Photo AI
I₃(s) → I₃(g) ΔH = +60 kJ mol⁻¹ I(g) → 2I(g) ΔH = +243 kJ mol⁻¹ I(g) + e⁻ → I⁻(g) ΔH = -349 kJ mol⁻¹ Which of the following would show the energy diagram for I₃(s) + 2e⁻ → 2I(g)? - Scottish Highers Chemistry - Question 17 - 2015
Question 17
I₃(s) → I₃(g) ΔH = +60 kJ mol⁻¹ I(g) → 2I(g) ΔH = +243 kJ mol⁻¹ I(g) + e⁻ → I⁻(g) ΔH = -349 kJ mol⁻¹ Which of the following would show the energy diagram for I₃(s) ... show full transcript
Worked Solution & Example Answer:I₃(s) → I₃(g) ΔH = +60 kJ mol⁻¹ I(g) → 2I(g) ΔH = +243 kJ mol⁻¹ I(g) + e⁻ → I⁻(g) ΔH = -349 kJ mol⁻¹ Which of the following would show the energy diagram for I₃(s) + 2e⁻ → 2I(g)? - Scottish Highers Chemistry - Question 17 - 2015
Step 1
Step 2
Combine Enthalpy Changes
Answer
Next, we combine the enthalpy changes in the context of the given process:
- From I₃(s) to I₃(g), we have +60 kJ.
- Convert I₃(g) to I(g) involves breaking it into I(g), which has a positive enthalpy change of +243 kJ.
- Keep in mind that we also need to account for two electrons reacting with I(g): the enthalpy change for this reaction involves -349 kJ.
Total Enthalpy Change = +(60) + +(243) + 2(-349) = 60 + 243 - 698 = -395 kJ
Step 3
Draw the Energy Diagram
Answer
The energy diagram should begin at the energy level corresponding to I₃(s), rise to I₃(g), then peak at the point where it is split into I(g), and finally drop again for the reaction with electrons. The final product will be 2I(g) at a lower energy level due to the significantly negative enthalpy change.
Options should be carefully compared against this logic. The correct diagram that represents these changes is option A.
Step 4