Ap Chemistry Notes 7-1 Kinetics and Rate Law – an Introduction

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AP CHEMISTRY NOTES 7-1 KINETICS AND RATE LAW – AN INTRODUCTION

CHEMICAL KINETICS – the study of rates of chemical reactions and the mechanisms by which they occur

FACTORS WHICH AFFECT REACTION RATES

1. Nature of the Reactants – more active substances react faster (ie. fluorine, etc.)

2. Surface Area – the more surface area of reactants exposed, the faster the reaction

3. Concentration of Reactants – the higher the concentration of reactants, the faster the reaction

4. Temperature – in general, the higher the temperature, the faster the reaction

5. Catalysts - involved in the reaction, but are unchanged themselves

RATE OF REACTION

Rate is measured by either a. a decrease in concentration of a reactant per unit time b. an increase in concentration of a product per unit time

Rate is always determined in terms of initial reaction rate:

A + 2B → C + D

Rate = - ∆A or Rate = ∆D ∆t ∆t

Relative rates can be determined from a balanced chemical equation using reciprocal coefficients:

4PH3 → P4 + 6H2

Initial Reaction Rate = - 1/4 ∆[PH3] = ∆[P4] = 1/6 ∆[H2] ∆t ∆t ∆t

EXAMPLE 1: For the reaction below,

A + 3B → 4C

what is the rate of ∆[B] with respect to ∆[C]? ∆t ∆t

Using this information, what would be the initial rate of change in [B] in Trial 2 below?

EXAMPLE 2: For the following reaction, what is ∆[NH3] with respect to ∆[H2]? ∆t ∆t

N2 + 3H2 → 2NH3

DIFFERENTIAL RATE LAWS

*General form:

Reaction: aA + bB → xX

General Rate Law (or Rate Expression): Rate = k [A]m [B]n

Where k = rate constant (for a specific rxn at a specific To) [A] = concentration of reactant A (mol/L) [B] = concentration of reactant B (mol/L) m = order of reaction with respect to reactant A n = order of reaction with respect to reactant B

NOTE: * These are initial rates, so the concentration of products is negligible * “m” and “n” can only be determined experimentally

EXAMPLE: What is the general rate law (or rate expression) for the reaction in Example 2 above?

AP CHEMISTRY NOTES 7-2 RATE LAWS AND THE ORDER OF A REACTION

RATE ORDER

The order with respect to one of the reactants is equal to the power to which the concentration of that reaction is raised in the rate equation.

Zero Order Changing the concentration of the reactant has no effect on the rate (ie. change in concentration = no change in rate) Not very common

Rate = k [A]o or Rate = k

First Order Doubling the concentration of the reactant doubles the rate (ie. change in concentration = change in rate) Very common

Rate = k [A]1 or Rate = k [A]

Second Order Doubling the concentration of the reactant quadruples the change in rate (ie. change in concentration = change2 in rate)

Common, especially in gas phase reactions

Rate = k [A]2

In a nutshell, for the general rate law Rate = k [A]m [B]n

If m = 0 The reaction is zero order with respect to [A] m = 1 The reaction is first order with respect to [A] m = 2 The reaction is second order with respect to [A]

n = 0 The reaction is zero order with respect to [B] n = 1 The reaction is first order with respect to [B] n = 2 The reaction is second order with respect to [B]

*Adding the orders of each reactant gives the overall order of the reaction

*If no other information is given, assume the units for rate are M/s or mol/L.s or mol . L-1 . s-1

EXAMPLE: Consider the reaction: 2A + 3B → 2C + D

The following information about this reaction was obtained experimentally:

Experiment # Initial [A] Initial [B] Initial Rate of Formation of C 1 0.10 M 0.10 M 2.0 x 10-3 M/s 2 0.10 M 0.30 M 6.0 x 10-3 M/s 3 0.30 M 0.30 M 5.4 x 10-2 M/s 4 0.25 M 0.15 M ?

a. What is the general rate law, or rate expression, for this reaction?

b. Give the differential rate law for this reaction.

By Inspection:

*In Experiments __(#)___ and __(#)__, the concentration of (RCO) was held constant while the

concentration of (RCH) was (doubled, tripled, etc...). This caused the rate of the reaction to (double, triple, increase by a factor of “x”, etc…).

Therefore the order with respect to (RCH) is _____. *

Using this information, the differential rate law for this reaction is: ______

By Calculation:

c. The overall rate order of the reaction is: _____

d. Find the rate constant, k, for this reaction (including units).

e. Calculate the initial rate of formation of [C] in Experiment 4.

