Presentation on theme: "Chemical Kinetics The area of chemistry that concerns reaction rates and reaction mechanisms."— Presentation transcript:
1 Chemical KineticsThe area of chemistry that concerns reaction rates and reaction mechanisms.
2 Reaction RateThe change in concentration of a reactant or product per unit of time
3 Reaction Rates: 1. Can measure disappearance of reactants 2NO2(g) 2NO(g) + O2(g)Reaction Rates:1. Can measuredisappearance ofreactants2. Can measureappearance ofproducts3. Are proportionalstoichiometrically
4 Reaction Rates: 4. Are equal to the slope tangent to that point 2NO2(g) 2NO(g) + O2(g)Reaction Rates:4. Are equal to theslope tangent tothat point5. Change as thereaction proceeds,if the rate isdependent uponconcentration[NO2]t
5 Rate LawsDifferential 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
6 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](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-620.5005.72 x 10-632.86 x 10-6411.4 x 10-6
7 Writing a Rate LawPart 1 – Determine the values for the exponents in the rate law:R = k[NO]x[Cl2]yExperiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-620.5005.72 x 10-632.86 x 10-641.14 x 10-5In experiment 1 and 2, [Cl2] is constant while [NO] doubles.The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO]2[Cl2]y
8 Writing a Rate LawPart 1 – Determine the values for the exponents in the rate law:R = k[NO]2[Cl2]yExperiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-620.5005.72 x 10-632.86 x 10-641.14 x 10-5In experiment 2 and 4, [NO] is constant while [Cl2] doubles.The rate doubles, so the reaction is first order with respect to [Cl2] R = k[NO]2[Cl2]
9 Writing a Rate LawPart 2 – Determine the value for k, the rate constant, by using any set of experimental data:R = k[NO]2[Cl2]Experiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-6
10 Writing a Rate LawPart 3 – Determine the overall order for the reaction.R = k[NO]2[Cl2]2+1= 3 The reaction is 3rd orderOverall order is the sum of the exponents, or orders, of the reactants
11 Determining Order with Concentration vs. Time data (the Integrated Rate Law)Zero Order:First Order:Second Order:
12 Solving an Integrated Rate Law Time (s)[H2O2] (mol/L)1.001200.913000.786000.5912000.3718000.2224000.1330000.08236000.050Problem: Find the integrated rate law and the value for the rate constant, kA graphing calculator with linear regression analysis greatly simplifies this process!!
13 Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4 Time (s)[H2O2]1.001200.913000.786000.5912000.3718000.2224000.1330000.08236000.050Regression results:y = ax + ba = x 10-4b = 0.841r2 =r =
14 Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4 Time (s)ln[H2O2]12030060012001800-1.5142400-2.043000-2.5013600-2.996Regression results:y = ax + ba = x 10-4b = -.005r2 =r =
15 Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460 Time (s)1/[H2O2]1.001201.09893001.28216001.694912002.702718004.545524007.6923300012.195360020.000Regression results:y = ax + ba =b =r2 =r =
16 And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order2. The (differential) rate law is:3. The integrated rate law is:4. But…what is the rate constant, k ?
17 Finding the Rate Constant, k Method #1: Calculate the slope from theTime vs. ln[H2O2] table.Time (s)ln[H2O2]12030060012001800-1.5142400-2.043000-2.5013600-2.996Now remember: k = -slopek = 8.32 x 10-4s-1
18 Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis.Regression results:y = ax + ba = x 10-4b = -.005r2 =r =Now remember: k = -slopek = 8.35 x 10-4s-1
19 Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0 Zero OrderFirst OrderSecond OrderRate LawRate = kRate = k[A]Rate = k[A]2Integrated Rate Law[A] = -kt + [A]0ln[A] = -kt + ln[A]0Plot that produces a straight line[A] versus tln[A] versus tRelationship of rate constant to slope of straight lineSlope = -kSlope = kHalf-Life
20 Reaction MechanismThe 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 reactionThe mechanism must agree with the experimentally determined rate law
21 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
22 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
23 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
24 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?
25 Collision ModelCollisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy).1.Colliding particles must be correctly oriented to one another in order to produce a reaction.2.
26 Factors Affecting Rate Increasing temperature always increases the rate of a reaction.Particles collide more frequentlyParticles collide more energeticallyIncreasing surface area increases the rate of a reactionIncreasing Concentration USUALLY increases the rate of a reactionPresence of Catalysts, which lower the activation energy by providing alternate pathways
29 The Arrhenius Equation k = rate constant at temperature TA = frequency factorEa = activation energyR = Gas constant, J/K·mol
30 The Arrhenius Equation, Rearranged Simplifies solving for Ea-Ea / R is the slope when (1/T) is plotted against ln(k)ln(A) is the y-interceptLinear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slopeEa = -R(slope)
31 CatalysisCatalyst: A substance that speeds up a reaction without being consumedEnzyme: 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.