CHEM Pharmacy Week 11: Kinetics – Arrhenius Equation

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

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:

Reaction Mechanism 2 NO2 (g) + F2 (g)  2 NO2F (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 NO2 + F2  NO2F + F slow Step 2 NO2 + F  NO2F fast 2 NO2 + F2  2 NO2F 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 = k2 [NO2][F]. reaction intermediate k1 k2

Rate-Determining Step
Overall reaction is: 2 NO2 (g) + F2 (g)  2 NO2F (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 F2  NO2F + 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 [NO2][F2]

Reaction Mechanism - Example 2
The following elementary steps constitute a proposed mechanism for a reaction: Step A + B X fast k1[A][B] =k-1[X] Step X + C Y slow Step Y D fast 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] = k1/k-1[A][B][C] = k [A] [B][C] k1 k-1 k2 k-2 k3 k-3

Reaction Mechanism - Example 3
For the reaction: NO CO  NO CO2 The experimentally determined rate equation is: rate = k[NO2]2 Show the rate expression is consistent with the mechanism: Step 1 2 NO N2O4 fast equilibrium Step 2 N2O4  NO NO3 slow Step 3 NO CO  NO2 + CO2 fast Overall k1 k-1 k2 k3

Reaction Mechanism – Example 4
A proposed mechanism for the reaction: 2NO(g) + Br2(g)  2NOBr(g) consists of two elementary reactions: NO Br2 NOBr2 fast equilibrium NOBr2 + NO  2NOBr slow Task: Confirm that this mechanism is consistent with the stoichiometry of the reaction and the observed rate law: Rate = k[NO]2[Br2]

The Oscillating Iodine Reaction
Three solutions are added together quickly: iodate IO3- in acid, malonic acid, and H2O2. The resultant solution oscillates in colour several times until it finally stops. The overall reaction is: IO H2O2 +CH2(CO2H) + H+ → ICH(CO2H)2 + 2 O2 + 3 H2O Demo 7.5 transition from colourless to golden brown to deep blue

Temperature vs. Rate Low temperature: slow reactions
Egg cooking time at 80oC ~ 30 min Bacterial growth at 4oC = slow High temperature: faster reactions Egg cooking time at 100oC ~ 5 min Bacterial growth at 30oC = fast In this lecture the effect of temperature and catalysts on the reaction rate will be examined The principal questions to be answered are: (1) Why do most reactions increase their rate with increasing temperature and (2) How do catalysts increase the reaction rate (eg., Pt catalyst in catalytic converters and enzymes)

Collision Theory 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

Activation Energy Activated complex (TS) Consider: A + B C + D
Potential Energy Reaction Progress Ea A + B C + D Activated complex (TS) Consider: A + B C + D For reaction to occur: colliding molecules must have energy greater than Ea Ea = activation energy Increasing temperature increases no. of molecules with energy greater than Ea Exo Endo The diagrams show potential energy of reactants and products as a function of the reaction ordinate (reaction progress) Graphs show potential energy changes as reactants converted to products (Could also use Free energy - represent same idea) The reaction is only assumed to occur if the reacting molecules collide together with enough kinetic energy to overcome the barrier to the fragmentation. This is termed the activation energy and is the minimum amount of energy that needs to be supplied to the system for the reaction to occur Exothermic reaction: if products have less energy than reactants, energy released due to reaction (highlight top diagram) Endothermic reaction: if products have more energy than reactants, energy consumed during reaction (highlight bottom diagram) In a true solution, molecules have a range of kinetic energies - a distribution. At low temperatures only a small fraction of these have an energy greater than Ea. As the temperature increases, thermal energy transformed to KE and a greater proportion of atoms have an energy greater than Ea - hence rate increases. There is an analogy of the above with the metastable and stable state examined in colloid stability. Here, the kinetic energy of the collision transforms the nature of the species (cf. ten pin falling - not transformed but position changes)

Orientation and Collision
NO2Cl + Cl· → NO2 + Cl2 LOOK THIS UP IN ATKINS – give proper definition Only a small fraction of the possible collision geometries can result in a reaction. Orientation is independent of T.

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

Reaction Rate Depends on Temperature
Chemiluminescent reaction The Cyalume stick glows more brightly in the hot water beaker because the reaction is faster. Demo 7.2 The glow lasts for longer in a cold beaker, because the reaction is slower.

The Arrhenius Equation
The Arrhenius equation describes the temperature dependence of the rate constant, k: k Ea k = A k ~ 0 Talk about what A is as well – related to number of collisions with the right energy that are in the right geometry, essentially

Rate and Temperature Rate constant, k Temperature
Consider the decomposition of H2O2: 2 H2O2(aq) H2O(l) + O2(g) k (s–1) T(K) 7.77 x 10– Rate increases exponentially with temperature. Rate constant, k Example is decomposition of hydrogen peroxide Will examine effect of a catalyst on the activation energy shortly The temperature quoted is in terms of K. Note: how to convert celsius into Kelvin (T+273) Temperature

Using Arrhenius Equation
To calculate Ea, rearrange the Arrhenius equation (k = A e – Ea / R T) as: ln k = ln A – Ea / R T If k1 and k2 are the rate constants at two temperatures T1 and T2 respectively, then we can also calculate Ea easily: ln k1 = ln A – Ea / R T1 ln k2 = ln A – Ea / R T2 And subtracting the two, the term (ln A) disappears, so we get:

Arrhenius Equation – Example 1
For the reaction: 2 NO2(g)  2NO(g) + O2(g) The rate constant k = 1.00 · s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are the values for A and k at 273 K? 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
(a) Find the value of A (independent of T): A = k eEa / R T = 1.00· s-1 · exp [ J mol-1 / (8.314 J mol-1 K –1· 300 K)] = 2.13 ·109 s-1 (three significant figures) Calculate the value of k at 273 K: k = 2.13 · 109 s-1 exp (– J mol-1) / (8.314 J mol-1 K –1·273 K) = 1.23 · s-1 (three significant figures) Calculate the temperature when k = 1.00· 10-11s-1 T = Ea / [R· ln (A/k)] = J mol-1 / (8.314·46.8) J mol-1 K = 285 K (three significant figures)

Distribution of Collision Energy
Increasing the temperature increases the number of collisions with sufficient energy to react, i.e. with energy > Ea. Blackman Figure 14.13

Energy Landscape in Chemical Reactions
A + B C + D Endothermic reaction A + B C + D Exothermic reaction Ea (forw) (rev) Activated state Ea (forw) Ea (rev) A + B Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. C + D Forward reaction is faster than reverse Reverse reaction is faster than forward

Arrhenius Equation – Example 2
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 C, what is the activation energy ? Answer: ln {(2.0 · 10-5)/ (7.32 · 10-5)} = - (Ea/8.314) (1/ /303.00) Ea = 96 kJ mol-1 (TWO SIGNIFICANT FIGURES)

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

Arrhenius Equation – Example 3
The rate constant of a particular reaction triples when the temperature is increased from 25 C to 35 C. Calculate the activation energy, Ea, for this reaction. ln (1/3) = - (Ea / 8.314)(1/ /308) = - Ea(1.310 x 10-5) Ea = 8.4·104 J mol-1 or 84 kJ mol-1 Note significant figures!

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