C h e m i c a l K i n e t i c s C H A P T E R 6 | 1 · _____requirements using stoichiometry or...

C h e m i c a l K i n e t i c s C H A P T E R 6 | 1

Dena K. Leggett, Ph.D., Allen High School, Allen, TX 2011-2012

Chapter 6 CHEMICAL KINETICS1

ZUMDAHL CH. 12 & 21.2& 21.4

1. Introduction 1.1. Three key questions about reactions

Will it go? ____________________________ How far will it go? ____________________________

(100 %) and ____________________________ (<100%) How fast will it go? ____________________________

1.2. COLLISION THEORY

1.2.1. Molecules must ____________________________

1.2.2. Molecules must collide with _______________________ (called the activation energy, Ea)

1.2.3. Molecules must collide with ________________________. 2. Reaction Mechanisms

2.1. Let’s consider the following reaction: 2NO2(g) + F2(g) 2NO2F(g)

2.1.1. A reaction like this depicts the _____________________and the _________________of a process. 2.1.2. It is helpful for studying the ____________________________and

____________________________requirements using stoichiometry or equilibrium. 2.1.3. We can use it to study the change in ____________________________ (enthalpy) and

____________________________ (entropy). 2.1.4. BUT: What is the probability that three molecules will collide

____________________________with the correct ____________________________and correct ____________________________is not very high.

2.1.5. How do these molecules interact with one another at a ____________________________level? How do changing concentrations and conditions alter the rate with which the reaction reaches completion?

1 Most figures from Chemistry 5

th ed. Zumdahl and Zumdahl.



2.2. A ____________________________ ____________________________ addresses these questions by suggesting series of ____________________________ by which collisions actually occur. Reactions mechanisms are typically arrived at in what is called a ____________________________method – they are theoretical but they rely on experimental (the empirical part) data for verification and direction.

2.2.1. There are two main criteria that a mechanism must meet to be considered plausible: 2.2.1.1. The sum of the elementary steps must lead to the final, overall reaction with

the ____________________________ ____________________________. 2.2.1.2. The mechanism must agree with an experimentally determined

____________________________ ____________________________– a mathematical equation that provides the mathematical relationship between the rate of a reaction and the molarity or partial pressure for each reactant.

MODEL: 2NO2(g) + F2(g) 2NO2F(g)

If we double [NO2] and the rate ____________________________, then there is a direct

proportionality: rateµ[NO2 ]1

If we double [NO2] and the rate ____________________________, then there is a squared or

quadratic proportionality: rateµ[NO2 ]2

If we double [NO2] and the rate ____________________________ ____________________________, then the

rate does not depend on [NO2] at all! rateµ[NO2 ]0

The powers in these mathematical relationships are called the “____________________________” of the reaction with respect to a particular reactant.

An order is given by an ____________________________ (or steps) not the overall ____________________________ ____________________________.

2.3. The rate law for an elementary step is simply the concentrations of reactions with an

order that matches their stoichiometry.

MOLECULARITY ELEMENTARY STEP RATE LAW EXPRESSION

ORDER(S) OVERALL ORDER

1 (Unimolecular) A Products rate = k[A] 1st wrt [A] 1st 2 (Bimolecular) A + A Products rate = k[A]2 2nd wrt [A] 2nd

A + B Products rate = k[A][B] 1st wrt [A] 1st wrt [B]

2nd

3 (Termolecular) A + A + A Products rate = k[A]3 3rd wrt [A] 3rd A + 2B Products rate = k[A][B]2 1st wrt [A]

2nd wrt [B]

3rd

A + B + C Products rate = k[A][B][C] 1st wrt [A] 1st wrt [B] 1st wrt [C]

3rd



2.4. At this level we will solve for only two types of mechanisms: 2.4.1. One step is much slower than all the rest and can thus be considered the

____________________________step. The overall rate cannot be any faster than the slowest step in a mechanism.

2.4.2. One step reaches a ____________________________. Since reverse

forward

eqk

kK we can use this to

help solve for the rate-law expression.

Lets do it 1. The rate law for the following reaction: 2NO2(g) + F2(g) 2NO2F(g) was experimentally determined to be: rate = k[NO2][F2]. Explain which of the following mechanisms is most plausible.

