Reaction Rate (aka – Chemical Kinetics)

Reaction Rate (aka – Chemical Kinetics) • Reaction rate– how fast reactants disappear and how fast product appears

Reaction Rate • Reaction Rate = ∆ [A] ∆ t • Example: CO(g) + NO2(g) CO2(g) + NO(g) - at t = 4.0 min, [CO2] = .12 mol/L - at t = 8.0 min, [CO2] = .24 mol/L - reaction rate = .24 mol/L - .12 mol/L 8.0 min – 4.0 min = 0.030 mol/L . min • Unit for reaction rate = conc. with some time unit • Products have a (+) rate • Reactants have a (-) rate

Kinetics – Techniques for Measuring Reaction Rates • If a gas is produced (ex: Zn + HCl) • Mass of rxn mixture before and during • Pressure or volume of gas • Acid – Base Reaction ( ex: Cl2 + H2O → HOCl + H+ + Cl-) • Titration (also other rxns, e.g. redox) • Measure pH • Conductivity (if there is a significant change in ion concentration) • Light Absorption ( ex: CV+ + OH-) • Rxns that ∆ from clear to solid or from dark ↔ clear • Clock Techniques (Iodine Clock) • Rxns w/ a very fast final step (virtually instant recognition of end)

Average vs. Instantaneous Rate Use figure 15.2 p. 508 in text

Reaction Rate and Stoichiometry For the reaction: 2 NO(g) + O2(g) → 2 NO2(g) • - ∆ [NO] = ∆ t • 2 -∆ [O2] = ∆ t • ∆ [NO2] ∆ t Example: Write the rate equation in terms of hydrogen for the formation of ammonia from nitrogen and hydrogen. 3H2(g) + N2(g)→ 2NH3(g) • - ∆ [H2] = ∆ t • - 3 ∆[N2] = ∆ t • 3/2 ∆ [NH3] ∆ t If the rate of formation of NH3 is 6.0 x 10-2 M/s what is the rate for the disappearance of hydrogen? - 9.0 x 10-2 M/s

Reaction Coordinate Diagram • Energy of products is lower than energy of reactants • energy lost, exothermic, -∆H • Energy of products is higher than energy of reactants • energy gained, endothermic, +∆H

Factors Affecting Reaction Rate • Nature of the Reactants (Rxn Mechanism) • Concentration (Differential and Integrated Rate laws 3. Particle Size (Surface Area) • more particles on the surface = more particles available for collisions • more collisions = more act. complex = more product • smaller particles give you more surface area 4. Agitation • this puts more liquid/gas particles in contact with the solid = ↑ collisions = ↑ act. complex = ↑ product

Factors Affecting Reaction Rate • 5. Pressure • ↑ pressure by ↓ volume – puts particles closer together = ↑ collisions = ↑ act. complex = ↑ product • All of these factors are similar, in terms of • explanation, to concentration!!! 6. Temperature • most effective at speeding up a reaction • ↑ temp. = ↑ KE (particles moving faster) • particles move faster leading to more collisions • the collisions are also harder • these harder collisions contain the needed energy to overcome the Ea • therefore the reaction rate will increase

Temperature (continued) What is this thing called? This area represents the number of particles with enough KE to be able to form the activated complex The effect of temperature on rate is exponential and is given by the Arrhenius Equation: k = Ae-Ea/RT or ln k = - Ea/RT + ln A A is the frequency factor and the rest represents the fraction of particles with minimum energy for reaction. Graphing ln k vs. 1/T gives a line with a slope of –Ea/R

Effect of Concentration on Rate • The larger the concentration of reactant molecules, the faster the reaction will go. • Increases the frequency of reactant molecule collisions • Different reactants have different effects on reaction rates • based on the orderof the reactant • Differential Rate Law (rate expression) – represents the rate of a reaction in terms of the concentration of reactants • Example • For: H2 + O2 H2O Rate = k[H2]2[O2] • k = rate constant, specific for a particular reaction and at certain temperature, large k = fast reaction

The Method of Initial Rates: The table below gives initial reaction rates for the reaction 2 NO(g) + O2(g) → 2 NO2(g) Initial concentrations (mol/L) Initial Rate (mol/L.s) Experiment 1. Determine the order for each reactant, the overall order and write the rate expression. 2. Calculate the value for the rate constant. 3. Determine the rate if the concentrations of NO and O2 are 0.50 and 0.75 mol/L, respectively.

