Ch. 13: Chemical KineticsCh. 13: Chemical Kinetics

Dr. Namphol Sinkaset

Chem 201: General Chemistry II

I. Chapter OutlineI. Chapter Outline

I. Introduction

II. The Rate of a Chemical Reaction

III. Reaction Rate Laws

IV. Integrated Rate Laws

V. Temperature and Rate

VI. Reaction Mechanisms

VII. Catalysis

I. IntroductionI. Introduction

• Some reactions are quick (explosions) while others are slow (rusting of iron).

• Knowing the rate of a reaction and what factors influence it allow chemists to plan accordingly.

• If we understand what contributes to the rate, we can control the reaction.

I. IntroductionI. Introduction

• Balanced equations only give net change.• Equations tell us nothing about how the

reaction happens.• One possibility for the above: simultaneous

collision of 6 molecules.• Unlikely, so reaction must occur in a series of

small steps that leads to final products.• We study mechanisms later in the chapter.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

II. Reaction RatesII. Reaction Rates

• Rates are generally change of something divided by change in time.

• Reaction rates are no different.• The rate of a reaction can be written

with respect to any compound in that reaction.

• However, there can only be one numerical value for a rate of reaction.

II. Average Rates of ReactionII. Average Rates of Reaction

H2(g) + I2(g) 2HI(g)

II. General Reaction RatesII. General Reaction Rates

aA + bB cC + dD

II. Some Rate DataII. Some Rate Data

• If we plot average rate data as a function of time, we see that the reaction rate constantly changes.

• Thus, rate depends on concentration of reactants!

III. Rate LawsIII. Rate Laws

• If the rate depends on concentration of reactants, then we should be able to write an equation.

• A rate law describes the mathematical relationship between the concentration of reactants and how fast the reaction occurs.

III. A Simple Rate LawIII. A Simple Rate Law

• Consider a decomposition reaction where A products

• If the reverse reaction is negligible, then the rate law is: Rate = k[A]n. k is called the rate constant n is called the reaction order

III. Reaction OrdersIII. Reaction Orders

• The reaction order, n, determines how the rate depends on the concentration of the reactant. For the previous reaction, if… n = 0, zero order, rate is independent of [A] n = 1, first order, rate is directly

proportional to [A] n = 2, second order, rate is proportional to

the square of the [A]

III. Reaction Orders and RateIII. Reaction Orders and Rate

• The rate law for the decomposition can then be either: Rate = k[A]0 = k Rate = k[A]1

Rate = k[A]2

• Each will have a different type of curve when graphed.

III. Determining OrdersIII. Determining Orders

• Reaction orders can only be determined by experiment!!

• Reaction orders are not related to the stoichiometry of a reaction!

• If reaction orders match a reaction’s stoichiometry, it is just a coincidence.

• Therefore, orders cannot be determined without experimental data!

III. Sure-fire MethodIII. Sure-fire Method

[A] (M) Initial Rate (M/s)

0.10 0.015

0.20 0.060

0.40 0.240

For the reaction, A Products, we have the following data:

III. More Complex ReactionsIII. More Complex Reactions

• What if we have a more complicated reaction like: aA + bB cC + dD?

• Writing the general rate law is easy. Simply include all reactants, each with its own order. Rate = k[A]m[B]n

• If there are more reactants, there are more terms in the rate law.

III. Example ReactionIII. Example Reaction

• 2H2(g) + 2NO(g) N2(g) + 2H2O(g)

• After looking at experimental data, the rate law was found to be Rate = k[H2][NO]2.

• We say the reaction is 1st order in H2, 2nd order in NO, and 3rd order overall.

• Note that Rate always has units of M/s, so the units on k will depend on the rate law.

• What are the units of k for the rate law above?

III. Steps for Finding Rate LawIII. Steps for Finding Rate Law

1) Pick two solutions where one reactant stays same, but another changes.

2) Write rate law for both w/ as much information as you have.

3) Ratio the two and solve for an order.

4) Repeat for another pair of solutions.

5) Use any reaction to get value of k.

III. Sample ProblemIII. Sample Problem

[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)

0.010 0.010 0.0035

0.020 0.010 0.0069

0.020 0.020 0.0098

0.040 0.040 0.027

Determine the complete rate law for the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) using the data below.

III. Sample ProblemIII. Sample Problem

[NO] (M) [H2] (M) Initial Rate (M/s)

0.10 0.10 0.00123

0.10 0.20 0.00246

0.20 0.10 0.00492

Sometimes, rate laws can be found by inspection. Determine the rate law for the reaction 2NO(g) + 2H2(g) N2(g) + H2O(g) using the data below.

