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Chemical Kinetics

Overview:

- Reaction Rates
- Stoichiometry, Conditions, Concentration
- Rate Equations
- Order
- Initial Rate
- Concentration vs. Time
- First Order Rxns.
- Second Order Rxns.
- Graphical Methods

Cont’d

- Molecular Theory
- Activation Energy
- Concentration
- Molecular Orientation
- Temperature
- Arrhenius Equation
- Reaction Mechanisms
- Elementary Steps, Reaction Order, Intermediates
- Catalysts

Reaction Rates

- What Affects Rates of Reactions?
- Concentration of the Reactants
- Temperature of Reaction
- Presence of a Catalyst
- Surface Area of Solid or Liquid Reactants

Rates

- for A B- D[A] = D[B] Dt DtRate of the disappearance of A is equal in magnitude but opposite in sign to the rate of the appearance of B

Average Rate--D mol (or concentration) over a period of time, Dt

- Instantaneous Rate-- slope of the tangent at a specific time, t
- Initial Rate-- instantaneous rate at t = 0

[M]

tangent at time, t

t

time

Stoichiometry

4PH3 => P4 + 6H2

- 1D[PH3] = + 1D[P4] = + 1D[H2] 4Dt 1Dt 6Dt

- D[PH3] = + 4D[P4] = + 2D[H2] Dt Dt 3Dt

General Relationship

Rate = - 1 D[A] = - 1 D[B] = + 1 D[C] = + 1 D[D] aD t b D t cD t dD taA + bB cC + dD

Conditions which affect rates

- Concentration
- concentration rate
- Temperature
- temperature rate
- Catalyst
- substance which increases rate but itself remains unchanged

Rate Equations:

- aA + bB xX rate lawrate = k[A]m[B]n
- m, n are orders of the reactants
- extent to which rate depends on concentration
- m + n = overall rxn order
- k is the rate constant for the reaction

Examples:

- 2N2O5 => 4NO2 + O2 rate = k[N2O5] 1st order
- 2NO + Cl2 => 2NOClrate = k[NO]2[Cl2] 3rd order
- 2NH3 => N2 + 3H2rate = k[NH3]0 = k0th order

Determination of Rate Equations:

Data for: A + B => C

Expt. # [A] [B] initial rate

1 0.10 0.10 4.02 0.10 0.20 4.03 0.20 0.10 16.0

rate = k[A]2[B]0 = k[A]2 k = 4.0 Ms-1 = 400 M-1s-1 (0.10)2 M2

Exponent Values Relative to DRate

Exponent Value [conc] rate

0 double same 1 double double 2 double x 4 3 double x 8 4 double x 16

Problem:

Data for: 2NO + H2 => N2O + H2O

Expt. # [NO] [H2] rate

1 6.4x10-3 2.2x10-3 2.6x10-52 12.8x10-3 2.2x10-3 1.0x10-43 6.4x10-3 4.5x10-3 5.0x10-5 rate = k[NO]2[H2]

Units of Rate Constants

- units of rates M/s
- units of rate constants will vary depending on order of rxn M = 1 (M)2 (M) for s sM2 rate = k [A]2 [B]
- rate constants are independent of the concentration

Concentration vs. Time1st and 2nd order integrated rate equations

- First Order: rate = - D[A] = k [A]D t
- ln [A]t = - kt A = reactant [A]0
- or ln [A]t - ln [A]0 = - kt

Problem:

The rate equation for the reaction of sucrose in water is, rate = k[C12H22O11]. After 2.57 h at 27°C, 5.00 g/L of sucrose has decreased to 4.50 g/L. Find k. C12H22O11(aq) + H2O(l) => 2C6H12O6

ln 4.50g/L = - k (2.57 h) 5.00g/L

k = 0.0410 h-1

Concentration vs. Time

- Second Order:rate = - D[A] = k[A]2Dt
- 1 - 1 = kt [A]t [A]0
- second order rxn with one reactant: rate = k [A]2

Problem:

Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. If the original concentration of NH4NCO is 0.458 mol/L and k = 0.0113 L/mol min, how much time elapses before the concentration is reduced to 0.300 mol/L?NH4NCO(aq) => (NH2)2CO(aq) rate = k[NH4NCO]

1 - 1 = (0.0113) t

(0.300) (0.458)

t = 102 min

Graphical Methods

- Equation for a Straight Line
- y = bx + a
- ln[A]t = - kt + ln[A]01st order
- 1 = kt + 1 2nd order[A]t [A]0

b = slopea = y interceptx = time

Problem:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a half-life of 4.1 hr. If you begin with 1.6 x 10-3 mol of SO2Cl2 in a flask, how many hours elapse before the quantity of SO2Cl2 has decreased to 2.00 x 10-4 mol?

SO2Cl2(g) => SO2(g) + Cl2(g)

Temperature Effects

- Rates typically increase with T increase
- Collisions between molecules increase
- Energy of collisions increase
- Even though only a small fraction ofcollisions lead to reaction
- Minimum Energy necessary for reactionis the Activation Energy

Molecular Theory(Collision Theory)

Activation Energy, Ea

DH reactionEa forward rxn.Ea reverse rxn.