The table below gives initial reaction rates for the reaction 2 ClO2(aq) + 2 OH-(aq) → ClO3-(aq) + ClO2-(aq) + H2O(l) Initial concentrations (mol/L) Initial Rate (mol/L.s) Experiment Perform the same operations as in 1 – 3 on the previous slide, except use 0.050 M for ClO2 and 0.075 M for OH- to calculate the rate. Answers: 1. rate = k[ClO2]2[OH-]; 2. 75 L2/mol2.s 3. 0.014 mol/L . s

Integrated Rate Laws • In order to determine the rate law for a reaction from a set of data consisting of concentration (or the values of some function of concentration) versus time, make three graphs. • [A] versus t (linear for a zero order reaction) • ln [A] versus t (linear for a 1st order reaction) • 1 / [A] versus t (linear for a 2nd order reaction) • The graph that is linear indicates the order of the reaction with respect to A. Then, you can choose the correct rate equation:

For a zero order reaction,as shown in the following figure, the plot of [A] versus time is a straight line with k = -slope of the line. Other graphs are curved for a zero order reaction. For a zero order reaction: rate = k (k = - slope of line)

For a first order reaction,as shown in the following figure, the plot of the logrithm of [A] versus time is a straight line with k = - slope of the line. Other graphs are curved for a first order reaction. For a 1st order reaction: rate = k[A] (k = - slope of line) For a 1st order reaction, rate = k[A] (k = - slope of line)

For a second order reaction, as shown in the following figure, the plot of 1/[A] versus time is a straight line with k = slope of the line. Other graphs are curved for a second order reaction. For a 2nd order reaction: rate = k[A]2 (k = slope of line)

The Common Integrated Rate Laws: For a zero order reaction: A → products , rate = k The integrated rate law is [A] = -kt + [A]0 For a first order reaction: A → products , rate = k[A]The integrated rate law is ln [A] = -kt + ln [A]0 For a second order reaction: 2A → products, or A + B →products (when [A] = [B]) , rate = k[A]2The integrated rate law is 1/[A] = kt + 1/[A]0

Using the Integrated Rate Laws How Long Does it Take? To determine t, the time required for the initial concentration of a reactant to be reduced to some final value, we need to know: The initial concentration, [Ao]. The final concentration, [A]. The order of the reaction or enough information to determine it. The rate constant, k, for the reaction or enough information to determine it. Substitute this information into the integrated rate law for a reaction with this order and solve for t.

Using the Integrated Rate Laws How Much Remains After a Given Time? To determine [A], the concentration of a reactant remaining after some time, t, we need to know: The initial concentration, [Ao]. The length of time the reaction ran, t. The order of the reaction or enough information to determine it. The rate constant, k, for the reaction or enough information to determine it. Substitute this information into the integrated rate law for a reaction with this order and solve for [A].

Using the Integrated Rate Laws What Concentration Was Present Initially? To determine [Ao], the initial concentration of a reactant, we need to know: The final concentration, [A]. The length of time, t, the reaction ran to reach the final concentration. The order of the reaction or enough information to determine it. The rate constant, k, for the reaction or enough information to determine it. Substitute this information into the integrated rate law for a reaction with this order and solve the equation for [Ao].

Sample Problem: • Hydrogen peroxide decomposes in a dilute NaOH solution in a first-order reaction: 2 H2O2(aq) → 2 H2O(l) + O2(g) Rate = k[H2O2] with k = 1.06 x 10-3 min-1 What is the fraction of H2O2 remaining after 100. min? What is the concentration of H2O2 after 100. min if the [initial] is 0.020 M?

Half-life (Time it takes to reach ½ [A]0) For first order kinetics the half-life for a reaction is given by the equation: t1/2 = ln 2/k = 0.693/k This will be by far the most common type of ½-life calculation you will have to do, solving for either k or the ½-life of a reaction. ½ - lives for nuclear reactions are also governed by 1st – order kinetics. Notice that since the value for ½-life is determined only by the rate constant then it also remains constant. For example: T1/2 for the decay of 14C to 14N is 5570 years. If a baby mammoth is discovered that has 12.5% of it’s original 14C concentration, how long ago did it die? Answer: 16,710 yrs.

Half-life (Time it takes to reach ½ [A]0) For 0-order reactions t1/2 = [A]0/2k For 2nd-order reactions t1/2 = 1/k[A]0 Note that unlike first order these 2 ½-lives depend on initial concentration. Since that will be different at the beginning of each ½-life, t1/2 is not constant as it is for 1st-order kinetics.