(1) 2NO2(g) + F2(g) 2NO2F(g) (2) NO2(g) + F2(g) NO2F(g) + F(g) fast NO2(g) + F(g) NO2F(g) slow (3) NO2(g) + F2(g) NO2F(g) + F(g) slow NO2(g) + F(g) NO2F(g) fast

2.5. Some Terminology

Intermediate: Temporary substance ____________________________and ____________________________in the mechanism. Cannot show up in the final rate law expression and cancels out when forming the final, balanced equation.

Transition state/activated complex: point in reaction progress in which the bond(s) that are going to form are partially formed, the bond(s) that are broken are partially broken and the reaction is committed to proceeding to product.

Catalyst: A substance that increases the rate of a reaction by either stabilizing the transition state or forming a new transition state. Present at the ____________________________and ____________________________of an experiment. Might show up in a rate law expression but ultimately cancels out when forming the final, balanced equation.

Lets do it 2. 2 The suggested mechanism for the reaction between peroxide and iodide ion is below. Write the overall reaction and identify specie(s) which is acting as a catalyst and a specie(s) which is acting as an intermediate.

Step 1: H2O2 + I─ H2O + IO─ slow Step 2: IO─ + H2O2 H2O + O2 + I─ fast

2 http://intro.chem.okstate.edu/HTML/SCH13.HTM



Lets do it 3. 3 Given the following reaction mechanism determine the overall reaction, write the rate law for the reaction, and indicate the presence of catalyst and/or intermediates.

Step 1: NO2(g) + NO2(g) NO(g) + NO3(g) slow Step 2: NO3(g) NO(g) + O2(g) fast

Lets do it 4. What is the rate law for the following reaction and its mechanism? 2HgCl2 + C2O42- → 2Cl- + 2CO2 + Hg2Cl2 (overall reaction)

HgCl2 + C2O42- ⇌ HgCl2C2O42- (Fast)

HgCl2C2O42- + C2O42- → Hg + 2C2O4Cl2- (Slow)

Hg + HgCl2 → Hg2Cl2 (Fast)

2C2O4Cl2- → C2O42- + 2Cl- + 2CO2 (Fast)

3. Determination of Rate

3 http://intro.chem.okstate.edu/HTML/SCH13.HTM

[reactants] decrease with

time [products]

increase with time

When rate levels off

(slope = zero) system is at equilibrium

The rate wrt to a species is the slope of the line and is NOT constant.



3.1. Measurement of rate

3.1.1. _______________ rate (the one we’ll use the most):

Make tangent to curve at t=0 Rate = slope of the tangent.

3.1.2. _______________ rate: Make a tangent to curve at a

designated time. Rate = slope of the tangent

3.1.3. _______________ rate: Make a line between two designated

points Rate = slope of the tangent

3.2. Relationship between rates measured

aA + bB cC + dD

-1

a

D[A]

Dt

æ

èç

ö

ø÷= -

1

b

D[B]

Dt

æ

èç

ö

ø÷=

1

c

D[C]

Dt

æ

èç

ö

ø÷=

1

d

D[D]

Dt

æ

èç

ö

ø÷

Lets do it 5. 4 Give the relative rates for the disappearance of reactants and formation of

products for the following reaction: 4PH3(g) P4(g) + 6H2(g)

Lets do it 6. The rate of the following reaction measured with respect to N2O5 was found to be -8.31 x 10-4 mol dm-3 min-1 .5 Predict the rate had it been measured with respect to NO2.

2 N2O5 4NO2 + O2

4. Experimental determination of reaction orders

We are now dealing with overall reactions, NOT elementary reactions Order must be determined from experimental data. It CANNOT be determined from the stoichiometry of the reaction.

