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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Average Bond Enthalpy Method

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Bond Energies

Enthalpy (H) is the energy stored in bonds.

Bond dissociation energy (BDE) is the energy needed to break a chemical bond.

ΔHrxn=[(∑nHbonds broken)−(∑nHbonds formed )]\Delta H_{rxn}=\left[\left(\sum_{ }^{ }nH_{bonds\ broken}\right)-\left(\sum_{_{ }}^{ }nH_{bonds\ formed\ }\right)\right]

Or (an easier way to think about it):

ΔHrxn=[(∑nBDEreactants)−[(∑nBDEproducts)]\boxed{\Delta Hrxn=[(\sum n BDE_{reactants})-[({\sum nBDE_{products})]}}

ΔHrxn=\Delta H_{rxn}= enthalpy change for the reaction (kJ/mol)
BDE = bond energy per mole of bonds (kJ/mol), always positive

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ΔHrxn=[(∑nBDEreactants)−[(∑nBDEproducts)]\boxed{\Delta Hrxn=[(\sum n BDE_{reactants})-[({\sum nBDE_{products})]}}


Photo by Rice University / CC BY

Example: Using Bond Dissociation Energies to Calculate Enthalpy of a Reaction

What is ΔHrxn for the following reaction?
2H2(g)+O2(g)→2H2O(g)2H_2(g) +O_2(g) \rightarrow 2H_2O(g)

Bond Bond Energy (kJ/mol)
H-H 436
O=O 499
O-H 463


ΔHrxn=[(∑nBDEreactants)−(∑nBDEproducts)]{\Delta Hrxn=[(\sum n BDE_{reactants})-{(\sum nBDE_{products})}}]

ΔHrxn=[(2x436 + 499) - (4x463)]
=1371 - 1852
=-481kJ

Practice: Calculating the Enthalpy of a Reaction Using Bond Energies

Use the table of average bond energies to calculate the ΔH{\small \Delta \text{H}} for the following reaction:

H−C≡C−H→H2C=CH2\text{H}-\text{C}\equiv\text{C}-\text{H}\rightarrow\text{H}_2\text{C}=\text{CH}_2

Bond‾Bond Energy (kJ/mol)‾H−H432H−C413H−I295I−I149C−I240C−C347C=C614C≡C839\begin{array}{cc} {\small \underline{\text{Bond}}}&{\small \underline{\text{Bond Energy (kJ/mol)}}}\\ \text{H}-\text{H} & 432 \\ \text{H}-\text{C} & 413\\ \text{H}-\text{I} & 295\\ \text{I}-\text{I} & 149\\ \text{C}-\text{I} & 240\\ \text{C}-\text{C} & 347\\ \text{C}=\text{C} & 614\\ \text{C}\equiv\text{C} & 839 \end{array}

Extra Practice
Bond Energy
What is the total bond energy for the following molecule?
C - H 413 kJ/mol
C - C 347 kJ/mol
Practice: Bond Energy
Calculate the BDE of N2 from the following reaction:
N2 (g) + 2H2 (g) → N2H4 (g); ∆Hr = +99 kJ / mol
H-H: BDE=436 kJ/mol
Bond Energy: Enthalpy of dissociation
Given the reaction: N2(g) + 3H2(g) → 2NH3(g), what is the bond dissociation energy for the nitrogen to hydrogen (N-H) bond? (bond dissociation energy of N≡N bond = 928 kJ/mol,bond dissociation energy of H-H bond = 436 kJ/mol, and Δ\DeltaHrxn = -92.0 kJ
Bond Energy
What is the total bond energy for this molecule?
C - H 413 kJ/mol
C - O 358 kJ/mol
Enthalpy of Reaction
Use the table below to answer the following questions
Estimate the Enthalpy of the following reaction
Bond Energy: Bond enthalpy

Use the table below to answer] the following questions:
Estimate the enthalpy of the following unbalanced reactions:
SO2(g)+C(s)→S(g)+CO2(g)SO_{2(g)}+C_(s)\to S_(g)+CO_{2(g)}
Use the table below to answer the following questions:
When the reaction below was performed the change in enthalpy was measured to be 474 kJ / mol. Estimate the strength of a S-Cl bond.
Use the table below to answer the following questions:
Estimate the enthalpy change of the following reaction:
Bond Energy
Consider the combustion reaction:
CH4 + 2O2 → CO2 + 2H2O
BEs in kJ / mol: C―H = 414, O=O = 498, C=O = 736, O―H = 464