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Chemical Dynamics Lecture5 <ul><li>Effect of Catalyst on Reaction Rate </li></ul><ul><li>Reaction Mechanisms </li></ul>
Effect of Catalyst on Reaction Rate <ul><li>A  catalyst  is a substance that speeds up a reaction  without being  consumed...
Effect of Catalyst on Reaction Rate <ul><li>Note from the figure that the  catalyst lowers  activation energy . Therefore ...
Effect of Catalyst on Reaction Rate Uncatalysed Reaction Catalised Reaction
Effect of Catalyst on Reaction Rate <ul><li>The figure on the left shows the fraction of molecules having sufficient energ...
More Problems (based on Reaction Profiles and Effect of Catalyst)   <ul><li>Past Final Exam-Q1 </li></ul><ul><li>(i). The ...
Past Final Exam-Q2 (continued) <ul><li>(ii). Re-draw the ‘Potential Energy’ versus the ‘Reaction Coordinate’ diagram from ...
Reaction Mechanisms <ul><li>Reaction Mechanisms: </li></ul><ul><li>Most reactions occur by a  series of steps called eleme...
Reaction Mechanisms <ul><li>The sum of the elementary steps must give the overall balanced equation for the reaction </li>...
Reaction Mechanisms <ul><li>From the above rate law, a mechanism could be proposed that consists of the following elementa...
Reaction Mechanisms <ul><li>Molecularity : Molecularity is defined as the number of reactant molecules (species) in an ele...
Reaction Mechanisms <ul><li>Note that, in the above mechanism both the elementary steps are  bimolecular reactions. </li><...
Reaction Mechanisms <ul><li>We must realize that one of the elementary steps will be  slowest step.  The products of a rea...
Reaction Mechanisms <ul><li>Also, from the experimental rate law, we can see that the  rate does not depend on the concent...
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Chemical Dynamics Lecture5 Wk2

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  1. 1. Chemical Dynamics Lecture5 <ul><li>Effect of Catalyst on Reaction Rate </li></ul><ul><li>Reaction Mechanisms </li></ul>
  2. 2. Effect of Catalyst on Reaction Rate <ul><li>A catalyst is a substance that speeds up a reaction without being consumed in in the reaction. Catalyst allows reactions occur via alternate mechanisms with lower activation energy (see the diagram below). </li></ul><ul><li> </li></ul>(Uncatalysed Pathway – high E a ) (Catalised Pathway - low Ea )
  3. 3. Effect of Catalyst on Reaction Rate <ul><li>Note from the figure that the catalyst lowers activation energy . Therefore the fraction of molecules having sufficient energy (E a ) will increase when an appropriate catalyst is added to a reaction mixture. </li></ul><ul><li>Hence, the rate increase in the presence of a catalyst. This can be understood from the following figures (one of the figures shows the fraction of molecules having sufficient energy in the absence of catalyst and the other in the presence of a catalyst). </li></ul>
  4. 4. Effect of Catalyst on Reaction Rate Uncatalysed Reaction Catalised Reaction
  5. 5. Effect of Catalyst on Reaction Rate <ul><li>The figure on the left shows the fraction of molecules having sufficient energy in the absence of catalyst and that on the right in the presence of a catalyst (see slide 4). </li></ul><ul><li>Clearly, the fraction of molecules having sufficient energy (Ea) increases in the presence of a catalyst and hence reaction rate increases . </li></ul>
  6. 6. More Problems (based on Reaction Profiles and Effect of Catalyst) <ul><li>Past Final Exam-Q1 </li></ul><ul><li>(i). The reaction: A + B  C + D has an activation energy for forward reaction of E a = 60 kJ mol ˉ¹, and ∆H = 30 kJ mol ˉ¹. Draw a plot of the ‘Potential Energy’ (on the y-axis) versus the ‘Reaction Coordinate’ (on the X-axis), showing the ‘products’, ’reactants’, ‘activated complex’, E a and ∆H. </li></ul><ul><li>(Try this one. This is similar to the one drawn in the Lecture4. Only the difference is that, you need to show the values). </li></ul><ul><li>(ii). Re-draw the ‘Potential Energy’ versus the ‘Reaction Coordinate’ diagram from part (i) above, and show on it how the presence of a catalyst would affect the plot </li></ul><ul><li>Past Final Exam-Q2 </li></ul><ul><li> (i). The reaction: A + B  C + D has an activation energy for forward reaction of E a = 60 kJ mol ˉ¹, and ∆H = -30 kJ mol ˉ¹. Draw a plot of the ‘Potential Energy’ (on the y-axis) versus the ‘Reaction Coordinate’ (on the X-axis), showing the ‘products’, ’reactants’, ‘activated complex’, E a and ∆H. </li></ul>
