General Chemistry M R NaimiJamal Faculty of Chemistry

General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

Contents 14 -1 The Rate of a Chemical Reaction 14 -2 Measuring Reaction Rates 14 -3 Effect of Concentration on Reaction Rates: The Rate Law 14 -4 Zero-Order Reactions 14 -5 First-Order Reactions 14 -6 Second-Order Reactions 14 -7 Reaction Kinetics: A Summary

Contents 14 -8 Theoretical Models for Chemical Kinetics 14 -9 The Effect of Temperature on Reaction Rates 14 -10 Reaction Mechanisms 14 -11 Catalysis Focus On Combustion and Explosions

14 -1 The Rate of a Chemical Reaction • Rate of change of concentration with time. 2 Fe 3+(aq) + Sn 2+ → 2 Fe 2+(aq) + Sn 4+(aq) t = 38. 5 s Δt = 38. 5 s [Fe 2+] = 0. 0010 M Δ[Fe 2+] = (0. 0010 – 0) M Rate of formation of Fe 2+= Δ[Fe 2+] Δt = 0. 0010 M 38. 5 s = 2. 6 x 10 -5 M s-1

Rates of Chemical Reaction 2 Fe 3+(aq) + Sn 2+ → 2 Fe 2+(aq) + Sn 4+(aq) 1 Δ[Fe 3+] Δ[Sn 4+] 1 Δ[Fe 2+] = = Δt 2 Δt Δt 2

General Rate of Reaction a. A+b. B→c. C+d. D Rate of reaction = rate of disappearance of reactants 1 Δ[B] 1 Δ[A] ==b Δt a Δt = rate of appearance of products 1 Δ[D] 1 Δ[C] = = d Δt c Δt

14 -2 Measuring Reaction Rates H 2 O 2(aq) → H 2 O(l) + ½ O 2(g) 2 Mn. O 4 -(aq) + 5 H 2 O 2(aq) + 6 H+ → 2 Mn 2+ + 8 H 2 O(l) + 5 O 2(g) Experimental set-up for determining the rate of decomposition of H 2 O 2. Oxygen gas given off by the reaction mixture is trapped, and its volume is measured in the gas buret. The amount of H 2 O 2 consumed and the remaining concentration of H 2 O 2 can be calculated from the measured volume of O 2(g).

Example: Determining and Using an Initial Rate of Reaction. H 2 O 2(aq) → H 2 O(l) + ½ O 2(g) Initial rate: -(-2. 32 M / 1360 s) = 1. 7 x 10 -3 M s-1 Rate = -Δ[H 2 O 2] Δt

Example: What is the concentration at 100 s? [H 2 O 2]i = 2. 32 M Rate = 1. 7 x 10 -3 M s-1 = - Δ[H 2 O 2] Δt -Δ[H 2 O 2] = -([H 2 O 2]f - [H 2 O 2]i) = 1. 7 x 10 -3 M s-1 x Δt [H 2 O 2]100 s – 2. 32 M = -1. 7 x 10 -3 M s-1 x 100 s [H 2 O 2]100 s = 2. 32 M - 0. 17 M = 2. 17 M

14 -3 Effect of Concentration on Reaction Rates: The Rate Law a A + b B …. → g G + h H …. Rate of reaction = k [A]m[B]n …. Rate constant = k Overall order of reaction = m + n + ….

Example: Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respect to Hg. Cl 2 and C 2 O 22 - and also the overall order of the reaction.

Example: Notice that concentration changes between reactions are by a factor of 2. Write and take ratios of rate laws taking this into account.

Example: R 3 = k [Hg. Cl 2]3 m[C 2 O 42 -]3 n R 2 = k[Hg. Cl 2]2 m[C 2 O 42 -]2 n k (0. 105)m [C 2 O 42 -]2 n R 2 = R 3 k (0. 052)m [C 2 O 42 -]3 n -5 7. 1 x 10 R 2 = 2 m = 3. 5 x 10 -5 R 3 2 m = 2. 0 therefore m = 1. 0

