Chemistry 2016 Jamb Examination

Chemistry 2016 Jamb Examination
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Question 41

      H  H   H    H
       |    |     |     |
H - C - C - C - C - Cl
       |     |    |     |
      H  OH H   Cl

The functional groups present in the compound above are
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The given structure shows the presence of an \( \text{OH} \) group (hydroxyl group) attached to one of the carbon atoms and \( \text{Cl} \) atoms (chloro substituents) on the terminal and another carbon atom.

There are no carbon–carbon double bonds, so it is not an alkene.

Hence, the compound contains hydroxyl and chloro functional groups.
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Question 42

Energy vs Reaction progress

In the diagram above, X is the
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In an energy profile diagram, the curve shows the progress of a reaction against the energy change.

The highest point on the curve represents the activated complex, while the energy difference between the reactants and this highest point is called the activation energy (X).\[ E_a = E_{\text{activated complex}} - E_{\text{reactants}} \] Hence, X is the activation energy.
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Question 43

Potential energy vs reaction coordinate

The diagram represents
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In the potential energy diagram, the products have lower potential energy than the reactants. This means energy is released to the surroundings.

The difference in energy between reactants and products represents the enthalpy change (\( \Delta H \)), which is negative for exothermic reactions:
\[ \Delta H = E_{\text{products}} - E_{\text{reactants}} < 0 \]
Hence, the diagram represents an exothermic reaction.
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Question 44

PH vs cm3 of 0.1 moldm-3

The curve depicts titration between a strong acid of pH
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The titration curve starts at a very low pH (indicating a strong acid) and ends at a high pH (indicating a strong base).

The steep and nearly vertical jump around the equivalence point at pH 7 is characteristic of strong acid–strong base titrations.
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Thus, the curve represents titration between a strong acid and strong base.
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Question 45

ideal gas can be represented by

From the diagram above, an ideal gas can be represented by
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An ideal gas obeys Boyle’s law perfectly, which states that \( PV = \text{constant} \).

This relationship produces a smooth rectangular hyperbola when \( P \) is plotted against \( V \).

Among the curves shown, L represents the ideal gas because it shows a perfect hyperbolic relationship between \( P \) and \( V \).

Hence, the ideal gas is represented by L.
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Question 46

CH4(g) + Cl2(g) → CH2Cl(s) + HCl(g)

The major factor that influences the rate of the reaction above is
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The reaction between methane and chlorine occurs through a free radical substitution mechanism initiated by light energy (photochemical reaction).

The UV light provides the energy to dissociate \( \text{Cl}_2 \) molecules into chlorine radicals:
\[ \text{Cl}_2 \xrightarrow{hv} 2\text{Cl} \cdot \]
Hence, light is the major factor influencing the rate of this reaction.
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Question 47

N2O4(aq) ⇌ 2NO2(g), \( \Delta H = +\text{ve} \)

In the reaction above, an increase in temperature will
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The reaction is endothermic since \( \Delta H = +\text{ve} \).

According to Le Chatelier’s principle, increasing temperature favors the endothermic direction (forward reaction).

This increases the concentration of \( \text{NO}_2 \) and raises the equilibrium constant \( K \):
\[ \text{N}_2\text{O}_4 (aq) \rightleftharpoons 2\text{NO}_2 (g) \]\[ \text{Forward reaction is endothermic.} \]
Therefore, an increase in temperature will increase the value of the equilibrium constant.
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Question 48

MnO4-(aq) + Y + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l).

In the equation above, Y is
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The balanced redox reaction between permanganate ion and iron(II) in acidic medium is:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
Hence, Y = 8H+(aq).
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Question 49

           H
             |
CH3 — C — COOH
             |
           OH
The compound above exhibits
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Geometric (cis–trans) isomerism occurs due to restricted rotation around a double bond or ring structure with different substituents on each carbon. Since the compound in question meets this condition, it exhibits geometric isomerism.
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Question 50

n monosaccharide ⇌ PQ polysaccharide − n water.

In the process above, P and Q respectively represent
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When monosaccharides combine to form polysaccharides, they undergo polymerization by the condensation process, releasing water molecules:
\[ n\, \text{C}_6\text{H}_{12}\text{O}_6 \rightarrow (\text{C}_6\text{H}_{10}\text{O}_5)_n + n\,\text{H}_2\text{O} \]Therefore, P = Polymerization and Q = Condensation.
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Chemistry 2016 Jamb Examination
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