  7. 7. Past Final Exam-Q2 (continued) <ul><li>(ii). Re-draw the ‘Potential Energy’ versus the ‘Reaction Coordinate’ diagram from part (i) above, and show on it how the presence of a catalyst would affect the plot. </li></ul><ul><li>(Note the difference of sign for ∆H between the two past exam problems Q1 and Q2. The products will be of lower energy than the reactants in Q2. In Q1, it will be opposite. Try this one before tutorial and discuss). </li></ul><ul><li>These problems will be solved in the lecture and more of this kind of problems will be solved in the workshops. Please take workshops seriously. </li></ul>
  8. 8. Reaction Mechanisms <ul><li>Reaction Mechanisms: </li></ul><ul><li>Most reactions occur by a series of steps called elementary steps </li></ul><ul><li>To properly understand a reaction we must know these elementary steps and hence know the mechanism </li></ul><ul><li>One of the purposes of studying kinetics is to learn about the reaction mechanism </li></ul><ul><li>A reaction mechanism is a series of elementary steps which must satisfy following two requirements: </li></ul>
  9. 9. Reaction Mechanisms <ul><li>The sum of the elementary steps must give the overall balanced equation for the reaction </li></ul><ul><li>The mechanism must agree with the experimentally determined rate law </li></ul><ul><li>In order to understand the reaction mechanism, let us consider the elementary steps involved in the reaction, </li></ul><ul><li>NO 2 (g) + CO (g)  NO (g) + CO 2 (g) </li></ul><ul><li>Remember that the balanced equation does not give direct information about the mechanism. In fact, the mechanism needs to be determined from experimental rate law. For the above reaction, it is given that the experimentally determined rate law is, </li></ul><ul><li>Rate = k [NO 2 ] ² (1) </li></ul>
  10. 10. Reaction Mechanisms <ul><li>From the above rate law, a mechanism could be proposed that consists of the following elementary steps, </li></ul><ul><li>Step1 : NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) </li></ul><ul><li>Step2 : NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) </li></ul><ul><li>k1 and k2 are the rate constants of the step1 and step2 respectively. In this mechanism NO 3 is an intermediate . </li></ul><ul><li>A reaction intermediate is a species that is neither reactant nor the product, that is formed and consumed in the reaction sequence. Note that the sum of the elementary steps above gives us the overall balanced equation (satisfies first requirement). </li></ul>
  11. 11. Reaction Mechanisms <ul><li>Molecularity : Molecularity is defined as the number of reactant molecules (species) in an elementary step. </li></ul><ul><li>An elementary step involving one reacting molecule is called a unimolecular reaction </li></ul><ul><li>An elementary step involving two reacting molecule is called a bimolecular reaction </li></ul><ul><li>An elementary step involving three reacting molecule is called a termolecular reaction </li></ul><ul><li>Termolecular elementary steps are rare as the probability of three molecules colliding simultaneously is very small. </li></ul>
  12. 12. Reaction Mechanisms <ul><li>Note that, in the above mechanism both the elementary steps are bimolecular reactions. </li></ul><ul><li>we can write the rate laws for both the elementary steps as follows (this is done purely based on the proposed mechanism): </li></ul><ul><li>Step1 : Rate = k 1 [NO 2 ] [NO 2 ] = k 1 [NO 2 ] ² (2) </li></ul><ul><li>Step2 : Rate = k 2 [NO 3 ] [CO] (3) </li></ul><ul><li>The second criteria is to see if the mechanism meets the second requirement (i.e. the mechanism must agree with the experimental rate law). For doing this a new concept known as the rate- determining step (RDS) needs to be defined, which is done as follows. </li></ul>
  13. 13. Reaction Mechanisms <ul><li>We must realize that one of the elementary steps will be slowest step. The products of a reaction can never be produced faster than the rate of the slowest step. Therefore, the slowest step of the elementary reactions determines the rate of the overall reaction and hence is referred to as the rate-determining step (RDS) </li></ul><ul><li>Therefore, the rate-determining step can be identified by comparing the rate laws of the elementary steps with that of the experimentally observed rate law. For the above reaction, we see that the experimentally determined rate law is identical to the rate law for step1. Therefore, step1 is the RDS . </li></ul>
  14. 14. Reaction Mechanisms <ul><li>Also, from the experimental rate law, we can see that the rate does not depend on the concentration of CO </li></ul><ul><li>As two requirements are satisfied, the proposed mechanism may be the plausible and may be correct mechanism. Absolute confirmation of the mechanism can only be proved by the method of detection of reaction intermediates (such as spectroscopic method) </li></ul>
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