Unit 7: Kinetics

Rate Laws from Mechanisms and Relative Rates

· Chemical Kinetics is the study of:

· The rate at which reactants are converted to products during the course of a chemical reaction

· The factors, which include temperature, pressure, concentration, catalyst and surface area that affect the rate of a chemical reaction

· The sequence of steps, or the mechanism, which we believe occurs when reactants are converted to products

· Four factors that affect the rates of reactions are

1.

2.

3.

4.

· For a chemical reaction its rate, or rate of reaction, is expressed in terms of how fast the concentration of a substance changes in the course of a chemical reaction.

· Rate = ∆[product]/time

· Rate = -∆[reactant]/time

· [] = concentration in molarity

· Relative Rates

· Rates of products and reactants differ only in the – sign

· Rates must be positive, but the change in reactants is negative, so you must account for this and have the negative sign in the equation.

· Consider the following reaction

2N2O5(g)→2NO2(g) + O2(g)

· The rate can be expressed in the following ways

Rate = -∆[ N2O5] = 1 ∆[ NO2] = 2 [O2]

∆t 2 ∆t∆t

1. Given the hypothetical reaction:

2A + nB → qC + rD

And if: -∆[ A] = 0.050 mol/Ls

∆t

-∆[ B] = 0.150 mol/Ls

∆t

∆[ C] = 0.0750 mol/Ls

∆t

∆[ D] = 0.0250 mol/Ls

∆t

a. What are the coefficients of n, q and r?

· Rate Law Equation

· Rate = k[A]x[B]y

· k = rate constant in the mathematical equation (not a given)

· exponents x and y = order of reaction with respect to each substance. Combination of x + y represents the overall order of the reaction

· rate = initial rate = the fastest rate of the reaction and occurs at the very beginning of the reaction. At this point there are few competing reactions.

2. Consider the following reaction and mechanisms for this reaction:

H2O2 + 2I- + 2H+ → I2 + 2H2O

H2O2 + I- → HOI + OH-slow

HOI + I- → I2 + OH-fast

2OH + 2H+ → 2H2Ofast

a. What species are intermediates in the above mechanism? Any catalysts?

b. Sketch the energy profile for the three step mechanism and clearly label the rate determining step in your diagram

c. What rate law would be expected if this mechanism is correct? That is, will it be zero-order, first-order or second-order with respect to H2O2? Zero-order, first-order or second-order with respect to I-? Explain briefly.

d. What rate law would be expected if the second step in this reaction mechanism is the slow step?

3. HOBr + HBr → Br2 + H2O

HOBr + HBr → Br2 + H2O

HBr + O2 → HOOBr(slow)

HOOBr + HBr → 2HOBr

a. What is the overall reaction of the above mechanism?

b. What species are intermediates in the above mechanism? Any catalysts?

4. The rate law for the following reaction

2NO2(g) + F2(g) → 2NO2F(g)

was experimentally determined to be

rate = k[NO2]1[F2]1

Which of the following mechanism is the most reasonable? Explain your reasoning.

a. 2NO2(g) + F2(g) → 2NO2F(g)

b. NO2(g) + F2(g) → NO2F(g) + F(g)(fast)

NO2(g) + F(g) → NO2F(g)(slow)

c. NO2(g) + F2(g) → NO2F(g) + F(g)(slow)

NO2(g) + F(g) → NO2F(g)(fast)

5. When a certain substance’s concentration is doubled, the initial rate of a reaction proceeds to eight times as fast. What is the order with respect to this substance?

HW 7A pg 593-599 #23-28, 61, 62

Method of Initial Rates To Determine Rate Law (2nd way to find rate law)

· Rate by definition is the -∆[reactant]/time

· Concentration of reacting species must be followed over time

· Two or more experiments, each with different initial concentrations are required

· After collecting experimental data, the initial rate is determined

· The order of the reaction with respect to a particular reactant is obtained by comparing the ratio of the initial concentrations of the reactant with the ratio of their initial rates

6. The following initial rate data were collected for the reaction.

2NO2(g) + F2 (g) → 2NO2F (g)

Exp[NO2] [F2]Initial Rate (M/sec)

10.04820.0318 1.90 x 10-3

20.01200.0315 4.69 x 10-4

30.04800.127 7.57 x 10-3

a. Determine the reaction order for NO2 and F2

b. Determine the overall order of the reaction

c. Write the specific rate law for the reaction

d. Determine the rate constant for the reaction (include units)

7. Consider the reaction and the following initial rate data.

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

ExpPNO(mmHg)PH2(mmHg)Initial Rate(mmHg/s)

1 400 1500.66

2 400 3001.34

3 150 4000.25

4 300 4001.03

a. Determine the reaction order for NO and H2

b. Determine the overall order of the reaction

c. Write the specific rate law for the reaction

d. Determine the rate constant for the reaction (include units)

8. Consider the reaction and the following initial rate data.

A + 3B → products

Exp [A] [B] [C]Rate of formation of product (M/s)

11.05 x 10-22.50 x 10-24.00 x 10-31.74 x 10-4

28.71 x 10-22.50 x 10-24.00 x 10-31.74 x 10-4

32.10 x 10-22.10 x 10-22.10 x 10-21.34 x 10-3

44.20 x 10-22.10 x 10-24.20 x 10-27.58 x 10-3

a. Determine the reaction order for A, B and C.