4 Chemistry and Chemical Reactivity Kotz and Purcell, p. 610

5 http://www.newi.ac.uk/buckleyc/react.htm

Negative sign before rates measured wrt to reactants: remember [reactants] decrease

with time

2NO2 --> 2NO + O2

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 10 20 30 40 50

Time (min)

Co

nc

en

tra

tio

n (

M)



DIFFERENTIAL ORDER MATHMATICAL EQUATION

INTEGRATED ORDER MATHMATICAL EQUATION

Use when the question involves RATE and CONCENTRATION or PARTIAL PRESSURE

ZERO Rate = k Use when the question involves TIME and CONCENTRATION or PARTIAL PRESSURE

ZERO 0AktA

1ST Rate = k[A] 1ST

0lnln AktA

2ND Rate = k[A]2 2ND 0

11

Akt

A

4.1. DIFFERENTIAL RATE LAW EXPRESSION Goal is to find data where we have one concentration that is the independent variable and

the rest are dependent variables. Concentration is typically measured in molarity or partial pressures. Time can be measured in minutes or seconds (although there are reactions that take

centuries to occur!)6 Lets do it 7. 7 Determine the rate law expression (including the value of k) for the reaction, 2 NO(g) + 2 H 2(g) N2(g) + 2 H2O(g), using the following data. Exp. # PNO (mm Hg) PH2 (mm Hg) Initial Rate (mm Hg/s) 1 400 150 0.66 2 400 300 1.34 3 150 400 0.25 4 300 400 1.03

a) Order with respect to NO:

b) Order with respect to H2:

c) Overall order:

d) Temperature independent rate law expression

6 http://departments.oxy.edu/tops/Kinetics/kineticsprepost.htm

7 Data from http://intro.chem.okstate.edu/HTML/SCH13.HTM



e) Determination of k with units

f) Temperature dependent rate law expression Lets do it 8. 8 The acid catalyzed bromination of acetone (propanone) is as follows:

CH3COCH3 + Br2 CH3COCH2Br + H+ + Br─

Exp. # [CH3COCH3] [Br2] [H+] Initial rate (M/sec) 1 0.30 0.05 0.05 5.7 x 10-5 2 0.30 0.10 0.05 5.7 x 10-5

3 0.30 0.05 0.10 12.0 x 10-5 4 0.40 0.05 0.20 31.0 x 10-5 5 0.40 0.05 0.05 7.6 x 10-5

a) Deduce the rate expression for the reaction and give the order with respect to each reactant as well as the overall order.

b) What is the numerical value for k, including units, at this temperature?

c) If [H+] is maintained at 0.050 M while [CH3COCH3] and [Br2] are both 0.10 M, what is the initial rate of the reaction?

d) Calculate the impact on the rate if [CH3COCH3] was doubled and [H+] is tripled.

8 Data from Kotz and Purcell, p. 651



4.2. The Integrated Rate Law: Time v. Concentration zero order 1st order 2nd order Integrated rate law 0AktA 0lnln AktA

0

11

Akt

A

Graphical analysis: straight line graphs

[A] vs time straight line, negative slope

ln[A] vs time straight line, negative slope

1/[A] vs time straight line, positive slope

Determination of rate constant k

Slope = -k Slope = -k Slope = k

4.2.1. PSEUDO-ORDER: The graphical method for determining order only works if you have one reactant. If you have more than one reactant you set up an experiment so there is APPARENTLY only one reactant

Let A + B C, rate = k[A]x[B]y In order to determine the order with respect to “A” set up exp: [B] >>>[A] [B] is effectively constant over time (loss is negligible) k’ = k[B]y pseudo rate law: rate = k’[A]x

4.2.2. GRAPHICAL ANALYSIS

Lets do it 9. 9 Use your graphing calculator to graph the following data to determine the order with respect to NO2. Calculate the value of k from the graph that appropriately gives the order.

2NO2(g) 2NO(g) + O2(g)

Time(sec) [NO2] (M) Time(sec) [NO2] (M) 0 0.0100 150 0.0055 25 0.0088 175 0.0051 50 0.0079 200 0.0048 75 0.0071 250 0.0042 100 0.0065 300 0.0038

9 Data from http://intro.chem.okstate.edu/HTML/SCH13.HTM



[NO2] vs. time ln[NO2] vs. time 1/[NO2] vs. time Correlation Coefficient

Slope

Value for k

Lets do it 10. 10 The decomposition of NOCl(g) : 2NOCl(g) 2NO(g) + Cl2(g) is a second order

reaction with a rate constant of 0.0480 M-1·sec-1 at 200 ºC. In an experiment at 200 ºC, the initial concentration of NOCl was 0.400 M.

(a) What is the concentration of NOCl after 15.0 min have elapsed?

