JAMB Chemistry Past Questions & Answers - Page 345

1,721.

A catalyst speeds up the rate of chemical reaction by

A.

taking part in the reaction

B.

increasing the activation energy of the reaction

C.

lowering the activation energy of the reaction

D.

increasing the heat content

Correct answer is C

Catalyst lowers the activation energy of reaction and hence speeds up the rate of chemical reaction.

1,722.

Which of the following reaction is endothermic?

A.

C(s) + O2(g) → CO2(g)

B.

CaO(s) + H2O(l) → Ca(OH)2(s)

C.

C(s) + H2O(g) → CO(g) + H2(g)

D.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Correct answer is C

Water Gas is a mixture of carbon monoxide and hydrogen produced from synthesis gas.\(H_2O + C --> H_2 + CO\) (H=+131KJ/mol)
An Endothermic process is any process which requires or absorbs thermal energy from its surroundings, usually in the form of heat.in Endothermic reaction ∆H is positive.
Therefore it is Endothermic

1,723.

\(NH_{3(g)} + HCl_{(g)} → NH_4Cl_{(s)}\)
The entropy change in the system above is

A.

positive

B.

zero

C.

negative

D.

indeterminate

Correct answer is C

\(NH_{3(g)} + HCl_{(g)} → NH_4Cl_{(s)}\)
Reactant = 2 moles of gaseous molecules
Product = 1 mole of gaseous molecule
Entropy change = 1 - 2 = -1 (negative)

1,724.

If a given quantity of electricity liberates 0.65g of Zn2+, what amount of Hg2+ would be liberated by the same quantity of current?
[Zn = 65, Hg = 201]

A.

1.00g

B.

2.01g

C.

4.02g

D.

8.04g

Correct answer is B

At the cathode, we have:
\(Zn^{2+} + 2e^- → Zn_{(s)}\)
Normally, we need
2F → 65g
x → 0.65
x = 0.02F
Let us check how much mercury will be deposited by 0.02F of electricity
\(Hg^{2+} + 2e^-(aq) → Hg_{(s)}\)
2F → 201g
∴ 0.02F = x
x = 2.01g

1,725.

In recharging a lead-acid accumulator, the reaction at the cathode can be represented as

A.

Pb2+(aq) + SO2-4(aq) →PbSO4(s)

B.

Pb2+(aq) + 2e- →Pb(s)

C.

Pb2+(aq) + 2H2O(l) → PbO2(s) + 4H+(aq) + 4e-

D.

Pb(s) → Pb2+(aq) + 2e-

Correct answer is B

In lead acid ,lead dissociates at the cathode Pb2+(aq) + 2e- →Pb(s)