Chemistry 213: Course Outline Spring 2016

Chemical Kinetics

Factors that Affect Reaction Rates Kinetics => study of how fast chemical reactions occur. Four important factors which affect rates of reactions: reactant concentration, temperature, action of catalysts, and surface area. Goal: to understand chemical reactions at the molecular level (Mechanisms)

Reaction Rates Speed of a reaction is measured by the change in concentration with time. For a reaction A  B

Reaction Rates For the reaction A  B there are two ways of measuring rate: the speed at which the products appear (i.e. change in moles of B per unit time), or the speed at which the reactants disappear (i.e. the change in moles of A per unit time).

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Reaction Rates Change of Rate with Time Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional. Consider: C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Reaction Rates Change of Rate with Time C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) calculate average rate in terms of the disappearance of C4H9Cl. Units for average rate: mol/L·s or mol L-1 s-1 or M/s or M s-1 . The average rate decreases with time. Plot [C4H9Cl] versus time. The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. Instantaneous rate is different from average rate. We usually call the instantaneous rate the rate.

C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Reaction Rates Reaction Rate and Stoichiometry For the reaction C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) we know In general for aA + bB  cC + dD

Concentration and Rate In general rates increase as concentrations increase. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)

Concentration and Rate For the reaction NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) we note as [NH4+] doubles with [NO2-] constant, the rate doubles, as [NO2-] doubles with [NH4+] constant, the rate doubles, We conclude rate  [NH4+][NO2-]. Rate law: The constant k is the rate constant.

Concentration and Rate Exponents in the Rate Law For a general reaction with rate law we say the reaction is mth order in reactant 1 and nth order in reactant 2. The overall order of reaction is m + n + …. A reaction can be zeroth order if m, n, … are zero. Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry.

Using Initial Rates to Determine Rate Laws Given data: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) Expt. # [NO] / M [H2] / M Rate / M s-1 1 0.10 1.23x10-3 2 0.20 2.46x10-3 3 4.92x10-3 Determine Rate Law for the reaction. i.e. Rate = k [NO]x [H2]y ; Find x , y , and k .

Given data: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) Expt. # [NO] / M [H2] / M Rate / M s-1 1 0.10 1.23x10-3 2 0.20 2.46x10-3 3 4.92x10-3

Given data: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) Expt. # [NO] / M [H2] / M Rate / M s-1 1 0.10 1.23x10-3 2 0.20 2.46x10-3 3 4.92x10-3

The Change of Concentration with Time First-Order Reactions Goal: convert rate law into a convenient equation to give concentrations as a function of time. For a first-order reaction, the rate doubles as the concentration of a reactant doubles. Plot [C4H9Cl] versus time. Simulation

The Change of Concentration with Time First-Order Reactions

The Change of Concentration with Time First-Order Reactions The first-order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 yr-1 . A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/cm3 of water. Assume that the average temperature of the lake is 12oC. (A) What is the concentration of the insecticide on June 1 of the following year? (B) How long would it take for the concentration of the insecticide to drop to 3.0x10-7 g/cm3 ?

The first-order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 yr-1 . A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/cm3 of water. Assume that the average temperature of the lake is 12oC. (A) What is the concentration of the insecticide on June 1 of the following year? (B) How long would it take for the concentration of the insecticide to drop to 3.0x10-7 g/cm3 ?

(B) How long would it take for the concentration of the insecticide to drop to 3.0x10-7 g/cm3 ?

The Change of Concentration with Time Second-Order Reactions For a second-order reaction with just one reactant A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0 For a second-order reaction, a plot of ln[A]t vs. t is not linear.

The Change of Concentration with Time Second-Order Reactions

The Change of Concentration with Time Second-Order Reactions The NO2 reaction has a rate constant of 0.543 M-1 s-1 . If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 hr ?

The NO2 reaction has a rate constant of 0. 543 M-1 s-1 The NO2 reaction has a rate constant of 0.543 M-1 s-1 . If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 hr ?

The Change of Concentration with Time Zeroth-Order Reactions For a zeroth-order reaction with just one reactant A plot of [A]t versus t is a straight line with slope -k and intercept [A]0 Applicable to catalysis on metal surfaces.

The Change of Concentration with Time Zeroth-Order Reactions A zeroth-order reaction has a rate constant of 1.1x10-7 M s-1 . The reaction began with a reactant concentration of 0.0200 M . What is the fraction of reactant concentration remaining after 45.0 hr ?

