Chemical Kinetics. SCH4U: Grade 12 Chemistry. Unit Mind Map. Chemical Kinetics. Chemical kinetics? Why do we study this? Who makes use of chemical kinetics?. Chemical Kinetics. Chemical kinetics studying how we can make reactions go faster or slower Why do we study this?

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Chemical Kinetics

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Chemical Kinetics • Chemical kinetics? • Why do we study this? • Who makes use of chemical kinetics?

Chemical Kinetics • Chemical kinetics studying how we can make reactions go faster or slower • Why do we study this? • Who makes use of chemical kinetics?

Chemical Kinetics • Chemical kinetics studying how we can make reactions go faster or slower • Why should we study this? • Economics effects Materials are expensive • Developing pharmaceutical drugs • Who makes use of chemical kinetics?

Chemical Kinetics • Chemical kinetics studying how we can make reactions go faster or slower • Why should we study this? • Economics effects Materials are expensive • Developing pharmaceutical drugs • Who makes use of chemical kinetics? • Biologists metabolic rxns, food digestion, bone regeneration • Automobile engineers rate of rusting, decrease pollutants • Agriculture slow down food ripening

Reaction Rates • What is a chemical reaction rate? • Possible formula for measuring rate?

Reaction Rates • What is a chemical reaction rate? • measure of how fast reactants are used up or how fast products are produced • Possible formula for measuring rate?

Reaction Rates • What is the reaction rate? • measure of how fast reactants are used up or how fast products are produced • Possible formula for measuring rate? • Rate = Δ Concentration/Δ Time

Sample problem N2(g) + 3H2 (g) 2NH3(g) • What is the average rate of production of ammonia for the system if the concentration is 3.5mol/L after 1.0min and 6.2mol/L after 4.0 min?

Sample problem N2(g) + 3H2 (g) 2NH3(g) • What is the average rate of production of ammonia for the system if the concentration is 3.5mol/L after 1.0min and 6.2mol/L after 4.0 min? • ANSWER: 0.9mol/(L*min)

Measuring reaction rates Graphically • Average rate of reaction take the slope of the secant of the line • Instantaneous rate of reaction take the slope of the tangent of the line

Measuring reaction rates Graphically • Average rate of reaction take the slope of the secant of the line • Draw a secant line between two points • Calculate the slope • Slope = Rise/Run • = Δconcentration/Δ time

Measuring reaction rates Graphically • Instantaneous rate of reaction take the slope of the tangent of the line Draw a tangent line to the graph Calculate the slope of the tangent line Slope = Rise/Run = Δconcentration/Δ time

Measuring reaction rates • What are some factors that we can measure experimentally to determine the rate of reaction ?

Reaction rates and stoichiometry CO(g) + NO2(g) CO2(g) + NO(g) In this reaction, the ratio of CO to NO is 1:1 Therefore, the disappearance of COis the same as the production of NO Rate = - Δ [CO]/Δt = + Δ[NO]/Δt

Reaction rates and stoichiometry Suppose the ratio is NOT 1:1? Example, H2(g) + I2(g) 2HI (g) 2 mols of HI are produced for every 1 mol H2 used Rate = The rate at which H2 is used up is only half of which HI is produced

Sample Problem • IO3-(aq) + 5I-(aq) + 6H+(aq) 3I2(aq) + 3H2O (l) The rate of consumption of iodate ions (IO3-) is determined experimentally to be 3.0 x 10-5mol/(L*s). What are the rates of reaction or the other reactants and products in this reaction? • Complete p.364 #3, 4 • Homework: practice problem package

Collision Theory • Why do reactions occur the way that they do? VS

Collision Theory • Collision theory reactions can only occur if: • There is a collision between molecules

Collision Theory • Collision theory reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way

Collision Theory • Collision theory reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way 3. enough energy is provided to break the chemical bonds that hold molecules together (Activation energy)

Activation Energy • Activation Energy minimum potential energy the system needs to overcome for the molecules to react

Factors That Affect Rate of Reaction • What are some of the factors that may speed up or slow down a chemical reaction? • Chemical nature of the reactant • Concentration of reactants • Surface area of reactants • Temperature • Catalysts • Get into groups, brainstorm and then explain these factors to the rest of class in a creative way, making use of collision theory

