Presentation on theme: "CHAPTER 15 Chemical Kinetics. KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. Reaction."— Presentation transcript:
KINETICS — 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 rates initial rate average rate instantaneous rate
The 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:
The Rate of Reaction Mathematically, the rate of a reaction can be written as: aA (g) + bB (g) cC (g) + dD (g)
The Rate of Reaction The 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.
The Rate of Reaction Important terminology for kinetics. The order of a reaction can be expressed in terms of either: 1each reactant in the reaction or 2the overall reaction. Order for the overall reaction is the sum of the orders for each reactant in the reaction. For example:
In general, for a A + b B --> x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction order can be 0, 1, 2 or fractions must be determined by experiment!
The Rate of Reaction Given 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 4. 2 2 = 4 –If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2) 2 = 1/4
Factors That Affect Reaction Rates There are several factors that can influence the rate of a reaction: 1. The nature of the reactants. 2. The concentration of the reactants. 3. The temperature of the reaction. 4. The presence of a catalyst. We will look at each factor individually.
Nature of Reactants This 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.
Nature of Reactants Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.
Nature of Reactants The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.
Nature of Reactants However, Mg reacts with steam rapidly to liberate H 2 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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Concentrations 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) Products A B B A B 4 different possible A-B collisions 6 different possible A-B collisions 9 different possible A-B collisions
Concentrations of Reactants: The Rate-Law Expression Example 16-1: The following rate data were obtained at 25 o C for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A (g) + B (g) 3 C (g) Experiment Number Initial [A] (M) Initial [B] (M) Initial rate of formation of C (M/s) 10.10 2.0 x 10 -4 20.200.304.0 x 10 -4 18.104.22.168 x 10 -4
Concentrations of Reactants: The Rate-Law Expression
The following data were obtained for the following reaction at 25 o C. 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) Experiment Initial [A] (M) Initial [B] (M) Initial [C] (M) Initial rate of formation of D (M/s) 10.200.10 2.0 x 10 -4 20.200.300.206.0 x 10 -4 22.214.171.1242.0 x 10 -4 40.600.300.401.8 x 10 -3
Concentrations 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. Experiment Initial Rate (M/s) Initial [A] (M) Initial [B] (M) 14.0 x 10 -3 0.200.050 21.6 x 10 -2 ?0.050 33.2 x 10 -2 0.40?
Concentrations of Reactants: The Rate-Law Expression
Concentration 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 1 st order in the reactant and 1 st order overall.
Concentration vs. Time: The Integrated Rate Equation An example of a reaction that is 1 st order in the reactant and 1 st order overall is: a A products This is a common reaction type for many chemical reactions and all simple radioactive decays. Two examples of this type are: 2 N 2 O 5(g) 2 N 2 O 4(g) + O 2(g) 238 U 234 Th + 4 He
Concentration vs. Time: The Integrated Rate Equation 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. The integrated rate equation for first order reactions is:
Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t. Define the half-life, t 1/2, 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.
Concentration vs. Time: The Integrated Rate Equation At time t = t 1/2, the expression becomes:
Concentration 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 1000 0 C. Calculate the half life of cyclopropane at 1000 0 C.
Concentration 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.
Concentration vs. Time: The Integrated Rate Equation The half-life for the following first order reaction is 688 hours at 1000 0 C. Calculate the specific rate constant, k, at 1000 0 C and the amount of a 3.0 g sample of CS 2 that remains after 48 hours. CS 2(g) CS (g) + S (g)
Concentration 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.
Temperature: The Arrhenius Equation Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (E a ), and (3) the specific rate constant (k). If the Arrhenius equation is written for two temperatures, T 2 and T 1 with T 2 >T 1.
Temperature: The Arrhenius Equation 1.Subtract one equation from the other. 2.Rearrange and solve for ln k 2 /k 1.
Temperature: The Arrhenius Equation Consider the rate of a reaction for which E a =50 kJ/mol, at 20 o C (293 K) and at 30 o C (303 K). –How much do the two rates differ?
Temperature: The Arrhenius Equation For reactions that have an E a 50 kJ/mol, the rate approximately doubles for a 10 0 C rise in temperature, near room temperature. Consider: 2 ICl (g) + H 2(g) I 2(g) + 2 HCl (g) The rate-law expression is known to be R=k[ICl][H 2 ].
Catalysts 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.
General rate expression Differential rate law (with respect to concentration) Rate = k [A] m [B] n [C] p If zero order Rate = k[A]0 = k first order Rate = k[A]1 = k[A] second order Rate = k[A] 2 I ntegrated rate law (with respect to time) If zero order [A] 0 - [A] = ak t first order second order Kinetics Initial rate Instantaneous rate Average rate Rate of reaction Order of reaction Overall order of reaction Catalyst Four factors that affect the rate of reaction nature of reactant concentration temperature presence of a catalyst [A] vs tln [A] vs t1/[A] vs t [A] ln [A] 1/[A] ttt
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55. 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 happens to the rate? 87.The following statements relate to the reaction with the following rate law: Rate = k[H 2 ][I 2 ] H 2(g) + I 2(g) → 2 HI (g) Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. a)The reaction must occur in a single step b)This is a second-order reaction overall c)Raising the temperature will cause the value of k to decrease d)Raising the temperature lowers the activation energy for this reaction e)If the concentration of both reactants are doubled, the rate will double f)Adding a catalyst in the reaction will cause the initial rate to decrease
89.Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. a)The rate determining elementary step in a reaction is the slowest step in a mechanism b)It is possible to change the rate constant by changing the temperature c)As a reaction proceeds at constant temperature, the rate remains constant d)A reaction that is third order overall must involve more than one step e)Reactions are faster at a higher temperature because activation energies are lower f)Rate increase with increasing concentration of reactants because there are more collisions between reactant molecules g)At higher temperatures a larger fraction of molecules have enough energy to get over the activation energy barrier h)Catalyzed and uncatalyzed reactions have identical mechanisms