2025 A Level H2 Chemistry Paper 1 Solutions - Questions 11 to 20
Question 11
Answer: D
Topic: Chemical Equilibria
Explanation:
*This same question was tested in 2017 A Levels H2 Chemistry Paper 1 Question 14.
As equimolar amounts of reactants are used and the mole ratio of reactants to products are all 1 to 1, the reactants H2O and CO concentrations will remain the same, and products H2 and CO2 concentrations will remain the same.
When concentration of H2 is greater than concentration of H2O, there are more products than reactants and equilibrium constant K will be greater than 1.
Therefore Gibbs free energy given by ΔG = - RTlnK will be negative, which corresponds to points 3 and 4 only.
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Question 12
Answer: A
Topic: Intro to Organic Chemistry
Explanation:
Empirical formula is the simplest ratio of all elements present in the compound. Based on the skeletal structures given we need to determine the total number of carbon, hydrogen and oxygen and convert the molecular formula to empirical formula for compound X and compare with those for molecules 1 to 3.
Molecular and empirical formula for X are C8H12O2 and C4H6O respectively.
Molecular and empirical formula for 1 are C8H12O2 and C4H6O respectively (same as X).
Molecular and empirical formula for 2 are C4H6O and C4H6O respectively (same as X).
Molecular and empirical formula for 3 are C16H24O4 and C4H6O respectively (same as X).
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Question 13
Answer: B
Topic: Solubility Product
Explanation:
Concentration of a saturated solution of zinc (II) hydroxide is just finding the solubility of the solution (x).
The concentration of Zn2+ and OH- in a saturated solution is x and 2x respectively.
These concentrations can be substituted into the Ksp expression to solve for x = 1.79 x 10-5 mol dm-3.
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Question 14
Answer: C
Topic: Halogenoalkanes
Explanation:
Step one involves elimination of HBr from 1-bromopropane to form propene.
Step two involves electrophilic addition of HBr to form 2-bromopropane. According to Markovnikov rule, H will be added to terminal alkene carbon CH2 with more hydrogen to form the major product.
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Question 15
Answer: D
Topic: Introduction to Organic Chemistry
Explanation:
We have to permutate systematically to find all structural isomers for P (C4H8O2), given P can be a carboxylic acid or ester.
For acid functional group, all the 4 carbons are together and the 2 chain isomers are butanoic acid (1) and 2-methylpropanoic acid (2).
For ester functional group, the 4 carbons have to be split between the acid and alcohol portion of the ester.
a. For 3 carbons for acid + 1 carbon for alcohol, only methyl propanoate (3) can be formed.
b. For 2 carbons for acid + 2 carbon for alcohol, only ethyl ethanoate (4) is possible.
c. For 1 carbon for acid + 3 carbons for alcohol, propanyl methanoate (5) and 1-methylethyl methanoate (6) are possible.
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Question 16
Answer: C
Topic: Kinetics, Hydrocarbons
Explanation:
CO is neutral so will not contribute to acid rain formation.
CO2 is acidic but dissolves in rain water to form weak acid H2CO3 so will not contribute to acid rain formation.
NO2 can dissolve in rain water to form HNO2 and HNO3 hence forms acid rain. It also catalyses the oxidation of SO2 to SO3 and both sulfur oxides are contributors to acid rain as well. NO2 also contributes to production of photochemical smog giving smog its brown colour.
SO2 dissolves in rain water to form H2SO3 hence contributes to acid rain formation but does not produce smog.
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Question 17
Answer: D
Topic: Halogenoalkanes
Explanation:
Student 1 is wrong as the organic compound might not contain any halogens at all, ie neither halogenoalkane nor halogenoarene, therefore no ppt can be formed.
Student 2 is wrong as the organic compound may be a fluoroalkane (halogenoalkane) but the strong C-F bond cannot be broken for ppt to be formed. Also the conditions of warm silver nitrate might not be drastic enough to cause nucleophilic substitution of halogenoalkanes hence no halides are present to form ppt.
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Question 18
Answer: C
Topic: Organic Chemistry Mechanism
Explanation:
A. Methane undergoes free radical substitution and behaves as neutral species in the reaction.
B. Iodomethane undergoes nucleophilic substitution and behaves as electrophile.
C. Ethene undergoes electrophilic addition and behaves as nucleophile.
D. Propanone undergoes nucleophilic addition and behaves as electrophile.
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Question 19
Answer: B
Topic: Alkanes
Explanation:
Reaction K must be initiation step where Cl-Cl bond breaks homolytically to form 2 chlorine atoms, hence the mass is halved.
Reaction L must be termination step where 2 chlorine radicals or 2 ethyl radicals forms Cl-Cl or C-C bond, hence the mass doubles.
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Question 20
Answer: A
Topic: Benzene
Explanation:
We can work backwards to deduce the structure of alkyl group. 2 benzene must react with 1,1-dichloroethane to form product G.
Lewis acid catalyst required for electrophilic substitution of benzene must be AlCl3 since Al is electron deficient and can function as an electron pair acceptor.
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