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Chem 112 - Exam 1 (Kinetics) Study Guide

by: Christopher Cooke

Chem 112 - Exam 1 (Kinetics) Study Guide CHEM 112

Marketplace > Pennsylvania State University > Chemistry > CHEM 112 > Chem 112 Exam 1 Kinetics Study Guide
Christopher Cooke
Penn State
GPA 3.5

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This is a comprehensive study guide for the first exam in Chem 112 this semester, covering all aspects of Kinetics as discussed in class.
Chemical Principles II
Dr. Raymond Shaak
Study Guide
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This 6 page Study Guide was uploaded by Christopher Cooke on Friday September 9, 2016. The Study Guide belongs to CHEM 112 at Pennsylvania State University taught by Dr. Raymond Shaak in Fall 2016. Since its upload, it has received 213 views. For similar materials see Chemical Principles II in Chemistry at Pennsylvania State University.


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Date Created: 09/09/16
Chem 112 – Exam 1 Study Guide Kinetics Key Concepts for the Exam  Rates of chemical reactions define behavior, reactivity, properties, and application.  Express, define and report reaction rates.  Evaluate reaction rates. Collision Theory  Collisions between molecules will lead to chemical reactions o Not all collisions will necessarily lead to a chemical reaction o Collision frequency, orientation, and chemical reactivity determine whether or not a collision will cause a reaction.  Heat, catalysts, concentration, and phase will influence the reaction rate. o Rate α # effective c/llisions time o Rate α [x]; rate = k[x] o k = the proportionality or rate constant  Only molecules with sufficient (activation or threshold) energy (denoted as Ea) can react Interpreting Reaction Rate Graphs  General Principles o The point where the graph begins is the energy level of the reactants. o The point where the graph ends is the energy level of the products. o Any peak(s) in the graph denote(s) a transition state. o Any trough(s) in the graph (not at the very end) denote the presence of intermediates, or any compounds that appear during the reaction and are phased out by the end of the reaction. They are neither reactants nor products.  Intermediates occur during the steps that together will form an overall reaction. For example, consider 2NO(g) + O (g)2 2NO. The “steps” may be: o NO + NO  N O 2 2 o N 2 2 O  2NO 2  Here, N 2 2s the intermediate. o The highest peak in the graph denotes which step in the reaction is the rate-limiting step, or the slowest step in the reaction. This step can be used to determine the rate and the overall order of the reaction. o The difference between the rate-limiting step and the reactants on the graph is the activation or threshold energy (E ). a o The difference between the products and the reactants denote ΔE, the change in energy during the reaction.  If ΔE is positive, the reaction is endothermic.  If ΔE is negative, the reaction is exothermic.  Which reaction is exothermic, and which is endothermic?  Correctly label this graph. Is it exothermic or endothermic? How many transition states are there? How many intermediates are there, if any? What is ΔE? What is E a Which is the rate-limiting step? The Arrhenius Equation  Remember: ΔE and reaction rate are NOT CORRELATED.  The linearized Arrhenius equation is as follows, in y = mx+b form: o ln(k) = (-E aR)(1/T) + ln(A)  If there is more than one k value (rate constant), then the formula is: o ln(k 1k 2 = (E aR)(1/T –21/T ) 1  Activation energy is typically ~100kJ/mol (half as strong as an atomic bond) Reaction rate  Rate = Δ[X]/Δt – units are meters per second (m/s)  To fully express the rate of a given formula, defined as: αA + ßB = γC + δD, where the reactants are A and B, the products are C and D, and α, ß, γ and δ represent their coefficients,  The rate law can be expressed as: - (ΔA/αΔt) = -(ΔB/ßΔt) = (ΔC/γΔt) = (ΔD/δΔt); o all reaction rates are positive Rate Law  Rate = k[x] o Recall: αA + ßB = γC + δD o Extrapolated, Rate = -Δ[A]/Δt = k[A] [B] ; this is the generic rate law  Rate is always positive and measured in m/s  x: order with respect to A  y: order with respect to B o x and y are NOT the same as α and ß  x + y: overall order of the reaction ***Keep in mind for the following three sections: It is impossible to tell if a reaction is elementary from looking at the equation. If a reaction is elementary, it will be specified. If otherwise, it will not be mentioned. Do not assume a reaction is elementary just by looking at it!