Chapter 14 – Chemical Kinetics The rate of a chemical reaction is the speed at which products or formed and reactants broken down. There factors that affect the rate of a chemical reaction are temperature, concentration, state of reactants, and the presence of a catalyst. A rate law can be constructed to predict the rate at which products form or reactants disappear. Some chemical reactions happen in several steps called a reaction mechanism. The slowest step of a reaction mechanism is called the rate-limiting step. A catalyst is a factor that can speed up a chemical reaction. Biological catalysts are called enzymes that are an important part of biochemical reactions within living organisms.

Chemical Kinetics Factors that affect rates of chemical reactions o Collision Theory – In order for two reactants to form products, they must collide with enough kinetic energy and in the correct orientation.

Chemical Kinetics Factors that affect rates of chemical reactions

Chemical Kinetics Factors that affect rates of chemical reactions o Concentration – The more abundant the reactants are, the more successful collisions to form products.

Chemical Kinetics Factors that affect rates of chemical reactions o Temperature – The more kinetic energy the reactants possess, the more likely a collision will result in a successful reaction.

Chemical Kinetics Factors that affect rates of chemical reactions o Catalyst – A substance the helps the reactants collide in the correct orientation. o They are not consumed in a chemical reaction.

Chemical Kinetics Factors that affect rates of chemical reactions o Inhibitors – Compete with binding sites on a catalyst that slow, or even stop, a chemical reaction form occurring.

Chemical Kinetics Rates of chemical reactions o The speed at which a chemical reaction occurs is called its ‘reaction rate’. o The rate of a chemical reaction is measured as the change in concentration of reactant or products per unit of time. Rate = Δ Concentration / Time = Molarity / Time = M / s ** In chemistry [ ] means the molar concentration **

Chemical Kinetics Rates of chemical reactions o A flask is charged with mol of A and allowed to react to form B according to the hypothetical gas- phase reaction: A (g)  B (g). a.) Calculate the number of mole of B at each time in the table. Time (s) Moles of A

Chemical Kinetics Rates of chemical reactions o A flask is charged with mol of A and allowed to react to form B according to the hypothetical gas- phase reaction: A (g)  B (g). b.) Calculate the average rate of disappearance of A for each 40 s interval in units of mol / s. Time (s) Moles of A

Chemical Kinetics Rates of chemical reactions o A flask is charged with mol of A and allowed to react to form B according to the hypothetical gas- phase reaction: A (g)  B (g). c.) What additional information would be needed to calculate the rate in units of concentration per time? Time (s) Moles of A

Chemical Kinetics Rates of chemical reactions o Instantaneous Rate versus Average Rate of Reactions  Instantaneous Rate – Refers to the rate of a chemical reaction at a specific time.  Average Rate – Refers to the rate of a reaction over an interval of time.  Which rate do you think is usually higher?

Chemical Kinetics Rates of chemical reactions o The role of coefficients on reaction rates aA + bB  cC + dD In terms of ‘A’, the rate of reaction will be Rate = - 1 Δ[A] a Δt

Chemical Kinetics Rates of chemical reactions o The role of coefficients on reaction rates aA + bB  cC + dD In terms of ‘B’, the rate of reaction will be Rate = - 1 Δ[B] b Δt

Chemical Kinetics Rates of chemical reactions o The role of coefficients on reaction rates aA + bB  cC + dD In terms of ‘C’, the rate of reaction will be Rate = 1 Δ[C] c Δt

Chemical Kinetics Rates of chemical reactions o The role of coefficients on reaction rates aA + bB  cC + dD In terms of ‘D’, the rate of reaction will be Rate = 1 Δ[D] d Δt

Chemical Kinetics Rates of chemical reactions o The decomposition of N 2 O 5 proceeds according to the following reaction: 2N 2 O 5(g)  4NO 2(g) + O 2(g) If the rate of decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2 x M/s, what is the rate of appearance of (a) NO 2 and (b) O 2 ?

Chemical Kinetics Rates Laws of Chemical Reactions o The rate of a chemical reaction cannot be determined without conducting an experiment. o The initial concentration of reactants are measured and the rate of the reaction is determined. X + Y  Z TrialInitial [X]Initial [Y]Rate (M/s) x x x x x x 10 -7

Chemical Kinetics Rates Laws of Chemical Reactions aA + bB  cC + dD Rate = k [A] m [B] n k = rate constant [A] = initial molar concentration of A [B] = initial molar concentration of B m = reaction order exponent n = reaction order exponent

Chemical Kinetics Rates Laws of Chemical Reactions NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) Notice that when the concentration of NH 4 + is doubled, and NO 2 - is held constant, the rate doubles. Notice that when the concentration of NO 2 - is doubled, and NH 4 + is held constant, the rate doubles. TrialInitial [NH 4 + ]Initial [NO 2 - ]Rate (M/s) x x x x x x 10 -7

