chemistry Unit 5.4

Topic 5.4: Chemical kinetics II recall that rates of reaction may be expressed by empirical rate equations of the form: rate = k[A]m[B]n, where m and n are 0, 1 or 2 define the terms rate constant and order of reaction and understand that these are experimentally determined the concept of molecularity is not required deduce rate equations from given experimental initial rate data recall that reactions with a lar ge activation energy will have a small rate constant students will be expected to be familiar with the Arrhenius equation but not to recall it understand that many reactions take place in several steps, one of which will be the rate-determining step understand that it is sometimes possible to deduce information regarding the mechanism of a chemical reaction from kinetic data understand that many reactions proceed through a transition state select and describe a suitable experimental technique for fol lowing a given reaction present and interpret the results of kinetic measurements in graphical form define the term half-life and recall that this is constant for any given first-order reaction. questions requiring a knowledge of the products of the radioactive decay will not be asked Rate equations A + B Products are 0, 1 or 2)

If the rate of reaction depends upon the conc. of A and B:

m

Reaction Rate = k[A] [B]

n

(m and n

Rate constant and order of reaction m n If the reaction Rate = k[A] [B] then reaction is of order m with respect to A and of order n with respect to B. The overall order of  reaction is (m+n). The proportionality constant k is called the rate constant for the reaction. The rate constant and order o f reaction are experimentally determined. Rate equations from experimental data m Rate equations are of the form rate = k[A] where k is a proportionality constant. A graph of rate of reaction against m [concentration] is plotted and the gradient of the graph will will give you the constant of the reaction k. k. A data table may yield a rate equation. E.g. A reacts with B to form C. From the table below below find the rate equation equation and calculate calculate the rate constant. constant.

experiment

[A]/mol -3 dm

[B]/mol -3 dm

initial rate/ mol -3 -1 dm s

1

1.00

1.00

4.00

2

2.00

1.00

8.00

3

1.00

2.00

16.0

In experiment 1 and 2 doubling [A] multiplies rate by 2 so rate proportional to [A] 2 2 In experiment 1 and 3 doubling [B] multiplies rate by 2 so rate proportional to [B] 2 so rate = k[A][B] . 2 -3 -3 2 -3 -1 k = [A][B]  /rate = 1.00mol dm * (1.00mol dm ) /4.00mol dm s -3 k = 0.250 mol dm s Activation energy and rate constant Some bonds in a molecule must break before it can react and form new bonds. Energy is needed to break these bonds is called the activation energy. Reactant molecules must must be given enough energy to pass the activation energy barrier to react. The activation energy and the rate constant are linked by the Arrhenius equation. -Eact/RT k=Ae where k=rate constant, constant, e = the base of natural logarithms, A is a constant constant for any given reaction, Eact = the activation energy, energy, R = the gas constant, constant, T = the temperature temperature in K. The Arrhenius equation shows that the r ate constant (k) decreases if the activation energy (Eact) increases. A reaction will have a small rate constant if it has a large activation energy. The activation energy for a reaction can be calculated as follows. -Eact/RT ln k = ln Ae -Eact/RT ln k = ln A + ln e ln k = ln A - Eact/RT log k=log A - Eact/2.3RT log k=log A - Eact/2.3R * 1/T If k is calculated for different values of T the n a plot of log k against 1/T gives a line of gradient = - Eact/2.3R. The rate determining step in a reaction Reactions often occur in several steps, one of which, the rate determining slow step, is likely to control the overall rate o f reaction. e.g. for an S N1 reaction two steps are involved + RX -------> R + X step 1 slow + R + OH -----> ROH step 2 fast

The rate depends on the slow step 1. rate = k[RX] first order E.G. RX=(CH3)3CBr For an SN2 reaction there is a rate determining slow step involving two species RX + OH ------> HO--R--X rate = k[RX][OH ] second order E.G. RX = CH 3Br

Mechanisms and kinetic data The mechanism for a reaction can be proposed with help from kinetic data but some speculation is needed. 1. The rate equation gives us information about what reacts in the rate determining step. 2. Sensible products must be suggested for the rate determining step. 3. If more molecules of reactant remain and more product molecules are still to be formed more s teps must be proposed. e.g. What is the mechanism for the following reaction CH3CH2I + NH3 -----> CH3CH2NH2 + HI if rate = k[CH k[CH3CH2I] CH3CH2I is a halogenoalkane so likely likely to take part in a nucleophilic substitution. substitution. NH3 is a nucleophile because of its lone pair of  electrons on the nitrogen atom. The rate equation shows that that it is first order so the slow step in the mechanism mechanism must involve only CH3CH2-I. +

-

CH3CH2---I -----> CH3CH2 + I A nucleophilic attack by ammonia is now possible in a fast step; +

