.5 g of Si3N4 has a greater percentage by mass of Si than 6.1 g of SiO2

(b) Standard Clean 1 (SC1) is an aqueous cleaning agent used to remove microscopic impurities from silicon chips during their manufacture. Organic contaminants can be removed by an extremely powerful oxidising mixture called piranha solution, named for its

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violent reaction with skin. It is formed by mixing pure sulphuric acid with 30% H2O2 in approximately a 4:1 ratio. Peroxomonosulphuric acid or Caro’s acid, H2SO5, is believed to be formed in high yield.

(i) Draw the displayed formula for Caro’s acid.

(ii) State the oxidation numbers of the sulphur atoms in sulphuric and Caro’s acid. Hence, state, if any, the reaction undergone when sulphuric acid is converted to Caro’s acid.

H2SO4: +6, H2SO5: +8. Oxidation..................................................................................................................................................

(iii) Write balanced half-equations showing the conversion of sulphuric acid to Caro’s acid and hydrogen peroxide to water.

H2O + H2SO4 H2SO5 + 2H+ + 2e-

H2O2 + 2H+ + 2e- 2H2O.................................................................................................................................................

(iv) Hence, write down the overall ionic equation showing the formation of Piranha solution.

H2O2 + H2SO4 H2SO5 + H2O.................................................................................................................................................

(c) Aluminium is used as a metal conductor on the surface of silicon chips due to its low electrical resistance to maximise the speed at which electrical signals are passed. However, aluminium suffers from electromigration – it diffuses into the underlying silicon film when a voltage is applied across the chip.

The corrosion of aluminium in chips is also a potential problem since aluminium is highly reactive to moist air, but luckily an oxide layer is formed almost instantaneously. This coating prevents further reaction unless exposed to a corrosion catalyst like ‘chlorine’, as shown in the equations below.

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)

AlCl3(aq) + 3H2O(l) Al(OH)3(s) + 3HCl(aq)

(i) Describe briefly what would happen if a few drops of a strong alkali are added to the aluminium connections of a silicon chip.

The aluminium will gradually dissolve, liberating hydrogen gas during the process due to complex ion formation.

2Al + 2OH- + 10H2O 2Al(OH)4(H2O)2- + 3H2

.................................................................................................................................................

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(ii) Explain why it is valid to mention that hydrogen chloride (‘chlorine’) acts as a corrosion catalyst for aluminium.

HCl is consumed during the process and is regenerated at the end of the process. .................................................................................................................................................

(iii) Corrosion of aluminium can also be caused by virtually every anion including carbonate, nitrate, sulphate and fluoride.

State the electronic feature shared by sulphate, carbonate and nitrate ions, but not by fluoride ions, explaining your reasoning by drawing appropriate structures.

Resonance / delocalisation effect .................................................................................................................................................

Chemistry Notes and Questions Chapter 20 14 The type of bonding between two elements can be rationalised and even

predicted using a van Arkel triangle. The triangle is based on electronegativity values. Difference in electronegativity is plotted along the y-axis and average electronegativity is plotted along the x-axis.

(a) What is meant by the term electronegativity?

It refers to the ability to attract bonding / valence electrons .................................................................................................................................................

(b) State and explain the trend in electronegativity

(i) across a period from left to right,

increases with increasing nuclear charge / same shielding but more protons.................................................................................................................................................

(ii) down a group.

decreases / gets smaller as electrons further from the nucleus / more shielding.................................................................................................................................................

(c) Use the electronegativity values quoted below to plot on the template van Arkel triangle below each of the following compounds. Label your points with the formulae.

(i) titanium boride, TiB2

(ii) silicon carbide, SiC(iii) gallium antimonide, GaSb

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(d) What is the type of bonding present at each of these bonding extremes, labelled A, B and C on the triangle?

A ...............metallic..............

B ...............covalent..............

C ...............ionic..............

Cambridge Pre-U Specimen Paper15 Ethyl ethanoate may be hydrolysed by treatment with aqueous sodium

hydroxide.

(a) Write an equation for this reaction, using structural formulae.

CH3CH2COOCH2CH3 + NaOH CH3CH2COO-Na+ + CH3CH2OH.................................................................................................................................................

(b) The rate equation for the reaction is found to be first order with respect to both the concentration of hydroxide ion and the concentration of the ester.

(i) Write down the rate equation for the reaction, where the rate constant is k2.

Rate = k2 [OH-][ester].................................................................................................................................................

(ii) State the units of the rate constant k2.

mol-1 dm3 s-1

.................................................................................................................................................

During the reaction, the concentrations of both the hydroxide ion and the ester change. However, if the initial concentration of the ester, [ester]0, is much larger than that of the hydroxide ion, the concentration of the ester remains essentially constant as the hydroxide ion is used up during the reaction. The

17

reaction is then said to follow pseudo-first order kinetics.

rate of reaction = k1 [OH–]

where k1 is the pseudo-first order rate constant and k1 = k2 [ester]

In a typical experiment, 10.0 cm3 of 0.5 mol dm–3 aqueous ethyl ethanoate was added to 10 cm3 of 0.005 mol dm–3 NaOH, all solutions being maintained at 0°C. The change in concentration of the hydroxide ion over the course of the reaction was followed by monitoring the change in conductivity of the solution. The data obtained is shown in the table.

(c)(i) From the data given in the table, calculate suitable values to enable you to plot a graph to obtain the pseudo-first order rate constant, k1. Enter your values in the blank column of the table. (A useful equation is provided in the Data Booklet.)

ln ([OH–]0/[OH–]t) calculated (not log10)

(ii) By plotting a suitable graph on the grid below, determine the pseudo-first order rate constant, k1.

18

graph of ln[OH-]0/[OH–]t on y-axis against time on x-axis, all points correctly plotted with straight line through origin (e.c.f. on (i))

k1 = 5.7 x 10–3 s–1

(iii) What is the concentration of ester at time t = 0?

0.5/2 = 0.25 mol dm–3

.................................................................................................................................................

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(iv) Hence, calculate the value of the second order rate constant, k2.

k2 = 5.7 x 10–3 /0.25 = 0.0228 s–1 mol–1 dm3

.................................................................................................................................................

(v) How would you expect the gradient of your graph to change if the initial concentration of the ester was doubled?

no change/stays the same/same gradient.................................................................................................................................................

End of Paper

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