14.1 expressing the reaction rate -...
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February 181 Expressing Reaction Rates
14.1 Expressing the Reaction Rate
Differential Rate and Integrated Rate Laws
Fred Omega GarcesChemistry 201Miramar College
February 182 Expressing Reaction Rates
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
February 183 Expressing Reaction Rates
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.
February 184 Expressing Reaction Rates
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.
February 185 Expressing Reaction Rates
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
February 186 Expressing Reaction Rates
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
February 187 Expressing Reaction Rates
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
February 188 Expressing Reaction Rates
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
February 1810 Expressing Reaction Rates
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
February 1811 Expressing Reaction Rates
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.
February 1812 Expressing Reaction Rates
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)
February 1813 Expressing Reaction Rates
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
February 1814 Expressing Reaction Rates
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.
February 1815 Expressing Reaction Rates
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.
February 1816 Expressing Reaction Rates
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
February 1817 Expressing Reaction Rates
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
February 1818 Expressing Reaction Rates
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
February 1819 Expressing Reaction Rates
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==
February 1820 Expressing Reaction Rates
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.
==
==
February 1821 Expressing Reaction Rates
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]
February 1822 Expressing Reaction Rates
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
February 1823 Expressing Reaction Rates
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
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