Post on 12-Jan-2016
RATE LAW
Experiment #4
What are we doing in this experiment?
Measure the rate of a chemical reaction betweenpotassium iodide (KI) and hydrogen peroxide(H2O2) and use this information to obtain the orderand rate constant of the reaction.
What is rate?
Rate refers to speed. In this experiment we are trying to measure the speed or the rate of a chemical reaction.
What is rate of a reaction?
It describes how fast reactants are used upand how fast products are formed.
What is rate of a reaction?
Change in concentration of the reactant or product with change in time
Mathematically expressed,
12
12
ttCC
timeionConcentrat
Rate
change Square bracketsdenotes concentration
Average and instantaneous rate
time
ionConcentratRate
If (time) is large----- Average rate
If (time) is small ----- instantaneous rate
Large and small can be defined based on the time scale of the reaction
Unit of rate
time
ionConcentratRate
timeofUnit
ionConcentratofUnitRateofUnit
ondsorutesMolarity
RateofUnitsecmin
ondsMolarity
orutes
MolarityRateofUnit
secmin
Unit of rate
ondsMolarity
orutes
MolarityRateofUnit
secmin
1 molLLmol
Litersmoles
Molarity
sLmol
orLmol
RateofUnit
min
1111 min smolLormolLRateofUnit
11minmin
MsorMsM
orM
RateofUnit
Rate of a reaction
A + 3B 2C+ 2D
time
ionConcentratRate
tD
tC
tB
tA
Rate
21
21
31
11
- sign signifiesdecrease in concentration
+ sign signifiesdecrease in concentration
A + 3B 2C+ 2D
tD
tC
tB
tA
Rate
21
21
31
11
Rate of consumption of B is 3 times the rate of consumption of A
Rate of formation of C is 2 times the rate of consumption of A
Rate of formation of D is 2 times the rate of consumption of A
Rate of formation of C is equal to the rate of formation of D
Rate of consumption of B is 1.5 times the rate of formation of C and D
These fractions makes sure that the rates are equal
Conclusion I
If want to measure the rate of a reaction we can follow the decrease in concentration of a reactant or increase in the concentration of a product overa period of time.
Rate lawThe rate of a homogenous reaction at anyinstant is proportional to the product of molarconcentrations of the reactants, each molarity raised to some power or exponent that hasto be found by experiment.
A + B products
According to the above state, its rate can be expressed as
nm BARate
Instantaneous rate
Rate law
nm BARate
nm BAkRate
k is called the rate constant
This equation is called the rate law
m and n are the order of the reaction due to A and B respectively
Overall order of the reaction = m + n
RATE CONSTANTS HAVE UNITSTHAT DEPEND ON THE ORDER
OF THE REACTION
Rate constant of a first order reaction:
A products
1AkRate
1ARate
k
1ARate
k
11minmin
MsorMsM
orM
RateofUnit
1}{ AofUnitRateofUnit
kofUnit
sMsM
MsM
kofUnit 11
Rate constant of a second order reaction:
A products
2AkRate
2ARate
k
2}{ AofUnitRateofUnit
kofUnit
LsmolsmolLMsMs
MMsM
kofUnit 1
1122
11)(
1)(
Rate constant of a third order reaction:
A products
3AkRate
3ARate
k
3}{ AofUnitRateofUnit
kofUnit
122233
11)(
1)(
sLmolsMMs
MMsM
kofUnit
sLmolkofUnit 22
Conclusion II
A + B productsFor any reaction,
nm BAkRate
Rate law is given by,
k is called the rate constantm and n are the order of the reaction due to A and B respectively. The orders have to be determined experimentally.
Overall order of the reaction = m + n
Conclusion II
A + B productsFor any reaction,
The order of the reaction due to each reactant is usually not the same as the stoichiometric coefficients of the reactants.
The order of the reaction due to each reactant could be the same as the stoichiometric coefficients of the reactants, only if experimentally proved so.
