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Chapter 19: Chemical Kinetics, Lecture notes of Chemical Kinetics

Analyze this in terms of the overall reaction order. A first order reaction depends on the concentration of one of the reactants raised to the first power. We ...

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Chapter 19: Chemical Kinetics
Key topics:
Reaction Rates
Factors That Affect Reaction Rates
Reaction Mechanisms
How fast does a reaction occur?
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Red paint from the pond in Van Gogh's Wheat Stack Under a Cloudy Sky has degraded to grey/white over time.

Chapter 19: Chemical Kinetics

Key topics: Reaction Rates Factors That Affect Reaction Rates Reaction Mechanisms How fast does a reaction occur? e.g. , many famous painting are changing color…

(from ibchem.com) Also see page 434. Collision Theory of Chemical Reactions A very simple picture of a chemical reaction between two molecules is that

  1. They must collide with enough energy to react
  2. They must collide in the correct orientation to react The collision (kinetic energy) can be transferred into vibrational energy, which can lead to bond breaking. This is the first step in a chemical reaction. The minimum energy required to initiate a reaction is called the activation energy , Ea. If the collision is < Ea, it is ineffective and the reactants simply bounce off each other. Temperature has a large effect on having energy > Ea.

The transition state (activated complex) is a transient species of high energy that falls apart to generate the products of the reaction. Transition state theory (TST), also known as Activated Complex Theory, is a more sophisticated theory that has two main assumptions

  1. The transition state is in equilibrium with the reactions.
  2. One (special) vibrational motion results in the breakup of the transition state into the products. Measuring Reaction Progress & Expressing Reaction Rate We can express the reaction rate as either: o Rate of disappearance of reactants (decrease or – ve) o Rate of appearance of products (increase or +ve)

Reaction progress with time. The rate changes over time. Reaction progress can be monitored using a spectrometer. We can follow a color change, conductance change, pressure change, or whatever is convenient. Consider the reaction A → B. average rate = [A] t = [B] t

Value of k? k = rate / [Br 2 ]

at t = 50 s, k =

5

M s

1

0. 0101 M

3

s

1

at t = 250 s, k =

5

M s

1

0. 00500 M

3

s

1 (same within experimental error) Consider the reaction A → 2B. rate = [A] t = 1 2 [B] t

In general for the reaction aA^ +^ bB^!^ cC^ +^ dD

rate =

a

[A]

t

b

[B]

t

c

[C]

t

d

[D]

t

e.g. , Write the balanced equation corresponding to the following rate expression rate = 1 2 [N 2 O 5 ] t = 1 2 [N 2 ] t = 1 5 [O 2 ] t Solution:

  • ve sign indicates reactants and +ve sign indicates products. So this reaction should be: 2N^2 O^5!^ 2N^2 + 5O^2 We should double check that the reaction is balanced. e.g. , Consider the following unbalanced equation: A + B → C. When C is being formed at the rate of 0.086 M /s, A is being consumed at a rate of 0.172 M /s, and B at 0.258 M /s. Balance the equation from this information.

Solution: We are told that [C] t = 0. 086 , [A] t = 0. 172 , [B] t = 0. 258 If we write the balanced equation a A + b B → c C, we know that

a

[A]

t

b

[B]

t

c

[C]

t

or

a

b

c

If we divide each expression by 0.086 we get 2 a = 3 b = 1 c Therefore a = 2, b = 3, and c = 1. Dependence of Reaction Rate on Reactant Concentration In the bromine reaction we considered earlier,

Br 2 (aq) + HCOOH(aq)! 2Br

(aq) + 2H

(aq) + CO 2 (g)

we determined the rate = k [Br 2 ]. In terms of the reactant concentrations, we write

rate = k[Br 2 ]

x

+ [HCOOH]

y with x = 1 and y = 0. The exponents (x and y) are empirically determined. Without experimental data we can’t simply predict their values.

In general, for aA^ +^ bB^!^ cC^ +^ dD, the rate law is

rate = k[A] x [B] y . Ø x = 0: the reaction is 0 th order with respect to A. Ø x = 1: the reaction is 1 st order with respect to A. Ø x = 2: the reaction is 2 nd order with respect to A. The overall reaction order is the sum of the exponents, so the bromine reaction is 1 st order overall (1 + 0 = 1).

Then we can determine k from any single exp.

k =

rate

[F 2 ][ClO 2 ]

3

M s

1

[0. 20 M ][0. 010 M ]

= 1. 2 M

1

s

1 The rate has units of M s

  • 1 , so k has units as follows: Note: Ø rate law defined in terms of reactants only Ø exponents are usually positive whole numbers or zero § can be negative (inhibitor) § can be a fraction (won’t see in this course)

e.g. , For the reaction A + B → 2C, the rate = k [A] 2 with k = 1. × 10

  • 2 M - 1 s - 1 . Fill in the missing table entries: Experiment [A] ( M ) [B] ( M ) Initial rate ( M /s) 1 0.013 0.250 (^) 2.20 × 10
  • 6 2 0.026 0. 3 0.500 (^) 2.20 × 10
  • 6 Solution: Blue entry: the rate is the same as exp. 1, so [A] must be the same as exp. 1 since this is what the rate depends on. Green entry: since [A] doubles from exp. 1, the rate must be 2 2 = 4 times the rate of exp. 1 = 8.80 × 10
  • 6 . Dependence of Reaction Concentration on Time How do the reactant (and product) concentrations change with time? Analyze this in terms of the overall reaction order. First Order Reactions A first order reaction depends on the concentration of one of the reactants raised to the first power. We will write such a reaction as aA → products. rate = 1 a [A] t = k[A] ) [A] t = (ak)[A] = k ⇤ [A] where we have redefined the rate constant to include the stoichiometric coefficient. Now we change Δ to d (finite change to infinitesimal change) and just write k for k *

d[A]

dt

= k[A] )

d[A]

