2011 JC2 H1 Chemistry Prelim Examination Paper 2

Meridian Junior College

2011 JC2 H1 Chemistry Prelim Examination Paper 2

Section A (Suggested Answers)

Answer all the questions in this section in the spaces provided.

1(a)Define the term relative atomic mass.

[1]

Relative atomic mass of an element is the average mass of its atom in the isotopic mixture, compared to 1/12 the mass of an atom of carbon-12.

(b)Chlorine consists of two stable isotopes: 35Cl and 37Cl.The percentage abundance of these two isotopes are 75.78% and 24.22% respectively.

Calculate the relative atomic mass of chlorine to four significant figures.

[1]

Ar = = 35.48

(c)Chlorine reacts with aqueous sodium hydroxide differently in hot and cold conditions. The equation for the reaction under hot conditions is given below.

3Cl2 + 6NaOH  5NaCl + NaClO3 + 3H2O

(i)Explain, in terms of changes in oxidation number, why this is a redox reaction.

The oxidation number of chlorine decreased from 0 Cl2 to –1 in NaCland increased from 0 in Cl2 to +5 in NaClO3.

(ii)What is unusual about this reaction?

[2]

Chlorine has undergone disproportionation.

(d)Explain the relative values of the first ionisation energies of sulfur and chlorine.

[3]

From S to Cl, increase in nuclear charge outweighs the negligibleincrease in shielding effect(sincevalence electrons in the same outer shell are ineffective in shielding one another ).

Effective nuclear chargeincreaseshence stronger electrostatic force of attractionbetween nucleus and valence electrons and more energy is required to remove the valence electron from Cl.

(e)The chlorides of the elements sodium to phosphorus all dissolve in or react with water.

(i)Describe, with the aid of an equation, the reaction between sodium and chlorine.

Na + ½ Cl2 NaCl

A yellow flame will be seen.

(ii)Describe, with the aid of an equation, the reaction beween silicon tetrachloride and water and state the pH of the solution formed.

[4]

SiCl4 + 2H2O  SiO2 + 4HCl

White fumes are formed.

pH of solution = 1

[Total: 11]

2Alkenes are very useful compounds and can be used as fuels and in the manufacture of a wide variety of polymers. The following reactions involve the formation of some alkenes.

only 1 product formed

(a)State the conditions needed to form A and B and the type of reaction.

Conditions:UV light

Type of reaction:Substitution

[1]

(b)Suggest why the experimental yields of A and B will be poor.

[1]

Due to multiple substitution of H atoms or due to random substitution of H atoms (hence giving another product).

(c)Suggest a structural formula for A and B.

[2]

A:B:

(d)The flow chart below involves the reaction of pent-4-en-1-ol.

(i)Draw the structural formulae for compounds C and D.

C:D:

(ii)State the reagents and conditions for Step I – IV in the spaces provided.

[6]

[Total: 10]

3(a)Define the term standard enthalpy change of neutralisation.

[1]

It is the energyreleased when an acid and a base react to form onemole of water at 298K and 1 atm.

(b)The table below shows data regarding the neutralisation reactions of various acids and bases.

(i)Explain as fully as you can the above data.

The first two reactions involve strong acids and strong bases and thus release the same amount of energy.

The last reaction involves a weak acid and strong base and since the weak acid only partially dissociates, some of the energy released during neutralisation is absorbed to further dissociate it, thus the enthalpy change of neutralization is less exothermic.

(ii)25.0 cm3 of a 1.00 mol dm-3 solution of HNO3 was added to
50.0 cm3 of a 1.00 mol dm-3 solution of KOH. Calculate the increase in temperature of the solution.

[5]

HNO3 KOH hence KOH is in excess.

No. of moles of H2O formed = no. of moles of HNO3 = 1 x 0.025

= 0.0250

| Hn x nH2O | = mcT

| (–57 x 1000) x 0.0250 | = (25 + 50) x 4.18 x T

T = 4.55C

(d)The hydrogen halides produce acidic solutions when dissolved in water.