(b) How many seconds will it take for the concentration of NOCl(g) to drop to 0.150 M? Lets do it 11. 11 C4H8 decomposes according to the following equation; C4H8(g) 2C2H4(g). The

rate constant for the decomposition is 6.07 x 10-10 sec-1 at 25 ºC.

a) What is the order of the reaction?

b) How long would it take for 1.00 % of a sample of C4H8 to decompose at 25 ºC and 1 atm?

10

Problem from http://intro.chem.okstate.edu/HTML/SCH13.HTM 11

http://intro.chem.okstate.edu/HTML/SCH13.HTM



4.2.3. KINETICS AND STOICHIOMETRY – If you know the beginning and ending amounts of one species, you can do stoichiometry to find the amount of another species.

Determine the amount used: [conc]initial – [conc]final for reactants OR Determine the amount formed: [conc]final – [conc]initial for products Multiply by the magic mole ratio to convert to new species Add this amount if the new species is a product, subtract it if it is a reactant. Using a RICE (reaction, initial, change, end) table might help!

Lets do it 12. Use the data in U do it 9 to determine the amount of oxygen formed after 50 seconds.

4.3. Half-life Calculations

The half-life (t1/2) is the time required for half of an amount to react. It is frequently used in radiological dating (specifically carbon dating). In the following example, the half-life is determined graphically to be 100 seconds. NOTE: in this case the half-life was independent of [N2O5]0. It was 100 sec whether one started with 0.100 M or 0.0250 M. From the summary table we can see that the only t1/2 that is independent of [A]0 is 1st order, therefore this reaction must be 1st order wrt [N2O5] zero order 1st order 2nd order Half-life

k

At

2

0

21

kt

693.0

21

021

1

Akt



Lets do it 13. 12The decomposition of H2O2 to H2O and O2 follows first order kinetics with a rate constant of 0.0410 min-1 at a particular temperature: H2O2(l) 2H2O(l) + O2(g). What is the half-life for this reaction?

Lets do it 14. 13 The second-order decomposition of nitrous oxide, N2O, has a half-life of 75.0 min at 900 K when the initial concentration of N2O is 2.00 x 10-2 M.

a. What is the concentration of nitrous oxide after 150 minutes?

b. How long will it take for 40.0 % of the sample to decompose?

Lets do it 15. 14 32P is used to treat some diseases of the bone. Its half-life is 14 days. Find the time it would take for a sample of 32P to decay from an activity of 10,000 counts per minute to 8,500 counts per minute. Assume first order kinetics.

12

http://intro.chem.okstate.edu/HTML/SCH13.HTM 13


http://chem.lapeer.org/Chem1Docs/HalflifeWorksheet.html



4.4. Radiological Dating: Radioactive decay follows first order kinetics. Since t1/2 is constant, the value can be used to estimate the age of objects. For once-living objects, carbon-14 is typically used. While alive, the carbon cycle ensures a relatively constant ratio of C-14/C-12. Once dead, no more carbon is taken up and the current level of C-14 begins to decay. Carbon-14 decay is not valid for objects that are recently dead, or older than ~40,000 years.

Lets do it 16. 15 At one time on the Historic Tour in the Mammoth Cave park in Kentucky an

Native American, found in the caves was on display. Because of the cave conditions the Indian's body was well preserved. It was estimated that the Native American was about 145 pounds and died at the age of 45 from a falling rock that crushed his spinal column. Determine the approximate year of the Indian's death from the following data collected from the activity of 14C. Assume the initial activity of 14C in his body was 15.7 cpm per gram of 14C, the current activity (circa 1950) was 11.76 cpm per gram of 14C, and the half life of 14C is 5730 years.

15

http://chem.lapeer.org/Chem1Docs/HalflifeWorksheet.html



5. Temperature

5.1. Theory

REMEMBER: Temperature is a measure of the AVERAGE kinetic energy of molecules. The actual distribution of kinetic energy is described by a Boltzmann distribution. Notice the following:

The peak of the lower temperature is at a lower energy than the higher temperature peak.

The higher temperature peak is broader and the maximum height is lower than the lower T peak

The “tail” of the higher T curve is higher than the lower T curve

More molecules, and hence more collisions, have E > Ea therefore the rate is faster.

5.2. Mathematics

5.2.1. The Arrhenius equation describes the temperature dependence of rate. Note it is the rate constant, k, that is temperature dependent.