*** Rate Law for Unimolecular Elementary Reactions  Unimolecular Elementary Reaction: A  products (e.g. 2H O  22 + O ) 2 2  Rate law: k[A] , x=1  The units for k in a unimolecular elementary reaction are M/s Rate Law for Bimolecular Elementary Reactions  Bimolecular Elementary Reaction: A + B  products (2H + O  2H2O) 2 2 x y  Rate law: k[A] [B] , x=y=1; overall order=2  Can also be: 2A  products (2O + O 2 2O )2 3 2  Rate law: k[A]  The units for k in a bimolecular elementary reaction are M s ( / )1 -1 1Ms Rate Law for Termolecular Elementary Reactions  Termolecular Elementary Reaction: A+B+C  products x y z  Rate law: k[A] [B] [C] , x=y=z=1; overall order=3  Can also be: A+2B, 2A+B, or 3A  products x 2 2 x 3  Rates: k[A] [B] , k[A] [B] , or k[A]  The units for k in a termolecular elementary reaction are M s (1/M s)-1 2 Rate-Determining Steps for Non-Elementary Reactions  Non-elementary reactions can be broken down into elementary steps.  The slowest step is the rate-determining step.  Given a series of steps: o Write out the slowest step using the guidelines above. o If there is an intermediate, go back to the first step and define its concentration in terms of the reactants. o Replace the intermediate with its reactant-defined form, so that only reactants appear in the final rate. (See – Day 6 Notes for clarification if you are confused.) The Method of Initial Rates  Given a table of initial rates and concentrations, we may be able to determine the rate law expression for a given reaction.  In a given reaction, if the concentration of [X] is altered and the initial rate is similarly altered (compared to some sort of baseline), we can extrapolate the rate law expression. o For example, if doubling [X] doubles the rate, the reaction is first order with respect to [X]. o If doubling [X] quadruples the rate, the reaction is second order with respect to [X]. o If doubling [X] has no obvious effect on the rate, the reaction is of no order with respect to [X] (its exponent, then, would be zero in the rate expression, meaning it can be left off). Integrated Rate Laws  For a first order reaction, the integrated rate law can be depicted as follows: o ln[A] – ln [A o = ln [A]/[A ] o -kt Linearized Graph of a First-Order Reaction using the Integrated Rate Law  For a second-order reaction, the integrated rate law can be depicted as follows: o 1/[A] – 1/[A o = kt Linearized Graph of a Second-Order Reaction using the Integrated Rate Law  Half-life for a first-order reaction o Half-life: time it takes for the concentration of a reactant to reach ½ of its initial value  ln(½) = -kt ½  t½= 0.693/k o For a first-order reaction, half-life does not depend on concentration  Half-life for a second-order reaction o 1/[A] = kt + 1/[A o  t½; let [A] = ½[Ao] o 2/[Ao] = kt½+ 1/[A o o 1/[Ao] = kt ½ t =½1/k[A ] o Catalysts and Catalysis  Catalysts change the speed of reactions without essentially changing their own form in the process – they do not react during the course of the reaction. o May change E orathe mechanism of the reaction, but will not affect ΔE.  Homogeneous catalysts are in the same phase as the reactants (solid, liquid or gas).  Heterogeneous catalysts are in different phases from the reactants.  Enzymes are very specific biological catalysts that accelerate and control reaction sites. o Reactants are known as substrates and the binding regions are known as active sites. o The active sites are very particularly formed such that only very specific compounds may attach to them and undergo chemical reactions. ***Do the homework problems on ANGEL for extra practice! They will help you immensely as you prepare for this exam.***


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