Energy

Reactant

Product

Reaction Progress

Activation Energy

- Activation Energy varies greatly
- almost zero to hundreds of kJ
- size of Ea affects reaction rates
- Concentration
- more molecules, more collisions
- Molecular Orientation
- collisions must occur “sterically”

The Arrhenius Equation

- increase temperature, inc. reaction rates
- rxn rates are a to energy, collisions, temp. & orient
- k = Ae-Ea/RT

k = rxn rate constant

A = frequency of collisions

-Ea/RT = fraction of molecules with energy necessary for reaction

Graphical Determination of Ea

rearrange eqtn to give straight-line eqtn

y = bx + a

ln k = -Ea1 + ln A RT

slope = -Ea/R

ln k

1/T

Problem:

Data for the following rxn are listed in the table. Calculate Ea graphically, calculate A and find k at 311 K.

Mn(CO)5(CH3CN)+ + NC5H5 => Mn(CO)5(NC5H5)+ + CH3CN

ln k k, min-1 T (K) 1/T x 10-3

-3.20 0.0409 298 3.35

-2.50 0.0818 308 3.25

-1.85 0.157 318 3.14

slope = -6373 = -Ea/REa = (-6373)(-8.31 x 10-3 kJ/K mol) = 53.0 kJ

y intercept = 18.19 = ln A A = 8.0 x 10 7

-3.20 -2.50 -1.85

ln k

k = 0.0985 min-1

3.14 3.25 3.35 x 10-3

1/T

Problem:

The energy of activation for

C4H8(g) => 2C2H4(g)

is 260 kJ/mol at 800 K and k = 0.0315 sec

Find k at 850 K.

ln k2 = - Ea (1/T2 - 1/T1) k1 R

k at 850 K = 0.314 sec-1

Reaction Mechanisms

- Elementary Step
- equation describing a single molecular event
- Molecularity
- unimolecular
- bimolecular
- termolecular
- 2O3 => 3O2

(1) O3 => O2 + O unimolecular(2) O3 + O => 2 O2 bimolecular

Rate Equations

- Molecularity Rate Law

unimolecular rate = k[A] bimolecular rate = k[A][B] bimolecular rate = k[A]2 termolecular rate = k[A]2[B]

- notice that molecularity for an elementary step is the same as the order

2O3 => 3O2

O3 => O2 + O rate = k[O3]

O3 + O => 2O2 rate = k’[O3][O]

2O3 + O => 3O2 + O

O is an intermediate

Problem:

- Write the rate equation and give the molecularity of the following elementary steps:

NO(g) + NO3(g) => 2NO2(g)

rate = k[NO][NO3] bimolecular

(CH3)3CBr(aq) => (CH3)3C+(aq)+ Br-(aq)

rate = k[(CH3)3CBr] unimolecular

Mechanisms and Rate Equations

rate determining step is the slow step --the overall rate is limited by the ratedetermining step

step 1 NO2 + F2 => FNO2 + F rate = k1[NO2][F2] k1 slow

step 2 NO2 + F => FNO2 rate = k2[NO2][F] k2 fast

overall 2NO2 + F2 => 2FNO2 rate = k1[NO2][F2]

Problem:

- Given the following reaction and rate law:NO2(g) + CO(g) => CO2(g) + NO(g)rate = k[NO2]2
- Does the reaction occur in a single step?
- Given the two mechanisms, which is most likely:NO2 + NO2 =>NO3 + NO NO2 => NO + ONO3 + CO => NO2 + CO2 CO + O => CO2

k2

Reaction Mechanisms & Equilibria2O3(g) 3O2(g) overall rxn

1: O3(g) O2(g) + O(g) fastequil. rate1 = k1[O3] rate2 = k2[O2][O]

2: O(g) + O3(g) 2O2(g) slow rate3 = k3[O][O3]

rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities

k3

Substitution Method

at equilibrium k1[O3] = k2[O2][O]

rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2]

rate3 = k3k1 [O3]2 or k2 [O2]

overall rate = k’ [O3]2 [O2]

substitute

Cont’d

rate1 = k1 [OCl -][H2O] =

rate 2 = k2 [HOCl][OH -]

[HOCl] = k1[OCl -][H2O] k2[OH -]

rate 3 = k3 [HOCl][I -]

rate 3 = k3k1[OCl -][H2O][I -] k2 [OH -]

overall rate = k’ [OCl -][I -][OH -]

solvent

Catalyst

- Facilitates the progress of a reaction bylowering the overall activation energy
- homogeneous
- heterogeneous

Ea

DHrxn

Energy

Reaction Progress

catalysts are used in an early rxn step but regenerated in a later rxn step

O3(g)<=> O2(g) + O(g)

O(g) + O3(g) => 2O2(g)

Catalyzed Reaction:

Step 1: Cl(g) + O3(g) + O(g) => ClO(g) + O2(g) + O(g)

Step 2: ClO(g) + O2(g) +O(g) => Cl(g) + 2O2(g)

Overall rxn: O3(g) + O(g) => 2O2(g)

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