2011 JC1 H2 Chemistry Promotional Examination Paper 2 Corrections SECTION A 1 In car accidents, the activation of an airbag is often able to save a life when it cushions impact on the passenger’s head. The expansion of the airbag is caused by the electronic detonation ofasodium compound,NaNx, that triggersa decomposition reaction, producing only sodium metal and nitrogen gas. A student was given 1.00 g of the sodium compound. He heated the solid on a weighing balance until a constant mass reading of 0.354 g was obtained. The gas produced was collected in a container under roomconditions. (a)
Find the empirical formula of the sodium compound, NaNx. (Assume that all products formed do not take part in further reactions.)
Comment [LHYC1]: Question is poorly done in general. Students show a lack of understanding of mole concept and ideal gas theories.
[2]
Mass / g Amount / mol
. .
Na 0.354 = 1.539 × 10-2
Simplest mole ratio 1 The empirical formula of the compound is NaN3.
(b)
N 0.646 . = 4.614 × 10-2 .
3
Write a balanced chemical equation for the decomposition and hence calculate the volume (in cm3) of nitrogen gas collected. (Assume that molar volume under room conditions is 24.0 dm3.) [2] 2NaN3 (s) → 2Na (s) + 3N2 (g) Amount of N2 = × 1.539 × 10-2 = 2.309 × 10-2mol Volume of N2 = 2.309 × 10-2 × 24.0 = 0.554 dm3 = 554 cm3
(c)
The nitrogen gas collected is compressed to halfthe volume calculated in (b) and heated from 20 oCto 40oC. Calculate the pressure exerted by the gas on the container. [2] =
Comment [LHYC2]: A common mistake made was to find the amount of N2, compare it to the amount of Na and obtain a mole ratio of 2:3, and using that mole ratio for the empirical formula. Some students simply wrote an equation assuming one mole of N2 is formed, then balanced the equation.
Comment [LHYC3]: Students commonly wrote the Na (s) as Na+ (s) instead. Comment [LHYC4]: Students found the amount of N atoms instead of the amount of N2. Comment [LHYC5]: A very common mistake was to leave the answer in dm3. Question asks for answer in cm3.
= p2 = 216000 Pa
(d)
In reality, the pressure exerted by the gas on the container was measured to be lower than the calculated value in (c).State two reasons why this discrepancy was observed. [2] The particles of nitrogen gas, a non-ideal gas, have significant volume compared to that of the container and there are significant intermolecular forces of attractions between the particlesand between the particles and the wall of the container. Total 8 marks
Comment [LHYC6]: Students gave a wide range of irrelevant answers including heat loss to surroundings, average values used, not a pure ionic compound, gas leaked from container, LCP, etc. Many students stated that the assumption for the calculations above is that the gas is ideal, but did not proceed to elaborate on how it is non-ideal. Students who caught on to the correct concepts commonly left out the key word, “particles” (“gas has significant volume compared to container” not acceptable), and the key phrase, “and between particles and the walls of the container”.
2
The table below shows some thermochemical data: ∆Hθ / kJ mol-1 Ca (s)→ Ca2+(aq) + 2e-
+347
Ca(s) + S(s) + 2O2(g) → CaSO4(s)
-1435
S(s) +2O2(g) + 2e-→ SO42-(aq)
-907
(a)
Use the data in the table above to construct an energy cycle and hence calculate the standard enthalpy change of the following reaction: Ca(NO3)2 (aq) + H2SO4 (aq) → CaSO4 (s) + 2HNO3 (aq) [3] ∆Hθrxn Ca(NO3)2 (aq) + H2SO4 (aq)
CaSO4 (s) + 2HNO3 (aq)
-907
-1435
Ca2+(aq) + 2e- + S (s) + 2O2 (g) + 2HNO3 (aq)
+347 Ca(s) + S(s) + 2O2(g) + 2HNO3 (aq) Comment [LHYC8]: Students are unable to form an energy cycle that is balanced. HNO3 and number of electrons are stumbling blocks. A common careless mistake is the leaving out of state symbols.
