The “Speed” of the Reaction Or Reaction Rates Chemical Kinetics The “Speed” of the Reaction Or Reaction Rates

Graphical Analysis of Rates Reaction Kinetics Reactants Products Rate of Change Initial rates Method Reaction Order Average Instantaneous Integrated Rate Forms of Rate Law Rate Law Graphical Analysis of Rates Half-Life

Reactants  What happens to the reactants in a reaction? Map What happens to the reactants in a reaction? If we measure the concentration of reactants as a reaction proceeds, what would the graph look like? Or How does the concentration vary with time? Is it linear? Exponential? Random? Do all reactions only move forward? Assume for now there is no reverse reaction Or the reverse reaction proceeds so slowly, we are willing to ignore it graph

 Products Where do the products come from? Map  Products Where do the products come from? If we measure the concentration of the products from the first second the reaction starts, how would the concentration vary with time? graph

Rate of Change [A] = the molarity of A In the reaction A  B Map Rate of Change In the reaction A  B How does the [A] vary with time? Or What is the rate of the reaction? Rate = [A]time2 – [A]time1 Time 2 – Time 1 [A] = the molarity of A

Rate of Change The symbol delta, , means “the change in” So the reaction rate can be written Rate = [A]t2 – [A]t1 or time 2 – time 1 Rate = [A] t graph

Average Rate of Change The concentration varies as time goes on. Because we use up reactants We can calculate an average rate of product consumption over a period of time. Rate is like velocity of a reaction. the rate of change of meters vs rate of change of [A] Instead of meters per second, it is concentration per second. To calculate speed/velocity, divide the distance traveled by the time it took. Meters = m2 – m1 =  m t2 – t1 t sec graph

Average Rate of Change What is the change in concentration? (how far did it go? If you are calculating speed) [A] @ time2 – [A] @time1 = [A] How much time has elapsed? [A] @ time2 – [A] @time1 = [A] t2 – t1 t graph

Instantaneous Rate An average rate describes what reaction rate over a time, but does not tell us the rate at any particular moment. The rate at any moment is the instantaneous rate

Map Instantaneous Rate If we take the average rate over a period of time and continuously make the time period smaller When the time period is infinitesimally small, you approach the instantaneous rate Graphically, it is the slope of the tangent line at the instant. That’s why graphing programs have that tangent line function! Rates are important in bio, physics and chem.

(differential) Rate Law Map (differential) Rate Law Expresses how rate depends on concentration. Rate = - [Reactant] = k [reactant]n t k is the rate constant The bigger the k value, the faster the reaction The smaller the k value, ….? n = the order of the reaction and must be determined experimentally.

Reaction Rates Reaction rates are considered positive Rateinstantaneous = kinstantaneous = - slope of tangent line Rateaverage = k average = - ([A]2 - [A]1) t2-t1 So the rate constant, k, is always negative Rate = - [Product] = k [Product]n t Assuming no reverse reaction!

Back to: Reactant/product Average Rate Reaction Rate Instantaneous Rate Product Formation Average Rate of Change From 0 to 300 s = 0.01 –0.0038 = 0.000021 M/s 300 s

2NO2  2NO + O2 SPECTROSCOPY Let’s consider the above reaction How can we measure the rate? What data do we need? Measure the time Measure the concentration We will take advantage of color in our lab If we are measuring light, we are doing….. SPECTROSCOPY

Spectrometer One frequency Of light Many frequencies of light Source Monochrometer, LED Or Filter Sample Detector The sample absorbs the light. The detector determines how much.

Beer’s Law Beer’s law states that the amount of light absorbed depends on: The material molar absorbtivity (physical property) How much is there? molarity And how big the sample holder The light spends more “time” in contact with a longer sample

Spectrometer One frequency Of light Many frequencies of light Source Monochrometer, LED Or Filter Sample Detector The sample absorbs the light. The detector determines how much.

Spectroscopy Assume the concentration is directly proportional to absorbance of light The more stuff there is that absorbs the light the less light that goes through …. or More light is absorbed Beer’s Law a = e l c = k M a = absorbtivity e = molar absorbtivity (physical property) l = length of light path c = molarity or the solution Molarity and absorbtivity Are directly proportional

in the first 50 secs and the last 50 secs 2NO2  2NO + O2 Compare the [NO2] in the first 50 secs and the last 50 secs Time [NO2] [NO] [O2] 0.0100 50 0.0079 0.021 0.0011 100 0.0065 0.0035 0.0018 150 0.0055 0.0045 0.0023 200 0.0048 0.0052 0.0026 250 0.0043 0.0057 0.0029 300 0.0038 0.0062 0.0031 350 0.0034 0.0066 0.0033 400 0.0069 Why does the rate slow down?

