Transcript 16 Kinetics (AHL) - slider-dpchemistry-11

DP Chemistry
R. Slider
Rate Equation
Recall that the rate of a reaction is a measure
of the change in concentration of a reactant,
R, (or product, P) over time.
Notice the initial rate is
measured from the graph
Units for rate
Δ[𝑅]
𝑅𝑎𝑡𝑒 = −
Δ𝑡
Δ[𝑃]
𝑅𝑎𝑡𝑒 =
Δ𝑡
The rate of disappearance of the reactants is equal to the rate of appearance of the
products for a 1:1 molar ratio. This can be seen graphically above
Measuring reaction rates
H2(g) + I2(g)  2HI(g)
The rate of reaction changes as the
reaction proceeds.
This can be seen by the change in
the gradient of the curves on the
graph.
We can collect information about
rate at different times by measuring
the gradient at different points on
the graph.
Notice in the above graph, the formation of HI is initially
occurring at a faster rate than the disappearance of reactants
as seen by the steeper gradient. Can you explain why?
Reaction Order
When the concentration of a
particular reactant is changed,
the rate of reaction may also
change
This relates to reaction order
which can only be determined
experimentally
Consider the following reaction
for which we have a measured
initial rate for A and B:
A + B  products
We can alter concentrations
and measure the change in rate
Scenario 1: First order
Doubling [A] doubles the rate of
reaction. This means:
Rate α [A]
Or
Rate = k[A]
where k is the rate constant which is
dependent upon temperature and use
of a catalyst.
This reaction is “first order with respect
to A”
Scenario 2: Second order
Doubling [A] increases the rate 4 times.
This means:
Rate α [A]2
Or
Rate = k[A]2
This reaction is “second order with
respect to A”
Reaction Order
When the concentration of a
particular reactant is changed,
the rate of reaction may also
change
This relates to reaction order
which can only be determined
experimentally
Consider the following reaction
for which we have a measured
initial rate:
Scenario 3: Zero order
Doubling [A] does not change the rate
of reaction. This means:
Rate α [A]0
Or
Rate = k[A]0
This reaction is “zero order with
respect to A”
Scenario 4: Overall reaction order
Doubling [A] increases the rate 4 times
and doubling [B] doubles the rate. This
means:
Rate α [A]2[B]
This is the rate
Or
expression for the
Rate = k[A]2[B]
reaction
A + B  products
We can alter concentrations
and measure the change in rate
This reaction is “second order with
respect to A and first order with
respect to B”
Overall reaction order = 2 + 1 = 3
Reaction Order Summary
For a general reaction: R  P,
Rate α [R]m
Order of
Reaction with
respect to
reactant, R
Index value Rate
(value of m) equation
Experimental observation
Zero
0
Rate=k[R]0
No change in rate with a change
in [R]
One
1
Rate=k[R]1
Rate and [R] are directly
proportional – as [R] is doubled or
tripled, so is the rate
Two
2
Rate=k[R]2
Rate is directly proportional to
[R]2 – doubling [R] will increase
the rate fourfold
Rate Constant (k)
What units?
The units of the rate constant is
dependent upon the overall rate
expression.
To determine the units for the
rate constant, simply solve for k
and derive the units.
Source: Chemistry for use with the IB Diploma Program, Derry et. al.
A small value for k is an indication of a slow rate of reaction whereas a large
value is indicative of a fast reaction rate.
Rate expression
So, now we know the rate expression has a general
form that looks something like this:
For the reaction A + B  products the rate expression
looks like:
Order of reaction
with respect to B
Order of reaction
with respect to A
Rate =
k [A]m[B]n
rate in mol dm-3 s-1
rate constant
concentrations in
mol dm-3
Be sure to practise
solving problems
involving this rate
expression to solve for:
•Rate
•Rate constant
•Unknown concentration
Determining the rate expression
Procedure:
1. Measure the initial rate for a series of reactant concentrations
2. Change one of the concentrations keeping the other constant and measure the rate
again
3. Change the concentration of the one previously kept constant, and measure the initial
rate again.
4. Repeat this procedure until all reactants have been changed and enough data is
obtained
You try:
Look at the data to the right and determine the rate
expression and the rate constant for a reaction that has
3 reactants, A, B and C.
Experiment
[A] / mol [B] / mol [C] / mol
dm-3
dm-3
dm-3
Rate/ mol dm-3 s-1
1
0.1
0.1
0.1
6.2 x 10-4
2
0.1
0.2
0.1
1.2 x 10-3
3
0.1
0.1
0.2
6.2 x 10-4
4
0.2
0.1
0.2
2.5 x 10-3
Determining the rate expression
Procedure:
1. Measure the initial rate for a series of reactant concentrations
2. Change one of the concentrations keeping the other constant and measure the rate
again
3. Change the concentration of the one previously kept constant, and measure the initial
rate again.
