# Chapter 17 Reaction Kinetics 17-1 The Reaction Process.

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Chapter 17 Reaction Kinetics 17-1 The Reaction Process

How did you meet? Can you remember the first time you ever made a friend? What had to happen before the friendship could begin? Eye Contact Mutual Friend Accidentally Bumped into each other

Collision Theory In order for a reaction to occur particles must collide in: 1.A specific orientation and 2.with enough energy

Activation Energy The amount of energy required for a reaction to occur

Activation Energy Activation energy - the amount of energy the particles must have when they collide to force a reaction to occur. Activation Energy Products Reactants

Reaction Pathways The products have less energy than the reactants. The rxn released energy (heat) = exothermic ∆H will be negative since energy has left the system

Reaction Pathways The products have more energy than the reactants. The rxn absorbed energy (heat) = endothermic ∆H will be positive since energy has been added to the system

Practice Draw and label the energy diagram for a reaction in which ΔE = 30 kJ/mol, E a = 40 kJ/mol. Place reactants at energy level zero. Indicate determined values of ΔE forward, ΔE reverse & E a ’

Reaction Mechanisms Step-by-step sequence of rxns in order to obtain a final product Proposed Mechanism for Ozone Depletion via Free Chlorine Atoms Created by Decomposition of CFCs Step 1) Cl + O 3 → ClO + O 2 Step 2) 2 ClO → ClOOCl Step 3) ClOOCl → ClOO + Cl Step 4) ClOO → Cl + O 2 Proposed Mechanism for Ozone Depletion via Free Chlorine Atoms Created by Decomposition of CFCs Step 1) Cl + O 3 → ClO + O 2 Step 2) 2 ClO → ClOOCl Step 3) ClOOCl → ClOO + Cl Step 4) ClOO → Cl + O 2

Mechanisms Intermediates overall rxn

Mechanisms Slow Fast Fast Fast Rate Determining Step overall rxn

Catalysts vs. Intermediates overall rxn Catalysts appear 1 st as a reactant and then as a product during a mechanism. Intermediates appear 1 st as a product and then as a reactant during a mechanism.

Chapter 17 Reaction Kinetics 17-2 Reaction Rate

How can we increase the rate of a reaction? 1.Increase Surface Area 2.Increase Temperature 3.Increase Concentration 4.Increase in Pressure 5.Add a Catalyst

Surface Area Increase the surface area allows for a greater chance for effective collision

Temperature An increase in temperature will cause particles to move at a higher velocity resulting in more effective collisions

Concentration An increase in concentration will also cause an increase in the chance that effective collisions will occur

Pressure Increasing the pressure of a gas system will cause more frequent collisions

Catalysts Adding a catalyst lowers the amount of activation energy required

Catalysts Reactants Catalyst

Slow Rate Determining Step Rate Laws Rate = k[HBr][O 2 ] An equation that relates the rxn rate and the concentration of reactants

Rate Laws If no mechanism is given, then… 2H 2 + 2NO  N 2 + 2H 2 O Rate = k[H 2 ] 2 [NO] 2

Rate Orders 0, 1 st and 2 nd order rates Order is dependent upon what will yield a straight line 0 order 2 nd order ln [reactants] [reactants] 1 st order 1/[reactants]

Rate Orders 1 st order: reaction rate is directly proportional to the concentration of that reactant 2 nd order: reaction rate is directly proportional to the square of that reactant 0 order: rate is not dependant on the concentration of that reactant, as long as it is present. For Individual Components:

Rate Orders Overall reaction orders is equal to the sum of the reactant orders. Always determined experimentally! For Overall Order:

Calculating for k A + 2B  C Rate = k[A][B] 2 ExperimentInitial [A]Initial [B]Rate of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min What is the value of k, the rate constant?

Calculating for k Rate = k[A][B] 2 ExperimentInitial [A]Initial [B]Rate of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min 2.0 x 10 -4 = k[0.20][0.20] 2 2.0 x 10 -4 = k(0.008) k = 2.50 x 10 -2 min -1 M -2

Practice 1. In a study of the following reaction: 2Mn 2 O 7(aq) → 4Mn (s) + 7O 2(g) When the manganese heptoxide concentration was changed from 7.5 x 10 -5 M to 1.5 x 10 -4 M, the rate increased from 1.2 x 10 -4 to 4.8 x 10 -4. Write the rate law for the reaction. 2. For the reaction: A + B → C When the initial concentration of A was doubled from 0.100 M to 0.200 M, the rate changed from 4.0 x 10 -5 to 16.0 x 10 -5. Write the rate law & determine the rate constant for this reaction. Rate = k[Mn 2 O 7 ] 2 Rate = k[A] 2 Constant = 4.0 x 10 -3 M/s

More Practice 3. The following reaction is first order: CH 3 NC (g) → CH 3 CN (g) The rate of this reaction is 1.3 x 10 -4 M/s when the reactant concentration is 0.040 M. Predict the rate when [CH 3 NC] = 0.025. 4. The following reaction is first order: (CH 2 ) 3(g) → CH 2 CHCH 3 (g) What change in reaction rate would you expect if the pressure of the reactant is doubled? New Rate = 8.1 x 10 -5 M/s An increase by a factor of 2

Even More Practice 5. The rate law for a single step reaction that forms one product, C is R = k[A][B] 2. Write the balanced reaction of A & B to form C. 6. The rate law of a reaction is found to be R = k[X] 3. By what factor does the rate increase if the concentration of X is tripled? 7. The rate of reaction, involving 2 reactants, X & Z, is found to double when the concentration of X is doubled, and to quadruple when the concentration of Z is doubled. Write the rate law for this reaction. A + 2B → C R = k[X][Z] 2 The rate will increase by a factor of 27