14.1 expressing the reaction rate -...

February 181 Expressing Reaction Rates

14.1 Expressing the Reaction Rate

Differential Rate and Integrated Rate Laws

Fred Omega GarcesChemistry 201Miramar College


OutlineKinetics: Intro

Factors influencing Reaction Rates

Rates & Stoic RelationsExpressing reaction Rates

Rate Law and componentsDifferential Rate Law

Initial Rate Method

Rate Laws and Half-Lives 0th Order 1st Order2nd Order

Kinetic Molecular Theory

PreExponential FactorCollision FrequencySteric Factor

Activation EnergyArrhenius EquationReaction Coordinate Diagram

Catalyst factor

Maxwell-Boltzmann Diagram

Reaction Mechanism & Chapman cycle

Elementary Step

Rate Determining Step


Thermodynamics Vs. KineticsThermoChem: Will a reaction occur ?

Kinetics: If so, how fast ?

• Iron rusting; DG -,

but rate of rusting depends on reaction conditions.

4 Fe (s) + 3 O2(g) + xH2O(g) D 2Fe2O3•XH2O (s)

Whether it occurs overnight or over many years, the reaction condition influence how fast it occurs.

• Conversion of diamond to graphite is

thermodynamically favorable (DG -).

C (diamond) D C (graphite) DG = -2.84 kJ/mol

Kinetics makes this reaction nearly impossible.


Factors Influencing Reaction Rates

Factors Affecting RatesConcentration of Reactant - Molecules must collide in order to react.

More reactants means the faster rxn. Note: rate µ concentration

Physical state - Molecules must mix to form product. The more finely a reactant chemical is divided, the greater its surface area per unit mass, the more contact it makes with other reactants the greater the speed of the reaction. Stirring and mixing helps in this process. Note: rate µ collision frequency

Temperature- Molecules must collide with enough energy to react. Note: rate µ collision energy (energy is temperature dependent)

Catalyst- The pathway in which the reactant convert to product influence the reactivity.


Rates ; Another word for Speed

Rate - Change / time (A positive quantity)

Speed - D distance / D time = D x / DtimeWages - D money / D time = D $ / DtimeReaction - D conc. / D time = D [A] / Dtime

Differential Rate Law; An expression that relates the rate to the concentration,D [A] with respect to time.

General Form: D [A]D t


Instantaneous Vs. Average rates(Integrated Form)

Consider: Reactant ® ProductExperiment will monitor concentration of reactant (product) as

a function of time.

rate= - [D c] = - 0.45-0.60 = + 0.15D t 30 - 15 15

rate (Avg.) = - 0.0010 M s-1

rate= - [D c] = - 0.40-0.67 = + 0.27D t 40 - 10 30

rate (Avg.) = - 0.0090 M s-1

rate = - dc = Tangent atdt


Stoichiometry Relationship: Example

H2 (g) + I2 (g) ® 2 HI (g)

Rate = - D [H2] = - D [I2] ? + D [HI] Rxn D t Dt D t

- D [H2] = + D [HI] ? - 1 mol H2D t Dt 2 mol HI

- D [H2] = + D [HI]D t 2 Dt

Rate of reaction does not depend on which specie is monitored during the reaction.

Rate = - D [H2] = - D [I2] = + D [HI] Rxn D t D t 2 Dt

Disappearance Appearance


Stoichiometry Rate Relationship

aA + bB ® cC+ dD

Rate = - D [A] = - D [B ] = + D [C ] = + D [D ] Rxn a Dt b Dt c Dt d Dt

Rate = - D [H2] = - D [O2 ] = + D [H2O ] Rxn 2 D t Dt 2 D t

2H2 + O2 ® 2 H2O

2O3 ® 3 O2

Rate = - D [O3] = + D [O2 ]Rxn 2 D t 3 Dt

10 mol/min

-5 mol/min -5 mol/min 5 mol/min


Reaction Rate RelationshipRates of a reaction in terms of any specie in the

reaction: aA +bB g pP + rRCoefficient- The coefficients from the balance equation must be

taken into account when calculating rates (speed of rxn).Stoichiometry- Although mole ratio from overall stoichiometry does

not relate to the order of the rate or the Rate Law directly (to be discussed later), it does provide the relationship of the rate of reactant consumption and product generation.