Dilute solutions of HF behave as weak acids whereas dilute solutions of HCl, HBr, and HI behave as strong acids.

(i)Explain what is meant by aweak acid.

A weak acid dissociates partially and donates protons to another substance.

(ii)By considering relevant data from the Data Booklet, suggest an explanation for the difference in acidity between HF and the rest of the hydrogen halides.

[3]

Since an acid donates protons, in order for the hydrogen halides to do so, the H–F bond must be broken. Due to the very high bond energy of the H – F bond, it is hard for the bond to breakand thus it only dissociates partially.

Bond energy of H–F = 562 kJ mol-1; H–Cl = 431 kJ mol-1

H–Br = 366 kJ mol-1; H–I = 299 kJ mol-1

[Total: 9]

4Oceans play an essential part in the cycling of many chemicals throughout Earth as approximately 70% of the planet’s surface is covered by sea. The oceans have a surface area of about 3.6 x 108 km2 and an average depth of 3.7 km and they transport enormous quantities of energy and materials.

In fact, gold compounds are present in solution but their concentration is very low (estimated to be at most 1 x 10-11 g dm-3) and Fritz Haber,a brilliant German chemist, wanted to extract gold from the sea so that his country’s war debts could be paid off.

When water evaporates, intermolecular attractions must be overcome and this requires energy. The enthalpy change of vaporisation, Hvap, is a measure of this energy. The Hvap of water, propanone, and hexane are +2260, +520, and +330 kJ kg-1 respectively. This is important because if our present rain were replaced by the same mass of ‘propanone-rain’, it would require four times more rain to fall in order to keep the temperature of the environment to be the same!

Oceans also play an important part in absorbing the excess carbon dioxide we release into the atmosphere from the combustion of fuels. This is done through a series of equilibrium reactions as shown:

CO2(g)CO2(aq)

CO2(aq) + H2O(l) H+(aq) + HCO3-(l)

HCO3-(aq)H+(aq) + CO32-(aq)

Due to these reactions, water in the oceans produce a mixture containing mainly HCO3- ions and CO32- ions.

Many marine organisms build protective shells composed of insoluble calcium carbonate and thus chemistry and marine life together make up a very efficient CO2 removal system and help keep the composition of our atmosphere constant.

(a)(i)Calculate the estimated volume of the oceans in km3.

Volume of oceans = (3.6 x 108) x 3.7 = 1.33 x 109 km3

(ii)What is this estimated volume in dm3?

1 km = 1000 m = 10000 dm

1 km3 = (10000)3 dm3 = 1012 dm3

Volume of oceans = (1.33 x 109) x 1012 = 1.33 x 1021 dm3

(iii)Calculate the estimated maximum value for the total mass of gold thought to be present in the oceans today.

[3]

Maximum total mass of gold = (1.33 x 1021) x 10-11

= 1.33 x 1010 g

(b)Account for the differences in the Hvap between propanone and
hexane.

[2]

Both have simple molecular structures but propanone has stronger permanent dipole – permanent dipole attractions between its molecules and hexane has weaker induced dipole – induced dipole attractions between its molecules hence more energy is needed to overcome the stronger intermolecular forces of propanone, hence it has a higher Hvap than hexane.

(c)(i)The water in oceans can be considered to be an acidic buffer solution. Explain what is meant by an acidic buffer solution and write an equation to show what would happen when acid is added to the buffer solution formed from the water in oceans.

It is a solution containing a weak acid and its conjugate base and it can resist pH change when SMALL amounts of acid or base are added to it.

CO32- + H+ HCO3-

(ii)Suggest, with the help of the relevant reactions described in the passage, how the actions of certain marine organisms can cause the oceans to continuously remove carbon dioxide from the atmosphere.

