Chapter 15: Phenomena. Chapter 15 Chemical Kinetics. Reaction Rates. Reaction Rates R P. Reaction Rates. Rate Laws

Similar documents
AP Chemistry--Chapter 12: Chemical Kinetics

2) Of the following questions, which ones are thermodynamic, rather than kinetic concepts?

Chapter 12. Chemical Kinetics

Chapter 13 Homework Answers

Chapter 14 Homework Answers

CHEMICAL KINETICS: 1. Rate Order Rate law Rate constant Half-life Temperature Dependence

Chapter 14 Chemical Kinetics

Chapter 15: Phenomena

Chapter 14 Chemical Kinetics

Northern Arizona University Exam #1. Section 2, Spring 2006 February 17, 2006

Math 333 Problem Set #2 Solution 14 February 2003

Advanced Organic Chemistry

CHEMICAL KINETICS Rate Order Rate law Rate constant Half-life Molecularity Elementary. Complex Temperature dependence, Steady-state Approximation

A First Course on Kinetics and Reaction Engineering. Class 19 on Unit 18

KINEMATICS IN ONE DIMENSION

Solutions to Assignment 1

Math 2214 Solution Test 1A Spring 2016

04. Kinetics of a second order reaction

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

CHAPTER 12 DIRECT CURRENT CIRCUITS

5.2. The Natural Logarithm. Solution

4.5 Constant Acceleration

CHAPTER 16 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS

APPM 2360 Homework Solutions, Due June 10

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

Chapter 7: Solving Trig Equations

Math 2214 Solution Test 1B Fall 2017

KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

04. Kinetics of a second order reaction

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

1.6. Slopes of Tangents and Instantaneous Rate of Change

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

Math 10B: Mock Mid II. April 13, 2016

המחלקה : ביולוגיה מולקולרית הפקולטה למדעי הטבע

Reaction Order Molecularity. Rate laws, Reaction Orders. Determining Reaction Order. Determining Reaction Order. Determining Reaction Order

6.003 Homework 1. Problems. Due at the beginning of recitation on Wednesday, February 10, 2010.

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010

Vehicle Arrival Models : Headway

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

Some Basic Information about M-S-D Systems

2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

1. VELOCITY AND ACCELERATION

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

Homework 2 Solutions

5.1 - Logarithms and Their Properties

4.6 One Dimensional Kinematics and Integration

d = ½(v o + v f) t distance = ½ (initial velocity + final velocity) time

EXERCISES FOR SECTION 1.5

Physics 20 Lesson 5 Graphical Analysis Acceleration

Phys1112: DC and RC circuits

Predator - Prey Model Trajectories and the nonlinear conservation law

Math 334 Test 1 KEY Spring 2010 Section: 001. Instructor: Scott Glasgow Dates: May 10 and 11.

Sub Module 2.6. Measurement of transient temperature

2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes

Chapter 2. First Order Scalar Equations

Physics 127b: Statistical Mechanics. Fokker-Planck Equation. Time Evolution

13.3 Term structure models

Motion along a Straight Line

Suggested Problem Solutions Associated with Homework #5

Math 116 Second Midterm March 21, 2016

IB Physics Kinematics Worksheet

= ( ) ) or a system of differential equations with continuous parametrization (T = R

( ) a system of differential equations with continuous parametrization ( T = R + These look like, respectively:

AP Calculus BC Chapter 10 Part 1 AP Exam Problems

Today: Graphing. Note: I hope this joke will be funnier (or at least make you roll your eyes and say ugh ) after class. v (miles per hour ) Time

The equation to any straight line can be expressed in the form:

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

EE650R: Reliability Physics of Nanoelectronic Devices Lecture 9:

Sterilization D Values

!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

Displacement ( x) x x x

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

Exam 1 Solutions. 1 Question 1. February 10, Part (A) 1.2 Part (B) To find equilibrium solutions, set P (t) = C = dp

Final Spring 2007

where the coordinate X (t) describes the system motion. X has its origin at the system static equilibrium position (SEP).

