Chemical Kinetics Honors Unit 11 Chemical Kinetics Chemical

Chemical Kinetics Honors Unit 11

Chemical Kinetics § Chemical equations do not give us information on how fast a reaction goes from reactants to products. § We can use thermodynamics to tell if a reaction is product – or reactant – favored but how do we know about the speed?

Chemical Kinetics § KINETICS = The study of reaction rates and the mechanism (the way the reaction proceeds) ***Only kinetics will tell us how fast the reaction happens!

Reaction Rate n

Rate of Reaction n Reaction rate = change in concentration of a reactant or product over time

Types of Rates 1) Initial rate = rate at “time zero” 2) Average rate = the rate over a given time interval 3) Instantaneous rate = the slope of the tangent line at a given point

Types of Rates

Expressing a Rate

Collision Theory of Reactants § Reactions occur when molecules collide to exchange or rearrange atoms § Effective collisions occur when molecules have correct energy and orientation

Factors Affecting Rates 1. Concentration and physical state of reactants and products o A Rate Law relates the rate of the reaction to the concentration of the reactants 2. Temperature 3. Catalysts o Catalysts are substances that speed up a reaction but are unchanged by the reaction

Writing Rate Laws For a. A + b. B c. C + d. D The rate law is: Rate = k[A]m[B]n § k is the rate constant § The exponents must be determined by performing an experiment. § They are NOT derived from the stoichiometry coefficients in an overall chemical equation!!

Rate Laws & Orders of Reactions Rate Law for a reaction: Rate = k[A]m[B]n[C]p The exponents m, n, and p: § Are the reaction order § Can be 0, 1, 2, or fractions (may be other whole numbers in fictional examples) § Must be determined by experimentation Overall Order = sum of m, n, and p

Interpreting Rate Laws n n n For reaction a. A products If m = 1 (1 st order) m 1 Rate = k[A] Rate = k [A] If [A] doubles, then the rate doubles (goes up by a factor of 2) If m = 2 (2 nd order) Rate = k [A]2 If [A] doubles, then rate quadruples (increases rate by a factor of 4) If m = 0 (zero order) Rate = k [A]0 If [A] doubles, rate does not change!

Rate Constant, k Relates rate and concentration at a given temperature. Overall Order 0 1 Units of k M time-1 Time-1 2 3 M-1 Time-1 M-2 Time-1 General formula for units of k: M(1 - overall order) time-1

Example #1 (part a): Rate Law Problem The initial rate of decomposition of acetaldehyde, CH 3 CHO, was measured at a series of different concentrations and at a constant temperature. CH 3 CHO(g) CH 4(g) + CO(g) [CH 3 CHO] (mol/L) 0. 162 0. 195 0. 273 0. 410 0. 518 Rate (mol/L*min) 3. 15 4. 56 8. 94 20. 2 35. 2 (a) Using the data, determine the order of the reaction; that is, determine the value of m in the equation Rate = k[CH 3 CHO]m

Strategy You are looking at how the concentration affects the rate so compare the two in a proportion! Use the equation: Pick any two points from the given data!

Example #1: CH 3 CHO(g) CH 4(g) + CO(g) CH 3 CHO (mol/L) 0. 162 0. 195 0. 273 0. 410 0. 518 Rate (mol/L*min) 3. 15 4. 56 8. 94 20. 2 35. 2 Rate = k[CH 3 CHO]m

Significance of Rate Laws Rate of rxn = k[CH 3 CHO]2 Here the rate goes up by FOUR when the initial concentration doubles. Therefore, the reaction is SECOND order overall.

Example #1 (part b & c): Using the same set of data from part a, and knowing the order of the reaction, determine: b) the value of the rate constant, k (w/ units!) c) the rate of the reaction when [CH 3 CHO] = 0. 452 mol/L Strategy: ¨ Use any set of data to find k. ¨ Solve for rate using k, rate order equation, and given concentration.

Example #1 (part b & c):

Example #2: The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800 o. C. 2 NO(g) + 2 H 2(g) N 2(g) + 2 H 2 O(g) (a) Determine the order of the reaction with respect to both reactants and write the rate law (b) calculate the value of the rate constant, and (c) determine the rate of formation of product when [NO]=0. 0024 M and [H 2]=0. 0042 M. Strategy: Choose two experiments where the concentration of one reactant is constant and the other is changing! Ø Solve for m and n separately Ø

Example #2: Experiment 1 2 3 4 5 6 [NO] 0. 0010 0. 0020 0. 0030 0. 0040 [H 2] 0. 0040 0. 0010 0. 0020 0. 0030 0. 12 0. 48 1. 08 0. 48 0. 96 1. 44

