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**منابع کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی**

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**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. slides

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**A B time D[A] rate = - Dt D[B] rate = Dt 911213.....56 slides**

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Reaction Rates Reaction rate = change in concentration of a reactant or product with time. Three “types” of rates initial rate average rate instantaneous rate slides

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**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 slides

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2NO22NO+O2 slides

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**Concentrations & Rates**

0.3 M HCl 6 M HCl Mg(s) HCl(aq) ---> MgCl2(aq) + H2(g) slides

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**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 slides

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**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 ________ slides

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**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. slides

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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. slides

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**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. slides

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**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. slides

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**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+. slides

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**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 slides

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**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: slides

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**[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 slides

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**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 slides

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**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 slides 13.3

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**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! slides

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**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 slides

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**Collision Theory O3(g) + NO(g) → O2(g) + NO2(g) 10 31 Collisin/Lit.S**

Reactions require (A) geometry (B) Activation energy slides

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**Three possible collision orientations-- a & b produce reactions, **

c does not. slides

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**Collision Theory TEMPERATURE 10 c0 T 100-300% RATE**

25 c0 T 35 c 2% COLLISIN EFFECTIVE COLLISIONS slides

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**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. slides

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**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. slides

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**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 slides

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**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 slides

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**ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A 300400C0 Rate20000 Times **

k = A • e( -Ea/RT ) ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A 300400C0 Rate20000 Times 400500C0 Rate400 Times Ea=60KJ/mol 300310C0 Rate2 Times Ea=260KJ/mol 300310C0 Rate25 Times slides

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**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 slides

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**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. slides

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**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. slides

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**Mechanisms 2NO+F22NOF R=K[NO][F2] 1=NO+F2NOF+F R1=K1[NO][F2]**

Rate limiting step(RDS) R=R1 2=NO+F NOF R2=K2[NO][F] slides

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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-] slides

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**SN2 OH- +CH3Br CH3OH + Br- R=K[CH3Br][OH-] 911213.....56 slides**

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**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 slides

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**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 slides

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**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. slides

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**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) 2ClOCl2(g)+O2(g) Ea=140KJ/mol Cl2 Catal. slides

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**2)Heterogeneous Catalysis**

C2H4 + H2 C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption slides

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**Enzyme Catalysis 13.6 911213.....56 slides**

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**2N2 + 3O2 Catalytic Converters CO2 + H2O catalytic converter catalytic**

CO + Unburned Hydrocarbons + O2 catalytic converter 2N2 + 3O2 2NO + 2NO2 slides 13.6

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Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois.

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