Collision Theory of Kinetics • Kineticsis the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. • In order for a reaction to take place, the reacting molecules must collide into each other. • Once molecules collide they may react together or they may not, depending on two factors - • Whether the collision has enough energy to "break the bonds holding reactant molecules together"; • Whether the reacting molecules collide in the proper orientation for new bonds to form.

Effective Collisions • Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. • The higher the frequency of effective collisions the faster the reaction rate. • When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex • It is a transition state between reactant and product • It has a very short lifetime (10-13 s) • Has to form for product to be formed

Activated Complex • The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, Ea • This is the minimum amount of energy that particles must have in order to react. • The larger the activation energy, the slower the reaction • The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat. • Different reactions have different activated complexes and therefore different activation energies

Reaction Mechanisms Nearly all reactions involving molecules occur in a series of steps. For example, the reaction between Cl2 (g) + CH4 (g) Cl2(g) + CH4(g) CH3Cl(g) + HCl(g) • Cl2 Cl + Cl (fast) • Cl + CH4  CH3Cl + H (slow) • H + Cl  HCl (very fast) • individual steps = elementary steps • all steps together = reaction mechanism • the slowest step determines the rate of the reaction • called the rate determining step • eliminating the intermediates allows us to write the balanced equation of the mechanism • Intermediates – product in one step, reactant in another • Cl, Cl, H in the reaction above

Reaction Mechanisms • The rate law of an elementary step can be written from its molecularity (# of reactant molecules; order). • A → products Unimolecular rate = k [A] • A + A → products Bimolecular rate = k [A]2 • A + B → products Bimolecular rate = k [A][B] • A + B + C → products Termolecular (rare) For a mechanism to be considered viable, must meet 2 criteria: • Elementary steps must give overall balanced equation. • Must agree with experimentally determined rate law.

Reaction Mechanisms • 2NO2(g) + F2(g) → 2NO2 F(g); rate = k [NO2][F] The proposed mechanism for the above reaction is given below. Determine if the mechanism is viable. NO2(g) + F2(g) → NO2 F(g) + F(g) slow NO2(g) + F(g) → NO2 F(g) fast K1 K2 The intermediate is F and if they are crossed out and we add the 2 NO2 molecules on the left and the 2 NO2F molecules on the right, we do get the correct balanced equation. Since the rate determining step is bimolecular and involves the collision of 1 NO2 and 1 F2 molecule the rate law would be rate = k [NO2][F] which is also correct, so the mechanism could be the correct one.

Reaction Mechanisms A two-step mechanism has been suggested for the reaction between iodine monochlorideand hydrogen gas: rate = k [H2][ICl] step 1: H2(g) + ICl(g) → HI(g) + HCl(g) (slow) step 2: HI(g) + ICl(g) → I2(g) + HCl(g) (fast) Write the equation for the overall reaction. Identify any reaction intermediates. What is the molecularity of each elementary step? Write the rate law predicted by this mechanism. Is the one given above a possible mechanism for this reaction? What are the orders in each reactant and the overall order for the reaction?

Reaction Mechanisms Answers to previous slide: H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g); HI(g); 2 (bimolecular) Rate = k [H2][ICl]; yes; 1, 2 A somewhat more complicated situation arises if there is an intermediate in the rate determining step. This may be solved in there is a fast equilibrium step in the mechanism which produces the intermediate. Example: 2 NO(g) + O2(g) → 2 NO2(g); rate = k[NO][H2] NO2(g) + O2(g) ↔ NO3 (g) (fast, equilibrium) NO(g) + NO3(g) → 2 NO2 (g) (slow)

Reaction Mechanisms Solution:

Reaction Mechanisms Try this one: 2 NO + 2 H2 → 2 H2O + N2 rate = k[NO]2[H2] Mechanism 1: NO + H2 → N + H2O (slow) N + NO → N2O (fast) N2O + H2 → N2 + H2 O (fast) K1 K2 K3 No? Try this one: Work? Mechanism 2: NO + H2  N + H2O (fast) N + NO → N2O (slow) N2O + H2 → N2 + H2 O (fast) K1 K-1 K2 K3

Using a Catalyst A catalyst • is a substance that speeds up a reaction, but isn’t used up in the reaction • provides a “different pathway” that requires lower Ea • lower Ea = more collisions having the proper amount of energy = ↑ act. complex = ↑ product

Reaction Rate (aka – Chemical Kinetics)