1. Chemical Energy
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2. Enthalpy 07/12/2018 I can understand what is meant by “enthalpy”.
I can give a definition of enthalpy of combustion.
I can use specific heat capacity, mass and temperature to calculate the enthalpy change for a reaction.
I can work out the enthalpy of combustion from practical experiments.

3. Exothermic and Endothermic reactions
REVISION
P.E.
Endothermic
P.E. Ea
-H Ea
+ H
Exothermic
Exothermic reactions give out energy
Endothermic reactions take in energy from the surroundings

4. Enthalpy Chemical energy
Enthalpy (H) is a measure of the energy stored in a chemical. 
In the previous sections we have seen potential energy diagrams during exothermic and endothermic reactions.
It is important to recognise we cannot measure the enthalpy of a chemical, however we can measure the enthalpy change in a reaction.
The enthalpy of combustion is the heat energy given out when 1 mole of fuel burns completely in oxygen.
The enthalpy of combustion of methane can be represented by the equation
CH4(g) + O2 (g)  CO2(g) H2 O(l)
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5. Combustion equations
For each of the following write a balanced equation to represent the complete combustion.
Propane
Cyclohexane
Hydrogen
Glucose
Ethanol
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6. Calculating Enthalpy Changes
REVISION
The energy change of a reaction can be determined by measuring the temperature change a reaction causes in a known mass of water.
The following equation can then be used to calculate the change in heat energy:
Where,
Eh = heat energy gained or lost by the water (kJ)
c = specific heat capacity of water = 4.18 kJ kg-1 oC-1
m = mass of water (kg)
∆T = change in temperature of the water (oC)
Eh = c m ∆T

7. Enthalpy of Combustion
The enthalpy of combustion is the quantity of energy released when 1 mole of a substance completely burns in oxygen.

8. Enthalpy of combustion
The enthalpy of combustion of a substance is the amount of energy given out when one mole of a substance burns in excess oxygen.
Worked example 1.
0.19 g of methanol, CH3OH, is burned and the heat energy given out increased the temperature of 100g of water from 22oC to 32oC. Calculate the enthalpy of combustion of methanol.
Use DH = cmDT
DH = x x 10
DH = kJ
c = 4.18 kJ kg-1 oC-1)
m = 0.1 kg
DT = 10o
NOTE: The change of sign and units
0.19 g  kJ
32 g  32/0.19 x = 704 kJ
Enthalpy of combustion of methanol is –704 kJ mol-1.

9. DH = cmDT 1 mole = 46g of ethanol 0.25 g  6.897 kJ
0.25g of ethanol, C2H5OH, was burned and the heat given out raised the temperature of 500 cm3 of water from 20.1oC to 23.4oC.
DH = cmDT
1 mole = 46g of ethanol
0.25 g  kJ
DH = x 0.5 x 3.3
46g  46/0.25 x
= kJ mol-1
= kJ
0.1 moles of methane was burned and the energy given out raised the temperature of 200cm3 of water from 18oC to 28.6oC. Calculate the enthalpy of combustion of methane.
DH = cmDT
0.1 mol  kJ
DH = x 0.2 x 10.6
1mol  1/0..1 x
= kJ mol-1.
= kJ

10. Worked example
0.22g of propane was used to heat 200cm3 of water at 20oC. Use the enthalpy of combustion of propane in the data book to calculate the final temperature of the water.
From the data booklet 1 mole ( 44g) of propane DH = kJmol-1
By proportion:
44g propane  2219kJ
0.22g propane  /44 x = kJ
Rearrange DH = -cmDT to find DT
DT = DH /cm
DT = 11.1 / (4.18 x 0.2)
DT = oC
Final water temperature =
= 33.3oC

