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.

14.1

• 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.
14.1

14.2 Reaction Rates

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

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.

14.2

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

14.2

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

14.2

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)

14.3

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)

M2

14.3

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

14.3

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

1
2 3

0.10
0.10 x 1 0.20

0.10
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

14.3

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

14.4

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.

14.4

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

300.0

0.00380

263

14.4

14.5 Temperature & Rate

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

14.5

• 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

14.5

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

14.5

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

14.5

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

RT

14.5

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

14.6

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)

14.6

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.

14.7

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

Enzymes

• 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 (; ….

14.7

Homework Page 619 21 & 27 Page 621 45