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

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

1 CHEM Pharmacy Week 11: Kinetics – Arrhenius Equation Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone:

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 ISBN:

3 Lecture 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 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 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 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 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 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 H 2 O 2 +CH 2 (CO 2 H) + H + → ICH(CO 2 H) O H 2 O

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

16 Lecture 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 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 For the reaction: 2 NO 2 (g)  2NO(g) + O 2 (g) The rate constant k = 1.00 · 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· s -1 ? Method: Make use of, and rearrange k = A e – Ea / R T Arrhenius Equation – Example 1

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

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

23 Lecture 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 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/ /308) = - E a (1.310 x ) E a = 8.4·10 4 J mol -1 or 84 kJ mol -1 Arrhenius Equation – Example 3

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