Chemical kinematics, Thesis of Biochemistry

This topic cuts across the molecular reaction taking place in the biochemistry of living systems and the rates and conditions in which they occur

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2017/2018

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Chapter 14
Chemical Kinetics
Learning goals and key skills:
Understand the factors that affect the rate of chemical reactions
Determine the rate of reaction given time and concentration
Relate the rate of formation of products and the rate of disappearance of reactants given the
balanced chemical equation for the reaction.
Understand the form and meaning of a rate law including the ideas of reaction order and rate
constant.
Determine the rate law and rate constant for a reaction from a series of experiments given
the measured rates for various concentrations of reactants.
Use the integrated form of a rate law to determine the concentration of a reactant at a given
time.
Explain how the activation energy affects a rate and be able to use the Arrhenius Equation.
Predict a rate law for a reaction having multistep mechanism given the individual steps in the
mechanism.
Explain how a catalyst works.
C
(diamond)
C
(graphite)
G°
rxn
= -2.84 kJ spontaneous!
C
(graphite)
+ O
2 (g)
CO
2 (g)
G°
rxn
= -394.4 kJ spontaneous!
A reaction may be thermodynamically
favored but not kinetically favored.
Kinetics studies the rates of a chemical
process.
The study of kinetics gives insights into the
reaction mechanism (i.e., how a reaction
occurs).
Chemical Kinetics Reaction Rates
Rates of reactions can be determined
by monitoring the change in
concentration of either reactants or
products as a function of time.
reaction rate: number of atoms or
molecules that react in a given time
for a reaction to proceed,
–contact is necessary between reactants
–contact must lead to breaking of bonds
(need sufficient energy)
–the reactants must have proper
orientation
Collision model theory
Measured from the change in concentration of
reactants or products per unit time.
Reaction rates are affected by
concentration of the reactants
physical state of the reactants, i.e., surface
area/particle size
temperature
catalysts (and inhibitors)
Reaction rates
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Chapter 14

Chemical Kinetics

Learning goals and key skills:

 Understand the factors that affect the rate of chemical reactions  Determine the rate of reaction given time and concentration  Relate the rate of formation of products and the rate of disappearance of reactants given the balanced chemical equation for the reaction.  Understand the form and meaning of a rate law including the ideas of reaction order and rate constant.  Determine the rate law and rate constant for a reaction from a series of experiments given the measured rates for various concentrations of reactants.  Use the integrated form of a rate law to determine the concentration of a reactant at a given time.  Explain how the activation energy affects a rate and be able to use the Arrhenius Equation.  Predict a rate law for a reaction having multistep mechanism given the individual steps in the mechanism.  Explain how a catalyst works.

C (^) (diamond) → C (^) (graphite) ∆G°rxn = -2.84 kJ spontaneous!

C (^) (graphite) + O2 (g) → CO2 (g) ∆G°rxn = -394.4 kJ spontaneous!

  • A reaction may be thermodynamically favored but not kinetically favored.
  • Kinetics studies the rates of a chemical process.
  • The study of kinetics gives insights into the reaction mechanism (i.e., how a reaction occurs).

Chemical Kinetics

Reaction Rates

Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

  • reaction rate: number of atoms or

molecules that react in a given time

  • for a reaction to proceed, –contact is necessary between reactants –contact must lead to breaking of bonds (need sufficient energy) –the reactants must have proper orientation

Collision model theory

Measured from the change in concentration of reactants or products per unit time.

Reaction rates are affected by

  • concentration of the reactants
  • physical state of the reactants, i.e. , surface area/particle size
  • temperature
  • catalysts (and inhibitors)

Reaction rates

Reaction Rates

  • Note that the average rate decreases as the reaction proceeds.
  • This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq )

Reaction Rates

  • A plot of concentration versus time for this reaction yields a curve like this.
  • The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq )

Reaction rates

rate = change in conc. / change in time

aA + bB → cC + dD

The rate can be related to the concentration of the reactants or products. Reactants will have a negative sign. Don’t forget to use the stoichiometric coefficients.

Example

Ammonia reacts with oxygen to form nitrogen monoxide and steam. Relate the reactant and product reaction rates.

Rate laws

Rate laws show the relationship between the reaction rate and the concentration of reactants (and sometimes products).

aA + bB cC + dD

rate equation:

Rate = R = k [A]x^ [B]y

x and y are experimentally determined. They are not the stoichiometric coefficients!

Rate laws

Rate = R = k [A]x^ [B]y

k is the rate constant for this reaction.

x = order of the reaction with respect to the concentration of reactant A

y = order of the reaction with respect to the concentration of reactant B

x+y = overall reaction order or (total) order of the reaction.

Half-life, t1/

ln

[ ]

[ ]

A

A

kt

t 0

For a 1 = −

st (^) order

The half-life, t1/2 is the time needed for the concentration of a reactant to decrease to half of its initial value.

[A]t1/2 = ½[A] 0

t

A A k

A A k k k k

t

1 2

0

(^12) 0 0

1 2 2 0 693 /

ln [ ] [ ] ln [ ] [ ] ln^ ln. = − = − = − = =

Example

The 1st^ order rate constant for the decomposition of N 2 O 5 to NO 2 and O 2 at 70 °C is 6.82 × 10 -3^ s-1. What is the half-life of this reaction at 70 °C?

