Question 3
(a)
- Draw the structure of the sixth member of the alkenes.
- Calculate the relative molecular mass of the sixth member of the alkenes.
- State one difference between cracking and reforming in the petroleum industry. [ H = 1, C = 12 ] [5 marks]
(b)(i) Define the term enthalpy of neutralization.
(ii) Describe briefly how the enthalpy of neutralization of the reaction of dilute hydrochloric acid and aqueous potassium hydroxide could be determined.
[9 marks]
(c) An electrochemical cell is constructed with copper and silver electrodes.
- (i)State which of the electrodes will be the:
- anode;
- cathode.
- Give the reason for your answer in 3(c)(i).
- State the type of reaction occurring at each electrode.
- Write a balanced equation for the overall cell reaction. [7marsks]
(d)(i) Name the compound formed when iron is exposed to moist air for a long time.
(ii) Write a balanced chemical equation for the reaction in 3 (d)(i).
(iii) Name one ore of iron.
Observation
This question was not satisfactorily answered by the candidates.
In part (a), majority of the candidates wrongly identified hexane as the sixth member of the alkene series instead of heptene.
In addition to this, majority of the candidates could not differentiate between cracking and reforming in the petroleum industry.
In part (b), majority of the candidates defined enthalpy of neutralization but could not describe how the enthalpy of neutralization could be determined.
In part (c), few candidates understood the principle of electrochemical cell.
In part (d), majority of the candidates could not name the compound formed when iron is exposed to moist air for a long time.
The expected answers include:
(iii).
CRACKING |
REFORMING |
Involves breaking large molecules of petroleum fractions into smaller molecules. |
Involves re-arrangement of atoms in molecules of petroleum fraction to obtain branched and cyclic hydrocarbons |
Used to increase the quantity of petrol. |
Used to improve the quality of petrol. |
Can be achieved thermally or catalytically. |
Occurs in presence of a catalyst. |
(b) (i) Enthalpy of neutralization is the heat evolved when an acid reacts with a base to form one mole of water / when one mole of H+ from an acid reacts with one mole of OH- from a base to form (one mole of) water.
(ii)Equimolar solutions of hydrochloric acid and potassium hydroxide are prepared separately. A known volume of the acid is placed in a calorimeter/polystyrene beaker and the temperature recorded.
The same volume of the potassium hydroxide solution at the same temperature Is poured into the calorimeter and the mixture stirred gently. The maximum
temperature of the mixture is recorded.
Heat lost in the reaction = heat gain by the solution formed.
OR
H neutralization = mass of solution x specific heat capacity of water
x change in temperature.
[9 marks]
OR
Prepare equimolar solutions of HCl and NaOH separately.
Measure equal volume of acid and alkali into two separate beakers.
Take temperature of each solution i.e. t1 and t2.
Find average of the two temperatures.
t 1 + t 2 = t 3
2
Pour the acid into a calorimeter.
Quickly add the NaOH to the acid and stir very well.
Record the highest temperature attained t4
Heat lost in the reaction = t4 – t3
ΔH neutralization = mass of solution x specific heat capacity of water x change in temperature
(c) (i) I. copper electrode
II. silver electrode
(ii) Silver has a more positive standard electrode potential than copper.
Copper is more electropositive than silver / copper is higher in the
electrochemical series.
(iii) Anode - oxidation ( occurs at the copper electrode)
Cathode - reduction (occurs at the silver electrode)
(iv) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
(d) (i) Hydrated iron (III) oxide
(ii) 4Fe(s) + 3O2(g) + 2xH2O(l)→ 2Fe2O3.xH2O
Accept 4Fe(s) + 3O2 (g) + 2H2O → 2Fe2O3.H2O
(iii) - haematite
- magnetite
- siderite /spathic iron ore
- limonite
- iron pyrites