Homework and Test Date

The practical assessment on rates of reaction should be complete by wed 5th Oct

Revise for test on rates of reaction on mon 3rd oct

Homework

Look at the kinetics power points, check that you understand all the aspects of the kinetics topic

Make sure that your investigation is properly planned.

Arrhenius calculations

The Arrhenius equation can be used to calculate the missing value of: A, K or Ea. Remember to convert Ea values to J/mol as they are usually in KJ/mol. Therefore you need to be consistent with the gas constant which is: 8.314 JK^-1 mol^-1. T must be in Kelvin. The other tip is that when you are calculating an Ea value, use the natural log form: lnK = lnA - Ea/RT.

Arrhenius Equation Calculations

The Arrhenius Expression

The rate constant is only constant for a fixed temperature. It increases exponentially as the temperature increases. Arrhenius came up with the following relationship:

k= A exp ^-Ea/RT       

This is often shown in logarithmic form so that it is the form of y=mx + c form and a graph of lnk against 1/T can be plotted which has a slope of -Ea/R.

Look at page 123 in The Chemistry Course Companion (Geoff Neuss) and the following Chemguide link is useful.

The Arrhenius Equation

Possible Graphs For Rate Experiments

Often concentration of a reactant against time is plotted or rate against the concentration of a reactant is plotted. You need to know and understand how each of the graphs would look like for: 0, 1st and 2nd order with respect to reactants concentration. Don't get confused between the two graphs; you need to look carefully at the axis.

See if you can plot and explain each of the possibilities?

Rates Investigation and Homework

To find the value of k, you just to insert the values from a particular experiment making sure that the concentrations are raised to the appropriate powers according to the order with respect to each reactant. Then just solve for k. To work out the units you simply do the same but with all the units inserted into the rate expression.

Rate investigation.

Please have the details of the investigation that you want to carry out ready. You must know how you are going to follow the rate, what your independent and dependent variables are and how you are going to control the other varaibles. I need to make sure that all the apparatus is ready, so please email your list of apparatus and chemicals.

The Rate Expression and the link with Mechanisms

The Rate Expression and Orders of Reaction

Simplistically one would imagine that doubling the concentration of a reactant would double the rate of a reaction. This seems logical; there will be double the number of collisions. However, although this is often the case it is not always true. Indeed, sometimes doubling the concentration of a reactant could have no affect on the rate and sometimes it will increase the rate by more than double, for example the rate could be quadrupled.

How can this be so? The answer lies in the fact that reactions usually take place by a series of steps and one of the steps (the slowest one (rate determining step RDS)) will control the rate. Only reactants involved in this step will affect the rate.  For reactants that are not involved in the RDS their concentration will have no affect on the rate. The order with respect to this reactant will be zero and this reactant will not appear in the rate expression. If a reactant is involved once in the RDS then doubling its concentration will cause the rate to double. It is said to be first order with respect to this reactant. The rate expression would be rate ≈ [conc] or Rate = k [conc]. It could be the case that a reactant is involved twice in the RDS, in which case, doubling its concentration will cause the rate to go quadruple i.e. go up by 4. The rate expression would be: Rate ≈[conc]²  or  Rate = k[conc]². K = the rate constant and the powers are the orders for each of the reactants. The overall order is simply the addition of all of the powers in the rate expression.

For example for the rate expression:  Rate = k  [BrO₃^-] [Br^-] [H⁺]²

What is the order with respect to each reactant?

What is the overall order?

What units will k have in this rate expression.  Note that the units of k will vary depending on the expression.

 

Remember that the rate expression can only be determined experimentally by following the rate and changing one reactant at a time. The stoichiometry of a reaction tells you nothing about the rate expression or possible mechanism.

Deducing  a Mechanism from the Rate Expression

This can be quite complex. However, simplistically if a reactant has zero order then it is not in the RDS but must be in a faster step. If a reactant is first order then it is directly or indirectly involved once in the RDS. If a reactant is second order then it’s either directly or indirectly involved twice in the RDS. Indirectly means that it could be involved with producing an intermediate, which is then involved in the RDS.

It can be fun proposing mechanisms that fit with the rate expression. There may well be several possibilities.

 The following links are useful:

Orders and Mechanisms

 Virtual Experiment

 Orders of reaction

Homework

Write about the uses of catalysts. Please cover, The Haber process, The Contact process, catalytic convertors in car exhausts, plus others of your choice. You should also be ready to discuss orders of reaction, the rate expression and how this links to mechanisms.

