Chapter 14_ Chemical Kinetics

Document Sample
Chapter 14_ Chemical Kinetics Powered By Docstoc
					Chapter 14: Chemical Kinetics

14.1 Factors that Affect Reaction Rates

Factors that affect Reaction Rates
• The physical state of the reactants- Reactions that involve solids proceed faster if the area of the solid is increased. • The concentrations of the reactants-As concentration increases, collision rate increases, leading to increased rates.


• The temperature at which the reaction occurs-the rates of chemical reactions increase as temperature increases. • The presence of a catalyst- catalysts are agents that increase reaction rates without being used up.

14.2 Reaction Rates

Reaction Rates
• Reaction rate (M/s): change in concentration is measured in Molarity divided by seconds.

D[A] rate = Dt rate = D[B] Dt

D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt
Because [A] decreases with time, D[A] is negative.


C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
D[C4H9Cl] [C4H9Cl]final – [C4H9Cl]initial •average rate = =Dt tfinal - tinitial •Instantaneous rate: rate at a particular moment in the reaction

slope of tangent

slope of tangent


Practice Problem #1
• For the reaction figure given; calculate the average rate at which A disappears over the time interval from 20 s to 40 s.

Solving the problem
average rate average rate average
D[A] = - Dt D[A] = - Dt

[A]final – [A]initial tfinal - tinitial = - 0.30M – 0.54M 40 s – 20 s

D[A] rate = - Dt

1.2 x 10 -2 M/s


14.3 Concentration & Rate

• • • •

Rate=k[NH4+][NO2-] Rate= k[A]m[B]n 5.4 x 10-7 M/s = k(0.0100 M)(0.200 M) K = 5.4 x 10-7M/s = 2.7 x 10-4 M-1 s -1 (0.0100M)(0.200M)


Exponents in rate law & Units of Rate constants
• Rate=k [reactant 1]m[reactant 2]n • Units of rate=(units of rate constant) (units of concentration)2
• Units of rate constant= units of rate = M/s = M -1 s-1

(units of concentration)



Practice problem #2
• 2N2O5(g)  4NO2(g) + O2(g) Rate= k[N2O5] 2. a. Using the reaction above what is the overall reaction order ? b. What is the units of the rate constant? ANSWERS: • The rate law of the reaction is 1st order
• Units of rate constant = units of rate = M/s = s-1 (units of concentration) M


Using Initial Rates to determine Rate Laws
The following data was measured for the reaction of nitric acid with hydrogen: 2NO(g) + 2H2(g)  2H2O(g)
Experiment Number [NO] (M) [H2] (M) Initial Rate (M/s)

Practice problem # 3

2 3

0.10 x 1 0.20

0.20 0.10 x2

1.23 x 10-3
2.46 x 10-3 x 2 4.92 x 10-3

(a) Determine the rate law for the reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.50 M and [H2] = 0.150


Solving problem # 3
• Rate=k[NO]m[H2]n • 21 = 2 thus, n=1


14.4 The Change of concentration with time

• First order reaction- the rate depends on the concentration of a single reactant raised to the first power. Consider the process in which methyl isonitrile is converted to acetonitrile.


Second-Order Reactions
• Second Order Reaction- the rate depends on the reactant concentration raised to the second power or on the concentrations of two different reactant, each raised to the first power. A graph of 1/[NO2] vs. t
gives this plot.
[NO2], M 0.01000 0.00787 0.00649 0.00481 Time (s) 0.0 50.0 100.0 200.0 1/[NO2] 100 127 154 208





14.5 Temperature & Rate

• Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent.


• The collision Model is based on the kinetic molecular theory: molecules must collide to react, increasing the temperature increases the number of collisions. • The orientation factor Cl + NOCl  NO + Cl2


• Activation Energy there is a minimum amount of energy required for reaction: the activation energy, Ea.


• reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. It shows the energy of the reactants and products (and, therefore, DE).


The Arrhenius Equation
Svante Arrhenius developed a mathematical relationship between k and Ea: k is the rate constant, Ea is the activation energy, r is the gas constant (8.314 J/ mol-k) Ea and T is the absolute temperature 

k  Ae



14.6 Reaction Mechanisms

• Reaction mechanism- the process by which a reaction occurs. • Elementary Reactions
Overall reaction NO2 (g) + CO (g) NO3 (g) +CO2 (g) Elementary Reaction : NO2 (g) + NO2 (g)  NO3 (g) + NO (g) Elementary Reaction : NO3 (g) + CO (g)  NO2 (g) + CO2 (g) NO2(g) + NO2(g) + NO3(g) + CO(g)NO2(g) + NO3(g) + NO (g) + CO2(g)

Because NO3 is neither a reactant or a product in the overall reaction it is formed in one elementary reaction and consumed in the next which makes it the intermediate


Practice Problem # 4
It has been proposed that the conversion of Ozone with O2 proceeds by a two-step mechanism. a) Write the equation for the overall reaction. b) Identify the intermediate O2(g)  O2(g) + O (g) O3(g) + O (g)  2O2 (g)


Solving the problem
O2(g)  O2(g) + O (g) O3(g) + O (g)  2O2 (g) Adding the two elementary reactions gives: 2O3 (g) + O (g)  3O2 (g) + O (g) a) 2O3 (g) 3O2 b) the intermediate is O because it is neither an original reactant nor the final, but it is formed in the first step and consumed in the next. 14.6

Rate determining step
• The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally.
Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast)

Overall: NO2(g) + CO (g) + CO2 (g)
Step one is the rate determining 14.6

14.7 Catalysis

• Homogeneous catalysis- it is present in the same phase as the reacting molecules. • Heterogeneous catalyst- exists in a different from the reactant molecules usually as solid in contact with either gaseous reactants or with reactants in liquid solution.


One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. 14.7


• Enzymes are catalysts in biological systems. • The substrate fits into the active site of the enzyme much like a key fits into a lock. 14.7

Practice Problem # 5
Make up your own problem that includes anything that you learned about rate laws… You will have 5 minutes to create and solve your own problem then you MUST turn in your paper to me. (: Just kidding (; ….


Homework Page 619 21 & 27 Page 621 45