KINETICS - Grand Junction High School

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KINETICS
What is Kinetics?

Kinetics is the study of the speed of reactions and is largely an
experimental science. Some general qualitative ideas about reaction
speed may be developed, but accurate quantitative relationships require
that experimental data be collected.

For a chemical reaction to occur, there must be a collision between
reactant particles. That collision is necessary to transfer kinetic energy
needed to break reactant chemical bonds and reform product bonds. If
the collision does not provide enough energy, no reaction will occur. The
collision must also take place with the proper orientation at the correct
place on the molecule, called the reactive site.
Five factors that affect rates of
chemical reactions:

1. Nature of the reactants: Large, complex molecules tend to react more slowly than
smaller ones because statistically there is a greater chance of collisions occurring
somewhere else on the molecule, rather than at the reactive site.

2. An increase in temperature increases the kinetic energy, providing more energy
transfer and speeding up the number of collisions.

3. The higher the concentration of reactants, the greater the chance of collision and the
greater the reaction rate. In gases, the pressure is directly related to the concentration;
the greater the pressure, the greater the rate of collision.

4. Reactants in the same physical state react more readily with one another. Gases and
liquids tend to react faster than solids because of increased surface area.

5. Catalysts are substances that speed up the reaction rate without affecting the
reaction. Catalysts work by reducing the activation energy needed to start a reaction.
Orientation
Click
Types of Rate Laws

Differential rate law shows how the rate is dependent on the concentration.

Integrated rate law shows how the concentrations of species in the reaction
depend on time.

Our rate laws will involve only concentrations of reactants.

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
Rates of Reaction
(3:28)

The rate (speed) of a reaction is related to the change in concentration of
either a reactant or product over time.

For the reaction: 2A + B  C + 3D

As the reaction proceeds, the concentrations of A and B will decrease and the
concentrations of C and D will increase. Thus, the rate can be expressed as:
1 Δ[A]
Rate = - 2
Δt = -
Δ[B]
Δt
Δ[C]
=
Δt
=
1
Δ[D]
3
Δt
The negative signs for the reactants indicate their concentrations are decreasing with
time. The brackets indicate concentration (molarity).
Since the rate of reaction decreases during the course of the reaction, the above
calculation will indicate the average rate of reaction over a given time frame, or more
commonly, as the initial reaction rate—the rate of reaction at the instant the reactants
are mixed.
Instantaneous rate of reaction: N2O5  2NO2 + ½ O2

The instantaneous rate of reaction is the
rate at any point in time and is
represented by the instantaneous slope
of the curve at that point.

You can determine the instantaneous
rate by calculating the slope of the
tangent to the curve at the point of
interest.

I.R.= - Δ[N2O5] = .0019 M = 1.9 x 10-5
ΔT

100 s
I.R.= + 1 Δ[NO2 ] = .0037 M = 1.9 x 10-5
2 ΔT
100 s
Rate equation (law)

Rate of reaction may depend on reactant concentration, product concentration (rare)
and temperature.

For the general reaction, aA + bB  cC + dD where the lower case letters are the
coefficient and the upper-case letters are the reactants, the rate law is written: Rate = k
[A]m [B]n …

k is the rate constant—a constant for each chemical reaction at a given temperature.

The exponents m and n are called the orders of reaction. They indicate what effect a
change in concentration of that reactant species will have on the reaction rate.

Example: if m=1 and n=2, that means that if the concentration of reactant A is doubled,
then the rate will also double ( [2]1 = 2), and if the concentration of reactant B is doubled,
then the rate will increase fourfold ( [2]2 = 4). We say that it is first order with respect to A
and second order with respect to B. If the concentration of a reactant is doubled and
that has no effect on the rate of reaction, then the reaction is zero order with respect to
that reactant ([2]0 = 1).

Many times the overall order of reaction is calculated; it is simply the sum of the individual
coefficients, third order in this example.

Rate = k[A][B]2 where the exponent 1 is understood
Reaction Rates

The reaction rate of a chemical reaction is defined as the change in
concentration of a reactant or product per unit of time:
Concentration of A at time t2 – concentration of A at time t1

Rate =
t2 – t1
Δ[A]

Therefore, Rate =
Δt

It is important to realize that the rate law (the rate, rate constant, and
orders of reaction) is determined experimentally. Do not use the balanced
chemical equation to determine the rate law.

Ways to determine rate of reaction can be measured in many ways
including taking the slope of the concentration versus time plot for the
reaction.

