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Is a study of how fast chemical reactions occur.

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Presentation on theme: "Is a study of how fast chemical reactions occur."— Presentation transcript:

1 Is a study of how fast chemical reactions occur.
Chemical Kinetics Is a study of how fast chemical reactions occur.

2 Factors affecting rates of reactions. . . .
Concentration of reactants. Temperature of reactions. Presence or absence of a catalyst. Surface area of solid or liquid reactants or catalysts.

3 Reaction Rates. . .the speed of a chemical reaction.
Speed of a reaction is defined as the change per unit time. Often determined by measuring the change in concentration of a reactant or product with time. Is called its reaction rate.

4 If A B Change in the number of moles of B Average rate =
Change in time Change in the number of moles of B = moles B final – moles B initial G. = 1.00molA AF.= average rate R. = t=0 A=1.00mol B=0mol t=10min A=.74mol B=.26mol t=20min A=.54mol B=.46molB

5 Two ways of measuring rate: appearance of B per unit of time
Average rate = D moles B Dt Average rate = .26 mol – 0 mol 10min – 0 min = mol/min Two ways of measuring rate: appearance of B per unit of time disappearance of A per unit of time OR Average rate = -D moles A Dt

6 Hints. . . The coefficients of the balanced chemical equation tell you relative rates. Rate of disappearance of reactants is given a negative sign.

7 Nitrogen monoxide reacts with chlorine gas to form the gaseous substance nitrosyl chloride (NOCl) according to the following equation: 2NO(g) + Cl2(g) NOCl(g) At the beginning of the reaction, the concentration of chlorine gas is mol/L after 30.0s, the chlorine concentration has decreased to mol/L. Calculate the average over time in terms of the disappearance of chlorine.

8 Rates in terms of Concentration
Determine rate by monitoring change in concentration. (of reactant or product) Of reactant or product. Most useful unit is molarity. Since volume is constant: Molarity and moles are directly proportional.

9 2NO2(g) 2NO(g) + O2(g)


11 C4H9Cl(aq) +H2O(l) C4H9OH(aq) +HCl(aq)
Calculate the average rate in terms of disappearance of C4H9Cl Units for average rate are mol/L.s or M/s Average rate decreases with time. If we plot [C4H9Cl] vs. time Instantaneous rate = the rate at any instant in time. Slope of the straight line tangent to the curve at that instant. Different from average rate.

12 Rate of appearance of C4H9OH must equal the disappearance of C4H9Cl
Reaction Rates and Stoichiometry C4H9Cl(aq) +H2O(l) C4H9OH(aq) +HCl(aq) Rate of appearance of C4H9OH must equal the disappearance of C4H9Cl - D [C4H9Cl] = D [C4H9OH ] Rate = Dt Dt

13 2HI(g) H2(g) + I2(g) OR aA + bB cC +dD Rate = 1 D[HI] D[H2] D[I2]
Dt Dt Dt * - = OR aA + bB cC +dD Rate = 1 D[A] D[B] D[C] D[D] a Dt b Dt c Dt d Dt - =

14 Rate vs. Concentration Rate:
Increases when reactant concentration increases. Decreases as the concentration of reactants is reduced.

15 Reaction order. . . If. . . Must be determined experimentally.
Rate= k[reactant 1]m[reactant 2]n m and n are reaction orders. Overall reaction order is the sum of the reaction orders. If no number is written, remember it equals 1. Must be determined experimentally.

16 Reaction Order. . .

17 Crystal Violet

18 Reaction Order. . .

19 Units of Rate Constants. . .
Depend on the overall reaction order. If a reaction is second order: Unit of rate=(unit of rate constant)(unit of concentration)2 Therefore. . . unit of the rate constant = (units of rate) = M/s = M-1 s-1 (Units of concentration)2 M2

20 Using Initial Rates To Determine Rate Laws
Observe the effect of changing initial concentrations: Zero order in a reactant. . . Changing concentration of that reactant will have no effect on rate. First order. . . Doubling concentration will cause rate to double. Second order. . . Doubling concentration will result in a 22increase in rate. Tripling concentration results in a 32 increase in rate. nth order . . . If doubling concentration causes a 2n increase in rate.

21 Zero order reaction

22 Hints. . . The sum of the exponents is called the reaction order.
From the experimental data, determine how rate varies with varying concentration of one while the concentration of the other is held constant. When solving for rate constant k, pick any equation though it is best to average them.

23 The Change in Concentration with Time (convert the rate law into a convenient equation that gives concentration as a function of time) First order, the rate doubles as the concentration of a reactant doubles. Therefore: Rate = -D[A]/Dt = k[A] After integrating: ln([A]t/[A]0) = -kt A plot of ln[A]t vs. t is a straight line with slope –k and intercept ln[A]0

24 Half Life. . .

25 Half Life. . .t 1/2 Is the time required for the concentration of a reactant to decrease to half its original value. t 1/2 is the time required for [A]0 to reach ½[A]0. The half life of a first order reaction is independent of the initial concentration of the reactant.