In our previous example: 2 NO2(g) ®. 2 NO(g) + O2(g)

Reactant NO2 , and product NO, have coefficient of 2, but product O2 has coefficient of 1. This leads to the production of NO having the same rate as NO2 consumption, but the production of O2 being only half as fast. Re

acti

on P

rogr

ess t = 1

t = 2

t = 3

t = 4


In Class Exercise

Acrylonitrile is produced from propene, ammonia and oxygen by the following reaction: 2C3H6 + 2NH3 + 3O2 g 2CH2CHCN + 6H2O

1. Relate the rates of reaction of starting materials and products.2. Draw a Concentration Vs. Time diagram for the reaction.3. If 120 moles of H2O is produced in 1-min. How many moles of propene

and ammonia are consumed in 1-hr. Assume 1-L vessel.4. If 19.20 grams of Oxygen is consumed in 1-min, how long (min) will it

take to produce 100 g of acrylonitrile. Assume 1-L vessel.

Strategy: The rates for different species participating in a chemical reaction are related by the stoichiometric coefficients of the balanced chemical equation.

In Class activity: Number heads - Break in to groups of 4 and answer the following question. I will call one of the members to explain answer in 10 min.


Summary of reaction rateThe average reaction rate is the change in reactant (or product) concentration in a certain time increment. The rate slows down however as reactants are used up. The instantaneous rate at time t is obtained from the slope of the tangent on a concentration-time curve at time t. The method of initial rates is generally used to avoid complications of back reaction.

The next step is to solve the differential rate law which leads to the integrated rate law. (This requires Calculus)


Solutions to the Differential Rate Law*

Differential Form of the Rate Law -Expression that provides rate dependence on time.

Also referred to as simply the Rate Law

aA + bB ® mM+ nN

rate = K [A]x [B]y [C]w

C = catalyst x, y and w - order of reaction

x-th order in A: y-th order in B: w-th order in C, (the catalyst)x, & y is not related to the coefficient of the balance equation; it is

experimentally determinedK - rate constant

* Zumdahl


Methods of Initial RatesIn our discussion so far, the reverse reaction has not been considered: 2 NO2(g) ® 2 NO(g) + O2(g)

Suppose 2 NO(g) + O2(g) ® 2 NO2(g also occurred?

In determining the concentration of species in a reaction, the formation of products may interfere with the measurements.

This can be avoided if the reaction is studied under conditions in which the reverse reaction makes negligible contribution. This is the Initial Rates Methods

Methods of Initial Rates: Rates of the reaction depends only on the concentration of the reactants.


Initial Rate MethodologyDetermining the Order of a Reaction

• Several experiments are carried out in which the initial concentration of all substances are known but are varied in the different trials

• The reaction is allowed to proceed for a short period (~3% of completion) such that the average rate during this time of the reaction is close to the instantaneous rate.

• The result of the analysis yields the Differential Rate Law (or Rate Law)

• The experiment is such that the rate is a function of only the reactants in the reaction.


Consider the combination reaction of NO and O2 to produce NO2 :2 NO(g) + O2(g) ® 2 NO2 (g)

Determination of the Rate Law (via Methods of Initial Rates)

Initial Concentrations (mol/ L) Initial Rate

Experiment [NO] [O2] (mol/L•s)1 0.020 0.010 0.028 2 0.020 0.020 0.0573 0.020 0.040 0.1144 0.040 0.020 0.2275 0.010 0.020 0.014

Based on these data, what is the rate law? What is the value of the rate constant, K?

Rate Law Determination


Experiments are chosen such that the concentration of all except one of the reactant does not change. If there are two reactants in the reaction then a minimum of three experiments is required. Three reactants require a minimum of four experiments.

The two experiments that are selected are divided between each other in the general formula:

Expt 1 = Rate1 = K [A1]x [B1]y [C1]z …Expt 2 = Rate2 = K [A2]x [B2]y [C2]z …

The reaction order is determined by solving the ratios.

