The following notes were written for the previous IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

Kinetics (hl)

16.1 - The Rate Expression

Rate = k[A]m[B]n

where: k is a constant, and m, n show the orders of the reaction with respect to each reactant.

It is an experimentally determined equation in that the information (n,m,k etc) can only be found through experimentation and not through theoretical considerations. The rate equation shows the relationship between the speed of a reaction and the concentrations of the individual reactants. Once the orders are found then they provide information regarding the mechanism of the specific reaction.

The overall order of reaction is the total of m+n above.

k is the rate constant. This gives a measure of how fast the reaction proceeds. External factors such as temperature, pressure, particle size and catalysts affect the value of the rate constant.

Graphing results

If a graph of reaction concentration against time is plotted a curve is obtained as the reactant is used up in the course of the reaction. The rate of any reaction is at its greatest at the beginning (time=0). As the reactants are used up the rate decreases.

If a graph of rate against time is plotted then the shape obtained will depend on the overall order of the reaction.

  • 0th order graph:- straight flat line
  • 1st order graph:- straight line
  • All other orders: curve

If a curve is obtained then further mathematical treatment of the results is necessary.

as: Rate = k[A]m[B]n

if [B] is kept constant, then: log Rate = log k' + m log[A]

And a plot of log Rate against log [A] will give a straight line of gradient m (the graph has the form y = mx + c)

keeping [A] constant and treating the results of rate when [B] varies will allow a similar determination of the order with respect to B. Once the two orders are ascertained then the rate constant k can be found.

Half Life

The half life is the time taken for the concentration of reactants to reach half of its original value. For most reactions the half life changes as the reaction procedes but this is not the case for first order reactions where the half life is constant. (short half life= fast rate)

Using the half life to find the rate constant

The rate constant can be found from a concentration/time graph by taking a point, finding its concentration, then finding a point on the graph which corresponds to half this concentration. The half life is the time between these two points. The half life is also equal to ln2/k where k is the rate constant (equation given in data book).


Solving the rate equation by inspection

If a series of experimental results are obtained for rate at different reactant concentrations the change in rate can be ascertained for those reactions when the concentration of one of the reactants is kept constant while the other reactant concentration is changed.

Example

experiment concentration of A concentration of B Rate
1 0.1M 0.1M 6 x 103
2 0.2M 0.1M 1.2 x 104
3 0.4M 0.1M 2.4 x 104
4 0.1M 0.2M 6 x 103
5 0.2M 0.2M 1.2 x 104

In experiments 1,2 and 3 the concentration of A changes while the concentration of B is kept constant. This means that the rate equation can be written as:

Rate = k'[A]m

For these three experiments. If we inspect the rate of experiments 1,2 and 3 we see that as the concentration of A is doubled so the rate doubles. In other words the order of reaction must be 1 so that whatever happens to the concentration must also happen in equal amounts to the rate.

Similar inspection of experiments 1 and 4 ( or 2 and 5) show that while A is kept constant there is no effect on the rate when the concentration of B is changed. The order with respect to B must be 0.

The orders can now be substituted into the rate equation:

Rate = k[A]1[B]0

To obtain a value for the rate constant we simple substitute the values for one of the experiments above using the newly determined orders. (choose the experiment with the simplest numbers - in this case experiment 1)

Values from experiment 1

6 x 103 = k x [0.1]1 x [0.1]0

k = 6 x 103 / 0.1

k = 6 x 104

Solving the rate equation - test yourself

More information about the rate expression  

16.2 - Reaction mechanism

The mechanism of a reaction is a series of reactions between the particles of a reaction that eventually lead to the final products. A reaction may have many steps in the mechanism.

Rate determining step The slowest step in a reaction. It determines the rate of the overall reaction. (note 1)
Molecularity The number of particles reacting in the rate determining step of a reaction.
Activated complex As two particles collide (with sufficient energy to react and in the correct orientation) they form an intermediate called the activated complex...not literally a chemical substance, but an intermediate in which the bonds are in the process of being broken and formed.
The order of the reaction This gives information about the particles involved in the rate determining step (which is one step in the mechanism). For example, if two of one type of particle are colliding, the order with respect to that particle will be 2 (and zero to any others).
More information about reaction mechanisms  

16.3 - Activation energy

Arrhenius equation:

k = Ae(-Ea/RT) (data book)

where:

  • A is a constant related to the number, orientation and frequency of collisions occurring between the particles in the reaction.

  • k is the rate constant

  • R is the universal gas constant

  • T is the absolute temperature

Determination of the activation energy from practical results

If rates experiments are carried out at different temperatures the results can be plotted on a graph to obtain a value for the activation energy for a specific reaction.

  k = Ae(-Ea/RT)  
     
therefore: ln k = lnA - Ea/RT  

A plot of natural log of k against (1/T) will give a straight line of gradient - Ea/R


Enthalpy level diagrams

These show a curve representing the path between the reactants and products in terms of energy. Energy is shown on the y-axis. The reactants and products are of different chemical energies and the curve goes between the two levels (the reactants and products) having an energy maximum between them (with the distance between the highest point and the reactants being equal to the activation energy of the forward reaction).

An energy graph showing an exothermic reaction

Reactions only occur when the reacting particles have energy greater than the activation energy and are able to get over the activation energy barrier.


Catalysis

Catalysts provide an alternative mechanism with lower activation energy. This means that a greater number of collisions will be successful and the reaction procedes at a faster rate

Homogeneous catalysts -- catalysts in the same state (phase -- ie solid, liquid or gas) as the reactants. Hetrogeneous catalysts -- catalysts in a different phase (usually a solid) from the reactants.

Homogeneous catalysts operate by reacting with the reactants and eventually producing a reaction pathway of lower activation energy (and also being regenerated at the end of this process). Hetrogeneous catalysts provide a reactive site on which an activated complex forms, weakening the bonds and increasing the rate of collisions thus increasing the rate of reaction.

Examples of catalysed processes

Process catalyst used type
Contact (sulphur trioxide) Vanadium pentoxide heterogenous
Haber (ammonia) Iron heterogenous
Hydrogenation of alkenes Nickel heterogenous
Polymerisation Pt (Zeigler Natta) heterogenous
Polymerisation Oxygen homogeneous

catalysts

More information about activation energy  


Resources

Solving the rate equation - test yourself


Useful links


Notes:

1. Why is the slowest step of a mechanism said to be rate determining?

It acts like a bottleneck in that it prevents the other processes from reaching the end of the reaction and in this way determines the overall rate. It is similar to the idea of a car journey that has to pass through an area of roadworks. The car will be slowed right down by the roadworks and this is therefore the crucial factor in determining the overall time of the journey. return



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