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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 > 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

2 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)

3 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 = J/K.mol

4 b. Example: 2 N 2 O 5 4 NO 2 + O 2 Ea? 4. Reminder about the Dilution Equation Example: 5 ml of 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) x x x x x x x x x x

5 5.Today’s Reactions: KIO 3 + NaHSO 3 rate = k[IO 3 - ] n [HSO 3 - ] m IO HSO > I - + 3SO H + fast IO I - + 6H > 3I H 2 O fast I HSO H 2 O > 3I - + SO H + slow Starch + I > 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) = [ M]/73s = 1.72 x 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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