Download presentation

Presentation is loading. Please wait.

Published byMaxwell Geels Modified over 2 years ago

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 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
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 = 8.3145 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 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
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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on forward rate agreement quote Download ppt on historical monuments in india Download ppt on heritage of india Download ppt on android operating system Download ppt on mind controlled robotic arms manufacturing Ppt on asian continent animals Ppt on l&t finance Ppt on rc phase shift oscillator Ppt on combination of resistances eve Ppt on edge detection filter