)( RTEa

Aek

Ea = activation energy in J R = ideal gas constant 8.314 A = frequency factor T = temperature in Kelvin

5.2.2. This equation is most helpful for finding the activation energy for a reaction. If we take the natural log of both sides we get the equation:

)ln(1

)ln( ATR

Ek a

y = m x + b A graph of ln(k) vs. 1/T yields a straight line with slope = (─Ea/R)



Lets do it 17. 16The following rate data was obtained at different temperatures for the reaction:

O3(g) + NO(g) O2(g) + NO2(g)

Temp (K) k (M-1s-1) 600 0.28 650 0.22 700 1.30 750 6.00 800 23.0

(a) What is the order of the reaction?

(b) Use your graphing calculator to determine the activation energy for the reaction.

5.2.3. We can also derive an equation that relates the rate constant at two temperatures.

121

2 11ln

TTR

E

k

k a

Lets do it 18. 17 At 300 ºC the rate constant for the reaction: is 2.41 x 10-10 sec-1. At 400 ºC the rate constant is 1.16 x 10-6 sec-1. Calculate the activation energy for the reaction.

Lets do it 19. The decomposition of NOCl to form nitrogen monoxide and chlorine has a rate

constant of 9.3 x 10─5 L mol─1 s─1 at 100oC. The activation energy for this reaction is 99 KJ mol─1. At what temperature is k=1.0 x 10─3 L mol─1 s─1?18

16



Question 14.75 Chemistry: The Study of Matter and its Changes by Brady and Holum.



6. Catalysis

Provides an ____________________________pathway that has a lower activation energy than the uncatalyzed reaction.

Either ____________________________the activated complex or forms a ____________________________ activated complex with a lower potential energy.

Speeds up the reaction without being __________________________ in the process.

Biological catalysts are called ____________________________ ____________________________Catalysis: The catalyst is a different phase than the reactants. Often the catalyst is a solid onto which the reactants adsorb. See Website: http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ethylene8.html ____________________________Catalysis: The catalyst is a same phase than the reactants.

7. Summary table – Collision theory and kinetics Collision Theory Factors that affect the rate Formulas Molecules must collide for reaction to occur

Temperature: an increase in temperature increases the frequency of collision and thus the rate.

Arrhenius equation:

)( RTEa

Aek

“A” = z*p where z is a measure of collision frequency

Concentration: an increase in concentration increases the frequency of collision and thus the rate. NOTE: the rate does not always depend on the concentration of reactants.

Rate vs. Concentration: rate law expressions Time vs. Concentration: Integrated rate law expression

Molecules must collide with sufficient energy for a reaction to occur. This energy is called the Activation Energy, Ea

Temperature: an increase in temperature means molecules/atoms have more energy on average and thus more collisions will have energy > Ea.

Arrhenius equation:

)( RTEa

Aek

“A” = z*p where p is a steric factor

See website for Ea animation: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/flash.mhtml

Notice that the curve is the same, but the activation energy is lower, therefore more effective collisions!



Collision Theory Factors that affect the rate Formulas Catalyst: a catalyst is a substance

NOT used up in a reaction. It provides an alternative pathway with a lower activation energy.

Molecules must collide with the correct orientation.

Surface Area: Large solids are less exposed than ground solids which in turn are less exposed than aqueous and gases.

Arrhenius equation:

)( RTEa

Aek

“A” = z*p where p is a steric factor, a number < 1 that reflects the fraction of collisions with effective orientation

See website for orientation animation: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/flash.mhtml

8. Terminology Order of reaction:

Number that represents the effect of concentration on the rate. Order for each species in the reaction (including product in some cases!) for your level will be

whole numbers 0, 1, 2, and maybe 3. Overall order: sum of the individual orders.

Differential rate law (rate law expression): We think of rate as [A]/ t. It is really a differential equation d[A]/dt. Don’t panic! We won’t be

doing calculus – we’ll just take advantage of the magic of mathematics. If the data in a problem is rate and concentration you will use these equations.

Integrated rate law: This is the equation that results after calculus is performed. If the data in a problem is time and concentration you will use these equations.

Elementary Reaction: Reaction occurs exactly as written. The order can actually be determined from the stoichiometry of

an elementary reaction. Molecularity: A term that indicates the number of molecules involved in an elementary reaction.