11. 0.1g of methanol, CH3OH, was burned and the heat given out used to raise the temperature of 500 cm3 of water at 21oC.
Use the enthalpy of combustion of methanol in the data booklet to calculate the final temperature of the water.
From the data booklet burning 1 mole, 32g, of methanol DH = -727 kJ
32g of methanol  727 kJ
0.1 g of methanol  0..1/32 x 727
= kJ
Rearrange DH = cmDT to find DT
DT = DH /cm
DT = 2.27 / (4.18 x 0.5)
DT = 5.4 oC
Final water temperature =
= 26.4oC

12. 0.2g of methane, CH4, was burned and the heat given out used to raise the temperature of 250 cm3 of water. Use the enthalpy of combustion of methane in the data booklet to calculate the temperature rise of the water.
From the data booklet burning 1 mole, 16g, of methane DH = -891 kJ
16g of methane 891 kJ
0.2 g of methanol  0.2/16 x 891
= kJ
Rearrange DH = cmDT to find DT
DT = DH /cm
DT = / (4.18 x 0.25)
DT = oC

13. Enthalpy of combustion
The heat energy released when alcohols burn can be measured
The enthalpy of combustion of a substance is the amount of energy given out when one mole of a substance burns in excess oxygen.
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14. Enthalpy of combustion
Procedure
1. Weigh the spirit burner (already containing ethanol) with its cap on and record its mass. (The cap should be kept on to cut down the loss of ethanol through evaporation)
2. Using the measuring cylinder, measure out 100 cm3 of water into the copper can.
3. Set up the apparatus as directed by your teacher/lecturer.
4. Measure and record the temperature of the water.
Remove the cap from the spirit burner and immediately light the burner.
Slowly and continuously stir the water with the thermometer. When the temperature has risen by about 10 °C, recap the spirit burner and measure and record the maximum temperature of the water.
7. Reweigh the spirit burner and record its mass.
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15. The heat energy released on burning 0.25 g of ethanol = 5.225 kJ
CALCULATION
Suppose 0.25 g of ethanol had been burned and the temperature of the water had risen by 12.5 °C.
The heat energy gained by the water (Eh) is calculated using the formula:
Eh = c m DT
Eh = 4.18 x 0.10 x 12.5
= kJ
We assume that the heat energy released by the burning ethanol is gained only by the water.
The heat energy released on burning 0.25 g of ethanol = kJ
Ethanol: CH3CH2OH Mass of 1 mole = 2(12) + 6(1) = 46 g
We can now calculate the heat energy released on burning 1 mole of ethanol.
0.25g  kJ
46g  (46 x 5.225) / 0.25
= kJmol-1
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16. Sources of inaccuracy Heat loss to surroundings
Ignore heat rise of calorimeter
Incomplete combustion
Possible loss of fuel by evaporation from wick
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17. Commercial ‘bomb’ calorimeters
The calorimeter is heated electrically. Energy required to heat the entire apparatus by 1 0C is calculated.
In what ways are the inaccuracies of our class experiment eliminated?
The burning fuel is supplied with oxygen to encourage complete combustion
The combustion chamber is entirely surrounded so there is no heat loss to the surroundings.
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18. Enthalpy of Solution
The enthalpy of solution is the quantity of energy released or absorbed when mole of solute is dissolved in water.

19. Enthalpy of solution
The enthalpy of solution of a substance is the energy change when one mole of a substance dissolves fully in water.
The enthalpy of solution for NaOH can be represented by the equation
NaOH(s)  Na+(aq) OH-(aq
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20. 1 mole of ammonium chloride = 53.5g 5g  1.38 kJ
The enthalpy of solution of a substance is the energy change when one mole of a substance dissolves in water.
Worked example :
5g of ammonium chloride, NH4Cl, is completely dissolved in 100cm3 of water. The water temperature falls from 21oC to 17.7oC.
Use DH = cmDT
DH = x x 3.3
DH = kJ
c = kJ kg-1 oC-1)
m = 0.1 kg
DT = -3.3oC
1 mole of ammonium chloride = 53.5g
5g  1.38 kJ
So g  53.5/5 x = kJ mol-1
Note:- As the temperature falls the reaction is endothermic (takes in heat) this is shown by the positive value for the enthalpy change.

