a(i) Write balanced chemical equations for the production of lead(II) chloride by:
I. direct combination of constituent elements;
II. double decomposition;
III. displacement reaction.
(ii) State one condition that would increase the rate of reaction for each of the reactions stated in 3a(i).
I. ........................... II. ......................... III. .......................
(b) Consider the following Chemical reaction in equilibrium:
2SO\(_2\)(g) + O\(_2\)(g) ⇌ 2SO\(_3\)\(_(g)\) ; ΔH = positive
(i) State two conditions that would increase the yield of SO\(_3\)\(_(g)\)
(ii) If 60cm\(^3\) of SO\(_2\) reacts with 40cm\(^3\) of O\(_2\), determine the volume of the residual gas, assuming that the reaction goes to completion in a cylinder.
(iii) What is the change in the oxidation number of sulphur in the reaction in 3(b) ?
(c) Using structural diagrams, explain why cyclopropane is not an isomer of n-propane.
(d) A structural organic compound X containing three carbon atoms reacted with butanoic acid in the presence of dilute H\(_2\)SO\(_4\) to form a sweet smelling compound Y.
(i) Name compounds X and Y.
(ii) State the functional group present in X.
(iii) Write an equation for the reaction between X and butanoic acid.
a(i) Production of Lead(II)Chloride by:
I. direct combination : Pb + Cl\(_2\) → PbCl\(_2\)
II. double decomposition: Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq)
III. displacement reaction: Pb + CuCl₂ → PbCl₂ + Cu.
(ii) Condition that would increase the rate of reaction for:
I. direct decomposition: increase the temperature, concentration of reactants, or introduce a catalyst.
II.double decomposition: increase the concentration of reactants, and/or increase the surface area of solid reactants.
III. displacement reaction: increase the temperature.
b(i) Conditions that would increase the yield of SO\(_3\)\(_(g)\)
- Increase in temperature,
- Increase in pressure,
- Removing the product formed
(ii) 2SO\(_2\)(g) + O\(_2\)(g) ⇌ 2SO\(_3\)\(_(g)\)
Reacting ratio: 2 moles : 1 mole
Total Reacting Volume: 60cm\(^3\) : 40cm\(^3\)
Actual reacted Volume: 60cm\(^3\) : 30cm\(^3\) → (Following the ratio from the balanced chemical equation)
Residual Volume: (Total reacting volume - Actual reacted volume)
For SO\(_2\): 60 - 60 = 0 ⇒ SO\(_2\) was completely used up in the reaction, hence it is called the limiting reactant.
For O\(_2\) : 40 - 30 = 10cm\(^3\) → Volume of the residual gas is 10cm\(^3\).
(iii) Change in the oxidation number of sulphur in the reaction: 2SO\(_2\)(g) + O\(_2\)(g) ⇌ 2SO\(_3\)\(_(g)\)
i.e From SO\(_2\)(g) → SO\(_3\)\(_(g)\)
S + O\(_2\)= 0 → S + O\(_3\) = 0
S + (-2)2 = 0 → S + (-2)3 = 0
S - 4 =0 → S - 6 = 0
S = + 4 → S = + 6
Change in oxidation state is from +4 to +6.
(c) cyclopropane n-propane
CH\(_2\) H H H
/ \ I I I
CH\(_2\) — CH\(_2\) H —C — C — C — H
( C\(_3\)H\(_6\)) I I I
H H H ( C\(_3\)H\(_8\))
For any two or more compounds to be isomers, their molecular formula must be the same while their structural formular would be different from each other. In this case, the molecular formula of the two compounds are not the same. i.e C\(_3\)H\(_6\) and C\(_3\)H\(_8\) are not the same, hence cyclopropane cannot be said to be an isomer of n-propane.
d(i) Compound X is Propanol while Compound Y is Propylbutanoate
(ii) The functional group present in X is - OH (hydroxyl)
(iii) Propanol + Butanoic acid → Propyl butanoate + Water
CH\(_3\)CH\(_2\)CH\(_2\)OH + CH\(_3\)CH\(_2\)CH\(_2\)COOH → CH\(_3\)CH\(_2\)CH\(_2\)COOCH\(_2\)CH\(_2\)CH\(_3\) + H\(_2\)O
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