2KINETICS— the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.Reaction rate = change in concentration of a reactant or product with time.Three “types” of ratesinitial rateaverage rateinstantaneous rate
3The Rate of Reaction Consider the hypothetical reaction, A(g) + B(g) C(g) + D(g)equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:
4aA(g) + bB(g) cC(g) + dD(g) The Rate of ReactionMathematically, the rate of a reaction can be written as:aA(g) + bB(g) cC(g) + dD(g)
5The Rate of ReactionThe rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance.[A] is the concentration of A in molarity or moles/L.k is the specific rate constant.k is an important quantity in this chapter.
6The Rate of Reaction Important terminology for kinetics. The order of a reaction can be expressed in terms of either:each reactant in the reaction orthe overall reaction.Order for the overall reaction is the sum of the orders for each reactant in the reaction.For example:
7In general, fora A + b B --> x X with a catalyst CRate = k [A]m[B]n[C]pThe exponents m, n, and p• are the reaction order• can be 0, 1, 2 or fractions• must be determined by experiment!
8The Rate of ReactionGiven the following one step reaction and its rate-law expression.Remember, the rate expression would have to be experimentally determined.Because it is a second order rate-law expression:If the [A] is doubled the rate of the reaction will increase by a factor of = 4If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4
9Factors That Affect Reaction Rates There are several factors that can influence the rate of a reaction:The nature of the reactants.The concentration of the reactants.The temperature of the reaction.The presence of a catalyst.We will look at each factor individually.
10Nature of ReactantsThis is a very broad category that encompasses the different reacting properties of substances.For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.
11Nature of ReactantsCalcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.
12Nature of ReactantsThe reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.
13Nature of ReactantsHowever, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide.The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.
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15Concentrations of Reactants: The Rate-Law Expression This is a simplified representation of the effect of different numbers of molecules in the same volume.The increase in the molecule numbers is indicative of an increase in concentration.A(g) + B (g) ProductsA BBA BA BA B6 different possible A-B collisions9 different possible A-B collisions4 different possible A-B collisions
16Concentrations of Reactants: The Rate-Law Expression Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? A(g) + B(g) ® 3 C(g)ExperimentNumberInitial [A](M)Initial [B]Initial rate of formation of C (M/s)10.102.0 x 10-420.200.304.0 x 10-43
17Concentrations of Reactants: The Rate-Law Expression
18Concentrations of Reactants: The Rate-Law Expression The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction?2 A(g) + B(g) + 2 C(g) ® 3 D(g) + 2 E(g)ExperimentInitial [A](M)Initial [B]Initial [C]Initial rate of formation of D (M/s)10.200.102.0 x 10-420.306.0 x 10-4340.600.401.8 x 10-3
19Concentrations of Reactants: The Rate-Law Expression
20Concentrations of Reactants: The Rate-Law Expression Consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks.ExperimentInitial Rate(M/s)Initial [A](M)Initial [B]14.0 x 10-30.200.05021.6 x 10-2?33.2 x 10-20.40
21Concentrations of Reactants: The Rate-Law Expression
22Concentration vs. Time: The Integrated Rate Equation The integrated rate equation relates time and concentration for chemical and nuclear reactions.From the integrated rate equation we can predict the amount of product that is produced in a given amount of time.Initially we will look at the integrated rate equation for first order reactions.These reactions are 1st order in the reactant and 1st order overall.
23Concentration vs. Time: The Integrated Rate Equation An example of a reaction that is 1st order in the reactant and 1st order overall is:a A ® productsThis is a common reaction type for many chemical reactions and all simple radioactive decays.Two examples of this type are:2 N2O5(g) 2 N2O4(g) + O2(g)238U 234Th + 4He
24Concentration vs. Time: The Integrated Rate Equation The integrated rate equation for first order reactions is:where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.k = specific rate constant. t = time elapsed since beginning of reaction.a = stoichiometric coefficient of A in balanced overall equation.
25Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t.Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.
26Concentration vs. Time: The Integrated Rate Equation At time t = t1/2, the expression becomes:
27Concentration vs. Time: The Integrated Rate Equation Cyclopropane, an anesthetic, decomposes to propene according to the following equation.The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.
28Concentration vs. Time: The Integrated Rate Equation Refer to Previous Example: How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds?The integrated rate laws can be used for any unit that represents moles or concentration.In this example we will use grams rather than mol/L.
29Concentration vs. Time: The Integrated Rate Equation The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.CS2(g) ® CS(g) + S(g)
30Concentration vs. Time: The Integrated Rate Equation For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is:Where:[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.k = specific rate constant. t = time elapsed since beginning of reaction.a = stoichiometric coefficient of A in balanced overall equation.
31Temperature: The Arrhenius Equation Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (Ea), and (3) the specific rate constant (k).If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.
32Temperature: The Arrhenius Equation Subtract one equation from the other.Rearrange and solve for ln k2/k1.
33Temperature: The Arrhenius Equation Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K).How much do the two rates differ?
34Temperature: The Arrhenius Equation For reactions that have an Ea»50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature.Consider:2 ICl(g) + H2(g) ® I2(g) + 2 HCl(g)The rate-law expression is known to be R=k[ICl][H2].
35Catalysts Homogeneous catalysts exist in same phase as the reactants. Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.Homogeneous catalysts exist in same phase as the reactants.Heterogeneous catalysts exist in different phases than the reactants.Catalysts are often solids.
36General rate expression Differential rate law (with respect to concentration)Rate = k [A]m[B]n[C]pIf zero order Rate = k[A]0 = kfirst order Rate = k[A]1 = k[A]second order Rate = k[A]2Integrated rate law (with respect to time)If zero order [A]0 - [A] = ak tfirst ordersecond orderKineticsInitial rateInstantaneous rateAverage rateRate of reactionOrder of reactionOverall order of reactionCatalystFour factors that affect therate of reactionnature of reactantconcentrationtemperaturepresence of a catalyst[A] vs tln [A] vs t1/[A] vs t[A]ln [A]1/[A]ttt
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4555. A reaction has the following experimental rate equation: Rate = k[A]2[B]. If the concentration of A is doubled and the concentration of B is halved what happensto the rate?The following statements relate to the reaction with the following rate law:Rate = k[H2][I2] H2(g) + I2(g) → 2 HI(g)Determine which of the following statements are true. If a statement is false, indicate why it is incorrect.The reaction must occur in a single stepThis is a second-order reaction overallRaising the temperature will cause the value of k to decreaseRaising the temperature lowers the activation energy for this reactionIf the concentration of both reactants are doubled, the rate will doubleAdding a catalyst in the reaction will cause the initial rate to decrease
46Describe each of the following statements as true or false Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct.The rate determining elementary step in a reaction is the slowest step in a mechanismIt is possible to change the rate constant by changing the temperatureAs a reaction proceeds at constant temperature, the rate remains constantA reaction that is third order overall must involve more than one stepReactions are faster at a higher temperature because activation energies are lowerRate increase with increasing concentration of reactants because there are more collisions between reactant moleculesAt higher temperatures a larger fraction of molecules have enough energy to get over the activation energy barrierCatalyzed and uncatalyzed reactions have identical mechanisms