By Hess’ Law, ∆Hθrxn = -(-907) – (+347)+ (- 1435) = – 875 kJmol-1 (b)
(i)
Explainwhat is meant by the lattice energy of calcium sulfate. Lattice energy is the energy evolved when 1 mol of solid CaSO4 is formed from its constituent gaseous ionsOR Ca2+ (g) + SO42- (g) → CaSO4 (s)
(ii)
Comment [LHYC7]: Students who attempted this question with memory of definitions and explanations for solubility managed to obtain the latter 2 marks. Overall, surprisingly better done than expected.
Calcium sulfate is sparingly soluble in water while magnesium sulfate is soluble. Explain, in terms of bonding and structure whymagnesiumsulfate is soluble in water. [2] Magnesiumsulfate is soluble in water as favourable ion-dipole interactions can be formed, which releases energy to break the giant ionic lattice structure. Total 5 marks
Comment [LHYC9]: “energy change” was accepted in place of “energy evolved”, but students should be told to use the latter in future. Students commonly left out the word “solid” and other parts of the definition.
Comment [LHYC10]: Second marking point commonly left out. Some students cited magnesium sulphate as having “intermolecular forces of attraction”, a major misconception.
3
Ethene reacts with bromine in tetrachloromethane to form 1,2-dibromoethane (CH2BrCH2Br)as shown by the equation: CH2=CH2 + Br2→ CH2BrCH2Br To determine the orders of reaction with respect to ethene and bromine, ethene and bromine were first dissolved separately in tetrachloromethane. Various volumes of these solutions and tetrachloromethane were then mixed and the time taken for the brown colour of bromine to disappear was recorded. The results are shown in the table below:
1
Volume of ethene solution/ cm3 20
Volume of bromine solution/ cm3 20
2
12
20
8
25
3
20
10
10
25
Experiment
(a)
Volume of Time taken for tetrachloromethane/ colour of bromine cm3 to disappear/ s 0 15
4 40 20 20 t4 With reference to Experiments 1 to 3, explain why varying volumes of tetrachloromethane were used. [1] To keep the total volume constant so that the concentration of ethene or bromine used is directly proportional to the volume used. Alternatively: [ethene] α volume of ethene and [bromine] α volume of bromine.
(b)
The relationship between the rate of reaction with the time taken for the colour of bromine to V disappearandthe volume of bromine used is given as shown:Rate∝ Br2 t Using the experimental data, deduce the order of reaction with respect to ethane and bromine. [2] Comparing Experiments 1 and 3, [] [ !] " !] "
#
$
#
$
= [] [
[] []"
= [] []"
= % y ≈ 2
Order of reaction w.r.t. bromine is 2. Comparing Experiments 1 and 2,
#
$
#
$
=
[] [ !] " [] [ !] "
[] []"
= [] []"
= & x = 1
Order of reaction w.r.t. ethene is 1.
Comment [KSH11]: Question is very badly done. Students did not understand that tetrachloromethane is a solvent and that it’s role is to keep the total volume constant such that concentration is proportional to volume and mentioned that it was to determine the order of reaction wrt tetrachloromethane. In addition, many students mentioned that the total volume was constant but did not elaborate on why total volume was constant.
Comment [KSH12]: Many students who attempted the question were unable to use the given relationship to determine the order of reaction wrt ethane and bromine. Students used 1/t to calculate the rate instead. Some students concluded the order of reaction without relevant working and marks were not awarded.
(c)
Suggest a value for t4, the time taken for the brown colour of bromine to disappear in Experiment 4. [2] Comparing Experiments 1 and 4, '
#/)# *'
#/'#
$
= =
[]' [ !]' [] [ !]
Comment [KSH13]: Students did not recognise that [ethene] and [Br2] is affected by total volume and that they need to make use of the given relationship of rate ∝ VBr/t to determine the time taken.
'# # )# )#
# # +'#,+'#,
+ ,+ ,
t4= 30s (d)
The following graph shows the second ionisation energy trend of successive elements from Period 3.
(i)
With the aid of the Data Booklet , account for the decrease in 2nd ionisation energy from element B to element C. Element B is Al, electronic configuration: 1s22s22p63s23p1; Element C is Si, electronic configuration: 1s22s22p63s23p2 For 2nd IE: Al+ Al2++e ; Si+ Si2+ +e + electronic configuration of Al : 1s22s22p63s2electronic configuration of Si+: 1s22s22p63s23p1
Comment [KSH14]: Students need to identify the elements or electronic configuration correctly in order to explain the decrease in 2nd IE. Many students identified the wrong elements or gave the wrong electronic configuration and mentioned inter-electron repulsion between electrons in the same orbitals led to the decrease in 2nd IE.