4. Repeat this procedure until all reactants have been changed and enough data is
obtained
You try:
Look at the data to the right and determine the rate
expression and the rate constant for a reaction that has
3 reactants, A, B and C.
Exp’t 2: Doubles [B], which doubles the rate
Exp’t 3: Doubles [C], which has no effect on the rate
Exp’t 4: Doubles [A], which quadruples the rate
Therefore, A is second order, B is first order and C is
zero order and the rate expression is:
Rate = k[A]2[B]
(note [C]0 = 1 so is not included in the rate expression)
Experiment
[A] / mol [B] / mol [C] / mol
dm-3
dm-3
dm-3
Rate/ mol dm-3 s-1
1
0.1
0.1
0.1
6.2 x 10-4
2
0.1
0.2
0.1
1.2 x 10-3
3
0.1
0.1
0.2
6.2 x 10-4
4
0.2
0.1
0.2
2.5 x 10-3
Solving for k,
6.2 x 10-4 = k [0.1]2[0.1]1
k = 6.2 x 10-1 dm6 mol-2 s-1
Determining reaction order
Experiment
We need to be able to measure the change in
concentration of either a reactant or a product such
as a gas being produced.
Then we can change the concentration of reactants
one by one to determine how they affect the reaction
rate.
Concentration (vol here) vs. Time graph
This graph allows us to calculate instantaneous rate
information. The gradient equals the rate.
Rate vs. Concentration graph (not shown)
Plotting rates vs. changes in concentration allows us
to easily determine the order of the reaction by
analysing their shape.
Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm
Reaction order graphs - zero
Rate vs. Concentration
For a zero order reaction, rate is constant with
changes in concentration
A  products , rate = k
Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm
Concentration vs. Time
Because the rate is constant, [A] will decrease
by the same amount every second and the
gradient is constant and negative (-k)
Also, the time it takes for half of the reactants
to disappear (1/2 life), decreases with reduced
concentration
Reaction order graphs - first
Rate vs. Concentration
For a first order reaction, the rate
increases in proportion to changes in
concentration
A  products , rate = k[A]
Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm
Concentration vs. Time
The rate will decrease every second and
the gradient will become less negative as
[A] decreases
Also, the 1/2 life remains constant with
reduction in concentration
Reaction order graphs - second
Rate vs. Concentration
For a second order reaction, the rate
increases exponentially with increases
in concentration (like y = x2)
A + B  products , if [A] = [B]
rate = k[A]2
Concentration vs. Time
The rate will decrease every second
and the gradient will become less
negative, but more dramatically than
first order reactions
Also, the 1/2 life increases with
reduction in concentration
Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm
Exercise
The reaction between NO (g) and Cl2(g) has been studied at 50 °C, recording the initial rate of
formation of NOCl(g) for the initial concentrations of reactants as shown in the table .
NO(g) + ½Cl2(g)  NOCl(g)
1.
2.
3.
4.
5.
6.
7.
[NO(g)]
(mol dm-3)
[Cl2(g)]
(mol dm-3)
Initial Rate
(mol dm-3 s-1)
0.250
0.250
1.43 x 10-6
0.250
0.500
2.86 x 10-6
0.500
0.500
11.4 x 10-6
The order of reaction with respect to NO(g) is:
The order of reaction with respect to Cl2(g) is:
The overall order of reaction is:
The value of the rate constant, k, at 50 °C is:
mol-2 dm6 s-1
The rate of formation of NOCl when [NO(g)] and [Cl2(g)] are both equal to 0.110 mol dm-3 is:
The rate at which NO is reacting, at the instant when Cl2 is reacting at the rate of 2.21 x 10-7 mol dm-3 s-1 is:
The rate at which NOCl is forming, at the instant when Cl2 is reacting at the rate of 2.21 x 10-7 mol dm-3 s-1 is:
Exercise answers
The reaction between NO (g) and Cl2(g) has been studied at 50 °C, recording the initial rate of
formation of NOCl(g) for the initial concentrations of reactants as shown in the table.
NO(g) + ½Cl2(g)  NOCl(g)
1.
2.
3.
4.
5.
6.
7.