+

NH3 + CH3CH2 -----> CH3CH2NH3 a final fast step might be loss of hydrogen ion; +

CH3CH2H2N----H -----> CH3CH2NH2 + H

+

For the reaction 2ICl(g) + H2(g) ----> 2HCl(g) + I2(g) Experiments show that rate = k[ICl(g)][H2(g)] so the rate determining step involves 1 molecule of ICl and one of H 2 ICl(g) + H2(g) ----> products possible products are HCl because it is a product of the overall reaction and HI because the elements hydrogen and iodine are left over. ICl(g) + H2(g) ----> HCl(g) + HI(g) HI(g) slow step check full equation to see what is unaccounted for one molecule of ICl still has to react so ICl(g) + HI(g) ----> HCl(g) + I2(g) fast step The kinetic data has led to a 2 step mechanism for the reaction. The transition state When particles collide the breaking breaking of bonds and the formation of new bonds bonds may take place at the same time. Half way through the process an intermediate called the transition state is formed. The transition state has more energy than the the reactants. reactants. The transition state has more energy than t he products. e.g. for the reaction between a hydrogen molecule and a chlorine radical. . . H-H + Cl H- - - H - - - Cl ----> H + H-Cl reactants energy = E transition state energy > E products energy < E for exothermic reaction Experimental techniques for following reactions Reactions can be followed by * sampling, quenching and titrating (suitable for acid base reactions) * measuring gas volumes (suitable for reactions involving a change in gas volume) * polarimetry (suitable for r eactions involving optically active substances) * measuring conductivity (suitable for reactions producing o r consuming ions) * colorimetry (suitable for reactions involving coloured substances) * dilatometry (suitable for reactions involving r eactions in which liquids change volumes) Presenting and interpreting kinetics graphs From a graph of concentration against time the initial slope of the graph is the initial rate of reaction. A graph of rate against concentration will be a straight line through the origin if the rate is pr oportional to the concentration. Half-life The time taken for the reaction to go to half completion is called the half -life of the reaction t1/2. The half life of a 1st ord er reaction is independent of the initial initial concentration. 1st order reactions have a constant half life. t 1/2 = 0.69/k where k = the rate constant for the reaction 3. (a) Explain what is meant “Rate of reaction” and “Overall order of a reaction”



Rate of increase/decrease/change in

concentration of reactants/products with time 

Sum of the powers to which the

concentrations are raised in the rate equation OR number of species involved in (up to and including) the rate determining step OR sum of partial orders if illustrated with a general rate equation (b) gases A and B react according to the equation

A + 3B

AB3

(i) State the order of reaction with respect to each of the reactants 1 + 3: double [A], doubles rate so order 1

Expt Expt

1 + 2: double [B], four x rate so order

2

OR

keeping [B] constant doubles rate so order

Double [A]

1

Double [B] keeping [A] constant four x rate so order 2 (ii) Write the rate equation for the reaction between A and B. Rate = k [A] [B]

2

(iii) Use the experimental data from Experiment 1 to calculate the rate constant, including units

=

(iv) Suggest a possible mechanism for the reaction between A and B, leading to the formation of AB 3. Identify the ratedetermining step.

OR

OR

(c) The rate constant, k , for the reaction in (b) was measured at different temperatures. Plot a graph of log 10 k against 1/T

(ii) The Arrhenius equation can be written

where

Calculate the gradient of the graph and hence calculate the value of activation energy, E a.

= -5750 (K) (+)5750 x 2.30 x 8.31= (+)110 kJ mol -1 2. Two gases, A and B, react according to the equation A(g) + 2B(g) AB2(g) A series of kinetics experiments performed at constant temperature gave the following results:

Ea =

→

(a) (i) Calculate, showing your working, the order of reaction with respect to A and to B. Expt 1 + 2: as [B] doubles rate x4 so second order (wrt B) [A] constant rate x4 so second order (wrt B) Expt 1 + 3: as [A] doubles rate x2 so first order (wrt A) [B] constant rate x2 so first order (wrt A) 

OR

As [B] doubles with



OR

As [A] doubles with

(ii) Write the rate equation for the reaction. 2 rate = k [A] [B] (iii) Calculate the rate constant, k , for the reaction in experiment 3 , stating its units. 2 -2 6 -1 0.000195/0.1x0.1 = 0.195mol dm s

k=

(b) (i) Explain, in terms of collision theory, why the rate of r eaction increases with an increase in temperature. Increasing T means molecules have/collide with greater energy so a greater proportion /more of the molecules collide with/have E > E a /the activation energy so a greater proportion of the collisions are successful OR more of the collisions are successful/more successful collisions in a given time  



(ii) Suggest, with an explanation, the least number of steps which is likely to exist in the mechanism of the reaction between A and B. (at least) two steps Simultaneous collision of   

three particles is unlikely OR valid mechanism e.g. A+B → AB fast AB + B → AB2 slow OR A+B → AB slow AB + B →AB2 fast (c) The variation of the rate constant, k , with change in temperature is given by the

