1. Lesson 5 – Hess’ law and enthalpy cycles
Module 3
Lesson 5 – Hess’ law and enthalpy cycles

2. Objectives
Must Recall Hess’ Law Should Interpret an enthalpy cycle using Hess’ Law Could Construct an enthalpy cycle and use it to calculate an enthalpy change of reaction from combustion data

3. Starter - Test What are standard conditions? Define ΔHc Define ΔHf
Define ΔHr
Describe the following equations as ΔHc, ΔHr or ΔHf
3C(s) + 4H2(g) C3H8(g)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
C2H4(g) + H2(g)  C2H6(g)
2C2H6(l) + 7O2(g)  4CO2(g) + 6H2O(l)
Write the equation for ΔHc for H2
CH3OH C

4. Starter - Answers
Standard conditions are 298K, 100 kPa and 1M for solutions. All substances should be in their standard states
The standard enthalpy change of combustion ΔHc is the enthalpy change that takes place when one mole of a substance reacts completely with oxygen under standard conditions, all reactants and products being in their standard states.
The standard enthalpy change of formation ΔHf is the enthalpy change that takes place when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.
The standard enthalpy change of reaction ΔHr is the enthalpy change that accompanies a reaction in the molar quantities expressed in a chemical equation under standard conditions, all reactants and products being in their standard states.

5. Starter - Answers Describe the following equations as ΔHc, ΔHr or ΔHf
3C(s) + 4H2(g) C3H8(g) ΔHf
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHc
C2H4(g) + H2(g)  C2H6(g) ΔHr
2C2H6(l) + 7O2(g)  4CO2(g) + 6H2O(l) ΔHr (because 2mols are shown to be burnt)
Write the equation for ΔHc for
H2(g) + ½ O2(g)  H2O(l)
CH3OH(l) + O2(g)  CO2(g) + H2O(l)
C(s) + O2(g)  CO2(g)

6. Measuring enthalpy changes
Using a calorimeter for a combustion reaction or or other reaction can often give the ΔHr directly (think about the spirit burners and copper sulphate experiments you have done).
Bond enthalpies can also be used to estimate ΔHr
However it may not always be possible to measure the enthalpy change of a reaction directly.

7. Problems There may be: a high activation energy a slow reaction rate
more than one reaction taking place
Take for example 3C(s) + 4H2(g)  C3H8(g)
This is virtually impossible to measure directly - think of the number of compounds of hydrogen and carbon that could form!

8. Hess’ Law ΔH(Route A) = ΔH(Route B) – ΔH(Route C) A B C
Hess’ law states that, if a reaction can take place by more than one route and the initial and final conditions are the same, the total enthalpy change is the same.
Reactants
Products
Intermediate A B C
ΔH(Route A) = ΔH(Route B) – ΔH(Route C)
Draw an enthalpy cycle – then if you follow the direction of an arrow then ADD. If opposite to the direction of the arrow then SUBTRACT.

9. Using combustion data ΔHr = Σ ΔHc(reactants) – Σ ΔHc(products) ΔHr
The combustion of reactants and combustion of products is a useful way to complete the cycle in a way that can be experimentally determined.
ΔHr = Σ ΔHc(reactants) – Σ ΔHc(products)
Reactants
Products
Combustion products
ΔHr
Σ ΔHc(reactants)
Σ ΔHc(products)
Draw an enthalpy cycle – then if you follow the direction of an arrow then ADD. If opposite to the direction of the arrow then SUBTRACT.

10. Example
You are provided with the following enthalpy changes of combustion
Determine the enthalpy change for the following reaction:
4C(s) + 5H2(g)  C4H10(g)
Substance
C(s)
H2(g)
C4H10(g)
C2H5OH(l)
ΔHc / kJmol-1
-394
-286
-2877
-1367

11. ΔHf = (4 x -394) + (5 x -286) – (-2877) = -129 kJmol-1
Substance
C(s)
H2(g)
C4H10(g)
C2H5OH(l)
ΔHc / kJmol-1
-394
-286
-2877
-1367
4C(s) + 5H2 (g)
C4H10(g)
Combustion products
ΔHf
4ΔHc(C)+ 5ΔHc(H2)
ΔHc(C4H10)
ΔHf = (4 x -394) + (5 x -286) – (-2877) = -129 kJmol-1

12. Example
You are provided with the following enthalpy changes of combustion
Determine the enthalpy change for the following reaction:
2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l)
Substance
C(s)
H2(g)
C4H10(g)
C2H5OH(l)
ΔHc / kJmol-1
-394
-286
-2877
-1367

13. ΔHr = (2 x -394) + (3 x -286) – (-1367) = -279 kJmol-1
Substance
C(s)
H2(g)
C4H10(g)
C2H5OH(l)
ΔHc / kJmol-1
-394
-286
-2877
-1367
ΔHr
2C(s) + 3H2 (g) + ½ O2(g)
C2H5OH(l)
ΔHc(C2H5OH)
2ΔHc(C)+ 3ΔHc(H2)
Combustion products
ΔHr = (2 x -394) + (3 x -286) – (-1367) = -279 kJmol-1

14. Examination question

16. Mark scheme

17. Mark scheme