A Level Chemistry
CIE A Level Chemistry Solved Past Paper May/June 2020 P23

CIE A Level Chemistry Solved Past Paper May/June 2020 P23

1 ai) Barium metal burns with pale green flame and forms white solid BaO.

bi) Thermal decomposition of group II carbonates give metal oxides and CO2 and when metal oxides react with water, we get metal hydroxides.

ciii) Ga2O3 + HCl ——-> GaCl3 + H2O

Ga2O3 + NaOH ——–> NaGaO2 + H2O

2 ciii) For Cu & Cr atoms, orbitals are stable if they are either half or completely filled. This means distribution of electrons has to be in such a way that s orbital gets 1 or 2 electrons, p orbital gets 3 or 6 and d orbital gets 5 or 10 electrons. Cu can’t have its configuration ending at 4s2 3d9 but on 4s1 3d10/3d10 4s1. For other atoms, lower energy level fills completely before it shifts to a higher energy level.

di) First we should determine the number of moles of sodium thiosulfate in titration and divide that by 2 to get moles of I2 in titration since the mole ratio of I2 to sodium thiosulfate is 1:2. Next it says that 250cm3 of I2 which was produced in reaction 1, out of that only 25cm3 of I2 was separated for titration: This means that moles of I2 which we calculated initially were for 25cm3 of I2, we should get moles of I2 for 250cm3 because 250cm3 of I2 was produced in reaction 1. Moving ahead, mole ratio of CuSO4 to I2 is 2:1 therefore we should multiply the number of moles of 250cm3 of I2 by 2.

ii) 10.68g has both unsaturated CuSO4 and water while 7.98g has got only unsaturated CuSO4. This means we have 10.68-7.98=2.7g of water in CuSO4•xH2O. Now that we have mass of unsaturated CuSO4 and mass of water in CuSO4•xH2O, we can convert them to moles and get mole ratio to find the value of x.

3 aii) Adding catalyst to a reaction lowers activation energy, this means now more molecules of sucrose and water will have energy greater than reaction’s activation energy. As a result rate of reaction increases. We can illustrate this concept on energy level diagram where x-axis denotes energy and activation energy shifts to the left in presence of catalyst. You must show both the activation energy before and after catalyst was added and shade the region under graph which indicates number of molecules that have achieved energy higher than activation energy.

ciii) Since in this Q, enthalpy of combustion values are to be used for calculating enthalpy of reaction, following should be your Hess’ cycle:

Hess’ cycle for finding enthalpy of reaction using enthalpy of combustion

Although oxygen molecules were not there, I added them to both sides of equation for combustion of reactants and products. There are 2 routes arriving on the same products i.e CO2 and water. One is direct which is reactants——>combustion products and the other is indirect route. Using Hess’ law, you can now equate all enthalpies and get enthalpy of reaction.

4b) This free radical substitution reaction between hexane and bromine divides into 3 steps known as initiation, propagation & termination.

Initiation step involves homolytic fission of halogen molecule Br2—–> Br + Br

Propagation step is reaction between one of bromine radicals and hexane from our reactants which forms a new radical and one of our products Br + C6H14 ——> HBr + C6H13 followed by hexane radical reacting with bromine from our reactants forming one more bromine radical and final product C6H13 + Br2 ——–> Br + C6H13Br.

Termination step involves reaction within radicals like Br + Br ———-> Br2, Br + C6H13 ——–> C6H13Br & C6H13 + C6H13 ——-> C12H26

ci) If your product has -diol group, this pretty much means that cold dilute acidified KMnO4 was used with unsaturated carbon chain. Position of -diol group also indicates position of double bond between carbon atoms.

5 ai) Alcohol——–> alkene+water (dehydration)

ii) Position of double bond tells us the position of hydroxyl group.

iii) Although C and D have exact same methyl groups attached to each carbon atom and double bond’s position is also the same, C has 2 methyl groups attached on the top which makes it a cis-isomer. On the other hand D has 2 methyl groups attached across C=C bond and that makes it a trans-isomer. E is not showing stereoisomerism.

b) Reagents which are used in this Q can tell us what functional groups are in G and H. Metal carbonate like Na2CO3 can tell us whether the structure has -COOH group if bubbles form. Metal like Na can identify if a structure has -OH or -COOH group if colourless gas bubbles form and finally alkaline aqueous iodine can differentiate secondary alcohols(-OH), ketones(-CO) and exclusively the only aldehyde ethanal(containing CH3CO group) from other substances if yellow precipitate forms.

iii) It is said that G which is a carboxylic acid doesn’t show positive result with alkaline aqueous iodine, this means that G doesn’t have CH3CO group. Hence G can’t be ethanoic acid but methanoic acid and H is ethanol.

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