#### Transcript Exp 17 - PowerPoint

```Exp. 17 – video
(time: 36:06 minutes)
Exp. 17: Kinetics: Determination of the
order of a reaction
Chemical Kinetics – is the study of rates of
chemical reactions. The rate of a chemical
reaction describes how fast a reaction proceeds.
(basically: how quickly are reactants consumed
and products produced)
Exp 17 experiment:
2 H2O2 (aq)
I 2H2O (l) + O2 (g)
Decomposition of hydrogen peroxide catalyzed by
iodide ion
General expression for this reaction (rate law):
Rate = k [H2O2]x [I-]y
Rate = k [H2O2]x [I-]y
Note: brackets typically refers to concentration in M.
k is the specific rate constant and is related to a particular
rxn and temperature
x and y are referred to as the order of the reactant; it
describes how the reactant concentration affects the rate of
the reaction. Values are typically a positive integer but not
always.
orders determined experimentally not by stoichiometry of
balanced equation
1) A + A  A2
2) A2 + B  C
2A + B  C
elementary (slow)
elementary (fast)
molecular
Order of a reactant is determined by the effect changing the
reactant conc has on rate
Change conc of reactant
and rate remains same
=
0 order (must be present)
20 x rate = rate
30 x rate = rate
Change conc of reactant
and rate changes same
=
Change conc of reactant
and rate changes to the sq
of change
=
1st order (linear)
21 x rate = 2rate
31 x rate = 3rate
2nd order (square)
22 x rate = 4rate
32 x rate = 9rate
3rd order?
23 x rate = 8 rate
Overall order of reaction equals the sum of all the
orders.
x + y + … = overall order
How do we determine the order? Collect data
carefully with a well designed experiment
Ex.
A + B  P
rate = k [A]x [B]y
Exp 1
[A]
1M
[B]
1M
rate
1M/s
Exp 2
1M
2M
2M/s
Exp 3
2M
1M
8M/s
Two ways:
1.) inspection
Compare exp1/exp2, [B] doubles and rate doubles, linear effect
y= 1: 1st order
Compare exp1/exp3, [A] doubles and rate is eight-fold; cube effect
x= 3: 3rd order
Rate = k [A]3 [B]
Ex.
A + B  P
[A]
1M
1M
2M
Exp 1
Exp 2
Exp 3
Find x:
exp3
exp1
rate = k [A]x [B]y
[B]
1M
2M
1M
rate
1M/s
2M/s
8M/s
rate3 = k[A3] x [B3] y
rate1 = k[A1] x [B1] y
8M/s = k[2M] x [1M] y
1M/s = k[1M] x [1M] y
8 = 2x
log 8 = log 2x = x log 2
x = log 8 = 3
log 2
2.) initial rate method
Solving for k; we will use exp 3 data, but you can use any set
rate3 = k[A3] x [B3] y
8M/s = k[2M] 3 [1M]
8 M/s
(2M)3(1M)
8 M/s
(8M3)(1M)
1 /s
M3
=k
=k
=k
M: 1 – 3 – 1 = – 3
M0.25
= M-1.10
0.50
0.85
M
M
M: 0.25 – 0.50 – 0.85 = – 1.10
k = 1 M-3 s-1
rate = 1 M-3 s-1 [A]3 [B]
overall order = 3 + 1 = 4th
In experiment 17, rate of H2O2 decomposed will be
determined by plotting volume of O2 gas generated vs. time.
Y axis
X axis
We will do 3 different experiments but only once each
Exp 1
KI, mL H2O, mL H2O2, mL rate, mL/min
10.00 15.00
5.00
slope1
Exp 2
20.00
5.00
5.00
slope2
Exp 3
10.00
10.00
10.00
slope3
Note: each group only needs 50 mL of KI and H2O2 will be
given out by the TA as needed.
Pg 117 describes how the experiment will be conducted
Important points:
- levels of buret and drying tube must be equal for readings
- you can dump excess water out
- make sure all air bubbles are out
- check for leaks
- Add H2O2 just before you are ready for exp to begin
-wait 1 – 2 mL before call time “0 min”
note: we are following the change in volume over a
particular change in time; therefore, it doesn’t matter when
call time zero.
Plot cumulative volume, mL
vs
time, min
time,
min
mL
Cumulative vol.,
mL
0.00
2.00
0.00
1.00
4.10
2.10
2.00
6.15
4.15
3.00
8.30
6.30
4.00
10.45
8.45
5.00
12.30
10.30
Graphing:
-Must have a descriptive title
-Label both axis with units
-Large graph over majority of page, select axis increments
which allows for this
Graph 30 blocks available on x axis and 5 minutes of data:
5 min
30 blocks
=
0.16 min
block

0.20 min
block
-Legend explaining data
-Best line, not connecting the dots, and do not force
through zero
Slope of each line gives the rate for that experiment.
Slope= rise
run
= D y = D mL = y2-y1 = rate D mL
Dx
D time x2 – x1
D min
Note: pick points on best line, not data points
In this experiment, we are using volume instead of
concentration in our rate unit. Notice in the experiment that
the total volume is held constant to 30 mL in every
experiment (volume of water changes to assure this).
This means that because the way the experiment is
designed that the original conc. of KI / H2O2 and total
volume cancel out leaving the volume of solution the only
variable.
Conc of KI in experiment 1
(0.100 M KI) (10.00 mL)
30.00 mL
= conc KI1
Conc of KI in experiment 2
(0.100 M KI) (20.00 mL)
30.00 mL
When compare exp 1
exp 2
= conc KI2
what happens?
KI2 = 2 x KI1
because of vol
Therefore, in this experiment D mL α D conc and the slope
of the line equals the rate
Overall goal of experiment is to report the rate law
expression for the decomposition of H2O2 :
Rate = k (H2O2, mL)x (I-, mL)y
Must give k with units, x, y, and overall order
```