Ch6: Bond Energy and Energy at Molecular Level

Energy Changes in Chemical Reactions (Sub-Microscopic Perspective)

  • Chemists must look at the "sub-microscopic" (molecular) picture to understand why reactions give off or absorb energy.
  • KEY QUESTION often asked for any exothermic reaction: “Where does the released energy actually come from?”
    • Answer lies not in macroscopic heat flow, but in the making and breaking of chemical bonds.

Two Fundamental Processes in Every Ordinary (Non-Nuclear) Reaction

  • 1️⃣ Bond Breakage
    • Always an endothermic step (energy must be absorbed).
    • Symbolically: E_{\text{break}}>0 (positive sign).
  • 2️⃣ Bond Formation
    • Always an exothermic step (energy is released).
    • Symbolically: E_{\text{form}}<0 (negative sign).
  • Net energy change ((\Delta E) or (\Delta H) under constant pressure) depends on numerical difference between these two totals: ΔE=∑E<em>break+∑E</em>form\Delta E = \sum E<em>{\text{break}} + \sum E</em>{\text{form}}
    • Negative (\Delta E): reaction is exothermic.
    • Positive (\Delta E): reaction is endothermic.
  • Practical consequence: If more (or “stronger”) bonds are formed than are broken, reaction releases heat; if the reverse is true, it absorbs heat.

Definition & Sign Convention for Bond Energy

  • Bond Energy (BE): “Energy required to break 1 mole of a given bond in the gas phase.”
    • Units: kJ    mol−1\text{kJ \; mol}^{-1}
    • Same magnitude is released when the identical bond is made (but with opposite sign).
    • Example: If breaking an X–X bond needs +200 kJ mol−1+200\,\text{kJ mol}^{-1}, then forming X–X releases −200 kJ mol−1-200\,\text{kJ mol}^{-1}.

Qualitative Trends in Bond Energies (Table Snippets Discussed)

  • Not all bonds have the same strength; numerical values differ by bond order and by element pair.
  • Representative values cited in class:
    • H–H (single): ≈436 kJ mol−1\approx 436\,\text{kJ mol}^{-1} (table value shown only implicitly)
    • C–C (single): 356 kJ mol−1356\,\text{kJ mol}^{-1}
    • C–C (double): 602 kJ mol−1602\,\text{kJ mol}^{-1}
    • C≡C (triple): 837 kJ mol−1837\,\text{kJ mol}^{-1} – highest for C–C series.
    • C–H (single): 416 kJ mol−1416\,\text{kJ mol}^{-1}
    • O=O (double in O₂): 498 kJ mol−1498\,\text{kJ mol}^{-1}
    • C=O (double in CO₂): 803 kJ mol−1803\,\text{kJ mol}^{-1}
    • O–H (single in H₂O): 467 kJ mol−1467\,\text{kJ mol}^{-1}
  • Strength trend for the same atomic pair:
    \text{triple} > \text{double} > \text{single} (more shared electron pairs ⇒ stronger bond ⇒ higher BE).

Worked Example: Heat of Combustion of Propane (C₃H₈)

1. Balanced Chemical Equation

C<em>3H</em>8+5 O<em>2→3 CO</em>2+4 H2O\mathrm{C<em>3H</em>8 + 5\,O<em>2 \rightarrow 3\,CO</em>2 + 4\,H_2O}

2. Draw Structural Formulas (Needed to Count Bonds)

  • Propane: CH₃–CH₂–CH₃
    • 2 C–C single bonds
    • 8 C–H single bonds
  • O₂: O=O (double); 5 molecules ⇒ 5 O=O bonds.
  • CO₂: O=C=O; each molecule has 2 C=O double bonds; 3 molecules ⇒ 6 C=O bonds.
  • H₂O: H–O–H; each molecule has 2 O–H single bonds; 4 molecules ⇒ 8 O–H bonds.

3. Categorize by Process

Reactants (bonds broken, ++)Products (bonds formed, −-)
2 × C–C (+2×356)(+2\times356)6 × C=O (−6×803)(−6\times803)
8 × C–H (+8×416)(+8\times416)8 × O–H (−8×467)(−8\times467)
5 × O=O (+5×498)(+5\times498)

4. Numerical Totals

  • Energy required to break bonds (reactants):
    E<em>break=2(356)+8(416)+5(498)E<em>{\text{break}} = 2(356) + 8(416) + 5(498)E</em>break=712+3328+2490=6530 kJ mol−1E</em>{\text{break}} = 712 + 3328 + 2490 = 6530\,\text{kJ mol}^{-1}
  • Energy released on forming bonds (products):
    E<em>form=−[6(803)+8(467)]E<em>{\text{form}} = -\big[6(803) + 8(467)\big]E</em>form=−[4818+3736]=−8554 kJ mol−1E</em>{\text{form}} = -\big[4818 + 3736\big] = -8554\,\text{kJ mol}^{-1}

5. Net Heat of Combustion

ΔE=E<em>break+E</em>form=6530+(−8554)\Delta E = E<em>{\text{break}} + E</em>{\text{form}} = 6530 + ( -8554 )
ΔE=−2024 kJ mol−1\boxed{\Delta E = -2024\,\text{kJ mol}^{-1}}

  • Negative sign confirms the reaction is exothermic.
  • Same procedure (count bonds → insert average BE values → sum) applies to any combustion or other reaction once structural formulas are known.

6. Conceptual Interpretation

  • Although 6.53 MJ are absorbed to rip apart the original C–C, C–H, and O=O bonds, the newly formed C=O and O–H bonds release 8.55 MJ.
  • The excess release (≈2.0 MJ) is the heat we observe during propane burning.

Connections & Practical Notes

  • Relates directly to Hess’s Law: total enthalpy change depends only on initial & final states; bond-energy approach is one practical application.
  • Uses average bond energies, so computed (\Delta E) is approximate; experimental heats of combustion are typically a few percent different.
  • If calculated under constant pressure conditions, result is reported as ΔHcombustion\Delta H_{\text{combustion}}, but the numerical procedures are analogous.
  • Real-world relevance:
    • Guides fuel choice/efficiency.
    • Helps design endothermic vs. exothermic industrial processes.

Quick ‘Checklist’ for Bond-Energy Calculations

  1. Balance the reaction.
  2. Draw full structural/Lewis formulas for all species.
  3. On reactant side, count each unique bond type → multiply by tabulated BE; treat as positive.
  4. On product side, do the same → treat as negative.
  5. Add reactant and product totals to obtain net (\Delta E).
    • Negative = exothermic
    • Positive = endothermic
  6. Interpret magnitude & sign; link back to physical observation (heat released/absorbed).

Key Takeaways

  • Energy changes in chemical reactions are fully traceable to bond energetics.
  • Breaking bonds costs energy; making bonds pays it back—often with interest.
  • Average bond energy tables are powerful predictive tools, even though they are approximations.
  • Propane’s combustion liberates about 2.0 MJ mol−12.0\,\text{MJ mol}^{-1} due to stronger C=O & O–H bonds replacing weaker C–C, C–H, and O=O bonds.