Notes on Jones/Atkins Chapter 13: The Rates of Reactions
(Handout 13.2: 2/3/01)

• Themodynamics only tells you IF a reaction CAN occur.
• Chemical Kinetics answers the questions:
  • How fast is the reaction; when will it be over?
  • Can I make the reaction go faster or slower?
  • Does the speed of the reaction tell us anything about how the reaction occurs?
• Factors which afect the speed or rate of a reaction include:
  • concentration
  • temperature
  • presence of a catalyst
  • physical state of reactants
• These factors are controlled from the macroscopic realm.

• The rate of a reaction is the change in concentration with respect to time: Æc/Æt.
• The rate of a reaction can be a measure of the loss of reactants or the gain of products. Most rates decrease with time.
• There are other types of rates measured as tangent lines:
  • the initial rate
  • the instantaneous rate
• When the rate is divided by the concentration of the reactant raised to the power m, we get a “rate constant”.
• This algebraic expression can be rearranged to give the rate law: Rate = k[A]^m[B]ⁿ... where m and n are reaction “orders” and k is the rate constant.

• Rate Laws
  • Zero Order: Rate = k, it is not affected by concentration
  • First Order: Rate = k[A]¹, rate decreases as concentration decreases
  • Second Order: Rate = k[A]² or Rate = k[A][B]
• The order of a reaction and the stoichiometry of the balanced equation ARE NOT related. Example: 2 NH₃ --> N₂ + 3 H₂ is a zero order reaction.
• The order of a reaction can only be determined by experiment. Reaction orders can be fractions or negative (3/2, -1, etc.)
• Some methods for determining reaction orders include:
  • integrated rate laws
  • initial rate comparisons

• Collect a series of initial rate measurements in which one concentration is changed while the others are fixed. Example:
  • Exp. Initial Conc CO Initial Conc. Cl₂ Initial Rate
    1 0.12 0.20 0.121
    2 0.24 0.20 0.241
    3 0.24 0.40 0.682
• Make a ratio of rate laws and simplify the ratio to calculate an isolated concentration effect:
  • Rate 2/Rate 1 = 0.682/0.241 = [0.40]^m/[0.20]^m
  • 2.83 = 2^m
• Use logarithms and algebra to solve for the reaction order of the isolated reactant.
  • log(2.83) = log(2) * m
  • m = log(2.83)/log (2) = 1.5

Integrated Rate Laws

• Allows us to calculate the concentration, [A]_t, of the reactants or products at any time, t, and determine the order of a reaction.
• To find the order of a reactant, make a plot for each order to find which gives the best straight line.
• Half-life is the time needed for the concentration to half of its previous value.
• Zero Order Reactions
  • A plot of [A]_t will be linear
• First Order Reactions
  • [A]_t = [A]₀e^-kt or ln[A]_t = ln[A]₀ -kt
  • A plot of ln[A]_t vs. t gives a straight line with a slope of -k and a y-intercept of [A]₀.
  • Half-life: t_1/2 = ln 2/k
• Second Order Reactions
  • 1/[A]_t = 1/[A]₀ + kt
  • A plot of 1/[A]_t vs. t gives a straight line with a slope of k and a y-intercept of 1/[A]₀.
  • Half-life: t_1/2 = 1/k[A]₀

• Molecules in gases, liquids and solutions are in constant and random motion and therefore posses kinetic energy.
• When molecules collide, the translational energy is transferred between each of the bodies and is also transferred in more energetic vibrations and rotations.
• If a molecule absorbs energy from a collision so that its vibrational energy is increased to the point at which a bond breaks, a reaction occurs.
• The potential energy required break the bond is called the “activation energy” or E*.
• As temperature increases the number of molecules with enough energy to overcome E* increases and the reaction rate increases.

• The “atom assembly” at the activation energy is called the “activated complex”.
• The activation energy is NOT the same as ÆH and is always endothermic.
• For similar reactions at a given temperature, a higher E* means a slower rate of reaction.
• As the temperature is increased, more molecules have enough energy to overcome E* and the rate increases.
• As concentration increases, there are more reactant collisions. The chances of having a collision provide the needed potential energy to overcome E* are greater and the rate increases.

• Most reactions are a series of “elementary steps”.
• A sequence of elementary steps is a called a “mechanism”. A molecule only appearing within a mechanism is an called an “intermediate”.
• The “molecularity” of an elementary step is the number of colliding particles.
• For an elementary step, the molecularity is directly related to the rate law for that step.
• The slowest elementary step is “the rate-limiting step”.
• In a mechanism the overall rate and rate law is determined by the rate limiting step and those that precede it.

• Consider the reaction 2 N₂O₅ ---> 4 NO₂ + O₂
• The rate law for this reaction is: Rate = k[N₂O₅]¹. Is this consistent with the reaction stoichiometry?
• A proposed mechanism for this reaction is:
  • 2 N₂O₅ Ú 2 NO₂ + 2 NO₃ (fast)
  • NO₂ + NO₃ ---> NO + O₂ + NO₂ (slow)
  • NO₃ + NO ---> 2 NO₂ (fast)
• Show that the steps in the mechanism sum to give the balanced reaction equation.
• Show that the mechanism is supported by the rate law.
• Identify the intermediates in the reaction.

• Show that the mechanism is supported by the rate law.
  • Rate Limiting Step: Rate = k[NO₂][NO₃]
  • Combine this with the step(s) that feed(s) it. Make a rate law ratio for the forward and reverse reaction.
  • Rate forward/Rate reverse = (k_f/k_r)[N₂O₅]²/[NO₂]²[NO₃]²
  • Rate forward=Rate reverse so Rate_f/Rate_r = 1.
  • So, 1 = (k_f/k_r)[N₂O₅]²/[NO₂]²[NO₃]²
  • Simplify by taking the square root of both sides to give:
    • 1 = (k_f/k_r)^1/2[N₂O₅]/[NO₂][NO₃]
  • Solve for the concentrations in the rate limiting step:
    • [NO₂][NO₃] = (k_f/k_r)^1/2[N₂O₅]
  • Substitute the right term into its rate law:
    • Rate = k[NO₂][NO₃] = k(k_f/k_r)^1/2[N₂O₅]
• Identify the intermediates in the reaction.
  • NO and NO₃

• A catalyst speeds up a reaction without being consumed.
• It DOES NOT lower E* but rather provides an alternate reaction pathway with a lower E*.
• The catalyst may be included in the reaction stoichiometry or placed over the reaction arrow.
• A catalyst makes a kinetically-stable, product favored reaction unstable.
• Example: 2 H₂O₂ ---> 2 H₂O + O₂
  • ÆG° = -206 kJ; ÆH° = -189 kJ; E* = +76 kJ
  • If I^- is added then E* = + 57 kJ
  • H₂O₂ + I^- ---> IO^- + H₂O
  • H₂O₂ + IO^- ---> I^- + H₂O + O₂

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