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Chem 112 - Week 3 Notes

by: Christopher Cooke

Chem 112 - Week 3 Notes CHEM 112

Christopher Cooke
Penn State
GPA 3.5

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Notes from week 3 of Chem 112, covering first-order and second-order reaction rate laws, integrated rate laws, the method of initial rates, and catalysts/catalysis.
Chemical Principles II
Dr. Raymond Shaak
Class Notes
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This 4 page Class Notes was uploaded by Christopher Cooke on Friday September 9, 2016. The Class Notes belongs to CHEM 112 at Pennsylvania State University taught by Dr. Raymond Shaak in Fall 2016. Since its upload, it has received 29 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 9-7-16 What we have so far  Reaction with reactant coefficients α and ß  αA + ßB = γC + δD  Overall rate expression: Rate = -ΔA/αΔt = -ΔB/ßΔt  Rate laws relate exponents to orders of elementary reactions that are part of mechanisms o Not necessarily observable or able to be directly measured  Rate = k[A]x[B]y  Key: combine this knowledge to experimentally validate this proof Acquiring useful experimental data  Method of initial rates o Why?  Reactants decrease over time, products increase  A+B  Products  Vary [A]o and [B]o – keep one constant, vary the other, then vice versa  Determine initial rates – measure the change in concentration change over short periods of time  Observe the effects of the concentration changes on the reaction rate  Analyze the data  Integrated rate laws o A  Products x  Rate = -ΔA/αΔt = k[A]  After some derivation:  ln[A] – ln [A o = ln [A]/[A o = -kt, assuming a first order reaction. o Concentrations of compounds involved in chemical reactions change over time, as does the rate o Consider the rate as a function of concentration as Δt = 0 Half-life of a first-order reaction  Half-life: time it takes for the concentration of a reactant to reach ½ of its initial value o ln(½) = -kt ½  t½= 0.693/k  For a first-order reaction, half-life does not depend on concentration Assume a second-order reaction; A  products 2  Rate = k[A] = -Δ[A]/Δt  After some derivation:  1/[A] – 1/[A ]o= kt Half-life of a second-order reaction  1/[A] = kt + 1/[A ] o t ;½let [A] = ½[A ] o  2/[A o = kt ½ 1/[A ] o  1/[A o = kt ½ t =½1/k[A ] o REMEMBER: Units of k can tell you the order of the reaction. --------------------------------------------------------------------------------------------------------------------- --------------------- Chem 112 9-9-16 Speeding up reactions  Recall the video of the H O 2e2ction in the first class. o Each reaction is the same, but all the reaction rates are different. o How does this happen?  Water-Gas-Shift reaction o CO + H O2 CO + H2 2 o Same reaction, but different pathways. o How does this happen? Introduction to catalysis  Catalysts change the speed of reactions without essentially changing their own form in the process. o Does not change during the reaction; does not react o Tends to lower E or change the mechanism of the reaction, but does a not change thermodynamic state functions.  ΔE does not change. o Can also cause intermediates to appear Which species is the catalyst?  2Cl + 2O  3ClO + 2O 2  2ClO + hv  2Cl + 2O  O + O  O 2  Cl is the catalyst; it goes in and comes out unchanged  Overall Reaction: 2O  3O 2 Homogeneous catalysis  Catalysts are in the same phase as the reactants  2H 2 (2)  2H O 2l) + O (g2 o Adding in I (liquid) o I (l) + 2H2O2(l)  2H 2 (l) + O (2) + I (l) - o What does I do to this reaction?  Color change  Implies an intermediate was formed  IO intermediate decomposes to form I , whic2 reacts with - - I to form I 3 which causes the color change  Speeds up change from liquid to gas  Bubbles/foam  Steam (heat)  exothermic  Back to colorless at the end of the reaction Heterogeneous catalysts  Catalysts are in a different phase than the reactants  2H 2 (2)  2H O2(l) + O (2) o Add MnO 2 o MnO (2) +2H O (2)2 2H O (l2 + O (g) 2 MnO (s) 2 o What does MnO do 2o this reaction?  Steam (heat)  exothermic reaction  No color change  no intermediates Enzyme catalysis  Enzymes are very specific biological catalysts o Proteins o Accelerate and control reaction rates o Reactants are called substrates and the binding regions are called active sites  Large “blob” of amino acids arranged in such a way where certain molecules fit very precisely inside the active site of the enzyme  How much does lowering the activation energy really affect reaction rates? o By a factor of >100,000!


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