Formation of Products 2NO2  2NO + O2 For every two NO2 consumed Rate of Consumption NO2 = Rate Formation NO Rate = k[NO2 ] = - k[NO] Because For every two NO2 consumed two NO formed

Formation of Products 2NO2  2NO + O2 For every two NO2 consumed Rate of Consumption NO2 = 2 x Rate Formation O2 Rate = k[NO2 ] = - k/2 [O2] Because For every two NO2 consumed one O2 formed

Compare the Instantaneous Rates At any moment in time [NO2] = - [NO] = 2 - [O2] t t t Or k [NO2] = - k [NO] = - k/2 [O2] graph

Form of the Rate Law For aA + bB  cC +dD Rate = k [A]n [B]m Where k is the rate constant n = order of reactant A m = order of reactant B n and m must be determined experimentally n +m = order of the reaction

Experimental Order the order in the integrated rate law Rate = - [Reactant] = k [Reactant]n t n = 0, zero order n = 1, first order n = 2, second order Determine order

Order of Reaction A + B → C Rate = k[A]n [B]m (n + m) = order of the reaction = 1 unimolecular =2 bimolecular =3 trimolecular This means how many particles are involved in the rate determining step

Method of Initial Rates A series of experiments are run to determine the order of a reactant. The reaction rate at the beginning of the reaction and the concentration are measured These are evaluated to determine the order of each reactant and the overall reaction order

If you plot the concentration versus time of [N2O5], you can determine the rate at 0.90M and 0.45M. What is the rate law for this reaction? Rate = k [N2O5]n n = the order. It is determined experimentally.

2N2O5(soln)  4NO2(soln) + O2(g) At 45C, O2 bubbles out of solution, so only the forward reaction occurs. Data [N2O5] Rate ( mol/l • s) 0.90M 5.4 x 10-4 0 45M 2.7 x 10-4 The concentration is halved, so the rate is halved

2N2O5(soln)  4NO2(soln) + O2(g) Rate = k [N2O5]n 5.4 x 10-4 = k [0.90]n 2.7 x 10-4 = k [0 45]n after algebra 2 = (2)n n = 1 which is determined by the experiment Rate = k [N2O5]1

Method of Initial Rates Map Method of Initial Rates Measure the rate of reaction as close to t = 0 as you can get. This is the initial rate. Vary the concentration Compare the initial rates.

NH4+ + NO2-  N2 + 2H2O Rate = k[NH4+1]n [NO2-1]m How can we determine n and m? (order) Run a series of reactions under identical conditions. Varying only the concentration of one reactant. Compare the results and determine the order of each reactant Order

NH4+ + NO2-  N2 + 2H2O Experiment [NH4]+ Initial [NO2]- Initial Rate Mol/L ·s 1 0.001M 0.0050 M 1.35 x 10-7 2 0.010 M 2.70 x 10-7 3 0.002M 0.010M 5.40 x 10-7

NH4+ + NO2-  N2 + 2H2O Compare one reaction to the next 1.35 x 10-7 = k(.001)n(0.050)m 2.70 x 10-7 = k (0.001)n(0.010)m Exp [NH4]+ Initial [NO2]- Initial Rate Mol/L ·s 1 0.001M 0.0050 M 1.35 x 10-7 2 0.010 M 2.70 x 10-7 3 0.002M 0.010M 5.40 x 10-7 Form

1.35 x 10-7 = k(0.001)n(0.0050)m 2.70 x 10-7 k (0.001)n(0.010)m 1.35 x 10-7 = (0.0050)m 2.70 x 10-7 (0.010)m 1/2 = (1/2)m m = 1 In order to find n, we can do the same type of math with the second set of reactions

NH4+ + NO2-  N2 + 2H2O Compare one reaction to the next 2.70 x 10-7 = k (0.001)n(0.010)m 5.40 x 10-7 = k(.002)n(0.010)m Exp [NH4]+ Initial [NO2]- Initial Rate Mol/L ·s 1 0.001M 0.0050 M 1.35 x 10-7 2 0.010 M 2.70 x 10-7 3 0.002M 0.010M 5.40 x 10-7

