RATES OF REACTION SUROVIEC SPRING 2014 Chapter 13

I. Reaction Rates Chemical Kinetics: the investigation of the rate at which reactions occur Rate: how fast a quantity changes with time

I. Reaction Rates Rate = change in quantity / time elapsed

A. Average Rate Vs. Instantaneous Rate

B. Writing Rate Reactions Consider the rate reaction: 2NO 2 (g)  2NO (g) + O 2 (g)

General form aA + bB  cC + dD

Example Give the relative rates of disappearance of the reactants and formation of products for each of the following reactions:  2O 3 (g)  3O 2 (g)  2HOF (g)  2HF (g) + O 2 (g)

Example A  2B Rate increases or decreases as there is less A to react How is rate of change of [A] related to rate of change of [B]? Find rate of change of [A] for time internal from 10.0 to 20.0 s What is the instantaneous rate when [B] = 0.750M time[B] (mol/L) Rate (  [B]/  t)

Example N 2 (g) + 3H 2 (g)  2NH 3 (g) What are the relative rates? If –  [H 2 ]/  t = 4.5X10 -4 M/min, What is –  [N 2 ]/  t? What is  [NH 3 ]/  t?

II. Rate Laws Why look at initial rates? Ex. 2N 2 O 5  4NO 2 + O 2 [N 2 O 5 ]Rate (M/s)

General Form of the Rate Law aA + bB  cC + dD Rate = k[A] x [B] y

A. Method of Initial Rates Initial rate of reaction is measured with several different starting amount of reactants

Example The reaction between ozone and nitrogen dioxide at 231 K is first order in both [NO 2 ] and [O 3 ]. 2 NO 2 (g) + O 3 (g)  N 2 O 5 (g) + O 2 (g) 1. Write the rate equation for the reaction 2. If the concentration of NO 3 is tripled, what is the change in the reaction rates? 3. What is the effect on the reaction rate if the concentration of the O 3 is halved?

Example 2NO(g) + O 2 (g)  2NO 2 (g) 1. Determine the order of the reaction for each reactant 2. Write the rate equation for the reaction 3. Calculate the rate constant 4. Calculate the rate in M/s at the instant when [NO] = M and [O 2 ] = M [NO] (mol/L)[O 2 ] (mol/L)Rate of [NO] disappearing (mol/L. s) X X X 10 -5

Example A + B  C What is the rate equation? Run #[A] i [B] i Initial Rate (M/s) X X X10-2

Base 10 log and Natural Log Base 10 log: 10 2 = 100  log 100 = _______ 10 3 = 1000  log 1000 = _______ log 45 = 1.65  =_________ log x = y  10 y =x Natural Log: base is e = ….. e 1 = ln( ) = 1 ln(e)=_______ ln(45) = 3.81  e 3.81 =_________ ln x = y  e y =x

Important properties of both log and ln log ab = log a + log b log a/b = log a – log b log a b = b(log a)

Example Rate = k[A] 2 [B] x

III. Concentration-time relationships Useful to know  How long a reaction is going to proceed to reach concentration of interest for product or reactant  After a given time what are the concentrations of the reactants and the products?

A. 1 st order reactions A  B Rate = k[A] Integrated rate eqn for 1 st order reactions: Useful for 3 reasons  Can calculate k if we know [A]/[A] o  if k and [A] o are known then we can determine the amount of material expected after time t ([A])  If k is known than after time t we can calculate the fraction [A]/[A] o

B. 2 nd order reactions Rate = k[A] 2 Integrated rate eqn for 2 nd order reactions: 1/[A] = kt + 1/[A] o OR 1/[A] – 1[A] o = kt

Example Ammonium cyanate, NH 4 NCO, rearranges in water to give urea, (NH 2 ) 2 CO. NH 4 NCO (aq)  (NH 2 ) 2 CO (aq) The rate equation for this process is: Rate = k [NH 4 NCO] 2 where k = L/mol. min. If the original concentration of NH 4 NCO in solution is 0.229M how long will it take for the concentration of decrease to 0.180M?

time (min)[sucrose]ln [sucrose]1/[sucrose] Example Sucrose breaks down to form glucose and fructose. C 12 H 22 O 11 (aq) +H 2 O (l)  2C 6 H 12 O 6 (aq) 1.Plot ln sucrose versus time and 1/[sucrose] versus time. Find the order of the reaction 2.Write the rate constant for the reaction and calculate k 3.Find concentration of sucrose After 175 min

C. Zero Order Reactions Rate =k[A] 0 Integrated rate eqn for 0 th order reactions: [A] o – [A] = kt

Table 15-1, p.719

D. Half-Life Time required for half of a reactant to be consumed 1. 1 st order

Fig. 15-9, p.720

Example The rate equation for the decomposition of producing NO 2 and O 2 is –  [ N 2 O 5 ]/  t = k[N 2 O 5 ] The value of k = 5.0 X s -1 for the reaction at a known temperature. 1. Calculate the half-life of N 2 O 5 2. How long does it take for the N 2 O 5 concentration to drop to 1/10 th of its original value?

2. 2 nd order and 0 th order A  B Rate = k[A] 2 A  B Rate = k

IV. Activation Energy NO 2 (g) + F 2 (g)  FNO 2 (g) + F (g) What makes this reaction occur? What makes this reaction not occur?

A. Collision Theory Reactant Molecules must collide with each other Reactant molecules must collide with sufficient energy Reactant molecules must have proper orientation 2 questions: What makes a reaction go faster? What does sufficient energy mean?

A. What makes a reaction go faster? 1.2.

B. What does sufficient energy mean? RXN: AB + C  A + BC Mechanism:

C. How can we measure E a ? From collision theory Rate of reaction = (collision freq)X(fraction with proper orientation)X (fraction of collisions with enough energy) k = Ae -Ea/RT Arrhenuis equation

Example When heated to a high temperature, cyclobutane, C 4 H 8, decomposes to ethylene. C 4 H 8 (g)  2C 2 H 4 (g) The activation energy, E a for this reaction is 260 kJ/mol. At 800K the ate constant k = s -1. Determine the value of k at 850K.

D. What is a catalyst? A substance added to a reaction to speed it up, but not consumed in the reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2 ---> 2 H 2 O + O 2 Uncatalyzed reaction Catalyzed reaction

V. Reaction mechanisms AB + C  A + BC Possible mechanisms

V. Reaction Mechanisms Molecularity Elementary step

A. Mechanisms and Rate Laws 1. For net reactions, no simple relationship can be assumed between reaction coefficients and reactant orders 2. For each elementary step reactant coefficients DO equal reactant orders

A. Mechanisms and Rate Laws 3. The rate of a net reaction is essentially that of the slowest step (the rate determining step)

A. Mechanisms and Rate Laws 4. When a fast step proceeds a slow step the fast step can be assumed to be at equilibrium RXN: H 2 (g) + CO (g)  H 2 CO (g) Step 1. H 2 (g)  2 H (g) fast Step 2. H (g)  HCO (g) slow Step 3. H (g)  H 2 CO (g) fast

A. Mechanisms and Rate Laws 5. A catalyst can provide a new mechanistic pathway for a reaction 2Ce 4+ (aq) + Tl + (aq)  2Ce 3+ (aq) + Tl 3+ (aq) Mechanism Rate equation Add Mn 2+ catalyst

In General 1. Measure reaction rates 1. Through initial rates 2. Graphically concentraion vs. time 2. Formulate the rate law 3. Postulate the mechanism 1. Cannot prove, can only disprove