EXAMPLE: For the reaction 2NO + H2 → N2O + H2O

the following information was obtained experimentally:

Initial Pressure Initial Pressure Initial Rate of Pressure Experiment # NO H2 Decrease 1 0.150 atm 0.400 atm 0.020 atm/min 2 0.075 atm 0.400 atm 0.005 atm/min 3 0.150 atm 0.200 atm 0.010 atm/min

Determine:

a. The general rate law:

b. The differential rate law for the reaction:

By Inspection:

Using this information, the differential rate law for this reaction is: ______

By Calculation:

c. The overall rate order is: ______

d. The value of the rate constant, k:

EXAMPLE: A particular reaction has a rate law of: Rate = k[A]2[B] If a reaction were carried out in which the [A] was doubled, while [B] was tripled, what would be the effect on the rate?

Note: Units for k if the reaction is first order overall (assume time in “s”): s-1 second order overall: L . mol-1 . s-1 or atm-1 . s-1 third order overall: L2 . mol-2 . s-1 or atm-2 . s-1 zero order overall: mol . L-1 . s-1 or atm . s-1

AP CHEMISTRY NOTES 7-3 GRAPHICAL ANALYSIS: RATE ORDER

DETERMINATION OF THE RATE OF A REACTION

Graphical representation of the rate of the disappearance of the reactant and the rate of appearance of the product in a chemical reaction:

The instantaneous rate of the reaction with respect to a reactant or product can be determined by drawing a line tangent to the curve of “concentration vs. time” at a given time. The slope of the line tangent to the curve at time “t” is equal to the rate.

EXAMPLE: Determine the instantaneous rate of the reactant in the graph above at 1.2 seconds.

Graphs of Zero, First, and Second Order Reactions:

Graph of a zero order reaction :

Graph of a first order reaction:

Graph of a second order reaction:

DETERMINATION OF RATE ORDER

In order to determine the order with respect to a given reactant [A], 3 graphs must be prepared:

[A] vs. time ln[A] vs. time [A]-1 vs. time

For a given set of data, obtaining a straight line for one of the above conditions indicates the order of that reactant as follows: If [A] vs. time is a straight line - zero order - Rate = k[A]0 or Rate = k If ln[A] vs. time is a straight line - first order - Rate = k[A]1 or Rate = k[A] If [A]-1 vs. time is a straight line - second order - Rate = k[A]2

In order to determine the rate constant, k, for the reaction, the slope of the straight line is determined: k = І slope l

In addition, the half-life of the reaction can be determined, as well as the integrated rate law – both based on the order of the reaction determined. The following table is a summary for the kinetics of a reaction in the form of: aA → products

Differential Plot for a Relationship of Integrated Rate Order Half-life Rate Law straight line “k” to slope Rate Law

zero Rate = k [A] vs. t k = І slope l [A] - [A]o = -kt t1/2 = [A]o / 2k first Rate = k [A] ln [A] vs. t k = І slope l ln [A] - ln[A]o = -kt t1/2 = 0.693/k 2 -1 second Rate = k [A] [A] vs. t k = І slope l 1/[A] - 1/[A]o = kt t1/2 = 1 / k[A]o

- - EXAMPLE: For the reaction (CH3)3CBr + OH → (CH3)3COH + Br the following data were obtained experimentally: TIME (s) [(CH3)3CBr] 0 0.100 30 0.074 60 0.055 90 0.041

Using graphical means, it was determined that the graph of ln[(CH3Br)] vs. time is a straight line with a slope of -0.00991. Determine the following:

the rate law: the rate constant, k:

the integrated rate law: the half-life:

the amount of reactant left after 100 seconds have elapsed:

EXAMPLE: For the reaction A → B + C the following data were obtained experimentally:

TIME (min) [A] 0 0.030 5 0.025 10 0.020 15 0.015

Using graphical means, it was determined that the graph of [(CH3Br)] vs. time is a straight line with a slope of -0.00100. Determine the following:

the rate law: the rate constant, k:

the integrated rate law: the half-life:

the amount of reactant left after 22 minutes have elapsed:

EXAMPLE: For the reaction X → Y the following data were obtained experimentally:

TIME (hr) Pressure of X (atm) 0 3.0 0.5 2.5 1.0 2.1 1.5 1.9

-1 Using graphical means, it was determined that the graph of the (PX) vs. time is a straight line with a slope of 0.131. Determine the following:

the rate law: the rate constant, k:

the integrated rate law: the half-life:

the amount of reactant left after 1.25 hours have elapsed:

AP CHEMISTRY NOTES 7-4 GRAPHICAL ANALYSIS: ACTIVATION ENERGY

DETERMINATION OF ACTIVIATION ENERGY

Activation energy can be determined using the Arrhenius Equation:

푬 ퟏ 퐥퐧 (풌) = − ( 풂) ( ) + 퐥퐧 (퐀) 푹 푻

( y ) = ( a ) ( x ) + ( b)

Where k = the rate constant Ea = activation energy (J/mol) R = 8.3145 J/K.mol T = temperature (K) A = frequency factor*

*the “frequency factor” is the frequency of collisions with correct geometry and is given in units of L/mol.s

Notice that a graph of ln (k) vs. 1/T (“y” and “x” in the above equation) will yield a slope (“a”) equal to _ Ea R So: 퐄 퐒퐥퐨퐩퐞 = − ( a) 퐑

or: E a = - R (slope)

Keep in mind:

If Ea is smaller, “k” is larger, so the reaction is faster. If Ea is larger, “k” is smaller, so the reaction is slower.

EXAMPLE: For the reaction 2N2O5 → 4NO2 + O2

the following data were obtained:

k (s-1) T (oC) 2.0 x 10-5 20 7.3 x 10-5 30 2.7 x 10-4 40 9.1 x 10-4 50 2.9 x 10-3 60

Calculate the activation energy (Ea) for this reaction.

*Another form of the Arrhenius Equation is as follows:

퐤 퐄 ퟏ ퟏ 퐥퐧 ( ퟐ) = 퐚 ( − ) 퐤ퟏ 퐑 퐓ퟏ 퐓ퟐ

This allows us to calculate activation energy when “k” values are known at only two temperatures.

EXAMPLE: The colorless gas dinitrogen tetroxide decomposes to the brown gas nitrogen dioxide in a first order reaction with a value of “k” equal to 4.5 x 103 s-1 at 275 K. If “k” is later found to have a value of 1.00 x 104 s-1 at 283 K, what is the activation energy?

AP CHEMISTRY NOTES 7-5 REACTION MECHANISMS

Chemical Reactions occur by a series of steps called the “reaction mechanism” – the sequence of bond- making and bond-breaking steps that occurs during the conversion of reactants to products in a chemical reaction. elementary step - a basic step in a reaction mechanism;

The rate law of an elementary step can be written directly from its molecularity (unlike more complex reactions whose rate laws must be determined experimentally). molecularity – the number of species that must collide to produce the reaction indicated by that step in the reaction mechanisms

Elementary Step Molecularity Rate Expression A → products unimolecular rate = k[A] A + B → products bimolecular rate = k[A][B] A + A → products bimolecular rate = k[A]2 2A + B → products termolecular rate = k[A]2[B]

Note: A unimolecular step is always first order overall A bimolecular step is always second order overall A termolecular step is always third order overall (and is quite rare)

Consider the complex reaction between nitrogen dioxide and carbon monoxide:

NO2 + CO → NO + CO2

2 From experimental evidence, the rate law for this reaction is: Rate = k[NO2]

The reaction mechanism is thought to occur in the following elementary steps:

NO2 + NO2 → NO3 + NO

NO3 + CO → NO2 + CO2

Notice the presence of the intermediate NO3 in the reaction mechanism

In order for this to be a valid mechanism, three requirements must be met:

1. The sum of the elementary steps must give the overall balanced equation for the reaction

2. The mechanism must agree with the experimentally-determined rate law

3. The rate law must not include an intermediate species – one which appears in the reaction mechanism, but not in the overall equation

In order to determine whether requirement #2 has been met, another term must be introduced:

rate determining step – the slowest step in a reaction mechanism

The overall rate can be no faster than the rate-determining step.

If the rate-determining step (slow step) is the first step in the reaction mechanism, the rate law is determined solely from the molecularity of this step:

NO2 + NO2 → NO3 + NO (slow)

NO3 + CO → NO2 + CO2 (fast)

The rate law can be determined by using the molecularity of the elementary steps up to and including the slow step.

EXAMPLE: Find the overall reaction and the rate law for the following mechanism:

I2 ↔ 2 I (fast) I + H2 ↔ H2I (fast) H2I + I → 2 HI (slow)

EXAMPLE: The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in the gas phase.

Step 1 Cl2 ↔ 2 Cl fast equilibrium

Step 2 CH4 + Cl → CH3 + HCl slow

Step 3 CH3 + Cl2 → CH3Cl + Cl fast

Step 4 CH3Cl + Cl → CH2Cl2 + H fast

Step 5 H + Cl → HCl fast

a) In the mechanism, is CH3Cl a catalyst, or is it an intermediate? Justify your answer.

b) Identify the order of the reaction with respect to each of the following according to the mechanism. In each case, justify your answer.

i) CH4

ii) Cl2

c) Give the rate law for this reaction.