Once the orders of reaction have been determined, it is easy to calculate
the rate constant (k).
2NO(g) + O2(g)  2NO2(g)
Experiment
Initial
NO
Initial
O2
Rate of
NO2
Formation
M/s
1
0.01
0.01
0.05
2
0.02
0.01
0.20
3
0.01
0.02
0.10

If the numbers are simple, you can
reason out the orders of reaction. From
exp. 1 to exp. 2, [NO] is doubled and
[O2] stays the same. The rate increased
fourfold. This means the reaction is
second order with respect to NO. In exp.
1 and 3, the [O2] was doubled, [NO]
remained the same and the rate
doubled. Therefore, the reaction is first
order with respect to O2 and the rate
equation can be written as: Rate =
k[NO]2[O2]

The rate constant can be determined by
substituting the values of the
concentrations of NO and O2 from any
of the experiments into the rate
equation above and solve for k.
Types of Reactions

Zero-order Reaction: rate = k[A]0

The concentration of the reactant decreases linearly with time and the slope of
the line is constant indicating constant rate.

Sublimation is an example of a zero rate reaction. As one layer of particles
sublimes, the next layer is exposed and the number of particles available to
sublime remains constant.
First order reaction

The rate of the reaction is directly proportional to the concentration of the
reactant.

Rate = k[A]1

The rate slows down as the reaction proceeds because the concentration
of the reactants decreases.
The rate of the reaction
is proportional to the square of the
Second-order
Reaction

concentration of the reactant.

Rate = k[A]2

For a second-order reaction, the rate is even more sensitive to the
reactant concentration. That’s why the curve (rate) flattens out
more quickly than it does for a first-order reaction.
Determining the Order of a Reaction

The order can ONLY be determined by experimentation so you will need
data!

Rate = k[A]m

Therefore: k = rate
[A]m
[A] (M)
Initial rate (M/s)
0.10
0.015
0.20
0.030
0.40
0.060
[A] (M)
Initial rate (M/s)
0.10
0.15
0.20
0.15
0.40
0.15
[A] (M)
Initial rate (M/s)
0.10
0.015
0.20
0.060
0.40
0.0240
What if the numbers aren’t obvious?

If you are unsure about how the initial rate is changing with the initial
reactant concentration, or if the numbers are not as obvious as the ones in
the preceding examples, you can substitute any two initial concentrations
and the corresponding initial rates into a ratio of the rate laws to
determine order.

rate 2 = k[A]n

rate 1
k[A]n

The reaction AB has been experimentally determined to be firs order.
The initial rate is 0.0100M/s at an initial concentration of A of 0.100M. What
is the initial rate of [A] = 0.500M?

rate 2

0.0100M/s

= .0500M/s
= 0.500M
0.100M
Reaction order for multiple reactants

aA + bB  cC + dD

rate = k[A]m [B]n

The overall order is the sum of m + n.

For example, the reaction between hydrogen and iodine has been
experimentally determined to be first order with respect to hydrogen and
first order for iodine.

H2 + I2  2HI

Thus it is second order overall.
rate = k[H2]1[I2]1
NO2(g)+ CO(g)  NO(g) + CO2(g)

From the data determine:

The rate law for the reaction

The rate constant (k) for the reaction.
[NO2] (M)
Experiment
[CO] (M)
Initial rate
(M/s)
1
0.10
0.10
0.0021
2
0.20
0.10
0.0082
3
0.20
0.20
0.0083
4
0.40
0.10
0.033

Between the first two experiments the NO2 concentration doubles
and the initial rate quadruples suggesting NO2 is second-order.

Between 2nd and 3rd experiments, the NO2 stays constant, the CO
doubles yet the initial rate remains constant which means CO is zeroorder.

Overall rate = 2 + 0 = 2

Rate = k[NO2]2 [CO]0 = k[NO2]2

k = rate
[NO2]2
k = 0.0021 M/s
(0.10 M)2
k = 0.21
M· s
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)

From the data, determine:

The rate law for the reaction

The rate
constant (k)
for the
reaction[Cl ] (M)
Experiment
[CHCl
3] (M)
2
Initial rate (M/s)
1
0.010
0.010
0.0035
2
0.020
0.010
0.0069
3
0.020
0.020
0.0138
4
0.040
0.040
0.027
Algebraically

rate2 = k[A]m[B]n

rate1

1.38 x 10-2 = k[0.020][0.020]n

6.9 x 10-3

2= 2n

log 2= n · log 2

n = log 2
k[A]m[B]n
** log xn = n log x**
log 2
n= 1
k[0.020][0.010]n

Solve for k:

Rate = k[A]m[B]n

k = rate
[A]m[B]n
k = .0035 M/s
[.010M][.010M]
k = 35 M -1 s-1
The Integrated Rate Law: The
Dependence of Concentration on Time

The integrated rate law for a chemical reaction is a relationship between the concentrations of
the reactants and time.