Example #2: Experiment 1 2 4 5 [NO] 0. 0010 0. 0020 0. 0040 [H 2] 0. 0040 0. 0010 0. 0020 0. 12 0. 48 0. 96

Example #2: Experiment 1 2 4 5 [NO] 0. 0010 0. 0020 0. 0040 [H 2] 0. 0040 0. 0010 0. 0020 0. 12 0. 48 0. 96

Example #3: The initial rate of a reaction A + B C was measured with the results below. Write the rate law, the value of the rate constant (with units), and the rate of reaction when [A] = 0. 050 M and [B] = 0. 100 M. Experiment 1 2 3 [A] (M) [B] (M) 0. 1 0. 2 0. 1 Initial Rate (M/s) 4. 0 x 10 -5 16. 0 x 10 -5

Example #3:

Potential Energy Diagrams n Molecules need a minimum amount of energy for a reaction to take place. ¨ Activation energy (Ea) – the minimum amount of energy that the reacting species must possess to undergo a specific reaction n Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products. ¨ Formation depends on the activation energy & the correct geometry (orientation)

Potential Energy Diagrams

Potential Energy Diagrams

Potential Energy Diagrams

Catalyzed Pathway Catalysts lower activation energy!!!

Reaction Mechanisms Mechanism – how reactants are converted to products at the molecular level Most reactions DO NOT occur in a single step! They occur as a series of elementary steps (a single step in a reaction).

Rate Determining Step Rate determining step – the slowest step in a reaction COCl 2 (g) COCl (g) + COCl 2 (g) COCl (g) + Cl 2 (g) 2 COCl (g) 2 CO (g) + 2 Cl (g) Cl 2 (g) fast slow fast

Rate Determining Step COCl 2 (g) COCl (g) + COCl 2 (g) COCl (g) + Cl 2 (g) 2 COCl (g) 2 CO (g) + 2 Cl (g) Cl 2 (g) 2 COCl 2 (g) 2 Cl 2 (g) + 2 CO (g) Adding elementary steps gives the net (or overall) reaction! fast slow fast

Intermediates § Intermediates are produced in one elementary step but reacted in another NO (g) + O 3 (g) NO 2 (g) + O 2 (g) NO 2 (g) + O (g) NO (g) + O 2 (g) O 3 (g) + O (g) 2 O 2 (g)

Catalysts n Catalyst – a reactant in an elementary step but unchanged at the end of the reaction ¨ A substance that speeds up the reaction but is not permanently changed by the reaction ¨ Both an original reactant and a final product NO (g) + O 3 (g) NO 2 (g) + O 2 (g) NO 2 (g) + O (g) NO (g) + O 2 (g) O 3 (g) + O (g) 2 O 2 (g)

Example #4 Cl 2 (g) 2 Cl (g) n Fast Cl (g) + CHCl 3 (g) CCl 3 (g) + HCl (g) Slow CCl 3 (g) + Cl (g) CCl 4 (g) Fast Identify: ¨ The rate determining step ¨ The overall (net) reaction ¨ The identity of any intermediates ¨ The identity of any catalysts

Example #5 n H 2 O 2(aq) + I 1 -(aq) H 2 O(l) + IO 1 -(aq) Slow H 2 O 2(aq) + IO 1 -(aq) H 2 O(l) + O 2(g) + I 1 - (aq) Fast Identify: ¨ The rate determining step ¨ The overall (net) reaction ¨ The identity of any intermediates ¨ The identity of any catalysts

Example #6 n O 3 (g) + Cl (g) O 2 (g) + Cl. O (g) Slow Cl. O (g) + O (g) Cl (g) + O 2 (g) Fast Identify: ¨ The rate determining step ¨ The overall (net) reaction ¨ The identity of any intermediates ¨ The identity of any catalysts

Pb 2+ (aq) + 2 Cl- (aq) Pb. Cl 2 (s) Chemical Equilibrium UNIT 11 (PART 2)

What is equilibrium? Definition (dictionary. com): a state of rest or balance due to the equal action of opposing forces Chemical Equilibrium: A process where a forward and reverse reaction occur at equal rates ***Not all chemical reactions are reversible!!!

General Characteristics of Equilibrium 1. DYNAMIC (in constant motion) 2. REVERSIBLE 3. Can be approached from either direction (reaction can run in the forward direction or the reverse direction) Pink to blue Co(H 2 O)6 Cl 2 Co(H 2 O)4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O)4 Cl 2 + 2 H 2 O Co(H 2 O)6 Cl 2 Video Clip - Cobalt Chloride Complex Equilibrium

Characteristics of Dynamic Equilibrium Fe 3+ + SCN- Fe. SCN 2+ + After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.