21. Calculations for you to try.
g of ammonium nitrate, NH4NO3, is dissolved in 200cm3 of water. The temperature of the water falls from 20oC to 17.1oC. Calculate the enthalphy of solution.
When 0.1 mol of a compound dissolves in 100cm3 of water the temperature of the water rises from 19oC to 22.4oC . Calculate the enthalpy of solution of the compound.
3. The enthalpy of solution of potassium chloride, KCl, is kJ mol-1. What will be the temperature change when 14.9g of potassium chloride is dissolved in 150cm3 of water? 21

22. ANSWERS
8g of ammonium nitrate, NH4NO3, is dissolved in 200cm3 of water. The temperature of the water falls from 20oC to 17.1oC.
DH = x x 2.9
DH = kJ
1 mole of ammonium nitrate = 80g
8g  2.42 kJ
So 80g  80/8 x = kJ mol-1.
When 0.1 mol of a compound dissolves in 100cm3 of water the temperature of the water rises from 19oC to 22.4oC . Calculate the enthalpy of solution of the compound.
DH = x x 3.4
DH = kJ
0.1 mol  kJ
So 1 mol  1/ x = kJ mol-1.

23. ANSWERS
The enthalpy of solution of potassium chloride, KCl, is kJ mol-1.
What will be the temperature change when 14.9g of potassium chloride is dissolved in 150cm3 of water?
Use proportion to find the energy change (DH) for 14.9g of potassium chloride dissolving.
74.5g (1 mol)  kJ
So g  /74.5 x = kJ
Rearranging DH = cmDT
Gives DT = DH / cm
DT = 3.35 / (4.18 x 0.15)
= oC Decrease
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24. Enthalpy of Neutralisation
The Enthalpy of neutralisation is the quantity of energy released when 1 mole of water is formed from its ions.

25. Enthalpy of neutralisation
The enthalpy of neutralisation of a substance is the amount of energy given out when one mole of water is formed in a neutralisation reaction.
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26. 100cm3 of 1 mol l-1 hydrochloric acid, HCl, was mixed with 100 cm3 of 1 mol -1 sodium hydroxide, NaOH, and the temperature rose by 6.2oC.
Use DH = cmDT
DH = x x 6.2
DH = kJ
c = 4.18 kJ kg-1 oC-1
m = 0.2 kg
DT = 6.2oC
The equation for the reaction is: HCl + NaOH  NaCl H2O
Number of moles of acid / alkali used = C x V = 1 x = moles
So number of moles of water formed = 0.1
0.1 mole  kJ
So 1 mole  1/0.1 x = kJ mol-1.