The unit of the rate constant changes with the orderof the reaction.
Example problem
A + B productsFor any reaction,
Concentration-Rate Data for the hypothetical reaction A + B Products
Initial concentration
Exp [A] [B] Initial rate of formation of P(mol L-1) (mol L-1) (mol L-1s-1)
1 0.10 0.10 0.202 0.20 0.10 0.40
3 0.30 0.10 0.604 0.30 0.20 2.405 0.30 0.30 5.40
Example problemWhat are the rate law, orders and rate constant of the reaction?
Solution:
The given reaction is,
A + B productsSince we do not know the rate law for the reaction, we have to start with a general rate law. The general rate law is given by
nm BAkRate
Where k is the unknown rate constant that we have to find.
Where m, n are the unknown orders of A and B respectively, that we have to determine based on the experimental data.
Initial concentration
Exp [A] [B] Initial rate of formation of P(mol L-1) (mol L-1) (mol L-1s-1)
1 0.10 0.10 0.202 0.20 0.10 0.40
nm BAkRate
nmk 10.010.020.0
Substituting the concentrations and rate in the general rate law
------------- (1)
nmk 10.020.040.0 ------------- (2)
nmk 10.010.020.0 ------------- (1)
nmk 10.020.040.0 ------------- (2)
Dividing equation (1) by equation (2)
nm
nm
kk
10.020.010.010.0
40.020.0
1
2 m
20.010.0
21
m
21
21
m
21
21
m
21
21
1
Since the bases are equal, for left hand side (LHS) of theequation to be equal to the right hand side (RHS) of the equation, the exponents must be equal.
Method 1
Base Base
Exponent
1m
m
21
21
m
21
21
1Method 2
m5.05.0 1
Take log on both sides of the equation
m5.0log5.0log 1
aba b loglog Applying
5.0log5.0log)1( m
5.0log5.0log)1( m
m1Initial concentration
Exp [A] [B] Initial rate of formation of P(mol L-1) (mol L-1) (mol L-1s-1)
3 0.30 0.10 0.604 0.30 0.20 2.40
nm BAkRate
nmk 10.030.060.0
Substituting the concentrations and rate in the general rate law
------------- (3)
nmk 20.030.040.2 ------------- (4)
nmk 10.030.060.0 ------------- (3)
nmk 20.030.040.2 ------------- (4)
Dividing equation (3) by equation (4)
nm
nm
kk
20.030.010.030.0
40.260.0
1
4n
20.010.0
41
n
21
41
n
21
41
n
21
21
2
Since the bases are equal, for left hand side (LHS) of theequation to be equal to the right hand side (RHS) of the equation, the exponents must be equal.
Method 1
Base Base
Exponent
2n
m
21
21
m
21
21
1Method 2
We know that the LHS of the equation must be equal to RHS.
So we can ask ourselves this question:How many times do I have to multiply half toitself to obtain one-fourth?
The answer is 2 times
2n
n
21
41
n
21
41
Method 3
n5.025.0
Take log on both sides of the equation
n5.0log25.0log
aba b loglog Applying
)301.0(602.0 n 5.0log25.0log n
n2
Now we have, 2n1m
So the rate law becomes,
21 BAkRate
Now we can pick any one of the 5 experiments and plug-in in the above equation, the values of concentration and rate and obtain the value of rate constant for this reaction.
The overall order of the reaction= m + n = 1 + 2 = 3
)301.0(602.0 m
Let us say, we pick experiment 5Initial concentration
Exp [A] [B] Initial rate of formation of P(mol L-1) (mol L-1) (mol L-1s-1)
5 0.30 0.30 5.40
21 BAkRate
211111 30.030.040.5 molLmolLksmolL
kmolLmolL
smolL
2111
11
30.030.0
40.5
kmolLmolL
smolL
2111
11
30.030.0
40.5
kLmolmolL
smolL
221
11
90.030.0
40.5
kLmolmolL
smolL
221
11
90.030.040.5
kLmol
s
22
1
20
kLmols 22120
Our experiment
2 I-(aq) + H2O2(aq) + 2 H+(aq) → I2(aq) + 2 H2O(l)
hnmHOHIkRate 22
There are 3 concentrations that change in thisexperiment. We can simplify the experiment bykeeping one of the concentrations a constant. This means that we will never vary the concentration of that particular molecule in allour experiments.
