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Chpt 12 - Chemical KineticsReaction RatesRate LawsReaction MechanismsCollision TheoryCatalysisHW set1: Chpt 12 - pg. 580-592, # 22, 23, 28 Due Dec. 13HW set2: Chpt 12 # 39, 48, 53, 56, 58, 60 due Dec. 17HW set3: Chpt 12 #62, 69, 72 due Dec. 19Reaction Rate ? Change in concentration of a reactant or product per unit time.

[A] means concentration of A in mol/L; A is the reactant or product being considered.

Decomposition of NO2 Graph

Instantaneous RateValue of the rate at a particular time.Can be obtained by computing the slope of a line tangent to the curve at that point.2NO2 --> 2NO + O2 Stoichiometry of Rates

2NO2 --> 2NO + O2The rate of comsumption (disappearance) of NO2 is same as rate of production of NO and twice the rate of production of O2. Calculate ave. rate from data

Rate LawShows how the rate depends on the concentrations of reactants.For the decomposition of nitrogen dioxide:2NO2(g) 2NO(g) + O2(g)

Rate = k[NO2]n:k = rate constantn = order of the reactantRate Law (cont)Rate = k[NO2]nThe 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.Types of Rate LawsDifferential 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.Rate Laws (cont)typically consider reactions when the reverse reaction is unimportant, so our rate laws involve only [reactants].differential and integrated rate laws for a given reaction are related in a welldefined way, can use either rate law.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.Determining the form of the Rate LawDetermine experimentally the power to which each reactant concentration must be raised in the rate law.Method of Initial RatesThe 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.Initial Rates examplesPg. 549 Table 12.4 discuss data

Pg. 550 Table 12.5 discuss dataOverall Reaction OrderThe sum of the exponents in the reaction rate equation.

Rate = k[A]n[B]m

Overall reaction order = n + m

k = rate constant[A] = concentration of reactant A[B] = concentration of reactant BIntegrated Rate Laws - zero first & second order

Take actual data for each reactant and plot it k is slope of line [A] vs t or ln[A] vs t or 1/[A] vs t straight line says Oo, 1o, 2ofor calculations at any time t t = time, [A]o = initial conc, [A] = conc at t, k = rate constantTable 12.6Which Order?

A plot of [A] vs timeWhich order?

Which order?

Exercise of Reaction OrderConsider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 x 102 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is:

Zero order First order Second order4.7 M3.7 M2.0 MIntegrated rate from a graphExcel file AP Chem Chpt12problems.xlsFind it on wiki spaceHalf-life determinationHalf-life equation provided in Table 12.6 determined from integrated form. Can calculate yourself using the concentration when half remains as [A]=[A]o/2 which is half original conc and do the math. Answer has time related to k and [A]o Reaction MechanismMost 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.Mechanism example

A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CONO2(g) + CO(g) NO(g) + CO2(g)MechanismsUnimolecular 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.Elementary Steps (Molecularity)

Mechanism SummaryThe sum of the elementary steps must give the overall balanced equation for the reaction.The mechanism must agree with the experimentally determined rate law.Rate determining stepA 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.Mechanism Ex. Decomposition of N2O52N2O5(g) 4NO2(g) + O2(g)

Step 1: 2( N2O5 NO2 + NO3 )(fast) Step 2: NO2 + NO3 NO + O2 + NO2(slow)Step 3: NO3 + NO 2NO2 (fast)Does this satisfy overall balanced eqn?Elementary ReactionsIf you know that it is an elementary reaction, then the rate law just uses the coefficients from the balanced equation.If you have the mechanism in elementary steps you can figure out a plausible rate lawReaction Mechanism Ex.The reaction A + 2B C has the following proposed mechanism:A + B D(fast equilibrium)D + B C(slow)

Write the rate law for this mechanism.

rate = k[A][B]2What if the fast and slow were reversed?Collision TheoryMolecules must collide to react.Main Factors:Activation energy, EaTemperatureMolecular orientationsReaction Diagram

Arrhenius Model of rxnsHow fast a reaction happens is related by energy

Number of collisions with Ea = (total collisions) x e-Ea/RTwhere R is gas constant and T is temp (K)

But this value was still too high for observed rates so total collisions must not all react (orientation effect)so k = zpe-Ea/RT with z = collision frequency and p = steric factor (orientation)Arrhenius Equation

A=frequency factor (zp)Ea=activation energyR=gas constant (8.3145 J/Kmol)T =temperature (in K)Linear Form Arrhenius Eqn

Slope = ?Ex. Activation Energy, EaChemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25C to 35C?

Ea = 53 kJCatalystsA substance that speeds up a reaction without being consumed itself.Provides a new pathway for the reaction with a lower activation energy.Rxn Diagram for catalyzed

The catalyst provides a new pathway which has a lower Ea

It does not lower the Ea for the original rxnEffective collisions graphic

Catalyzed Rxn has a lower Ea, so many more collisions are effective at making the product(s) Heterogeneous CatalystsMost often involves gaseous reactants being adsorbed on the surface of a solid catalyst.Adsorption collection of one substance on the surface of another substance.

Heterogeneous Catalysis ProcessAdsorption and activation of the reactants.Migration of the adsorbed reactants on the surface.Reaction of the adsorbed substances.Escape, or desorption, of the products.