Chapter 12 - Chemical Kinetics

¡Supera tus tareas y exámenes ahora con Quizwiz!

First-Order (Contd....)

- Time required for a reactant to reach half its original concentration - Half-Life: t1/2 = 0.693/k k = rate constant - Half-life does not depend on the concentration of reactants.

Rate-Determining Step

- A reaction is only as fast as its slowest step. - The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.

Catalyst

- A substance that speeds up a reaction without being consumed itself. - Provides a new pathway for the reaction with a lower activation energy.

Rate Laws: A Summary

- Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. - Because the differential and integrated rate laws for a given reaction are related in a well-defined way, the experimental determination of either of the rate laws is sufficient. - Experimental convenience usually dictates which type of rate law is determined experimentally. - Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

For Reactants to Form Products

- Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). - Relative orientation of the reactants must allow formation of any new bonds necessary to produce products.

Determining the form of Rate Law

- Determine experimentally the power to which each reactant concentration must be raised in the rate law.

Types of Rate Laws

- Differential Rate Law (rate law) - shows how the rate of a reaction depends on concentrations. - Integrated Rate Law - shows how the concentrations of species in the reaction depend on time.

Activation Energy, Ea

- Energy that must be overcome to produce a chemical reaction.

Heterogeneous Catalyst Heterogeneous Catalyst

- Heterogeneous Catalyst - Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. - Adsorption - collection of one substance on the surface of another substance. - Adsorption and activation of the reactants. - Migration of the adsorbed reactants on the surface. - Reaction of the adsorbed substances. Escape, or desorption, of the products - Homogeneous Catalyst - Exists in the same phase as the reacting molecules. - Enzymes are nature's catalysts.

Collision Model

- Molecules must collide to react. - Main Factors: - Activation energy, Ea - Temperature - Molecular orientations

Reaction Mechanism

- Most chemical reactions occur by a series of elementary steps. - An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction.

Rate Law

- Review the pic - The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. - The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

Reaction Mechanism Requirements

- The sum of the elementary steps must give the overall balanced equation for the reaction. - The mechanism must agree with the experimentally determined rate law.

Method of Initial Rates

- The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. - Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. - The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

Elementary Steps (Molecularity)

- Unimolecular - reaction involving one molecule; first order. - Bimolecular - reaction involving the collision of two species; second order. - Termolecular - reaction involving the collision of three species; third order. Very rare.

Instantaneous Rate

- Value of the rate at a particular time. - Can be obtained by computing the slope of a line tangent to the curve at that point.

Reaction Rate

Change in concentration of a reactant or product per unit time.

Exercise

Ea = 53 kJ ln(2) = (Ea / 8.3145 J/K·mol)[(1/298 K) - (1/308 K)]

How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs?

For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent).

How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why?

The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism).

Exercise

a) 4.7 M [A] = -(1.0×10^-2)(30.0) + 5.0 b) 3.7 M ln[A] = -(1.0×10^-2)(30.0) + ln(5.0) c) 2.0 M (1 / [A]) = (1.0×10^-2)(30.0) + (1 / 5.0)

Exercise

a) rate = k[A]^2 We know this is second order because the second half-life is double the preceding one. b) k = 8.0 x 10^-3 M^-1 min^-1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10^-3 M^-1 min^-1)(525 min) + (1 / 5.0 M)

A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?

ln(0.65) = -k(55) + ln(1) k = 7.8 x 10^-3 min^-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem.

Concept Check

rate = k[A][B]^2 The sum of the elementary steps give the overall balanced equation for the reaction.


Conjuntos de estudio relacionados

BIO 121 Chapter 7: Membrane Structure and Function-Study Guide

View Set

UNIT 3: LESSON 3 MOBILE APPLICATION

View Set

subscriber and data management quiz MC

View Set

NURS (FUNDAMENTAL): Ch 21 NCLEX Teacher and Counselor

View Set

Chapter 3: Perception and Communication

View Set