Kept Constantby the useof buffer
Determination 1:
Iodide Solution:
Peroxide Solution:
10 mL 0.3 M KI
30 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
123 mL distilled water
Determination 2:
Iodide Solution:
Peroxide Solution:
20 mL 0.3 M KI
30 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
113 mL distilled water
Determination 3:
Iodide Solution:
Peroxide Solution:
30 mL 0.3 M KI
30 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
103 mL distilled water
Determination 3:
Iodide Solution:
Peroxide Solution:
30 mL 0.3 M KI
30 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
123 mL distilled water
Determination 4:
Iodide Solution:
Peroxide Solution:
30 mL 0.3 M KI
50 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
83 mL distilled water
Determination 5:
Iodide Solution:
Peroxide Solution:
30 mL 0.3 M KI
70 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
63 mL distilled water
Example problem
A + B productsFor any reaction,
Concentration-Rate Data for the hypothetical reaction A + B Products
Initial concentration
Exp [A] [B] Initial rate of formation of P(mol L-1) (mol L-1) (mol L-1s-1)
1 0.10 0.10 0.202 0.20 0.10 0.40
3 0.30 0.10 0.604 0.30 0.20 2.405 0.30 0.30 5.40
hnmHOHIkRate 22
hHkk '
nmOHIkRate 22
'
Determining m:
nmOHIkRate 22
'
Taking natural log on both sides of the equation
}{)( 22
' nmOHIkLnRateLn
}{)( 22
' nmOHIkLnRateLn
)()()()(Re cLnbLnaLnabcLnmember
nmOHLnILnkLnRateLn 22
' )()(
)()(Re axLnaLnmember x
22
' )()( OHnLnImLnkLnRateLn
When determining m, the rates are measured by varying [I-] and keeping [H2O2] constant. So, nLn[H2O2] remains a constant say C1.
22
' )()( OHnLnImLnkLnRateLn
1
' )()( CImLnkLnRateLn
Since K’ is a constant, Ln(K’) is also a constant.We can sum Ln(k’) and C1 and assign a new constant C’
')( CImLnRateLn
[I-] Rate Ln (Rate) Ln [I-]
[I1]
[I2]
[I3]
Rate1 Ln [I1]
Rate2 Ln [I2]
Rate3 Ln [I3]
Ln (Rate1)
Ln (Rate2)
Ln (Rate3)
Ln [I-]
Ln
(Rat
e)
Ln [I-]p
Best-fit
line
x
x
p
q
x
y
Equation of the best-fit line:
Y=mX + z
m = slope =
12
12
XXYY
pq
pq
ILnILn
RateLnRateLn
][][
)()(
Ln [I-]2,Ln (Rate2)
Ln [I-]3,Ln (Rate3)
Ln
(R
ate p
)L
n (
Rat
e q)
Ln [I-]1,Ln (Rate1)
Ln [I-]qIntercept, Z
')( CImLnRateLn
ZXmY
Slope = order = mDetermining n:
22
' )()( OHnLnImLnkLnRateLn
When determining n, the rates are measured by varying [H2O2] and keeping [I-] constant. So, mLn[I-] remains a constant say C2.
