# 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 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:

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

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. 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

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

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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