Unimolecular: one molecule involved in step – always 1st order for the elementary step. Bimolecular: two molecules involved in step – always 2nd order for the elementary step. Termolecular: Very rare since it involves three molecules simultaneously colliding!

Overall Reaction: Sum of all the elementary steps in a reaction. The order CANNOT be determined from the

stoichiometry of an overall reaction. We will use experimental data to determine the order. We don’t really go any more in depth than Pre-AP so this should be review!

Mechanism: Proposed series of elementary steps that agrees with experimental data and sums up to provide the

overall reaction stoichiometry (ie the overall balanced equation). Half-life

Time required for half of the original amount to react



9. Summary Table – concentration effects

zero order 1st order 2nd order Differential rate law kAkrate

0 Akrate 2Akrate

Integrated rate law 0AktA 0lnln AktA

0

11

Akt

A

Graphical analysis: straight line graphs

[A] vs time straight line, negative slope

ln[A] vs time straight line, negative slope

1/[A] vs time straight line, positive slope

Determination of rate constant k

Slope = -k Slope = -k Slope = k

Half-life k

At

2

0

21

kt

693.0

21

021

1

Akt

NOTE: [A]0 is the concentration at time = zero, ie the beginning of the reaction.



Selected AP Problems – Copyright College Board – for face to face teaching only

1998 Answer the following questions regarding the kinetics of chemical reactions.

a) The diagram below at right shows the energy pathway for the reaction O3 + NO --> NO2 + O2.

Clearly label the following directly on the diagram.

i. The activation energy (Ea) for the forward reaction

ii. The enthalpy change (ΔH) for the reaction b) The reaction 2 N2O5 4 NO2 + O2 is first order with

respect to N2O5. i. Using the axes at right, complete the graph that

represents the change in [N2O5] over time as the reaction proceeds.

ii. Describe how the graph in (i) could be used to find the reaction rate at a given time, t.

iii. Considering the rate law and the graph in (i), describe how the value of the rate constant, k, could be determined.

iv. If more N2O5 were added to the reaction mixture at constant temperature, what would be the effect on the rate constant, k? Explain.

c) Data for the chemical reaction 2A B + C were collected by measuring the concentration

of A at 10-minute intervals for 80 minutes. The following graphs were generated from analysis of data.

Use the information in the graphs above to answer the following. (i) Write the rate-law expression for the reaction. Justify your answer. (ii) Describe how to determine the value of the rate constant for the reaction.



1997 2 A + B C + D

The following results were obtained when the reaction represented above was studied at 25 °C

Experiment Initial

[A] Initial

[B]

Initial Rate of Formation

of C (mol L−1 min−1)

1 0.25 0.75 4.3 x 10−4

2 0.75 0.75 1.3 x 10−3

3 1.50 1.50 5.3 x 10−3

4 1.75 ?? 8.0 x 10−3

a) Determine the order of the reaction with respect to A and B. Justify your answer. b) Write the rate law for the reaction. Calculate the value of the rate constant, specifying units. c) Determine the initial rate of change of [A] in Experiment 3. d) Determine the initial value of [B] in Experiment 4. e) Identify which of the reaction mechanisms represented below is consistent with the rate law developed in part (b). Justify your choice.

1 A + B C + M Fast

M + A D Slow

2 B M Fast equilibrium

M + A C + X Slow

A + X D Fast

3 A + B M Fast equilibrium

M + AC + X Slow

X D Fast



1996 The reaction between NO and H2 is believed to occur in the following three-step process.

NO + NO N2O2 (fast equilibrium) N2O2 + H2 N2O + H2O (slow) N2O + H2 N2 + H2O (fast)

(a) Write a balanced equation for the overall reaction. (b) Identify the intermediates in the reaction. Explain your reasoning. (c) From the mechanism represented above, a student correctly deduces that the rate law for the reaction is rate = k[NO]2[H2]. The student then concludes that (1) the reaction is third-order and (2) the mechanism involves the simultaneous collision of two NO molecules and an H2 molecule. Are conclusions (1) and (2) correct? Explain. (d) Explain why an increase in temperature increases the rate constant, k, given the rate law in (c).

C h e m i c a l K i n e t i c s C H A P T E R 6 | 1 · _____requirements using stoichiometry or...

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