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Chapter 4Kinetics: Rates and Mechanisms of Chemical Reactions

16-1

Kinetics: Rates and Mechanisms of Chemical Reactions16.1 Factors That Influence Reaction Rate 16.2 Expressing the Reaction Rate

16.3 The Rate Law and Its Components16.4 Integrated Rate Laws: Concentration Changes over Time 16.5 The Effect of Temperature on Reaction Rate

16.6 Explaining the Effects of Concentration and Temperature 16.7 Reaction Mechanisms: Steps in the Overall Reaction16.8 Catalysis: Speeding Up a Chemical Reaction

16-2

Figure 16.1

Reaction rate: the central focus of chemical kinetics.

16-3

Figure 16.2

The wide range of reaction rates.

16-4

Factors That Influence Reaction Rate

Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants.

Four factors can be controlled during the reaction: 1. Concentration - molecules must collide to react; 2. 3. 4. Physical state - molecules must mix to collide; Temperature - molecules must collide with enough energy to react; The use of a catalyst.

16-5

Figure 16.3

The effect of surface area on reaction rate.

16-6

Figure 16.4

Collision energy and reaction rate.

16-7

Expressing the Reaction Ratereaction rate - changes in the concentrations of reactants or products per unit time reactant concentrations decrease while product concentrations increase for

A

B=-

Rate of reaction = -

change in concentration of A change in time

conc A2 - conc A1t2 - t1

-

(conc A) t

16-8

Table 16.1 Concentration of O3 at Various Times in its Reaction with C2H4 at 303 KC2H4(g) + O3(g) C2H4 O(g) + O2(g)

Time (s)

Concentration of O3 (mol/L) 3.20x10-5 2.42x10-5 1.95x10-5 1.63x10-5 1.40x10-5 1.23x10-5 1.10x10-5

-

(conc A) t

0.010.0 20.0 30.0 40.0 50.0 60.0

16-9

Figure 16.5

The concentration of O3 vs. time during its reaction with C2H4.C2H4(g) + O3(g) C2H4 O(g) + O2(g)

rate = [C2H4] t

=

-

[O3] t

16-10

Figure 16.6

Plots of [C2H4] and [O2] vs. time.

Tools of the Laboratory

16-11

In general, for the reaction aA + bB cC + dD

rate =

-

1 a

[A]t

= -

1 b

[B]t

= +

1 c

[C]t

= +

1 d

[D]t

The numerical value of the rate depends upon the substance that serves as the reference. The rest is relative to the balanced chemical equation.

16-12

Sample Problem 16.1 PROBLEM:

Expressing Rate in Terms of Changes in Concentration with Time

Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by earthbound engines in the near future. 2H2(g) + O2(g) 2H2O(g)

(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.

(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H 2O] increasing? PLAN: Choose [O2] as a point of reference since its coefficient is 1. For every molecule of O2 which disappears, 2 molecules of H2 disappear and 2 molecules of H2O appear, so [O2] is disappearing at half the rate of change of H2 and H2O.SOLUTION:

(a) (b) [O2] t

rate = -

1 2

[H2] t

=-

[O2] t t

1 =+ 2

[H2O] t = 0.46 mol/L*s

= - 0.23 mol/L*s = + 1 2

[H2O]

[H2O]t

16-13

Sample Problem 16.2 Determining Reaction Order from Rate Laws PROBLEM: For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. 2NO2(g); rate = k[NO]2[O2] CH4(g) + CO(g); rate = k[CH3CHO]3/2

(a) 2NO(g) + O2(g) (b) CH3CHO(g)

(c) H2O2(aq) + 3I-(aq) + 2H+(aq)

I3-(aq) + 2H2O(l); rate = k[H2O2][I-]

PLAN: Look at the rate law and not the coefficients of the chemical reaction. SOLUTION:

(a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall.(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall. (c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+, while being 2nd order overall.

