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by: Leslie Pike

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# Chem 222 Week 3 notes Chem 222

Leslie Pike
WKU
GPA 3.9

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Notes over Chapter 13 and kinematics equations
COURSE
College Chemistry 2
PROF.
Darwin Dahl
TYPE
Class Notes
PAGES
2
WORDS
CONCEPTS
Chemistry
KARMA
25 ?

## Popular in Chemistry

This 2 page Class Notes was uploaded by Leslie Pike on Friday February 12, 2016. The Class Notes belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 15 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.

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Date Created: 02/12/16
Chem 222 notes Week 3 Chapter 13: Kinematics Kinetics deal with how fast a reaction takes place. Thermodynamics deal with whether or not a reaction will occur. Kinetics and thermodynamics are not related to each other. The rate of reaction is the change in molarity with respect to time, dM/dt. For a reaction A  2B, the rate expression for the rate of reaction would be: Rate= -Δ[A]/Δt Rate= Δ[B]/2Δt Reactants have a negative sign in front of them, and the expression is divided by any coefficient that may be in front of a reactant or product. Four things affect rate: 1. Reactant concentration 2. Presence of a catalyst 3. Temperature 4. Surface area A rate law describes the rate of reaction in terms of the concentration of the reactants. RATE LAW IS DETERMINED EXPERIMENTALLY. A sample rate law for an equation: 2A + 3B  5C Rate=k[A] [B] y Notice that C, a product, is not a part of this rate law. Notice that the balancing coefficients in front of A and B do not necessarily correspond to X and Y. X and Y are usually small integer values, such as 0, 1, or 2. The order of a reaction is the sum of the exponents. Say X is 2 and Y is 1, the reaction above would be a third-order reaction. First-order Reactions: For a first order equation A  B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k[A]. A is raised to the first power because the sum of all exponents must equal 1 since this a first-order reaction, and there is only one exponent so it must be 1. The equation for determining the concentration after a set amount of time for a first-order reaction is: Ln[A] tLn[A] -0t The expression for the half-life is: t1/2n(2)/k. This half life does not change with respect to time. Second-order Reactions: For a second order equation A  B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k[A] . A is raised to the second power because the sum of all exponents must equal 2 since this a second-order reaction, and there is only one exponent so it must be 2. The equation for determining the concentration after a set amount of time for a second-order reaction is: 1/[A] t1/[A] +0t The expression for the half-life is: t =1/2[A] . Th0s half-life increases with respect to time. Zero-order Reactions: For a zero order equation A  B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k. A is raised to the zero power because the sum of all exponents must equal 0 since this a zero-order reaction. Anything raised to the 0 power is 1; therefore, it disappears out of the equation. The equation for determining the concentration after a set amount of time for a zero-order reaction is: [A]t=[A] 0kt The expression for the half-life is: t =1/2 /2k.0This half-life decreases with respect to time. Calculating the Rate Constant, k: K=Zfp, where  Z=collision frequency  F=fraction of collisions with minimum energy  P= orientation factor Ln(k)=-E /aT + Ln(A), where A is the frequency factor and R=8.314J/(mol*K)

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