1. Title: Lesson 5 Reaction Mechanisms Learning Objectives: – Understand what a reaction mechanism is – Understand the relationship between rate equations and reaction mechanisms This will explain how things can be ‘zero’ order – Learn to identify possible reaction mechanisms from suitable rate data.

2. Main Menu Recap  What is the order of reaction with respect to NO 2 (g) and F 2 (g) given the following rate data at a certain temperature? [NO 2 (g)] / mol dm –3 [F 2 (g)] / mol dm –3 Rate / mol dm –3 min –1 0.10.20.1 0.2 0.4 0.10.40.2 Order with respect to NO 2 (g)Order with respect to F 2 (g) A. first B. firstsecond C. secondfirst D. second

3. Main Menu Reaction Mechanisms Explained…with ping pong balls  The task:  As a class seated in a big circle, you need to see how quickly you can pass 10 ping pong balls around the circle.  Most of you can use your hands  One of you must use THE SPATULA OF DOOM  The rules:  If the person you are trying to pass to already has a ball, hold on to it until they have given it away  Each person can only have one ball at a time

4. Main Menu Reflecting on the silly ping-pong balls thing  What were the main factors that influenced how fast the balls could be passed around?  Was each step in the reaction the same speed?  This is a polite way of asking if someone found this activity ‘challenging’  Picture a million ping-pong balls being passed around the circle, which person would have the biggest impact on how long it took to pass them all around.  How much can the people present after the slow one influence the rate?

5. Main Menu Eating Elephants…  How do you eat an elephant?  One mouthful at a time!  Chemical reactions are similar.

6. Main Menu Reaction Mechanisms - Example 1  Most reactions happen by a series of small steps, and this series is called the ‘reaction mechanism’  The individual steps (elementary steps) can take place at very different rates and are not usually observed directly.  For example: NO 2 (g) + CO(g)  NO(g) + CO 2 (g)  The (theoretical) steps in the mechanism are:  Step 1: NO 2 + NO 2 (g)  NO(g) + NO 3 (g) (slow)  Step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g)(fast)  Overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g)  Note:  NO 3 is an intermediate something that is made in one step and used up in another  The left and right sides of the mechanism cancel out to give you the overall equation.

7. Main Menu The rate determining step is the slowest step in the reaction mechanism  When city planners are trying to improve the flow of traffic through a particular area, they will look at where the traffic is moving most slowly as this will be the place that limits the overall traffic flow.  For chemical reactions with several steps, the overall rate will be determined by the slowest step in the sequence.  Products can only appear as fast as the products of this elementary step.

8. Main Menu Potential energy level profiles The two maxima represent the transition states. Minimum between the maxima represents the intermediate species after the first step. The energy of the transition state for the first step is higher compared to the second step, hence is will be the slowest step. The activation energy for the overall reaction is equal to the activation energy of the RDS. Catalysts usually provide an alternate pathway for the RDS that has a lower activation energy

9. Main Menu Rate expression for an overall reaction is determined by the reaction mechanism  RDS is an elementary step, a single molecular event, its rate law comes directly from its molecularity.  Molecularity describes the number of particles involved in a single step.  The rate law for the equation in the RDS must contain the concentration of the reactants raised to the power of it’s co-efficient.  NOTE: This relationship only exists for elementary step equations, not overall reactions! (See lesson 3 to recap!)

10. Main Menu A note on molecularity…  Molecularity describes the number of particles involved in a single step:  Unimolecular: one molecule involved: A  B +C  Bimolecular: two molecules involved A+A  C A+B  C  Termolecular: three molecules involved (extremely rare, you will not encounter these at IB level).

11. Experiment shows that each of the following reactions has the rate equation shown. 1.Hydrolysing a Primary Haloalkane RCH 2 Br + OH -  RCH 2 OH + Br - Rate = k[RCH 2 Br][OH - ] 2.Hydrolysing a Tertiary Haloalkane R 3 CBr + OH -  R 3 COH + Br - Rate = k[R 3 CBr] 3.Nitrating Benzene using a c. H 2 SO 4 Catalyst C 6 H 6 + HNO 3  C 6 H 5 NO 2 + H 2 O Rate = k[HNO 3 ] 4.Iodination of Propanone in Acid Solution CH 3 COCH 3 + I 2  CH 3 COCH 2 I + H + + I - Rate = k[CH 3 COCH 3 ] [H + ] RCH 2 Br and OH - both take part in RDS of mechanism Only R 3 CBr (not OH - ) takes part in RDS of mechanism Only HNO 3 takes part in the RDS of mechanism, not H 2 SO 4 or C 6 H 6 CH 3 COCH 3 and H + both take part in RDS of mechanism but not I 2 Note: In example 4, H + influences rate but is not a reactant  H + is a catalyst

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13. What about if the RDS is not the first step in the mechanism? These examples show why the order of the reaction w.r.t each reactant is not linked to their co-efficients in the overall reaction equation. If a reaction is zero order w.r.t to that reactant (concentration will not affect the rate) it means it does not take part in the RDS. If a zero order reactant does appear in the rate equation, then the reactant or something derived from it must take part in the RDS.

