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Titanium(IV) chloride can be made from titanium(IV) oxide as shown in the equation - AQA - A-Level Chemistry - Question 5 - 2017 - Paper 1

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Titanium(IV) chloride can be made from titanium(IV) oxide as shown in the equation. TiO₂(s) + 2C(s) + 2Cl₂(g) → 2CO(g) + TiCl₄(l) ΔH⁼ -60.0 kJ mol⁻¹ Some entropy ... show full transcript

Worked Solution & Example Answer:Titanium(IV) chloride can be made from titanium(IV) oxide as shown in the equation - AQA - A-Level Chemistry - Question 5 - 2017 - Paper 1

Step 1

Calculate ΔS for the reaction

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Answer

We start by calculating the change in entropy (ΔS) using the formula:

ΔS=SproductsSreactantsΔS = S_{products} - S_{reactants}

Where:

  • For the products (2 CO(g) + TiCl₄(l)):
    • 2 × 198 J K⁻¹ mol⁻¹ (for CO) + 253 J K⁻¹ mol⁻¹ (for TiCl₄)
  • For the reactants (TiO₂(s) + 2C(s) + 2Cl₂(g)):
    • 50.2 J K⁻¹ mol⁻¹ (for TiO₂) + 2 × 5.7 J K⁻¹ mol⁻¹ (for C) + 2 × 223 J K⁻¹ mol⁻¹ (for Cl₂)

Calculating:

  • Products: 2 × 198 + 253 = 649 J K⁻¹ mol⁻¹
  • Reactants: 50.2 + 2 × 5.7 + 2 × 223 = 507.6 J K⁻¹ mol⁻¹

Thus:

ΔS=649507.6=141.4JK1mol1ΔS = 649 - 507.6 = 141.4 \, J \, K^{-1} \, mol^{-1}

Step 2

Calculate Gibbs free-energy change (ΔG)

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Answer

To calculate the Gibbs free energy change (ΔG), we use the equation:

ΔG=ΔHTΔSΔG = ΔH - TΔS

Where:

  • ΔH = -60.0 kJ mol⁻¹
  • T = 989 °C = 989 + 273.15 K = 1262.15 K
  • ΔS = 141.4 J K⁻¹ mol⁻¹ = 0.1414 kJ K⁻¹ mol⁻¹

Substituting these values:

ΔG=60.0(1262.15K)(0.1414kJK1mol1)ΔG = -60.0 - (1262.15 \, K)(0.1414 \, kJ \, K^{-1} mol^{-1})

Calculating:

ΔG=60.0178.3=238.3kJmol1ΔG = -60.0 - 178.3 = -238.3 \, kJ \, mol^{-1}

Step 3

Explain feasibility of the reaction

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Answer

The reaction is feasible since the calculated Gibbs free energy change (ΔG) is negative:

ΔG=238.3kJmol1ΔG = -238.3 \, kJ \, mol^{-1}

A negative ΔG indicates that the reaction can occur spontaneously under the given conditions.

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