Reaction Rates The term reaction rate refers to how fast a product is formed or a reactant is consumed. Consider the following generic reaction: aA(g) + bB(g) cC(g) + dD(g) 1 a Δ[A] Δt 1 b Δ[B] Δt Rateave = - = - 1 c Δ[C] Δt 1 d Δ[D] Δt = =

Rates are always positive quantities. The concentration of reactants do not change at the same rate when coefficients are different. Consider the following specific example: N2(g) + 2O2(g) 2NO2(g) Δ[N2] Δt 1 2 Δ[O2] Δt Rateave = - = - 1 Δ[NO2] M = = 2 Δt s

Instantaneous Rate vs Average Rate The following graph shows the decomposition of N2 as a function of time. There are two different rates shown in the graph. The instantaneous rate is the rate of change (Δ) at a specific instant of time, i.e. 40 s.

In a non-calculus environment, this is determined by calculating the slope of a tangent line containing that point. The average rate is computed over a time interval necessitating using two different times. Δ[N2] Δt Δ[N2] Δt 0.30 M – 0.55 M Rateinst Rateinst = = - - = = - 60. s – 30. s Rateinst = 8.3 x 10-3 M s-1

This is the average rate from 20. s to 60. s. Δ[N2] Δt 0.30 M – 0.67 M Rateavs = - = - 60. s – 20. s Rateave = 9.2 x 10-3 M s-1

[N2] vs Time average rate instantaneous rate [N2] Δ[N2] Δt Time (s)

Factors Affecting Reaction Rates Chemical reactions involve the breaking of bonds (endothermic) and the forming of bonds (exothermic). The factors affecting reaction rates: The physical state of reactant rates. When reactants are in different phases (states), the reaction is limited to their area of contact.

The concentration of reactants. As the concentration of reactants increase, the rate of reaction increases. Temperature As the temperature is increased, molecules have more kinetic energy resulting in higher reaction rates. Presence of a catalyst. Increases the rate of reaction without being consumed.

Rate Laws Consider the following generic reaction: aA(g) + bB(g) cC(g) + dD(g) The rate law is written: Rate = k[A]x[B]y k is called the rate constant and is temperature dependent. The exponents x and y are called reaction orders.

The overall reaction order is the sum of the orders with respect to each reactant. The reaction orders in a rate law indicate how the rate is affected by the concentration of the reactants. The effect of reactant concentration can not be predicted from the balanced equation. only be determined empirically!

The rate of a reaction will decrease as the reaction proceeds because the reactants are being consumed. A very important distinction needs to be made at this point: The rate of a reaction depends on concentration but k, the rate constant, does not! The rate constant and the rate of reaction are affected by temperature and the presence of a catalyst.

Reaction Rate and Stoichiometry How is the rate of disappearance of N2 and O2 related to the rate of appearance of NO2 in the reaction shown below. N2(g) + 2O2(g) 2NO2(g) Δ[N2] Δt 1 2 Δ[O2] Δt Rate = - = - = 1 2 Δ[NO2] = Δt

(b) If the rate of decomposition of N2 at an instant is 3.7 x 10-6 M•s-1, what is the rate of disappearance of O2 and the rate of appearance of NO2? Δ[N2] Δt Rate = - = 3.7 x 10-6 M•s-1 Δ[O2] Δt = 2 × 3.7 x 10-6 M•s-1 = 7.4 x 10-6 M/s Δ[NO2] Δt = 2 × 3.7 x 10-6 M•s-1 = 7.4 x 10-6 M/s

Determining Rate Laws The following data was measured for the reaction of nitrogen and oxygen to form nitrogen(IV) oxide. N2(g) + 2O2(g) 2NO2(g)

Trial [N2] (M) [O2] Init Rate (M/s) 1 0.12 1.19×10-3 2 0.23 2.44×10-3 0.22 4.92×10-3

Determine the rate law for this reaction. k[N2]x[O2]y = Rate 1 2.44 × 10-3 M/s 1.19 × 10-3 M/s = k(0.12)x(0.23)y k(0.12)x(0.12)y 2 = 2y y = 1

