Chapter 13.1-13.3. Earlier, rate was shown to be a change in concentration compared to a change in time. This is also know as a. slope

Rate LawsChapter 13.1-13.3

Earlier, rate was shown to be a change in concentration compared to a change in time.

This is also know as a .

Rate

slope

Rate, the speed of a reaction, can be expressed as a function of concentration

The term “function” meaning, of course, an equation…

The Rate Law

Rate = k [Reactant 1]x [Reactant 2]y [Reactant 3]z

Rate Constant

Concentrationsof

Reactants

Orders

Brackets = Molarity

k is the rate constant, a function of the particular reaction ◦ depends on strength of bonds which must be

broken (Activation Energy)◦ depends on presence of catalyst◦ depends on temperature

The Rate Law

[Reactants] is the concentration of one of the reactants◦ notice [Products] do not effect the rate ◦ some reactants may not effect rate, and so they

do not show up in the rate law (these are said to have an “order of zero”)

The Rate Law

the order of a reactant is the exponent x, which is called the “order with respect to that Reactant” ◦ the higher the order wrt R1, the greater the

influence of [R1] on the rate ◦ a reactant which does not effect rate is said to be

“0th order” ◦ orders do not come from the balanced equation;

they must be determined experimentally by studying the reaction mechanism

◦ orders can be fractional or even negative

The Rate Law

example Hydrogen peroxide oxidizes iodide ion in acidic solution:

H2O2(aq) + 3 I–(aq) + 2 H+(aq) I3–(aq) + 2 H2O(l)

The rate law for this reaction isRate = k [H2O2] [I–]

What is the order of the reaction with respect to each reactant species?

What is the overall order?

The Rate Law

1st order WRT H2O2

1st order WRT I–

1 + 1 = 22nd order overall

Notice that the rate law covers all of the factors which effect rate. ◦ concentration◦ temperature◦ pressure/volume of gases◦ catalyst◦ surface area of reactants

The Rate Law

So if the rate law can not be determined by the coefficients from the balanced equation, how can the orders be determined?

Method of Initial Rates Or

Graphical Analysis of Data

The Rate Law

In the Method of Initial Rates, a reaction is run several times as the concentrations of the reactants are changed and the initial rate is measured each time

to solve for the orders…1. make sure you have at least two trials for which

only [x] has changed2. comparing the initial rates of these two trials will

yield the order wrt x3. repeat for each of the other reactants

Determining the Rate Law using Method of Initial Rates


Practice2 NO+2 H2 N2+2 H2O

Rate = k[NO]x[H2]y

to solve for the orders…1. make sure you have at

least two trials for which only [NO] has changed

Exp.

[NO]o [H2]o Rate (mol/L•sec)

1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11



Rate = k[NO]x[H2]y

After picking the experiments, plug the information into the reaction’s generic rate law as a ratio:

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

12.0x10 11

108.0x10 11

k 0.020 x0.010 y

k 0.060 x0.010 y


The order WRT [NO] = 2


Rate = k[NO]x[H2]y

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

12.0x10 11

108.0x10 11

k 0.020 x0.010 y

k 0.060 x0.010 y

1

9

1

3

x

2 x



Rate = k[NO]2[H2]y

to solve for the orders…1. make sure you have at

least two trials for which only [H2] has changed

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11



Rate = k[NO]2[H2]y

After picking the experiments, plug the information into the reaction’s generic rate law as a ratio:

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

3.0x10 11

9.0x10 11

k 0.010 20.010 y

k 0.010 20.030 y


The order WRT [H2] = 1


Rate = k[NO]2[H2]y

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

3.0x10 11

9.0x10 11

k 0.010 20.010 y

k 0.010 20.030 y

1

3

1

3

x

1 x



Rate = k[NO]2[H2]1

The rate law so far looks like…

Rate = k[NO]2[H2]1

The Overall Order is…2 + 1 = 3Ex

p.[NO]o [H2]o Rate

(mol/L•sec)1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11



Rate = k[NO]2[H2]1

to solve for the value of k…1. pick any trial, plug in

the initial [ ]’s and the orders, and solve for k

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11


What about units???


