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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) ΔHf = -60.0 kJ mol⁻¹ Some entrop... 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

To find the change in entropy, ΔS, we can use the equation:

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

Calculating for the given reaction:

ΔS=[2(198)+253][50.2+2(5.70)+2(223)]ΔS = [2(198) + 253] - [50.2 + 2(5.70) + 2(223)]

Calculating the values:

  • For products: 2(198)+253=396+253=6492(198) + 253 = 396 + 253 = 649
  • For reactants: 50.2+2(5.70)+2(223)=50.2+11.4+446=507.650.2 + 2(5.70) + 2(223) = 50.2 + 11.4 + 446 = 507.6

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

The Gibbs free-energy change is calculated using the following equation:

ΔG=ΔHTΔSΔG = ΔH - TΔS

Where:

  • ΔH = -60.0 kJ/mol = -60000 J/mol (convert to Joules)
  • T = 989 °C = 1262 K (convert to Kelvin by adding 273)
  • ΔS = 141.4 J K⁻¹ mol⁻¹ (calculated previously)

Now substitute the values into the equation:

ΔG=60000(1262)(141.4)ΔG = -60000 - (1262)(141.4)

Calculating:

=60000178,880.8=238,880.8J/mol= -60000 - 178,880.8 = -238,880.8 \, J/mol

Converting to kJ:

ΔG=238.9kJ/molΔG = -238.9 \, kJ/mol (to three significant figures)

Step 3

Explain feasibility of the reaction

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Answer

The reaction is feasible since the Gibbs free-energy change, ΔG, is negative. A negative ΔG value indicates that the reaction will proceed spontaneously under the given conditions. Therefore, this reaction is thermodynamically feasible.

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