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Reaction Kinetics

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• 1. CHEMISTRY II SCI 4211 Prepared by Hartini Hamzah 1
• 2. REACTION KINETICS 2
• 3. Reaction Rate Defined Reaction rate: changes in a concentration of a product or a reactant per unit time. [] Reaction rate = —— t concentration [] change [] t t 3
• 4. Expressing reaction rates For a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation. Note the expression and reasons for their relations for the reaction 2 NO + O2 (g) = 2 NO2 (g) [O2] 1 [NO] 1 [NO2] Reaction rate = – ——— = – — ———— = — ——— t 2 t 2 t Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for 2 N2O5 = 4 NO2 + O2 How can the rate expression be unique and universal? 4
• 5. Calculating reaction rate The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for 2 N2O5 = 4 NO2 + O2 Solution: (0.93 – 1.24)e-2 – 0.31e-2 M Decomposition rate of N2O5 = – ———————— = – —————— 1200 – 600 600 s = 5.2e-6 M s-1. Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1. Be able to do this type problems The reaction rates are expressed in 3 forms 5
• 6. Determine Reaction Rates To measure reaction rate, we measure the concentration of either a reactant or product at several time intervals. The concentrations are measured using spectroscopic method or pressure (for a gas). For example, the total pressure increases for the reaction: 2 N2O5 (g) 4 NO2 (g) + O2(g) Because 5 moles of gas products are produced from 2 moles of gas reactants. For the reaction CaCO3 (s) barometer CaO(s) + CO2 (g) The increase in gas pressure is entirely due to CO2 formed. 6
• 7. Rate Laws  An equation that relates the rxn rate and the concentration of reactants Rate Determining Step Slow Rate = k[HBr][O2]
• 8. Rate Laws  If no mechanism is given, then… 2H2 + 2NO  N2 + 2H2O Rate = k[H2]2[NO]2
• 9. Rate Orders  0, 1st and 2nd order rates  Order is dependent upon what will yield a straight line 1st order ln [reactants] [reactants] 0 order 1/[reactants] 2nd order
• 10. Rate Orders  1st order: reaction rate is directly proportional to the concentration of that reactant  2nd order: reaction rate is directly proportional to the square of that reactant  0 order: rate is not dependant on the concentration of that reactant, as long as it is present. For Individual Components:
• 11. Rate Orders  Overall reaction orders is equal to the sum of the reactant orders.  Always determined experimentally! For Overall Order:
• 12. Calculating for k A + 2B  C Rate = k[A][B]2 Experiment Initial [A] Initial [B] Rate of Formation of C 1 0.20 M 0.20 M 2.0 x 10-4 M/min 2 0.20 M 0.40 M 8.0 x 10-4 M/min 3 0.40 M 0.40 M 1.6 x 10-3 M/min What is the value of k, the rate constant?