First order Integrated Rate Law is where the rate is directly proportional to the concentration of
A: rate = k[A]

-Δ[A]
ΔT
= k[A]
ln[A]t = -kt + ln[A]0 or ln [A]t = - kt
[A]0
Where [A]t is the concentration of A at any time t, k is the rate constant, and [A]0 is the initial
concentration of A.
Notice that the integrated rate law has the form of an equation for a straight line: y = mx + b

For a first-order reaction, a plot of the natural log of the reactant
concentration as a function of time yields a straight line with a slope of –k
and a y-intercept of ln[A]0. The slope may be negative but the rate
constant (k) is always positive.
SO2Cl2(g)  SO2(g) + Cl2(g)
The concentration of SO2Cl2 was monitored at a fixed
temperature as a function of time during the
decomposition reaction, and the following data was
tabulated.
Show that the reaction is first order, and determine the
rate constant for the reaction.
Time (s)
[SO2Cl2] (M)
Time (s)
[SO2Cl2] (M)
0
.100
800
.0793
100
.0971
900
.0770
200
.0944
1000
.0748
300
.0917
1100
.0727
400
.0890
1200
.0706
500
.0865
1300
.0686
600
.0840
1400
.0666
700
.0816
1500
.0647

If the plot is linear, then it is a first-order reaction.

To obtain the rate constant, fit the data to a line. The slope of the line will
be equal to –k.

Since the slope of the best fitting line is -2.90 x 10-4s-1, the rate constant is
+2.90 x 10-4s-1.

Now use your graph to predict the concentration at 1900 seconds.
Practice
SO2Cl2(g)  SO2(g) + Cl2(g)

Using the rate constant from the last problem, (+2.90 x 10-4 s-1) if the firstorder reaction is carried out at the same temperature, and the initial
concentration of SO2Cl2 is 0.0225 M, what will the SO2Cl2 concentration be
after 865 s?

Given: k = +2.90 x 10-4 s-1

[SO2Cl2]0 = 0.0225 M

Find: [SO2Cl2] at t = 865 s

Use first –order integrated rate law.

Equation: ln[A] = -kt + ln[A]0

Solution:

ln [SO2Cl2]t = -kt + ln[SO2Cl2]0

ln[SO2Cl2]t = -(2.90 x 10-4 s-1)865s + ln(0.0225)

ln[SO2Cl2]t = -0.251 - 3.79

[SO2Cl2]t = e-4.04

= 0.0175M
Second-Order Integrated Rate Law

Rate = k[A]2

Since rate = -Δ[A] = k[A]2
Δt
The above can be integrated to obtain:
1/[A]t = kt + 1/[A]0
The second-order integrated rate law is also in the form of an equation for a
straight line: y = mx + b
NO2(g)  NO(g) + O(g)

Time (s)
Time (s) as a[NO
The concentration
of NO2 is at[NO
a fixed
function
2] (M)temperature
2] (M) of time.
Show by graphical
analysis that
the reaction
and that it is
0
0.01000
500 is not first order
0.00440
second order. Determine the rate constant for the reaction.
50
0.00887
550
0.00416
100
0.00797
600
0.00395
150
0.00723
650
0.00376
200
0.00662
700
0.00359
250
0.00611
750
0.00343
300
0.00567
800
0.00329
350
0.00528
850
0.00316
400
0.00495
900
0.00303
450
0.00466
950
0.00292

The plot should NOT be linear so it is not a first order.

Next prepare a graph of 1/[NO2] versus time.

0.05 M/s = k(0.01 M)2(0.01 M)

k = 0.05 M/s ÷ (0.01 M)2(0.01 M)

k = 5 x 104/M2 s

**In choosing experiments to compare, choose two in which the
concentration of only one reactant has changed while the other has
remained constant.

Go to page 8 in your packet.
Keywords and Equations
ln[A]t – ln[A]0 = -kt (first order)
1
-
[A]t
1

T = time (seconds)

Ea = activation energy

K = rate constant

A = frequency factor

Gas constant, R = 8.314 J/ mol K
[A]0 = kt (second order)
ln 2 = 0.693
t1/2 =
k
k
-E a
ln k =
RT + ln A

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