Let’s solve this step by step.
We are given the reaction:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)
and told that
ΔH° = -2220 kJ for this reaction (this is the standard enthalpy of combustion of propane).
We are also told to use bond energies from a table — but since no table is provided, we’ll assume standard average bond energies commonly used in textbooks. Here are typical values (in kJ/mol):
- C–C: 347
- C–H: 413
- O=O: 498
- C=O (in CO₂): 799 *(note: CO₂ has two C=O bonds per molecule)*
- O–H: 463
*(Note: These are approximate averages; actual values may vary slightly depending on source, but these are standard for such problems.)*
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Step 1: Understand what ΔH° means here
The given ΔH° = -2220 kJ is the
actual experimental value for the combustion of 1 mole of propane.
But the problem asks us to calculate ΔH° using
bond energies, and then compare it to the actual value to find the error.
So we will compute ΔH° using bond energies via:
>
ΔH° ≈ Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)
Because breaking bonds requires energy (positive), forming bonds releases energy (negative).
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Step 2: Count bonds broken (reactants)
Reactants:
1 mol C₃H₈ and
5 mol O₂
#### In C₃H₈ (propane):
Structure: CH₃–CH₂–CH
Bonds:
- 2 C–C bonds
- 8 C–H bonds
→ Energy to break = 2×(C–C) + 8×(C–H)
= 2×347 + 8×413
= 694 + 3304 =
3998 kJ
#### In 5 O₂:
Each O₂ has 1 O=O bond → 5 × O=O
= 5 × 498 =
2490 kJ
Total energy absorbed (bonds broken) = 3998 + 2490 =
6488 kJ
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Step 3: Count bonds formed (products)
Products:
3 mol CO₂ and
4 mol H₂O(l)
#### In 3 CO₂:
Each CO₂ has 2 C=O bonds → 3 molecules × 2 = 6 C=O bonds
Energy released = 6 × 799 =
4794 kJ
#### In 4 H₂O:
Each H₂O has 2 O–H bonds → 4 molecules × 2 = 8 O–H bonds
Energy released = 8 × 463 =
3704 kJ
Total energy released (bonds formed) = 4794 + 3704 =
8498 kJ
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Step 4: Calculate ΔH° using bond energies
ΔH° = (energy to break bonds) – (energy released when bonds form)
= 6488 – 8498 =
–2010 kJ
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Step 5: Compare with actual value and find percent error
Actual ΔH° = –2220 kJ
Calculated ΔH° = –2010 kJ
Error = |calculated – actual| / |actual| × 100%
= |–2010 – (–2220)| / 2220 × 100%
= |210| / 2220 × 100%
≈
9.46%
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Final Answer:
–2010 kJ, 9.46%
Parent Tip: Review the logic above to help your child master the concept of enthalpy calculations worksheet.