4

Value of slope = -1.2 ×10 Multiply by -8.31 Divide by 1000 to give 104 (kJ -1 mol )  

Arrhenius equation: where A is a constant. In a series of experiments performed at various temperatures T to determine the rate constant, k , for the decomposition of a gas X, a graph of ln k against 1/ T  T gave a straight line of slope - Ea/R 



Use the graph to calculate the value of the activation energy, in kJ mol – 1, for the  – 1. decomposition of X. The value of the gas constant R = 8.31 J K – 1 mol – 

4. 2-bromo-2-methylbutane reacts with aqueous sodium hydroxide in a substitution reaction

(a) The rate of reaction can be followed by measuring the concentration of 2-bro mo-2-methylbutane at various times. In one such experiment, a known amount of 2-bromo-2-methylbutane was added to a large excess of aqueous sodium hydroxide. The following results were obtained

(i) Plot a graph of the concentration of 2-bromo-2-methylbutane on the y (vertical) axis against time on the x (horizontal) axis.

(ii) Show TWO successive half-life measurements on your graph and write their values below. nd 1min) 2 half life 15min (± 1min)



st

1 half life 15min (±



(iii) What is the order of reaction with respect to 2-bro mo-2-methylbutane? Give a reason for your answer. st 1 order t is constant ½

(b) When the reaction is repeated using equal concentrations of 2-bromo-2-methylbutane and aqueous sodium hydroxide, the same results are obtained. (i) What is the order of reaction with respect to hydroxide ions? Zero (ii) Write the rate equation for the reaction. Rate = k [2-bromo-2-methylbutane] (iii) Write a mechanism for the reaction which is consistent with your rate equation

(c) The reaction between 2-bromobutane, C2H5CHBrCH3 and NaOH(aq) proceeds by the same mechanism as in (b)(iii). Use the mechanism to explain why the reaction of a single optical isomer of 2-bromobutane produces a mixture that is no longer optically active.



The intermediate carbocation

is



Equal attack from either side



planar Therefore racemic mixture produced (i) Overall order of reaction 4. (a) Define the terms sum of the powers to which the concentration (terms) are raised in the rate equation  / number of species involved up to and and including the rate determining step (in the the reaction mechanism) 

(ii) Rate constant



constant (of proportionality) in the rate equation / numerically = rate

-3

when all concs 1 mol dm  /correct example (b) In a kinetic study of the reaction

At a certain temperature the following data were obtained:

 – ions. With reasons. (i) State the order of reaction with respect to CH3I and with respect to OH – ions. Both orders 1 Double concentration of one while other is constant and the rate doubles OR refer to two specific experiments (ii) Write the rate equation for the reaction. 



-

rate = k[CH I][OH ] 3

(iii) Calculate the value of the rate constant for this reaction, stating its units.

e.g.

k

-

= rate/[CH I][OH ] 3

so

k

= 1 x 10

–3

mol

–1

3

dm s

-1

(c) The reaction CH3CH2Br + OH –  CH3CH2OH + Br –  has an SN2 mechanism that proceeds through a transition state. (i) Draw the mechanism, showing the structure of the transition state.

(ii) Draw a reaction profile for this exothermic reaction. Show the energy level of the transition state o n the profile

1. (a) Consider the following table, which shows data for the reaction between reactants A and B.

(i) Define the term order of reaction. The sum of the powers to which the concentrations are raised in the rate equation (ii) Determine, giving reasons, the orders of reaction with respect to A and B. Hence write the rate equation for the reaction 1st order because rate halves as [A] halves in expt. 1 → 2 or [B] constant 2 2nd order because rate quadruples / increases by 2 as [B] doubles in expt. 2 → 3 or [A] constant (iii) Calculate a value for the rate constant and give its units. -2 6 -1 k = 0.008mol dm s (iv) State how, if at all, the value of the rate constant would change if the temperature were increased. (k) increases (b) (i) Draw a Maxwell-Boltzmann curve for a sample of a gas

(iii) The rate of a reaction can also be increased by raising the temperature. Describe how the Maxwell-Boltzmann curve at a higher temperature differs from the curve you have drawn in (i). Peak ( 

more ) to the right Peak lower (iv) Transition metals are important industrial industrial catalysts. Identify an industrial process involving a transition metal metal catalyst and name the catalyst used. Explain why many transition metals and their compounds are successful catalysts. 

OR

Manufacture of ammonia Iron Hydrogenation of oils

Ni/Pt/Pd

OR Manufacture of H2 from CH4 Ni Explanation: uses d orbitals to bond with reactants( at active sites) (v) How do the rate constants for the catalysed and uncatalysed reactions compare? bigger/Higher OR uncatalysed k lower

Catalysed k