IV. Concentration and TimeIV. Concentration and Time

• Study and elucidation of rate laws allow the prediction of when a reaction will end.

• An integrated rate law for a chemical reaction is a relationship between the concentrations of reactants and time.

• Integrated rate laws depend on the order of the reaction; thus, we examine each separately.

• We will only consider reactions with one reactant.

IV. 1IV. 1stst Order Integrated Rate Law Order Integrated Rate Law

IV. 1IV. 1stst Order Integrated Rate Law Order Integrated Rate Law

• Notice this equation is in y = mx + b form.

• A plot of ln[A] vs. t for a 1st order reaction yields a straight line with m = -k and b = ln[A]0.

IV. 2IV. 2ndnd Order Integrated Rate Law Order Integrated Rate Law

IV. 2IV. 2ndnd Order Integrated Rate Law Order Integrated Rate Law

• Again, this equation is in y = mx + b form.

• A plot of 1/[A] vs. t yields a straight line with slope equal to k and y-intercept equal to 1/[A]0.

IV. Zero Order Integrated Rate LawIV. Zero Order Integrated Rate Law

IV. Zero Order Integrated Rate LawIV. Zero Order Integrated Rate Law

• Yet again in y = mx + b form!

• Plot of [A] vs. t results in a straight line with slope equal to -k and b = [A]0.

IV. Reaction Half LivesIV. Reaction Half Lives

• The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to half its initial value.

• Half life equations depend on the order of the reaction.

IV. 1IV. 1stst Order Reaction Half Life Order Reaction Half Life

IV. 1IV. 1stst Order Reaction Half Life Order Reaction Half Life

• Notice that the half life doesn’t depend on reactant concentration!

• Unique for 1st order.• The half life for a 1st order reaction is

constant.

IV. 1IV. 1stst Order Half Lives Order Half Lives

IV. 2IV. 2ndnd Order Reaction Half Life Order Reaction Half Life

IV. 2IV. 2ndnd Order Reaction Half Life Order Reaction Half Life

• For 2nd order, the half life depends on initial concentration.

• As concentration decreases, half life gets longer and longer.

IV. Zero Order Reaction Half LifeIV. Zero Order Reaction Half Life

IV. Zero Order Reaction Half LifeIV. Zero Order Reaction Half Life

• We see that for zero order reactions, the half life depends on concentration as well.

V. Temperature and RateV. Temperature and Rate

• In general, rates of reaction are highly sensitive to temperature.

• If Rate = k[A]n, where does the temperature factor in?

• It’s in the constant k!

• Generally, increasing temperature increases k.

V. The Arrhenius EquationV. The Arrhenius Equation

• Note that R is the gas constant, and T is temperature in kelvin.

V. Parameters of Arrhenius Eqn.V. Parameters of Arrhenius Eqn.

• We can describe the physical meanings of the aspects of the Arrhenius equation by considering a specific reaction.

V. Activation EnergyV. Activation Energy

• To get to product state, reactant must go through high-energy activated complex, or transition state.

• Even though reaction is exo overall, it must go through an endo step.

• Higher Ea means slower reaction.

V. Frequency FactorV. Frequency Factor• The frequency factor represents the number of

approaches to the activation barrier per unit time.

• For this reaction, it represents how often the NC part of the molecule vibrates.

• Note that not all approaches result in reaction due to not having enough energy.

• A frequency factor of 109/s means that there are 109 vibrations per second of the NC group.

V. Exponential FactorV. Exponential Factor

• The exponential factor is a number between 0 and 1 that represents the fraction of molecules that successfully react upon approach.

• An exponential factor of 10-7 means that 1 out of every 107 molecules has enough energy to cross the energy barrier.

V. Exponential Factor & TempV. Exponential Factor & Temp

• Since exponential factor = e-Ea/RT, temperature has a huge influence.

• As T 0, the factor goes to 0, and as T ∞, the factor goes to 1.

• Thus, higher temperatures mean more successful approaches because the molecules have more energy to overcome the activation barrier.

V. Finding A and EV. Finding A and Eaa

V. Arrhenius PlotsV. Arrhenius Plots

• If we have kinetic data at various temperatures, we can plot ln k vs. 1/T.

• We should get a straight line with m = -Ea/R and b = ln A.

V. Two-Point FormV. Two-Point Form

V. Sample ProblemV. Sample Problem

• The decomposition of HI has rate constants of k = 0.079 1/M·s at 508 °C and k = 0.24 1/M·s at 540 °C. What is the activation energy of this reaction in kJ/mole?