chapter 12 chemical kinetics - lebanon high school 6 kinetics part 1.pdfthe rate law • the rate...

Chapter 12

Chemical Kinetics

The Rate of a Chemical Reaction

Section 12.1

Reaction Rates

Copyright © Cengage Learning. All rights reserved 3

Reaction Rate

� Change in concentration of a reactant or product

per unit time.

[A] means concentration of A in mol/L; A is the

reactant or product being considered.

[ ]

2 1

2 1

concentration of A at time concentration of A at time Rate =

A=

−

−

∆

∆

t t

t t

t

Section 12.1

Reaction Rates


The Decomposition of Nitrogen Dioxide

Section 12.1

Reaction Rates


Instantaneous Rate

� Value of the rate at a particular time.

� Can be obtained by computing the slope of a line

tangent to the curve at that point.

Reaction Rate and Stoichiometry

What do coefficients have to do with rate?

• If there are coefficients, then change in the number of molecules of one substance is a multiple of the change in the number of molecules of another.

• To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient.

• So, for the reaction:

aA + bB � cC + dD

© 2014 Pearson Education, Inc.

Reaction Rate and Stoichiometry

H2 (g) + I2 (g) → 2 HI(g)

– For the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made.


Practice Problems:

• Be able to relate the overall reaction rate to the

change in concentration of a particular reactant or

product

• Be able to relate the change concentration of one

reactant or product to anther using their respective

coefficients. Watch out for the sign convention!


Practice Problems:

Consider the reaction:

4PH3 (g) � P4 (g) + 6H2 (g)

If in a certain experiment, over a specific time

period, 0.0048 moles of PH3 are consumed in a 2.0 L

container each second of reaction, what are the rates

of production of P4 and H2, respectively?


Practice Problems:

Consider the reaction:

4PH3 (g) � P4 (g) + 6H2 (g)

If in a certain experiment, over a specific time period 0.0048 moles of PH3+

are consumed in a 2.0 L container each second of reaction, what are the rates

of production of P4 and H2, respectively?

Rate = s

L)mol/2.00.048(][3 −−

=∆

∆−

t

PH = 2.4 ×

310

−mol/L•s

t∆

]PH[∆

4

1

t∆

]P[∆ 34 −= = 2.4 × 3

10−

/4 = 6.0 × 4

10−

mol/L•s

t∆

]PH[∆

4

6

t∆

]H[∆ 32 −= = 6(2.4 × 3

10−

)/4 = 3.6 × 3

10−

mol/L•s

The Rate Law

• The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well.

• The rate law must be determined experimentally!

• The rate of a reaction is directly proportional to the concentration of each reactant raised to a power.

• For the reaction aA + bB →→→→ products the rate law would have the form given below.

– n and m are called the orders for each reactant.

– k is called the rate constant.

Section 12.2

Rate Laws: An Introduction

The Rate Law (aka differential rate law)

� Relates Rate to Reactant Concentrations.(remember, the rate changes; it depends on concentration!)

� For the decomposition of nitrogen dioxide:

2NO2(g) → 2NO(g) + O2(g)

Rate = k[NO2]n:

� k = rate constant

� n = order of the reactant


Section 12.2


Rate Law

Rate = k[NO2]n

� The concentrations of the products do not appear in the

rate law because the reaction rate is being studied

under conditions where the reverse reaction does not

contribute to the overall rate.


Section 12.2


Rate Law

Rate = k[NO2]n

� The value of the exponent n must be determined by

experiment; it cannot be written from the balanced

equation.


Rate = k[A]n

� If a reaction is zero order, the rate of the reaction is

always the same.

� Doubling [A] will have no effect on the reaction rate.

� If a reaction is first order, the rate is directly

proportional to the reactant concentration.

� Doubling [A] will double the rate of the reaction.

� If a reaction is second order, the rate is directly

proportional to the square of the reactant

concentration.

� Doubling [A] will quadruple the rate of the reaction.

Section 12.2


Types of Rate Laws

�


Section 12.2


Rate Laws: Which one is more useful?

� Because the differential and integrated rate laws for a

given reaction are related in a well–defined way, the

experimental determination of either of the rate laws is

sufficient.


Section 12.2


Rate Laws: Which one is more useful?

