Syllabus Edition

First teaching 2023

First exams 2025

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Bond Enthalpy Calculations (SL IB Chemistry)

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Alexandra

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Alexandra

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Chemistry

Bond Enthalpy Calculations

  • In 1840, the Russian chemist Germain Hess formulated a law which went on to be known as Hess’s Law
  • This went on to form the basis of one of the laws of thermodynamics. The first law of thermodynamics relates to the Law of Conservation of Energy
  • It is sometimes expressed in the following form:

Energy cannot be created or destroyed, it can only change form

  • This means that in a closed system, the total amount of energy present is always constant
  • Hess’s law can be used to calculate the standard enthalpy change of a reaction from known standard enthalpy changes
  • Hess’s Law states that:

"The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same."

  • This means that whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same

Diagram to show Hess's Law

Hess's law

The diagram above illustrates Hess’ Law: the enthalpy change of the direct route, going from reactants (A+B) to product (C) is equal to the enthalpy change of the indirect routes

  • Hess’ Law is used to calculate enthalpy changes which can’t be found experimentally using calorimetry, eg:

3C (s) + 4H2 (g) → C3H8(g)

  • ΔHf (propane) can’t be found experimentally as hydrogen and carbon don’t react under standard conditions

Calculating ΔHr from ΔHf using Hess’s Law energy cycles

  • You can see the relationships on the following diagram:

Diagram to show Hess's Law 

hesss-law-cycle

The enthalpy change from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route)

 

  • The products can be directly formed from the elements = ΔH2

OR

  • The products can be indirectly formed from the elements = ΔH1 + ΔHr
  • Equation

ΔH2 = ΔH1 + ΔHr

Therefore for energy to be conserved,

ΔHr = ΔH2 – ΔH1

Average bond energy

  • Bond energies are affected by other atoms in the molecule (the environment)
  • Therefore, an average of a number of the same type of bond but in different environments is calculated
  • This bond energy is known as the average bond energy and is defined as
    • 'The energy needed to break one mole of bonds in a gaseous molecule averaged over similar compounds

average-bond-enthalpies

Average bond enthalpy of C-H in methane

  • The average bond enthalpy of C-H is found by taking the bond dissociation enthalpy for the whole molecule and dividing it by the number of C-H bonds
  • The first C-H bond is easier to break than the second as the remaining hydrogens are pulled more closely to the carbon
  • However, since it is impossible to measure the energy of each C-H bond an average is taken
  • This value is also compared with a range of similar compounds to obtain an accepted value for the average bond enthalpy

Bond enthalpy calculations

  • Bond energies are used to find the ΔHr of a reaction when this cannot be done experimentally
  • The process is a step-by-step summation of the bond enthalpies of the all the molecules present finishing with this formula:

w5wvYaa9_equation-for-calculating-standard-enthalpy-change-using-bond-energies

Formula for calculating the standard enthalpy change of reaction using bond energies

  • These two worked examples show how to lay out your calculation

Worked example

You do not need to learn Hess's Law word for word as it is not a syllabus requirement, but you do need to understand the principle as it provides the foundation for all the problem solving in Chemical Energetics

Worked example

The complete combustion of ethyne, C2H2 , is shown in the equation below:

2C2H2 (g)  + 5 O2 (g)      2H2O (g) + 4CO2 (g)

Using the average bond enthalpies given in the table, what is the enthalpy of combustion of ethyne?

Bond Average Bond Energy (kJ mol-1)
Cnegative signH 414
Cidentical toC 839
OequalsO 498
CequalsO 804
OminusH 463
Onegative signC 358

Answer:

  • Step 1: The enthalpy of combustion is the enthalpy change when one mole of a substance reacts in excess oxygen to produce water and carbon dioxide

   The chemical reaction should be therefore simplified such that only one mole of ethyne reacts in excess oxygen:

      H-C≡C-H + 2 ½ O=O → H-O-H + 2O=C=O

  • Step 2: Set out the calculation as a balance sheet as shown below:
Bonds broken  (kJ mol-1) Bonds formed  (kJ mol-1)
2 x CH = 2 x 414= 828
1 x CC= 1 x 839= 839
2½ OO = 2½  x 498 = 1245
2 x OH= 2 x 463 = 926
4 x CO =  4 x 804 = 3216
Total = +2912 Total= -4142

ΔHr = enthalpy change for bonds broken + enthalpy change for bonds formed

      = (+2912 kJ mol-1) + (- 4142 kJ mol-1)

      = -1230 kJ mol-1

Exam Tip

The key to success in bond enthalpy calculations is to be very careful when accounting for every bond present. Always draw out the full displayed structures of the molecules so you don't miss any of the bonds.

Watch out for coefficients in the balanced equations as students often miss those, forget to multiply them by the bond enthalpies and get the answer wrong!

It is super important to show your steps because bond enthalpy calculations often carry 3 marks, 2 of which could be for workings if you get the final answer wrong

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Alexandra

Author: Alexandra

Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.