Math 115 Final Exam December 14, 2017

Math Final Exam Solutions

SPH3U1 Lesson 03 Kinematics

Math 116 Practice for Exam 2

Notes 8B Day 1 Doubling Time

4.1 - Logarithms and Their Properties

MA Study Guide #1

The average rate of change between two points on a function is d t

8. Basic RL and RC Circuits

ENV 6015 Solution to Mixing Problem Set

AP CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Unsteady Mass- Transfer Models

Section 4.4 Logarithmic Properties

Y 0.4Y 0.45Y Y to a proper ARMA specification.

Physics 221 Fall 2008 Homework #2 Solutions Ch. 2 Due Tues, Sept 9, 2008

Transcription:

Chaper 5: Phenomena Phenomena: The reacion (aq) + B(aq) C(aq) was sudied a wo differen emperaures (98 K and 35 K). For each emperaure he reacion was sared by puing differen concenraions of he 3 species ha ake par in he reacion ino an oherwise empy conainer. The reacion rae was hen measured. Wha paerns do you noice abou he reacion raes? Can reacion raes be prediced? Daa Taken a 98 K Daa Taken a 35 K Exp. [] [B] [C] Rae Exp. [] [B] [C] Rae M M M M M M M M M M M M 3 M M M 3 M M M 4 M M M 4 M M M 5 M M M 5 M M M 6 M 3 M M 6 M 3 M M 7 M M M 4 7 M M M 8 8 4 M M M 6 8 4 M M M 3 9 M M M 4 9 M M M 8.5 M 6 M 3 M.5 M 6 M 3 M Big Idea: The raes of chemical reacions are described by simple expressions ha allow us o predic he composiion of a reacion mixure a anyime. These expressions also sugges he seps in which he reacions akes place. Chaper 5 Chemical Kineics o Thermo Review o Reacion Raes o o Concenraion and o o Explaining Reacion Rae Facors Reacion Raes Reacion Raes Caalys: subsance ha increases he reacion rae wihou being consumed in he reacion. Homogeneous Caalys: caalys ha is in he same phase as he reacans. Heerogeneous Caalys: caalys ha is in a differen phase han he reacans. Reacion Raes: The change in concenraion of one of he reacans or producs divided by he ime inerval over which he change akes place. verage Rae of Consumpion of R: R P verage Rae of Producion of P: Noe: Raes are always posiive, herefore, since he reacans are consumed, a negaive sign mus be added o make he rae posiive. Unique verage Rae (UR) 3 4 Reacion Raes Insananeous Rae of Reacion The bes approximaion o he rae a a single insan is obained by drawing a line angen o he plo of he concenraion agains ime. The slope of he angen line is called he insananeous rae of he reacion. Rae Law: n equaion expressing he insananeous reacion rae in erms of he concenraions, a any insan, of he subsances aking par in he reacion. Noe: k is he rae consan and x,y, are he orders of reacion. Noe: This form of he rae law is called he differenial rae law. Noe: The unis of rae are always, herefore, he unis of k will differ depending on he overall reacion order. 5 6

Things o know abou he rae law: How o find he unis of k Sep : Deermine he order of he Rae laws can conain producs, reacans, caalyss bu usually only saring maerial. reacion Rae laws do no conain inermediaes. Examples for rae=k[][b]: Rae laws can ONLY be deermined Overall order experimenally. Sep : Subrac from he overall order Orders do NOT correlae wih coefficiens in Examples for rae=k[][b]: balanced equaion! Orders can be an inegers, zeros, fracions, Sep 3: Find posiives, OR negaives! # Each species has is own individual reacion Examples for rae=k[][b]: order. The overall reacion order is he sum of he individual orders found in he reacion. 7 8 Order h s nd Deermine he rae law: + B + C D ime Iniial Concenraion [] 4 6 Rae.3.6.9 3 3 3 4 6 4 6 4 6 Rae Law Experimen [] o (M) [B] o (M) [C] o (M) Iniial Rae ( 6 4 6 8 3 3 4 4 6 ) General Rae Law: ) Order wih respec o : 3) Order wih respec o B: 4) Order wih respec o C: 9 Deermine he rae law and k for + B C Experimen [] o (M) [B] o (M) Iniial Rae (.5..74...888 3.5.. Deermining Order (long way) Sep : Find wo experimens in which he concenraions of everyhing, excep one species, is held consan. Sep : Divide he rae laws for hese wo experimens by each oher. Noe: This will cancel ou k and all oher variables excep for he order ha you are rying o deermine. Mah Noe: Sep 3: Solve for order. Mah Noe: I is someimes useful o ake he log of boh sides of he equaion. The log(x y )=ylog(x).

