Question

using bond dissociation energies, determine the approximate enthalpy of reaction for the following reaction:
2 ch₃ch₂oh (l) + 7 o₂ (g) → 4 co₂ (g) + 6 h₂o (g)
note that bond dissociation energies can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 4 sig figs, but do not include units in your answer.

question 6
0.5 pts
which of the following chemical equations represents the enthalpy of formation, δh_f °, of c₂h₅oh (l)?
○ c₂h₅oh (l) → 2 c (graphite) + 1/2 o₂ (g) + 3 h₂ (g)
○ 2 c₂h₅oh (l) + 7 o₂ (g) → 4 co₂ (g) + 6 h₂o (g)
○ 2 c (graphite) + 1/2 o₂ (g) + 3 h₂ (g) → c₂h₅oh (l)
○ 2 c₂h₅oh (l) → 4 c (graphite) + o₂ (g) + 6 h₂ (g)
○ 4 c (graphite) + o₂ (g) + 6 h₂ (g) → 2 c₂h₅oh (l)

Explanation:

First Problem (Enthalpy of Reaction using Bond Dissociation Energies)

Step1: Identify Bonds in Reactants

For \( 2 \, \text{CH}_3\text{CH}_2\text{OH} \) (liquid, but we consider gaseous bonds for BDE; assume vaporized or use BDE for liquid if available). Let's break down the bonds:

• Each \( \text{CH}_3\text{CH}_2\text{OH} \) has: \( 5 \, \text{C-H} \), \( 1 \, \text{C-C} \), \( 1 \, \text{C-O} \), \( 1 \, \text{O-H} \).
• For 2 molecules: \( 2 \times (5 \, \text{C-H} + 1 \, \text{C-C} + 1 \, \text{C-O} + 1 \, \text{O-H}) = 10 \, \text{C-H} \), \( 2 \, \text{C-C} \), \( 2 \, \text{C-O} \), \( 2 \, \text{O-H} \).
• \( 7 \, \text{O}_2 \) has \( 7 \, \text{O=O} \) bonds.

Step2: Identify Bonds in Products

• \( 4 \, \text{CO}_2 \): Each \( \text{CO}_2 \) has \( 2 \, \text{C=O} \) (double bonds), so \( 4 \times 2 = 8 \, \text{C=O} \).
• \( 6 \, \text{H}_2\text{O} \): Each \( \text{H}_2\text{O} \) has \( 2 \, \text{O-H} \) bonds, so \( 6 \times 2 = 12 \, \text{O-H} \).

Step3: Use BDE Formula (\( \Delta H = \sum \text{BDE}_{\text{reactants}} - \sum \text{BDE}_{\text{products}} \))

From standard BDE tables (values in kJ/mol):

• \( \text{C-H} = 413 \), \( \text{C-C} = 348 \), \( \text{C-O} = 358 \), \( \text{O-H} = 463 \), \( \text{O=O} = 498 \), \( \text{C=O} = 799 \) (for \( \text{CO}_2 \), \( \text{C=O} \) is 799 kJ/mol).

Calculate Reactant BDE:

• \( 10 \times 413 = 4130 \) (C-H)
• \( 2 \times 348 = 696 \) (C-C)
• \( 2 \times 358 = 716 \) (C-O)
• \( 2 \times 463 = 926 \) (O-H)
• \( 7 \times 498 = 3486 \) (O=O)

Total Reactant BDE: \( 4130 + 696 + 716 + 926 + 3486 = 9954 \)

Calculate Product BDE:

• \( 8 \times 799 = 6392 \) (C=O)
• \( 12 \times 463 = 5556 \) (O-H)

Total Product BDE: \( 6392 + 5556 = 11948 \)

Calculate \( \Delta H \):

\( \Delta H = 9954 - 11948 = -1994 \) (Wait, but let's check again—maybe I missed bonds or used wrong BDE. Wait, \( \text{CH}_3\text{CH}_2\text{OH} \) structure: \( \text{CH}_3-\text{CH}_2-\text{OH} \). So per molecule: \( 3 \, \text{C-H} \) (CH3), \( 2 \, \text{C-H} \) (CH2), 1 C-C, 1 C-O, 1 O-H. So 5 C-H, correct. For 2 molecules: 10 C-H, 2 C-C, 2 C-O, 2 O-H. O2: 7 O=O. Products: 4 CO2 (each 2 C=O), 6 H2O (each 2 O-H). Wait, maybe BDE for \( \text{CO}_2 \) is 2799 (since each CO2 has two C=O bonds, so per CO2, 2799? Wait no, \( \text{CO}_2 \) is O=C=O, so two C=O bonds, so per CO2, 2799. So 4 CO2: 42799 = 6392, correct. H2O: 62463 = 5556, correct. Reactants: 2 ethanol: 2(5413 + 348 + 358 + 463) + 7498. Let's recalculate ethanol BDE: 5413=2065, +348=2413, +358=2771, +463=3234. For 2 ethanol: 23234=6468. O2: 7498=3486. Total reactant: 6468 + 3486 = 9954. Products: 42799=6392, 62*463=5556. Total product: 6392 + 5556 = 11948. So \( \Delta H = 9954 - 11948 = -1994 \). But wait, maybe the ethanol is liquid, so we need to consider vaporization? But the problem says "using bond dissociation energies", which are for gaseous species. Maybe the given reaction has \( \text{CH}_3\text{CH}_2\text{OH} \) as gas? The problem says (l), but maybe it's a typo. Alternatively, maybe I used wrong BDE values. Let's check standard enthalpy of combustion of ethanol. The actual enthalpy of combustion of ethanol (liquid) is about -1367 kJ/mol, so for 2 mol: -2734 kJ. But our calculation is -1994, which is different. Maybe BDE values are approximate. Alternatively, maybe I messed up the number of bonds. Let's re-express ethanol: \( \text{C}_2\text{H}_5\text{OH} \) is \( \text{CH}_3\text{CH}_2\text{OH} \), so bonds:

• C-H: 5 (3 in CH3, 2 in CH2)
• C-C: 1
• C-O: 1
• O-H: 1

So per molecule: 5 C-H, 1 C-C, 1 C-O, 1 O-H. Correct. For 2 molecules: 10 C-H, 2 C-C, 2 C-O, 2 O-…

Enthalpy of formation (\( \Delta H_f^\circ \)) is the enthalpy change when 1 mole of a compound is formed from its most stable elemental forms (C: graphite, H: \( \text{H}_2 \) gas, O: \( \text{O}_2 \) gas) under standard conditions.

• The reaction must have 1 mole of \( \text{C}_2\text{H}_5\text{OH} (l) \) as product, and reactants as elements in their standard states.
• Analyze options:
• Option 1: Decomposition (product is elements, reverse of formation) → No.
• Option 2: Combustion (ethanol reacts with O2 to form CO2 and H2O) → No.
• Option 3: \( 2 \, \text{C (graphite)} + \frac{1}{2} \, \text{O}_2 (g) + 3 \, \text{H}_2 (g)

ightarrow \text{C}_2\text{H}_5\text{OH} (l) \) → This forms 1 mole of ethanol from C (graphite), \( \text{H}_2 \) (g), and \( \text{O}_2 \) (g) (standard states).

• Option 4: Decomposition of 2 ethanol → No.
• Option 5: Formation of 2 ethanol → No (formation is for 1 mole).

Thus, the correct option is the third one.

Answer:

(First Problem: Approximate, assuming BDE values)
-1994 (or -2000, but based on calculation with given O2=7, though unbalanced, but following problem’s numbers)