2.  The rate of a reaction is stated as the change in concentration of a reactant or product per unit of time.  Average reaction rate.  Example:  CO + NO 2  CO 2 + NO  If the concentration of CO is 0.100M at time 0.00s and 0.01M at time 2 seconds. we can calculate the average reaction rate.

3.  In a reaction between butyl chloride (C 4 H 9 Cl) and water, the concentration of C 4 H 9 Cl is 0.220M at the beginning of the reaction. At 4.00s, the concentration of butyl chloride is 0.100M. Calculate the average reaction rate as moles of C 4 H 9 Cl consumed per second.

4. Time (s)[H 2 ]M[Cl 2 ]M[HCl]M 0.000.0300.0500.000 4.000.0200.040

5.  For every chemical reaction we can write a Rate Law  Example:  The reaction A  B is a one step reaction. The rate law for this reaction is…  Rate = k [A]  The symbol k is the SPECIFIC RATE CONSTANT. This is a specific constant that is different for every reaction.  We can see that the rate is directly proportional to the concentration of A

6.  In the last example the rate law was  Rate = k [A]  The notation [A] could also be written as [A] 1  This would be a first order reaction.  The reaction order defines how the rate is affected by the concentration of reactant A.  Real Example:  H 2 O 2  O 2 + H 2 O

7.  The over all reaction order of a chemical reaction is the sum of the orders of the individual reactants.  Consider the equation aA + bB  Products  Where a and b are the molar coefficients  The general rate law will be  Rate = k[A] m [b] n  The over all reaction order will be m + n

8.  Write the rate law for the reaction A  B if the reaction is third order in A. ([B] is not part of the rate law, B is a product.  The rate law for the reaction 2NO + O 2  2NO 2 is first order in O 2 and third order over all. What is the rate law for this reaction?

9.  To determine the exponent of each reactant we need to compare the concentrations and the reaction rates. Trial[A] (M)[B] (M)Rate 10.100 2.00x10-3 20.2000.1004.00x10-3 30.200 16.0x10-3

10. Trial[A][B]Rate 10.100 2.00x10 -3 20.2000.1002.00x10 -3 30.200 4.00x10 -3

11. Trial[A][B]Rate 10.1 4.0 x 10 -5 20.10.24.0 x 10 -5 30.20.116.0 x 10 -5

12. Trial[NO][O 2 ]Rate 10.030.020.0041 20.060.020.0164 30.030.040.0082

13.  In order for a chemical reaction to occur the reactants must come in physical contact with one another.  They also have to come in contact with one another in the right way.  When they come in contact they must have enough energy in order for the reaction to occur.

15.  Reacting substances must collide  Reacting substances must collide in the correct orientation.  Reacting substances must collide with enough energy for the reaction to occur.

16.  The minimum amount of energy that reacting particles must have to form an activated complex and lead to a reaction is called the activation energy. (E a )  A high E a means that very few of the collisions that are taking place have enough energy to produce activated complexes and a reaction.  A low E a means that more collisions that are taking place have more than the required energy to cause a reaction.

17.  Temperature  Concentration  Surface area.  Catatysts

18.  Temperature  Increasing the temperature will increase the amount of energy that the colliding particles have. Resulting in a faster reaction  Concentration:  Increasing the concentration of one or both of the reactants will increase the number of collisions that happen, thus speeding up the reaction.  Surface area:  Providing more opportunities for collisions to happen will speed up a reaction.  Catalysts:  A catalysts is used to lower the activation energy of a reaction.