The rae law for he following reacion NO(g) +O (g) NO (g) was experimenally found o be in he form rae=k[no] x [O ] y I was also found ha when he NO concenraion was doubled, he rae of he reacion increases by a facor of 4. In addiion, when boh he O and he NO concenraion were doubled, he rae increases by a facor of 8. Wha is he reacion order of O? a) h b) s c) nd d) 3 rd e) None of he above Deermine he rae law and k for + B C Experimen [] o (M) [B] o (M) Iniial Rae ( 5.. 5. 5... Pseudo Order Reacion: reacion in which he rae law can be simplified because all bu one of he species have virually consan concenraions. 3 4 Zero Order Inegraed Rae Law Firs Order Inegraed Rae Law ) Rae k dx x cons d ) k 5) k 3) 4) d k d k k ) Rae k d ) k 3) d k 4) d k dx x 5) ln ln ln x cons ln k k ln 5 6 Second Order Inegraed Rae Law ) Rae k dx cons x x d ) k 5) k 3) 4) d k d k k The rae law for +B C + D was only found o be dependen on. Using he following daa deermine he rae law and k. (s) [] (M) 5.4..5 4..77 6..5 7 8 3

(s) [] 6 5 4 3 [] (M) 5.4..5 4..77 6..5 ln[] - - ln[].7 -.9 -.6-3. []^- 5 5 [] - (M - ).9 6.7 3 9 Calculae he concenraion of N O afer he firs order decomposiion: N O(g) N (g) + O (g). The rae of decomposiion of N O = k[n O]. The reacion has coninued a 78ºC for. ms, and he iniial concenraion of N O was. M and k = 3.4. -3 5 4 6 8-4 4 6 8 4 6 8 9 Half Life: i akes for he concenraion o drop o half he iniial amoun s Order h Order k k k k ln k ln k ln k ln ln ln k ln ln ln k ln k nd Order k k k Reacion Mechanism: The seps by which a reacion akes place. Elemenary Reacion: One of he reacion seps in an overall reacion. Noe: RTE LWS CN BE DETERMINED FROM ELEMENTRY RECTIONS. Noe: In order o be a valid mechanism he sum of he elemenary reacions mus equal he overall reacion. 3 4 4

Mechanism : Elemenary Reacion(s): Mechanism : Elemenary Reacion(s): Reacion Inermediae: species ha plays a role in he reacion bu does no appear in he reacion. Moleculariy: The number of reacan molecules/aoms aking par in an elemenary reacion. Noe: Differen elemenary reacions usually give differen raes. Noe: In order o be a valid mechanism he rae law derived from he elemenary reacions mus mach he experimenally deermined rae law. Examples: Moleculariy of Moleculariy of 5 6 Use he experimenally found rae laws given below o deermine which reacion is mos likely o occur in a single sep. Experimenally Found a) NO (g)+f (g)no F(g) b) H (g)+br (g)hbr(g) c) NO(g)+O (g)no (g) + O (g) d) NO (g)+co(g) NO(g) + CO (g) Consider he following hypoheical reacion: + B E. The mechanism for his reacion is: () + B C (slow) () B + C D (fas) (3) D E (fas) The rae law consisen wih his mechanism is: a) rae=k[][b] b) rae=k[] [B] c) rae=k[] d) rae=k[][b] e) None of he above 7 8 Relaing he Rae and Equilibrium Consans The reacion NO + Cl NOCl was experimenally found o have he rae law: rae = k[cl ][NO]. Which mechanism could no be he correc mechanism? +B C Equilibrium (Rae forward = Rae reverse ) a) Cl Cl (fas equilibrium) Cl + NO NOCl (slow) b) NO + Cl NOCl (slow) NOCl NOCl 3 +NOCl (fas equilibrium) NOCl 3 NOCl + Cl (fas equilibrium) c) NO + Cl NOCl d) NO N O (fas equilibrium) N O + Cl NOCl (slow) e) ll of he above are possible mechanisms Equilibrium Consan C k K B k k K k Rae Consan 9 3 5

