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CHEM1612 - Pharmacy Week 11: Kinetics – Arrhenius Equation Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196

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Presentation on theme: "CHEM1612 - Pharmacy Week 11: Kinetics – Arrhenius Equation Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196"— Presentation transcript:

1 CHEM1612 - Pharmacy Week 11: Kinetics – Arrhenius Equation Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au

2 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd. 2008 ISBN: 9 78047081 0866

3 Lecture 32 - 3 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g)reaction (1) A reaction mechanism is a series of elementary reactions (or steps) that add up to give a detailed description of a chemical reaction Step 1 NO 2 + F 2  NO 2 F + Fslow Step 2NO 2 + F  NO 2 Ffast 2 NO 2 +F 2  2 NO 2 F Overall An elementary reaction is not made up of simpler steps. For elementary reactions the stoichiometric coefficients are equal to the exponents in the rate law: e.g. for step 2 the rate law is rate = k 2 [NO 2 ][F]. Reaction Mechanism reaction intermediate k1k1 k2k2

4 Lecture 32 - 4 Overall reaction is: 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) A rate-determining (or rate-limiting) step is an elementary reaction that is the slowest step in the mechanism. e.g. Step 1:NO 2 + F 2  NO 2 F+ F(slow) The exponents in the overall rate law (overall reaction) are the same as the stoichiometric coefficients of the species involved in the rate-limiting elementary process (if no intermediates are involved; otherwise their concentrations need to be expressed in terms of the reactants used), in this case : For the overall reaction (1) Rate = k [NO 2 ][F 2 ] Rate-Determining Step

5 Lecture 32 - 5 Reaction Mechanism - Example 2 The following elementary steps constitute a proposed mechanism for a reaction: Step 1 A + B Xfastk 1 [A][B] =k -1 [X] Step 2 X + C Yslow Step 3 YDfast Overall reaction? A + B + C D What are the reaction intermediates? X and Y Show the mechanism is consistent with the rate law: rate = k [A] [B][C] Rate = k [X][C] = k 1 /k -1 [A][B][C] = k [A] [B][C] k 1 k -1 k 2 k -2 k 3 k -3

6 Lecture 32 - 6 For the reaction: NO 2 + CO  NO + CO 2 The experimentally determined rate equation is: rate = k[NO 2 ] 2 Show the rate expression is consistent with the mechanism: Step 12 NO 2 N 2 O 4 fast equilibrium Step 2 N 2 O 4  NO + NO 3 slow Step 3 NO 3 + CO  NO 2 + CO 2 fast Overall Reaction Mechanism - Example 3 k1k1 k2k2 k3k3 k -1

7 Lecture 32 - 7 A proposed mechanism for the reaction: 2NO(g) + Br 2 (g)  2NOBr(g) consists of two elementary reactions: NO + Br 2 NOBr 2 fast equilibrium NOBr 2 + NO  2NOBrslow Task: Confirm that this mechanism is consistent with the stoichiometry of the reaction and the observed rate law: Rate = k[NO] 2 [Br 2 ] Reaction Mechanism – Example 4

8 Lecture 32 - 8 The Oscillating Iodine Reaction Three solutions are added together quickly: iodate IO 3 - in acid, malonic acid, and H 2 O 2. The resultant solution oscillates in colour several times until it finally stops. The overall reaction is: IO 3 - + 2 H 2 O 2 +CH 2 (CO 2 H) + H + → ICH(CO 2 H) 2 + 2 O 2 + 3 H 2 O

9 Lecture 32 - 9 Low temperature: slow reactions Egg cooking time at 80 o C ~ 30 min Bacterial growth at 4 o C = slow High temperature High temperature: faster reactions Egg cooking time at 100 o C ~ 5 min Bacterial growth at 30 o C = fast Temperature vs. Rate

10 Lecture 32 - 10 A + B → C Reactants must collide to react - Orientation and energy factors slow down the reaction Not all collisions produce a reaction - Need enough energy Not all collisions are effective, they need a particular orientation - Orientation does not depend on T Collision Theory

11 Lecture 32 - 11 Consider: A + BC + D For reaction to occur: colliding molecules must have energy greater than E a E a = activation energy Increasing temperature increases no. of molecules with energy greater than E a Potential Energy Reaction Progress Potential Energy EaEa EaEa A + B C + D Activated complex (TS) Exo Endo Activation Energy

