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The Rate of Chemical Reactions 1.Rate Laws a.For generic reaction: aA + bB cC + dD b. Rate = k[A] x [B] y [Units of Rate always = M/s = mol/L s] c.Details 1.Reactants decrease (-); Products increase (+) 2.For simple forward only reactions: ONLY REACTANTS IN RATE LAW 3.Exponents (x and y) in the rate law must be determined experimentally N 2 O 4 -------> 2 NO 2 Rate = k[N 2 O 4 ] 1 (first order) 3 NO -------> N 2 O + NO 2 Rate = k[NO] 2 (second order) 4.Properties of the rate constant “k” Specific to each reaction and changes with Temperature Units of “k” depend on what order the reaction is

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2.Method of Initial Rates a.Method for determining the “order” of the reactants (exponents in rate law) b.First order reactants = [A] 1 Double the reactant concentration -------> Doubles the rate of reaction c.Second order reactants = [A] 2 Double the reactant concentration -------> quadruples the rate of reaction d.Zero order reactants = [A] 0 (changing concentration has no effect on rate)

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3. Temperature, Reaction Rates, and the Arrhenius Equation a.Taking the natural log of each side gives us another form of the equation that gives a linear plot. lnk vs. 1/T gives straight line with slope = -E a /R and intercept = ln(A) k = rate constant A = frequency factor (combines z and p) E a = activation energy T = temperature in Kelvins R = gas constant = 8.3145 J/K.mol

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b. Example: 2 N 2 O 5 4 NO 2 + O 2 Ea? 4. Reminder about the Dilution Equation Example: 5 ml of 0.015 M KIO 3 is added to 2.5 ml HSO 3 - and 7.5 ml H 2 O T( o C)T(K)1/T(K)k (s -1 )ln(k) 202933.41x10 -3 2.0x10 -5 -10.82 303033.30x10 -3 7.3x10 -5 -9.53 403133.19x10 -3 2.7x10 -4 -8.22 503233.10x10 -3 9.1x10 -4 -7.00 603333.00x10 -3 2.9x10 -3 -5.84

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5.Today’s Reactions: KIO 3 + NaHSO 3 rate = k[IO 3 - ] n [HSO 3 - ] m IO 3 - + 3HSO 3 - -------> I - + 3SO 4 2- + 3H + fast IO 3 - + 8I - + 6H + -------> 3I 3 - + 3H 2 O fast I 3 - + HSO 3 - + H 2 O -------> 3I - + SO 4 - + 3H + slow Starch + I 3 - -------> Colored Complex a)The “slow” reaction occurs until all of the HSO 3 - is gone, then Blue color forms b)We will determine the rate of the “slow reaction” by timing how long it takes for the Blue Complex to appear. c)Rate = ([HSO 3 - ] initial )/(time till Blue) = [0.0125 M]/73s = 1.72 x 10 -4 mol/Ls 6.Reminder: a)Experiments 1-5 at different concentrations give us Rate Law and k b)Experiments 6-9 at different temperatures give us E a

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