Chemical Reaction Rates

Kinetics • • • • • • • Chemical Reaction Rates Factors Affecting Reaction Rates Rate Laws Integrated Rate Laws Collision Theory Reaction Mechanisms Catalysts What is Kinetics? • Kinetics studies the factors that can affect reaction rates and investigates the mechanism (i.e. the pathway) a chemical reaction proceeds. • The speed of a chemical reaction is referred to as its reaction rate. • The rate measures how fast the reaction uses up reactants or makes new products. • Rate = - (D[reactant]/D time) • It is about a change (D) in quantity over in a given period of time. • Example: • Driving a car: • The speed you drive your car is a rate. It describes the distance your car travels (miles) for a given period of time (1 hour). • The rate your car travels has units of mi/hr. • Chemical reaction: • It measures how much the concentration (molarity) of a reactant decreases in a given period of time. • For reactants, a negative sign is placed in front of the definition. Rate Types • Instantaneous Rate is the change in concentration at any particular time. • On a graph it is the slope at one point of a curve. • First derivative of the function. • Average Rate is the change in concentrations at any particular time period. • It is the linear approximation of a curve. • The larger the time interval, the more the average rate deviates from the instantaneous rate. Reference: www.shodor.org Factors Affecting Reaction Rates • Chemical nature of the reacting substances: • Particle size: • Molecules tend to react faster than large molecules. • Kinetic energy (½ mv2) • Ions react faster than molecules. • No bonds need to be broken. • Phase: • Gases tend to react faster than liquids, which react faster than solids. • Powdered solids are more reactive than solid clumps. • Greater surface area. • Chemical reactivity(e.g., the activity series of metals). Factors Affecting Reaction Rates • Concentration: • More reactant particles present (e.g. concentrated) the faster the reaction. • Increase chance/frequency of particle contact. • Concentration of a solutions depends on the solute to solution ratio (molarity). • Increase pressure, specifically for reactants in the gaseous state: • The concentration of reactants in the gaseous state depend on the partial pressure of the gas. • Higher pressure → higher concentration. • Temperature: • In most cases, an increase in temperature will increase the chemical reaction rate. • In general, for each 10.0 °C rise in temperature, the speed of the reaction doubles. • The mathematical relationship between temperature and the rate of a reaction is the Arrhenius equation. −Ea/RT k = A𝑒 Factors Affecting Reaction Rates • Catalysts: • Substances that affect the rate of a reaction without being consumed. • A catalyst affect a reaction’s rate by lowering the energy (Ea) of activation. • Most catalysts (e.g. positive) speed up a reaction, however there are catalysts that will slow down (e.g. negative) a reaction rate. • Types of Catalysts: • Homogeneous: the catalyst and reactants are present in same phase. • Example: Biological systems – cellular reactions. • Heterogeneous: the catalyst and reactants are present in different phase. • Example: Automobile catalytic converter. Reference: openstax Homogeneous: catalyst is in the same phase (state) as reactants. (a) According to the lock-and-key model, the shape of an enzyme’s active site is a perfect fit for the substrate. Heterogeneous: catalyst is in a different phase (state) than the reactants. (a) (b) (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate. (c) (d) Reference: openstax Catalysis of the reaction C2H4 + H2 ⟶ C2H6 by nickel. Hydrogen is adsorbed on the surface, breaking the H–H bonds & forming Ni–H bonds. Ethylene is adsorbed on the surface, breaking the π-bond & forming Ni–C bonds. Atoms diffuse across the surface and form new C–H bonds when they collide. C2H6 molecules escape from the nickel surface, since they are not strongly attracted to nickel. Rate Laws • Mathematical expressions describing the relationship between reaction rate and reactant concentration. • Rate of reaction (M/time) = k [A]n • Where: k is the rate constant & n is the reaction order for reactant A. • The rate constant (k) is independent of concentration with its numerical value increasing with temperature. • A reaction’s rate constant (k) and orders (n) must be determined EXPERIMENTALLY. What is a rate order? • Rate orders (i.e. the exponents in a rate law) describe the effects reactant concentrations have on the rate of reaction. Rate = k [A]n[B]o[C]p • The exponents n, o & p represents the reaction order. • Values can be positive or negative whole number integers (e.g. 1-,0, 1, 2….) or fractions. • The order values are NOT the stoichiometric coefficients from the chemical reaction. • The overall order of reaction is the sum of the individual reaction orders. • Example: If the rate law for a reaction is Rate = k [A]3[B]3 then the overall order of for the reaction is 6th order. Rate Orders: Zero • A reactant with a ZERO rate order has a reaction rate that is INDEPENDENT of that reactant’s concentration. • Over time the reactant’s concentration decreases linearly over time. • Rate = k [A]0 or Rate = k • Graphic representation: A reactant having a zero order – its data when plotted will yield a straight line graph: [A] vs time. Reference: www.purdue.chem.edu Rate Orders: First • A reactant with a FIRST rate order has a reaction rate that is DEPENDENT/DIRECTLY on the reactant’s concentration. • Rate = k [A]1 • Graphic representation: A reactant having a first order – its data when plotted will yield a straight line graph: ln[A] vs time. Reference: www.purdue.chem.edu Rate Orders: Second • A reactant with a SECOND order rate has a reaction rate that is PROPORTIAL to the reactant’s concentration. • Rate = k [A]2 • Graphic representation: A reactant having a second order – its data when plotted will yield a straight line graph: (1/[A] )vs time. • Over to time, the concentration of the reactant decreases linearly. Reference: www.purdue.chem.edu The Rate Constant, k and Its Units • The rate constant (k) units are determined by the overall rate order for the reaction as described by its rate law. • Example: • Rate law: Rate = k [A]1[B]2 • The overall rate order for the above is 3 • Reaction rate units are concentration per change in time. • Example: • Rate (Molarity/hour) • Rate (milligram/second). • The units for the rate constant, k, vary according to the overall rate order of the reaction. • Example: Rate law: Rate = k [A]1[B]2 • Rate (mol/L s) = k (mol/L)3 • ‘k’ units have to be such that both sides of the rate law will have the same units. • The units of k for this rate law are (L2/mol2 s) or M-2 s-1 • (mol/L s) = (L2/mol2 s) (mol3/L3) rate = k [M]3 Problem: An experiment was conducted, and it was found that reactant A was third order, reactant B was zero order and reactant C has a rate order of -1. The unit of time was in seconds and the concentration of the reactants was molarity. 1. Write a rate law for this reaction. 2. Determine the overall rate order for this reaction. 3. Determine the units for the rate constant, k. Rate orders: Experimentally Determined • Rate orders and the overall rate order of a reaction can ONLY be determined through experimentation (i.e. data collected from an experiment. • They CANNOT be determined stoichiometrically. • To determine a reactant’s rate order from experimental data: • Inspection • Algebraically • Graphically Rate Orders by Inspection • Using two sets of data, comparing a reactant concentration to its corresponding reaction rates when a reactant’s concentration is doubled while the other reactant concentrations and rates are held constant. • Doubled reactant’s concentration and: • No change in the reaction’s rate is observed, then the order is said to be ZERO order. • A doubling of the reaction’s rate occurs, then the order is said to be 1st order. • A quadrupling of the reaction’s rate, then the order is said to be 2nd order. Problem: When heated, acetaldehyde gas decompose to produces methane gas and carbon monoxide gas. CH3CHO (g) → CH4 (g) + CO (g) From the data below: 1. Determine the rate law by inspection for the decomposition of acetaldehyde. 2. Determine the units for the rate constant, k. Exp. Run 1 2 3 [CH3CHO] 1.75 x 10-3 3.50 x 10-3 7.00 x 10-3 Rate (mol/L•sec) 2.06 x 10-11 8.24 x 10-11 3.30 x 10-10 Rate Orders: Algebraically • Using two sets of data, comparing a reactant concentration to its corresponding reaction rates while keeping the other reactants’ concentration and rates constant. • Example: Run 1: Run 3: 0.001 M/s = k1 [0.010]x [0.20]y [0.0050]z 0.004 M/s = k3 [0.010]x [0.30]y [0.0050]z Problem: Given the following information, determine the: 1. 2. 3. 