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ASChemistryCIERevision Notes5. Chemical Energetics5.1 Enthalpy Change, ΔHEnthalpy & Bond Energies

Enthalpy & Bond Energies (CIE AS Chemistry)

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Enthalpy & Bond Energies

  • During a reaction, enthalpy changes take place because bonds are being broken and formed
  • Energy (in the form of heat) is needed to overcome attractive forces between atoms
  • Bond breaking is therefore endothermic
  • Energy is released from the reaction to the surroundings (in the form of heat) when new bonds are formed
  • Bond forming is therefore exothermic

Making and breaking bonds 

Chemical Energetics Bond Breaking and Forming, downloadable AS & A Level Chemistry revision notes

Breaking bonds requires energy from the surroundings and making bonds releases energy to the surroundings

  • If more energy is required to break bonds than energy is released when new bonds are formed, the reaction is endothermic
  • If more energy is released when new bonds are formed than energy is required to break bonds, the reaction is exothermic
  • In reality, only some bonds in the reactants are broken and then new ones are formed

Enthalpy Calculations

Exact bond energy

  • The amount of energy required to break one mole of a specific covalent bond in the gas phase is called the bond dissociation energy
  • Bond dissociation energy (E) is also known as exact bond energy or bond enthalpy
  • The type of bond broken is put in brackets after E
    • Eg. EE(H-H) is the bond energy of a mole of single bonds between two hydrogen atoms

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
  • Since bond energies cannot be determined directly, enthalpy cycles are used to calculate the average bond energy

How bond energies are affected by other atoms in the molecule

Chemical Energetics Exact and Average Bond Energies, downloadable AS & A Level Chemistry revision notes

Bond energies are affected by other atoms in the molecule

Calculating enthalpy change from bond energies

  • Bond energies are used to find the ΔHrꝋ of a reaction when this cannot be done experimentally
    • E.g. the Haber Process
  • The equation to calculate the standard enthalpy change of reaction using bond energies is:

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

Worked example

Calculate the enthalpy change of reaction the Haber process reaction.

The relevant bond energies are given in the table.

Bond Average bond energy
kJ mol-1
Nbegin mathsize 16px style identical to end styleN 945
H–H 436
N–H 391

Answer

  • Step 1: The chemical equation for the Haber process is:
    • N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
  • Step 2: Set out the calculation as a balance sheet:
Bonds broken
kJ mol-1		 Bonds formed
kJ mol-1

1 x Nidentical toN = 1 x 945 = 945

3 x H–H = 3 x 436 = 1308

6 x N–H = 6 x 391 = 2346
Total = +2253 Total = –2346
    • Note: Values for bonds broken are positive (endothermic) and values for bonds formed are negative (exothermic)
  • Step 3: Calculate the standard enthalpy of reaction:
    • ΔHθr = enthalpy change for bonds broken + enthalpy change for bonds formed
    • ΔHθr = (+2253 kJ mol-1) + (–2346 kJ mol-1)
    • ΔHθr = –93 kJ mol-1

Worked example

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

2H–Cidentical toC–H + 5O=O → 2H–O–H + 4O=C=O

Calculate, using the average bond enthalpies given in the table, the enthalpy of combustion of ethyne.

Bond Average bond energy
kJ mol-1
C–H 414
Cidentical toC 839
O=O 498
C=O 804
O–H 463
O–C 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:
Bonds broken
kJ mol-1		 Bonds formed
kJ mol-1

1 x Cidentical toC = 1 x 839 = 839

2 x C–H = 2 x 414 = 828

2½ x O=O = 2½ x 498 = 1245 

2 x O–H = 2 x 463 = 926

4 x C=O = 4 x 804 = 3216
Total = +2912 Total = –4142
    • Note: Values for bonds broken are positive (endothermic) and values for bonds formed are negative (exothermic)
  • Step 3: Calculate the standard enthalpy of reaction:
    • ΔHθr = enthalpy change for bonds broken + enthalpy change for bonds formed
    • ΔHθr = (+2912 kJ mol-1) + (–4142kJ mol-1)
    • ΔHθr = –1230 kJ mol-1 

Measuring enthalpy changes

  • Calorimetry is a technique used to measure changes in enthalpy of chemical reactions
  • A calorimeter can be made up of a polystyrene drinking cup, a vacuum flask or metal can

 Example calorimetry equipment

Chemical Energetics Calorimeter, downloadable AS & A Level Chemistry revision notes

A polystyrene cup can act as a calorimeter to find enthalpy changes in a chemical reaction

  • The energy needed to increase the temperature of 1 g of a substance by 1 oC is called the specific heat capacity (c) of the liquid
  • The specific heat capacity of water is 4.18 J g-1 oC-1
  • The energy transferred as heat can be calculated using the following equation:

q = m x c x ΔT

    • q = the heat transferred, J
    • m = the mass of water, g
    • c = the specific heat capacity, J g-1 oC-1 or J g-1 K-1 
    • T = the temperature change, oC

Worked example

In a calorimetric experiment, 2.50 g of methane is burnt in excess oxygen. 

30% of the energy released during the combustion is absorbed by 500 g of water, causing the temperature to rise from 25 oC to 68 oC.

The specific heat capacity of water is 4.18 J g-1 K-1.

Calculate the total energy released per gram of methane burnt.

Answer

  • Step 1: Gather the necessary values for the q = m x c x ΔT equation:
    • m (of water) = 500 g
    • c (of water) = 4.18 J g-1 K-1
    • ΔT (of water) = 68 oC - 25 oC = 43 oC = 43 K
      • The change in temperature in oC is equal to the change in temperature in K
  • Step 2: Complete the calculation:
    • q = 500 x 4.18 x 43
    • q = 89 870 J
  • Step 3:  Conversion from 30% to 100%
    • The value of q calculated is only 30% of the total energy released by 2.50 g of methane
    • Total energy x 0.3 = 89 870 J
    • Total energy = 299 567 J
  • Step 4: Calculate the energy released by 1.00 g of methane
    • The energy calculated is released by 2.50 g of methane
    • Energy released by 1.00 g of methane = fraction numerator 299567 over denominator 2.50 end fraction
    • Energy released by 1.00 g of methane = 119 827 J
  • Step 5: Convert the answer from J to kJ:
    • 119827 over 1000 = 120 kJ (to 3 significant figures)

Examiner Tip

  • When new bonds are formed the amount of energy released is equal to the amount of energy absorbed when the same bonds are broken
  • For example:

O2 (g) → 2O (g)   E (O=O) = +498 kJ mol-1

2O (g) → O2 (g)   E (O=O) –498 kJ mol-1

  • Aqueous solutions of acid, alkalis and salts are assumed to be largely water so you can just use the m and c values of water when calculating the energy transferred
  • To then calculate any changes in enthalpy per mole of a reactant or product the following relationship can be used:

ΔH = –m x c x ΔT

  • When there is a rise in temperature, the value for ΔH becomes negative suggesting that the reaction is exothermic
  • When there is a fall in temperature, the value for ΔH becomes positive suggesting that the reaction is endothermic
  • Also, remember that ΔT is the same in oC and K!

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    Caroline

    Author: Caroline

    Expertise: Physics Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.