{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture 06_Ch15b

# Lecture 06_Ch15b - Summary of Rate Laws(Table 15.6 Key...

This preview shows pages 1–10. Sign up to view the full content.

Summary of Rate Laws Key assumption so far: forward reactions only . This assumption is generally good over short times. (Table 15.6)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Click to edit Master subtitle style Crystal Violet Demo (CH 3 ) 2 N C N(CH 3 ) 2 N(CH 3 ) 2 (CH 3 ) 2 N C N(CH 3 ) 2 N(CH 3 ) 2 OH + + OH -
Integrated Rate Law for Reactions with More than 1 Reactant Example: aA + bB + cC products Overall reaction order = n + m + p How do we solve for all of these unknowns? rxn order w.r.t. A = n w.r.t. B = m w.r.t. C = p Rate = k [A]n [B]m [C]p

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Method of Initial Rates Rate1 = k [A]1n [B]0m [C]0p Rate2 = k [A]2n [B]0m [C]0p Rate2 [A]2 Rate1 [A]1 = n Solve for n: Rate2 [A]2 Rate1 [A]1 = n ln ln Rate2 [A]2 Rate1 [A]1 ln ln n =
Suppose we used a large excess of [B]0 and [C]0 i.e. [B]0 >> [A]0 and [C]0 >> [A]0 This simplifies the rate law, which becomes pseudo-first order : Rate = k ' [A]n where k ' = k [B]0 [C]0 We can then solve for k ' and n in the usual way Determination of k ' at several different [B]0 and [C]0 allows determination of m , p , and k

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
What have we learned about kinetics so far? (2) Reaction rates are expressed quantitatively by rate constants and rate orders , both of which must be determined experimentally. (1) Reaction kinetics are described by rate laws (4) Second-order reactions show a curved decrease of the reactant concentration with time; the half-life is inversely related to the initial reactant concentration (5) Zero-order reactions show a linear decrease of the reactant concentration with time; the half-life is directly related to the initial reactant concentration (3) First order reactions show an exponential decrease of the reactant concentration with time; the half-life is independent of the initial reactant concentration
What have we learned about kinetics? (1) Rate: change in concentration per unit of time (2) Rate expression: for aA + bB cC + dD rate = -d [A] = -d [B] = d [C] = d [D] a*dt b*dt c*dt d*dt (3) Differential rate law: rate=k[A]n[B]m , where k, m, and n have to be experimentally determined (4) Order of reaction w.r.t. A: 1st: rate = k[A]; 2nd: rate = k[A]2; zero: rate = k (5) Integrated rate law: 1st: plot of ln[A] vs t is linear ln[A] = -kt + ln[A]o 2nd: plot of 1/[A] vs t is linear 1/[A] = kt + 1/[A] o zero: plot of [A] vs t is linear [A] = -kt + [A] o (6) Integrated rate law for reactions with more than 1 reactant: Pick one reactant to change and hold the others constant (use large excess). Combine rate constant and constant concentrations into k’, the pseudo-rate constant, simplifying the rate law with respect to the chosen reactant.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter #15 – Chemical Kinetics 15.1) Reaction Rates 15.2) Rate Laws: Introduction 15.3) Determining the Form of the Rate Law 15.4) Integrated Rate Law 15.5) Rate Laws: Summary 15.6) Reaction Mechanisms 15.7) The Steady-State Approximation 15.8) A Model for Chemical Kinetics 15.9) Catalysis
Why do we study reaction kinetics?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern