Chapter 13 Chemical Kinetics CHEMISTRY

Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: –reactant concentration, –temperature, –action of catalysts, and –surface area. Goal: to understand chemical reactions at the molecular level. Factors that Affect Reaction Rates

Change of Rate with Time Consider: C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Reaction Rates

Change of Rate with Time C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) –We can calculate the average rate in terms of the disappearance of C 4 H 9 Cl. –The units for average rate are mol/L·s or M/s. –The average rate decreases with time. –We plot [C 4 H 9 Cl] versus time. –The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. –Instantaneous rate is different from average rate. –We usually call the instantaneous rate the rate. Reaction Rates

For the reaction A  B there are two ways of measuring rate: –the speed at which the products appear (i.e. change in moles of B per unit time), or –the speed at which the reactants disappear (i.e. the change in moles of A per unit time). Reaction Rates

Reaction Rate and Stoichiometry In general for: aA + bB  cC + dD Reaction Rates

Rate is (-) if reagent is consumed. Rate is (+) if compound is produced. Rate will ultimately be (+) because change in concentration will be negative. Two (-)’s become (+).

Take Note: Rate must ALWAYS be a positive value!

Rate Comparison 2NO 2  2NO + O 2 *If main interest is on the consumption of starting reagent, then: Rate = -  [NO 2 ] =      t  t  t Since 2 NO 2 molecules are consumed for every O 2

Take Note! Since the direction of equilibrium changes as more product is produced, rates have to be determined as soon as the experiment has begun.

2N 2 O 5 4NO 2 + O 2 [N 2 O 5 ] (mol/L)Time (sec)

Sample Problem A. How is the rate at which ozone (O 3 ) disappears related to the rate at which O 2 appears in the reaction: 2 O 3 (g)  3 O 2 (g)? B. If the rate at which O 2 appears,  [O 2 ]/  t, is 6.0 x M/s at a particular instant, at what rate is O 3 disappearing at this same time, -  [O 3 ]/  t?

Answers A. “Related to” means compare, so write the rate expression comparing the compounds. B. 4.0 x M/s

Sample Problem The decomposition of N 2 O 5 proceeds according to the following equation: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) If the rate of the 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, (b) O 2 ?

Differential Rate Law - is a rate law that expresses how rate is dependent on concentration Example: Rate = k[A] n

Differential First Order Rate Law First Order Reaction – Rate dependent on concentration – If concentration of starting reagent was doubled, rate of production of compounds would also double

Using Initial Rates to Determines Rate Laws A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. A reaction is first order if doubling the concentration causes the rate to double. A reacting is nth order if doubling the concentration causes an 2 n increase in rate. Note that the rate constant does not depend on concentration. Concentration and Rate

Differential Rate Law For single reactants: A  C Rate = k[A] n For 2 or more reactants: A + B  C Rate = k[A] n [B] m Rate = k[A] n [B] m [C] p

Exponents in the Rate Law For a general reaction with rate law we say the reaction is mth order in reactant 1 and nth order in reactant 2. The overall order of reaction is m + n + …. A reaction can be zeroth order if m, n, … are zero. Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. Concentration and Rate

In general rates increase as concentrations increase. NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) Concentration and Rate

Rate law: The constant k is the rate constant. Concentration and Rate

Experimental Data Expt.[NH 4 ] initial [NO 2 - ] initial Initial Rate M M1.35 x M0.010 M2.70 x M0.010 M5.40 x 10 -7

Problem NH 4 + +NO 2 - N 2 +2H 2 O Give the general rate law equation for rxn. Derive rate order. Derive general rate order. Solve for the rate constant k.

To Determine the Orders of the Reaction (n, m, p, etc….) 1. Write Rate law equation. 2. Get ratio of 2 rate laws from successive experiments. Ratio = rate Expt.2 = k 2 [NH 4 + ] n [NO 2 - ] m rate Expt.1 k 1 [NH 4 + ] n [NO 2 - ] m 3. Derive reaction order. 4. Derive overall reaction order. 5. Use one set of values, plug in the orders and solve for the rate constant.

Experimental Data Expt.[NH 4 ] initial [NO 2 - ] initial Initial Rate M M1.35 x M0.010 M2.70 x M0.010 M5.40 x 10 -7

Experiment Number [A] (mol·L -1 )[B] (mol·L -1 )Initial Rate (mol·L -1 ·s -1 ) x x x A + B  C (a) Determine the differential rate law (b) Calculate the rate constant (c) Calculate the rate when [A]=0.050 mol·L -1 and [B]=0.100 mol·L -1

Use the data in table 12.5 to determine 1) The orders for all three reactants 2) The overall reaction order 3) The value of the rate constant

Experiment Number [NO] (mol·L -1 )[H 2 ] (mol·L -1 )Initial Rate (mol·L -1 ·s -1 ) x x x (a) Determine the differential rate law (b) Calculate the rate constant (c) Calculate the rate when [NO]=0.050 mol·L -1 and [H 2 ]=0.150 mol·L -1 2NO(g) + 2H 2 (g)  N 2 (g) + 2H 2 O(g)

Consider the general reaction aA + bB  cC and the following average rate data over some time period Δt: Determine a set of possible coefficients to balance this general reaction. Sample Problem:.

Problem Reaction: A + B  C obeys the rate law: Rate = k[A] 2 [B]. A. If [A] is doubled (keeping B constant), how will rate change? B. Will rate constant k change? Explain. C. What are the reaction orders for A & B? D. What are the units of the rate constant?

You now know that…. The rate expression correlates consumption of reactant to production of product. For a reaction: 3A  2B - 1  [A] = 1  [B] 3  t2  t The differential rate law allows you to correlate rate with concentration based on the format: Rate = k [A] n

You also know that… 1. Rate of consumption of reactant decreases over time because the concentration of reactant decreases. Lower concentration equates to lower rate. 2. If a graph of concentration vs. time were constructed, the graph is not a straight line

How can we make the line straight? What is the relationship between concentration and time?

By graphing concentration vs. 1/time?

The Integrated Rate Law makes this possible!

Integrated Rate Law Expresses the dependence of concentration on time

Integrated Rate Laws Zero Order: [A] t = -kt + [A] o First Order:ln[A] t = -kt + ln[A] o Second Order: 1 = kt + 1 [A] t [A] o where [A] o is the initial concentration and [A] t is the final concentration.

Integrated First-Order Rate Law ln[A] t = -kt + ln[A] 0 Eqn. shows [concn] as a function of time Gives straight-line plot since equation is of the form y = mx + b

Zero Order Reactions A plot of [A] t versus t is a straight line with slope -k and intercept [A] 0. The Change of Concentration with Time

First Order Reactions A plot of ln[A] t versus t is a straight line with slope -k and intercept ln[A] 0. The Change of Concentration with Time

Second Order Reactions A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0. The Change of Concentration with Time

Activation Energy How does a molecule gain enough energy to overcome the activation energy barrier? From kinetic molecular theory, we know that as temperature increases, the total kinetic energy increases. We can show the fraction of molecules, f, with energy equal to or greater than E a is where R is the gas constant (8.314 J/mol·K). Temperature and Rate

Determining the Activation Energy If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: From the above equation, a plot of ln k versus 1/T will have slope of –E a /R and intercept of ln A. Temperature and Rate

Determining the Activation Energy If we do not have a lot of data, then we recognize Temperature and Rate

Elementary Steps Elementary step: any process that occurs in a single step. Molecularity: the number of molecules present in an elementary step. –Unimolecular: one molecule in the elementary step, –Bimolecular: two molecules in the elementary step, and –Termolecular: three molecules in the elementary step. It is not common to see termolecular processes (statistically improbable). Reaction Mechanisms

Rate Laws for Elementary Steps Reaction Mechanisms

Rate Laws for Elementary Steps The rate law of an elementary step is determined by its molecularity: –Unimolecular processes are first order, –Bimolecular processes are second order, and –Termolecular processes are third order. Rate Laws for Multistep Mechanisms Rate-determining step: is the slowest of the elementary steps. Reaction Mechanisms