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2. The End!!!!!!!
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4. منابع کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی
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5. Chemical Kinetics Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
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6. A B time D[A] rate = - Dt D[B] rate = Dt 911213.....56 slides

7. Reaction Rates
Reaction rate = change in concentration of a reactant or product with time.
Three “types” of rates
initial rate
average rate
instantaneous rate
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8. slope of tangent slope of tangent slope of tangent
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time
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9. 2NO22NO+O2
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12. Concentrations & Rates
0.3 M HCl
6 M HCl
Mg(s) HCl(aq) ---> MgCl2(aq) + H2(g)
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14. Reaction Order What is the rate expression for aA + bB → cC + dD
Rate = k[A]x[B]y
where x=1 and y=2.5?
Rate = k[A][B]2.5
What is the reaction order?
First in A, 2.5 in B
Overall reaction order?
= 3.5
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15. Interpreting Rate Laws
Rate = k [A]m[B]n[C]p
If m = 1, rxn. is 1st order in A
Rate = k [A]1
If [A] doubles, then rate goes up by factor of __
If m = 2, rxn. is 2nd order in A.
Rate = k [A]2
Doubling [A] increases rate by ________
If m = 0, rxn. is zero order.
Rate = k [A]0
If [A] doubles, rate ________
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16. Deriving Rate Laws Derive rate law and k for
CH3CHO(g) → CH4(g) + CO(g)
from experimental data for rate of
disappearance of CH3CHO
Rate of rxn = k [CH3CHO]2 ????
Therefore, we say this reaction is _____ order.
Here the rate goes up by _____ when initial conc. doubles.
Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec)
Now determine the value of k. Use expt. #3 data— 1,2,3 or 4
R= k [CH3CHO] mol/L•s = k (0.30 mol/L)2
k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CH3CHO] at same T.
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17. Expression of R &K
A series of four experiments was run at different concentrations, and the initial rates of X formation were determined.
From the following data, obtain the reaction orders with respect to A, B, and H+.
Calculate the numerical value of the rate constant.
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18. Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
Comparing Experiment 1 and Experiment 2, you see that when the
A concentration doubles (with other concentrations constant),
the rate doubles.
This implies a first-order dependence with respect to A.
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19. Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
4.60 x 10-6
Exp. 4
Comparing Experiment 1 and Experiment 3, you see that when the
B concentration doubles (with other concentrations constant),
the rate 4 times.
This implies a second-order dependence with respect to B.
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20. Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
4.60 x 10-6
Exp. 4
Comparing Experiment 1 and Experiment 4, you see that when
the H+ concentration doubles (with other concentrations constant),
the rate is the same.
This implies a zero-order dependence with respect to H+.
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21. Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
4.60 x 10-6
Exp. 4
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22. Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
4.60 x 10-6
Exp. 4
You can now calculate the rate constant by substituting values from
any of the experiments. Using Experiment 1 you obtain:
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24. [A] = [A]0 - kt Zero-Order Reactions
D[A] Dt
= k -
A product
rate =
rate = k [A]0 = k
rate
[A]0
k =
=RATE= M/s
[A] = [A]0 - kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ =
[A]0 2k
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25. First-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]
rate
[A]
M/s M =
k =
= 1/s or s-1
D[A] Dt
= k [A] -
t½ = t when [A] = [A]0/2
[A] is the concentration of A at any time t
t½ =
0.693 k
[A]0 is the concentration of A at time t=0
[A] = [A]0exp(-kt)
ln[A] = ln[A]0 - kt
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26. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
rate
[A]2
M/s M2 =
D[A] Dt
= k [A]2 -
k =
= 1/M•s
[A] is the concentration of A at any time t 1
[A] =
[A]0
+ kt
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ = 1
k[A]0
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13.3

