Kinetics Part II: Rate Laws

1
Kinetics Part IIRate Laws Order of Reaction

• Dr. C. Yau
• Fall 2009

Based on Brady Senese 5/eChap. 13 Sec 3
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Rate of Rxn vs. Rate Law
Rate of reaction is based on one component
(reactant or product) of the reaction
disappearance of a reactant or formation of a
product. Rate law is a rate expression that
includes all reactants. LEARN THESE TERMS SO YOU
KNOW WHAT IS BEING ASKED FOR!!
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The Rate Law Depends On The Concentrations Used

• Rate kreactantorder
• k is a reaction rate constant, a measure of time
  efficiency
• High values of k mean high efficiency
• k must be determined experimentally
• Each experiment has its own rate law
• Rate law must be determined experimentally

4
A B products

• Rate k AmBn
• where m and n are the "orders of reaction" and
  are found by experiment, NOT based on the
  coefficients of the chemical equation,
• and k is the "rate constant."
• This expression is called the "rate law."

5
H2SeO3 6I- 4H Se 2I3- 3H2O

• Rate 5.0x105 L5mol-5 s-1 H2SeO3 I-3H2
• 5.0x105 mol-5 s-1 is the rate constant (k).
• We speak of the reaction as being
• first order with respect to H2SeO3,
• third order with respect to I-,
• second order with respect to H, and
• the overall order of reaction is 6 (sum of all
  the orders).
• Learn this terminology!
• What is the unit of Rate in the equation shown
  above?

Do Practice Exercises 7 8 on p.529.
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What is a rate law used for? Rate changes with
concentrations. The rate law allows us to
determine the rate for various concentrations of
the reactants.

• Example
• The rate law for the reaction 2A B?3C is
• Rate 0.045M-1s-1 AB
• If the concentration of A is 0.2M and that of B
  is 0.3M, what will be the reaction rate?

rate 0.045 M-1 s-1 0.2M0.3M
rate 0.0027 M/s
Do Practice Exercises 5 6 p.526
7

• Chlorine Dioxide, ClO2, is a reddish-yellow gas
  that is soluble in water. In basic solution it
  gives ClO3- and ClO2- ions.
• ClO2(aq) OH?(aq) ?? ClO3?(aq) ClO2?(aq) H2O
  (l)
• The rate law is RatekClO22OH-. What is the
  value of the rate constant given that when
  ClO20.060M, OH- 0.030M, the reaction rate
  is 0.0248 M/s
• 0.02 M-1 /s
• 0.02 M/s
• 0.02 s-
• None of these

2.3x102 M-2 s-1
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Orders

• indicate the degree of resistance to reaction- a
  high order suggests a reactant that is slow to
  react,
• are indicated for each reactant,
• the overall reaction order is the sum of
  individual reactant orders
• may be negative, fractional or integers, but in
  this course we will usually encounter positive
  integers
• must be determined from experimental data

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Determining The Rate Law

• Run reaction under the same conditions, varying
  only the concentrations of reactants (not the
  temperature, etc.)
• A ratio of rate laws for each experiment allows
  us to determine the orders of each reactant
• The rate law is unique to temperature and
  concentration conditions

10
Use Rate Laws To Determine Orders
2NO(g) O2(g) ? 2NO2(g)

• Select 2 rate laws that vary in concentration for
  only one of the substances (NO).

Hint Write the fractions with the larger R on
top.
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Use Rate Laws To Determine Orders
2NO(g) O2(g) ? 2NO2(g)

• Next choose 2 rate laws where the concentration
  for the other component
• (O2) changes.

x2, y 1 so.. Rate k NO2O2
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Determining The Value Of k

• Finally we can solve for k. Use any rate law and
  the orders that we have determined.

rate kNO2O2
0.048M/s k 0.015M20.015M
1.4104 M-2s-1 k
Do Example 13.4 p.531, Exercises 9 10 p.531.
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Determine The Rate Law From Given Data
z0
x1
y1
ratekAB Note that changing the concentration
of C had no effect on the rate. We say it is
zero order with respect to C.
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Effect of Order of Rxn on Rate

• Consider Rate kAn
• If n 0, change in conc has no effect on rate.
• If n 1, Rate kA1 and when conc is 2x,
• rate is 2x.
• If n 2, Rate kA2 and when conc is 2x,
• rate is 4x
• If n 2, when conc is tripled, rate is ?
• rate is 9x
• If n 3, and conc is doubled, rate is?
• rate is 8x

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What is the rate law? Rate kA?B? Rate
kA1B2
16

• For the following data, determine the order of
  NO2 in the reaction at 25
• 2 NO2(g) F2(g)? 2 NO2F(g)

Do Examples13.4, 13.5, Exercise 11, 12 p.532-533

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