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16 Kinetics (AHL). DP Chemistry R. Slider. Rate Equation. Notice the initial rate is measured from the graph. Recall that the rate of a reaction is a measure of the change in concentration of a reactant, R, (or product, P) over time. Units for rate.

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16 Kinetics (AHL)

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16 kinetics ahl

16 Kinetics (AHL)

DP Chemistry

R. Slider


Rate equation

Rate Equation

Notice the initial rate is measured from the graph

Recall that the rate of a reaction is a measure of the change in concentration of a reactant, R, (or product, P) over time.

Units for rate

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

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

Reaction Order

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”

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

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”

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


Reaction order1

Reaction Order

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”

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

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]

Or

Rate = k[A]2[B]

This reaction is “second order with respect to A and first order with respect to B”

Overall reaction order = 2 + 1 = 3

Consider the following reaction for which we have a measured initial rate:

A + B  products

We can alter concentrations and measure the change in rate

This is the rate expression for the reaction


Reaction order summary

Reaction Order Summary

For a general reaction: R  P, Rate α [R]m


Rate constant k

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

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:

Rate = k [A]m[B]n

Order of reaction with respect to B

Order of reaction with respect to A

  • Be sure to practise solving problems involving this rate expression to solve for:

  • Rate

  • Rate constant

  • Unknown concentration

rate in mol dm-3 s-1

rate constant

concentrations in

mol dm-3


Determining the rate expression

Determining the rate expression

Procedure:

Measure the initial rate for a series of reactant concentrations

Change one of the concentrations keeping the other constant and measure the rate again

Change the concentration of the one previously kept constant, and measure the initial rate again.

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.


Determining the rate expression1

Determining the rate expression

Procedure:

Measure the initial rate for a series of reactant concentrations

Change one of the concentrations keeping the other constant and measure the rate again

Change the concentration of the one previously kept constant, and measure the initial rate again.

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)

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

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 vs. Time graph

This graph allows us to calculate instantaneous rate information. The gradient equals the rate.

Rate vs. Concentration graph

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

Reaction order graphs - zero

Rate vs. Concentration

For a zero order reaction, rate is constant with changes in concentration

A  products   ,    rate =  k

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

Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm


Reaction order graphs first

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]

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

Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm


Reaction order graphs second

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

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)

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

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)

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 10-5mol-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

Graphing Exercise

Plot this data to determine the order of the reaction:

2NOBr(g)  2NO(g) + Br2 (g)


Reaction mechanisms

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 intermediatespecies, 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

Molecularity

This is a description of each elementary step, indicating the number of reacting particles. Quite simply:

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

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

Reaction Mechanism example 2

This is a reaction between 2-bromo-2-methylpropane 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 and bromine.

The second step is likely to be fast due to the strong attraction between the positive carbon and negative hydroxide.

The slowest step is known as the rate determining step because the rate can only be as fast as the slowest step.

Notice that adding the two elementary steps together gives the overall balanced equation for the reaction

Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.html


Reaction mechanism example 21

Reaction Mechanism example 2

This is a reaction between 2-bromo-2-methylpropane 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

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.

The reaction occurs all at once due to the partial positive charge on the carbon atom

Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.html


Exercise1

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 answers1

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


Exercise2

Exercise

  • If the reaction 2 NO2 + 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 answers2

Exercise answers

  • If the reaction 2 NO2 + F2 = 2 NO2F follows the mechanism,

  • Step 1. NO2 + F2 = NO2F + F (slow)Step 2. NO2 + F = NO2F (fast)

  • Rate = k[NO2][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

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 Maxwell-Boltzmann 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 energy1

Activation Energy

Mathematically, the fraction, f, of particles Ea

f =

and the fraction is proportional to the rate constant, k.

f α k

Now, all we need to do is understand this proportionality…


Arrhenius equation

Arrhenius Equation

The fraction of particles

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)

Activation energy

Arrhenius constant

rate constant

Temperature in Kelvin

Mathematical quantity (2.303)

Gas constant (8.314 J K-1mol-1 )

Therefore, since f αk…

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 equation1

Arrhenius Equation

  • The negative sign on the exponent means that:

  • As Ea increases, k decreases

  • As T increases, k increases

  • So…

Activation energy

Arrhenius constant

rate constant

Temperature in Kelvin

Mathematical quantity (2.303)

The rate constant will increase with increasing temperature and decrease with decreasing temperature

Gas constant (8.314 J K-1mol-1 )


Arrhenius equation integrated form

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

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.


Exercise3

Exercise

  • What happens to the fraction of particles that are able to react when we raise the temperature from 200C to 300C, assuming the activation energy is 50 kJ mol-1?


Exercise4

Exercise

  • What happens to the fraction of particles that are able to react when we raise the temperature from 200C to 300C, assuming the activation energy is 50 kJ mol-1?

= 1.21 x 10-9

= 2.38 x 10-9

1.

2.

This shows that an increase in 100C has effectively doubled the fraction of particles that can react.


Determining ea using a graph

Determining Ea using a graph

We can use the integrated form of the Arrhenius Equation to determine the activation energy.

Recall,

This can be rewritten in the form of an equation for a straight line, y = mx + b

Where:

y = ln k, y intercept = ln A

x =

gradient =


Determining e a using a graph

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…


17 kinetics ahl

17 Kinetics (AHL)


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