The 3p electron in Si or Si+ (C or C+) is further away from the nucleus. (or the 3p1 electron experience a weaker electrostatic force of attraction from the nucleus.) Less energy is required to remove the electron resulting in the decrease of 2nd IE from B to C (ii)
Sketch the trend of electrical conductivity for element A to D. Electrical Conductivity
[3]
element A
B
C
D Total 8 marks
Comment [KSH15]: Most students were unable to sketch the correct trend of electrical conductivity. This may be because they did not identify the elements A to D and thus unable to know the electrical conductivity of the elements.
4
Nitrates are commonly used in the making of pyrotechnics. Hydroxylamine nitrate, NH3OHNO3, is one of them and it can also be used as a rocket propellant. Hydroxylamine nitrate undergoes catalytic decomposition according to the following equation: 4NH3OHNO3 (s) → 3N2O (g) + 7H2O (l) + 2HNO3 (l) The standard enthalpy change of reaction, ΔHrxnθ, is -526.6 kJ mol-1 and the standard entropy change, ΔSθ, is +180 J mol– 1K-1. (a)
Explain the significance of the sign of ΔSθ. [1]
Comment [ECH(C16]: Most students give the definition of ΔSθ instead of explaining the sign in reference to the chemical reaction in the question.
The positive sign for ΔSθ implies that there is an increase in the number of gaseous molecules/particles, ie. Δn = +3. Or A change in state from solid reactants to liquid and gaseous products. (b)
Determine the value of the standard Gibbs free energy, ΔGθ, for the reaction.
[1]
ΔGθ = (-526.6) – 298 (180/1000)= -580 kJ mol–1 (c)
Comment on the effect of high temperatures on the feasibility of the reaction. [2] The reaction will be feasible at all temperatures./ Temperature has no effect on feasibility of reaction. ΔGθ = ΔHθ - TΔSθ = -ve – (+ve)( +ve) = -ve ΔGθwill be negative at all temperatures and the reaction will be spontaneous.
(d)
Draw the dot-and cross diagram of N2O.[1]
Comment [ECH(C17]: Most students forget to convert ΔSθto kJ mol-1 K-1 and substitute temperature as ‘293 K’ and ‘ 273 K’ instead. Comment [ECH(C18]: Some students interpret as ‘kinetic feasibility’ instead of ’thermal feasibility’, hence use Ea to explain. The above reaction is not reversible so do not cite LCP! Most students compare the magnitude of ΔHθvs.TΔSθ when it is not necessary.
Comment [ECH(C19]: Do not draw circles to represent electron shelves, use other shapes (besides dots and crosses) to represent electrons and ‘→’ to represent dative bond.
(e)
The plots of pV/RT against p for one mole of an ideal gas at 300K is given below. pV/RT
Ideal Gas 1.0
p (i)
Show on the same axes how one mole of N2O gaswill behave at the same temperature of 300K.
(ii)
On the same axes, draw a curve to representN2O gas when it is cooled to 50K.
(iii)
On the same axes show how one mole of N2 gaswill behave at the same temperature of 300K. Explain your answer. [4] pV/RT
(ii) N2O (50 K) (i)N2O (300 K)
(iii) N2 (300 K)
1.0
Ideal Gas (300K)
p
Or (ii) N2O (50 K) pV/RT (i)N2O (300 K)
(iii) N2 (300 K)
1.0
p (iii)
N2O and N2 both have simple molecular structure. N2 has weaker temporary dipole-dipole interaction as compared to the permanent dipole-dipole interaction in N2O Or N2O has more electrons/ higher Mr than N2 hence experiences more extensive Van der Waals’ forces of attractions. N2moleculeswill tend towards ideal gas behaviour. Total 9 marks
Comment [ECH(C20]: Many students compare N2 at 300 K with N2O at 50 K. They should compare with N2O at 300 K since both gases are at the same temperature! N2O cannot formed hydrogen bonds. Do not compare the size of N2 and N2O in relation to the size of the container as the focus of the questions is about changes in temperature and not pressure.