Chemical Kinetics Rates Laws of Chemical Reactions NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) Rate = k [NH 4 + ] [NO 2 - ] Since NH 4 + and NO 2 - affect the rate proportionally, we do not need to raise to concentration of the ions to any power. The order of both reactants is 1. TrialInitial [NH 4 + ]Initial [NO 2 - ]Rate (M/s) x x x x x x 10 -7

Chemical Kinetics Rates Laws of Chemical Reactions NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) Rate = k [NH 4 + ] [NO 2 - ] The overall order of this reaction is the sum of the two orders of the reactants. (1+1 = 2) This is second order reaction overall. TrialInitial [NH 4 + ]Initial [NO 2 - ]Rate (M/s) x x x x x x 10 -7

Chemical Kinetics Rates Laws of Chemical Reactions Example 1: The experimentally determined rate law for the reaction 2NO (g) + 2H 2(g)  N 2(g) + 2H 2 O (l) is rate = k[NO] 2 [H 2 ]. a.) What are the reaction orders in this rate law? b.) Does doubling the concentration of NO have the same effect on the rate as doubling the concentration of H 2 ?

Chemical Kinetics Rates Laws of Chemical Reactions o Units of the rate constant (k) Units of rate constant = units of rate = M/s = M -1. s -1 (units of concentration) 2 M 2

Chemical Kinetics Rates Laws of Chemical Reactions o Example 2: The initial rate of reaction A + B  C was measured for several different starting concentrations of A and B, and the results are as follows; a.) Determine the rate law for this reaction. Trial[A][B]Initial Rate (M/s) x x x 10 -5

Chemical Kinetics Rates Laws of Chemical Reactions o Example 2: The initial rate of reaction A + B  C was measured for several different starting concentrations of A and B, and the results are as follows; b.) Determine the rate constant. Trial[A][B]Initial Rate (M/s) x x x 10 -5

Chemical Kinetics Rates Laws of Chemical Reactions o Example 2: The initial rate of reaction A + B  C was measured for several different starting concentrations of A and B, and the results are as follows; b.) Determine the rate of reaction when [A] = M and [B] = M Trial[A][B]Initial Rate (M/s) x x x 10 -5

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws o An integrated rate law allows use to calculate the rate of a chemical reaction during a specific time interval. integrate rate law (first-order reactions) : ln [A] t = -kt or ln [A] t - ln [A] 0 = -kt [A] 0 ln = natural logarithm [A] t = concentration of A a specific time [A] 0 = initial concentration of A k = rate law constant t = time interval

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Law ln [A] t - ln [A] 0 = -kt rearrange into ln [A] t = -k t + ln [A] 0 y = m. x + b ln = natural logarithm [A] t = concentration of A a specific time [A] 0 = initial concentration of A k = rate law constant t = time interval

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws – The graph on the left shows [A] versus time. The graph in the middle shows the ln [A] versus time. A linear relationship of ln [A] proves that it is a first-order reaction.

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws o Example: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y -1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1 st, leading to a concentration of 5.0 x g/cm 3. Assuming that the lake remains at 12°C: a.) What is the concentration of the insecticide in the lake on June 1 st of the following year?

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws o Example 1: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y -1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1 st, leading to a concentration of 5.0 x g/cm3. Assuming that the lake remains at 12°C: a.) What is the concentration of the insecticide in the lake on June 1 st of the following year?

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws o Example 1: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45y -1 at 12°C. A quantity of this insecticide is seeps into a lake on June 1 st, leading to a concentration of 5.0 x g/cm 3. Assuming that the lake remains at 12°C: b.) How long will it take for the concentration of the insecticide in the lake to drop to 3.0 x g/cm 3 ?

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws o Example 2: The decomposition of dimethyl ether, (CH 3 ) 2 O, at 510°C is a first-order process with a rate constant of 6.8 x s -1 : (CH 3 ) 2 O (g)  CH 4(g) + H 2(g) + CO (g) If the initial pressure of (CH 3 ) 2 O is 135 torr, what is the pressure after 1420 s?

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws (second-order reactions) 1 = kt + 1 [A] t [A] 0 [A] 0 = initial concentration [A] t = concentration at time t k = rate law constant t = time

Chemical Kinetics Rates Laws of Chemical Reactions o Integrated Rate Laws (second-order reactions) Example: The following is a second-order reaction 2NO 2(g)  2NO (g) + O 2(g) k = M -1 s -1 If the initial concentration of NO 2 in a closed vessel is M, What is the remaining concentration after 0.500h?