2.70 x 10-7 = k (0.001)n(0.010)m 5.40 x 10-7 k(.002)n(0.010)m 0.5 = (0.5)n n = 1 n + m = order of the reaction 1 + 1 = 2 or second order Form

Review Method of initial rate In the form Rate = k[A]n[B]m Where k is the rate constant n, m = the order of the reactant The order is determined experimentally Rate law is important so we can gain an insight into the individual steps of the reaction

The Integrated Rate Law Map The Integrated Rate Law Expresses how concentrations depend on time Depends on the order of the reaction Remember Rate = k[A]n[B]m Order = n + m Integrated Rate law takes the form by “integrating” the rate function. (calculus used to determine) The value of n and m change the order of the reaction The form of the integrated rate depends on the value of n You get a different equation for zero, first and second order equations.

Reaction Order Order of the reaction determines or affects our calculations. Zero order indicates the use of a catalyst or enzyme. The surface area of catalyst is the rate determining factor. First or second order is more typical (of college problems)

Integrated Law - Zero Order Rate = - [A] = k t Set up the differential equation d[A] = -kt Integral of 1 with respect to A is [A]

Integrated Rate Law – First Order Map Integrated Rate Law – First Order Rate = [A] = k [A] n t If n = 1, this is a first order reaction. If we “integrate” this equation we get a new form. Ln[A] = -kt + ln[A0] where A0 is the initial concentration

Why? If Rate = - [A] = k [A] 1 t Then you set up the differential equation: d[A] = -kdt [A] Integral of 1/[A] with respect to [A] is the ln[A].

Integrated Rate Law ln[A] = -kt + ln[A]0 The equation shows the [A] depends on time If you know k and A0, you can calculate the concentration at any time. Is in the form y = mx +b Y = ln[A] m = -k b = ln[A]0 Can be rewritten ln( [A]0/[A] ) = kt This equation is only good for first order reactions!

First Order Reaction [N2O5] Time (s) 0.1000 0 0.0707 50 0.0500 100 0.1000 0 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 Ln[N2O5] Time (s)

Zero First Second Rate Law Integrated Rate Law Line Slope = Half-life Rate = K[A]0 Rate = K[A]1 Rate = K[A]2 Integrated Rate Law [A] = -kt + [A]0 Ln[A] = -kt +ln[A]0 1 = kt + 1 [A] [A]0 Line [A] vs t ln[A] vs t 1 vs t [A] Slope = - k k Half-life t1/2 = [A]0 2k t1/2 = 0.693 T1/2 = 1 k[A]0 Graph Data Map

Given the Reaction 2C4H6  C8H12 And the data [C4H6] mol/L Time (± 1 s) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200

Equations 2C4H6  C8H12

Graphical Analysis Map Data ___1___ [C4H6] Ln [C4H6]

Experimental Derivation of Reaction Order Arrange data in the form 1/[A] or ln [A] or [A] Plot the data vs time Choose the straight line y = mx + b Determine the k value from the slope Graphical rate laws 1/[A] = kt + b → 2nd ln[A] = kt + b → 1st [A] = kt + b → zero

Half-life The time it takes 1/2 of the reactant to be consumed This can be determined Graphically Calculate from the integrated rate law

Half-Life Graphical Determination

Half-Life Algebraic Determination Map Zero First Second Half-life t1/2 = [A]0 2k t1/2 = 0.693 k T1/2 = 1 k[A]0 Equations are derived from the Integrated Rate Laws.

As temperature increases, the rate increases. Temperature and Rate Most reactions speed up as temperature increases. (EX. food spoils faster when not refrigerated.) As temperature increases, the rate increases. Typically the rate doubles for every 10ºC increase in temperature.

Why? Temperature and Rate Since the rate law has no temperature term in it, the rate constant must depend on temperature. Why? The temperature effect is quite dramatic.

Temperature and Rate The Collision Model Observations: rates of reactions are affected by concentration and temperature. In order for a reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

Temperature and Rate The Collision Model The more molecules present, the greater the probability of effective collisions and the faster the rate. The higher the temperature, the more energy available to the molecules and the faster the rate.

