Rates of Chemical Reactions CHEMICAL KINETICS

The rate of a reaction is measured by looking at the change in concentration over time. RATES OF CHEMICAL REACTIONS Finding the slope will give you the reaction rate at any given time.

What is the average rate between 10 and 30 seconds? What is the average rate between 30 and 40 seconds?

RELATIVE RATES

What are the relative rates of reaction for the following reaction? 2 NOCl → 2 NO + Cl 2 PRACTICE

Page 675 Exercise 15.2 Page 712 2, 4, 6 ASSIGNMENT

How will increasing temperature affect the rate? Why? How will increasing concentration affect the rate? Why? AFFECTING REACTION RATES

Catalysts increase the reaction rate by decreasing the activation energy(energy needed to start reaction). CATALYSTS

A rate law is an equation that relates concentration to the rate of reaction. For the equation: N 2 O 5 → 2 NO 2 + ½ O 2 The rate law is: rate = k[N 2 O 5 ] k is the rate constant. We can calculate its value. The rate will have units of mol/L·time RATE LAWS

For the general equation a A + b B → x X, the rate law is rate = k[A] m [B] n m and n have to be determined from data. The data will tell you how the rate changes when concentrations of each reactant change. Order of a reaction is related to the exponents in the rate law. In the rate law, rate = k[NO] 2 [Cl 2 ], the reaction is second order for NO, first order for Cl 2, and third order overall. DETERMINING RATE LAWS

Determine the rate equation and the value of k. CO + NO 2 → CO 2 + NO EXAMPLE [CO], mol/L[NO 2 ], mol/LRate, mol/L· h 5.10 X X X X X X X X X X X X X X X 10 -8

Using the rate equation, find the rate of reaction when [CO] = 3.8 X10 -4 mol/L and [NO 2 ] = 0.650X10 -4 mol/L. EXAMPLE

Page 683 Exercise 15.3 and 15.4 Page 713 8, 10, 12, 14 ASSIGNMENT

CONCENTRATION AND TIME

Cyclopropane will rearrange to propene in a first order process. If the initial concentration is mol/L, how many hours would elapse if the concentration is mol/L? k = 2.42 h -1 EXAMPLE

Hydrogen peroxide decomposes in a first order reaction. H 2 O 2 → H 2 O + O 2, k = 1.06 X min -1 What fraction of hydrogen peroxide remains after 100. min? What is the concentration after 100. min if the initial concentration was mol/L? PRACTICE

Page 686 Exercise 15.5 and 15.6 Page , 19 ASSIGNMENT

SECOND-ORDER REACTIONS

The decomposition of HI is a second order reaction and k = 30. L/mol· min. How much time does it take for the concentration to drop from mol/L to mol/L? PRACTICE

ZERO-ORDER REACTIONS

Regardless of the order, we should be able to create a straight line if we graph the right things. If we are not sure of the order, if the graph the three possibilities, we can determine the order. DETERMINING ORDER FROM A GRAPH

Zero Order First Order Second Order

ASSIGNMENT Page 687 Exercise 15.7 Page 689 Exercise 15.8 Page , 28, 30

HALF-LIFE

Sucrose, C 12 H 22 O 11 decomposes to fructose and glucose. If k = h -1, what is the half life? How long will it take for 87.5% of the sample to decompose? PRACTICE

Radon-222 gas has a half-life of 3.8 days. If there are 4.0X10 13 atoms per liter initially, how many per liter will remain after 30 days? EXAMPLE

Reacting molecules must collide. The molecules must have enough energy The molecules must collide in the correct way. COLLISION THEORY

If the concentration increases, the reaction rate will increase. If there are more particles present, more collision will occur. EFFECT OF CONCENTRATION ON RATE

The activation energy is the amount of energy required to start a reaction. On an energy diagram, it is the amount of energy from the reactants to the highest point. What is the activation energy for the diagram? ACTIVATION ENERGY

Page 692 Exercise 15.9 Page , 24 ASSIGNMENT

Increasing the temperature of a sample is really increasing the energy of the particles. That means more particles have the required energy to react. In addition, more collisions will occur due to the increased speed of the particles. EFFECT OF TEMPERATURE ON RATE

ARRHENIUS EQUATION

Calculate the activation energy for the decomposition of HI if k 1 = 2.15X10 -8 at 650 K and k 2 = 2.39X10 -7 at 700K.

Catalysts speed up reaction rates be lowering activation energy. Even though they are not part of the equation, they are sometimes included in rate laws. Reaction take place in steps, and a catalyst takes part in certain steps, but is canceled out in the end. EFFECT OF CATALYST ON RATE

ASSIGNMENT Page 699 Exercise Page and 38

If a reaction takes place in steps, we call each step an elementary step. Br 2 + NO → Br 2 NOStep 1 Br 2 NO + NO → 2 BrNOStep 2 Br NO → 2 BrNOOverall Reaction Each step has its own activation energy and k. Any substance that is not part of the overall reaction is a reaction intermediate. What is the reaction intermediate above? ELEMENTARY STEPS

Unimolecular- reactants of step include one molecule Bimolecular- reactants of step include two molecules, which could be two of the same molecule or two unique molecules Termolecular- reactants include three molecules MOLECULARITY OF ELEMENTARY STEPS

3 ClO - (aq) → ClO 3 - (aq) + 2Cl - (aq) The above reaction takes place in two steps: ClO - (aq) + ClO - (aq) → ClO 2 - (aq) + Cl - (aq) ClO 2 - (aq) + ClO - (aq) → ClO 3 - (aq) + Cl - (aq) What is the molecularity of each step? Write the rate equation for each step. Show that the sum of the two steps gives the net reaction. PRACTICE

When a reaction has multiple steps, the slow step will determine the reaction rate. For the reaction 2 NO 2 (g) + F 2 (g) → 2 FNO 2 (g), the rate law is: rate = k[NO 2 ][F 2 ] The reaction takes place in two steps: NO 2 (g) + F 2 (g) → FNO 2 (g) + F(g) NO 2 (g) + F(g) → FNO 2 (g) Which step is slow? Which is fast? RATE-DETERMINING STEP

Page 705 Exercise Page 707 Exercise Page , 42 ASSIGNMENT

The rate law for an overall reaction can’t include an intermediate. This happens if the second step is slow. For the reaction 2 NO(g) + O 2 (g) → 2 NO 2 (g) the steps are: – NO(g) + O 2 (g) ⇌ OONO(g)fast, equilibrium – NO(g) + OONO(g) → 2 NO 2 (g)slow The rate law for the rate-determining step is rate = k 2 [NO][OONO] The forward and reverse rates of step one are equal. k 1 [NO][O 2 ]=k -1 [OONO], so [OONO]=k 1 /k -1 [NO][O 2 ] Substituting for [OONO], rate = k 1 k 2 /k -1 [NO] 2 [O 2 ]. This is the same as you would find experimentally. MECHANISMS WITH AN EQUILIBRIUM STEP

Another mechanism for 2 NO(g) + O 2 (g) → 2 NO 2 (g) is: NO(g) + NO(g) ⇌ N 2 O 2 (g)fast N 2 O 2 (g) + O 2 (g) → 2 NO 2 (g)slow Show this mechanism leads to the same rate law. EXAMPLE

Page 710 Exercise Page (skip b), 46, 48, 52, 54, 62, 78, 80, 83 ASSIGNMENT