5
(a)
Formic acid, also known as methanoic acid, is the simplest carboxylic acid. It occurs naturally in the venom of bee and ant stings. Formic acid dissociates in water as shown: HCOOH (aq) HCOO- (aq) + H+ (aq) In an experiment, 25.0 cm3 of 0.1 mol dm-3 formic acid was titrated against 0.05 mol dm3 of aqueous sodium hydroxide. [Ka of formic acid = 1.8 x 10-4 mol dm-3] (i)
At a certain point in the titration, a solution of maximum buffering capacity can be obtained. Calculate the pH of this point. At half-equivalence point, pH = pKa pH = -log10(1.8 x 10-4)= 3.74
(ii)
Calculate the concentration of the formic acid when 15.00 cm3 of aqueous sodium hydroxide has been added. Hence, calculate the pH of the mixture. [4] HCOOH (aq) HCOO–(aq) + H+ (aq) – + HCOO Na (aq) →HCOO–(aq) + Na+(aq)
Amount of HCOOH before mixing = × 0.1 = 2.5 x 10-3 mol
Comment [TEY21]: This part is well done. Most students are able to recognise that at maximum buffering capacity, pH = pKa. However, there are some students who think that at this point, the solution is that of a weak acid.
Comment [TEY22]: Many students do notknow that the mixture is a buffer. The weak acid is in the conical flask and as it has only been partially neutralised, some weak acid remains and some salt would have been formed. A number of students mistook the mixture to be a salt solution or a solution of weak acid.
Amount of NaOH added = × 0.05 = 7.5 x 10-4 mol HCOOH (aq) + NaOH (aq) → HCOO-Na+ (aq) + H2O (l) Amount of HCOOH reacted = 7.5 x 10-4 mol Amount of HCOO-Na+ formed = 7.5 x 10-4 mol New volume after mixing = 25 + 15 = 40 cm3 New [HCOOH] =
-. ×0 1 –-. ×0' 1 .
Comment [TEY23]: Some students did not realise that some of the formic acid has reacted with the sodium hydroxide added and used the original amount of acid / amount of acid reacted to find its concentration.
= 0.0438mol dm-3
. × 0' = 0.01875 mol dm-3 . 0 [3455 ] .6 = pKa + log10[34553]= –log10(1.8 × 10-4) + log10. =
New [HCOO-Na+] = pH
3.38
Comment [TEY24]: Note that since the question asked for boththe concentration of the weak acid and the pH of the mixture, both answers have to be calculated and left to 3 sig. fig. Comment [TEY25]: Note the formula! Some students remembered the buffer formula wrongly.
(b)
A saturated solution of iron(II) hydroxide has a pH of 8.76 at 25°C. (i) Write an expression for the solubility product of iron(II) hydroxide.
Comment [TEY26]: This is a very easy question. However, some students do not know that solubility product refers to Ksp and wrote the dissociation equation instead! Some students included the solid in the expression!
Fe2+ (aq) + 2OH- (aq)
Fe(OH)2(s) Ksp = [Fe2+][OH-]2 (ii)
Calculate the value of the Ksp of iron(II) hydroxide, stating its units. pOH = 14 – 8.76 = 5.24 pOH = -lg [OH-] [OH-] = 10-5.24 =5.754 x 10-6 mol dm-3 Let the solubility of Fe(OH)2 be x mol dm–3. Fe(OH)2(s) ∆ in [ ] :
Fe2+ (aq) + 2OH- (aq) -x +x +2x
Ksp = [Fe2+][OH-]2 =(5.754 x 10-6÷2)(5.754 x 10-6)2 = 9.53 x10-17 mol3 dm-9 (iii)
Explain, with the aid of an equation, the decrease in solubility of iron(II)hydroxide in the presence of 0.10 mol dm-3 of aqueous iron(II) sulphate. [5] Fe(OH)2 (s) Fe2+ (aq) + 2OH– (aq) ----- (1) 2+ FeSO4 (aq) →Fe (aq)+SO42– (aq) ----- (2) Due to thecommon ion effect, by Le Chatelier’s Principle, the position of equilibrium in (1) will shift left to reduce [Fe2+]. Hence, solubility of Fe(OH)2 will be reduced. Total 10 marks
TOTAL for SECTION A: 40 Marks
Comment [TEY27]: Most students know that they need to find [OH-] from the pH given. However, many students do not know that they need to divide by 2 in order to find the [Fe2+] at saturation. Comment [TEY28]: Some students forgot to write down the units which can easily earn them one mark! Comment [TEY29]: This question does not require a quantitative explanation. Students who attempted to give a quantitative explanation failed to fully understand what is happening and to correctly prove that the solubility of Fe(OH)2 is decreased in the presence of FeSO4. Comment [TEY30]: Many students did not get the mark. This was due to the following reasons: (i) no/wrong state symbols (ii) wrong type of arrow (iii) wrong /no equation. Comment [TEY31]: For the explanation, it is important for students to include that the reason the position of equilibrium will shift left is to decrease [Fe2+]. Many answers omitted this point and was hence, not awarded the mark. A number of students also missed out the word ‘position’ despite numerous reminders. Some students used Ksp to explain. Students need to take note that Ksp will not change as temperature did not change! For this examination, the explanation mark was awarded as long as the equation is written in parts (i) – (iii). However, it should be noted that in the future, the correct equation should be written together with the explanation before the explanation can be awarded any marks.
