Chapter 14 Chemical Kinetics

Review Section of Chapter 14 Test Net Ionic Equations

Reaction Rate The rate of a chemical reaction is measured as the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time.

Reaction Rate The rate of a chemical reaction is measured as the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time. ∆[ ] ∆time Rate =

Reaction Rate The rate of a chemical reaction is measured as the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time. ∆[ ] ∆time Rate = What units would we use for rate?

Reaction Rate The rate of a chemical reaction is measured as the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time. 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g) ∆[ ] ∆time Rate =

Reaction Rate The rate of a chemical reaction is measured as the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time. 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g) ∆[ ] ∆time Rate = How could the rate be expressed for this reaction in terms of H 2 O 2 ?

2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g)

What is the rate of the reaction from 0s to 2.16 x 10 4 s?

2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g) What is the average rate of appearance of O 2 from 0s to 2.16 x 10 4 s? 1.16 x mol O 2 L -1 s -1

General Rate of Reaction a A + b B → c C + d D Rate of reaction = rate of disappearance of reactants We can use the coefficients in the equation to compare the reaction rates for all the substances in the reaction. Rate of reaction = rate of appearance (formation) of products or

15-1 The Rate of a Chemical Reaction Rate is change of concentration with time. 2 Fe 3+ (aq) + Sn 2+ (aq) → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s [Fe 2+ ] = M Rate of formation of Fe 2+ = = = 2.6 x M s -1 Δ[Fe 2+ ] ΔtΔt M 38.5 s ∆t = 38.5 s ∆ [Fe 2+ ] = ( – 0) M

Rates of Chemical Reaction 2 Fe 3+ (aq) + Sn 2+ (aq) → 2 Fe 2+ (aq) + Sn 4+ (aq) Rate of formation of Fe 2+ = 2.6 x mol L -1 s -1 What is the rate of formation of Sn 4+ ? What is the rate of disappearance of Fe 3+ ? 1.3 x mol Sn 4+ L -1 s x mol Fe 3+ L -1 s -1

What does the slope of the line represent?

What is the concentration at 100s for the reaction: 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g)? Given: Δ[H 2 O 2 ] = (1.7 x M s -1 ) (∆t) Rate = 1.7 x M s -1 ΔtΔt = Δ[H 2 O 2 ] ∆[H 2 O 2 ] = (1.7 x M s -1 )(100 s) = 0.17M = 2.15 M = 2.32 M M [H 2 O 2 ] 100 s [H 2 O 2 ] i = 2.32 M

What does it mean when the rate of a reaction reaches zero? For a normal reaction it means that one or more of the reactants are used up and the reaction has stopped. For a reversible reaction it means that the reaction has reached equilibrium.

Factors Affecting Reaction Rates 1.The nature of the reacting substances.

Factors Affecting Reaction Rates 2.The state of subdivision of the reacting substances.

Lycopodium Powder

Factors Affecting Reaction Rates 3. The temperature of the reacting substances.

Factors Affecting Reaction Rates 4. The concentration of the reacting substances. Air (21% oxygen) 100% oxygen

Factors Affecting Reaction Rates 5.The presence of a catalyst. Catalysts speed up reactions but are left unchanged by the reaction.

The Rate Law a A + b B …. → g G + h H …. Rate = k [A] m [B] n …. Rate constant = k (k is constant for a particular reaction at a specific temperature) Order of A = mOrder of B = n Overall order of reaction = m + n + ….

Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

After finding the trials to compare: A reactant is zero order if the change in concentration of that reactant produces no effect on the rate. A reaction is first order if doubling the concentration of that reactant causes the rate to double. A reactant is nth order if doubling the concentration of that reactant causes an 2 n increase in rate. Note that the rate constant does not depend on concentration. Concentration and Rate Summary

Use the data provided to write the rate law and indicate the order of the reaction with respect to HgCl 2 and C 2 O 4 2- and also the overall order of the reaction.

First determine the order of HgCl 2

Next determine the order of C 2 O 4 2-

Now write the rate law and determine the order of the reaction.

Calculate the rate constant “k” and its units. Initial rate of disappearance HgCl 2 mol L -1 min -1

What is the average rate of disappearance of C 2 O 4 2- in trial 1? Initial rate of disappearance HgCl 2 mol L -1 min -1

Use the data provided to write the rate law and indicate the order of the reaction with respect to NO 2 and CO (support your answers). Also give the overall order of the reaction.

Calculate the rate constant “k” and its units.

What is the average rate of disappearance of CO in trial 2?

Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

Two Factors -Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). -Orientation of reactants must allow formation of new bonds.

2HI → H 2 + 2I

Concentration and Collision Theory Why does an increase in concentration cause an increase in reaction rate?

Concentration and Collision Theory Why does an increase in concentration cause an increase in reaction rate? Increasing the concentration increases the number of collisions and therefore there are more collisions leading to product.

Temperature and Collision Theory Why does a temperature increase cause the reaction rate to increase?

Temperature and Collision Theory Why does a temperature increase cause the reaction rate to increase? At higher temperatures there are more collisions and a greater percentage of the collisions have the energy necessary to create a successful collision.

Activation Energy The activation energy is the minimum amount of energy necessary for a reaction to occur.

Temperature and Activation Energy (E a ) Figure (Page 432)

Activation Energy The activation energy can also be thought of as the energy necessary to form an activated complex during a collision between reactants.

Transition State Theory The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

The activated complex is a transition state between reactants and products where old bonds have begun to break and new bonds have started to form. It cannot be isolated.