[NO(g)]
(mol dm-3)
[Cl2(g)]
(mol dm-3)
Initial Rate
(mol dm-3 s-1)
0.250
0.250
1.43 x 10-6
0.250
0.500
2.86 x 10-6
0.500
0.500
11.4 x 10-6
The order of reaction with respect to NO(g) is: 2
The order of reaction with respect to Cl2(g) is: 1
The overall order of reaction is: 3
The value of the rate constant, k, at 50 °C is: 9.17 x 10-5 mol-2 dm6 s-1
The rate of formation of NOCl when [NO(g)] and [Cl2(g)] are both equal to 0.110 mol dm-3 is: 1.22 x 10-7 mol dm-3 s-1
The rate at which NO is reacting, at the instant when Cl2 is reacting at the rate of 2.21 x 10-7 mol dm-3 s-1 is: 4.42 x
10-7 mol dm-3 s-1 (half of Cl2 as seen in the reaction ratio)
The rate at which NOCl is forming, at the instant when Cl2 is reacting at the rate of 2.21 x 10-7 mol dm-3 s-1 is: 4.42 x
10-7 mol dm-3 s-1 (same as the disappearance of NO in #6)
Graphing Exercise
Plot this data to determine the order of the reaction:
2NOBr(g)  2NO(g) + Br2 (g)
Reaction Mechanisms
Most reactions occur in more than one step
because it is rare that more than two individual
particles will simultaneously collide and
successfully react.
Reactions often go through
intermediate species, which means a
reaction may go through multiple
steps before reaching the final
products.
These possible multi-step pathways
are known as reaction mechanisms. It
is important to note that there may be
more than one possible mechanism.
Each step in the reaction mechanism is
known as the elementary step or
elementary process.
Source:
http://www.talktalk.co.uk/reference/encyclopaedia/hutchinson/m0030471.html
Molecularity
This is a description of each elementary step, indicating the number of
reacting particles. Quite simply:
Number of
reacting particles
Molecularity
Example
Rate/order
1
Unimolecular
CuCO3  CO2 + CuO
k[CuCO3]/first
2
Bimolecular
NO2 + CO  NO + CO2
k[NO2][CO]/second
3
Termolecular
2CO + O2  2CO2
k[CO]2[O2]/third
Unimolecular and bimolecular are by far the most common. Single step termolecular
reactions are very rare and no examples of higher molecularity are known.
Reaction Mechanism example 1
Consider this reaction:
2 NO(g) + O2 → 2 NO2
This reaction does not occur in a single step,
however, but rather through two steps.
Step 1: 2 NO → N2O2
Step 2: N2O2 + O2 → 2 NO2
Overall,
Step 1: 2 NO → N2O2
Step 2: N2O2 + O2 → 2 NO2
Overall: 2 NO(g) + O2 → 2 NO2
Reaction Mechanism example 2
This is a reaction between 2-bromo-2methylpropane and the hydroxide ions from
sodium hydroxide solution:
This shows a two-step mechanism.
The first step shows electrons being
transferred to the Br forming two ions.
This is slow due to strong bonds between
the
carbon
andadding
bromine.
Notice
that
the two
elementary steps together
The
second
step is likely
to be fast due to
gives
the overall
balanced
the strong attraction between the positive
equation for the reaction
carbon and negative hydroxide.
Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.html
The slowest step is known as the rate
determining step because the rate can
only be as fast as the slowest step.
Reaction Mechanism example 2
This is a reaction between 2-bromo-2methylpropane and the hydroxide ions from
sodium hydroxide solution:
Experimentally:
We find that the overall rate expression is:
Rate = k[(CH3)3CBr]
Notice that the [OH-] has no effect on the
rate. This supports the assumption that
the first step is slow.
If the second step were also slow,
increasing the [OH-] would have an effect
on the rate.
Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.html
The rate determining step must have a rate expression that matches the rate expression of
the overall reaction. The molecularity of this step is equal to the overall order of reaction.
Reaction Mechanism example 3
This is a reaction between bromo-ethane and
the hydroxide ions. Similar species, but
different mechanism.
Experimentally:
We find that the overall rate expression is:
Rate = k[(CH3)3CBr][OH-]
Notice that in this example the [OH-] has
an effect on the rate. This supports the
assumption that the reaction occurs in one
step.
Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.html
The reaction occurs all at once due to the
partial positive charge on the carbon atom
Exercise
Step 1:
2 NO → N2O2
Step 2:
N2O2 + H2 → N2O + H2O
Step 3:
N2O + H2 → N2 + H2O
For this reaction find the following:
the overall balanced equation
any reaction intermediates
Exercise answers
Step 1:
2 NO → N2O2
Step 2:
N2O2 + H2 → N2O + H2O
Step 3:
N2O + H2 → N2 + H2O
For this reaction find the following:
the overall balanced equation
any reaction intermediates
Net Reaction:
2 NO + 2 H2 → N2 + 2 H2O
To identify the reaction intermediates, look for substances that first appear on
the product side of the equation, but then appear in the next step as a reactant. In
this example there are two reaction intermediates - N2O2 and N2O
Exercise
If the reaction 2 NO 2 + F2  2 NO2F follows the mechanism,
Step 1. NO2 + F2  NO2F + F (slow)
Step 2. NO2 + F  NO2F (fast)
Work out the rate expression
What is the order of the overall reaction
Exercise answers
If the reaction 2 NO 2 + F2  2 NO2F follows the mechanism,
Step 1. NO2 + F2  NO2F + F (slow)
Step 2. NO2 + F  NO2F (fast)
Rate = k[NO 2][F2]
..because the rate expression of the slowest step is the same as the
overall rate expression. Therefore the overall order of reaction is 2.