26 Second Order Reactions
Rate depends on the reactant concentration to the second power or on the concentration of two reactants to the first power. A plot of 1/[A]t vs. t is a straight line with slope k and intercept 1/[A]0. A plot of ln[A]t vs. t is not linear. t 1/2= 1/k[A]0 Rate=k[A]2 or =k[A][B]

27 Temperature and Rate Most reactions speed up as temperature increases.
Reflecting temperature and rate in a rate expression: Rate law has no temperature term, so rate constant must depend on temperature. Consider first order reaction: CH3NC CH3CN As temperature increases from 190C-250C the rate constant increases. Approximately doubling with each 10C increase in temperature. Use light sticks in hot and cold water to demonstrate

28 Collision Model. . . Based on kinetic molecular theory.
In order for molecules to react, they must collide. The greater the number of collisions, the faster the rate. Rate should increase with an increase in concentration of reactants. Higher temperature, more energy,more collisions. Rate should increase with temperature increase. Not all collisions lead to products. Molecules must collide in the correct orientation with enough energy to form products.

29 Activation Energy. . . Arrhenius: molecules must possess a minimum amount of energy to react. Bonds must be broken in the reactants. Molecules with too little kinetic energy do not react when they collide. Activation Energy, Ea: is the minimum energy required to initiate a chemical reaction. Use laser disk clip Side B chapter 11

30 Reaction Mechanism. . . Is the process by which the reaction occurs.
Provide a picture of which bonds are broken and formed during the course of a reaction.

31 Single Step Reactions. . . A+B 2C Rforward = kforward [A][B]
If reversible: Rreverse=kreverse[C]2 Only if the reaction occurs in one step.occurs at the molecular level just as it is written in the chemical equation.

32 Elementary Steps. . . Any processes that occur in a single step.
Number of molecules present in an elementary step gives the molecularity of the elementary step: Uni-, bi-, ter- Multistep reactions: consist of a sequence of elementary steps. Elementary steps add to give the balanced chemical equation. Multistep reactions will include intermediates. Appear in an elementary step, but are not products or reactants. Formed in one elementary step and consumed in another. Not found in the balanced equation for the overall reaction.

33 Rate Laws of Elementary Steps
Determine the overall rate law for the reaction. Rate law of elementary step is determined by its molecularity. Uni- first order Bi- second order Ter- third order

34 Rate Laws of Multistep Mechanisms
One step is slower than others. Slow step limits the overall reaction rate. This is the rate determining step. This step governs the rate law for the overall reaction.


36 NO2(g)+CO(g) NO(g)+CO2(g)
Experimentally derived rate law: Rate= k1[NO2]2 Mechanism for the reaction: Step 1:NO2(g)+NO2(g) NO3(g) + NO(g) slow Step 2: NO3(g)+CO(g) NO2(g) + CO2 fast NO3 is an intermediate If k2>>k1, then the overall rate depends on first step. Rate = k1[NO2]2 supports but does not prove mechanism. k1 k2

37 2NO2(g)+ F 2(g) 2NO2F(g) A proposed mechanism:
Step 1. NO2+F2 NO2F+F (slow) Step 2. F+NO2 NO2F (fast) Identify the rate determining step, and write an acceptable rate law. ANSWER: Combine the two steps, the intermediate F cancels out and we are left with the original equation. The first step is the slow step and is considered the rate determining step. R=k[NO2][F2]

38 X + 2Y XY2 This reaction occurs by a single step mechanism. Write the rate law for this reaction, then determine the following effects: a)doubling the concentration of X b)doubling the concentration of Y c) using one third the concentration of Y

39 Answer: Single step therefore the rate law can be written from the equation. The rate will vary directly with the concentration of X, which has the implied coefficient of 1 in the equation. The rate will vary directly with the square of the concentration of Y,which has a coefficient of 2: R=k[X][Y]2

40 a. double the rate: R=k[2X][Y]2
b. increase the rate four fold: R=k[X][2Y]2 c. Reduce the rate to one ninth of its original value: R=k[X][1/3Y]2

41 Mechanisms with an Initial Fast Step
Consider: 2NO(g)+ Br2(g) 2NOBr(g) Experimentally determined rate law is: Rate = k[NO]2[Br2] Step 1: NO(g) + Br2(g) NOBr2(g) fast Step 2: NOBr2+NO(g) NOBr(g) slow k1 K-1 k2


43 Theoretical rate law based on step 2: rate=k2[NOBr2][NO]
Problem: depends on the concentration of an intermediate species. Intermediates unstable, have low/unknown concentrations. Find a way to remove term from rate law. Assuming equilibrium in step 1, express [NOBr2] in terms of NOBr and Br2.

44 By definition of equilibrium, forward rate equals the reverse rate:
k1[NO][Br2] = k-1[NOBr2] Therefore, the overall rate law becomes: Rate = k2 [NO][Br2][NO]=k[NO]2[Br2] Note that the final rate law is consistent with the experimentally observed rate law. k1 k-1

45 Catalysis

46 Catalysts. . . Change the rate of a chemical equation without itself undergoing a permanent chemical change in the process. Two types: Homogeneous: Present in the same phase as reacting molecules. Lower activation energy for the reaction. Increase the number of effective collisions. From the Arrenhenius equation, catalysts increase k by increasing A or decreasing Ea Usually provides a completely different mechanism for the reaction. If an intermediate is added, the activation energy for both steps must be lower than the activation energy for the uncatalyzed reaction.

47 Heterogeneous catalysts. . .
Exists in a different phase than reactants. Many industrial catalysts are heterogeneous. Adsorption is the binding of reactant molecules to the catalyst surface. Happens due to the high reactivity of atoms or ions on the surface of the solid. Use laser disk Side A chapter 8


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