27. Calculations for you to try.
cm3 of 0.5 mol l-1 hydrochloric acid. HCl, was reacted with 400 cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by 6.4oC . Calculate the enthalpy of neutralisation.
cm3 of 0.5 mol l-1 sulphuric acid. H2SO4, was reacted with 500 cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by 2.1oC. Calculate the enthalpy of neutralisation.
cm3 of 0.5 mol l-1 NaOH is neutralised by 100cm3 of 0.5 mol l-1 HCl. Given that the enthalpy of neutralisation is 57.3 kJ mol-1, calculate the temperature rise.
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28. 400 cm3 of 0. 5 mol l-1 hydrochloric acid
400 cm3 of 0.5 mol l-1 hydrochloric acid. HCl, was reacted with 400 cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by 6.4oC Calculate the enthalpy of neutralisation.
Use DH = cmDT
DH = x x 6.4
DH = kJ
The equation for the reaction is HCl + KOH  KCl H2O
Number of moles of acid / alkali used = C x V (in litres)
= x = moles
So number of moles of water formed = 0.2
0.2 mole  kJ
So 1 mole  /0.2 x = kJ mol-1.
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29. 250 cm3 of 0. 5 mol l-1 sulphuric acid
250 cm3 of 0.5 mol l-1 sulphuric acid. H2SO4, was reacted with 500 cm3 of 0.5 mol l -1 potassium hydroxide and the temperature rose by 2.1oC. Calculate the enthalpy of neutralisation.
Use DH = cmDT
DH = x x 2.1
DH = kJ
The equation for the reaction is:
H2SO KOH  K2SO H2O
1 mole of acid reacts with 2 moles of alkali to form 2 mole of water.
Number of moles of acid used = x =
Number of moles of alkali used = x = 0.25
So number of moles of water formed = 0.25
0.25 mole  kJ
So 1 mole  /0.25 x = kJ mol-1.
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30. 100cm3 of 0. 5 mol l-1 NaOH is neutralised by 100cm3 of 0
100cm3 of 0.5 mol l-1 NaOH is neutralised by 100cm3 of 0.5 mol l-1 HCl. Given that the enthalpy of neutralisation is 57.3 kJ mol-1, calculate the temperature rise.
The equation for the reaction is: HCl + NaOH  NaCl H2O
1 mole of acid reacts with 1 mole of alkali to form 1 mole of water.
Number of moles of acid used = x = 0.05
Number of moles of alkali used = x = 0.05
So number of moles of water formed = 0.05
1 mol  kJ
So mol  /1 x = kJ DH
-cm
DT = = = 3.4oC
2.865
4.18 x 0.2
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31. Enthalpy I can understand what is meant by “enthalpy”.
I can give a definition of enthalpy of combustion.
I can use specific heat capacity, mass and temperature to calculate the enthalpy change for a reaction.
I can work out the enthalpy of combustion from practical experiments.
Hess’s Law

32. Hess’s Law 07/12/2018 State Hess’s Law Applying Hess’s Law:-
 I can use enthalpies of combustion, given in the data book, to calculate enthalpies of formation.
 I can calculate the overall enthalpy change for a reaction using given enthalpy data.

33. Hess’s Law Hess’s Law states that:
the enthalpy change for a reaction is always the same no matter what route was taken to get from reactants to products or
The total energy for the overall reaction is the sum of the energies of all the other steps
This is useful if we want to know the enthalpy change for a reaction which is impractical or impossible to do directly.

34. The aim of this experiment is to verify Hess’s Law using the following example.
Solid potassium hydroxide can be converted into potassium chloride solution using two different routes as shown in the diagram below.
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35. According to Hess’s Law :  H1 =  H2a +  H2b
Route 1 : Solid potassium hydroxide is added to dilute hydrochloric acid to produce potassium chloride solution. KOH (s) + HCl (aq)  KCl (aq) + H2O (l) H1 Route 2 :The first step is to dissolve solid potassium hydroxide in water. KOH (s) + H2O (l)  KOH (aq)  H2a The potassium hydroxide solution is then added to hydrochloric acid to form potassium chloride solution. KOH (aq) + HCl (l)  KCl + H2O (l)  H2b
According to Hess’s Law :  H1 =  H2a +  H2b
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36. Enthalpy
We know how to measure & calculate the enthalpies of combustion, solution and neutralisation.
We did this indirectly by heating water during the reaction and using the equation
DH= cmDT
Then working out the enthalpy change (energy change for one mole).

37. Hess’s Law
The temperature changes in each step are measured and cmDT used to find the energy released (temperature increased)
The mass of the pellets are measured and the moles of KOH used calculated using
Moles = mass/formula mass
Typically the calculations yield results for route 1 and route 2 which are about 3 or 4 kjmol-1 different
Consistent with experimental error

38. Title: Hess’s Law
Aim of the experiment: To confirm Hess’s Law.
Procedure: Use equations to describe the two routes whereby you converted solid potassium hydroxide into potassium chloride solution and label them with appropriate ΔH values.