Activation Energy
Recall that the activation
energy, Ea, is the energy
required for particles to react.
We have also discussed that
raising the temperature
increases the number of
particles that are able to react
as seen in the MaxwellBoltzmann distribution curves
We have also stated that the rate constant, k, is affected by
temperature and catalysts. So, how can we quantify this?
The rate constant is proportional to the fraction of particles that have
energies ≥ Ea, represented by the shaded part of the graph.
Activation Energy
Mathematically, the fraction, f,
of particles ≥ Ea
𝑠ℎ𝑎𝑑𝑒𝑑 𝑎𝑟𝑒𝑎
f = 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒
and the fraction is proportional
to the rate constant, k.
fαk
Now, all we need to do is understand this proportionality…
Arrhenius Equation
The fraction of particles ≥ 𝐸𝑎 𝑖𝑠
𝑓=𝑒
−𝐸𝑎
Arrhenius
constant
𝑅𝑇
Where:
R = gas constant 8.314 J K-1 mol-1
T = temp in K
e = 2.718 (base for ln)
Ea = activation energy (must be
in J mol-1 to cancel)
Therefore, since f α k…
Activation
energy
𝑘 = 𝐴𝑒
−𝐸𝑎
𝑅𝑇
rate constant
Mathematical
quantity
(2.718)
Temperature
in Kelvin
Gas constant
(8.314 J K-1
mol-1 )
Note: the Arrhenius constant, A, is based on the probability of correct
molecular orientation and frequency of collisions.
It is virtually constant over a wide temperature range for a particular
reaction. It is also called the frequency factor
Arrhenius Equation
The negative sign on the
exponent means that:
Activation
energy
Arrhenius
constant
• As Ea increases, k decreases
• As T increases, k increases
𝑘 = 𝐴𝑒
So…
−𝐸𝑎
𝑅𝑇
rate constant
The rate constant will increase with increasing
temperature and decrease with decreasing
temperature
Mathematical
quantity
(2.303)
Temperature
in Kelvin
Gas constant
(8.314 J K-1
mol-1 )
Arrhenius Equation – integrated form
The Arrhenius Equation is also
sometimes written in the so-called
integrated form.
This is just a rearrangement of the
equation to help us to:
• solve for temperature
• determine Ea using a graph
𝐸𝑎
𝑙𝑛𝑘 = 𝑙𝑛𝐴 −
𝑅𝑇
Where, ln is the natural logarithm (your scientific
calculator will have a button)
Also, no need to memorize either of these equations.
They are found in your Chemistry Data Booklet.
Exercise
What happens to the fraction of particles that are able to react when we
raise the temperature from 20 0C to 300C, assuming the activation energy is
50 kJ mol-1?
Exercise
What happens to the fraction of particles that are able to react when we
raise the temperature from 20 0C to 300C, assuming the activation energy is
50 kJ mol-1?
𝒆
1.
2.
𝑒
−50000𝐽 𝑚𝑜𝑙
𝑒
−𝑬𝒂
−
1
−50000𝐽 𝑚𝑜𝑙
(8.314)(293𝐾)
𝑹𝑻
= 1.21 x 10-9
−
1
(8.314)(303𝐾)
= 2.38 x 10-9
This shows that an increase in 100C has effectively
doubled the fraction of particles that can react.
Determining Ea using a graph
We can use the integrated form of
the Arrhenius Equation to
determine the activation energy.
Recall,
𝐸𝑎
𝑙𝑛𝑘 = 𝑙𝑛𝐴 −
𝑅𝑇
This can be rewritten in the form
of an equation for a straight line,
y = mx + b
𝐸𝑎
𝑙𝑛𝑘 = −
𝑅
1
+ 𝑙𝑛𝐴
𝑇
Where:
y = ln k, y intercept = ln A
1
x=𝑇
gradient = −
𝐸𝑎
𝑅
Determining Ea using a graph
So, a graph of ln k vs. 1/T will give a
straight line that will allow us to
determine the value of Ea for a
reaction from the gradient.
Now, practice some of these problems…