Chemical Kinetics

A brief note on Collision Theory All matter is made up of particles: atoms, molecules and ions. Kinetics is all about how chemicals react together. Chemicals can only react if their particles come into contact with each other. The particles in gases and liquids are in constant motion and consequently they will collide with each other. Millions of these collisions occur every second but not every one produces a new substance. The type of collision and how energetic it is will help to decide whether or not a reaction will happen. When two particles collide, ‘head on’ with sufficient speed (or energy) then a reaction takes place.

KINETICS The rate of a chemical reaction depends on a number of factors:- 1.Temperature 2.Concentration 3.Pressure 4.Particle size 5.Catalysts

CONCENTRATION In this section we will look at the effect of concentration on the rate of a chemical reaction. Initially it must be stated that [ X ] square brackets represents concentration with units mol dm -3 A typical graph of how concentration varies with time is shown on the next slide.

[X] mol dm -3 Time ‘s’ A B C The “initial rate” is measured by the initial slope of the graph i.e. Initial rate = slope = BC where units are mol dm -3 = mol dm -3 s -1 AC s In a chemical reaction as the products are formed so the reactants are used up. The graph shows conc. of products against time. The rate of the reaction is the time taken in seconds for a measured change in concentration of a reactant or product

RATE EQUATIONS A + B Products In the above the rate could be dependant on both the concentrations of A and B. i.e. Rate  [A] [B] or:- Rate = k [A] [B] where ‘k’ is a proportionality constant called the rate constant.

ORDERS OF REACTION The order of reaction tells us how the reaction rate is affected by changing the concentration of each of the products. To determine the “order of a reaction” we need to determine how the rate of the reaction varies with a change in the concentration. This can only be evaluated by carrying out experiments. A graph of this variation is shown on the next slide.

Rate (mol dm -3 s -1) Concentration ( mol dm -3) ZERO ORDER FIRST ORDER SECOND ORDER

Example: If we had a reaction with the equation: 2A + 3B + C D + 2E And you found from experiments that doubling the concentration of A had no effect on rate of reaction B increased the rate of reaction four times C doubled the rate of reaction The reaction is zero order with respect to A second order with respect to B first order with respect to C The overall order of the reaction is 3 – the sum of the individual orders( )

MATHEMATICAL EXPLANATION ZERO ORDER (n =0) Concentration Rate FIRST ORDER (n = 1) SECOND ORDER (n =2) [1] 0 1 [1] 1 1 [1] 2 1 [2] 0 1 [2] 1 2 [2] 2 4 [3] 0 1 [3] 1 3 [3] 2 9 This is based on X n where n = order Concentration Rate Zero order – change in concentration gives no change in rate First order – double the concentration and the rate doubles Second order – double the concentration and the rate increases by four

Rate equations Hence the rate equation would be written as:- Rate = k [A] a [B] b where a and b are the orders of reaction Overall order for the reaction In the case above it would be (a + b)

The rate constant Example Suppose we have a rate equation:- Rate = k [A] 1 [B] 1 Rearranging we get:- k = Rate [A] 1 [B] 1 Units for ‘k’ are important and the method of calculation is shown on the next slide.

Units of ‘k’ Write the units to match the equation and finally cancel down to a single set. i.e. mol dm -3 s -1 = s -1 mol dm -3 x mol dm -3 mol dm -3 Bringing the denominator to the top (i.e. change the charges of the powers):- Units of k are:- mol -1 dm 3 s -1

Problems Work out the units for k for the following:- 1. Rate = k [A] 0 [B] 1 2. Rate = k [A] 1 [B] 2 3. Rate = k [A] 2 [B] 2

ANSWERS 1. mol dm -3 s -1 = s -1 mol dm mol dm -3 s -1 = mol dm -3 s -1 = mol -1 dm 3 s -1 ( mol dm -3 ) 2 mol dm -3 x mol dm mol dm -3 s -1 = mol dm -3 s -1 (mol dm -3 ) 2 x (mol dm -3 ) 2 (mol dm -3 ) 4 = mol dm -3 s -1 = mol -3 dm 9 s -1 mol dm -3 x (mol dm -3 ) 3

CALCULATIONS The reaction given below gave results as shown in the table. A + B PRODUCTS EXPERIMENT NUMBER [A] (mol dm -3 ) [B] (mol dm -3 ) Initial rate (mol dm -3 s -1 ) Determine:- 1. the order with respect to B 2. the overall order 3. the value of ‘k’ with correct units

ANSWER From the table of results:- In experiments 1 and 2 concentration of A is constant but the concentration of B has doubled and the rate has not changed. Hence concentration of B has no effect so must be ZERO ORDER. In experiments 2 and 3 concentration of B is constant but the concentration of A has doubled causing the rate to double. Hence the reaction must be FIRST ORDER with respect to A. Therefore:-1. Order with respect to B is ZERO 2. Overall order is (0 + 1) = 1 (first order) 3.Rate = k [A] 1 [B] 0 which can be written as Rate = k [A] so k = Rate = = 0.32 s -1 [A] 0.01

PROBLEMS For the reaction:- 2NO (g) + O 2(g) 2NO 2(g) the following results were obtained:- Experiment number [NO] mol dm -3 [O 2 ] mol dm -3 Initial rate mol dm -3 s Determine the rate equation for the reaction 1.

ANSWER 1. In experiments 1,2 and 3 the concentration of O 2 gas is constant. When The concentration of NO is doubled the rate increases by a factor of four and when the concentration is tripled the rate increases by a factor of nine. Hence the reaction must be SECOND ORDER with respect to NO. In experiments 3 and 4 the concentration of NO is constant. Here the concentration of O 2 gas is tripled and the rate is also tripled. Hence the reaction is FIRST ORDER with respect to oxygen gas. Therefore the rate equation is:- Rate = k [NO] 2 [O 2 ] 1 (Actually power 1 is not needed.)

PROBLEM For the reaction:- 2H 2(g) + 2NO (g) 2H 2 O (g) + N 2(g ) the following results were obtained:- [H 2 ] x mol dm [NO] x mol dm -3 Initial rate x mol dm -3 s -1 (a) Determine the rate equation for the reaction (b) Calculate the value of the rate constant with correct units 2.

ANSWER 2. In the first three readings the concentration of NO is constant at 6.0 x mol dm -3. When the concentration of H 2 is doubled and tripled the rate is doubled and tripled respectively. Hence the reaction Must be FIRST ORDER with respect to H 2. If we look at the concentration of H 2 going from 1.5 x to 3.0 x mol dm- 3 it has doubled and since it is first order we would expect the rate to have doubled to 9.0 x mol dm -3 s -1. However, the concentration of NO has reduced by a half so if it were first order the rate would also reduce by half. In fact it has reduced by a factor of four which shows us that the reaction is SECOND ORDER with respect to NO.

ANSWER TO Q.2 continued (a) Hence the rate equation for this reaction is:- Rate = k [H 2 ] 1 [NO] 2 (b) Rearranging the equation we get:- k = Rate [H 2 ] 1 [NO] 2 = 1.5 x x x ( 6.0 x ) 2 = 1.5 x = 8.33 x 10 4 mol -2 dm 6 s x 10 -8