RATE LAW Experiment #4. What are we doing in this experiment? Measure the rate of a chemical...

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


kofUnit

LsmolsmolLMsMs

MMsM

kofUnit 1

1122

11)(

1)(

Rate constant of a third order reaction:

A products

3AkRate

3ARate

k


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


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


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.



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


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


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




Determination 3:

Iodide Solution:

Peroxide Solution:

30 mL 0.3 M KI

30 mL 0.1 M H2O2

10 mL 0.02 M Na2S2O3




Determination 3:

Iodide Solution:

Peroxide Solution:

30 mL 0.3 M KI

30 mL 0.1 M H2O2

10 mL 0.02 M Na2S2O3




Determination 4:

Iodide Solution:

Peroxide Solution:

30 mL 0.3 M KI

50 mL 0.1 M H2O2

10 mL 0.02 M Na2S2O3




Determination 5:

Iodide Solution:

Peroxide Solution:

30 mL 0.3 M KI

70 mL 0.1 M H2O2

10 mL 0.02 M Na2S2O3




Example problem


Concentration-Rate Data for the hypothetical reaction A + B Products



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


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


When determining n, the rates are measured by varying [H2O2] and keeping [I-] constant. So, mLn[I-] remains a constant say C2.

22


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?


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 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




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

RATE LAW Experiment #4. What are we doing in this experiment? Measure the rate of a chemical...

Documents

Transcript of RATE LAW Experiment #4. What are we doing in this experiment? Measure the rate of a chemical...