chemistry - mcquarrie 4e ch.17 - chemical kinetics: rate...

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CHEMISTRY - MCQUARRIE 4E

CH.17 - CHEMICAL KINETICS: RATE LAWS

CONCEPT: RATES OF CHEMICAL REACTIONS

____________________________ is the study of reaction rates, and tells us the change in concentrations of reactants or

products over a period of time.

Although a chemical equation can help us calculate the theoretical yield from reactants, it can’t tell us how fast it goes.

• Looking at a chemical reaction in the simplest way can be seen as ___________ breaking down to form

______________.

A BReaction :

0 Seconds 30 Seconds 60 Seconds 90 Seconds



CONCEPT: FACTORS INFLUENCING REACTION RATES

VS.

1. Concentration: Molecules must _________________ to react.

• Increasing the number of molecules in a container, increases their _______________ and thereby causes the rate

to increase.

A BReaction : A B+

2. Surface Area: The frequency of collisions increases with ____________________ surface area.

CH3CH2CH2CH3 VS.

H2C

H2C CH2

CH2

3. Temperature: Increasing the temperature increases the reaction rate by increasing the ___________ and

___________ of collisions.

4. Catalyst: A catalyst increases the rate of a reaction by ______________________ the energy of activation.



CONCEPT: TEMPERATURE AND RATE

Temperature can have a huge effect on the reaction rate. It affects the rate by affecting the rate constant, k.

• Increasing the temperature or ______________ amount of catalyst will ______________ the rate constant, k.

• The ______________ Equation relates both the temperature and the rate constant.

For a reaction to be successful the molecules must collide with sufficient ____________________ and the correct

____________________.

Conversion of the equation into its logarithmic form can help us find Ea when two rate constants or temperatures are given.

ln k2k1=−EaR

1T2−1T1

"

#$

%

&'

A plot-wise approach can also be used to convert the equation in order to find Ea.

ln A

slope = ΔyΔx

= −EaRln k

1T

k =Ae−Ea

RT

A =

Ea =

R =

T =

lnk = − EaR

"

#$

%

&'1T"

#$

%

&'+ lnA



PRACTICE: TEMPERATURE AND RATE (CALCULATIONS)

EXAMPLE: The reaction 2 HCl (g) H2 (g) + Cl2 (g) has Ea of 1.77 x 104 kJ/mol and a rate constant of 1.32 x 10-1

at 700 K. What is the rate constant at 685 K?

PRACTICE 1: If a first order reaction has a frequency factor of 3.98 x 1013 s-1 and Ea of 160 kJ, then calculate the rate

constant at 25oC.

PRACTICE 2: Generally, the slower the rate of the reaction then the ___________ the energy of activation (Ea) and the higher the temperature, the ____________ the value of rate constant, k.

a) larger, larger b) smaller, smaller c) smaller, larger d) larger, smaller



CONCEPT: RATE LAW

Although, chemical reactions can be reversible we will only look at the ___________ reaction for chemical reactions.

• By ignoring ___________ reactions then the rate depends only on __________ concentrations and ___________.

4 NO (g) + O2 (g) 2 N2O3 (g)

Rate Law = k [NO]x [O2]y k = ___________________________________

x & y = ___________________________________

Unless the reaction is classified as a ______________ step, a rate-determining step, then x & y must be calculated

experimentally.

EXAMPLE: For the following reaction, use the given rate law to determine the best answer for the reaction with respect to each reactant and the overall order.

H2O2 (aq) + 3 I – (aq) + 2 H+ (aq) I3 – + 2 H2O (l) Rate = k [H2O2]2 [I –]

a) H2O2 is 1st order, I – is 1st order, 2nd order overall.

b) H2O2 is 2nd order, I – is 1st order, 3nd order overall.

c) H2O2 is 0th order, I – is 1st order, H+ is 1st order, 3rd order overall.

d) H2O2 is 2nd order, I – is 1st order, H+ is 0th order, 3rd order overall.



PRACTICE: REACTION ORDERS

Answer each of the following question based on the following chemical reaction:

3 A (g) + B (g) + 2 C (g) D (g)

Experiment Initial [A] Initial [B] Initial [C] Initial Rate

1 0.0500 M 0.0500 M 0.0100 M 6.25 x 10-3

2 0.1000 M 0.0500 M 0.0100 M 1.25 x 10-2

3 0.1000 M 0.1000 M 0.0100 M 5.00 x 10-2

4 0.0500 M 0.0500 M 0.0200 M 6.25 x 10-3

EXAMPLE 1: Calculate the reaction order for reactant A.

PRACTICE: Calculate the reaction orders for reactants B and C.

EXAMPLE 2: Calculate the rate constant and the new rate for the given reaction if the the initial concentrations of [A] =

0.300 M, [B] = 0.150 M and [C] = 0.150 M.



PRACTICE: RATE CONSTANT & RATE

EXAMPLE 1: A certain chemical reaction has the given rate law:

Rate = k [A]3 [B]2 [C]-1

What are the units of the rate constant for the given reaction?

a) M-2 · s-1 b) M2 · s-1 c) M-4 · s-1 d) M-3 · s-1 e) M3 · s-1

EXAMPLE 2: The reaction of 3 A + B 2 C + D, was found to be: Rate = k [A]2 [B]3. How much would the rate

increase by if A were tripled while B were increased by half?

a) 0.50 b) 30.38 c) 0.75 d) 20.25 e) 1.125

PRACTICE: If the rate law for the following reaction is found to be the following: Rate = [Cl2 ] ⋅[HCCl3] . What are

the units for the rate constant, K?