4. rate orders for each of the reactants; the rate law for the chemical reaction; the overall rate order for the chemical reaction and the units for the rate constant Chemical reaction: 2 NO (g) + Cl2 (g) → 2 NOCl (g) Run # [NO] (mol/L) [Cl2] (mol/L) Rate (mol/L min) 1 0.10 0.10 0.00300 2 0.10 0.15 0.00450 3 0.15 0.10 0.00675 Integrated Rate Laws • Rate = k [A]x [B]y is a rate law that describes a reaction rate to its reactant concentrations. • Integrated rate laws relate the reactants’ concentration to time based on experiment data. • Using calculus, the differential rate law relates the amount of reactant or product present over a specified period of time. • Integrated rate laws: • Zero order: [A]t = −kt + [A]0 • First order: ln[A] = –kt + ln[A]o or • Second order: 1/[A]t = kt + 1/[A]0 ln {[A]t /[A]0} = - kt Summary of Rate Orders: Zero Zero Order: Independent of Concentration • Rate = k[A]0 = k [A]o • The integrated rate law : [A]t = −kt + [A]0 • The graph of [A] vs. time is straight line with slope = −k & y-intercept = [A]o [A] • The half life relationship is: t½ = [Ao]/2k • Zero-order reaction: the lower the initial concentration of the reactants, the shorter the half-life. Time (t) Summary of Rate Orders: First First Order: Dependent on Reactant Concentration • Rate = k[A]1 = k [A] • Integrated rate law: ln[A] = –kt + ln[A]o or ln {[A]t /[A]0} = - kt ln [A]o ln [A] • The graph of ln [A] vs. time is straight line with slope = −k & y-intercept = ln [A]o • The half life relationship is: • t½ = ln 2/k or • t½ = 0.693/k Time (t) Summary of Rate Orders: Second Second Order: Proportional to Reactant Concentration • Rate = k[A]2 • Integrated rate law: 1/[A]t = kt + 1/[A]0 (1/[A]) • The graph of (1/[A]) vs. time is straight line with slope = k & y-intercept = (1/[A]o ) • The half life relationship is: t½ = 1/(k[A0]) • In a second-order reaction, the half-life is inversely proportional to the initial concentration. • Increasing the initial concentration shortens the half-life. (1/[A]o) Time (t) Integrated Rate Laws & Half-life • A half-life is the amount of time required a substance (e.g. reactant) to decay to half its original amount. • Half-life is often associated with radioactive decay of elements. • Elemental radio active decay is a 1st order reaction. • For a zero-order reaction, the lower the initial concentration of the reactants, the shorter the half-life. • t1/2 = [A]o/2k • For a first-order reaction, the half-life is independent of the concentration. • t1/2 = ln(2)/k or 0.693/k • For a second-order reaction, the half-life is inversely proportional to the initial concentration, meaning increasing the initial concentration shortens the half-life. • t1/2 = 1/(k[A]o) Reference: Chem.libretexts.org Problem #1: Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6.0 hr and 73.0 hr, respectively. 1. Determine the rate constant for each of these radioactive isotopes. 2. What percent of the radioactivity would remain for each of the isotopes after 2 days? Useful information: ln {[A]t /[A]0} = - kt t½ = ln 2/k or t½ = 0.693/k Problem #2: The rate constant for the decomposition of cyclobutane to ethene at 500 °C is 9.2 × 10−3 s−1. C4H8 → 2 C2H4 1. What is the order of this reaction? 2. Determine how long it will take for 80.0% of a sample of C4H8 to decompose. 3. Determine the half-life (t1/2) of the reaction. Useful information: ln {[A]t /[A]0} = - kt t½ = ln 2/k or t½ = 0.693/k Problem #3: Given the following chemical reaction: 2C4H6 (g) ⟶ C8H12 (g) with a second order rate constant of 5.76 × 10−2 M-1 min-1. 1. The original concentration of C4H6 (g) was 0.200 M. Determine the concentration of C4H6 remaining after 20.0 minutes. 2. What is the half-life of this reaction? Useful information: 1/[A]t = kt + 1/[A]0 t½ = 1/(k[A0]) Radioactive Decay is 1st order Kinetics • The rate of decay associated with a radioactivity is s 1st order half-life. • It is the time required for half of an original quantity of an element to decay. • The shorter the half-life the faster the decay. • The decay can be particulate in nature (e.g. beta, alpha, proton, neutron, positron, elemental nuclei) or energy (e.g. gamma, x-rays). • Each isotope has its own radioactive decay half-life, which is constant and independent of any physical or chemical change the atom may undergo. • It can be calculated at any given moment by measuring the rate of decay of a known quantity using a radiation detector. t½ = 0.693/k ln (A)t = -kt + ln (A)to or ln (At/Ao) = -kt • The rate of radioactive change is not affected by temperature. • Meaning, radioactivity is not a chemical reaction! Problem: Calculate the mass of Pu-236 remaining after 5.00 years if the original mass of the sample was 1.35 mg. Half life is 2.86 years. Useful information: ln {[A]t /[A]0} = - kt t½ = ln 2/k or t½ = 0.693/k Problem: After 37.8 days, 1.05 x 10-2 g of Rn-222 was left. Determine the mass in grams of the original sample. t1/2 for Rn-222 is 