16-14

Table 16.2 Initial Rates for a Series of Experiments in the Reaction Between O2 and NO2NO(g) + O2(g) 2NO2(g)

Initial Reactant Concentrations (mol/L)

Experiment

O2

NO

Initial Rate (mol/L*s) 3.21x10-36.40x10-3 12.8x10-3

12 3

1.10x10-22.20x10-2 1.10x10-2

1.30x10-21.30x10-2 2.60x10-2

45

3.30x10-21.10x10-2

1.30x10-23.90x10-2

9.60x10-328.8x10-3

16-15

Determining Reaction OrdersUsing initial rates Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n

Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. rate 2 k [O2]2m[NO]2n [O2]2m [O2]1mm

=rate 1 k [O2]1m[NO]1n = 3.21x10-3 mol/L*s

=

=

[O2]2

m

[O2]1

6.40x10-3 mol/L*s

2.20x10-2 mol/L1.10x10-2 mol/L

;

2 = 2m

m=1

Do a similar calculation for the other reactant(s).

16-16

Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these isNO2(g) + CO(g) NO(g) + CO2(g)

rate = k[NO2]m[CO]n

Use the following data to determine the individual and overall reaction orders.Experiment Initial Rate (mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 0.0050 0.080 0.0050 0.10 0.40 0.10 0.10

1 23

0.100.20

PLAN:

Solve for each reactant using the general rate law using the method described previously.rate = k [NO2]m[CO]n

SOLUTION:

First, choose two experiments in which [CO] remains constant and the [NO2] varies.

16-17

Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data

rate 2

rate 10.080 0.0050

=

k [NO2]m2[CO]n2

k [NO20.40

]mm

1

[CO]n

=

[NO2] 2

m

1

[NO2] 1

The reaction is 2nd order in NO2.

=

;

16 = 4m and m = 2 [CO] 3[CO] 1 1 = 2n and n = 0n

0.10

rate 3rate 1 0.0050 =

k [NO2]m3[CO]n3k [NO2]m1 [CO]n1 0.20n

=

The reaction is zero order in CO.

0.0050

=

;

0.10rate = k [NO2]2[CO]0 = k [NO2]2

16-18

Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders

Overall Reaction Order 0

Units of k (t in seconds) mol/L*s (or mol L-1 s-1)

1

1/s (or s-1)

2

L/mol*s (or L mol -1 s-1)

3

L2 / mol2 *s (or L2 mol-2 s-1)

16-19

Integrated Rate Laws[A]

rate = -

t

= k [A]

first order rate equation[A]0 [A]t

ln[A]rate = t = k [A]2

= kt

ln [A]0 = kt + ln [A]t

second order rate equation

1 [A]t[A] rate = t = k [A]0

-

1 [A]0

= kt

1 [A]t

= kt

+

1 [A]0

zero order rate equation

[A]t - [A]0 = - kt

16-20

Sample Problem 16.5 Determining the Reactant Concentration at a Given Time PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s -1, to two molecules of ethylene (C2H4). (a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? PLAN: Find the [C4H8] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration.

SOLUTION:

ln(a)

[C4H8]0[C4H8]t

= kt ;

ln

2.00[C4H8]

= (87 s-1)(0.010 s)

[C4H8] = 0.83 mol/L(b) [C4H8]0 - [C4H8]t [C4H8]0 2.00 M - 0.87 M = 2.00 M

= 0.58

16-21

Figure 16.7

Integrated rate laws and reaction orders.

1/[A]t = kt + 1/[A]0

ln[A]t = -kt + ln[A]0

[A]t = -kt + [A]0

16-22

Figure 16.8

Graphical determination of the reaction order for the decomposition of N2O5.

16-23

Figure 16.9

A plot of [N2O5] vs. time for three half-lives. for a first-order process t1/2 =

ln 2k

=

0.693 k

16-24

Sample Problem 16.6 Using Molecular Scenes to Determine Half-Life PROBLEM: The 1st order process of compound A (red) converting to compound B (black) is depicted at 0.0 s and 30.0 s:

(c)

(a) Find the half-life, t1/2, of the reaction. (b) Calculate the rate constant, k. (c) Draw a scene that represents the reaction mixture at 2.00 min. PLAN: Number of spheres represents concentration. Half-life is constant for a 1st order reaction and the elapsed time when half the red spheres turn black. SOLUTION: (a) At t = 0, 8 A and 0 B. At t = 30.0 s, 6 A and 2 B. For 4 A and 4 B, t = 60.0 s = t1/2. (b) t1/2 = 0.693/k; k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10-2 s-1 (c) The 2.00 min is 120. s or two half-lives represented by the above scene.

16-25

Sample Problem 16.7 Determining the Half-Life of a First-Order Reaction PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its 60o bond angles allow poor orbital overlap, its bonds are weak. As a res