Lesson 2 • Review collision theory • Rate Laws • Group Practice problems • Individual practice problems • homework

Collision Theory • Collision theory reactions can only occur if: 1. There is a collision between molecules 2. the molecules are oriented in the correct way 3. enough energy is provided to break the chemical bonds that hold molecules together (Activation energy)

Factors That Affect Rate of Reaction • Chemical nature of the reactant • Concentration of reactants • Surface area of reactants • Temperature • Catalysts

The Rate Law • Mathematical relationship between reaction rate and factors that affect it • Determined empirically (experimentally) • Rate = k[X]m[Y]n • e.g. 2NO2 + F2 2NO2F The above reaction is 1st order with respect to NO2 and 2nd order with respect to F2. What is the rate law equation?

The Rate Law Answer: • e.g. 2NO2 + F2 2NO2F rate = k[NO2]1[F2]2

Steps to solve rate law problems eg. 2BrO3- (aq) + 5HSO3-(aq) Br2 (g) + 5SO42- (aq) + H2O (l) + 3H+(aq) Rate = k [BrO3-]m [HSO3-]n

Steps to solve rate law problems eg. 2BrO3- (aq) + 5HSO3-(aq) Br2 (g) + 5SO42- (aq) + H2O (l) + 3H+(aq) • Write out rate equation: rate = k [BrO3-]m [HSO3-]n • pick a trial where [BrO3-] changes but [HSO3-] stays constant= trial 1 and trial 2 • Using a ratio, determine the relationship between change in concentration and change in rate = (trial 1/trial2) = (4.0/2.0) = (1.6/0.8) = 2 = 2 4. Using the following chart, determine the rate order of [BrO3-]

Steps to solve rate law problems 5. If the concentration is doubled (2.0 to 4.0), the rate also doubles (0.80 to 1.60), therefore the rate order is 1 m=1

Steps to solve rate law problems 6. To find the rate order of the other reactant, follow the same steps but pick a trial where [HSO3-] changes but [BrO3-] stays constant= trial 2 and trial 3 = (6.0/3.0) = (0.8/0.2) = 2 = 4 7. Use the following table to determine the rate order

Steps to solve rate law problems 8. As the concentration doubles the rate was multiplied by 4, therefore rate order of HSO3- = 2 (n=2) 9. Plug values into rate equation rate = k [BrO3-]m [HSO3-]n and solve for k.

Example 1 From the data collected above, determine the rate law of the following equation: aA + bB products • Rate = k[A]m[B]n

Example 1 • aA + bB products • Rate = k[A]m[B]n • m = 0; n = 1; k = 0.6s-1 • Rate = 0.6s-1 [B]1

Example 2 From the data collected above, determine the rate law of the following equation: • 2NO2 + F2 2NO2F • Rate = k[A]m[B]n

Example 2 • 2NO2 + F2 2NO2F • Rate = k[A]m[B]n • m = 2, n = 0, k = 4 x 10-3M-1s-1 • Rate =4 x 10-3M-1s-1 [NO2]2

Team problem • Get into groups of 3 and obtain problem clues a) Determine the order with respect to each reactant b) Determine the overall order of reactionc) Write the rate expression for the reaction.d) Find the value of the rate constant, k.

Practice Problem • From the data collected above, determine the rate law for the following equation: IO3- (aq) + 5I-(aq) + 6H+(aq) 3I2(aq) + 3H2O (l)

Practice Problem • From the data collected above, determine the rate law for the following equation: IO3- (aq) + 5I-(aq) + 6H+(aq) 3I2(aq) + 3H2O (l) • m = 1; n = 1; o = 2; k = 5M-3s-1 • Rate = 5M-3s-1 [IO3-]1 [I-]1 [H+]2

Homework • P. 377 practice # 1, 2, 3, 6, • P. 415 # 15 a-e

Lesson 3 • Reaction mechanisms: • Ping pong activity • Assembly line • Elephant toothpaste demonstration • Practice problems

Reaction mechanisms • Rate Laws determined experimentally • Equation: Reactants products • This actually occurs in a series of steps called elementary steps • Analogy: Cooking takes place in several steps