. Rate 3 k[N2]x[O2]y = Rate 1 4.92 × 10-3 M/s k(0.22)x(0.12) = 1.19 × 10-3 M/s k(0.12)x(0.12) 4 = 2x x = 2 Rate = k[N2]2[O2]

(b) Calculate the rate constant. Rate = k[N2]2[O2] 1.19 × 10-3 Ms-1 = k(0.12 M)2(0.12 M) k = 0.69 M-2s-1 (c) Calculate the rate when [N2] = 0.047 M and [O2] = 0.16 M Rate = 0.69 M-2s-1 × (0.047 M)2 × 0.16 M Rate = 2.4 x 10-4 M/s

Integrated Rate Laws The second type of rate law is called the integrated rate law which shows the concentration as a function of time. There are three possible orders for a reaction, zeroth, first, or second.

Zeroth Order Kinetics The integrated rate law is given by: [A] = -kt + [A]0 A plot of [A] vs Time produces a straight line with a slope equal to the negative rate contant, -k.

The reaction rate is independent of concentration. The reaction rate is independent of time.

For 0th order kinetics: Rate Law: Rate = k Integrated Rate Law: [A] = -kt + [A]0 Plot needed for straight line: [A] vs T Slope: -k [A]0 Half-Life: t1/2 = 2k

First-Order Kinetics The integrated rate law for 1st order kinetics is given by: ln[A] = -kt + ln[A]0 A plot of ln[A] vs Time produces a straight line with a slope equal to the negative rate contant, -k.

The reaction rate is directly proportional to the concentration. The reaction rate decreases but not linearly with time.

A first-order reaction is one in which the rate depends on the concentration of a single reactant to the first power. aA(g) Product Δ[A] Δt Rate = - = k[A] Δ[A] = - k Δt [A] ln[A]t – ln[A]0 = -kt

ln[A]t = -kt + ln[A]0 . y = mx + b The slope of the straight line gives -k, the rate constant, and ln[A]0 is the y-intercept. For a first-order reaction, the half-life (the time required for one-half of a reactant to decompose) has a constant value. ln 2 k 0.693 k t1/2 = =

For a first-order reaction, the half-life is only dependent on k, the rate constant and remains the same throughout the reaction. The half-life is not affected by the initial concentration of the reactant.

For 1st order kinetics: Rate Law: Rate = k[A] Integrated Rate Law: ln[A] = -kt + ln[A]0 Plot needed for straight line: ln[A] vs T Slope: -k 0.693 Half-Life: t1/2 = k

Second-Order Kinetics The integrated rate law for 2nd order kinetics is given by: 1/[A] = kt + 1/[A]0 A plot of 1/[A] vs Time produces a straight line with a slope equal to the rate contant, k.

The reaction rate is directly proportional to the concentration. The reaction rate decreases but not linearly with time.

A second-order reaction is one in which the rate depends on the concentration of a single reactant concentration raised to the second power or concentrations of two different reactants, each raised to the first power. aA(g) Product or aA(g) + bB(g) Product Δ[A] Δt Δ[B] Δt Rate = - = - = k[A][B] or Δ[A]2 Δt Rate = - = k[A]2

The integrated rate law for a second-order reaction is given by: 1 1 [A]0 = k t + [A]t y = mx + b The slope of the straight line gives k, the rate constant, and is the y-intercept. 1 [A]0

. For a second-order reaction, the half-life (the time required for one-half of a reactant to decompose) is double the preceding one. 1 t1/2 = k[A]0

For 2nd order kinetics: Rate Law: Rate = k[A]2 Integrated Rate Law: Plot needed for straight line: Slope: k 1 1 1 1 1 [A]0 = kt + [A] [A] [A] 1 [A]0 vs T 1 Half-Life: t1/2 = k [A]0

Collision Theory Particles must collide for chemical reactions to occur. Not every collision leads to a reaction. For a reaction to occur, an “effective collision” must take place. An “effective collision” consists of two conditions.

The colliding particles must approach each other at the proper angle. v1 v2 H2 Cl2 v3 v4 HCl HCl

In addition, the colliding particles must have sufficient energy. At the point of impact, ΔKE = ΔPE. An elastic collision is assumed in which the law of conservation of mass-energy applies. The KE/molecule must be sufficient to break the H-H and the CI-CI in both hydrogen and chlorine.