Once the order for each reactant is determined the rate constant (K) is solved by using one of the experimental conditions.

…. but first,

GuideLines for Solving Rate Law


Exponential Operations:20 = 121 = 222 = 423 = 824 = 1625 = 3226 = 64

A Quick Math Review

Solving for exponents:Ax = BX • lnA = lnBX = lnB / lnA

i.e., 26 = 64

Try 2x = 64, solve for x

x = ln(64) / ln(2)= 2.77/.693= 6


Now consider our pervious problem:Initial Concentrations

(mol/ L) Initial RateExperiment [NO] [O2] (mol/L • s)1 0.020 0.010 0.028 2 0.020 0.020 0.0573 0.020 0.040 0.1144 0.040 0.020 0.2275 0.010 0.020 0.014

Rate Law Determination; Ratio of Experiments

Select experiments in which concentration of [NO] is constant but the

concentration of O2 is doubled: Experiments 1 and 2

Expt 2 Rate2 k [NO]x [O2]y 0.057 k [0.020]x [0.020]y

Expt 1 Rate1 k [NO]x[O2]y 0.028 k [0.020]x [0.010]y==


Rate Law Determination; Ratio of Experiments

Select experiments in which concentration of [NO] is constant but the concentration of O2 is doubled: Experiments 1 and 2

Expt 2 Rate2 k [NO]x [O2]y 0.057 k [0.020]x [0.020]y

Expt 1 Rate1 k [NO]x[O2]y 0.028 k [0.020]x [0.010]y

Result:

2 = 2y; y = 1 g 1st order in O2

Similarly, selecting experiments 2 and 5 leads to

Expt 2 Rate 2 k [NO]x [O2]y 0.057 k [0.020]x [0.020]y

Expt 5 Rate 5 k [NO]x[O2]y 0.014 k [0.010]x [0.020]y

This leads to: 4 = 2x; x = 2 g 2nd order in NO.

==

==


Rate Law; Solving for rate Constant

The general rate law is:

Rate law = k [NO]2 [O2]

the rate constant k is determine by selecting one of the experiments and solving the equation.

Consider experiment #1 :

Rate = 0.028 = k [0.020]2 [0.010]

k = 0.028 / (4•10-4) (0.010) = 7.1•103 M-2 s-1

Rate Law: Rate = 7.1•103 M-2s-1 [NO]2 [O2]


In Class ExerciseAcrylonitrile is produced from propene, ammonia and oxygen by the following reaction at 25�C: 2C3H6 + 2NH3 + 3O2 g 2CH2CHCN + 6H2O

The following data were obtained. (Units are mol/L)C3H6 NH3 O2 Initial Rates

1.00 e-4 2.00 e-2 2.00 e-2 1.66 e -7

2.00 e-4 2.00 e-2 2.00 e-2 3.33 e -7

3.00 e-4 2.00 e-2 2.00 e-2 4.99 e -7

1.00 e-4 4.00 e-2 2.00 e-2 6.66 e -7

1.00 e-4 1.00 e-2 2.00 e-2 0.42 e -7

1.00 e-4 2.00 e-2 4.00 e-2 13.2 e -7

1.00 e-4 1.00 e-2 4.00 e-2 3.36 e -7

Rate Law: Rate = 5.18 e5 M-5s-1 [C3H6]1 • [NH3]2 • [O2]3


Summary

Reaction rates depend on reactant concentrations, temperature, and the presence of a catalyst. The concentration dependence is given by the rat law, rate = k [A]m •[B]n, where k is the rate constant, m and n specify the reaction order with respect to reactants A and B, andm+n is the overall reaction order. The values of m and n must be determined by experiments; they can�t be deduced from the stoichiometry of the overall reaction.

The integrated rate law is a concentration-time equation that allows us to calculate concentrations at any time or the time required for an initial concentration to reach any particular value.

Answer last problem Rate = 5.25 e5 M-5s-1 [C3H6]1 • [NH3]2 • [O2]3

14.1 expressing the reaction rate -...

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