Chemical Kinetics Rates Laws of Chemical Reactions o Half-Life (t 1/2 ) – The time required for the concentration to fall to half of its original amount. [A] t1/2 = ½[A] 0 Calculating the the half life of a first-order reaction: t 1/2 = ln1/2 = k k

Chemical Kinetics Rates Laws of Chemical Reactions o Half-Life (t 1/2 ) – Example: Calculate the half-life for the decomposition of the insecticide used in the previous example (slide 31). How long does it take for the concentration of the insecticide to reach ¼ of its original value?

Chemical Kinetics Temperature and Rates of Chemical Reactions An increase in the temperature of a chemical reaction will also increase the rate constant, k.

Chemical Kinetics Temperature and Rates of Chemical Reactions An increase in the temperature is also an increase in the kinetic energy of the reactants. More kinetic energy = more velocity = more collisions = more successful collisions.

Chemical Kinetics Temperature and Rates of Chemical Reactions Not only do molecules need to collide with enough kinetic energy, but they also need to be in the correct orientation.

Chemical Kinetics Temperature and Rates of Chemical Reactions Activation energy (E a ) – The minimum amount of energy required to initiate a chemical reaction.

Chemical Kinetics Temperature and Rates of Chemical Reactions Activation energy (E a ) – The minimum amount of energy required to initiate a chemical reaction.

Chemical Kinetics Temperature and Rates of Chemical Reactions Energy diagrams

Chemical Kinetics Temperature and Rates of Chemical Reactions Activation Energy Arrhenius Equation – Incorporates the fraction of molecules that posses enough kinetic energy (E a ), the number of molecular collisions, and the fraction of reactants that are in the correct orientation. k = A e -Ea/RT k = reaction constant A e = frequency factor constant E a = activation energy R = gas law constant T = Temperature (K)

Chemical Kinetics Temperature and Rates of Chemical Reactions Determining Activation Energy ln k 1 = E a ( ) k 2 R ( T 2 T 1 ) K 1 = rate constant at T 1 K 2 = previously known rate constant at T 2 E a = activation energy T 1 = Temperature of reaction with k 1 T 2 = Temperature of reaction with k 2

Chemical Kinetics Reaction Mechanisms A detailed step-by-step explanation as to how a reaction occurs. Some reactions occur only as 1 step. Other reactions occur in several steps.

Chemical Kinetics Reaction Mechanisms Elementary reactions (1 step) NO (g) + O 3(g)  NO 2(g) + O 2(g) The rate of this reaction is only determined in this step.

Chemical Kinetics Reaction Mechanisms Multistep Reaction Mechanisms (2 or more steps) NO 2(g) + CO (g)  NO (g) + CO 2(g) This reaction is the result of two steps; NO 2(g) + NO 2(g)  NO 3(g) + NO (g) NO 3(g) + CO (g)  NO 2(g) + CO 2(g) The sum of these reactions gives us the overall reaction.

Chemical Kinetics Reaction Mechanisms Multistep Reaction Mechanisms (2 or more steps) NO 2(g) + NO 2(g)  NO 3(g) + NO (g) NO 3(g) + CO (g)  NO 2(g) + CO 2(g) 2NO 2(g) + NO 3(g) + CO (g)  NO 3(g) + NO (g) + NO 2(g) + CO 2(g) NO 2(g) + CO (g)  NO (g) + CO 2(g) NO 3(g) is called an intermediate, it is neither a reactant nor a product.

Chemical Kinetics Reaction Mechanisms Example: 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 a.) Is the proposed mechanism consistent with the overall reaction?

Chemical Kinetics Reaction Mechanisms Example: 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 b.) Identify any intermediate(s).

Chemical Kinetics Reaction Mechanisms Example: 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 b.) Identify the molecularity of each step of the mechanism.

Chemical Kinetics Reaction Mechanisms and Rate Determining Steps o Single step reactions – The rate of the reaction is determined by the speed of one step. o Multiple Step Reactions – The rate of the reaction will be determine by the slowest step (rate determining step). NO 2(g) + NO 2(g)  NO 3(g) + NO (g) (slow) NO 3(g) + CO (g)  NO 2(g) + CO 2(g) (fast) NO 2(g) + CO (g)  NO (g) + CO 2(g)

Chemical Kinetics Catalyts o A substance that speeds up a chemical reaction without being consumed as a reagent. o Catalysts speed up a chemical reaction by lowering the activation energy.

Chemical Kinetics Catalyts o Homogenous Catalyts – Catalysts in the same state as the reagents.

Chemical Kinetics Catalyts o Heterogeneous Catalyts – Catalysts that are not in the same state as the reagents.