Example: R 2 = k[Hg. Cl 2]21[C 2 O 42 -]2 n = k(0. 105)(0. 30)n R 1 = k[Hg. Cl 2]11[C 2 O 42 -]1 n = k(0. 105)(0. 15)n R 2 = R 1 k(0. 105)(0. 30)n k(0. 105)(0. 15)n -5 (0. 30)n 7. 1 x 10 n = = 2 (0. 15)n 1. 8 x 10 -5 2 n = 3. 98 therefore n = 2. 0 = 3. 94

Example: R = k [Hg. Cl 2] [C 2 O 42 -] 2 First order + Second order = Third Order

15 -4 Zero-Order Reactions A → products Rrxn = k [A]0 Rrxn = k [k] = mol L-1 s-1

Integrated Rate Law -Δ[A] Δt =k Move to the -d[A] infinitesimal dt =k And integrate from 0 to time t t [A]t -∫ d[A] [A]0 = ∫ k dt 0 -[A]t + [A]0 = kt [A]t = [A]0 - kt

15 -5 First-Order Reactions H 2 O 2(aq) → H 2 O(l) + ½ O 2(g) d[H 2 O 2 ] = -k [H 2 O 2] dt ; [k] = s-1 [A]t t d[H 2 O 2 ] = - ∫ k dt ∫ [H 2 O 2] [A] 0 0 ln [A]t [A]0 = -kt ln[A]t = -kt + ln[A]0

First-Order Reactions

Half-Life • t½ is the time taken for one-half of a reactant to be consumed. For a first order reaction: ½[A]0 ln = -kt½ [A]0 ln [A]t [A]0 = -kt ln 2 = kt½ ln 2 0. 693 t½ = = k k

Half-Life But. OOBut(g) → 2 CH 3 CO(g) + C 2 H 4(g)

Some Typical First-Order Processes Some typical first-order processes

15 -6 Second-Order Reactions • Rate law where sum of exponents m + n +… = 2 A → products d[A] dt [A]t ∫ [A]0 = -k[A]2 ; d[A] [A]2 t = - ∫ k dt 0 1 1 = kt + [A]t [A]0 [k] = M-1 s-1 = L mol-1 s-1

Second-Order Reaction 1 1 = kt + [A]t [A]0

Pseudo First-Order Reactions • Simplify the kinetics of complex reactions • Rate laws become easier to work with CH 3 CO 2 C 2 H 5 + H 2 O → CH 3 CO 2 H + C 2 H 5 OH • If the concentration of water does not change appreciably during the reaction. – Rate law appears to be first order • Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.

Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t. 2 nd order

15 -7 Reaction Kinetics: A Summary • Calculate the rate of a reaction from a known rate law using: Rate of reaction = k [A]m[B]n …. • Determine the instantaneous rate of the reaction by: Finding the slope of the tangent line of [A] vs t or, Evaluate –Δ[A]/Δt, with a short Δt interval.

Summary of Kinetics • Determine the order of reaction by: Using the method of initial rates Find the graph that yields a straight line Test for the half-life to find first order reactions Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.

Summary of Kinetics • Find the rate constant k by: Determining the slope of a straight line graph. Evaluating k with the integrated rate law. Measuring the half life of first-order reactions. • Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.

Activation Energy • For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s). • Activation Energy is: –The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Activation Energy

Kinetic Energy

Collision Theory • If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower. • As temperature increases, reaction rate increases. • Orientation of molecules may be important.

Collision Theory

Transition State Theory • The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

15 -9 Effect of Temperature on Reaction Rates • Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = Ae-Ea/RT ln k = -Ea 1 R T + ln A

Arrhenius Plot N 2 O 5(CCl 4) → N 2 O 4(CCl 4) + ½ O 2(g) -Ea R = -1. 2 x 104 K Ea = 1. 0 x 102 k. J mol-1

Arrhenius Equation k= ln k 2– ln k 1 = ln ln k = Ae-Ea/RT k 2 k 1 = -Ea 1 R T 2 Ea 1 1 R T 1 - + ln A - -Ea 1 R T 1 + ln A - ln A T 2 log k 2 k 1 = Ea 1 2. 3 R T 1 - 1 T 2

A Rate Determining Step

11 -5 Catalysis • Alternative reaction pathway of lower energy. • Homogeneous catalysis. – All species in the reaction are in solution. • Heterogeneous catalysis. – The catalyst is in the solid state. – Reactants from gas or solution phase are adsorbed. – Active sites on the catalytic surface are important.

11 -5 Catalysis