[5]

Because the marine organisms build shells made of insoluble calcium carbonate, this decreases the [CO32-]and hence by Le Chatelier’s Principle, the equilibrium position of the third reaction will shift right to increase [CO32-].

In doing so, the [HCO3-] decreases and hence equilibrium position of the second reaction will shift right to increase [HCO3-].

Hence [CO2(aq)] decreases and equilibrium position of the first reaction will shift right to increase [CO2(aq)] and thus removing CO2(g) from the atmosphere (since [CO2(g)] decreases).

[Total: 10]

1

Section B

Answer ALLquestions from this section on separate answer paper.

5Fritz Haber was a German chemist who received the Nobel Prize in Chemistry in 1918 for developing the Haber Process, which was important in helping to make explosives and kept the German army well supplied with ammunition during World War I. Despite Haber's high standing in German science, he was forced by the Nazis to leave the country after Adolf Hitler seized power due to his Jewish background.

The Haber Process involves the following equilibrium:

N2 (g) + 3H2 (g)2NH3 (g) ∆H = –92 kJ mol-1

(a)(i)State the temperature and pressure used in the Haber Process.

Moderate temperature of450ºC

High pressure of250 atm

(ii)Explain why the conditions you stated in (a)(i) are used.

[4]

By Le Chatelier’s Principle, to increase yield of NH3,

(1) a lower temperature would be preferred because a decrease in temperature will cause the equilibrium position to shift right towards the exothermic reaction to release heat

(2) a higher pressurewould be preferred because equilibrium position will shift right to reduce the number of moles of gas to decrease pressure.

However, if temperature is too low, the rate of reaction will be very slow hence this is not feasibleandif pressure is too high, thicker pipes need to be built to withstand higher pressure, leading tohigher costs.

(b)A mixture containing 0.5 moles each of N2 and H2 is allowed to reach equilibrium in a 3 dm3 vessel. The graph below shows how the number of moles of N2 and NH3 varies with time.

(i)Write an expression for the Kc of the reaction, stating its units.

Kc = units: mol-2 dm6

(ii)Use the given data to calculate a value for the Kc of the reaction.

[2]

From graph, at equilibrium,

no. of moles of N2 = 0.35 and no. of moles of NH3 = 0.3

Hence no. of moles of H2 = 0.5 – 0.45 = 0.05

Kc ==

=1.85 x 104 mol-2 dm6

(c)Ammonia is a weak base. A chemist starts with a 25 cm3 solution of 0.0010 mol dm-3 hydrochloric acid and he wishes to make a buffer solution by adding 25 cm3 of 0.0020 mol dm-3 of ammonia.

(i)Calculate the initial pH of the hydrochloric acid solution.

pH = – lg [0.0010] = 3.00

(ii)Draw asketch graph to show how the pH changes during this titration.Mark clearly where the end point occurs.

No. of moles of HCl = 0.001 x 0.025 = 2.50 x 10-5

No. of moles of NH3 needed = 2.50 x 10-5

Volume of NH3 needed to neutralise = = 12.5 cm3

(iii)Choose a suitable indicator from the table below for the above titration and explain your choice.

Aurin can be used for this strong acid–weak base titration since the end-point is within the pH range of the colour change of aurinorthe pH transition range of aurin lieswithin the range of rapid pH change (around 3 – 7) over the equivalence point.

(iv)Draw a dot-and-cross diagram for the salt produced during the above titration.

The salt produced is NH4+Cl-

(v)Would the salt produced or ammonia have a higher boiling point? Explain your reasoning.

[8]

The salt has a giant ionic structure with strong electrostatic forces of attraction between the oppositely charged ions and NH3has a simple molecular structure with weaker intermolecular hydrogen bonding.

Hence more energy is needed to overcome the forces in the saltand the salt would have a higher boiling point.