[A]

= k dt or

[A]

d[A] = k dt

First order kinetics; we obtain a straight line when the data is plotted on a logarithmic scale. From concentration vs. time data ( e.g. Lab 17 in Chem 1112) we would do the following:

ln[A]t

| {z }

y

= k

|{z}

m

t

|{z}

x

+ ln[A] 0

| {z }

b

Half-life: The half-life of a reaction (t1/2) is the time required for the reactant concentration to drop to half of its original value. We express this as

[A]t = [A]t

1 / 2

[A] 0

[A]t

1 / 2

[A] 0

Then the integrated rate law gives

ln

[A]t

1 / 2

[A] 0

= ln

= kt 1 / 2

First order kinetics; the concentration is halved every half-life. (from chemwiki.ucdavis.edu) ln 1 2 = ln 2 ⇡ 0 .693 so that t 1 / 2 =

  1. 693 k e.g. , Calculate the rate constant for the first order decay of 24 Na (t1/2 = 14.96 hours). Solution: 14 .96 h =
  2. 693 k ) k =
  3. 693 14 .96 h = 4. 63 ⇥ 10 2 h 1 ✓ ⇥ 24 h 1 day = 1.11 day 1 ◆

Second order kinetics: a plot of 1/[A] vs. time has slope k and intercept 1/[A] 0 To compute the half-life, we use

[A]t = [A]t 1 / 2 = [A] 0 / 2

[A] 0 / 2

= kt 1 / 2 +

[A] 0

) t 1 / 2 =

k[A] 0

e.g. , The reaction 2A → B is second order in A with k = 32 M

  • 1 s
  • 1 . (a) With [A] 0 = 0.0075 M , how long does it take for [A] to drop to 0.0018 M? Calculate the half-life of the reaction for [A] 0 = 0.0075 M and for [A] 0 = 0.0025 M.

Solution: 1 [A]t = kt +

[A] 0

) kt =

[A]t

[A] 0

) t =

k[A]t

k[A] 0 t = 1 (32 M ^1 s^1 )(0. 0018 M ) 1 (32 M ^1 s^1 )(0. 0075 M ) = 17.36 s 4 .17 s =13.2 s. For the half-life, t 1 / 2 = 1 (32 M ^1 s^1 )(0. 0075 M ) = 4.17 s t 1 / 2 = 1 (32 M ^1 s^1 )(0. 0025 M ) = 12.5 s Zeroth Order Reactions A zeroth order reaction is independent of reactant concentration: For aA → products, rate = k [A] 0 = k. Once again, we redefine the rate constant to include the stoichiometric coefficient and write

[A]

t

= k

Now we change Δ to d (finite change to infinitesimal change):

d[A]

dt

= k ) d[A] = k dt

Then we integrate from t^ = 0 to^ t^ =^ t

Z (^) [A] t [A] 0 d[A] = k Z (^) t 0 dt ) [A]t [A] 0 = kt or [A]t = kt + [A] 0

We can obtain the activation energy and the frequency factor from a plot of ln k versus 1/T. Dependence of Reaction Rate on Temperature Reactions almost always occur faster at higher temperature. Quantified by the Arrhenius Equation for the rate constant k = Ae

Ea/RT

A = frequency factor Ea = activation energy R = gas constant = 8.314 J K

  • 1 mol - 1 T = temperature (in K) Increase Ea : decrease k Increase T : increase k Take the logarithm of the Arrhenius Equation: ln k = ln A Ea RT (used ln xy = ln x + ln y)

Now, consider two different temperatures:

ln k 1 = ln A

Ea

RT 1

(1) and ln k 2 = ln A

Ea

RT 2

Subtract (1) – (2) to obtain ln k 1 k 2 = Ea R ✓ 1 T 2 1 T 1 ◆ e.g. , Determine Ea, and also k at 655 K, from the data: T (K) k (s

  • 1 ) 625 1.1 x 10
  • 4 635 1.5 x 10
  • 4 645 2.0 x 10
  • 4 Solution: From the first two data points, ln
  1. 1
  2. 5 = Ea R ✓ 1 635 K 1 625 K ◆ ) 0 .31 = (Ea/R)( 2. 52 ⇥ 10 5 K 1 ) Ea = ( 0 .31)(8.314 J K 1 mol 1 )/( 2. 52 ⇥ 10 5 K 1 ) = 102.3 kJ/mol Then, using T = 645 K and T = 655 K, we have ln

k

3 J mol 1 8 .314 J K^1 mol 1

655 K
645 K

) ln 2. 0 ⇥ 10 4 ln k = 0. 291 ) ln k = 8. 226 ) k = e 8. 226 = 2. 7 ⇥ 10 4