Part I: Reaction Rates: Change in Concentration with Time, halflives

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Part I Reaction Rates Change in
Concentration with Time, half-lives

• Tuesday, July 10th
• CHM 102

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What weve done so far with rates...

• Used rate laws to find relationships between
  concentrations, rates, and orders of reactions.
• Found average rates for reactions.
• Covered the 4 major things that affect rate
  (concentration, physical state, temperature, and
  catalysis).
• Now were going to look at integrated rate laws,
  which pull in time, and we can look at reaction
  rates and concentrations at any given time in a
  reaction!

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Whats an integrated rate law?

• Integrated rate laws relate time, concentration,
  and the rate constant k, with the rate.
• Youre not expected to be able to derive the
  integrated rate law from a basic rate law, so
  calculus is needed here!
• The integrated rate law is dependent upon the
  overall order of the reaction youre looking at!!

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First order integrated rate law

• 1st order integrated rate laws take the form
  lnAt -kt lnAo
• This rate law is typically used to solve for 3
  main things
• Reactant concentration at any given time.
• Time required for a fraction of reactant to be
  used up.
• Time required for a reactant concentration to
  fall below a certain level.

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Plots First order integrated rate law

• A plot of the 1st order integrated rate law
  should look like a line.
• A way to check if experimental data is 1st order
  is to plot the lnAt vs. t and see if it is
  linear!

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Practice 1st order integrated rate law

• The decomposition of dimethyl ether, (CH3)2O, at
  510oC is a 1st-order process with a rate constant
  of 6.810-4s-1
• (CH3)2O(g) ? CH4(g) H2(g) CO(g)
• If the initial pressure of (CH3)O is 135
  Torr, what is its partial pressure after 1420 s?
• So the same problem, but lets figure out how long
  it takes for 1/3 of the initial dimethyl ether to
  remain.

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Integrated rate laws and half-life

• Half-life is how long it takes for concentration
  to be reduced to 50 of where it started.
• Half-lives are important when calculating the
  presence of toxins, radiation, and many other
  important areas.
• For 1st order rate laws t1/2 0.693/k
• For 2nd order rate laws t1/2 1/(kAo)

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Practice 1st order integrated rate law and
half-life

• Youve got a pesticide that has run into a lake
  at a concentration of 5.010-7 g/cm3. The
  decomposition of the pesticide is 1st order with
  a rate constant of 1.45yr-1 at 12oC, which is the
  average temperature of the lake. Whats the
  half-life of t he pesticide? How long will it
  take for only 1/4th of the original pesticide
  concentration to remain?

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Part II Temperature and Rate, Mechanisms, and
Catalysis

• Tuesday, July 10th
• CHM 102

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Temperature and Rate

• Why does milk sour faster when its warm? Why
  does a glowstick glow brighter and burn out
  faster if you stick it in hot water?
• The ideas behind the collision model help to
  explain both the effects of temperature and
  concentration on the rates of chemical reactions.

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Collision Model

• The collision model centers around the idea that
  atoms and molecules must collide for chemical
  reactions to occur.
• The higher the energy per collision and the more
  collisions per unit time, the higher the rate for
  a chemical reaction.
• Collisions alone dont play the entire role in
  chemical reactions though!

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Collision Model Orientation Factor

• While collisions are necessary for reactions to
  occur, there also have to be orientations taken
  into account.
• In other words, a molecule may have to be facing
  a certain way for a collision to be successful in
  reacting.
• DEMO Hook and Eyelet Grab-hands

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Collision Model Activation energy

• The activation energy, Ea, is the minimum energy
  needed to initiate a chemical reaction.
• When atoms and molecules collide, the kinetic
  energy is put into bending, stretching and
  flexing, etc. the molecules.
• Upon collision, for the reaction to occur, the
  collision must produce the necessary Ea.

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Plots Activation Energy

• Activation energy can be thought of in terms of
  putt-putt.
• Lets think of the situation below

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Plots Activation Energy

• Moving aside from putt-putt, we can look at
  energy diagrams. Activation energy is related to
  rate constant by the Arrhenius equation k
  Ae(-Ea/RT)
• So as T goes up, k goes up, as Activation enegy
  goes up, k goes down.

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Practice Activation Energy

• If you take the natural log of both sides of the
  Arrhenius equation, after a bit of derivation,
  you get
• ln k (- Ea/R)(1/T) ln A
• Given the data table below, what is the value of
  Ea for the reaction (hint R 8.314 J/molK)?

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More Practice Activation Energy

• Since we know the activation energy is 160 kJ/mol
  now, we can figure out the rate constant at 553 K
  (hint check your units!) by using the equation
  below!.
• ln (k1/k2) ( Ea/R)(1/T2 1/T1)

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Mechanisms

• While a balanced chemical reaction (what youre
  used to) gives reactants and products, a
  mechanism gives the steps necessary to get from
  reactants to products.
• Think of it as making a loaded chilidog. You
  wouldnt put the chili, slaw, and mustard on the
  dog, then put it in the bun.
• The reactants of the dog, chili, and slaw also
  dont just poof together instantly. You
  probably put them on one at a time. This is
  similar to a mechanism.

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Reaction Mechanisms

• A reaction mechanism is composed of one or more
  elementary steps. Each single event or step in a
  reaction is called an elementary step.
• If the balanced chemical reaction is an
  elementary step, then the reaction is called an
  elementary reaction (or process).

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Reaction Mechanisms

• Mechanisms are broken down further into uni-,
  bi-, and termolecular steps.
• Unimolecular steps, only involve a single
  molecule in the event (such as a molecule
  rearranging itself).
• Bimolecular steps involve bringing together two
  molecules in a single event for an elementary
  step
• NO(g) O3(g) ? NO2(g) O2(g)

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Reaction Mechanisms

• For termolecular steps and beyond, the likelihood
  of these occuring becomes smaller and smaller.
• For even termolecular steps, there would have to
  be three molecules coming together to collide at
  the proper orientation with sufficient activation
  energy (recall collision theory!) for a single
  event.

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Lets look at ozone conversion!

• Ozone conversion is a multistep mechanism,
  meaning its mechanism consists of more than one
  elementary step
• O3(g) ? O2(g) O(g)
• O3(g) O(g) ? 2O2(g)
• Multistep mechanisms often have intermediates,
  which are species produced and consumed during
  the reaction, so they dont show up in the
  overall balanced equation!
• What is the molecularity of each elementary step?
  What is the overall equation? What are the
  intermediates?