Temperature and Rate The Collision Model Complication: Not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

Temperature and Rate Activation Energy

Temperature and Rate Activation Energy Arrhenius: molecules must possess a minimum amount of energy to react. In order to form products, bonds must be broken in the reactants. Bond breakage requires energy. Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

Temperature and Rate Activation Energy

Temperature and Rate Activation Energy Consider the rearrangement of acetonitrile: In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state. The energy required for the twisting and breaking of the reactant bonds is the activation energy, Ea. Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.

Temperature and Rate Activation Energy

Temperature and Rate Key Points for Activation Energy The change in energy, ∆E, for the reaction is the difference in energy between reactants (CH3NC) and products (CH3CN). The activation energy is the difference in energy between the reactants and the transition state. The reaction rate depends on Ea. The lower the Ea, the faster the reaction. Notice that if a forward reaction is exothermic (CH3NC  CH3CN), then the reverse reaction is endothermic (CH3CN  CH3NC).

Temperature and Rate Activation Energy Re-consider the reaction between Cl and NOCl: If the Cl effectively collides with the Cl end of NOCl, then the products are Cl2 and NO. If the Cl collided with the O of NOCl, then no products are formed. We need to quantify this effect.

Temperature and Rate As the temperature increases, more molecules have sufficient energy to have effective collisions and react. Lower temp Minimum energy needed for reaction, Ea Higher temp Fraction of molecules Kinetic energy

Temperature and Rate The Arrhenius Equation Arrhenius discovered most reaction rate data obeyed the equation: k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. A is called the frequency factor. It is a measure of the probability of a favorable collision. Both A and Ea are specific to a given reaction.

Temperature and Rate The Arrhenius Equation If you have a lot of data, you can determine Ea and A graphically by rearranging the Arrhenius equation: If you have only two data points, then you can use

Catalysis A catalyst changes the rate of a chemical reaction. There are two types of catalysts: homogeneous and heterogeneous. Homogeneous Catalysis The catalyst and reaction are in one phase. Hydrogen peroxide decomposes very slowly: 2H2O2(aq)  2H2O(l) + O2(g).

Catalysis Homogeneous Catalysis 2H2O2(aq)  2H2O(l) + O2(g). In the presence of the bromide ion, the decomposition occurs rapidly: 2Br-(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l). Br2(aq) + H2O2(aq)  2Br-(aq) + 2H+(aq) + O2(g). Br- is a catalyst because it can be recovered at the end of the reaction. Generally, catalysts operate by lowering the activation energy for a reaction.

Catalysis Homogeneous Catalysis

Catalysis Homogeneous Catalysis Catalysts can operate by increasing the number of effective collisions. From the Arrhenius equation: catalysts increase k by increasing A or decreasing Ea. A catalyst may add intermediates to the reaction. Ex: When Br- is added, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

Catalysis Heterogeneous Catalysis The catalyst is in a different phase than the reactants and products. First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) are very reactive.

Catalysis Heterogeneous Catalysis

C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol. Catalysis Heterogeneous Catalysis Consider the hydrogenation of ethylene: C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol. The reaction is slow in the absence of a catalyst. In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature. First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. The H-H bond breaks and the H atoms migrate about the metal surface.

C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol. Catalysis Heterogeneous Catalysis Consider the hydrogenation of ethylene: C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol. When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms. When C2H6 forms it desorbs from the surface. When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

Catalysis Enzymes Enzymes are biological catalysts. Most enzymes are protein molecules with large molecular masses (10,000 to 106 amu). Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. Substrates undergo reaction at the active site of an enzyme. A substrate locks into an enzyme and a fast reaction occurs.

Catalysis Enzymes

Catalysis Enzymes The products then move away from the enzyme. Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors). The number of events (turnover number) catalyzed is large for enzymes (103 - 107 per second).

Catalysis Nitrogen Fixation and Nitrogenase Nitrogen gas cannot be used in the soil for plants or animals. Nitrogen compounds, NO3, NO2-, and NO3- are used in the soil. The conversion between N2 and NH3 is a process with a high activation energy (the NN triple bond needs to be broken). An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.

Catalysis Nitrogen Fixation and Nitrogenase

Catalysis Nitrogen Fixation and Nitrogenase The fixed nitrogen (NO3, NO2-, and NO3-) is consumed by plants and then eaten by animals. Animal waste and dead plants are attacked by bacteria that break down the fixed nitrogen and produce N2 gas for the atmosphere.