Mark Scheme for SECTION B 6
(a)
Brass is a mixture of copper and zinc. It dissolves in nitric acid to give a mixture of Cu2+ (aq) and Zn2+ (aq) ions. 3Cu (s) + 2NO3- (aq) + 8H+ (aq) → 3Cu2+ (aq) + 2NO (g) + 4H2O (l) The copper ions, Cu2+, may be analysed by means of iodide and sodium thiosulfate. The zinc ions do not react during this analysis. 1.00 g of brass was dissolved in nitric acid and after boiling off oxides of nitrogen and neutralisation, excess potassium iodide, KI, was added to the Cu2+(aq) ions and white precipitate of copper(I) iodide was formed in iodine solution, I2 (aq). The iodine formed then reacted with 0.0100 mol of sodium thiosulfate. I2 (aq) + 2S2O32- (aq) → 2I- (aq) + S4O62- (aq) Construct an equation between the copper(II) ions and iodide. Hence, calculate the percentage by mass of copper in the brass. [2] 2Cu2+ (aq) + 4I- (aq) →2CuI (s) + I2 (aq) Amt of I2 produced = 0.0100 ÷ 2 = 0.00500 mol Amt of Cu2+ = 0.005 x 2 = 0.0100 mol Amt of Cu = 0.0100 mol
Comment [GXYS32]: Most students were not able to provide the correct equation (i.e. to form Cu+ (aq), instead of CuI (s)). However, it should not be the case since a similar question is found in Redox tutorial Q5. Some students also placed Cu2+ and I- as the products formed. This shows poor understanding of the question.
Mass of Cu in sample = 0.0100 x 63.5 = 0.635 g % by mass of Cu in brass = (0.635 ÷ 1) x 100% = 63.5% (b)
The sulfur dioxide and carbon dioxide mixture was subsequently passed through an industrial scrubber to separate the two gases. The sulfur dioxide gas obtained was then passed through excess oxygen, using vanadium(V) oxide as the catalyst at a temperature of 500°C in a reactor of 2 dm3. This is the key stage in the Contact process to produce sulfuric acid. 2SO2 (g) + O2 (g) (i)
2SO3(g)
Comment [GXYS33]: No e.c.f. was given as long as overall redox equation was not justifiable.
∆H = −197 kJ mol-1
Explain the considerations which lead to the temperature of 500oC being used.
At low temperature, according to Le Chatelier’s Principle, the position of equilibrium shifts to the right to release heat, favouring the exothermic reaction. Hence, a high yield of SO3 is obtained. However, too low a temperature will cause the reaction to be too slow, which makes the process uneconomical. Thus a moderate temperature of 500oC is adopted.
Comment [ec34]: A lot of students did not take into consideration the rate of reaction. They only wrote down equilibrium considerations.
(ii)
Write an expression for Kc for this reaction. [75 ]
] [5 ]
Kc = [75 (iii)
Assuming a 95% conversion of SO2 (g) into SO3 (g) was achieved, use your expression in (b)(i) to calculate the value for Kc when 4 mol of SO2 and 3 mol of O2 were allowed to reach equilibrium at 500°C. 2SO2 (g)
+ O2 (g)
2SO3 (g)
4
3
0
-3.8
–(3.8/2) = -1.9
+3.8
0.2
1.1
3.8
Initial amount /mol Change in amount/mol Equilibrium amount / mol Kc =
(iv)
.)
#. .
