Chem%20162-2011%20Lecture%205

Chem%20162-2011%20Lecture%205 - CHEMISTRY 162-2011 LECTURE...

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Chem 162-2011 Lecture 5 1 CHEMISTRY 162-2011 LECTURE 5 ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in EXAMS QUIZ
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Chem 162-2011 Lecture 5 2 PLAN FOR TODAY : COMPLETE LECTURE 4 CHAPTER 13 (cont.) - CHEMICAL KINETICS H&P 13.7-13.8 Collision theory Effect of temperature on reaction rates
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Chem 162-2011 Lecture 5 3 FORMULAS ET: Rate = speed. Using a diagram with Y being 100 lbs and 110 lbs, and days 0 to 5, show that the increase in weight = slope = (Y 2 – Y 1 )/(X 2 – X 1 ) = +2lb/day. Decrease in weight = -increase in weight = -slope = -(Y 2 – Y 1 )/(X 2 – X 1 ) = -2lb/day. ET: Point out that table in middle of page contains the key formulas in kinetics. Discuss these formulas at beginning of recitation, focusing on 1 o reaction. C 12 H 22 O 11 (sucrose) + H 2 O 2 C 6 H 12 O 6 (glucose) Rate of sucrose disappearance = -( [sucrose])/( time) = -([sucrose f ] - [sucrose i ])/(t f - t i ) Rate of glucose appearance = 2 x Rate of sucrose disappearance aA + bB cC + dD General rate of reaction = -(1/a)( [A]/ t) = -(1/b)( [B]/ t) = (1/c)( [C]/ t) = (1/d)( [D]/ t) Rate or a reaction may be written several ways. Reaction: 2A + 3B + C 2D Rate = -k[A] m [B] n [C] p C/ t = -k[A] m [B] n [C] p d[A]/dt = -k[A] m [B] n [C] p I n t e g r a t e d Reaction Differentiated rate law k Order Reaction Rate** RATE LAW*** (y = mx + b) Half-life * units 0 Avg Rate = -(C 2 -C 1 )/(t 2 -t 1 ) Rate = k[C] o =k [C] t = -kt + [C] o t 1/2 = [C] o /2k M 1 s -1 1 Avg Rate = -(C 2 -C 1 )/(t 2 -t 1 ) Rate = k[C] 1 ln[C] t = -kt + ln[C] o t 1/2 = 0.693/k M o s -1 2 Avg Rate = -(C 2 -C 1 )/(t 2 -t 1 ) Rate = k[C] 2 1/[C] t = kt + 1/[C] o t 1/2 = 1/(k[C] o ) M -1 s -1 ** Rate of appearance = +slope; rate of disappearance = -rate of appearance = -slope. *** Differentiated Rate Law may be for more than one component, e.g., Rate = k[C] 1 [D] 2 *Half-lives : For a zero order reaction, each successive half-life is ½ the time of the preceding one. For a first order reaction, each successive half-life is equal in time to the preceding one. For a second order reaction, each successive half-life is double time of the preceding one. Arrhenius equation: k = Ae (-Ea/RT) A = frequency factor = combination of steric factor and collisional frequency E a = energy of activation ln k 2 - ln k 1 = (-E a /RT 2 ) - (-E a /RT 1 ) ln k = -E a /RT + ln A ln (k 2 /k 1 ) = -(E a /R)[(1/T 2 ) - (1/T 1 )]
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Chem 162-2011 Lecture 5 4 LECTURE 4 CONTINUED PRIORITIES IN SOLVING KINETICS PROBLEMS FIND REACTION ORDER Use method of initial rates - If initial rates are available* *If initial rates are not available they can be calculated, but that is probably not worth the time on an exam. Evaluation of half-life data Identifying integrated rate law straight line graph - From actual graph - From description of graph Recognizing k units Plugging numbers into equations to get a constant “k”. (the “constant-constant” method) - Integrated rate law equation - Differentiated rate law equation - Half-life equation FIND THE RATE CONSTANT After identifying reaction order, plug numbers into - Differentiated rate law equation or - Integrated rate law equation or - Half-life equation Slope or negative slope of integrated rate law graph Note: Diff. rate law equation contains conc. and rate terms.
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This note was uploaded on 03/08/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Chem%20162-2011%20Lecture%205 - CHEMISTRY 162-2011 LECTURE...

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