Muli Sep Reacions Wih Unknown Speeds Sep : Wrie an expression for he rae of formaion of one of he final producs (someimes he produc of ineres is specified). If possible selec a produc ha is only in sep. Sep : Use he seady sae approximaion o solve for he concenraion of inermediaes. Sep 3: Plug back ino overall rae equaion. The rae consan for he second-order gasphase reacion HO(g) + H (g) H O(g) + H(g) varies wih he emperaure as shown here: Temperaure (K) Rae Consan ( 373. -9 473.8-8 573. -8 673 4.4-7 Deermine he acivaion energy. a).4-5 b) 4. 3 c) 4. 4 d) None of he above 3 3 Collision Theory (Gases Only) Rae = Collision Frequency Fracion wih Sufficien Energy How would we ge collision frequency Size of he molecules/aoms verage velociy of he molecules/aoms Concenraion of molecules/aoms Problem: lhough he rae has he righ form (rrhenius) i predics a larger reacion rae han is found. Soluion: Serics need o be aken ino accoun. How would we ge fracion wih sufficien energy? Bolzmann disribuion 4 E a = Minimum energy needed for reacion o occur. = Takes ino accoun number of collisions and serics. 33 34 Take way From Chaper 5 civaed Complex Theory (Soluions) Big Idea: The raes of chemical reacions are described by simple expressions ha allow us o predic he composiion of a reacion mixure a anyime. These expressions also sugges he seps in which he reacions akes place. Thermo Review Know he difference beween hermo and kineics Thermodynamics allows us o predic if a reacion will occur. Kineics allows us o predic how fas a reacion will occur. Be able o draw reacion coordinaes along wih labeling reacans, producs, inermediaes, ransiion saes, and acivaion energy. (8, 8, 8) Reacion Raes Be able o explain how a caalys can increase reacion rae. (84, 88) 35 36 6

Take way From Chaper 5 Take way From Chaper 5 Reacion Raes (Coninued) Be able o calculae he average rae of reacion of species given he average rae of reacion of anoher species. Be able o calculae he unique average rae of a reacion. Rae Law Know ha rae laws mus be deermined experimenally. Be able o deermine he order of a reacion and each individual species. Be able o calculae rae law from experimenal daa. (7, 8, 9,, ) Double concenraion and rae says he same: h order Double concenraion and rae doubles: s order Double concenraion and rae quadruples: nd order If none of hese use he mah rick Be able o deermine he unis of he rae consan (8, 4, 5) The unis of concenraion are consan (M) The unis of rae are consans Rae Law (Coninued) Know ha rae problems can be simplified when he concenraion of species is high and essenially unchanging (pseudo order reacions). (34, 37, 5) Be able o use he inegraed rae law o perform calculaions. (4, 45, 46, 48) h order: s order: ln nd order: Be able o idenify he order of he reacion by ploing daa. (8, 9, 3, 3, 33, 36) [] vs. linear h order, slope = -k ln[] vs. linear s order, slope= -k [] - vs. linear nd order, slope = k Be able o calculae half-life/ nd half-life/ec.. 37 38 Take way From Chaper 5 Take way From Chaper 5 (5) Be able o wrie a rae law of an elemenary reacion. (55) Be able o deermine he rae law of muli sep reacions. (56,57) Be able o eliminae inermediaes from muli sep reacions. Equilibrium (59,6,64) Seady sae approximaion (65,66) Concenraion Caalys Be able o draw poenial energy diagram of reacions Temperaure Know ha mos reacions follow rrhenius behavior. (68,73,74,65,75,77,78,) ln ln (coninued) Surface rea Know he ideas behind collision heory (gases). Know he ideas behind acivaed complex heory (soluions). 39 4 7