8 9
= 656.4 ≈656 mol-1 dm3
Some of the SO3 (g) formed was immediately removed from the reactor once equilibrium was established. Calculate the newequilibrium amount of SO3 (g) at 500°C if the amount of O2 (g) at the new equilibrium was 1.01 mol. 2SO2 (g) + O2 (g) 2SO3 (g) Initial amount 3.8-y 0.2 1.1 /mol Change in -(2 x 0.09) -0.09 +0.18 amount/mol = -0.18 Equilibrium 0.02 1.01 3.98-y amount / mol Since temperature remains constant, Kc = 656.4 mol-1 dm3. 2
3.98 − y 2 Kc = 2 0.02 1.01 2 2 2
3.98 − y 2 656.4 = 2 ( 0.01) ( 0.505) 0.3641 = 3.98 – y y = 3.6159 SO3 at equilibrium = 3.98 – 3.6159 = 0.364 mol
Comment [ec35]: Some students cannot differentiate between initial and final amount. Therefore, they are unable to draw the correct ICE table. A lot of students did not substitute concentrations in Kc expression. Instead they substitute amount (mol). DO REMEMBER to divide amount by volume before calculating Kc.
Comment [ec36]: A lot of students cannot draw the correct ICE table. They did not minus “y” from the initial amount.
(v)
(vi)
(c)
State and explain the impact on the equilibrium yield of SO3 (g) if more vanadium(V) oxide was added to the reacting system. No change on the equilibrium yield of SO3(g). Catalyst merely speeds up the rate at which equilibrium is achieved.
Comment [ec37]: Some students has the misconception that when rate increase, yield must increase which is wrong. Rate change but equilibrium yield is the same.
State and explain the impact on the equilibrium yield of SO3 (g) if the process was carried out at the temperature of 800oC. [11] The equilibrium yield of SO3(g) will decrease. When temperature is increased, the endothermic reaction is favoured so as to absorb the excess heat and position of equilibrium shifts left in order reduce the temperature.
Sulfuric acid, H2SO4,can behave as an acid, as an oxidising agent and as a dehydrating agent and is a central substance in the chemical industry. Its principal uses include lead-acid batteries for cars and other vehicles, ore processing, fertiliser manufacturing, oil refining, wastewater processing, and chemical synthesis. (i) 50 cm3 of 1 mol dm-3 of ethanoic acid, CH3COOH (aq), was mixed together with 25 cm3 of 1 mol dm-3 of sodium hydroxide, NaOH (aq). A small amount of sulfuric acid was then added to the mixture. With an aid of an equation, comment if there is any pH change. With 50 cm3 of 1 mol dm-3 of ethanoic acid, CH3COOH (aq) and 25 cm3 of 1 mol dm-3 of sodium hydroxide, NaOH (aq) Presence of Buffer Solution. The small amount of H2SO4 will be removed by the buffer system
(ii)
(d)
H+ + CH3COO- CH3COOH Thus, pH will remain fairly constant. The boiling point of pure sulfuric acid, at 270oC, is higher than that of SCl2. Explain, in terms of structure and bonding, why the boiling point of sulfuric acid is higher than that of SCl2. [5] Both have simple molecular structure. H2SO4 has intermolecular hydrogen bonds and SCl2 has intermolecular van der Waals’forces of attractions. More energyis required to overcome the stronger hydrogen bonds in H2SO4 than the weaker van der Waals’ forces of attraction in SCl2. Hence H2SO4 has a higher boiling point. Give the symbols (showing the proton number, nucleon number and charges) of the following two particles particle Q R
protons 17 18
neutrons 17 17
electrons 20 17 [2]
34 17
3−
Q
35 18
+
R
Total 20 marks
Comment [GXYS38]: Students were generally able to identify SCl2 as a simple molecule. However, many students identified H2SO4 as a giant ionic compound due to the conception of H+ and SO42- ions present during dissociation. Students should be able to identify that H2SO4 is in fact a simple molecule since it consists of non-metal atoms only. Comment [GXYS39]: Most students were not able to identify “intermolecular H-bonds” present in H2SO4. Most mentioned about the greater extensiveness of VDW forces due to polarity, or greater number of electrons. Comment [GXYS40]: Some students identified SCl2 to consist of temporary dipole-dipole interactions, which is not true. This is because they failed to identify that SCl2 is a polar molecule. Comment [GXYS41]: Phrasing for intermolecular forces needs to be improved. Many students failed to include keywords: “intermolecular”, or “between molecules”. Some students also had misconceptions about IMF – “between atoms” was mentioned. Comment [GXYS42]: Most students were able to identify the accurate nucleon and proton number, and charges. However, some failed to read the question carefully to express the answer in symbols.