Keep in mind that temperature is a measure of the KE/molecule. Some molecules will be traveling faster than others while the slower molecules will bounce off each other without reacting.

A less than ideal set of conditions for a collision. v1 v2 H2 Cl2 v3 v4 H2 Cl2

Factors Affecting Reaction Rates The bonding found in the reactants. Triple bonds are stronger than double bonds which are stronger than single bonds. The temperature of the system (reactants and products). The initial concentration of the reactants.

The amount of surface area when reactants are in more than one phase. The use of a catalyst to provide an alternate pathway.

Activation Energy In order to react, colliding molecules must have KE equal to or greater than a minimum value. This energy is called the activation energy (Ea) which varies from reaction to reaction. The Ea depends on the nature of the reaction and is independent of temperature and concentration.

The rate of a reaction depends on Ea. At a sufficiently high temperature, a greater number of molecules have a KE > Ea.

Maxwell-Boltzmann Distribution lower temperature higher temperature Ea

For the Maxwell-Boltzmann Distribution, only those molecules with energies in excess of Ea react. As shown on the graph, as the temperature is increased: The curve shifts to higher energies meaning more molecules to have energies greater than Ea. The curve gets broader and flatter but the area remains the same under the curve.

Energy Profile for Exothermic Rxn Minimum Energy for Reaction Ea Energy Content of Reactants ΔH = - Energy Content of Products

Energy Profile for Endothermic Rxn Minimum Energy for Reaction Ea Energy Content of Products ΔH = + Energy Content of Reactants

Arrhenius Equation The Arrhenius equation illustrates the dependence of the rate constant, k, on the frequency factor, A, the activation energy, Ea, the gas constant, R, and the absolute temperature, T. k = Ae-Ea/RT The frequency factor, A, is related to the frequency of collisions and the probability of these collisions being favorably oriented.

Taking the ln of both sides gives: ln k = ln A – Ea/RT Rearranging the equation above shows that the ln k is directly proportional to 1/T. ln k = -Ea/R • 1 T + ln A y = m x + b The equation also shows that the reaction rate decreases as the activation energy increases.

An unknown gas at 300.°C has a rate constant equal to 2.3 x 10-10 s-1 and at 350.°C the rate constant is determined to be 2.4 x 10-8 s-1. (a) Calculate Ea for this reaction. T1 = 300.°C = 573 K T2 = 350.°C = 623 K k1 = 2.3 x 10-10 s-1 k2 = 2.4 x 10-8 s-1 R = 8.31 J mol-1K-1 ln k2 k1 = Ea R [ 1 T1 - ] T2 1 T2

2.4 × 10-8 s-1 ln = 2.3 × 10-10 s-1 . Ea 1 573 K 1 623 K [ - ] 8.31 J mol-1 K-1 Ea = 2.7 × 105 J/mol

(b) Determine the rate constant at 425°C. k2 ln = 2.4 x 10-10 s-1 2.7 x 105 J mol-1 1 573 K 1 698 K [ - ] 8.31 J mol-1 K-1 k2 = 6.2 x 10-6 s-1

Reaction Mechanisms A reaction mechanism consists of a series of elementary steps indicating how reacting particles rearrange themselves to form products. Elementary steps are individual steps showing what molecules must collide with each other and the proper sequence of collisions. Most reactions occur in more than one step.

A reaction mechanism provides much more information than a balanced chemical equation. Intermediates are species that are formed and consumed between elementary steps which never appear in a balanced chemical equation. Not all elementary steps occur at the same rate and consequently are designated as fast or slow steps.

Each elementary step has a molecularity associated with it. An elementary step is referred to as a unimolecular step when one molecule decomposes or rearranges to form a product as shown below: A B A unimolecular reaction follows a first-order rate law, i.e. Rate = k[A]

An elementary step is referred to as a bimolecular step when two molecules collide to form a product as shown below: A + B C A bimolecular reaction follows a second-order rate law, i.e. Rate = k[A][B]

An elementary step is referred to as a termolecular step when three molecules collide to form a product as shown below: A + B + C D A termolecular reaction follows a third-order rate law, i.e. Rate = k[A][B][C] Termolecular reactions are very rare.