Rate = k[NO]2[H2]1

to solve for the value of k…1. pick any trial, plug in

the initial [ ]’s and the orders, and solve for k

12.0x10–11 = k[0.020]2[0.010]1

k = 3.0x10–5

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11



Rate = k[NO]2[H2]1

to solve for the units of k…

12.0x10–11 = k[0.020]2[0.010]1

k = 3.0x10–5 1/M2•s

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

M

sk M 2

M 1

M

M 2M 1

sk

1

M2sk



Rate = k[NO]2[H2]1

Final Rate Law…

Rate=(3.0x10–5 1/M2•s)[NO]2[H2]1

Exp.


1 0.010 0.010 3.0 x 10–11

2 0.010 0.030 9.0 x 10–11

3 0.020 0.010 12.0 x 10–11

4 0.010 0.050 15.0 x 10–11

5 0.060 0.010 108.0 x 10–11

Example2 Fe2+ + Cl2 2 Fe3+ + 2 Cl–


Exp [Fe2+] [Cl2] [H+] Rate (M/sec)

1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4

What is the reaction order with respect to Fe2+, Cl2, and H+?

What are the rate law and the relative rate constant?




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4


What are the rate law and the relative rate constant? Rate=k[Fe2+]x[Cl2]y[H+]z




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the order of Fe2+?




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the order of Fe2+?

0.0040

0.0020

x

2.0x10 5

1.0x10 5




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4


What are the rate law and the relative rate constant? Rate=k[Fe2+]1[Cl2]y[H+]z




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the order of Cl2?





How do you find the order of Cl2?Exp [Fe2+] [Cl2] [H+] Rate

(M/sec)1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4

0.0040

0.0020

y

2.0x10 5

1.0x10 5




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4


What are the rate law and the relative rate constant? Rate=k[Fe2+]1[Cl2]1[H+]z




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the order of H+?




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the order of H+?

0.5

1.0

z

2.0x10 5

1.0x10 5




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4


What are the rate law and the relative rate constant? Rate=k[Fe2+]1[Cl2]1[H+]–1




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the value for k?

1.0x10–5=k[.002]1[.002]1[1]–1

2.5=k




1 0.0020 0.0020 1.0 1.0x10–5

2 0.0040 0.0020 1.0 2.0x10–5

3 0.0020 0.0040 1.0 2.0x10–5

4 0.0040 0.0040 1.0 4.0x10–5

5 0.0020 0.0020 0.5 2.0x10–5

6 0.0020 0.0020 0.1 1.0x10–4



How do you find the unit for k?

M/s=k[M]1[M]1[M]–1

M1M–1M–1M1s–1=k

s–1=k

The second way, also experimental, to determine the rate law is based on the integration of the basic rate law.

Let us begin by introducing the integrated rate laws… ◦ Let us assume that…

the Reactant of interest is A the reaction which contains A has been run, allowing

the concentration of A to be measured as time changes

Determining the Rate Law using Graphical Analysis of Data

The Integrated Rate laws Let’s begin with the

simplest… Zeroth-Order Rate

Law:◦ Rate = k [A]0

◦ Rate = k The instantaneous

rate law:

By combining these two rates and rearranging them…

This results in the integrated rate law:◦ [A]t – [A]o = – kt

The Integrated Rate laws Let’s rearrange this

integrated rate law into something more useful… such as the form…

y = m x + b[A]t = – kt + [A]o

[A]

t

[A]o

Slope = –k

The Integrated Rate laws First-Order Rate Law:

◦ Rate = k [A]1

The instantaneous rate law:

By rearranging these…

This results in the integrated rate law:◦ ln [A]t – ln [A]o = – kt



y = m x + bln [A]t = – kt + ln [A]o

ln[A]

t

ln[A]o

Slope = –k

The Integrated Rate laws Second-Order Rate

Law:◦ Rate = k [A]2

The instantaneous rate law:

By rearranging these…

This results in the integrated rate law:1

A t

1

A o

kt



y = m x + b

t

Slope = +k

1

A t

kt 1

A o

1

A

1

A o

Documents

Chapter 13.1-13.3. Earlier, rate was shown to be a change in concentration compared to a change in time. This is also know as a. slope