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General Chemistry 2 (CHEM 215) Week 3 Class Notes

by: snufkin

General Chemistry 2 (CHEM 215) Week 3 Class Notes CHEM 215

Marketplace > San Francisco State University > Chemistry > CHEM 215 > General Chemistry 2 CHEM 215 Week 3 Class Notes
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Week 3 lecture notes from CHEM 215 (General Chemistry 2 course) covering half life, activation energy, and equilibrium.
General Chemistry 2
Russel Jensen
Class Notes
Chemistry, General Chemistry, ratemechanism, halflife, activationenergy, Equilibrium
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This 6 page Class Notes was uploaded by snufkin on Saturday September 17, 2016. The Class Notes belongs to CHEM 215 at San Francisco State University taught by Russel Jensen in Fall 2016. Since its upload, it has received 24 views. For similar materials see General Chemistry 2 in Chemistry at San Francisco State University.


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Date Created: 09/17/16
Chemistry Notes: Week 3    ❈ Explain the form and function of an integrated rate law    Integrated Rate Law:  ­Provides a concentration as a function of time  ­Predicts concentration at any time    Ex. SO​ Cl​  →   SO​  + Cl​    2​ 2​ 2​ 2 −4 What is [SO​ Cl2​] 2​ t= 2390s, given ​k=2.90× 10 s −1 and initial  ­­­­­­­­​ oncentration is 0.0225M?    rate= k[A]  rate M/s 1 k= [A] → M = s → 1st order    ­kt [SO​ 2​​2​ t​ [SO​ 2​​ 2​ o​​   ­(2.90  10 s  1) (2390s)  =0.0225M e​ =0.01125M      ❈ Define half life and carry out related calculations    ❈ Use integrated rate laws to calculate [A]​ , [A]​  , t​ t, ao​ fractional  ­­­​ mounts    Half Life:  ­The time required for the concentration of a reactant to be half its  ­­­­­­­­­​ nitial value          1 Order:  Rate Law:  Integrated Rate Law:  k Units  t 2   2nd  rate= k[A]​   1 1 1 1   [A]t =kt +  [A]o   Ms   k[A] o   1st  rate= k[A]  ln[A]​t​­kt + ln[A]​ o     or  1   0.693   [A]t= [A]​  e​   s k o​ M [A]o 0th  rate= k  [A]t= ­kt + [A]​ o s   2k       Ex. NO​  2​composes to O​  and NO2​t 500K in a second order process with  ­­­­­​t = 71s and [​NO​ ]​ 2​ o​00M. How long will it take for 85%  2 ­­­­­​completion?    [NO​ 2​ t​(1 ­ 0.85)[NO​ ]​2​ o  = 0.15 NO​ ]​2​ o      1 =kt +  1   [A] t [A]o Need to know k:  1 1 k=  t2[NO ]2 o =  71.5 × 1.00M   1 =0.014 Ms   1 1 [A] =kt +  [A]   t o 1 ­  1 ​ 0.014 1 ​  0.15[NO ]2 o 1.00M Ms 0.66M −1M −1 t=  −1 −1   0.014M s =4.0 × 10​  s or 400s  Arrhenius Equation:    k= Ae​ ­Ea/ RT R= 8.314 J/Kmol  A=frequency factor    Factors that affect reaction rate:  ­collision orientation  ­collision rate (concentration, temperature, kinetic energy)    How to quantify these factors:      Ea= Activation energy  ΔH= Enthalpy of reaction  Peak= Transition state    ­Reaction rate is not related to ΔH because of the activation energy  ­Activation energy is closely tied to temperature and rate constant k  ­Changing temperature, measuring k, calculating Ea is the experimentation    Activation Energy:  ­If the reaction progresses, ΔH is not related to how fast it progresses  k1 R × ln( ) k2 Ea=    1 − 1 T 2 T 1   Catalysts:  ­Lowers activation energy by stabilizing transition state (speeds up  ­­­­­­­­­​reaction)  ­Not consumed in reaction  ­Can also force correct orientation  ­Can form intermediates    ❈ Calculate activation energies using rate data at two temperatures    ❈ Describe the dynamic nature of equilibrium and shifts in equilibrium in  ­­­​ erms of rate​ forward​nd rate​ reverse​  ❈ Differentiate between reaction quotient, Q, and Keq    In Equilibrium:  ­Reactions go forward and backward at the same rate  (rate​ forward​ rate​reverse​ ­Constant reaction and product concentrations    In General: mA + nB ⇋ xC + yD    m​ n​ x​ y In equilibrium: K​f​A]​ [B]​  = K​ r​C]​ [D]​    kf [C] [D] y k​eq​ k ​ [A] [B] n  r K​eq​s unitless  K​eq​k​ c​when using concentration  k​eq​k​ p​hen using partial pressure    ***Do not include solvents (ex. H​ O(l)) or2​olids  Not in equilibrium:  ­It takes time to re­establish equilibrium after starting or perturbing a  ­­­­­­­­­r​ eaction    [C] [D] y Q=  [A] [B] n     Establishing equilibrium:    When K​  > eq​ reaction going forward  When K​  < eq​ reaction going backward  When K​  = Q : reaction in equilibrium  eq​   Using ICE Tables to Establish Equilibrium Quantitatively:    ICE= Initial, Change, Equilibrium    ­Fill in initial concentration for partial pressure  ­Fill in change with an unknown (use stoichiometry, negative  ­­­­­­­­­­­­­­­­​changes for disappearing reagents)  ­Sum the initial and change for equilibrium, solve for unknown    Ex. For the reaction 2NH​  ⇋ N​  + 33​  , t2​ initia2​[NH​ ] is 1.30M, and 3​e  ­­­­­­​equilibrium [NH​ ] is 3​5M. What is K​ ?  eq​   2NH​  ⇋3​​  + 2​​    2                 NH​   N​   H​    3 2 2   Initial [C]  1.30  0  0  Change  ­2x  x  3x  Equilibrium [C]  1.30 ­ 2x  0 + x  0 + 3x    Set equal to equilibrium concentration    1.30 ­ 2x = 0.50 0 + 0.40 0 + 3x  2x= 1.30 ­ 0.50 x= 0.40 x= 1.2  x= 0.40    3 [N2][H 2 (0.40)(1.2) K​eq​  [NH ] 2 ​ (0.50)2 ​  2.8  3


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