Determining a Rate Expression The slowest step in a mechanism is called the rate determining step and determines the rate of the overall reaction. To determine the rate expression from a mechanism do the following: The rate of the overall reaction is the same as the rate of the rate-determining step.

Write the rate expression for the slowest step. The rate expression must only include those species that appear in the balanced equation. Intermediates are species that are produced in one step and consumed in the next step. Intermediates can not appear in the rate expression because their concentrations are always small and undectable.

Determine the overall reaction and the rate expression for the suggested mechanism: NO2(g) + NO2(g) NO3(g) + NO(g) (slow) NO3(g) + CO(g) NO2(g) + CO2(g) (fast) NO2(g) + CO(g) NO(g) + CO2(g)

Note that the NO3 is an intermediate because it is produced in one step of the reaction and consumed in the next. Because the rate-determining step comes first, the intermediate NO3 cancels out. This is good because intermediates can not appear in the overall reaction. The rate expression is determined by using the rate-determining step and is given by: Rate = k[NO2][NO2] = k[NO2]2

The following reaction 2NO2(g) + O3(g) N2O5(g) + O2(g) has as its rate law, Rate = k[NO2][O3]. Which of the following mechanisms predicts a satisfactory rate law? Mechanism 1: NO2(g) + NO2(g) N2O2(g) + O2(g) (slow) N2O2(g) + O3(g) N2O5 (g) (fast)

Mechanism 2: NO2(g) + O3(g) NO3(g) + O2(g) (slow) NO3(g) + NO2(g) N2O5(g) (fast) For Mechanism 1: NO2(g) + NO2(g) N2O2(g) + O2(g) (slow) N2O2(g) + O3(g) N2O5 (g) (fast) 2NO2(g) + O3(g) N2O5(g) + O2(g)

The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct. However, the predicted rate law determined from the rate-determining step is: Rate = k[NO2]2 which does not agree with given rate law.

For Mechanism 2: NO2(g) + O3(g) NO3(g) + O2(g) (slow) NO3(g) + NO2(g) N2O5(g) (fast) 2NO2(g) + O3(g) N2O5(g) + O2(g) The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct.

The predicted rate law determined from the rate-determining step is: Rate = k[NO2][O3] which does agrees with given rate law. Mechanism 2 is consistent with the given rate law. When the intermediate is not a reactant in the rate-determining step, mechanisms are usually easier to work with.

When the intermediate is a reactant in the rate-determining step, mechanisms are more involved. Determine the overall reaction and the rate expression for the suggested mechanism: NO(g) + Br2(g) NOBr2(g) (fast) NOBr2(g) + NO(g) 2NOBr(g) (slow)

NO(g) + Br2(g) NOBr2(g) (fast) NOBr2(g) + NO(g) 2NOBr(g) (slow) The predicted rate law determined from the rate-determining step is: Rate = k[NOBr2][NO] which is incorrect because intermediates can not appear in the rate law. 2NO(g) + Br2(g) 2NOBr(g)

When this situation occurs, you must equate the forward and reverse rates obtained from the fast step as shown below. NO(g) + Br2(g) NOBr2(g) Ratef = Rater kf[NO][Br2] = kr[NOBr2] kf [NOBr2] = [NO][Br2] = k[NO][Br2] kr

Now you can use the rate law obtained originally but substitute for [NOBr2]. Rate = k[NOBr2][NO] [NOBr2] = k[NO][Br2] Rate = k[NO]2[Br2] There is no need to keep track of the individual rate constants because they would be incorporated into a single constant.

Catalysis A catalyst is a substance that increases the speed of a reaction without being consumed. Catalysts can be either homogeneous or heterogeneous depending if they are in the same phase as the reactants. The function of a catalyst is to provide an alternate pathway (mechanism) for a reaction which lowers the activation energy, Ea.

Reviewing the Arrhenius equation, shows that as activation energy, Ea, is made smaller, the rate constant ,k, increases. k = Ae-Ea/RT

For the given mechanism, identify the catalyst and the intermediate. O3 + Br O2 + BrO BrO + O Br + O2 Remember an intermediate is produced in one step and consumed in the other making BrO the intermediate. Remember a catalyst remains chemically unaltered during the reaction making the Br the catalyst.