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Chlorine can be produced by the reaction of hydrogen chloride with air using the Deacon process - Scottish Highers Chemistry - Question 9 - 2019
Question 9
Chlorine can be produced by the reaction of hydrogen chloride with air using the Deacon process. (a) (i) Using the potential energy diagram, determine the activatio... show full transcript
Worked Solution & Example Answer:Chlorine can be produced by the reaction of hydrogen chloride with air using the Deacon process - Scottish Highers Chemistry - Question 9 - 2019
Step 1
Using the potential energy diagram, determine the activation energy, in kJ mol⁻¹, for the forward reaction.
Answer
From the potential energy diagram, the activation energy for the forward reaction can be determined by identifying the energy difference between the reactants and the peak of the curve. According to the diagram, this value is approximately +220 kJ mol⁻¹.
Step 2
Explain why increasing the temperature in the Deacon process results in less chlorine being produced.
Answer
Increasing the temperature in the Deacon process favors the endothermic/reverse reaction. The forward reaction, which produces chlorine, is exothermic. As the temperature rises, the equilibrium shifts towards the endothermic direction to absorb the excess heat, leading to a decrease in the production of chlorine.
Step 3
Complete a labelled diagram to show an apparatus suitable for carrying out this preparation.
Answer
The apparatus should include a reaction flask containing concentrated hydrochloric acid connected to a delivery tube leading to a water bath. The potassium permanganate should be added in the flask. The collection vessel should be inverted over the water to collect insoluble chlorine gas. The delivery tube must be labeled accordingly.
Step 4
Calculate the enthalpy change, in kJ mol⁻¹, for this reaction using the following information.
Answer
Utilizing Hess's law, we can combine the equations to determine the overall enthalpy change for the reaction:
For the reaction: CCl₄(g) + CH₄(g) →
- CH₄(g) + 2HCl(g) → CH₃Cl(g) ΔH = -75 kJ mol⁻¹
- CH₄(g) + Cl₂(g) → CCl₄(g) ΔH = -98 kJ mol⁻¹
- 2H(g) + Cl₂(g) → 2HCl(g) ΔH = -92.1 kJ mol⁻¹
The overall change is:
ΔH = (-75) + (-98) + [ΔH of products - ΔH of reactants] = -391 kJ mol⁻¹.