Get Started with Reaction Kinetics

These notes are intended to provide an introduction to Rate of Reaction. Work through these, and then refer to other available materials to build your understanding of the concepts.

Answer the questions...

  1. What is meant by the 'rate' of a chemical reaction?
  2. How does the rate of a chemical reaction change as the reaction proceeds from start to finish?
  3. How can we follow experimentally the changing rate of a chemical reaction as it takes place?

A chemical reaction involves one or more reactants. We are interested in how the rate of reaction depends on each of these separately.

Two experimental methods which can be employed to investigate reaction rate are:

First of all, the Continuous method...

Consider the decomposition of dinitrogen pentoxide dissolved in liquid tetrachloromethane at 30 °C.

N2O5(sol)   ®   2NO2(sol) + ½O2(g)

The volume of O2(g) measured at a known temperature and pressure is related to the diminishing concentration of N2O5(sol) at intervals during the experiment. The following data are obtained:

[N2O5(sol)] (mol dm-3)Time, t (hours)  
1.400
1.051
0.782
0.603
0.454
0.335
0.236
0.187
0.148
0.109
0.0810
  1. Plot a graph of [N2O5(sol)] against Time (hours). This is a 'concentration against time' graph.

The gradient of a tangent to the curve at any instant of time (or concentration) is the rate at that instant.

  1. Calculate five of these, and then plot a 'rate against concentration' graph.

From this graph, decide how the rate of reaction is related to the concentration of N2O5(sol). If doubling the [N2O5(sol)] doubles the rate of reaction, then the reaction is said to be first order with respect to N2O5(sol).

If doubling the [N2O5(sol)] causes the reaction rate to quadruple, then the reaction is second order with respect to the reactant in question. If doubling the concentration of a reactant produces an 8 times increase in rate, then the reaction is third order with respect to the reactant concerned.

It might be that the concentration against time graph was a straight line, and consequently, the rate against concentration graph is a horizontal straight line. In this case, changing the concentration of the reactant has no effect on the reaction rate. The reaction is said to be zero order with respect to the reactant.

  1. What is the order of reaction with respect to the [N2O5(sol)]?
  2. Write a rate equation for the N2O5(sol) decomposition reaction.
  3. Now examine the concentration against time curve. Work out some repeating half-lives for this reaction. Write an appropriate statement about this observation.

If there are, for example, two reactants, then how is the continuous method applied experimentally? Clearly, it is necessary to keep each reactant concentration constant in turn whilst the other is investigated. This is achieved by having a large concentration of the reactant not being investigated, so that during the reaction its concentration changes negligibly, and so remains effectively constant.

Once you have found the order of reaction with respect to each reactant, the overall order of reaction can be obtained simply by adding together these values. A rate equation can be written for the reaction. For example,

2A(g) + B2(g)   ®   products

rate = k[A(g)]x[B2(g)]y

The idea is to find the values of x and y. It is important to remember that these values can be obtained only by experiment.

Now the Initial Rates method...

This method involves carrying out the experiment five or more times. Each reactant is investigated in turn by changing its concentration whilst keeping the other(s) unchanged. The time taken for the reaction to reach exactly the same stage (for example, the time taken for the same amount of a product to form) is measured for each experiment. A colour change, for example, might indicate this.

Here is a brief insight into this idea...

Imagine that one of the products is iodine, I2. If the same amount (volume and concentration) of sodium thiosulphate solution is present in the reaction mixture along with some starch solution, as the iodine forms it will react with the sodium thiosulphate so being removed from solution. At the instant when all of the sodium thiosulphate has reacted, the iodine then produced will immediately form a blue-black colour with the starch. The clock is stopped at this instant. Reactions of this type are often referred to as Clock Reactions.

Now try these examples to get the ideas of the initial rates method...

For the thermal decomposition of ethanal, CH3CHO

CH3CHO(g)   ®   CH4(g) + CO(g)

the following data at 800 K are given.

Initial [CH3CHO(g)] (mol dm-3)Initial Rate of decomposition of CH3CHO (mol dm-3 s-1)
0.10009.0 x 10-7
0.200036.0 x 10-7
0.300081.0 x 10-7
0.400014.4 x 10-6
  1. What is the order of reaction with respect to CH3CHO(g)? Write the rate equation for the reaction.
  2. Calculate the rate constant for the reaction at 800 K.
  3. Calculate the decomposition rate at 800 K at the instant when [CH3CHO(g)] is 0.250 mol dm-3.

Again, when two reactants participate in a reaction, the rate equation may be derived by keeping the concentration of one reactant constant while varying the concentration of the other.

The reaction

NO(g) + ½Cl2(g)   ®   NOCl(g)

has been studied at 50 °C.

Initial [NO(g)] (mol dm-3) Initial [Cl2(g)] (mol dm-3)Initial rate of formation of NOCl(g) (mol dm-3 s-1) 
0.2500.2501.43 x 10-6
0.2500.5002.86 x 10-6
0.5000.50011.4 x 10-6
  1. Work out a rate equation for this reaction.
  2. What is the overall order of the reaction?
  3. Calculate the rate constant for the reation at 50 °C.
  4. Calculate the rate of formation of NOCl when [NO] = [Cl2] = 0.110 mol dm-3.
  5. At the instant when Cl2 is reacting at 2.21 x 10-7 mol dm-3 s-1, what is the rate at which NO is reacting and NOCl is forming?