Q1 Combustion Of Ethanol. (A) The “Lewis Structure” Of The Molecule Ethanol Is Presented In Figure Q1.1. How Many Electron Bonding Pairs And Electron Lone Pairs Are Present In The Molecule? How Many Electrons Is This In Total? [2] H-C-0-0-H (B) (C) Figure Q1.1 Lewis Structure Of Ethanol Carbon (C) Has 4 Electrons On Its Valence Shell Whilst Hydrogen
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Q1 Combustion of ethanol. (a) The Lewis Structure of the molecule ethanol is presented in Figure Q1.1. How many electron bonding pairs and lone pairs are present in the molecule? How many electrons is this in total? Single Bond Energies (kJ/mol of bonds) H C N O S F Cl Br I H 436 C 413 346 N 391 305 163 O 463 358 201 146 S 347 272 F 565 485 In Table Q1.2, the total energy of the bonds in O2, CO2 and H2O are presented. Calculate the total energy, E°fo G) Calculate the standard change in entropy ?S° for this reaction at 25°C. (k) Explain why the entropy is positive.
The correct answer and explanation is:
Solution to Q1: Combustion of Ethanol
(a) Lewis Structure Analysis
The Lewis structure of ethanol (C₂H₅OH) consists of:
- Electron bonding pairs (covalent bonds):
- C-H bonds: 6 (each carbon is bonded to hydrogen)
- C-C bond: 1
- C-O bond: 1
- O-H bond: 1
- Total bonding pairs = 9
- Electron lone pairs:
- Oxygen (O) in the hydroxyl group has 2 lone pairs.
- Total number of valence electrons:
- Carbon (C) contributes 4 × 2 = 8 electrons
- Hydrogen (H) contributes 1 × 6 = 6 electrons
- Oxygen (O) contributes 6 electrons
- Total electrons = 20 valence electrons
(b) Bond Energy Calculation
The combustion reaction for ethanol: C2H5OH+3O2→2CO2+3H2OC_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O
Step 1: Breaking Bonds in Reactants
Bonds present in 1 molecule of ethanol:
- C-H: 5 × 412 kJ/mol = 2060 kJ
- C-C: 1 × 348 kJ/mol = 348 kJ
- C-O: 1 × 360 kJ/mol = 360 kJ
- O-H: 1 × 463 kJ/mol = 463 kJ
Bonds present in 3 molecules of O₂:
- O=O: 3 × 498 kJ/mol = 1494 kJ
Total energy required to break bonds: 2060+348+360+463+1494=4725 kJ2060 + 348 + 360 + 463 + 1494 = 4725 \text{ kJ}
Step 2: Forming Bonds in Products
Bonds present in 2 molecules of CO₂:
- C=O: 4 × 799 kJ/mol = 3196 kJ
Bonds present in 3 molecules of H₂O:
- O-H: 6 × 463 kJ/mol = 2778 kJ
Total energy released in forming bonds: 3196+2778=5974 kJ3196 + 2778 = 5974 \text{ kJ}
Step 3: Calculating Energy Change
ΔH=Bonds Broken−Bonds Formed\Delta H = \text{Bonds Broken} – \text{Bonds Formed} ΔH=4725−5974=−1249 kJ/mol\Delta H = 4725 – 5974 = -1249 \text{ kJ/mol}
(c) Standard Entropy Change (ΔS∘\Delta S^\circ) Calculation
Standard entropy values (S∘S^\circ) at 298 K:
| Substance | S∘S^\circ (J/mol·K) |
|---|---|
| C₂H₅OH (l) | 160.7 |
| O₂ (g) | 205.0 |
| CO₂ (g) | 213.7 |
| H₂O (g) | 188.8 |
Entropy change: ΔS∘=∑Sproducts∘−∑Sreactants∘\Delta S^\circ = \sum S^\circ_{\text{products}} – \sum S^\circ_{\text{reactants}} ΔS∘=[2(213.7)+3(188.8)]−[(160.7)+3(205.0)]\Delta S^\circ = \left[ 2(213.7) + 3(188.8) \right] – \left[ (160.7) + 3(205.0) \right] ΔS∘=(427.4+566.4)−(160.7+615)\Delta S^\circ = (427.4 + 566.4) – (160.7 + 615) ΔS∘=993.8−775.7\Delta S^\circ = 993.8 – 775.7 ΔS∘=218.1 J/mol\cdotpK\Delta S^\circ = 218.1 \text{ J/mol·K}
(d) Explanation for Positive Entropy
Entropy (ΔS∘\Delta S^\circ) is positive because:
- Phase Change: Ethanol is a liquid, while CO₂ and H₂O in the gaseous state contribute higher entropy.
- Molecular Freedom: The number of gaseous molecules increases from 3 O₂ molecules to 5 gaseous molecules (2 CO₂ and 3 H₂O).
- Dispersal of Energy: Combustion disperses energy and increases randomness.
Thus, entropy is positive because the system becomes more disordered after the reaction.
Generated Image: Lewis Structure of Ethanol
I’ll generate an image showing the Lewis structure of ethanol now.
Here is the Lewis structure of ethanol (C₂H₅OH) showing bonding pairs and lone pairs. Let me know if you need any modifications or further explanations!