(d)Compound E, C4H9Cl, reacts with hot aqueous NaOH to produce F, which gives a yellow precipitate when heated with iodine in aqueous NaOH. When E is heated with ammonia in a sealed container, compound G, C4H11N, is produced.

(i)Give the structures forE,F, andG and explain your reasoning clearly.

Eundergoes substitutionto give F.

E is a halogenoalkane and F is analcohol.

F undergoes oxidationwith hot iodine in NaOH(aq).

F has a group.

E undergoes substitutionto give G.

G is an amine.

F:

E:

G:

(ii)Write a balanced equation for the reaction of F when heated with iodine in aqueous NaOH.

[6]

CH3CH2CH(OH)CH3 + 4I2 + 6NaOH

CH3CH2COONa + CHI3 + 5NaI + 5H2O

[Total: 20]

6Pollution is the introduction of contaminants into a natural environment that causes harm to the ecosystem.

(a)Sulfur hexafluoride, SF6, is the most potent green house gas. SF6 gas must be removed during repairs since in the presence of sparks and oxygen, SF6 decomposes to give toxic sulfur oxyfluorides and sulfur tetrafluoride, SF4.

(i)Draw the structure of a molecule of sulfur tetrafluoride, SF4, showing the shape clearly.

(ii)State and explain the shape around the central atom in SF4.

There are 4 bond pairs and 1 lone pairaround the S atom.

To minimise repulsion, the 5 electron pairs are directed to the corners of a square pyramid.

Bond pair–lone pair repulsion > bond pair–bond pair repulsion

Hence shape is square pyramidal.

(iii)Sulfur hexafluoride, SF6, is sparingly soluble in water. Draw a diagram to show how SF6 interacts with water.

[5]

(b)Since the Industrial Revolution, emissions of sulfur dioxide and nitrogen oxides to the atmosphere have increased.The increase in sulfur dioxide results in production of acid rain which contains sulfuric acid.

(i)Assuming there is an average of 156 tonnes of rainfall annually and H2SO4 is present in 1.5 parts per million (ppm) of the annual rainfall, calculate the concentration of H2SO4, in mol dm-3, present in the annual rainfall.

[Given 1 ton = 1.0 × 106 g, Density of acid = 1.0 g cm-3]

Mass of H2SO4 = = 234 g

No. of moles of H2SO4 (in 156 x 106 cm3 of water) = = 2.385

[H2SO4] = x 1000 = 1.53 × 10-5 mol dm-3

(ii)A statue made of calcium carbonate is placed in the open. Assuming 0.1% of the annual rainfall came into contact with the statue, calculate the loss in mass of the statue in that year.

[4]

CaCO3 + H2SO4 CaSO4 + CO2 + H2O

No. of moles of H2SO4 that came into contact with statue

= × 2.385 = 2.385 × 10-3

CaCO3 ≡H2SO4

Mass of statue lost = (2.385 × 10-3) × 100.1 = 0.239 g

(c)The pH of soil affects the growth of plants and farmers add aluminium sulfate to decrease the pH of basic soil.

(i)Explain, with the aid of relevant equations, why an aqueous solution of aluminium sulfate is acidic.

Al2(SO4)3 undergoes hydration and slight hydrolysisin water.

Al2(SO4)3 (s) + 12 H2O (l)  2[Al(H2O)6]3+ (aq) + 3 SO42- (aq)

[Al(H2O)6]3+(aq)[Al(H2O)5(OH)]2+(aq) + H+(aq)

The high charge density of the Al3+ ion enables it to attract electrons away from one of its surrounding water molecules, thereby polarising and weakening the O–H bond which results in the release of a proton from one of the water molecules.

Soilpollution can occur when substances such as pesticides used by farmers leach into the soil.Pyrethra I and 1,3-indandione are two such pesticides and their structures are shown below.

(ii)State the number of geometrical isomers present in Pyretha I.

2

(iii)State the reagents and conditions you would use to distinguishPyrethra I and 1,3-indandione.