Cl2 (g) + HCCl3 (g) HCl (g) + CCl4 (g)

a) M3/2

s b) Ms c)

1M ⋅s d)

1M3/2 ⋅s e) M ⋅s



CONCEPT: REACTION PATHWAYS

A reaction ____________________ is a sequence of single reaction steps that add up to the overall chemical reaction. For example a possible chemical setup for the overall reaction:

A + 2 B E

might involve these steps:

1) A + B C

2) B + C D

3) D E

EXAMPLE 1: The elementary reaction, 2 NOBr (g) 2 NO + Br2, is an example of a _____________ reaction.

a) Unimolecular b) Dimolecular c) Bimolecular d) Tetramolecular

EXAMPLE 2: Answer the following questions to the following reaction:

Br + O2 + O BrO + O2 [FAST]

Br + BrO + O Br2O + O [SLOW]

a. Identify the intermediate(s).

b. Identify the catalyst(s).

c. What is the overall reaction?



CONCEPT: 1ST, 2ND or 0TH: WHAT’S YOUR ORDER?

If you were super observant you might have noticed that although we were talking about rate, the rate law didn’t include

___________ as a variable.

A BReaction :

The good thing is that the _______________ Rate Laws help to answer an important question in kinetics:

• “How long will it take x moles per liters of A to be consumed?”

Rate Laws Zero-Order First-Order Second-Order

Rate Equations [A]t = −kt +[A]0 ln[A]t = −kt + ln[A]0 1[A]t

= kt + 1[A]0

[A]

Time

[A]0

Slope = - k

ln[A

]

Time

ln[A]0

Slope = - k

1/[A

]

Time

1/[A]0

Slope = k

Half-Life t 12=[A]02k t 1

2=ln2k t 1

2=

1k[A]0



PRACTICE: INTEGRATED RATE LAWS (CALCULATIONS)

EXAMPLE 1: The oxidation of ethane follows a first order mechanism, with a very high rate constant of 32 s-1, to form H2O

and CO2 as products. If the initial [C2H6] is 4.12 M, what is the concentration after 1.12 x 10-3 minutes?

EXAMPLE 2: Iodine-123 is used to study thyroid gland function. This radioactive isotope breaks down in a first order process with a half-life of 8.50 hours at 800 K. How long will it take for the concentration of iodine-123 to be 74.1% complete?

PRACTICE 1: At 25oC, 2 NOBr (g) 2 NO (g) + Br2 (g). The rate of the reaction is found to be: rate = k [NOBr]2.

The constant at 25oC is 7.80 x 10-4 M-1 · s-1. If 0.550 moles of NOBr (g) is placed in a 5.0 L container, how long will take for the concentration to reach 0.063 moles of NOBr (g)?

PRACTICE 2: In a typical chemical reaction, nitrogen trioxide, NO3, reacts to produce nitrogen dioxide, NO2, and oxygen gas, O. 2 NO3 (g) 2 NO2 (g) + 2 O (g)

A plot of [NO3] vs. time is linear and the slope is equal to 0.183. If the initial concentration of NO3 is 0.930 M, how long will it take for the final concentration to reach 0.400 M?



PRACTICE: INTEGRATED RATE LAWS (CALCULATIONS 2)

EXAMPLE: Given the following graph for a second order reaction:

a) Calculate the frequency factor.

b) Calculate the energy of activation in (J/mol):

PRACTICE: The three plots were done based on a chemical reaction.

a. What is the rate constant of the reaction if it takes 21.2 minutes for the reaction to be 38.0% complete?



CONCEPT: HALF–LIFE Half-life is defined as the time it takes for half of the amount of a substance to decay in a certain amount of time.

• However, this is mainly true for a ______ order reaction.

Zero-Order First-Order Second-Order

t 12=[A]02k t 1

2=ln2k t 1

2=

1k[A]0



CONCEPT: INTEGRATED RATE LAWS Zeroth Order

• The concentration of reactants decreases in a uniform manner over time. • The rate of the reaction remains constant.

First–Order

• The concentration of reactants decreases in a logarithmic manner. More recognizable by a constant half-life. • The rate of the reaction decreases uniformly.

Second-Order

• The largest drop in concentration occurs initially followed by a decreasing loss of reactant. More recognizable by an increasing half-life

• The rate of the reaction decreases substantially before leveling off.

Time Concentration

[Reactants]

TimeRate

Time

[Products]

Time

Rate

Time

Time Concentration

[Reactants]

Time

Rate

Time

[Products]

TimeRate

Time

Time Concentration

[Reactants]

Time

Rate

Time

[Products]

Time

Rate

Time



chemistry - mcquarrie 4e ch.17 - chemical kinetics: rate...

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