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17. Mechanisms and Rates - Example 1  Looking at our previous example  NO 2 (g) + CO(g)  NO(g) + CO 2 (g)  Step 1: NO 2 (g) + NO 2 (g)  NO(g) + NO 3 (g) (slow)  Step 2: NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g)(fast)  If you think about it….  Changing the concentration of CO will not affect the rate  Because it is involved in a fast step after the Rate Determining Step (RDS), and so the only thing relevant to this step is how quickly the NO 3 can be made, and this is made by a step which is very slow.  Changing the concentration of NO 2 will affect the rate  Due to it being involved in the slow step  Since it appears twice in the slow step, changing it’s concentration will have double the impact

18. Main Menu Mechanisms and Rates - Example 2  2NO(g) + O 2 (g)  2NO 2 (g)  Step 1: 2NO(g)  N 2 O 2 (g)(fast)  Step 2: N 2 O 2 (g) +O 2 (g)  2NO 2 (g)(slow)  In this reaction:  Changing the concentration of O 2 will affect the rate  Because it is involved in the slow step.  Changing the concentration of NO will affect the rate  Because it will affect the rate at which N 2 O 2 is made, and this is needed in the slow step.  In summary:  Rate is affected by everything up to and including the slow step  Rate is not affect by anything after the slow step  For this reason, the slow step is called the rate determining step.

19. e.g. NO 2 + CO  NO + CO 2 can be shown to follow the following kinetics Rate = k[NO 2 ] 2  2 molecules of NO 2 only (not CO) are involved in the RDS. 2 mechanisms were proposed: (A) 2NO 2  N 2 O 4 (slow RDS) N 2 O 4 + CO  NO + NO 2 + CO 2 (fast) (B) 2NO 2  NO 3 + NO (slow RDS) NO 3 + CO  O 2 + CO 2 (fast) Both are consistent with the rate equation and both give the overall reaction when steps added but they cannot both be correct! Other experimental work shows NO 3 can be detected.  mechanism (B) is acceptable but only until further evidence contradicts it!

20. Main Menu How can we determine the reaction mechanism? Example 1: NO 2 (g) + CO(g)  NO(g) + CO 2 (g)  First we identify a series of possible reaction mechanisms:  From experimental data we find that the rate law is Rate = k[NO 2 ] 2  This tells us that the reactions at/before the RDS involve two NO 2 or things that can be made from it.  The only possibility that fits this is the first, with the first step as the RDS. Possibility 1: 2NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO(g) + CO 2 (g) Depending on which is RDS either: Rate = k[NO 2 ] 2 Rate = k[NO 2 ] 2 [CO] RDS most likely to be step 1. Possibility 2: NO 2  NO + O CO + O  CO 2 Depending on RDS either: Rate = k[NO 2 ] Rate = k[NO 2 ][CO] RDS most likely to be step 1.

21. FURTHER EXAMPLES: Suggest a possible mechanism for each of the following reactions which is consistent with the experimentally determined rate equation shown. 1. A + B  C + D where rate = k[A][B] 2. A + B  C where rate = k[A] A + B  [A-B]* RDS forms intermediate [A-B]*  C + D Fast step A  [A]* RDS forms intermediate [A]* + B  C Fast step

22. Q3 The reaction 2N 2 O 5  4NO 2 + O 2 is shown by experiment to follow the following rate equation: Rate = k [N 2 O 5 ] The mechanism is proposed to be Step 1N 2 O 5  NO 2 + NO 3 Step 2NO 2 + NO 3  NO + NO 2 + O 2 Step 3NO + NO 3  2NO 2 For this mechanism to be valid, which step must be the rate determining step? Step 1Step 2Step 3

23. Q4 The reaction H 2 O 2 + 2H 3 O + + 2I -  I 2 + 4H 2 O is considered to take place in 3 steps Step 1H 2 O 2 + I -  IO - + H 2 O SLOW Step 2IO - + H 3 O +  HIO + H 2 O FAST Step 3HIO + H 3 O + + I -  I 2 + 2H 2 O FAST For this mechanism to be acceptable, experiment must show the rate equation to be: Rate = K[H 2 O 2 ][H 3 O + ] 2 [I - ] 2 Rate = K[H 2 O 2 ][H 3 O + ][I - ] Rate = K[H 3 O + ][IO - ]Rate = K[HIO][H 3 O + ][I - ]Rate = K[H 2 O 2 ][I - ]

24. Main Menu Key Point  The order of reaction with respect to a reactant tells you the number of molecules of the reactant involved up to and including the rate determining step. https://www.tes.co.uk/teaching-resource/video- the-rate-determining-step-6354488

25. Main Menu How can we determine the reaction mechanism? Example 2: 2NO(g) + O 2 (g)  2NO 2 (g)  First we identify a series of possible reaction mechanisms:  From experimental data we find that the rate law is Rate = k[NO] 2 [O 2 ]  This tells us that the reactions at/before the RDS only at least two NO and one O 2.  The only possibility that fits this is the second, with the second step as the slow one Possibility 1: NO + O 2  NO 2 + O NO + O  NO 2 Step 1 would be slow so: Rate = k[NO][O 2 ] Possibility 2: 2NO  N 2 O 2 N 2 O 2 + O 2  2NO 2 Depending on RDS: Rate = k[NO] 2 Rate = k[NO] 2 [O 2 ] Possibility 3: NO  N + O NO + O  NO 2 N + O 2  NO 2 Step 1 likely to be slowest so: Rate = k[NO]

26. Main Menu Complete the reactions mechanisms worksheet...  Complete only questions:  1(a)+(b)  2(a)+(b)+(c)+(d)+(e)  3(a)+(b)+(c)+(d)  Leave the rest as it involves aspects of equilibrium which we haven’t covered yet...

27. Main Menu Review  Rate equations are determined by the mechanism of a reaction  We can use the rate equation to help us choose from possible candidate mechanisms  The order of the reaction with respect to each reactant tells you the number of times they are involved in the mechanism up to and including the RDS.