Determining the Activation Energy “The Arrhenius Equation” -Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). -Orientation of reactants must allow formation of new bonds.

Arrhenius Equation k = rate constant A = frequency factor E a = activation energy T = temperature R = ideal gas constant frequency factor: a value in the Arrhenius equation indicating how many collisions have the correct orientation to lead to products. A

Arrhenius Equation Determination of Activation Energy Graphical determination of activation energy (E a ). –plot the ln k on the y-axis. –Plot 1/T (use Kelvin temperature) on the x-axis.

Arrhenius Equation Determination of Activation Energy A plot of ln k versus 1/T (using Kelvin) will have: –slope of –Ea/R –y-intercept of the graph is = ln A.

Slope = - EaEa R ln A

x-axisy-axis

x 1,y 1 = 1.25 x 10 -3, x 2,y 2 = 1.78 x 10 -3,

For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant (k) and the slower rate.

Disregard R, T and A and focus on k and E a ln k is proportional to ‒ E a Because Ea is negative a higher activation energy results in a lower rate constant (k). Hidden Teacher Only Slide

2O 3  3O 2 -A chemical equation like the one above does not tell us how reactants become products - it is simply a summary of the overall reaction.

The reaction: 2O 3  3O 2 -Is proposed to occur through the two step process given below: O 3  O 2 + O O 3 + O  2O 2 This two step process is an example of a reaction mechanism

Reaction Mechanisms A reaction mechanism is a step-by-step description of a chemical reaction. Each step is called an elementary reaction.

Often Used Terms Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. Molecularity: the number of species that must collide to produce the reaction indicated by that step. Elementary Step: A step within a reaction mechanism whose rate law can be written from its molecularity.

Elementary Steps Molecularity: the number of molecules present in an elementary step. –Unimolecular: one molecule in the elementary step, –Bimolecular: two molecules in the elementary step, and –Termolecular: three molecules in the elementary step. It is not common to see termolecular processes (statistically improbable). Reaction Mechanisms

Rate Laws for Elementary Steps The rate law of an elementary step is determined by its molecularity: –Unimolecular processes are first order, –Bimolecular processes are second order, and –Termolecular processes are third order. Reaction Mechanisms

Rate Laws for Elementary Steps Reaction Mechanisms

The Rate Determining Step

Rate-Determining Step In a reaction mechanism, the rate determining step is the slowest step. It therefore determines the rate of reaction.

Reaction Mechanisms Reaction mechanisms must be consistent with: 1.Stoichiometry for the overall reaction. 2.The experimentally determined rate law.

Reaction mechanism must be consistent with the stoichiometry of the overall reaction. Is the mechanism below consistent with the overall reaction above? NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g)  NO(g) + CO 2 (g)

Determining the stoichiometry of a reaction mechanism. Page 439

The reaction mechanism must also support the rate law. Reaction Mechanisms

Rate Laws for Multistep Mechanisms with an initial fast step. Consider the reaction: 2NO(g) + Br 2 (g)  2NOBr(g) Reaction Mechanisms

Mechanisms with an Initial Fast Step 2NO(g) + Br 2 (g)  2NOBr(g) The experimentally determined rate law is Rate = k[NO] 2 [Br 2 ] Consider the following mechanism Reaction Mechanisms

The rate law is (based on Step 2): Rate = k 2 [NOBr 2 ][NO] The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable). NOBr 2 is an unstable intermediate, so we express the concentration of NOBr 2 in terms of NO and Br 2 Since there is an equilibrium in step 1 we have

By definition of equilibrium: Therefore, the overall rate law becomes Note the final rate law is consistent with the experimentally observed rate law.

Student Example: Determine the rate law for the reaction and the balanced equation given the mechanism below: 2NO ↔ N 2 O 2 fast N 2 O 2 + O 2 → 2NO 2 slow

Use the slide that follows to show the students the method for determining the slow step for each mechanism. It can only be used for students practicing how to relate rate law to the mechanism if you treat the first step of Mechanism B and C as an equilibrium and [H 2 O] is treated as a liquid and is therefore = 1.

Assume the rate law is: Rate = [H 2 O 2 ][H 3 O + ][I - ] Which step would be the rate – determining step? Page 439

Catalysts A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy, but which is left unchanged by the reaction.

What is the overall reaction? O 3  O 2 + O O 3 + O  2O 2

What is the overall reaction?

Identify the intermediates.

NO is a catalyst A homogeneous catalyst is of the same phase as the reacting substances. It lowers the activation energy by forming intermediates which allow the reaction to proceed by a different pathway.

Heterogeneous Catalysts A heterogeneous catalyst is of a different phase than the reacting substances. It lowers the activation energy by providing a surface on which the reaction can occur.

Inhibitor An inhibitor decreases the rate of a reaction. It often does this by rendering a catalyst ineffective. Catalyst “poisoning” occurs when a catalytic converter is exposed to exhaust containing substances that coat the working surfaces, encapsulating the catalyst so that it cannot contact and treat the exhaust. The most notable contaminant is lead, so vehicles equipped with catalytic converters can only be run on unleaded gasoline.

Inhibitor An inhibitor decreases the rate of a reaction.

Reaction Rate Lab – Part A Use different containers for Reaction Mixtures I and II. Don’t forget the starch.

Reaction Rate Lab – Part B In part B you will perform trial 1 at a temperature assigned to you by your instructor.

Reaction Rate Lab – Part C In part C you will perform trial 1 using a catalyst.