Chemical Kinetics

Kinetics Kinetics in chemistry is concerned with how quickly a reaction proceeds Factors that affect rate Physical state of the reactants Concentration of the reactants Temperature at which the reaction occurs The presence of a catalyst

Reaction Rates Reaction rates depend on the frequency of collisions between molecules Reaction rate = speed of a reaction (M/s) A → B Δ[B]/Δt = -Δ[A]/Δt Average Rate

Change of Rates with Time C 4 H 9 Cl (aq) + H 2 O (l) → C 4 H 9 OH (aq) + HCl (aq) What is happening to the rate as the reaction proceeds? Graphs of the data allow you to find the instantaneous rate

Reaction Rates and Stoichiometry Stoichiometry will affect the rates of disappearance and formation 2HI(g) → H 2 (g) + I 2 (g) -1/2Δ[HI]/Δt = Δ[H 2 ]/Δt = Δ[I 2 ]/Δt For any general reaction aA + bB → cC + dD

Practice How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O 3 (g) → 3O 2 (g) ? If the rate of appearance of oxygen is 6.0x M/s at a particular instant what is the value of the rate of disappearance of ozone at this same time? Answer: 4.0x M/s

Practice The decompostion of N2O5 proceeds according to the following equation: 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) If the rate of decomposition of dinitrogen pentoxide at a particular instant in a reaction vessel is 4.2x M/s, what is the rate of appearance of NO 2 and O 2 ? Answer: 8.4x10 -7 M/s, 2.1x10 -7 M/s

Beer's Law Spectroscopic methods are useful in seeing how concentration changes with time 2HI(g) → H 2 (g) + I 2 (g) A = abc A : absorbance a : molar absorptivity b : path length c : concentration

Concentration and Rate To determine the effect of concentration of rate, you can vary the concentration of reactants and monitor the change in initial rate NH 4 + (aq) + NO 2 - (aq) → N 2 (g) + 2H 2 O (l) What happens to the initial rate when the concentrations are changed?

Rate Law Rate law shows the rate of a reaction is related to the concentrations of the reactants Rate = k[NH 4 + ][NO 2 - ] For a general reaction: aA + bB → cC + dD Rate = k[A] m [B] n k = rate constant The magnitude of k is affected by changes in temp. If we know the rate law we can calculate k From exp. 1: r = 5.4x10 -7 M/s = k(0.0100M)(0.200M) k = 2.7x10 -4 M -1 ·s -1

Reaction Orders Rate = k[A] m [B] n m and n are reaction orders Rate = k[NH 4 + ][NO 2 - ] Each compound is 1 s t order but the overall order is 2 n d (just add the exponents) Reaction orders must be determined experimentally 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) : Rate = k[N 2 O 5 ] 2HI(g) → H 2 (g) + I 2 (g):Rate = k[H 2 ][I 2 ]

Units of Rate Constants Units of the rate constant depend on the overall order of the rate law What are the overall reaction orders and units of the rate constant for the following reactions? 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g)  Rate = k[N 2 O 5 ] 2HI(g) → H 2 (g) + I 2 (g)  Rate = k[H 2 ][I 2 ] CHCl 3 (g) + Cl 2 (g) → CCl 4 (g) + Hcl (g)  Rate = k[CHCl 3 ][Cl] 1/2

Using Initial Rates Observing the effect of changing the initial concentrations of the reactants on the initial rate allows us to determine reaction orders Exponents will commonly be 0, 1, 2 What effect will a reactant with a reaction order of 0 have on the reaction? 1? 2? Remember only the rate depends on concentration

Practice The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B. Using this data determine the rate law for the reaction; the magnitude of the rate constant; and the rate when [A] = 0.050M and [B] = 0.100M Answer: r = k [A]2, 4.0x M - 1 s - 1, r = 1.0x M/s

Practice The following data were measured for the reaction of nitric oxide with hydrogen: 2NO(g) + 2H 2 (g) → N 2 (g) + 2H 2 O(g) Determine the rate law for this reaction, the value of the rate constant and the rate when [NO] = 0.050M and [H 2 ] = 0.150M Answer: r = k[NO] 2 [H 2 ], k = 1.2 M - 2 s - 1, r = 4.5x M/s

Changing Concentration with Time Rate laws tell us how the rate of a reaction changes at a given temperature as concentration changes. Rate laws can be converted to tell us what the concentration of a substance is at any time during a reaction. There are two special cases and you must be able to use them.

First Order Reactions For a first order reaction that proceeds A → products the rate law is Rate = -Δ[A]/Δt = k[A] After some math magic (involving integration) ln[A] t – ln[A] 0 = -kt With some rearrangement we get something similar to y = mx + b ln[A] t = -kt + ln[A] 0 What would you graph to see if the reaction was first order?