A zeroth-order reaction has a rate constant of 1. 1x10-7 M s-1 A zeroth-order reaction has a rate constant of 1.1x10-7 M s-1 . The reaction began with a reactant concentration of 0.0200 M . What is the fraction of reactant concentration remaining after 45.0 hr ? Solution Key

The Change of Concentration with Time Half-Life Half-life is the time taken for the concentration of a reactant to drop to half its original value. For a first-order process, half life, t½ is the time taken for [A]0 to reach ½[A]0. Mathematically,

The Change of Concentration with Time Half-Life For a second-order reaction, half-life depends on the initial concentration: For a zeroth-order reaction:

Summary of Rate Laws First-Order Second-Order Zeroth-Order DRL k[A] k[A]2 k IRL [A]t = [A]oe-kt ln[A]t = -kt + ln[A]o 1/[A]t = kt + 1/[A]o [A]t = -kt + [A]o Linear Equation ln[A]t vs. t 1/[A]t vs. t [A]t vs. t Linear Plot Half-Life ln(2)/k 1/k[A]o [A]o/2k Units on k time-1 M-1 time-1 M time-1 m = -k b = ln[A]o m = k b = 1/[A]o m = -k b = [A]o

Temperature and Rate The Collision Model As temperature increases, the rate increases.

Temperature and Rate The Collision Model The collision model: in order for molecules to react they must collide. The greater the number of collisions the faster the rate. The more molecules present, the greater the probability of collision and the faster the rate. The higher the temperature, the more energy available to the molecules and the faster the rate. Complications: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

Temperature and Rate The Orientation Factor

Temperature and Rate Activation Energy Arrhenius: molecules must posses a minimum amount of energy to react. Why? In order to form products, bonds must be broken in the reactants. Bond breakage requires energy. Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

Temperature and Rate Activation Energy

Temperature and Rate The Arrhenius Equation Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. A is called the frequency factor. A is a measure of the probability of a favorable collision. Both A and Ea are specific to a given reaction.

Rate Constant Comparison at two different temperatures.

Temperature and Rate Determining the Activation Energy If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation: From the above equation, a plot of ln k versus 1/T will have slope of –Ea/R and intercept of ln A.

Temperature and Rate

Temperature and Rate Determining the Activation Energy If we do not have a lot of data, then we recognize Ea ~ 160 kJ/mol (previous slide of ln(k) versus 1/T plot)

Reaction Mechanisms The balanced chemical equation provides information about the beginning and end of reaction. The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. Elementary Steps Elementary step: any process that occurs in a single step.

Reaction Mechanisms Rate Laws for Elementary Steps

Catalysis A catalyst changes the rate of a chemical reaction. There are two types of catalyst: homogeneous, and heterogeneous. Chlorine atoms are catalysts for the destruction of ozone.

Catalysis

Catalysis Enzymes

Radioactivity Nuclear Equations Nucleons: particles in the nucleus: p+: proton n0: neutron. Mass number: the number of p+ + n0. Atomic number: the number of p+. Isotopes: have the same number of p+ and different numbers of n0. In nuclear equations, number of nucleons is conserved: 23892U  23490Th + 42He

Radioactivity Types of Radioactive Decay

Radioactivity Types of Radioactive Decay

Neutron-to-Proton Ratio The heavier the nucleus, the more neutrons are required for stability. The belt of stability deviates from a 1:1 neutron to proton ratio for high atomic mass.

Patterns of Nuclear Stability Radioactive Series For 238U, the first decay is to 234Th (-decay). The 234Th undergoes -emission to 234Pa and 234U. 234U undergoes -decay (several times) to 230Th, 226Ra, 222Rn, 218Po, and 214Pb. 214Pb undergoes -emission (twice) via 214Bi to 214Po which undergoes -decay to 210Pb. The 210Pb undergoes -emission to 210Bi and 210Po which decays () to the stable 206Pb.

Rates of Radioactive Decay 90Sr has a half-life of 28.8 yr. If 10 g of sample is present at t = 0, then 5.0 g is present after 28.8 years, 2.5 g after 57.6 years, etc. 90Sr decays as follows 9038Sr  9039Y + 0-1e Each isotope has a characteristic half-life. Half-lives are not affected by temperature, pressure or chemical composition. Natural radioisotopes tend to have longer half-lives than synthetic radioisotopes.

Rates of Radioactive Decay

Rates of Radioactive Decay Half-lives can range from fractions of a second to millions of years. Naturally occurring radioisotopes can be used to determine how old a sample is. This process is radioactive dating.

Rates of Radioactive Decay Dating Carbon-14 is used to determine the ages of organic compounds because half-lives are constant. We assume the ratio of 12C to 14C has been constant over time. For us to detect 14C the object must be less than 50,000 years old. The half-life of 14C is 5,730 years. It undergoes decay to 14N via -emission: 146C 147N + 0-1e

Rates of Radioactive Decay Calculations Based on Half Life Radioactive decay is a first order process: In radioactive decay the constant, k, is the decay constant. The rate of decay is called activity (disintegrations per unit time). If N0 is the initial number of nuclei and Nt is the number of nuclei at time t, then

Rates of Radioactive Decay Calculations Based on Half Life With the definition of half-life (the time taken for Nt = ½N0), we obtain A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 s-1 . The half-life of 14C is 5715 yr. What is the age of the archeological sample? [Answer: 2,229 yr]

A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 s-1 . The half-life of 14C is 5715 yr. What is the age of the archeological sample? [Answer: 2,229 yr] HW Key

Chemical Kinetics [A]t = [A]oe-kt [A]t = -kt + [A]o 1/[A]t = kt + 1/[A]t