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Chapter 36. Elementary Chemical Kinetics. Engel & Reid. Figure 36.1.
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Chapter 36 Elementary Chemical Kinetics Engel & Reid
Figure 36.1 Figure 36.1 Concentration as a function of time for the conversion of reactant A into product B. The concentration of A at time 0 is [A]0, and the concentration of B is zero. As the reaction proceeds, the loss of A results in the production of B.
36.2 reaction rate Example
36.3 rate laws Reaction order Rate law :R =k[A]a[B]b... Rate Constant • The rate of reaction is often found to be proportional to the molar concentrations of the reactants raised to a simple power. • It cannot be overemphasized that reaction orders have no relation to stoichiometric coefficients, and they are determined by experiment.
36.3 Rate laws 36.3.1 Measuring Reaction Rates
36.3.2 Determining Reaction Orders Strategy 1. Isolation method The reaction is performed with all species but one in excess. Under these conditions, only the concentration of one species will vary to a significant extent during the reaction. R = k’[B]b
36.3.2 Determining Reaction Orders Strategy 2. Method of initial rates The concentration of a single reactant is changed while holding all other concentrations constant, and the initial rate of the reaction is determined.
Example Problem 36.2 Using the following data for the reaction illustrated in equation 36.13. Determine the order of the reaction with respect to A and B, and the rate constant for the reaction
Example 36.2 Solution
36.3 Rate law • Determining the rate of a chemical reaction, experimentally • Chemical methods • Physical methods • Stopped-flow techniques • Flash photolysis techniques • Perturbation-relaxation methods
Figure 36.3 Figure 36.3 Schematic of a stopped-flow experiment. Two reactants are rapidly introduced into the mixing chamber by syringes. After mixing chamber, the reaction kinetics are monitored by observing the change in sample concentration versus time, in this example by measuring the absorption of light as a function of time after mixing.
Example 36.3 Example Problem 36.3 The decomposition of N2O5 is an important process in tropospheric chemistry. The half-life for the first order decomposition of this compound is 2.05×104 s. How long will it take for a sample of N2O5 to decay to 60% of its initial value? Solution
Example 36.3 Example Problem 36.4 Catbon-14 is a radioactive nucleus with half-life of 5760 years. Living matter exchange carbon with its surroundings (for example, through CO2) so that s constant level of C14 is maintained, corresponding to 15.3 decay events per minute. Once living matter has died, carbon contained in the matter is not exchanged with the surroundings, and the amount of C14 that remains in the dead material decreases with time due to radioactive decay. Consider a piece of fossilized wood that demonstrates 2.4 C decay events per minute. How old is the wood.
Example 36.4 Solution
Figure 36.6 Figure 36.6 Schematic representation of the numerical evaluation of a rate law.
Figure 36.7 Figure 36.7 Comparison of the numerical approximation method to the integrated rate law expression for a first-order reaction. The rate constant for the reaction is 0.1 m s-1. The time evolution in reactant concentration determined by the integrated rate law expression
36.7 Sequential First-Order Reaction At t = 0 Sequential First-Order Reaction
Figure 36.8a Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (a) kA=2kf =0.1 s-1.
Figure 36.8b Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (b) kA=8kf =0.4 s-1. Notice that both the maximal amount of I in a ddition to the time for the maximum is changed relative to the first channel.
Figure 36.8c Figure 36.8 Concentration profiles for a sequential reaction in which the reactant (A, blue line) from an intermediate (I, yellow) that undergoes subsequent decay to form the product (P, red line) where (c) kA=0.025kf=0.0125 s-1. In this case, very little intermediate is formed, and the maximum in [I] is delayed relative to the first two examples.
36.7 Sequential First-order Reaction Maximum Intermediate Concentration
Example problem 36.5 Example Problem 36.5 Determine the time at which [I] is at a maximum for kA = 2kI = 0.1 s-1. Solution
36.7 Sequential First-order Reaction Rate-Determining Steps
Figure 36.9a Figure 36.9 Rate-limiting step behavior in sequential reactions. (a) kA=20kf =1 s-1 such that the rate-limiting step is the decay of intermediate I. In this case, the reduction in [I] is reflected by the appearances of [P]. The time evolution of [P] predicted by the sequential mechanism is given by the yellow line, and the corresponding evolution assuming rate-limiting step behavior, [P]rl, is given by the red curve.
Figure 36.9b Figure 36.9 Rate-limiting step behavior in sequential reactions. (b) The opposite case from part (a) kA=0.04kf = 0.02 s-1 such that the rate-limiting step is the decay of reactant A.
36.7 Sequential First-order Reaction The steady-State Approximation This approximation is particularly god when the decay rate of the intermediate is greater than the rate of production so that the intermediates are present at very small concentrations during the reaction. In Steady-State approximation, the time derivative of intermediate concentrations is se to zero.
Figure 36.10 Figure 36.10 Concentration determined by numerical evolution of the sequential reaction scheme presented in Equation (36.44) where kA= 0.02 s-1 and kf = k2 = 0.2 s-1.
Figure 36.11 Figure 36.11 Comparison of the numerical and steady-state concentration profiles for the sequential reaction scheme presented in Equation (36.44) where kA= 0.02 s-1 and kf = k2 = 0.2 s-1. Curves corresponding to the steady-state approximation are indicated by the subscript ss.
36.8 Parallel Reactions Parallel Reactions
Figure 36.12 Figure 36.12 Concentration for a parallel reaction where kB = 2kC = 0.1 s-1.
36.8 Parallel Reactions Yield, F, is defined as the probability that given product will be formed by decay of the reactant.
Example Problem 36.7 Example Problem 36.7 In acidic conditions, benzyl peniciline (BP) undergoes the following paralll reaction: In the molecular structure, R1 and R2 indicate alkyl substitutions. In a solution where pH=3, the rat constants for the processes at 22 ℃are k1=7.0×10-4 s-1, k2=4.1×10-3 s-1, and k2=5.7×10-3 s-1. what is the yield for P1 formation?
Example Problem 36.7 Solution
36.9 Temperature Dependence of Rate Constants A frequency factor or Arrhenius pre-exponential factor Ea activation energy
Example Problem 36.8 Example Problem 36.8 The temperature dependence of the acid-catalyzed hydrolysis of penicillin (illustrated in Example problem 36.7) is investigated, and the dependece of k1 on temperature is iven in the following table. What is the activation eergy and Arrhenius preexponential factor for this branch of hydrolysis reaction?
Example Problem 36.8 Solution Draw a plot of ln (k1) versus 1/T
36.9 Temperature Dependence of Rate Constant Figure 36.13 A schematic drawing of the energy profile for a chemical reaction. Reactants must acquire sufficient energy to overcome the activation energy, Ea, for the reaction. The reaction coordinate represents the binding and geometry changes hat occur in the transformation of reactants into products.
36.10 Reversible Reactions and Equilibrium Reversible Reactions
36.10 Reversible Reactions and Equilibrium At equilibrium
36.10 Reversible Reactions and Equilibrium Figure 36.14 Reaction coordinate demonstrating the activation energy for reactants to form products, Ea, and the back reaction in which products form reactants, E’a
36.10 Reversible Reactions and Equilibrium Figure 36.15 Time-dependent concentrations in which both forward and back reactions exist between reactant A and product B. In this example, kA=2kB=0.06 s-1. Note that the concentration reach a constant value at longer times (t > teq) at which pint the reaction reaches equilibrium.