39. Results:

40. Calculation: Carry out a calculation to show the confirmation of Hess’s Law.

41. Calculation cont:

42. Conclusions:
The experiment has confirmed Hess’s Law with reasonable accuracy.

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46. 07/12/2018

48. Calculate a value for the enthalpy change involved in the formation of one mole of hydrogen peroxide from water (ΔH3).
The enthalpy change when hydrogen forms hydrogen peroxide is -188 kJ mol-1 and the enthalpy of combustion of hydrogen to form water is -286 kJ mol-1.

49. ?
-188kJmol
-286kJmol
H 1 =
H 2
H 3 +
H 3 =
H 1
H 2 -

50. The required equation can be built up from equations given
[1] CO + ½ O2
CO2 DH1=-283kjmol-1
[2] Cu + ½ O2
CuO DH2=-155kjmol-1
Find the enthalpy change for
[3] CuO + CO
CO2 + Cu DH3= ?

51. CuO + CO
CO2 + Cu DH3= ?
[1] CO + ½ O2
CO2 DH1=-283kjmol-1
[2] Cu + ½ O2
CuO DH2=-155kjmol-1
[2] Cu + ½ O2
CuO DH2=-155kjmol-1

52. Calculate the enthalpy for the reaction given that:
Examples to try
The thermite reaction is a displacement reaction used to join sections of railway. The equation for the reaction is:
2Al(s) + Fe2O3(s)
2Fe(s) + Al2O3(s)
Calculate the enthalpy for the reaction given that:
2Al(s) + 1 ½ O2(g)
Al2O3(s) DH1= -1600kjmol-1
Fe2O3(s) DH2= -820kjmol-1
2Fe(s) + 1 ½ O2(g)

53. Examples to try
Calculate the energy change when phosphorous (P) changes to P4 given:
4P2(RED)+ 5O2(g)
P4O10 DH1=-725.5kjmol-1
P4(YELLOW) + 5O2(g)
P4O10 DH2= -2390kjmol-1

54. Examples to try
Calculate the energy change when one mole of hydrogen peroxide decomposes to from water and oxygen given:
H2(g) + O2(g)
H2O2(l) DH1= -189kjmol-1
H2(g) + ½ O2(g)
H2O(l) DH2= -286kjmol-1

55. Harder Example 1
The enthalpy of combustion of carbon monoxide is -282kJ mol-1. Use this value and the enthalpy of combustion of carbon to calculate the enthalpy of formation of carbon monoxide. Target equation: C(s) + ½O2(g)  CO(g) ΔHf

56. Target equation: C(s) + ½O2(g)  CO(g) ΔHf
1. Combustion of CO:-
CO + ½O2  CO2 ΔH1= -282kJ mol-1
2. Combustion of C:-
C + O2  CO2 ΔH2 = -394kJ mol-1

57. Target equation: C(s) + ½O2(g)  CO(g) ΔHf
1. Combustion of CO:-
CO + ½O2  CO ΔH1= -282kJ mol-1
2. Combustion of C:-
C + O2  CO ΔH2 = -394kJ mol-
Therefore reverse equation 1
[-1] CO2  CO + ½O ΔH1 = 282kJ mol-
[+2] C O2  CO ΔH2 = -394kJ mol-
ADD EQNS C + ½O2  CO ΔHf = -112kJ mol-

58. Harder Example 2 Target equation:
Use the enthalpies of combustion of carbon, hydrogen and ethanol to calculate the enthalpy of formation of ethanol (C2H5OH).
Target equation:
2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHf

59. Target equation:
2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHf
1. C O2  CO ΔH1= -394kJ
2. H2 + ½ O2  H2O ΔH2 = -286kJ
3. C2H5OH + 3O22CO2 +3H2O ΔH3=-1367kJ

60. Reaction 1 (x2) 2C + 2O2  2CO2 ΔH1 = -788kJ Reaction 2 (x3) 3H2 + 1½O2 3H2O ΔH2 = -858kJ Reaction 3 (Reverse) 2CO2 + 3H2OC2H5OH + 3O2 ΔH3 = 1367kJ