� Experimental convenience usually dictates which type of

rate law is determined experimentally.

� Knowing the rate law for a reaction is important mainly

because we can usually infer the individual steps

involved in the reaction from the specific form of the

rate law.


Section 12.3

Determining the Form of the Rate Law

� Determine experimentally the power to which each

reactant concentration must be raised in the rate law.


Section 12.3


Method of Initial Rates

� The value of the initial rate is determined for each

experiment at the same value of t as close to t = 0 as

possible.

� Several experiments are carried out using different

initial concentrations of each of the reactants, and the

initial rate is determined for each run.

� The results are then compared to see how the initial

rate depends on the initial concentrations of each of the

reactants.


Section 12.3


Overall Reaction Order

� The sum of the exponents in the reaction rate equation.

Rate = k[A]n[B]m

Overall reaction order = n + m

k = rate constant

[A] = concentration of reactant A

[B] = concentration of reactant B


Section 12.3


How do exponents (orders) in rate laws compare to

coefficients in balanced equations?

Why?


CONCEPT CHECK!

Section 12.4

The Integrated Rate Law

First-Order

� Differential:

Rate = k[A]

� Integrated:

ln[A] = –kt + ln[A]o

[A] = concentration of A at time t

k = rate constant

t = time

[A]o = initial concentration of ACopyright © Cengage Learning. All rights reserved 24

Section 12.4


Plot of ln[N2O5] vs Time: 1st order


Section 12.4


First-Order Reactions and Half-Life

� Rate = k[A]

� Integrated:

ln[A] = –kt + ln[A]o

We can consider how long it would take for half of a

reactant to be consumed.

Rearrange this equation to solve for t when a concentration

[A] is halved. You will find that ln[A] - ln[A0] is equal to

0.693. This is how the half-life equation is derived.


Section 12.4


First-Order

� Time required for a reactant to reach half its original

concentration

� Half–Life:

k = rate constant

� Half–life of a 1st order reaction does not depend on the

concentration of reactants.


12

0.693 = t

k

Section 12.4


A first order reaction is 35% complete at the end of

55 minutes. What is the value of k?

k = 7.8 × 10–3 min–1


Section 12.4


Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time

Section 12.4


Second-Order

� Half–Life (2nd order) :

k = rate constant

[A]o = initial concentration of A

� For a 2nd order reaction, Half–life gets longer as the reaction

progresses and the concentration of reactants decrease.

� Each successive half–life is double the preceding one.


[ ]12

0

1 =

At

k

Section 12.4


For a reaction aA � Products,

[A]0 = 5.0 M, and the first two half-lives are 25 and 50

minutes, respectively.

a) Write the rate law for this reaction.

rate = k[A]2

b) Calculate k.

k = 8.0 × 10-3 M–1min–1

c) Calculate [A] at t = 525 minutes.

[A] = 0.23 MCopyright © Cengage Learning. All rights reserved 32

EXERCISE!

Section 12.4


Zero-Order

� Rate = k[A]0 = k

� Integrated:

[A] = –kt + [A]o


k = rate constant

t = time



Section 12.4


Plot of [A] vs Time


Section 12.4


Zero-Order

� Half–Life:

k = rate constant


� Half–life gets shorter as the reaction progresses and the

concentration of reactants decrease.


[ ]0

12

A =

2t

k

Section 12.4


Check your Royal Purple Papers

� What equations are given?

� Are they labeled?

� Which apply to first order reactions only?

� Which are second order?

� Which are zero order?


Section 12.4


How can you tell the difference among 0th, 1st, and 2nd

order rate laws from their graphs?


CONCEPT CHECK!

Section 12.4


Rate Laws


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Section 12.4


Summary of the Rate Laws


Important: Notice that only in the case of a first order reaction

is the half-life independent of [A].

Therefore, if the ½ life is constant, it’s 1st order.

Section 12.4


Consider the reaction aA � Products.

[A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate

for each case). Calculate [A] after 30.0 seconds have passed,

assuming the reaction is:

a) Zero order

b) First order

c) Second order


4.7 M

3.7 M

2.0 M

EXERCISE!

chapter 12 chemical kinetics - lebanon high school 6 kinetics part 1.pdfthe rate law • the rate...

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