Full Syllabus

Here's the link to the pdf file of the detailed syllabus. look at the S.Land H.L sections and check that you understand each part of the topic as we go through the syllabus. Ask about any parts that you are not clear about.

Full syllabus

Types of Catalyst

A useful link with some diagrams illustrating heterogeneous catalysis.

Types of Catalyst

Homework

Please look at the link to rates of reaction and make sure you understand the following: The collision theory, factors that affect the rate of reaction [temperature, concentration, surface area, catalysts and pressure (for gaseous systems)]. In addition you should be able to sketch enthalpy level diagrams showing the activation energy for exothermic and endothermic reactions. You should also be able to sketch the Maxwell-Boltzmann distribution curve for two different temperatures and be able it to use it to explain how the use of catalyst and increasing the temperature causes an increase in the rate of reaction.


Chem guide rates link Rates of reaction

You also need to think of an experiment that you can investigate a factor that you think will affect the rate. you need to think about all the variables and how you will measure the rate. Please look at the link to practical work to remind yourself of the criteria. This could possibly be used for all three skill areas.

Explaining the colour in complex ions

This chemguide site is worth looking at:

Colour in complexes

Hybrid Orbital Shapes

This is a great site which visually shows you the formation of the hybrid orbital shapes.

Hybrid Orbital Shapes

Homework

Please go through the rest of p2 and have ready any questions. Also please look at the kinetics chapter, you could also look at some good websites e.g chem guide.

Rates of reaction

Notes from going through the exam

Remember ionic oxides are basic (if they dissolve) and covalent oxides are acidic. One of the gases which leads to acid rain is sulphur dioxide which is a covalent oxide.

Ionic oxides contain the oxide ion which reacts with water to form the hydroxide ion.


The following chem guide link is good: Period Three Oxides

Why is SO3 a solid at room temp?

One of the questions stated that sulphur trioxide was a solid at room temp. This may seem strange as one would presume that there are just weak forces of attraction ( Van Der Waal's)  between the molecules.
The following Chem Guide link is useful:Chem Guide

Power points (knockhardy) www.knockhardy.org.uk/ppoints.htm

This site has some very useful powerpoints which will help with your revision and can be a useful tool.

Powerpoints

Practical Assessment

Practical (Internal) Assessment 

Criterion	marks available	marks
Design	two grades out of 6 are to be presented	12 marks
Data collection and processing	two grades out of 6 are to be presented	12 marks
Conclusions and Evaluation	two grades out of 6 are to be presented	12 marks
Manipulative skills	1 grade out of 6 will be presented	6 marks
Personal skills	1 grade out of 6 will be presented	6 marks
	Total =	48 marks

Every year a sample of the internal assment work is sent by the teacher in charge for external moderation. In this process a third part (the moderator) tries to decide whether the teacher has applied the criteria correctly in his/her assessment. More about moderation.

Final grades

The final grade awarded for an IB subject is from 1-7. The schools receive a breakdown of the grade achieved in each part of the exam.

The requirements to achieve the IB diploma are fairly complex and may be found here: IB diploma award requirements

Homework

Homework is to look through the rest of the MC questions and have questions ready about anything that you want me to go through.

Double and triple bonds

Double bond C=C

   

                                                                   Sp² Hybrid orbitals overlap

                                                                    Forming sigma bond

                                 Unhybridised p orbitals overlap

                                   To form a pi bond. Note that the

                                                                  pi Bond has a lobe above and

                                      below the Sigma bond. However,

                               it still Constitutes One bond.

                                        The C=C double bond is Therefore,

                              one sigma and one pi bond.

        

                 Triple bond C=_C

                                                                                  pz and py unhybridised

                                                                               orbitals overlap to form

                                                                                two pi bonds. 

 sp hybrid orbitals overlap to form                      The triple bond is therefore

 a sigma bond.                                                       Two pi bonds and one sigma bond.

Shapes of Hybrid Orbitals

Shapes of Hybrid Orbitals

Note: The hybridisation not only explains observed shapes but it also explains the expansion of the octet which sometimes occurs. For example SF₆  involves d²sp³  hybridisation, therefore, explaining the six bonding pairs.

Interactive Website

Please look at this site all the time. There is lots of great stuff and you can follow what we are doing and click on the related links. If you exchanged your facebook time for IB interactive chemistry time I'd be very happy.


IB Interactive Syllabus

Welcome back

I hope that you have all had fantastic holiday's and are eager to get back into the chemistry class.