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Drill

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Drill 5/21/08 1. What does each letter represent in the above energy diagram? 2. Is this reaction endothermic or exothermic? Reaction Rate Chapter 17, Section 2 ... – PowerPoint PPT presentation

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Title: Drill


1
Drill 5/21/08
1. What does each letter represent in the above
energy diagram? 2. Is this reaction endothermic
or exothermic?
2
Reaction Rate
  • Chapter 17, Section 2

3
What is Reaction Rate?
  • The change in concentration of reactants per time.

4
  • In order for a reaction to occur, there must be a
    COLLISION.
  • Collision Theory provides two requirements for
    reactants to yield products
  • 1. Reaction needs enough energy
  • 2. Colliding molecules must be oriented in a way
    to react

5
  • Reaction Rate changes are caused by changes in
    the amount (frequency) of collisions, changes in
    the orientation, and changes in the energy.

6
Factors that affect reaction rate
  • Nature of Reactants different atoms, molecules,
    or compounds react with each other in certain
    ways (ex. Na reacts vigorously with water)
  • Surface Area two reactants must be in contact
    with each other to react. Gases and dissolved
    particles collide freely, so react faster.
  • Temperature Increasing kinetic energy,
    increases motion collisions. Also increases
    energy.
  • Concentration Increasing conc., increases
    collisions, so increases reaction rate
  • Catalyst lower the activation energy so that
    reaction has enough energy.

7
Rate Laws
  • Equations that relates reaction rate and
    concentrations of reactions

8
R kAnBm
  • R reaction rate
  • k specific rate constant
  • A B MOLAR conc. of reactants
  • n m powers to which the conc. are raised or
    ORDERS

9
  • Orders must be determined experimentally you
    cant figure it out from the chemical formula.
    (Unless it is stated that it is a single or one
    step mechanism)
  • Ex 3NO ? N2O NO2 R kNO2
  • 2NO2 ? 2NO O2 R kNO22

10
  • Three experiments that have identical conditions
    were performed to measure the initial rate of the
    reaction
  • NO O2 ? NO2
  • Conditions were identical in the 3 experiments,
    except the concentrations of reactants varied.
    The results are as follows

Experiment NO O2 Rate (M/s)
1 0.20 0.20 1.50
2 0.40 0.20 3.00
3 0.20 0.40 1.50
11
  • First thing you need to do is figure out the
    general rate law R kAnBm
  • R kNOnO2m

12
  • Next, to figure out n m by comparing the
    rations of reactant concentration to the ratio of
    rates. To figure out the order for A use two
    experiments that keep B constant.

Experiment NO O2 Rate (M/s)
1 0.20 0.20 1.50
2 0.40 0.20 3.00
3 0.20 0.40 1.50
.40 M/.20 M 2 3.00 M/s / 1.50 m/s 2 If
the conc. doubles, than the rate doubles a 1 to
1 ratio. 1st Order
13
  • To figure out the order for B use two
    experiments that keep A constant.

Experiment NO O2 Rate (M/s)
1 0.20 0.20 1.50
2 0.40 0.20 3.00
3 0.20 0.40 1.50
.40 M/.20 M 2 1.50 M/s / 1.50 m/s 1 If
the conc. doubles, than the rate does not change
a 1 to 0 ratio. A order of zero!
14
  • R kNO1O20
  • Now pick one experiment and fill all the values
    in to solve for k.
  • k R / NO1O20
  • k 1.50 / .201.200 1.50 / .20
  • k 7.5

15
  • If a reaction is single-step then we CAN
    determine the order of concentrations from the
    balanced chemical equation. The coefficients
    tell you what power each reactant concentration
    should be raised to.
  • 2X Y ? X2Y
  • So R kX2Y

16
  • Do page 746 in the Blue Chemistry Book
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