• 13. Calculating for k Experiment Initial [A] Initial [B] Rate of Formation of C 1 0.20 M 0.20 M 2.0 x 10-4 M/min 2 0.20 M 0.40 M 8.0 x 10-4 M/min 3 0.40 M 0.40 M 1.6 x 10-3 M/min Rate = k[A][B]2 2.0 x 10-4 = k[0.20][0.20]2 2.0 x 10-4 = k(0.008) k = 2.50 x 10-2 min-1 M-2
• 14. Practice 1. In a study of the following reaction: 2Mn2O7(aq) → 4Mn(s) + 7O2(g) When the manganese heptoxide concentration was changed from 7.5 x 10-5 M to 1.5 x 10-4 M, the rate increased from 1.2 x 10-4 to 4.8 x 10-4. Write the rate law for the reaction. Rate = k[Mn2O7]2 2. For the reaction: A+B→C When the initial concentration of A was doubled from 0.100 M to 0.200 M, the rate changed from 4.0 x 10-5 to 16.0 x 10-5. Write the rate law & determine the rate constant for this reaction. Rate = k[A]2 Constant = 4.0 x 10-3 M/s
• 15. More Practice New Rate = 8.1 x 10-5 M/s 3. The following reaction is first order: CH3NC(g) → CH3CN(g) The rate of this reaction is 1.3 x 10-4 M/s when the reactant concentration is 0.040 M. Predict the rate when [CH3NC] = 0.025. 4. The following reaction is first order: (CH2)3(g) → CH2CHCH3 (g) What change in reaction rate would you expect if the pressure of the reactant is doubled? An increase by a factor of 2
• 16. Integrated Rate Laws concentrations as functions of time One reactant A decomposes in 1st or 2nd order rate law. Differential rate law Integrated rate law – d[A] / dt = k [A] = [A]o – k t d[A] – —— = k [A] dt [A] = [A]o e – k t or ln [A] = ln [A]o – k t d[A] – —— = k [A]2 dt 1 1 —— – —— = k t [A] [A]o [A] conc at t [A]o conc at t = 0 Describe, derive and apply the integrated rate laws Learn the strategy to determine rate-law 16
• 17. Concentration and time of 1st order reaction Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times. [A] ln[A ] ln [A] = ln [A]o – k t [A] = [A]o e – k t t½ t t 17
• 18. Half life & k of First Order Decomposition The time required for half of A to decompose is called half life t1/2. Since [A] = [A]o e – k t When t = t1/2, [A] = ½ [A]o Thus ln ½ [A]o = ln [A]o – k t1/2 or ln [A] = ln [A]o – k t – ln 2 = – k t1/2 k t1/2 = ln 2 = 0.693 relationship between k and t1/2 Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability. Evaluate t½ from k or k from t½ 18
• 19. st 1 order reaction calculation N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s. Solution next 19
• 20. st 1 order reaction calculation N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s. Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e – k t apply [A] = [A]o e– k t ln 0.9 = ln 1.0 – k 30 s – 0.1054 = 0 – k * 30 k = 0.00351 s – 1 t½ = 0.693 / k = 197 s apply k t ½ = ln 2 [A] = 1.0 e – 0.00351*500 = 0.173 Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 % After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed. After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed. Apply integrated rate law to solve problems 20
• 21. Molecularity of Elementary Reactions The total order of rate law in an elementary reaction is molecularity. The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react. A molecule decomposes by itself is a unimolecular reaction (step); two molecules collide and react is a bimolecular reaction (step); & three molecules collide and react is a termolecular reaction (step). O3 O2 + O rate = k [O3] NO2 + NO2 NO3 + NO rate = k [NO2]2 Br + Br + Ar Br2 + Ar* rate = k [Br]2[Ar] Caution: Derive rate laws this way only for elementary reactions. 21
• 22. The Arrhenius Equation  The dependence of the rate constant of a reaction on temperature can be expressed by the following equation known as Arrhenius equation. k=A e Ea RT Ea = activation energy k = rate constant R = gas constant T = absolute temperature e = the base of the natural logarithm scale A = quantity represents the collision frequency (frequency factor)
• 23. • The relation ship between the rate constant, k and temperature can be seen in the k vs T graph: -Ea∕ k = Ae RT 1/T (K-1)
• 24. By taking the natural logarithm of both sides. ln k = ln Ae -Ea/RT ln k = ln A – Ea/RT  The equation can take the form of a linear equation ln k =( y Ea ) ( 1 ) + ln A R T m x c 1 A plot of ln k versus gives a straight line T Slope m = Eaand intercept c with the y axis is ln A. R Video 4
• 25. Video 5
• 26. Graph Representation Of The Arrhenius Equation • Plotting a ln k vs 1/T graph would show a clearer relationship between k (Rate constant) and temperature ln k Ea 1 ( ) ln A R T Where, Ea = Activation Energy R = 8.314 Jmol-1K-1 T = Absolute Temp A = Collision freq. factor
• 27. Example 1 The rate constants for decomposition of acetaldehyde 2HI(g) H2(g) + I2(g) were measured at five different temperatures.The data are shown below. Plot ln k versus 1/T, and determine the activation energy (in kJ/mole) for the reaction. k (1/M s) T (K) 3.52 x 10-7 283 3.02 x 10-5 356 2.19 x 10-4 393 1.16 x 10-3 427 3.95 x 10-2 508
• 28. Solution We need to plot ln k on the y-axis versus 1/T on the x-axis. From the given data we obtain ln k 1/T (K-1) 14.860 1.80 x 10-3 10.408 1.59 x 10-3 8.426 1.50 x 10-3 6.759 1.43 x 10-3 3.231 1.28 x 10-3
• 29. ln k -2 -4 -6 -8 y -10 -12 -14 x -16 1.4 1.6 1.8 x 10-3 1/T(K-1) Plot of ln k versus 1/T. The slope of the line is calculated from two pairs of coordinates.
• 30. Slope = (-14.0) – (-3.9) (1.75– 1.3) x 10-3 K-1 = -2.24 x 104 K  Finally, calculate the activation energy from the slope:  The slope, m = – Ea/R. Ea = – R (m) = (–8.314 J K-1 mol-1) (–2.24 x 104 K) = 1.9 x 105 J mol-1 = 190 kJ mol-1
• 31.  An equation relating the rate constants k1 and k2 at temperatures T1 and T2 can be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known. ln k1 = ln A – Ea/RT1 ln k2 = ln A – Ea/RT2  Subtracting ln k2 from ln k1 gives ln k1 – ln k2 = Ea/R (1/T2 – 1/T1) Video 6
• 32. Activation Energy  Collisions only result in a reaction if the particles collide with enough energy to get the reaction started. This minimum energy required is called the activation energy for the reaction.  Only a fraction of the total population of a chemical species will be able to collide with enough energy to react (E ≥ Ea)
• 33. Maxwell-Boltzmann Distribution  In any system, the particles present will have a very wide range of energies. For gases, this can be shown on a graph called the MaxwellBoltzmann Distribution which is a plot of the number of particles having each particular energy.
• 34. The Maxwell-Boltzmann Distribution and Activation Energy  Notice that the large majority of the particles don't have enough energy to react when they collide.  To enable them to react we either have to change the shape of the curve, or move the activation energy further to the left.
• 35. M-B Distribution and Temperature  Increasing the temperature:  increases the average kinetic energy of the gas (slight shift to the right)  Increases the number of particles whose energy exceeds the activation energy
• 36. Catalysts and Activation Energy  To increase the rate of a reaction you need to increase the proportion of successful collisions.  Catalysts increase the rate of reaction by providing an alternative way for the reaction to happen which has a lower activation energy.
• 37. M-B Distribution and Catalysts
• 38. Catalysis Energy A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, Ea. It participates a reaction in forming an intermediate, but is regenerated. Uncatalyzed rxn Enzymes are marvelously selective catalysts. A catalyzed reaction, NO (catalyst) 2 SO2 (g) + O2 — 2 SO3 (g) via the mechanism i 2 NO + O2 2 NO2 ii NO2 + SO2 SO3 + NO Catalyzed rxn (3rd order) rxn 39
• 39. Homogenous vs. heterogeneous catalysts A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult. When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways. Platinum is often used to catalyze hydrogenation Catalytic converters reduce CO and NO emission. 40
• 40. Enzymes – selective catalysts Enzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites. Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction. After having reacted, the products P1 & P2 are released. Enzyme + Substrate ES ES (activated complex) P1 + P2 + E Enzymes are biological catalysts for biological systems. 41
• 41. Summary  Maxwell-Boltzmann Energy Distribution Curve  distribution of a population of a chemical by energy  Used to show effect of temperature and catalysts on the rate of a chemical reaction  Temperature  changes average kinetic energy of population  shifts and distorts the M-B Distribution curve  changes the proportion of particles with Ea.  Catalysts  provide alternative pathway with a lower the activation energy  Ea on M-B curve is shifted left  no effect on distribution curve itself
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