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