7
Tea light candles are a popular form of decoration during the Christmas season and are commonly made from alkanes. A student performed an experiment using tea light candles purchased from IKEA and The Body Shop. The calorimeter used in the experiment involving the IKEA candle was calibrated using the following formula: Q = C ΔT where Q is the heat energy transferred, C is the heat capacity of the calorimeter and ΔT is the temperature change. The student used the calibrated calorimeter to determine the standard enthalpy change of combustion,ΔHcθ,of The Body Shop candle. The information below shows the results of the experiment. Calibration of calorimeter using IKEA candle: -12 x 106 J mol-1 ΔHcθ Formula Initial mass of candle Final mass of candle Temperature change
C20H42 3.000 g 2.500 g 5 oC
Determination of ΔHcθ of The Body Shop candle: ? ΔHcθ Formula C22H46 Initial mass of candle 3.000 g Final mass of candle 2.400 g Temperature change 6 oC After the experiment, the student realised that he had forgotten to record the standard enthalpy change of combustion of The Body Shop candle. standard enthalpy change of combustion. (a) Define [1] The standard enthalpy change of combustion is the heat evolved when 1 mol of the substance is completely combusted under standard conditions.
Comment [YWC43]: Students wrote “heat” required which is wrong. Comment [YWC44]: Students left out the “standard condition” in their answer.
(b)
Using your practical knowledge, illustrate,with a clearly labelled diagram, the set-up for this experiment. You should include in your diagram the apparatus mentioned in this experiment and those commonly found in a school laboratory. [2]
Comment [YWC45]: Students did not mentioned or draw the water in the calorimeter. Students used Styrofoam cup to contain the water, not allowed as it will burn.
Thermometer
Thermometer is not included in the diagram.
Calorimeter with water
Tea Light Candle (c)
(i)
Using the instructions and the data collected from the experiment with the IKEA candle, calculate the heat capacity of the calorimeter. Leave your answer to 3 significant figures. ΔHcθ = −
Q n
-12 x 106 = −
Comment [YWC46]: Students got this formula wrong.
Q 3 − 2.5 282
Comment [YWC47]: Students did not subtract the mass.
Q = 2.1277 x 104 J Using Q = C ΔT 2.1277 x 104= C (5) C = 4255.3 J K-1= 4260 J K-1
Comment [YWC48]: Students added 273 to the temperature change factor. Comment [YWC49]: Unit is wrong.
(ii)
Hence, calculate the standard enthalpy change of combustion of The Body Shop candle. State one assumption you made in your calculations in c(i) and c(ii). [5] Q =C ΔT= 4255.32 x 6 = 25531.91 J
ΔHcθ = −
25531.91 = -13.2 x 106 J mol-1 0.60 12(22) + 46
Assumption: 100% efficiency / no heat lost to surrounding. (d)
Iron(III) oxide, Fe2O3, is produced from the oxidation of iron metal. It is one of the three main oxides of iron, the other two being FeO, which is rare and Fe3O4 which occurs naturally as the mineral magnetite. Standard enthalpy change of atomisation of iron = +414 kJ mol-1 First and second electron affinity of oxygen = +650 kJ mol-1 Standard enthalpy change of formation of iron(III) oxide = -823 kJ mol-1 (i)
With reference to the Data Booklet and given information, construct and label a BornHabercycle for the formation of iron(III) oxide from its elements. Comment [YWC50]: Energy and zero label is not present
Energy 2Fe3+ (g) + 3 O2- (g) st
3x 1 & 2 EA O
Electrons not balanced.
nd
Equations not balanced.
2Fe3+ (g) + 3 O (g) + 6e
2x 3rd IE Fe
If 1st EA and 2nd EA is used, students represented it wrongly on the Born Haber cycle.
2+
2Fe (g) + 3 O (g) +4e
2x 2nd IE Fe
2Fe+ (g) + 3 O (g) +2e
2x 1st IE Fe
2Fe (g) + 3 O (g)
3/2 x BE O
State symbols not included.