Examples of Chemical Equilibria Phase changes such as: H 2 O(s) H 2 O(liq)

Graphing Dynamic Equilibrium achieved when product and reactant concentrations remain constant!!

The Equilibrium Expression, Keq = equilibrium constant (for a given T) Brackets "[ ]" = concentration (molarity) "a, b, c, and d" = coefficients from balanced equation The "c" in Kc = concentration (Kc = a special Keq based on concentration)

There are two cases when a species is not shown in the equilibrium expression: #1: SOLIDS – (s) after the formula #2: pure LIQUIDS – (l) after the formula

Example #7 Write the equilibrium expression for the oxidationreduction reaction occurring between iron(III) chloride and tin(II) chloride: 2 Fe. Cl 3 (aq) + Sn. Cl 2 (aq) 2 Fe. Cl 2 (aq) + Sn. Cl 4 (aq)

Example #8 Write the equilibrium expression for the replacement of silver ions by copper: Cu (s) + 2 Ag+ (aq) Cu 2+ (aq) + 2 Ag (s)

Le Chatelier’s Principle “If a system at equilibrium is stressed, the system tends to shift its equilibrium position to counter the effect of the stress. ”

How does a “stress” influence equilibrium? The impact of addition of reactants on reaction rate

The Seesaw Analogy

“Stresses” to a System 1) 2) 3) 4) 5) Changes in amount of species Changes in pressure or volume Changes in temperature Adding an inert substance Adding a catalyst

“Stresses” (factors) that can cause changes at equilibrium 1) Changes in amount of species o Add reactant; system shifts to the RIGHT! (produces more products) o Add product; system shifts to the LEFT! (produces more reactants) o Remove reactant; system shifts to LEFT! o Remove product; system shifts to RIGHT!

Example #9: Predict the direction of shift based on the following concentration changes for the reaction: CH 4(g) + 2 S 2(g) CS 2(g) + 2 H 2 S(g) (A) Some S 2(g) is added. (B) Some CS 2(g) is added. (C) Some H 2 S(g) is removed. (D) Some argon gas (an inert gas) is added.

“Stresses” (factors) that can cause changes at equilibrium 2) Changes in pressure or volume o If P goes down (same as V goes up), system shifts to increased # of moles of gas o If P goes up (same as V goes down), system shifts to decreased # of moles of gas

Example #10: Predict the effect of increasing pressure (decreasing volume) on each of the following reactions. (a) CH 4(g) + 2 S 2(g) CS 2(g) + 2 H 2 S(g) (b) H 2(g) + Br 2(g) 2 HBr(g) (c) CO 2(g) + C(s) 2 CO(g) (d) PCl 5(g) PCl 3(g) + Cl 2(g)

“Stresses” (factors) that can cause changes at equilibrium 3) Changes in temperature o Write heat as a product (exothermic) or reactant (endothermic) o System shifts to get rid of added heat: o System shifts LEFT for exo. reactions as T goes up o System shifts RIGHT for endo. reactions as T goes up

Example #11: Predict the effect of increasing temperature on each of the following reactions: (a) CO (g) + 3 H 2 (g) CH 4 (g) + H 2 O (g) ΔH < 0 (b) CO 2(g) + C (s) 2 CO (g) ΔH > 0 (c) 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) EXO. (d) 2 H 2 O (g) 2 H 2(g) + O 2(g) ENDO.

“Stresses” (factors) that can cause changes at equilibrium 4) Adding an inert substance o If a substance is NOT in the reaction (or in the Keq expression) any changes will have NO EFFECT on equilibrium! Ex. ) If CO 2 is added to the system below, there would be no effect on the equilibrium. H 2 (g) + Br 2 (g) 2 HBr (g)

“Stresses” (factors) that can cause changes at equilibrium 5) Adding a catalyst o What do catalysts do? They increase the rate of the reaction! o Adding a catalyst will not affect equilibrium. It only changes the rate at which you reach equilibrium.

Example #12: How can the reaction below be shifted to the right? List all possibilities! CO(g) + 2 H 2 (g) <---> CH 3 OH (g)

Example #13: How can the reaction below be shifted to the right? This process (the Haber process) is used in the industry to produce ammonia. List all possibilities! N 2(g) + 3 H 2(g) 2 NH 3(g) ∆H = negative

Haber-Bosch Process for NH 3 N 2(g) + 3 H 2(g) 2 NH 3(g) + heat K = 3. 5 x 108 at 298 K