1. 16.1 Rate expression 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. 16.1.2 Deduce the rate expression for a reaction from experimental data. 16.1.3 Solve problems involving rate expression 16.1.4 Sketch, identify and analyze graphical representation for zero-, first- and second-order reactions.

2. Rate Law  Math expression to show how rate depends on concentration Rate = k[reactant 1] m [reactant 2] n …  m and n are called reaction orders. Their sum is called the overall reaction order.  k is the rate constant. It is specific to a reaction at a certain temperature.

3. Rate = k [A] m [B] n The exponents in a rate law must be determined by experiment. They are not derived from the stoichiometric coefficients in a chemical equation The values of exponents establish the order of a reaction for each species and overall order for the reaction The proportionality constant, k, is the rate constant and its value depends on the reaction, the temperature, and the presence or absence of a catalyst.

4. Chemical Equation showing stoichiometry: General expressions for the rate: Actual experimentally determined form of the rate law:

5. Sample Exercise 1: 2NO + 2H 2  N 2 + 2H 2 O is the reaction we’re studying, this is the data found during our experimentation Experiment Number Conc. of NO (M) Conc. of H 2 (M) Rate of N 2 forming (Ms -1 ) 10.2100.1220.0339 20.2100.2440.0678 3 0.4200.1220.1356

6. Study the data Exp #1 and #2, [NO] is unchanged, [H 2 ] is doubled and this causes the rate to double (0.0678/0.0339 =2), the H 2 ’s rate order is 1. Exp #1 and #3, [H 2 ] is unchanged, [NO] is doubled and this causes the rate to quadruple (0.1356/0.0339 =4), NO’s rate order is 2. Rate = k [NO] 2 [H 2 ] The overall rate order for this reaction is 3 (1+2)

7. Suppose… If when we ran exp #1 and #3, the rate didn’t change, what rate law would we expect?  Since it didn’t change when we doubled [NO], the rate order is 0, meaning the rate doesn’t depend on the concentration of NO at all, so rate= k[H 2 ] and the overall reaction order is 1.

8. Sample Exercise 2:  Use the kinetics data to write the rate law for the reaction. What overall reaction order is this? 2NO + O 2  2NO 2 Exp #[NO][O 2 ]Rate forming NO 2 (M/s) 10.015 0.048 20.0300.0150.192 30.0150.0300.096 40.030 0.384

9.  exp #1 and exp #2, [O 2 ] remained constant, where [NO] is doubled. The rate is quadrupled. The rate order for [NO] is 2.  Exp #1 and exp #3, [NO] remained constant, where [O2] is doubled. The rate is doubled. The rate order for [O2] is 1.  Overall reaction order (1+2) = 3

10. Sample Exercise 3 Rate data for the reaction: CH 3 Br + OH -  CH 3 OH + Br - Exp #CH 3 BrOH - Rate of forming CH 3 OH 10.200 0.015 20.4000.2000.030 30.400 0.060 Use the data to find the experimental rate law.

11. Distinctions Between Rate And The Rate Constant k  The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants.  The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants.  The rate and the rate constant have the same numerical values and units only in zero-order reactions.  For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.

12. Method Of Initial Rates Summary The effects of doubling one initial concentration: –For zero-order reactions, no effect on rate. –For first-order reactions, the rate doubles. –For second-order reactions, the rate quadruples. –For third-order reactions, the rate increases eightfold. The value of k for the reaction can be calculated.

13. Graphing conc. Vs time  Shows the effect of reactants being used up on the rate of reaction  If constant = zero order  Decreasing concentration is not affecting the rate

14. Graphing first order  If the reaction rate is halved when the concentration is halved, then the reaction is first order

15. Graphing second order  If halving the concentration causes the rate to decrease by a factor of 4, the reaction is second order  (1/2 ) 2 = ¼

16. Rate vs. conc graphs

17. First order  Conc. shows an exponential decrease, that is the time for the conc. to fall from its initial value to half its initial value, is equal to the time required for it to fall from half to one quarter to one eighth, etc…  This is known as half-life, t ½  Look at next slide for graphical representation

20. Radioactive decay  First order exponential decay  Half life is important and can be found from a graph or the equation  t ½ = ln2 k Example: if the rate constant of a first order reaction is 0.005 s-1, then the half life will be  t ½ = ln2 = 139 s 0.005 0.005 Now you can try to do #7 on handout