Reactions are either endothermic or exothermic – the heat transferred during a chemical reaction is either moving from the system to the surroundings or vice versa and is always the consequence of breaking bonds in the reactants and making new bonds in the products.
We can show this on an enthalpy profile diagram:
CH4(g) + 2O2(g) ⇾ CO2(g) + 2H2O(g) ΔrH = -890 kJ mol-1
- Ea is the activation energy – the energy we need to put in to start breaking bonds in the reactants (overcoming the electrostatic attraction between shared pairs of electrons and nuclei). This energy might come from the surroundings or from heating.
- once the reaction gets going in an exothermic reaction, energy is produced and the reaction becomes self-sustaining.
In exothermic reactions more energy is released making stronger bonds in the products than is used to break the bonds in the reactants, hence the fall in enthalpy as the reaction proceeds and a negative value for ΔrH, the overall enthalpy change of reaction.
CaCO3(s) ⇾ CaO(s) + CO2(g) ΔrH = +178 kJ mol-1
- in an endothermic reaction more energy is required to break strong bonds in the reactants than is released by making weaker bonds in the products, hence the overall rise in enthalpy over the course of the reaction and a positive value for ΔrH
Bond dissociation enthalpy is the enthalpy change per mole when a particular chemical bond is broken, with the substance being in the gas phase. It is always endothermic.
In reality, data tables always give us an average bond enthalpy value for a particular bond e.g. C-H or C=O …
Calculate ΔrH for the following reaction given average bond enthalpy data:
C2H6(g) + 3½ O2(g) ⇾ 2 CO2(g) + 3 H2O(g)
| Bond | Bond enthalpy / kJ mol-1 |
| C-C | 347 |
| C-H | 412 |
| O=O | 498 |
| C=O | 805 |
| O-H | 464 |
Step 1 is to ALWAYS draw out each molecule and all the bonds – you’d be amazed how easy it is to miss a bond out:
Step 2: add up the bond enthalpies for the all the bonds that are broken (in the reactants) and for all the bonds made (in the products)
- bonds broken: 347 + (6 x 412) + (3.5 x 498) = + 4562 kJ mol-1 (endothermic, positive value)
- bonds made: (4 x -805) + (6 x -464) = – 6004 kJ mol-1 (exothermic, negative value)
Step 3: add the two values together
ΔrH = (+ 4562) + (- 6004) = – 1442 kJ mol-1
Note that we have calculated a value for the enthalpy change of reaction, not the enthalpy change of combustion, because in the equation water has a gaseous state symbol, not a liquid state symbol (check the definitions!).
Practice questions
| Bond | Average bond enthalpy / kJ mol-1 |
| C-H | +413 |
| C=O | +805 |
| O=O | +498 |
| O-H | +464 |
| C-O | +358 |
| C-C | +347 |
| C=C | +610 |
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- Two oxides of nitrogen, N2O and NO, have endothermic standard enthalpies of formation.
(a) Explain, in terms of bond making and bond breaking, why the standard enthalpy of formation for NO is endothermic.
(b) Draw a fully labelled enthalpy profile diagram to show the standard enthalpy of formation of N2O. Include the reactants, products and relevant state symbols.
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- Ethene burns in oxygen to give carbon dioxide and water.
C2H4(g) + 3O2(g) ⇾ 2CO2(g) + 2H2O(g) ΔrH = -1318 kJ mol-1
(a) Explain, in terms of bond making and bond breaking, why the enthalpy change for this reaction is exothermic.
(b) Using the data in the table above, calculate the average bond enthalpy for a C=C bond in ethene.
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3. (a) Write an equation to represent the standard enthalpy of formation for ethanol, C2H5OH, including state symbols.
(b) Calculate the enthalpy of combustion (in kJ mol-1) for 1 mole of ethanol using the average bond enthalpy data in the table above.
C2H5OH(l) + 3O2(g) ⇾ 2CO2(g) + 3H2O(l)
(c) Draw a fully labelled enthalpy profile diagram to show the standard enthalpy of combustion of ethanol. Include the reactants, products and relevant state symbols.
(d) The data book value for the standard enthalpy change of combustion of ethanol is -1367 kJ mol-1. Give the main reason why this value is different from your answer to part (b).
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4. An expression has been derived using average bond enthalpy data to estimate ΔcH for an alkane. If the alkane has n carbon atoms, ΔcH = – (496n + 202) kJ mol-1.
(a) Using this expression, ΔcH for an alkane was estimated to be -6650 kJ mol-1.
Determine the molecular formula of the alkane.
(b) Suggest why a value for the enthalpy of combustion of a liquid alkane would be different from the value obtained from using the expression in part (a), other than the use of average bond enthalpies.
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5. Propan-2-ol, CH3CH(OH)CH3, can be made by hydrating propene, CH3CH=CH2. ΔrH for this hydration reaction is -50 kJ mol-1. Use this information, in addition to data from the table above, to calculate the average bond enthalpy for a C-C bond in these molecules.
Answers
- (a) ΔfH is endothermic because more energy is needed to break the bonds in the reactants (N2, O2) than is released making the new bonds ini the product (NO).
2. (a) The energy released when new bonds are made in the products (CO2,H2O) is greater than the energy needed to break the bonds in the reactants (C2H4, O2).
3. (a) 2C(s) + 3H2(g) + 1½O2(g) ⇾ C2H5OH(l)
(d) The value calculated above uses average bond enthalpies, which differ slightly from one compound to another in reality and are for gaseous molecules. ΔcH⦵ is for all substances in their standard states at 298K and 100kPa so ethanol is a liquid (as is water) and not a gas.
4. (a) ΔcH⦵ = – (496 n + 202) = -6650 kJ mol-1
496 n = 6650 – 202 = 6448
n = 13
Alkane formula = CnH2n+2 = C13H28
(b) ΔcH⦵ for an alkane is based on the balanced equation in which all substances are in their standard states at 298K and 100kPa, so the alkane is not a gas (as is the case when using average bond enthalpy data).
5.