3rd IE of Iron not mentioned in the Born Haber cycle.
∆Hlatt
2Fe (g) + 3/2 O2 (g)
2x∆Hat (Fe) 2Fe (s) + 3/2 O2 (g) 0 ∆Hf Fe2O3 (s)
(ii)
Hence, use the cycle in d(i) to calculate the lattice energy of Fe2O3(s). [3] By Hess Law, [2x ΔHat (Fe)] + [3/2 BE of O] + [2 x 1st IE Fe] + [2 x 2nd IE Fe]+ [2 x 3rd IE Fe] + [3 x 1st&2nd BE of O] + [ΔHlatt (Fe2O3)] = ΔHf (Fe2O3) (2x414) + (3/2 x496) + (2x762) + (2x1560) + (2x2960) + (3x650) + ΔHlatt (Fe2O3) = - 823 ΔHlattθ(Fe2O3) = -14909 kJ mol-1 = - 1.48 x 107 J mol-1
(e)
The following mechanism illustrates a reaction between reactants A and B. Step 1: A + B ⇌ C fast Step 2: C + B → D + F slow Prove that the overall order of this reaction is three. [2] From Reaction 1: [4]
Kc = [:][;] [C] = Kc[A][B] ----- (1) From reaction 2: Rate = k[C][B] ----- (2) Since C is an intermediate, it should not appear in the rate law. Substituting (1) into (2): Rate = k{Kc[A][B]} [B] Rate = k’[A][B]2 Therefore, the overall order of reaction is three.
Comment [S51]: Some serious misconceptions shown by majority of the students resulting in zero mark awarded: 1) deduce overall order from the overall equation instead of the rate determining step (slowest step); 2)express (totally meaningless) rate equation in term of products, D/F which cannot be controlled.
(f)
With the aid of an energy profile diagram,explain the following statement: “Diamond is energetically unstable but kinetically stable.”[2] Energy
HighEa Diamond
ΔH = -ve Graphite
Reaction Pathway The process of converting diamond to graphite is an exothermic reaction/diamond has higher energy level, hence diamond is energetically unstable, but the process required high activation energy resulting in kinetic stability.
Comment [S52]: This question deals with two concepts of energetics and kinetics. 1 mark is only awarded for correct explanation per concept if clearly supported by relevant labels on the energy profile diagram. No marks were given at all for any contradiction in concepts shown and when diamond was not indicated in the diagram and explanation. 1 mark is also deducted for wrong labels of energy profile diagram: common mistake is to label x-axis as time/s.
(g)
Iodination of propanone is done in aqueous acidic solution according to the equation: CH3COCH3 (aq) + I2 (aq) → CH3COCH2I (aq) + H+ (aq) + I- (aq) The rate of reaction was studied via a colorimetric method, in which the colour intensity of iodine was measured at regular time intervals. Three separate experiments were performed, in which the initial concentrations of iodine, propanone and acid were varied in turn, the other two being kept constant. The results are shown below in graphical form: Graph 1 [I2]/mol dm-3 0.004
0.002 [CH3COCH3] = 0.4 mol dm-3 [CH3COCH3] = 0.8 mol dm-3
[CH3COCH3] = 1.2 mol dm-3
Time/min 1
2
3
4
5
6
Graph 2 [I2]/mol dm-3 0.004
[H+] = 0.4 mol dm-3 0.002
[H+] = 0.8 mol dm-3
[H+] = 1.2 mol dm-3
Time/min 1
2
3
4
5
6
(i)
Using the graphs above, deduce the orders of reaction with respect to propanone, iodine and acid, respectively. Show all working clearly. Order of reaction with respect to: appropriate working Propanone = 1 Acid = 1 Iodine = 0
(ii)
Hence, [5]
write
down
the
rate
equation
for
this
reaction.
Rate = k[H+][CH3COCH3]
Total 20 marks
END OF CORRECTIONS
Comment [S53]: It is unfortunate that many students do not have the time to attempt this standard question. Among the answers given, few students could explain why it is zero order wrt iodine by simply stating that a straight line for concentration-time graph indicates constant rate which is independent of [I2]. Missing/wrong units for rate which is moldm-3min-1. Reminder that the correct concluding statement is zero or first order with respect to a particular reactant, NOT [reactant].
