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

Reaction Rates: 1. Can measure disappearance of reactants 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

Reaction Rates: 4. Are equal to the slope tangent to that point 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] 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] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 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]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 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]  R = k[NO]2[Cl2]y

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]2[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In 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]

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] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6

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: First Order: Second Order:

Solving an Integrated Rate Law Time (s) [H2O2] (mol/L) 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4 Time (s) [H2O2] 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Regression results: y = ax + b a = -2.64 x 10-4 b = 0.841 r2 = 0.8891 r = -0.9429

Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4 Time (s) ln[H2O2] 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999

Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460 Time (s) 1/[H2O2] 1.00 120 1.0989 300 1.2821 600 1.6949 1200 2.7027 1800 4.5455 2400 7.6923 3000 12.195 3600 20.000 Regression results: y = ax + b a = 0.00460 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: 3. The integrated rate law is: 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] 120 -0.0943 300 -0.2485 600 -0.5276 1200 -0.9943 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Now remember:  k = -slope k = 8.32 x 10-4s-1

Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999 Now remember:  k = -slope k = 8.35 x 10-4s-1

Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0 Zero Order First Order Second Order Rate Law Rate = k Rate = k[A] Rate = k[A]2 Integrated Rate Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 Plot that produces a straight line [A] versus t ln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

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 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.

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 k = rate constant at temperature T A = frequency factor Ea = activation energy R = Gas constant, 8.314 J/K·mol

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-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 adsorbed on a platinum surface Step #1: Adsorption and activation of the reactants.

Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide arranged prior to reacting Step #2: Migration of the adsorbed reactants on the surface.

Heterogeneous Catalysis Carbon dioxide and nitrogen form from previous molecules Step #3: Reaction of the adsorbed substances.

Heterogeneous Catalysis Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface Step #4: Escape, or desorption, of the products.