Practice The first order rate constant for the decomposition of a certain insecticide in water at 12°C is 1.45 yr - 1. A quantity of the insecticide is washed into a lake on June 1, leading to a concentration of 5.0x g/mL. What is the concentration of insecticide after 1 year? How long will it take for the concentration to drop to 3.0x g/mL? Answer: 1.2x g/mL, 0.35 years

Practice The decomposition of dimethyl ether, (CH 3 ) 2 O, at 510°C is a first order process with a rate constant of 6.8x s - 1 : (CH 3 ) 2 O(g) → CH 4 (g) + H 2 (g) + CO(g) If the initial pressure of dimethyl ether is 135 torr, what is the partial pressure after 1420s? Answer: 51 torr

Second Order Reactions For a reaction that proceeds A → products or A+B → products that are second order in just one reactant A: Rate = - Δ[A]/Δt = k[A] 2 After some calculus magic this becomes: 1/[A] t = kt + 1/[A] 0 If plotting 1/[A]t creates a straight line the reaction is second order

Practice The following data were obtained for the gas phase decomposition of nitrogen dioxide at 300°C, NO 2 (g) → NO(g) + ½ O 2 (g): Is the reaction first or second order in NO 2 ? What is k? If the initial concentration of NO 2 is M, what is the concentration after 0.500hr? Answer: 2 nd order r = k[NO 2 ] 2, k = M - 1 s - 1, 1.00x M

Half Life Half life (t ½ ) is the time it takes for the concentration of a reactant to drop to one half of its initial value Using algebra you can find the half life of a 1 st order reaction t ½ = 0.693/k Half life for a second order reaction depends on concetration t ½ = 1/k[A] 0

Temperature and Rate Rate of most chemical reactions increase as temperature increases Increasing temperature increases the rate constant and thus the rate Glow stick fun What effect did the ice and hot water have on the reaction?

Collision Model Molecule must collide with enough energy to react What does increasing temperature do? Collision Theory Orientation Factor Activation Energy  Ea

Activation Energy Higher activation energy the lower the rate Only a fraction of molecules have energy to generate products upon collision f = e -Ea/RT

Arrhenius Equation Rates depend on Fraction of molecules with an energy of Ea or greater Collisions per second Fraction of collision with proper orientation Arrhenius used these ideas to relate k and Ea k = Ae - E a / R T Taking the natural log of both sides ln k = (-Ea/R)T + ln A So we can graph ln k versus 1/T to find Ea We can relate the rate constants of a reaction at different temperatures ln(k 1 /k 2 ) = Ea/R (1/T 2 - 1/T 1 )

Practice The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures. From these data, calculate the activation energy for the reaction. What is the value of the rate constant at 430.0K? Answer: Ea = 160 kJ/mol, k = 1.0x s - 1

Reaction Mechanisms Reaction mechanisms show how a reaction occurs Elementary Steps NO(g) + O 3 (g) → NO 2 (g) + O 2 (g) Molecularity  Unimolecular  Bimolecular  Termolecular

Multistep Mechanisms Many reactions do not happen in just one step NO 2 (g) + CO(g) → NO(g) + CO 2 (g) NO 2 (g) + NO 2 (g) → NO 3 (g) + NO(g) NO 3 (g) + CO(g) → NO 2 (g) + CO 2 (g) Elementary steps must add up to give the overall reaction Intermediate

Practice It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps:  O 3 (g) → O 2 (g) + O(g)  O 3 (g) → O(g) + 2O 2 (g) Describe the molecularity if each step in this mechanism. Write the equation for the overall reaction. Identify any intermediates. For the reaction Mo(CO) 6 + P(CH 3 ) 3 → Mo(CO) 5 P(CH 3 ) 3 + CO The proposed mechanism is Mo(CO) 6 → Mo(CO) 5 + CO Mo(CO) 5 + P(CH 3 ) 3 → Mo(CO) 5 P(CH 3 ) 3 Is the proposed mechanism consistent with the equation for the overall reaction? Identify any intermediates.

Rate Laws for Elementary Steps Rate laws cannot normally be predicted from the coefficients of balanced equations. Why? For elementary steps the equation tells you the rate law. Rate law is determined by its molecularity

Practice If the following reaction occurs in a single elementary step, predict the rate law: H 2 (g) + Br 2 (g) → 2HBr(g) Consider the following reaction: 2NO(g) + Br 2 (g) → 2NOBr(g). Write the rate law for the reaction assuming it involves a single elementary step. Is a single elementary step likely for this reactipon?

Rate Laws for Multistep Mechanisms In multistep systems the rate laws is set by the slowest step Rate determining step Mechanisms with slow first step Step 1: NO 2 (g) + NO 2 (g) → NO 3 (g) + NO(g) (slow) Step 2: NO 3 (g) + CO(g) → NO 2 (g) + CO 2 (g) (fast) Overall: NO 2 (g) + CO(g) → NO(g) + CO 2 (g) Rate = k 1 [NO 2 ] 2

Mechanisms with Initial Fast Step You need to derive the rate law for a mechanism in which there is an intermediate. 2NO(g) + Br 2 (g) → 2NOBr(g) Rate = k[NO] 2 [Br 2 ] Possible Mechanism NO(g) + Br 2 (g) ↔ NOBr 2 (g)(fast) NOBr 2 (g) + NO(g) → 2NOBr(g) (slow) Algebra fun When a fast step precedes a slow one we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.

Practice Show that the following mechanism for the reaction producing NOBr also produces a rate law consistent with the experimentally observed one: Step 1: NO(g) + NO(g) ↔ N 2 O 2 (g) (fast) Step 2: N 2 O 2 (g) + Br 2 (g) → 2NOBr(g)(slow) The first step of a mechanism involving the reaction of bromine is: Br 2 (g) ↔ 2Br(g) (fast). What is the expression relating the concentration of Br(g) to that of Br 2 (g)

Catalysis What does a catalyst do? Types of catalysts Homogeneous – same phase as the reactants Heterogeneous – different phase from the reactants  Adsorption happens first Enzymes