Half-lives of radioactive elements

Radioactive decay follows first-order kinetics. Rate of decay of radioactive isotopes are given in terms of a half-life.

(^238) U → 234 Th + α 4.5 x 10 (^9) years (^14) C → 14 N + β 5730 years (^131) I → 131 Xe + β 8.05 days

Element 106 - seaborgium (^263) Sg 0.9 s

Other interesting half-lives

isotope half-life uses C-11 20.33 min PET scans Na-24 14.951 h cardiovascular system tracer K-40 1.248 × 109 yr dating of rocks Fe-59 44.495 d red blood cell tracer Co-60 5.2712 yr radiation therapy Tc-99 6.006 h biomedical imaging Ra-226 1600 yr radiation therapy Am-241 432.2 yr smoke detectors

Temperature and Rate

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

for a reaction to proceed,

  • contact is necessary between reactants
  • contact must lead to breaking of bonds (need sufficient energy)
  • the reactants must have proper orientation

Collision model theory

Ea, Activation energy

  • There is a minimum amount of energy required for reaction: the activation energy, Ea.
  • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

Reaction coordinate diagrams

Maxwell–Boltzmann Distributions

  • At any temperature there is a wide

distribution of kinetic energies. f = e-E a /RT

Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between k and Ea :

k = A e − Ea / RT

where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

ln k = - Ea/R ( 1 ) + ln A T

k = A e − Ea / RT

Example

For the reaction CO + NO 2 → CO 2 + NO k =0.220 M-1^ s-1^ at 650. K and k =1.30 M- s-1^ at 700. K. Show that the activation energy for this reaction is 134 kJ/mol.

Multistep Mechanisms

  • In a multistep process, one of the steps will be slower than all others.
  • The overall reaction cannot occur faster than this slowest, rate-determining step.

Mechanisms

Most reactions have more than one ELEMENTARY step. The NET reaction is the sum of all the elementary steps.

Mechanisms

Most rxns. involve a sequence of elementary steps. 2 I-^ + H 2 O 2 + 2 H+^ → I 2 + 2 H 2 O Rate = k [I-] [H 2 O 2 ] NOTE

  1. Rate law is determined from experiment.
  2. Order and stoichiometric coefficients are not necessarily the same!
  3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

Slow Initial Step

  • The rate law for this reaction is found experimentally to be Rate = k [NO 2 ]^2
  • CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
  • This suggests the reaction occurs in two steps.

NO 2 ( g ) + CO ( g ) → NO ( g ) + CO 2 ( g )

Slow Initial Step

  • A proposed mechanism for this reaction is

Step 1: NO 2 + NO 2 → NO 3 + NO (slow) Step 2: NO 3 + CO → NO 2 + CO 2 (fast)

  • The NO 3 intermediate is consumed in the second step.
  • As CO is not involved in the slow, rate-

determining step , it does not appear in the rate law.

Fast Initial Step

  • The rate law for this reaction is

found to be

Rate = k [NO]^2 [Br 2 ]

  • Because termolecular processes

are rare, this rate law suggests a

two-step mechanism.

2 NO ( g ) + Br 2 ( g ) → 2 NOBr ( g )

Fast Initial Step

A proposed mechanism is

Step 2: NOBr 2 + NO → 2 NOBr (slow)

Step 1 includes the forward and reverse reactions.

Step 1: NO + Br 2 NOBr 2 (fast) k 1 k- k 2

Fast Initial Step

  • The rate of the overall reaction depends upon the rate of the slow step.
  • The rate law for that (slow) step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?
  • NOBr 2 can react two ways:
    • With NO to form NOBr (step 2)
    • By decomposition to reform NO and Br 2 (step 1)
  • The reactants and products of the first step are in equilibrium with each other.
  • Therefore, Rate f = Rate r

Fast Initial Step

  • Because Rate f = Rate r ,

k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ]

  • Solving for [NOBr 2 ] gives us k 1 k

[NO] [Br 2 ] = [NOBr 2 ]

Substituting this expression for

[NOBr 2 ] in the rate law for the

rate-determining step gives

k 2 k 1 Rate = (^) k − [NO] [Br 2 ] [NO]

Rate = k [NO]^2 [Br 2 ]

Mechanisms

Rate of the reaction is controlled by the slow step — the rate-determining step (rds) or rate-limiting step. Elementary Step 1 is bimolecular and involves I-^ and H 2 O 2. Therefore, this predicts the rate law should be Rate ∝ [I-] [H 2 O 2 ] — as observed The species HOI and OH-^ are reaction intermediates.

2 I-^ + H 2 O 2 + 2 H+^ → I 2 + 2 H 2 O Rate = k [I-] [H 2 O 2 ]

Proposed Mechanism

Step 1 — slow H 2 O 2 + I-^ → HOI + OH- Step 2 — fast HOI + I-^ → I 2 + OH- Step 3 — fast 2 OH-^ + 2 H+^ → 2 H 2 O

Catalysts

  • Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
  • Catalysts change the mechanism by which the process occurs.

Iodine-Catalyzed Isomerization

of cis-2-Butene