16.1 Expressing the Reaction Rate Differential Rate and ...

16.1 Expressing the Reaction Rate Differential Rate and ...

13.1 Expressing the Reaction Rate Differential Rate and Integrated Rate Laws Fred Omega Garces Chemistry 201 Miramar College 1 Expressing Reaction Rates 2/9/20 Outline Kinetics: Intro Factors influencing Reaction Rates Rates & Stoic Relations Expressing reaction Rates Rate Law and components Differential Rate Law Initial Rate Method Rate Laws and Half-Lives 0th Order 1st Order 2nd Order Kinetic Molecular Theory PreExponential Factor Collision Frequency Steric Factor Activation Energy Arrhenius Equation Reaction Coordinate Diagram Catalyst factor Maxwell-Boltzmann Diagram

Reaction Mechanism & Chapman cycle Elementary Step Rate Determining Step 2 Expressing Reaction Rates 2/9/20 Thermodynamics Vs. Kinetics ThermoChem: Will a reaction occur ? Kinetics: If so, how fast ? Iron rusting; G -, but rate of rusting depends on reaction conditions. 4 Fe (s) + 3 O2(g) + xH2O(g) D 2Fe2O3XH2O (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 (G -). C (diamond) D C (graphite) G = -2.84 kJ/mol Kinetics makes this reaction nearly impossible. 3 Expressing Reaction Rates

2/9/20 Factors Influencing Reaction Rates Factors Affecting Rates Concentration of Reactant - Molecules must collide in order to react. More reactants means the faster rxn. Note: rate 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 collision frequency Temperature- Molecules must collide with enough energy to react. Note: rate collision energy (energy is temperature dependent) Catalyst- The pathway in which the reactant convert to product 4 influence the reactivity. Expressing Reaction Rates 2/9/20 Rates ; Another word for Speed Rate - Change / time (A positive quantity) Speed - distance / time = x / time Wages - money / time = $ / time Reaction - conc. / time = [A] / A] / time Differential Rate Law; An expression that

relates the rate to the concentration, [A] / A] with respect to time. General Form: [A] / A] t 5 Expressing Reaction Rates 2/9/20 Instantaneous Vs. Average rates (Integrated Form) Consider: Reactant Product Experiment will monitor concentration of reactant (product) as a function of time. rate = - [ c] = - 0.40-0.67 = + 0.27 rate (Avg.) t 40 - 10 = - 0.0090 M s-1 30 rate = - [ c] = - 0.45-0.60 = + 0.15 rate (Avg.)

t 30 - 15 = - 0.0010 M s-1 rate = - dc = 15 Tangent at dt 6 Expressing Reaction Rates 2/9/20 Stoichiometry Relationship: Example H2 (g) + I2 Rate 2 HI = - [H2] = - [I2] Rxn - (g) t [H2] = + [HI] t

t (g) ? + [HI] t t ? - 1 mol H2 2 mol HI Rate of reaction does not depend on which specie is monitored during the reaction. - [H2] = + [HI] t 2 t Disappearance Rate Rate Rxn Rxn 7 == Appearance - -[H [H22]] == - -[I[I22]] ==++[HI] [HI]

tt tt Expressing Reaction Rates 22t t 2/9/20 Stoichiometry Rate Relationship aA + bB Rate Rxn = - [A] = - [B ] a t cC+ dD = + [C ] = + [D ] b t c t d t 10 mol/min 2H2 + O2 2 H2O Rate Rxn = - [H2] = - [O2 ]

2mol/min t -5 t -5 mol/min = + [H2O ] 2t 5 mol/min 2O3 3 O2 Rate Rxn 8 = - [O3] = + [O2 ] 2t Expressing Reaction Rates 3 t 2/9/20 Methods of Initial Rates Methods of initial rates is used to determine the rxn rates: 2 NO2(g) 2 NO(g) + O2(g) Rate Law c * Rate Reaction = d[A] / NO2] / Rate dt Rxn

= - [NO2 ] = - [NO2] t=150 - [NO2] t=50 t 150s - 50 s Rate = - [NO2 ] = - 0.0052 - 0.0079 M Rxn Rate t 100 s = - [NO2 ] = - (-2.710-5) M t s Rxn Rate Rxn = - [NO ] = + 2.710-5 M t s = Rate Reaction s The rate of a reaction does not depend on which specie is measured.

