Spring 2024 A2 Sessions:
Session 1 – Accurate Enthalpy Change Calculation (PB)
Session 2 – Enthalpy and Born Haber Cycles (NS)
Session 3 – Iodine Clock Reaction – discussing order and mechanism (PB)
Session 4 – Approach to Unit 5 Assessment (NS)
Iodine Clock Reaction
Colorimetry
Clock Reaction
Timed event
Zn(s) + H2SO4(aq) 🡪 ZnSO4(aq) + H2(g)
2
Rate =
1__
Time
Unit =
s-1
Add S2O32-
S2O32- 🡪
I2 🡪
S4O62-
2
+ 2e-
2I-
+2e-
Half equations
10ml 0.005M Na2S2O3
n Na2S2O3
0.005 x 10/1000 = 5x 10-5
nI2 =
2.5 x 10-5
How much iodine?
5
2
1
3
4
Constant volume!!!
Change in volume is proportional to change in concentration
0
3
4
2
1
Constant volume = 50cm3
Concentration H2O2 @ 1cm3
n H2O2 = 1/1000 x 0.1= 1x10-4
C = 1x10-4 / (50/1000) = 2x10-3
volume | 1ml | 2ml | 3ml | 4ml | 5ml |
[H2O2] | 2x10-3 | | | | |
4x10-3
6x10-3
8x10-3
1.0x10-2
Graph shows that rate is proportional to concentration
Rate law =
k [H2O2]1
[I-]? [H+] ?
The graph
Order of H+ and I-
Rate law =
[I-]
[H+]0
[H2O2]
Look for intermediates
Find reactants
H2O2 + I- + H+ + H+ + I-
Find products
H20 + I2 + H20
Reaction
H2O2 + 2I- + 2H+ 🡪 2 H20 + I2
Change in [A] =
0.03/0.02 = x1.5
Effect on rate =
1.8/1.2 = x1.5
Rate = k [A]
Change in [B] =
0.25/0.10 = x2.5
Effect on rate =
11.25/1.80 = x 6.25
Square root of 6.25 = 2.5
[B]2
Mechanism
A + B 🡪 intermediate
Intermediate + B 🡪 product
fast
slow
Use the initial rate data to write a` rate law
Cl2 + CHCl3 🡪 HCl + CCl4
Rate = k
[Cl2] [CHCl3]
Balanced equation
Intermediates?
What is left
Rate =
Overall
equation
Intermediate?
O3 + O. 🡪 2 O2
Cl.
CATALYST
k [Cl.] [O3]
N2O5 🡪 2NO2 + 1/2 O2
N2O5 🡪 NO + NO3 + 1/2O2
NO + NO3 🡪 2NO2
Rate = k [N2O5]
Subsequent steps
Spring 2024 A2 Sessions:
Session 1 – Accurate Enthalpy Change Calculation (PB)
Session 2 – Enthalpy and Born Haber Cycles (NS)
Session 3 – Iodine Clock Reaction – discussing order and mechanism (PB)
Session 4 – Approach to Unit 5 Assessment (NS)