Br2 in CCl4, r.t.p., in the dark

(iv)Suggest a 2-step synthesis of Compound H starting from 1,3-indandione stating clearly the intermediates, reagents and conditions for each step.

Compound H

Step 1:LiAlH4 in dry ether, r.t.p.

Intermediate compound is

Step 2:PCl5, r.t.p.

(v)Draw the structural formulae of the compounds formed when Pyrethra I is treated with:

IH2 and Pt catalyst

IINaOH(aq), heat

[11]

and

[Total: 20]

7Nitrogen is an important component of our atmosphere as it makes up 78% by volume of Earth’s atmosphere. It is generally inert under standard conditions, causing difficulty for both organisms and industry to convert nitrogen into useful compounds.

(a)Suggest why nitrogen is generally inert under standard conditions.

[1]

A lot of energy is needed to break the strong NN bondhence there may not be enough energy under standard conditions to react.

(b)Describe the bonding that occurs in nitrogen in terms of the overlap
of orbitals. Draw diagrams to illustrate your answer.

[3]

The NN bond is made up of 1 sigma bond formed via the head on overlap of orbitals (sp hybridised)and 2 pi bonds formed via the side-on / sideways overlap of p orbitals.

Diagram showing head-on overlap:

Diagram showing side-on overlap:

(c)Nitric oxide, NO, can be formed from nitrogen, and the reaction between nitric oxide and oxygen is an important intermediate step in the industrial production of nitric acid. The equation for the reaction is given below.

2NO(g) + O2(g)  2NO2(g)

Given that the rate equation for the above reaction is rate = k[NO]2[O2] and that the reaction is endothermic,

(i)State the units of the rate constant, k.

mol-2 dm6 s-1

(ii)Sketch a graph of rate against concentration for:

INO

IIO2

(iii)Explain briefly why the rate of reaction would be expected to increase by increasing pressure.

[5]

Increasing pressure causes the number of gaseous reactant particles per unit volume to increase.

Hence, frequency of EFFECTIVE collisions increases and since rate of reaction frequency of EFFECTIVE collisions, rate of reaction increases.

(d)(i)Define bond energy.

Bond energy of X–Y bond is the average energy absorbed to break 1 mole of X–Y bonds in the gaseous state under standard conditions to form X and Y gaseous atoms.

(ii)Hydrazine, H2N–NH2(g), is another compound made from nitrogen and has been used as a rocket fuel because it reacts highlyexothermically with oxygen to form gaseous nitrogen and water vapour.

Use relevant data from the Data Booklet to calculate a value for theenthalpy change of the reaction.

[4]

N2H4 (g) + O2 (g)  N2 (g) + 2H2O (g)

H = [4BE(N–H) + BE(N–N) + BE(O=O)] – [BE(NN) + 4 BE(O–H)]

= [4(390) + 160 + 496] – [994 + 4(460)]

= – 618 kJ mol-1

(e)2,4-dintrophenylhydrazine is a derivative of hydrazine and has an important use in the field of organic chemistry.

(i)State the functional group that 2,4-dinitrophenyldrazine is used to test and the observations that could be made.

Carbonyl functional group;Orange ppt

Compound J has a molecular formula of C5H10O and gives a positive test with 2,4-dinitrophenylhydrazine. Compound J also gives a silver mirror with Tollens’ reagent.

(ii)State the number of structural isomers that can be compound Jbased on the above information and draw all these isomers.

4

(iii)CompoundJreacts with sodium borohydride in ethanol to give compound K, which does not react with hot concentrated sulfuric acid.

Suggest the structural formulae for compoundsJandK, explaining clearly your reasons.

[7]

J undergoes reduction with NaBH4 in ethanol to give an alcoholKwhich does not undergo eliminationwith conc. H2SO4.

 There must be no H atom bonded to the C atom adjacent to the C atom of the alcohol.

J:K:

[Total: 20]

1