22
' )()( OHnLnImLnkLnRateLn
Since K’ is a constant, Ln(K’) is also a constant.We can sum Ln(k’) and C2 and assign a new constant C’
'
22)( COHnLnRateLn
[H2O2] Rate Ln (Rate) Ln [H2O2]
[H2O2]1
[H2O2]2
[H2O2]3
Rate1 Ln [H2O2]1
Rate2
Rate3
Ln (Rate1)
Ln (Rate2)
Ln (Rate3)
222
' )()( OHnLnCkLnRateLn
Ln [H2O2]2
Ln [H2O2]3
Ln [H2O2]
Ln
(Rat
e)
Best-fit
line
x
x
p
q
x
y
Equation of the best-fit line:
Y=mX + z
m = slope =
12
12
XXYY
pq
pq
OHLnOHLn
RateLnRateLn
][][
)()(
2222
Ln
(R
ate p
)L
n (
Rat
e q)
Ln [H2O2]1,Ln (Rate1)
Intercept, Z
Ln [H2O2]2,Ln (Rate2)Ln [H2O2]3,Ln (Rate3)
Ln [H2O2]P Ln [H2O2]q
'
22)( COHnLnRateLn
ZXmY
Slope = order = n
How do we measure the rate of the reaction?
2 I-(aq) + H2O2(aq) + 2 H+(aq) → I2(aq) + 2 H2O(l)
Either by following the change in concentration of a reactant or product
In this reaction we follow the change inconcentration of I2 with time.
initialfinal
initialfinal
tt
IIRate
22
initialfinal
initialfinal
tt
IIRate
22
Since there is no I2 formed before the reaction begins, tinitial=0 and [I2]=0
final
final
final
final
t
I
t
IRate 22
0
0
)(
mindet
22 LV
IofmolesI
ationereachinsolutiontotal
final
How do we measure the rate of the reaction?
How to detect the formation of I2 ?
I2 on reaction with starch gives a blue black color
Is just adding starch enough to perform all the determinations?
No
Why not?
All the determinations will give a blue blackcolor instantaneously if both starch and I2 arein the same solution.
How do we work around the problem?
2 S2O3-2(aq) + I2(aq) → 2 I-(aq) + S4O6
-2(aq)
2 I-(aq) + H2O2(aq) + 2 H+(aq) → I2(aq) + 2 H2O(l)
2 moles of I- ≡ 1 mole of I2
2 moles of S2O32- ≡ 1 mole of I2
Moles of Na2S2O3 = 0.01 L × 0.02 M= 0.0002 moles
Moles of I2 = ½ × (Moles of Na2S2O3)= 0.0001 moles
We need to exceed the initial 0.0001 moles of I2
to be able to see the blue-black color.
The speed of formation of I2 depends on the[I-] and [H2O2]. Hence the blue-black color will form at different times depending on how quicklythe initial 0.001 moles of I2 are exceeded.
How do we work around the problem?
Determination 1:
Iodide Solution:
Peroxide Solution:
10 mL 0.3 M KI
30 mL 0.1 M H2O2
10 mL 0.02 M Na2S2O3
25 mL 0.5/0.5 M acetate buffer
2 mL starch solution
123 mL distilled water
Moles of I- = 0.01 L × 0.3 M= 0.003 moles
Moles of H2O2 = 0.03 L × 0.1 M= 0.003 moles
Moles of Na2S2O3 = 0.01 L × 0.02 M= 0.0002 moles
Moles of I2 = ½ × (Moles of Na2S2O3)= 0.0001 moles
Calculating concentrations
100 mL mark
100mL
100 mL mark
100mL
0.2 M of A 0.3 M of B
25 mL of A65 mL of B + 110 mL of H2O
What is the concentration of A and B in the final solution?
)(.ln LVAofmoles
Asototal
)()(
.ln LVLVM
Asototal
AA
Lmol
L
LLmol
A2.0
105)110.0065.0025.0(
025.02.03
M2105.2
100mL
0.2 M of A
)(.ln LVBofmoles
Bsototal
)()(
.ln LVLVM
Bsototal
BB
Lmol
L
LLmol
B2.0
1095.1)110.0065.0025.0(
065.03.0 2
M21075.9
100mL
0.3 M of B