IBDP SL Kinetics

1. Definitions
 Standard Enthalpy Change
 the change of enthalpy that accompanies the formation of 1
mole of a substance in its standard state from its constituent
elements in their standard states.
 Exothermic Rxn
 Typical examples; combustion, neutralisation, respiration
 Endothermic Rxn
 Typical examples; melting, evaporating, photosynthesis.

2. Signs 
 Exothermic
 Endothermic

3. Energy Level Diagrams

4. Exo and Endo Energy Level
Diagrams
http://www.docbrown.info/page03/3_51energy.htm#3.

5. Energetics
Calculation of Enthalpy
http://www.youtube.com/watch?v=lxTmei2yrBg

7. Experiments
 Enthalpy of Neutralisation of HCl with NaOH.
 Enthalpy of dissolution.
 Enthalpy of precipitation.

9. Define
 Enthalpy of combustion
 Enthalpy of formation
 Draw energy level diagram for the process of
evaporation.

10. Hess’s Law
 Students should be able to use simple enthalpy cycles
and enthalpy level diagrams and to manipulate
equations. Students will not be required to state Hess’s
law…..but I’ll tell you anyway 
 Energy can not be created or destroyed. It can only be
converted from one form to another – first law of
thermodynamics

11. Conventions
 -ve for exothermic
 +ve for endothermic
 If forward reaction is exo the reverse reaction is endo and of
identical magnitude.
 Hess’s Law states that the enthalpy change for a reaction is
independent of the route the reaction takes.
 The overall enthalpy change depends only on the initial and
final stages.
 Direct measurement of enthalpy is impossible.

12. Enthalpy types
 Combustion
 Energy released when one mole of a compound is burned
in excess oxygen.
 Formation
 Energy change when one mole of a compound is formed
under standard conditions from its constituent elements.
 Bond Enthalpies – we will discuss later 

13. Hess’ Law Defined
 Hess’ Law: H for a process involving the
transformation of reactants into products is not
dependent on pathway. Therefore, we can pick
any pathway to calculate H for a reaction.

14. Shortcuts  Learn Them 
 Enthalpies of formation :
 ΔH = Σproducts – Σ reactants
 Enthalpies of combustion :
 ΔH = Σreactants– Σ products
 Average Bond Enthalpies
 Δ H = [bonds broken] – [bonds made]

15. Hess’ Law: An Example

16. Using Hess’ Law
 When calculating H
N2 (g) + 2O2 (g) 2NO2 (g)
for a chemical
reaction as a single
step, we can use
2NO2 (g) combinations of
reactions as
q
“pathways” to
N2 (g) + 2O2 (g)
determine H for our
“single step” reaction.

17. Example (cont.)
 Our reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
• This reaction can also be carried out in two
steps:
N2 (g) + O2 (g) 2NO(g) H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) H = -112 kJ

18. Example (cont.)
 If we take the previous two reactions and add them, we
get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) H= 68 kJ

19. Changes in Enthalpy
 Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic

20. Example (cont.)
 Note the important things about this example,
the sum of H for the two reaction steps is
equal to the H for the reaction of interest.
 We can combine reactions of known H to
determine the H for the “combined” reaction.

21. Hess’ Law: Details
 Once can always reverse the direction of a reaction
when making a combined reaction. When you do
this, the sign of H changes.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
2NO2(g) N2(g) + 2O2(g) H = -68 kJ

22. Details (cont.)
 The magnitude of H is directly proportional
to the quantities involved (it is an “extensive”
quantity).
 As such, if the coefficients of a reaction are
multiplied by a constant, the value of H is
also multiplied by the same integer.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
N2(g) + 4O2(g) 4NO2(g) H = 136 kJ

23. Using Hess’ Law
 When trying to combine reactions to form a reaction of
interest, one usually works backwards from the reaction
of interest.
 Example:
What is H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)

24. Example (cont.)
3C (gr) + 4H2 (g) C3H8 (g) H=?
• You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ

25. Example (cont.)
 Step 1. Only reaction 1 has C (gr). Therefore, we will
multiply by 3 to get the correct amount of C (gr) with
respect to our final equation.
Initial:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
Final:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ

26. Example (cont.)
 Step 2. To get C3H8 on the product side of the
reaction, we need to reverse reaction 2.
Initial:
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220
Final:
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220

27. Example (cont.)
 Step 3: Add two “new” reactions together to see what
is left:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220
2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 k

28. Example (cont.)
 Step 4: Compare previous reaction to final reaction, and
determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038
kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286
kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4

29. Example (cont.)
 Step 4: Compare previous reaction to final reaction, and
determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038
kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144
kJ
3C (gr) + 4H2 (g) C3H8 (g)

30. Example (cont.)
• Step 4 (cont.):
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038
kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144
kJ
3C (gr) + 4H2 (g) C3H8 (g) H = -106 kJ

31. Changes in Enthalpy
 Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic

32. Another Example
 Calculate H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ

33. Another Example (cont.)
 Step 1: Only the first reaction contains the product of
interest (HCl). Therefore, reverse the reaction and
multiply by 2 to get stoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ

34. Another Example (cont.)
 Step 2. Need Cl2 as a reactant, therefore, add reaction
3 to result from step 1 and see what is left.
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
H = -277 kJ

35. Another Example (cont.)
 Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final
reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) H=?
Need to take middle reaction and reverse it

36. Another Example (cont.)
 Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final
reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
1
2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ
H2(g) + Cl2(g) 2HCl(g) H = -185 kJ

37. Changes in Enthalpy
 Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic