Chemical Kinetics The area of chemistry that concerns reaction rates and reaction mechanisms. Reaction Rate The change in concentration of a reactant or product per unit of time [ A] at timet2 [ A] at timet1 Rate t2 t1 [ A] Rate t Reaction Rates: 2NO2(g) 2NO(g) + O2(g) 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically Reaction Rates: 2NO2(g) 2NO(g) + O2(g) 4. Are equal to the slope tangent to that point 5. Change as the [NO2] reaction proceeds, t if the rate is dependent upon concentration [ NO2 ] constant t Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called “the rate law.” Integrated rate laws express (reveal) the relationship between concentration of reactants and time Writing a (differential) Rate Law Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: 2 NO(g) + Cl2(g) 2 NOCl(g) Experiment [NO] [Cl2] Rate (mol/L) (mol/L) Mol/L·s 1 0.250 0.250 1.43 x 10-6 2 0.500 0.250 5.72 x 10-6 3 0.250 0.500 2.86 x 10-6 4 0.500 0.500 11.4 x 10-6 Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]x[Cl2]1 2 y Experiment [NO] [Cl2] Rate (mol/L) (mol/L) Mol/L·s 1 0.250 0.250 1.43 x 10-6 2 0.500 0.250 5.72 x 10-6 3 0.250 0.500 2.86 x 10-6 4 0.500 0.500 1.14 x 10-5 In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO] In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2] Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: R = k[NO]2[Cl2] Experiment [NO] [Cl2] Rate (mol/L) (mol/L) Mol/L·s 1 0.250 0.250 1.43 x 10-6 2 mol 6 mol mol 1.43 x10 k 0.250 0.250 Ls L L 1.43 x106 mol L3 L2 k 9.15 x105 0.2503 L s mol 3 mol 2 s Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO]2[Cl2] 2 + 1 = 3 The reaction is 3rd order Overall order is the sum of the exponents, or orders, of the reactants Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: time vs. concentrationis linear First Order: timevs.ln(concentration) is linear 1 Second Order: time vs. is linear concentration Solving an Integrated Rate Law Time (s) [H2O2] (mol/L) Problem: Find the 0 1.00 integrated rate law 120 0.91 and the value for the 300 0.78 rate constant, k 600 0.59 1200 0.37 A graphing calculator 1800 0.22 with linear regression 2400 0.13 analysis greatly 3000 0.082 simplifies this process!! 3600 0.050 (Click here to download Rate Laws program for theTi-83 and Ti-84) Time vs. [H2O2] Time (s) [H2O2] 0 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 Regression results: 1800 0.22 2400 0.13 y = ax + b 3000 0.082 a = -2.64 x 10-4 3600 0.050 b = 0.841 r2 = 0.8891 r = -0.9429 Time vs. ln[H2O2] Time (s) ln[H2O2] 0 0 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 Regression results: 1800 -1.514 y = ax + b 2400 -2.04 a = -8.35 x 10-4 3000 -2.501 b = -.005 3600 -2.996 r2 = 0.99978 r = -0.9999 Time vs. 1/[H2O2] Time (s) 1/[H2O2] 0 1.00 120 1.0989 300 1.2821 600 1.6949 1200 2.7027 1800 4.5455 Regression results: 2400 7.6923 y = ax + b 3000 12.195 a = 0.00460 3600 20.000 b = -0.847 r2 = 0.8723 r = 0.9340 And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order 2. The (differential) rate law is: R k[ H 2O2 ] 3. The integrated rate law is: ln[ H 2O2 ] kt ln[ H 2O2 ]0 4. But…what is the rate constant, k ? Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[H2O2] table. Time (s) ln[H2O2] ln[ H 2O2 ] 2.996 0 0 slope t 3600 s 120 -0.0943 4 1 300 -0.2485 slope 8.32 x10 s 600 -0.5276 1200 -0.9943 Now remember: 1800 -1.514 ln[ H 2O2 ] kt ln[ H 2O2 ]0 2400 -2.04 3000 -2.501 k = -slope 3600 -2.996 k = 8.32 x 10-4s-1 Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Regression results: 4 1 slope a 8.35 x10 s y = ax + b a = -8.35 x 10-4 Now remember: b = -.005 ln[ H 2O2 ] kt ln[ H 2O2 ]0 r2 = 0.99978 r = -0.9999 k = -slope k = 8.35 x 10-4s-1 Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = k Rate = k[A] Rate = k[A]2 Integrated 1 1 Rate Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 kt [ A] [ A]0 1 Plot the [A] versus t ln[A] versus t versus t produces a [ A] straight line Relationship of rate constant Slope = -k Slope = -k Slope = k to slope of straight line Half-Life [ A]0 0.693 1 t1/ 2 t1/ 2 t1/ 2 2k k k[ A]0 Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step Identifying the Rate-Determining Step For the reaction: 2H2(g) + 2NO(g) N2(g) + 2H2O(g) The experimental rate law is: R = k[NO]2[H2] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g) Step #1 agrees with the experimental rate law Identifying Intermediates For the reaction: 2H2(g) + 2NO(g) N2(g) + 2H2O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g) 2H2(g) + 2NO(g) N2(g) + 2H2O(g) N2O(g) is an intermediate Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? Collision Model Collisions must have sufficient energy to 1. produce the reaction (must equal or exceed the activation energy). Colliding particles must be correctly 2. oriented to one another in order to produce a reaction. Factors Affecting Rate Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways Endothermic Reactions Exothermic Reactions The Arrhenius Equation Ea / RT k Ae k = rate constant at temperature T A = frequency factor Ea = activation energy R = Gas constant, 8.31451 J/K·mol The Arrhenius Equation, Rearranged Ea 1 ln(k ) ln( A) R T Simplifies solving for Ea -Ea / R is the slope when (1/T) is plotted against ln(k) ln(A) is the y-intercept Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope Ea = -R(slope) Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules. Lowering of Activation Energy by a Catalyst Catalysts Increase the Number of Effective Collisions Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide Step #1: adsorbed on a Adsorption and platinum surface activation of the reactants. Heterogeneous Catalysis Carbon monoxide and Step #2: nitrogen monoxide arranged prior to Migration of the reacting adsorbed reactants on the surface. Heterogeneous Catalysis Carbon dioxide and nitrogen form from Step #3: previous molecules Reaction of the adsorbed substances. Heterogeneous Catalysis Carbon dioxide and nitrogen gases escape Step #4: (desorb) from the platinum surface Escape, or desorption, of the products.