b. Determine the overall order of the reaction

c. Write the specific rate law for the reaction

d. Determine the rate constant for the reaction (include units)

9. Instantaneous Rate

a. The rate at any one point and time during the experiments.

b. To find it, find the slope of the curve at the time in question (the derivative)

c. The rate (slope of tangent line at any point) constantly decreases

i. Due to the concentration of the reactants constantly decreasing as the reaction proceeds

HW 7B pg 593-599 #30, 31, 33, 36

Integrated Rate Laws (3rd way to find rate law)

· For zero order reactions of the type A(g) → productswe can write the differential form of the rate law (useful in describing how the rate of disappearance of reactant is proportional to the concentration of the reactant) as rate = k[A]0

· When this relationship is integrated from t0 to t1 we obtain the relationship

· The ½ life of a chemical reaction is defined as the time required for the initial concentration, [A]0, to fall to half its value. This can be described using the mathematical equation

· For first order reactions of the type A(g) → productswe can write the differential form of the rate law (useful in describing how the rate of disappearance of reactant is proportional to the concentration of the reactant) as rate = k[A]1

· When this relationship is integrated from t0 to t1 we obtain the relationship

· The ½ life of a chemical reaction is defined as the time required for the initial concentration, [A]0, to fall to half its value. This can be described using the mathematical equation

· For second order reactions of the type A(g) → productswe can write the differential form of the rate law (useful in describing how the rate of disappearance of reactant is proportional to the concentration of the reactant) as rate = k[A]2

· When this relationship is integrated from t0 to t1 we obtain the relationship

· The ½ life of a chemical reaction is defined as the time required for the initial concentration, [A]0, to fall to half its value. This can be described using the mathematical equation

10. The decomposition of H2O2 to H2O and O2 follows first order kinetics with a rate constant of 0.0410 min-1 at a particular temperature. Calculate the [H2O2] after 10 min if [H2O2]0 od 0.200M.

H2O2(l) → 2H2O(l) + O2(g)

a. Calculate the [H2O2] after 10 min if [H2O2]0 od 0.200M.

b. How long would it take for half of the H2O2 to decompose?

11. The decomposition of N2O5 to O2 and NO2 follows first order kinetics. If a sample at 25°C with the initial concentration of N2O5 of 1.25 x 10-3 M falls to 1.02 x 10-3 M in 100. Minutes, calculate the rate constant for the reaction.

12. The decomposition of NOCl into NO and Cl2 is a second order reaction with a rate constant of 0.0480 M-1s-1 at 200°C. In an experiment, the initial concentration of NOCl was 0.400 M.

a. What is the concentration of NOCl after 15.0 minutes?

b. Calculate the half-life

13.

HW 7C pg 593-599 #37-41, 43, 45, 46, 53, 57

Catalysts and Activation Energy with the Arrhenius Equation

· How does temperature affect the reaction rate?

· Molecules have a range of energies that can be described by an energy distribution diagram

· Bell shaped curve and most of molecules have an average of 100 kJ/mol, although some have more and some have less.

· If the reaction has an Ea of 100 kJ/mol, then about half of the molecules will have enough enough energy to reach the activated complex, therefore half of collisions will be effective with respect to energy.

# molecules

# molecules

T = 400K T = 300K

energy

energy

· How does a catalyst affect the rate of the reaction?

· Best shown using progress of reaction diagrams

· Catalyst for a certain reaction lowers the Ea for that particular reaction from 118 kJ/mol to only 99 kJ/mol.

· Larger fraction of molecules will have the energy to reach the activated complex and will be able to react.

· Heterogeneous catalyst

· Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst.

· Adsorption – collection of one substance on the surface of another substance.

· Adsorption and activation of the reactants.

· Migration of the adsorbed reactants on the surface.

· Reaction of the adsorbed substances.

· Escape, or desorption, of the products.

· Homogeneous catalyst

· Exists in the same phase as the reacting molecules.

· Enzymes are nature’s catalysts.

· Activation Evergy, Ea

· Energy that must be overcome to produce a chemical reaction.

· Represents a measure of the energy barrier colliding molecules must surmount if they are to react rather than to recoil from one another

· It is assumed that every pair of molecules less than Ea will not react and every pair with greater energy than Ea and the proper orientation will react

· Collision Theory

· For every reaction to happen, two things must occur:

1. Molecules must collide with sufficient kinetic energy to cause them to reach the activation energy barrier.

2. Molecules must collide with the appropriate orientation

· Two factors summarized in the single temperature Arrhenius Equation

k = rate constant

A = frequency factor (how many collisions have the appropriate orientation)

e = energy of collisions

Ea = activation energy

R = gas constant (8.3145 J/K·mol)

T = temperature (in K)

· Linear Form of Arrhenius Equation

14. At 300°C the rate constant for the reaction is 2.41x10-10 s-1. At 400°C the rate constant is 1.16x10-6 s-1. Calculate the activation energy for the reaction.

15. If the activation energy for the decomposition of N2O5 is 1.0x102 kJ/mol, calculate the temperature change necessary to double the rate at room temperature.

HW 7D pg 593-599 #64, 65, 69, 75, 78, 81

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