3.80 days Useful information: ln {[A]t /[A]0} = - kt t½ = ln 2/k or t½ = 0.693/k Kinetic Collision Theory • For a reaction to take place, the reacting particles (e.g. ions, atoms, molecules) must encounter each other; must collide. • Types of collisions: • Ineffective • Effective: resulting in a temporary, high energy complex. • This high energy complex is referred to as an activated complex or transition state. • Factors that determine if collisions are effective: • Orientation: • A proper orientation results when the particles are aligned in such a way that the old bonds can break, and the new bonds can form. • Energy (Ea activation energy): • Does the collision possess enough energy to break the bonds holding reactant molecules together. • Frequency: • More collision, then the greater the probability of having effective collisions. Reference: openstax Collision Theory and the Rate of Reaction • Effective collisions possess enough energy for the reaction to proceed in a forward (products) direction via the formation of a transition state (i.e. high energy complex). • Factors which influence contact probability which lead to more effective collisions: • Phase (e.g. solid vs. liquid vs. gas). • Concentration (e.g. gas, liquids and aqueous phases) or partial pressure (e.g. gas phase) of reactant particles: • Except for zero-order reactions, the more particles present more likely hood there are more particles having sufficient kinetic energy for effective collisions (e.g. distribution is the same, just more particles having sufficient kinetic energy) and greater probability of contact. • Temperature: • Increase temperature results in an increase in the average kinetic energy of the reactant particles. • Have an increase in the number of particles having sufficient kinetic energy to overcome the reaction’s activation energy (Ea) of the reaction. • A minimum amount of kinetic energy is needed for a collision to be converted into enough potential energy to form the activated complex. Energy of Activation (Ea) & Arrhenius Relationship • Molecules need a minimum amount of energy to react; activation energy, Ea. Graphical of Exothermic Reaction • Changes in temperature result in a change in the reaction’s rate law constant (k); there is a direct correlation. • Arrhenius relationship relates reaction rate to activation −𝑬𝒂 energy. k = Ae 𝑹𝑻 • Frequency or Arrhenius constant (A) • Orientation factor • Collison frequency factor • T is the temperature in Kelvins. • Ea is the activation energy • R is the gas constant (8.314 J/(mol∙K)) Reference: openstax Arrhenius Equation & Graphical Representation Arrhenius Equation Graphical Representation • One set of data points • Two set of data points Reference: elte.prompt.hu The Connection Between an Arrhenius Graph & Energy Reaction Plot • The greater the slope of the line the larger the activation energy (Ea). • The higher the “hill” the larger the activation energy (Ea). Higher Ea Higher Ea Lower Ea Lower Ea Reference: openstax Reference: chem.libretexts.org Problem #1: Determine: 1. the Arrhenius factor and 2. the Ea in kJ for this reaction. Reference: keywordsuggests.com Chemical Mechanisms: Steps Behind the Chemical Reaction • Rate orders of reactants and reaction rates are determined from experimental data. • A rate law reflects all the chemistry, down to and including the slowest step, in a multistep reaction. • The rate law is derived from the SLOWEST ELEMENTARY step in reaction’s mechanism. • This step is referred to as the RATE DETERMING step. • A chemical equation is a summary statement of all mechanism steps. • A mechanism is a proposal as how reactants are converted to products at the molecular level. • A chemical reaction is a summary of a sequence of ELEMENTARY STEPS. • Intermediates are chemical species found in elementary steps that are used in subsequent mechanism steps. • They never appear in the chemical reaction or rate law. • Criteria for a mechanism to be valid; both conditions MUST be met. 1. The elementary steps must sum to the overall chemical reaction; 2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law. Problem: The chemical reaction for the decomposition of NO2Cl is 2 NO2Cl (g) → 2 NO2(g) + Cl2(g) A. Given following elementary steps of a proposed mechanism: 1. 2 NO2Cl(g) → ClO2(g) + N2O(g )+ ClO(g) fast 2. N2O(g) + ClO2(g) → NO2(g) + NOCl(g) fast 3. NOCl (g) + ClO (g)→ NO2 (g) + Cl2 (g ) slow B. Experimental data indicated that the rate law for this reaction was rate = k[NOCl]2 [ClO] I. II. Determine if this is a valid mechanism. Identify any intermediates.