Bond Energy Calculator

Calculate reaction enthalpy from bond energies

ΔH_rxn = Σ(Bonds Broken) - Σ(Bonds Formed)

Breaking bonds requires energy (+), forming bonds releases energy (-). The net difference is the enthalpy change.

Example Reactions:

Reactants (Bonds Broken) ⬆️ +Energy

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Products (Bonds Formed) ⬇️ -Energy

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Understanding Bond Energies

Bond Energy (Bond Dissociation Energy): The energy required to break one mole of bonds in gaseous molecules. Always a positive value (endothermic process).

  • Breaking bonds: Requires energy input (+)
  • Forming bonds: Releases energy (-)
  • Triple bonds stronger than double bonds stronger than single bonds
  • Average values: Bond energies are averages across many molecules

Key Concept: If more energy is released forming bonds than required to break them, the reaction is exothermic (ΔH < 0). Otherwise, it's endothermic (ΔH > 0).

Understanding Entropy

Entropy (S) is a measure of disorder or randomness in a system. The Second Law of Thermodynamics states that the entropy of the universe always increases in spontaneous processes.

ΔS° = Σ(S°products) - Σ(S°reactants)

  • S° = standard molar entropy
  • Units: J/(mol·K)
  • Measured at 25°C, 1 atm

Entropy Trends

  • ↑Gas > Liquid > Solid
  • ↑Higher temperature = higher entropy
  • ↑More moles of gas = higher entropy
  • ↑Larger, more complex molecules = higher entropy
  • ↑Dissolved solutes = higher entropy

Where It's Used

  • •Gibbs Free Energy: ΔG = ΔH - TΔS determines spontaneity
  • •Equilibrium: Understanding reaction direction
  • •Phase Changes: Melting, boiling, sublimation
  • •Chemical Engineering: Process optimization

Example Calculation

Problem:

Calculate ΔS° for: 2H₂(g) + O₂(g) → 2H₂O(l)

Step 1: Find S° values

S°[H₂(g)] = 130.7 J/(mol·K)
S°[O₂(g)] = 205.2 J/(mol·K)
S°[H₂O(l)] = 69.9 J/(mol·K)

Step 2: Calculate products sum

Σ(S°products) = 2 × 69.9
= 139.8 J/(mol·K)

Step 3: Calculate reactants sum

Σ(S°reactants) = 2(130.7) + 1(205.2)
= 261.4 + 205.2 = 466.6 J/(mol·K)

Step 4: Calculate ΔS°

ΔS° = 139.8 - 466.6
= -326.8 J/(mol·K)

Result:

ΔS° is negative because 3 moles of gas form 2 moles of liquid (decreased disorder).

Predicting ΔS° Sign

ChangeΔS° Sign
Gas moles increase+
Gas moles decrease-
Solid → Liquid → Gas+
Gas → Liquid → Solid-
Dissolving (usually)+
Simple → Complex molecules+

Quick Rule:

If Δngas > 0 (more gas moles produced), ΔS° is usually positive. If Δngas < 0 (fewer gas moles produced), ΔS° is usually negative.

Standard Molar Entropy Values (S° at 25°C, 1 atm)

SubstancePhaseS° [J/(mol·K)]Note
H₂gas130.7Diatomic gas
O₂gas205.2Diatomic gas
N₂gas191.6Diatomic gas
CO₂gas213.8Linear molecule
H₂Oliquid69.9Liquid water
H₂Ogas188.8Steam
NH₃gas192.8Ammonia
CH₄gas186.3Methane
C(graphite)solid5.7Very ordered solid
NaClsolid72.1Ionic solid

Entropy and Spontaneity

Second Law of Thermodynamics:

ΔSuniverse = ΔSsystem + ΔSsurroundings

For spontaneous processes: ΔSuniverse > 0

  • ✓Universe entropy always increases
  • ✓System entropy can decrease if surroundings entropy increases more

Gibbs Free Energy Connection:

ΔG = ΔH - TΔS

Spontaneous when ΔG < 0

ΔHΔSSpontaneous?
-+Always
+-Never
--Low T only
++High T only

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