12 Lecture 32 - 12 Only a small fraction of the possible collision geometries can result in a reaction. Orientation is independent of T. Orientation and Collision NO 2 Cl + Cl · → NO 2 + Cl 2

13 Lecture 32 - 13 The reaction constant k depends on several factors: k = Z·p· f Z·p = A (frequency factor) f = exp(– E a / RT ) where E a = activation energy, R gas constant, T temperature (K) Collision Theory Z: collision frequency p: orientation probability factor (the fraction of collisions with proper orientation) f: fraction of collisions with sufficient energy Potential Energy EaEa Reactants Products

14 Lecture 32 - 14 Reaction Rate Depends on Temperature Chemiluminescent reaction The Cyalume stick glows more brightly in the hot water beaker because the reaction is faster. The glow lasts for longer in a cold beaker, because the reaction is slower.

15 Lecture 32 - 15 The Arrhenius Equation k EaEa k = A k ~ 0 The Arrhenius equation describes the temperature dependence of the rate constant, k:

16 Lecture 32 - 16 Consider the decomposition of H 2 O 2 : 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) k (s –1 )T(K) 7.77 x 10 –6 298 Rate increases exponentially with temperature. Temperature Rate constant, k Rate and Temperature

17 Lecture 32 - 17 To calculate E a, rearrange the Arrhenius equation (k = A e – Ea / R T ) as: ln k = ln A – E a / R T If k 1 and k 2 are the rate constants at two temperatures T 1 and T 2 respectively, then we can also calculate E a easily: ln k 1 = ln A – E a / R T 1 ln k 2 = ln A – E a / R T 2 And subtracting the two, the term (ln A) disappears, so we get: Using Arrhenius Equation

18 Lecture 32 - 18 For the reaction: 2 NO 2 (g)  2NO(g) + O 2 (g) The rate constant k = 1.00 · 10 -10 s -1 at 300 K and the activation energy E a = 111 kJ mol -1. (a)What are the values for A and k at 273 K? (b)What is the value of T when k = 1.00· 10 -11 s -1 ? Method: Make use of, and rearrange k = A e – Ea / R T Arrhenius Equation – Example 1

19 Lecture 32 - 19 (a) Find the value of A (independent of T): A = k e Ea / R T = 1.00· 10 -10 s -1 · exp [111000 J mol -1 / (8.314 J mol -1 K –1 · 300 K)] = 2.13 ·10 9 s -1 (three significant figures) Calculate the value of k at 273 K: k = 2.13 · 10 9 s -1 exp (– 111000 J mol -1 ) / (8.314 J mol -1 K –1 ·273 K) = 1.23 · 10 -12 s -1 (three significant figures) (b)Calculate the temperature when k = 1.00· 10 -11 s -1 T= E a / [R· ln (A/k)] = 111000 J mol -1 / (8.314·46.8) J mol -1 K -1 = 285 K (three significant figures) Arrhenius Equation – Example 1

20 Lecture 32 - 20 Increasing the temperature increases the number of collisions with sufficient energy to react, i.e. with energy > E a. Distribution of Collision Energy Blackman Figure 14.13

21 Lecture 32 - 21 Energy Landscape in Chemical Reactions A + B C + D Activated state E a (forw) E a (rev) Exothermic reaction Endothermic reaction A + B C + D E a (forw) E a (rev) A + B C + D Forward reaction is faster than reverseReverse reaction is faster than forward Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

22 Lecture 32 - 22 If a reaction has a rate constant k of 2.0 · 10 -5 s -1 at 20.0  C and 7.32·10 -5 s -1 at 30.00  C, what is the activation energy ? Answer: ln {(2.0 · 10 -5 )/ (7.32 · 10 -5 )} = - (E a /8.314) (1/293 - 1/303.00) E a = 96 kJ mol -1(TWO SIGNIFICANT FIGURES) Arrhenius Equation – Example 2

23 Lecture 32 - 23 Determining E a Ideally you would require many more than two values to determine E a. ln k = ln A – E a / R T Plot lnk versus 1/T and get a line with a slope of –E a /R.

24 Lecture 32 - 24 The rate constant of a particular reaction triples when the temperature is increased from 25  C to 35  C. Calculate the activation energy, E a, for this reaction. ln (1/3) = - (E a / 8.314)(1/298 - 1/308) -1.099 = - E a (1.310 x 10 -5 ) E a = 8.4·10 4 J mol -1 or 84 kJ mol -1 Arrhenius Equation – Example 3


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