Chemical Kinetics *All of the v’s in this lecture were changed to r’s (if this is incorrect then the overlying textboxes can just be deleted)

The Rate of Reaction Chemical kinetics is concerned with the rate at which chemical reactions take place The rate of reaction is the change in concentration of a reactant or a product over time (in M/s)

The Rate of Reaction For the reaction A  B We want the rate of reaction to be positive, so for a reactant, we put a negative sign since

The Rate of Reaction For the reaction 2A  B : since A disappears twice as fast as B appears In general, for the reaction aA + bB  cC + dD Example: Write the rate expression for the following reaction: Solution:

The Rate of Reaction The rate of reaction can be determined from a graph of the concentration of a reactant or product versus time The rate of reaction at a particular time is given by the slope of the tangent at this point The rate of reaction decreases with time as the reactants are depleted

The Rate Law The rate law relates the rate of reaction to the concentrations of reactants and a proportionality constant (the rate constant) The effect of the concentration of a reactant is best determined by measuring the initial rate of reaction The rate of the inverse reaction (products  reactants) is negligible as there are not yet any products to react To determine the effect of the concentration of a reactant on the reaction rate, the concentrations of other reactants must be constant

The Rate Law eg.; F2(g) + 2 ClO2(g)  2 FClO2(g) [F2] (M) [ClO2] (M) Initial Rate (M/s) 0.10 0.010 1.2 x 10-3 0.10 0.040 4.8 x 10-3 0.20 0.010 2.4 x 10-3 If we keep [F2] constant, it is observed that the initial rate increases by a factor of four if [ClO2] increases by a factor of four If we keep [ClO2] constant, it is observed that the initial rate increases by a factor of two if [F2] increases by a factor of two

The Rate Law k is the rate constant for the reaction You can take any of the points from the empirical data to find the value of k Taking the first: This method is called the isolation method

The Rate Law In general, for the reaction N.B. most of the time: The sum of the exponents in the rate law (in this example, x + y) is the global order of the reaction In this example, the reaction is of order x for A and of order y for B

The Rate Law Example: We measured rates of the reaction A + 2 B  C at 25oC. Using this data, determine the rate law and the rate constant for this reaction. Experiment [A] initial [B] initial initial rate (M/s) 1 0.100 0.100 5.50 x 10-6 2 0.200 0.100 2.20 x 10-5 3 0.400 0.100 8.80 x 10-5 4 0.100 0.300 1.65 x 10-5 5 0.100 0.600 3.30 x 10-5

The Rate Law Solution: In the second and third experiments, the initial [B] is constant, and it is seen that the rate quadruples if we double [A]: In the fourth and fifth experiment, the initial [A] is constant, and it is seen that the rate doubles if we double [B]: Thus, the rate law is: N.B. We could have used other pairs of experiments too. To determine the value of k, we use the first experiment:

Concentration as a Function of Time Rate laws allow us to determine the concentrations of reactants at any moment in time For example, for a reaction of global first order that is first order with respect to A: where [A]o is the concentration of A at t=0 (which need not necessarily the beginning of the experiment)

Concentration as a Function of Time The formula is the integrated rate law for a first order reaction We can modify the formula This form of integrated rate law tells us that a graph of ln[A] versus time is a straight line with a slope of -k Experimentally, we determine k this way

Concentration as a Function of Time Example: The reaction 2A  B is first order with respect to A, and the rate constant is 2.8 x 10-2 s-1 at 80oC. How much time (in seconds) would it take for [A] to decrease from 0.88 M to 0.14 M?

Half-life The half-life of a reaction, t1/2, is the time required for the concentration of a reactant to decrease by half For a reaction of global first order: For a reaction of global first order, the half-life is independent of the initial concentration

Half-life Example: The half-life of a first order reaction is 84.1 min. Calculate the rate constant of the reaction. Solution: or

Half-life Example: 14C is a radioactive isotope that incorporates itself within living organisms. The half-life of 14C is 5730 years (The nuclear disintegration is a global first order reaction). If we find a piece of cloth produced from a previously living source that contains only 10% of the 14C that it would contain if it was living, what is the age of this piece of cloth?

The Effect of Temperature on the Rate Constant When the temperature increases, reaction rates increase, and thus the rate constants increase There are very few exceptions to this rule One exception is a reaction catalyzed by an enzyme (at high temperatures, the enzyme is denatured and will no longer function properly)

Activation Energy The activation energy is the energy barrier separating reactants from products (the reaction is either endothermic or exothermic) When the temperature is increased, the fraction of molecules with a kinetic energy greater than the activation energy increases Increasing the temperature increases the chance that a collision will have sufficient kinetic energy

The Arrhenius Equation The Arrhenius equation expresses the rate constant as: where Ea is the activation energy, R=8.3145 JK-1mol-1, T is the temperature in Kelvin, and A is the frequency factor (the frequency of collisions) The value of A does not vary significantly with varying temperatures A graph of ln k versus 1/T has a slope of -Ea/R

The Arrhenius Equation We can determine the value of the activation energy using the two rate constants, k1 and k2, at temperatures T1 and T2 We subtract the first equation from the second

The Arrhenius Equation Example: The rate constant of the (first order) reaction of methyl chloride (CH3Cl) with water to form methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 x 10-10 s-1 at 25oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol.

Reaction Mechanisms and Rate Laws The global reaction is often the sum of a series of simple reactions (elementary steps) The sequence of elementary steps that give the product is the reaction mechanism eg.; For the reaction 2 NO(g) + O2(g)  2 NO2(g) , the reaction mechanism is The sum of the two elementary steps is the global reaction In this example, N2O2 is an intermediate, i.e., a species that appears in the reaction mechanism but not in the overall balanced reaction An intermediate is produced and then consumed Step 1: Step 2:

Reaction Mechanisms and Rate Laws The number of molecules that react in an elementary step is the molecularity of that step A unimolecular step involves a single molecule A bimolecular step involves two molecules A trimolecular step involves three molecules A trimolecular step is rare because it is unlikely that three molecules will collide simultaneously The rate law for an elementary step is given by the stoichiometry of the step, i.e., For A  products, r = k[A] For A + B  products, r = k[A][B] For 2 A  products, r = k[A]2

Reaction Mechanisms and Rate Laws A mechanism can be deduced from experimental observations The reaction mechanism must obey two rules The sum of the elementary steps must correspond to the balanced equation of the global reaction The slowest elementary step of the series, the rate-determining step, should match that of the overall reaction The rate-determining step determines the rate of the overall reaction

Reaction Mechanisms and Rate Laws Take, for example, the reaction All alone, this reaction is slow However, with the presence of iodide ions, the reaction is fast and the rate law is The mechanism cannot simply be two molecules of H2O2 that collide and then react

Reaction Mechanisms and Rate Laws A possible mechanism is If step 1 is the rate-determining step, the rate law for the global reaction would be Such a rate law is observed experimentally, so the mechanism is plausible

Reaction Mechanisms and Rate Laws In the proposed mechanism, IO- is an intermediate since it is produced in the first step and consumed in the second In the proposed mechanism, I- is not an intermediate since it is present before the reaction occurs and is still present once the reaction is finished I- is a catalyst

Reaction Mechanisms and Rate Laws Example: We believe that the formation of NO and CO2 from NO2 and CO happens in two steps: The experimentally determined rate law is k[NO2]2. (a) Give the equation of the global reaction. (b) Indicate which species is an intermediate. (c) What can be said about the relative rates of Steps 1 and 2? Solution: (a) The sum of Steps 1 and 2 is: NO2 + CO  NO + CO2 (b) NO3 is an intermediate (c) Step 1 is the slow step since its kinetics correspond to that of the global reaction (the value of [CO] would be in the rate law if Step 2 was the rate-determining step)

The reaction A(aq) B(aq) is a first-order reaction with respect to A The reaction A(aq) B(aq) is a first-order reaction with respect to A. The half-life at 25.0oC is 875.3 s. If the activation energy of this reaction is 35.1 kJ/mol, what would the concentration of A be after 1000.0 s if we have an initial concentration of A of 0.500 M and temperature is held constant at 50.0oC?