* Integrated Rate Law; how the concentrations depend on time. 9 Expressing Reaction Rates 2/9/20 Reaction Rate Relationship Rates of a reaction in terms of any specie in the reaction: aA +bB g pP + rR Coefficient- 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. 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 10 Expressing Reaction Rates Reaction Progress In our previous example: 2 NO2(g) . 2 NO(g) + O2(g) Reactant NO2 , and product NO, have t=1 t=2 t=3 2/9/20

In Class Exercise 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. Acrylonitrile is produced from propene, ammonia and oxygen by the following reaction: 2C3H6 + 2NH3 + 3O2 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. 11 Expressing Reaction Rates 2/9/20 Summary of reaction rate

The 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) 12 Expressing Reaction Rates 2/9/20 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] / A]x [A] / B]y [A] / 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 determined K - rate constant * Zumdahl 13

Expressing Reaction Rates 2/9/20 Methods of Initial Rates In 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. 14 Expressing Reaction Rates 2/9/20 Initial Rate Methodology Determining 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. 15 Expressing Reaction Rates 2/9/20 Rate Law Determination 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 Experiment (mol/ L) [A] / NO] [A] / O2] (mol/Ls) 1 2 3 4 5 0.010 0.020 0.040 0.020 0.020 0.020 0.020 0.020 0.040 0.010 Initial Rate

0.028 0.057 0.114 0.227 0.014 Based on these data, what is the rate law? What is the value of the rate constant, K? 16 Expressing Reaction Rates 2/9/20 GuideLines for Solving Rate Law 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. 17 Expressing Reaction Rates 2/9/20 A Quick Math Review Exponential Operations: Solving for exponents: 20 = 1 X lnA = lnB 21 = 2 X = lnB / lnA Ax = B 22 = 4 i.e., 26 = 64 23 = 8 24 = 16 25 = 32 Try 2x = 64,

x = ln(64) / ln(2) = 2.77/.693 26 = 64 18 solve for x = 6 Expressing Reaction Rates 2/9/20 Rate Law Determination; Ratio of Experiments Now consider our pervious problem: Initial Concentrations (mol/ L) Initial Rate Experiment [NO] [O2] (mol/L s) 1 2 3 4 5 0.020 0.020 0.020 0.040 0.010 0.010

0.020 0.040 0.020 0.020 0.028 0.057 0.114 0.227 0.014 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 Rate1 k [NO]xExpressing [O2]y Reaction Rates0.028 = 0.057 = k [0.020]x [0.020]y 19 Expt 1

k [0.020]x [0.010]y 2/9/20 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 Expt 1 Rate1 k [NO]x[O2]y 0.057 = 0.028 k [0.020]x [0.020]y 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 [0.020]y Expt 5

Rate 2 k [NO]x [O2]y Rate 5 k [NO]x[O2]y = 0.057 = 0.014 k [0.020]x k [0.010]x [0.020]y 20 x Expressing Reaction Rates 2/9/20 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 / (410-4) (0.010) = 7.1103 M-2 s-1

Rate Law: Rate = 7.1103 M-2s-1 [A] / NO]2 [A] / O2] 21 Expressing Reaction Rates 2/9/20 In Class Exercise Acrylonitrile is produced from propene, ammonia and oxygen by the following reaction at 25C: 2C3H6 + 2NH3 + 3O2 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 22 Expressing Reaction Rates 2/9/20 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, and m+n is the overall reaction order. The values of m and n must be determined by experiments; they cant 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 23 Expressing Reaction Rates 2/9/20

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