61. 2C O2  2CO ΔH1 = -788kJ
3H ½O2 3H2O ΔH2 = -858kJ
2CO2 + 3H2O  C2H5OH + 3O2 ΔH3= 1367kJ
ADD
2C + 3H2 + ½O2  C2H5OH ΔHf= -297kJ

62. Hess’ Law Past Paper Questions

63. A B

64. C A

65. 2014 H
= −147 (kJ mol-1)
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66. 2015 H Q12
+ 20 kJ mol−1
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67. 2013 H Q 13
+206 kJ mol–1
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68. +34 kJ mol-1

69. Harder Past Paper 2011

70. -672kJ mol-1

71. Hess’s Law State Hess’s Law Applying Hess’s Law:-
 I can use enthalpies of combustion, given in the data book, to calculate enthalpies of formation.
 I can calculate the overall enthalpy change for a reaction using given enthalpy data.
Bond Enthalpy

72. Bond Enthalpies I understand what is meant by:- molar bond enthalpy
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Bond Enthalpies
I understand what is meant by:-
molar bond enthalpy
mean molar bond enthalpy
 I can use bond enthalpies to estimate the enthalpy change taking place for a gas phase reaction.

73. Bond Enthalpy
The enthalpy change for a reaction can also be found by looking at the energy needed to make and break the chemical bonds in gases.
This is known as the BOND ENTHALPY and is found on P10 of the Data Book.
Some are shown as MEAN (Average) BOND ENTHALPIES, as these bonds will be found in different molecules (eg a C-C bond in ethane will have a different energy than the C-C bond in butane).
The bond enthalpies quoted in the data book are the energies required to break 1 mole of a particular bond between a pair of atoms in the gaseous state.

74. Bond Breaking & Making
BREAKING – Energy is needed to be able to break a bond, so this is an ENDOTHERMIC process.
MAKING – Making is the opposite of breaking, so will be an EXOTHERMIC process (energy is released).

