[ Related Blog post: http://www.wewwchemistry.com/2013/09/the-mind-boggling-h2-chemistry ]

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Example of an Old H2 Chemistry Paper 3 Free-Response Question
from a School Preliminary Examination
(The question stem is highlighted in green.)
2007 Anglo-Chinese Junior College / Paper 3 / Question 1
The following experimental results were obtained from the hydrolysis of two
bromoalkanes at temperature T1, where the concentration of the solution is measured
in mol dm–3
and time taken for the experiment in seconds.
Order of reaction with respect toReaction Reactants
bromoalkane OH–
ion
A CH3CH2CH2Br + OH–
1 1
B (CH3)3CBr + OH–
1 0
(a) Using reaction A,
(i) write down the rate equation, giving the units of the rate constant.

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Proposed Solution
Rate equation : Rate = k[CH3CH2CH2Br][OH–
]
Units for k : mol–1
dm3
s–1
(ii) describe, and explain in molecular terms, how the rate of this reaction is
affected when it is repeated at a higher temperature T2. You should include a
reference to the Maxwell Boltzmann distribution in your answer. [6]
Proposed Solution
When temperature of a reaction increases,
 average kinetic energy of reacting molecules increases.
 Number of reactant particles, per unit volume, with energy at least equal to the
activation energy, Ea, increases.
 This increases the frequency of effective collisions.
 Since rate of reaction is proportional to the frequency of effective collisions, rate of
reaction increases.

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(b) Using reaction B,
(i) calculate the time taken for the bromoalkane to decrease to 6.25 % of its
original concentration, given that the half-life for this reaction is approximately
10.6 minutes.
Proposed Solution
Let n be the number of half-lives.
∴ Time taken for the bromoalkane to decrease to 6.25 % of its original concentration =
4 × 10.6 = 42.4 min

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(ii) state and explain the effect on the rate of this reaction when the
concentration of the hydroxide ion is doubled and the concentration of the
bromoalkane is halved simultaneously.
Proposed Solution
 Reaction is zero-order with respect to [OH–
].
⇒ Rate of reaction is unaffected by changes to [OH–
].
 Rate equation: Rate = k[(CH3)3CBr]
∴ Rate of reaction is halved.

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(iii) The following mechanism is proposed for this reaction:
Slow : (CH3)3CBr → (CH3)3C+
+ Br–
Fast : (CH3)3C+
+ OH–
→ (CH3)3COH
Draw a labelled enthalpy profile diagram for this proposed mechanism. [5]
Enthalpy,H
Reaction coordinate
(CH3)3CBr + OH–
(CH3)3COH+ Br–
(CH3)3C+ + Br – + OH–
Ea,2
Ea,1
!Hr

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(c) The following observations were obtained when two halogenoalkanes react
separately with aqueous silver nitrate.
Halogenalkane Observations after adding AgNO3 (aq)
CH3CH2CH2Br cream ppt forms after 2 minutes
CH3CH2CH2I yellow ppt forms almost immediately
Explain these observations as fully as you can. [3]
Proposed Solution
Reactions:
i CH3CH2CH2X + H2O → CH3CH2CH2OH + H+
+ X–
(where X = halogen)
ii Ag+
(aq) + X–
(aq) → AgX(s)
 In the nucleophilic substitution of halogenoalkanes, the slow step involves the
breaking of C–X.
⇒ Rate of this reaction depends on the strength of the C–X bond.

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 Since the C–Br bond is stronger than the C–I bond, the rate at which Br–
(Reaction i)
is released into solution to form cream precipitate of AgBr with Ag+
(Reaction ii) ions
is slower than the rate at which I–
is released into solution to form yellow precipitate
of AgI.
(d) Describe a simple chemical test that you could use to distinguish the following
compounds. You are to include reagents and conditions, observations and a
balanced equation(s) in each case.
(i)
and
I II

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Proposed Solution
Test : To separate samples of compounds I and II, add Br2(aq).
I II
Observation Reddish brown bromine is
decolourised, and a white
precipitate is formed.
Reddish brown bromine is not
decolourised, and a white
precipitate is not observed.
Equation:

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(ii)
and
III IV
[6]
[Total: 20]
Proposed Solution
Test : To separate samples of compounds III and IV, add KMnO4(aq) and H2SO4(aq)
and heat.
I II
Observation Purple solution of KMnO4 is
decolourised.
Purple solution of KMnO4 is not
decolourised.

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Equations:
Step 1: Hydrolysis of ester
!
Step 2: Oxidation of primary alcohol