Chemistry 2014 Jamb Examination

According to Le Chatelier’s principle, when the concentration of hydrogen ions increases, the system adjusts to counteract this change by shifting the equilibrium in the direction that consumes H⁺ ions. In this reaction, hydrogen ions appear on the product side, so adding more H⁺ will push the equilibrium backward, shifting it to the left. This results in the formation of more dichromate ions and fewer chromate ions. Consequently, the orange colour of dichromate solution becomes more intense as equilibrium shifts left.

Magnesium is an active metal that readily reacts with dilute tetraoxosulphate (VI) acid (H₂SO₄) to produce hydrogen gas and magnesium sulphate. The reaction proceeds because magnesium lies above hydrogen in the electrochemical series and can displace hydrogen from the acid:

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)
Less reactive metals like copper, lead, and gold do not react with dilute sulphuric acid under normal conditions.

Hydrogen chloride (HCl) is highly soluble in water, unlike gases such as oxygen or hydrogen. When water comes into contact with HCl gas, it rapidly dissolves to form hydrochloric acid. This solubility is due to strong hydrogen bonding and dipole-dipole interactions between HCl and water molecules. Therefore, water effectively removes HCl gas from a gaseous mixture.

In this displacement reaction, zinc acts as the reducing agent. Initially, zinc is in its elemental state with an oxidation number of 0. After the reaction, zinc forms Zn²⁺ in ZnSO₄, meaning its oxidation number increases from 0 to +2. This increase indicates oxidation, while copper ions are reduced from +2 to 0. The simultaneous oxidation and reduction show it is a redox process.

In this reaction, carbon is oxidised to carbon monoxide, meaning it loses electrons, while water vapour is reduced to hydrogen gas, meaning it gains electrons. The substance that accepts electrons (causes oxidation) is the oxidising agent. Hence, water (H₂O) acts as the oxidising agent because it accepts electrons from carbon during the reaction.

From Faraday’s first law of electrolysis:\[Q = n \times F \times z\]where:
\(Q\) = charge (C), \(n\) = number of moles of copper, \(F\) = Faraday constant, and \(z\) = number of electrons per ion (for Cu²⁺, \(z = 2\)).

Calculate moles of Cu\[n = \dfrac{10}{64} = 0.15625\; \text{mol}\] Calculate charge\[Q = 0.15625 \times 96500 \times 2 \]\[= 30156.25\; \text{C}\]

The reaction at the cathode for copper deposition is:\[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\]This shows that 2 moles of electrons (2 Faradays) are required to deposit 1 mole of copper. Therefore, for 0.25 mole of copper:\[F = 0.25 \times 2 = 0.5F\]Hence, 0.5 Faraday of electricity is required to produce 0.25 mole of copper.

Using the Gibbs free energy relation:\[\Delta G = \Delta H - T\Delta S\]Rearranging for \(\Delta H\):\[\Delta H = \Delta G + T\Delta S\]Given \(\Delta G = -899\; \text{J mol}^{-1}\), \(\Delta S = 10\; \text{J mol}^{-1}\text{K}^{-1}\), and \(T = 25°C = 298\;\text{K}\).

Substitute:\[\Delta H = -899 + (298 \times 10)\]\[\Delta H = -899 + 2980 = +2081\; \text{J mol}^{-1}\]Thus, the enthalpy change is +2081 J mol⁻¹.

Educating the public on the causes and effects of pollution promotes awareness and behavioural change, leading to long-term prevention strategies. Once individuals and industries understand the implications of their activities on the environment, they are more likely to adopt sustainable practices. Education also strengthens public support for environmental policies, ensuring collective responsibility in pollution control.

Litmus is a natural dye used as an acid-base indicator. In acidic conditions, it turns red, while in basic conditions it turns blue. In a neutral medium, such as pure water, there is no excess of H⁺ or OH⁻ ions, so litmus retains its intermediate colour, which is purple. This colour represents equilibrium between its acidic and basic forms.