75. Using bond enthalpies
∆H can be calculated from bond enthalpies using the equation
H =  H bonds broken +  H bonds made
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76. Worked Example
What is the enthalpy change when hydrogen is added to ethyne to produce ethane?
C2H2 (g) + 2H2 (g) C2H6 (g)
To answer this we must look at what types of bonds must be broken in the reactants and formed in the products.
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77. This will require energy to be put in.
C2H2 (g) + 2H2 (g)  C2H6 (g)
In this reaction, we must first break all the bonds inside the reactant molecules.
This will require energy to be put in.
Next, new bonds must be formed between the atoms in the product molecule.
This releases energy.
Potential energy
Reaction pathway
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78. To answer this question we can follow these steps.
What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)  C2H6 (g) C H H C H + H
To answer this question we can follow these steps.
Step One: Draw the full structural formulae of all the molecules from the equation. This will show exactly what bonds are involved.
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79. Make a list of all the bonds being broken in the reactants
What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)  C2H6 (g) + C H H C H C C H
Bond Breaking C
1 x C
1 x C H
2 x C H
2 x H
2 x
Step Two:
Make a list of all the bonds being broken in the reactants
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80. What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)  C2H6 (g) + C H
Bond Breaking
1 x C C
= 838
2 x C H
= 2 x 412
= 824
2 x H H
= 2 x 436
= 872
Step Three:
Fill in the values for the bond enthalpies from Page 10 of the data book.
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81. Repeat this process for the Bond Making steps.
What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)  C2H6 (g) + C H C H C
Bond Breaking
Bond Making
1 x C C
= 838 C
1 x
= 348
2 x C H
= 2 x 412
= 824 C H
6 x
= 6 x 412
= 2472
2 x H H
= 2 x 436
= 872
Step Four:
Repeat this process for the Bond Making steps.
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82. Remember: Bond breaking is an endothermic process
What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g)  C2H6 (g) + C H
Bond Breaking
Bond Making
1 x C C
= 838 C
1 x
= 348
2 x C H
= 2 x 412
= 824 C H
6 x
= 6 x 412
= 2472
2 x H H
= 2 x 436
= 872
Total put in = 2534 kJ
Step Five:
Calculate the total energy put in breaking bonds and total energy given out making new bonds.
Total given out = –2820 kJ
Remember: Bond breaking is an endothermic process
Remember: Bond making is an exothermic process
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83. Calculate the enthalpy change for the reaction. = 2534 – 2820 =
What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g) C2H6 (g) + C H
Bond Breaking
Bond Making
1 x C C
= 838 C
1 x
= 348
2 x C H
= 2 x 412
= 824 C H
6 x
= 6 x 412
= 2472
2 x H H
= 2 x 436
= 872
Total put in = 2534 kJ
Total given out = –2820 kJ ∆H =
2534 +
(–2820)
Step Six:
Calculate the enthalpy change for the reaction. =
2534 –
2820 =
– 286 kJ mol-1
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84. Bond Enthalpy Examples
Using bond enthalpies, calculate the enthalpy change in the
reaction shown below:
H2 (g) Cl2(g)  2 HCl(g)
Bond Breaking
Bond Making H
1 x
= 436 H Cl
2 x
= 2 x 432
= 864 Cl
1 x
= 243
Total put in = 679 kJ
Total given out = –864 kJ ∆H = +
679
(–864)
= – 185 kJ mol-1
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85. Bond Enthalpy Examples
CH4 (g) O2(g)  CO2 (g) H2O(g)
Bond Breaking C H
4 x
= 4 x 412 = 1648 O
1 x
= 498
Bond Making C O
2 x
= 2 x 743
= 1486 H
4 x
= 4 x = 1853
Total put in = 2146 kJ
Total given out = –3339 kJ
∆H = (–3339)
= – 1193 kJ mol-1
07/12/2018

86. Bond Enthalpy Examples
Calculate the enthalpy change for these reactions using bond
enthalpies:
1) C3H8 (g) O2(g)  CO2 (g) H2O(g)
2) C2H4 (g) + H2 (g)  C2H6(g)
3) N2 (g) H2 (g)  2 NH3(g)
4) Cl2(g)  2 Cl(g)
5) C2H2(g) H2 (g)  C2H6(g)
Answers:
– 1632 kJmol-1
-124 kJmol-1
-75 kJmol-1
+243 kJmol-1
-286 kJmol-1

87. Bond Enthalpy Examples: Answers
Calculate the enthalpy change for these reactions using bond
enthalpies:
1) C3H8 (g) O2(g)  CO2 (g) H2O(g)
Bond Breaking
Bond Making C O
6 x
= 6 x 743
= 4458 H
8 x
= 8 x = 3704
2 x C-C = 2 x 348 = 696
8 x C-H = 8 x 412 = 3344
5 x O=O = 5 x 498 = 2490
Total put in = 6530 kJ
Total given out = –8162 kJ
∆H = (–8162)
= – 1632 kJ mol-1

88. Bond Enthalpy Examples: Answers
Calculate the enthalpy change for these reactions using bond
enthalpies:
2) C2H4 (g) + H2 (g)  C2H6(g)
Bond Making
Bond Breaking
2 x C-C = 2 x 348 = 696
8 x C-H = 8 x 412 = 3344
5 x O=O = 5 x 498 = 2490
1 x C=C = 2 x 348 = 696
4 x C-H = 8 x 412 = 3344
1 x H-H = 5 x 498 = 2490
Total put in = 6530 kJ
Total given out = –8162 kJ
∆H = (–8162)
= – 1632 kJ mol-1

89. I understand what is meant by:-
Bond Enthalpies
I understand what is meant by:-
molar bond enthalpy
mean molar bond enthalpy
 I can use bond enthalpies to estimate the enthalpy change taking place for a gas phase reaction.