Rates of Reaction

Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how does a reaction proceed (reaction mechanism) and how fast (reaction rate). Reaction rate is the change in the concentration of a reactant or a product with time (Units: mol/L  s). R P rate = -  [R] tt rate =  [P] tt  [R] = change in [reactants] over a time period  t  [P] = change in [products] over a time period  t Because [R] decreases with time,  [R] is negative.

A B rate = -  [A] tt rate = [B][B] tt time

Reaction Rates & Stoich Gaseous dinitrogen pentoxide decomposes to form nitrogen dioxide and oxygen gas. Nitrogen dioxide is produced at a rate of 5.00 X mol/L·s. Determine the rate of decomposition of dinitrogen pentoxide and the rate of formation of oxygen. 2 N 2 O 5(g)  4 NO 2(g) + O 2(g)

Collision Theory In order for successful reactions to occur molecules must collide with: 1.The proper orientation 2.Sufficient energy

Factors that Affect the Rate of Reaction A) Temperature: By heating the mixture, you increase the Kinectic energy and increase the chances of reactions B) Concentration of Reactants: Increasing the concentration incresases the frequency of collisions C) Catalysts: Lowers the activation energy (minimum amount of energy required for a reaction to occur)

D) Surface Area of a Solid Reactant: Increased surface area increases the number of possible sites for a reaction to occur E) Pressure of gaseous reactions and products: Increasing the pressure, forces molecules closer together increasing the frequency of collisions

Activation Energy Activation Energy: The minimum collision energy required for a successful reaction to occur Maxwell Boltzmann Distribution Curve: kinetic energy vs. # of particles As temperature increases, KE increases and more particles will have sufficient energy to react

Potential Energy Diagrams When molecules collide, kinetic energy of the particles is converted to potential energy

Diagram Terminology Ea = activation energy ΔH = enthalpy change Transition State: point when reactant(s) are converted to product(s) Activated Complex: chemical species that exist at the transition state ▫Bonds partially broken and formed

Practice 1.It is a general rule that with a 10 o C temperature increase, most reaction rates will double. This is not a result of doubling the number of collisions. Explain. 1.Sketch a PE diagram for each of the following reactions. Include labels for the transition state. a.S(s) + O2(g)  SO2(g) ΔH = kJ b.b. Cl2 (g)  Cl(g) + Cl(g) ΔH = kJ

Catalysts Increases the rate of a chemical reaction without being consumed in the reaction process Lowers the activation energy so that a larger number of reactants have sufficient energy to react

Rate Law

aX + bY  products r=k[X] m [Y] n r=rate k=rate constant ▫specific to the reaction at a certain temperature

Order of Reaction r=k[X] m [Y] n Exponents (m and n) describe the relationship between the rate and the initial concentration Can only be determined empirically (experimentally) Ex: ▫2NO 2 + F 2  2NO 2 F ▫r=k[NO 2 ] 1 [F 2 ] 1 Exponents (m and n) do not have to be the same as the reaction coefficients The exponents are called order of reaction The order of reaction with respect to NO2 is 1 The order of reaction with respect to F2 is 1 The overall order of reaction: ▫sum of individual orders of reactants ▫(1 + 1) = 2

Determining Order of Reaction Ex: 2 X + 2 Y + 3 Z → product Given: r=k[X] 1 [Y] 2 [Z] 0 The order of [X] is 1 ▫If the initial conc of X is doubled, the rate will double (2 1 ) ▫If the initial conc of X is tripled, the rate will triple (3 1 ) The order of [Y] is 2 ▫If the initial conc of Y is doubled, the rate will multiply by 4 (2 2 ) ▫If the initial conc of Y is tripled the rate will multiply by 9 (3 2 ) The order of [Z] is 0 ▫If the initial conc of Z is double the rate will multiply by 1 (2 0 ), and so remain unchanged ▫The rate does not depend on Z The overall order of reaction is (1+2+0) = 3 r=k[X] 1 [Y] 2 [Z] 0 can be written r=k[X] 1 [Y] 2

Practice

Reaction Mechanism Many reactions don’t take place in only one step, but many Reaction Mechanism: A series of elementary steps that makes up an overall reaction Elementary Step: A step in a reaction mechanism that only involves one-, two-, or three-particle collisions

Rate Determining Step The slowest step in a reaction mechanism The step with the highest activation energy The overall reaction rate is controlled by the rate determining step Reaction intermediates: molecules formed as short-lived products in reaction mechanisms ▫More stable than transition state ▫Exist in the valleys (as oppose to the peaks) of a PE diagram

Example HBr(g) + O2 (go)  HOOBr (g) (slow) HOOBr(g) + HBr (g)  2 HOBr(g)(fast) 2HOBr(g) + 2HBr(g)  2H2O(g) + Br2(g) (fast) 4HBr(g) + O2(g)  2H2O(g) + 2Br(g) What step is the rate determining step? If we wanted to increase the rate of reaction, which reactant could we increase the concentration of? What is the rate equation? r=k[HBr][O2]

Practice

Reminder: Rate equation and Rate determining step Direct correlation between the exponents in the rate equation and the equation coefficients of the rate determining step Example: ▫r= k [HBr] 1 [O 2 ] 1 Rate determing step: ▫1 HBr + 1 O 2  reaction intermediate In general: If the rate equation is r = k [X] m [Y] n Rate determining step: mX + nY  reaction intermediate

Reminder: Three rules of reaction mechanisms 1.Each step must be elementary, involving no more than 3 reactants (usually only 1 or 2) 2.The slowest or rate-determining step must be consistent with the rate equation 3.The elementary steps must add up to the overall equation

Practice 2N2O5(g)  2N2O4(g) + O2 (g) a)What would the rate equation be if the reaction occurred in a single step? b)The actual experimentally derived rate equation is r=k[N 2 O 5 ] 1. What is the rate- determining step? c)Suggest the possible mechanism and indicate the slowest step.

Practice 2N2O5(g)  2N2O4(g) + O2 (g) a)What would the rate equation be if the reaction occurred in a single step? The exponent on the rate law would be the same as the coefficient on the reaction. ▫r=k[N 2 O 5 ] 2 b) The actual experimentally derived rate equation is r=k[N 2 O 5 ] 1. What is the rate-determining step? Because the coefficient on the reactant must be the same as the exponent in the rate equation, the rate-determining step must be ▫1N2O5 (g)  product or reaction intermediate c) Suggest the possible mechanism and indicate the slowest step. The only step we are sure of is the rate determining step, the others are guesses ▫N2O5  N2O4 + O (slow) ▫O + N2O5  N2O4 + O2 (fast)

Practice! The reaction of nitrogen dioxide with carbon monoxide produces nitrogen monoxide and carbon dioxide. NO 2(g) +CO (g)  NO (g) +CO 2(g) Rate experiments were performed on this reaction, providing the empirical data in Table 1.

1.Determine the rate law equation, along with the value of the rate constant, for this reaction. 2. Calculate the enthalpy change for the reaction (using heat of formation). 3. Represent the enthalpy change on a potential energy diagram, include the following additional information. Under SATP conditions the activation energy for the reaction is 45 kJ/mol. Include the changes of energy for the step(s) of the reaction mechanism on the energy diagram. Identify the activated complex(es)/transition state(s) and any reaction intermediate(s) for this reaction mechanism and show the corresponding position(s) on the energy diagram. Given Step 1: 2NO 2  NO 3 + NO Step 2: NO 3 + CO  NO 2 + CO 2