27. First step in evaluating rate data is to graphically interpret the order of rxn
Zeroth order: rate does not change with lower concentration
First, second orders:
Rate changes as a function of concentration
Graphs of different rates of reaction copy the chapter from Langmuir for them!
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28. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2
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29. Collision Theory O3(g) + NO(g) → O2(g) + NO2(g) 10 31 Collisin/Lit.S
Reactions require
(A) geometry
(B) Activation energy
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30. Three possible collision orientations-- a & b produce reactions,
c does not.
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31. Collision Theory TEMPERATURE 10 c0 T  100-300% RATE
25 c0 T  35 c  2%  COLLISIN
EFFECTIVE COLLISIONS
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33. ① with sufficient energy
The colliding molecules must have a total kinetic energy equal to or
greater than the activation energy, Ea.
Ea is the minimum energy of collision required for two molecules to initiate a chemical reaction.
It can be thought of as the hill in below. Regardless of whether the
elevation of the ground on the other side is lower than the original position,
there must be enough energy imparted to the golf ball to get it over
the hill.
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35. The Collision Theory of Chemical Kinetics
Activation Energy (Ea)- the minimum amount of energy required to initiate a chemical reaction.
Activated Complex (Transition State)- a temporary species formed by the reactant molecules as a result of the collision before they form the product.
Exothermic Reaction
Endothermic Reaction
Exothermic. Products are more stable than reactants and after product formation, there is release of heat.
endothermic.
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37. Temperature Dependence of the Rate Constant
(Arrhenius equation)
k = A • exp( -Ea/RT )
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
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38. Arrhenius Equation k = A • exp( -Ea/RT )
Can be arranged in the form of a straight line
ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A
Plot ln k vs. 1/T 
slope = -Ea/R
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39. ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A 300400C0 Rate20000 Times
k = A • e( -Ea/RT )
ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A
300400C0 Rate20000 Times
400500C0 Rate400 Times
Ea=60KJ/mol
300310C0  Rate2 Times
Ea=260KJ/mol
300310C0  Rate25 Times
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40. K in different T Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
Ln(K2/K1)=-Ea/R(1/T2-1/T1)
Log(K2/K1)=-Ea/2.303R(1/T2-1/T1)
Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
Log(K2/K1) =ΔH/2.303R(T2-T1)/T1T2
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42. Most rxns. involve a sequence of elementary steps.
Mechanisms
Rate = k [H2O2] [I-]
Most rxns. involve a sequence of elementary steps.
2 I- + H2O H+ ---> I H2O
Proposed Mechanism
Step 1 — slow HOOH + I- --> HOI + OH-
Step 2 — fast HOI + I- --> I2 + OH-
Step 3 — fast 2 OH H+ --> 2 H2O
2 I- + H2O H+ ---> I H2O
Rate = k [H2O2] [I-]
Rate can be no faster than slow step
RATE DETERMINING STEP,( rds.)
The species HOI and OH- are intermediates.
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43. Mechanisms NOTE Rate law comes from experiment
Order and stoichiometric coefficients not necessarily the same!
3. Rate law reflects all chemistry including the slowest step in multistep reaction.
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44. Mechanisms 2NO+F22NOF R=K[NO][F2] 1=NO+F2NOF+F R1=K1[NO][F2]
Rate limiting step(RDS) R=R1
2=NO+F NOF R2=K2[NO][F]
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45. SN1
OH- + (CH3)3CBr(CH3)3COH + Br- R=K[(CH3)3CBr] 1)(CH3)3CBr(CH3)3C+ +Br- R1=K1[(CH3)3CBr ] 2) (CH3)3C+ +OH- (CH3)3COH R2=K2[(CH3)3C+] +[OH-]
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46. SN2 OH- +CH3Br  CH3OH + Br- R=K[CH3Br][OH-] 911213.....56 slides

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48. Rate Laws andMechanisms
Proposed
NO2 + CO →NO(g) + CO2(g) Rate = k[NO2]2
Experimental
Two possible mechanisms
Two steps
is rational mechanism
Single step
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50. ratecatalyzed > rateuncatalyzed Ea k
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
k = A • exp( -Ea/RT )
ratecatalyzed > rateuncatalyzed Ea k
uncatalyzed
catalyzed
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51. Br - For the decomposition of hydrogen peroxide:
A catalyst(Br-) speeds up a reaction by lowering the activation energy.
It does this by providing a different mechanism by which the reaction
can occur. The energy profiles for both the catalyzed and uncatalyzed
reactions of decomposition of hydrogen peroxide are shown in Figure below.
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52. 1)Homogeneous Catalysis
N2O (g)  N2 (g) + O2 (g)
1) N2O (g)  N2 (g) + O (g)
O (g) + N2O (g)  N2(g)+ O2 (g)
Ea=240KJ/mol
Cl2(g)2Cl(g)
N2O(g)+Cl(g)N2(g)+ClO(g)
2ClOCl2(g)+O2(g)
Ea=140KJ/mol
Cl2 Catal.
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53. 2)Heterogeneous Catalysis
C2H4 + H2  C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption
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54. Enzyme Catalysis 13.6 911213.....56 slides
13.6

55. 2N2 + 3O2 Catalytic Converters CO2 + H2O catalytic converter catalytic
CO + Unburned Hydrocarbons + O2
catalytic
converter
2N2 + 3O2
2NO + 2NO2
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56. The End!!!!!!!
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