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Chapter 16
 The time taken for the disappearance of the
reactant or the appearance of the product .
Rate is a ratio as the amount of reactant
disappeared divided by the time.
 Average rate: The change in the
concentration divided by the total time
elapsed.
 Rate = amount reacted or produced/ time
interval
 units: g/s, mol/s, or %/s
Instantaneous rate: rate measured
between very short interval
Initial rate: instantaneous rate at the
beginning of an experiment
Page 592
Concept check
Rate depends on the concentration:
In some reactions doubling the
concentration doubles the rate of reaction.
In some doubling the reaction increases
the reaction four folds.This happens in the
decomposition of HI to form H 2 and I2.
An expression for the rate of a reaction as
a function of the concentration of one or
more of the reactants.
Rate=k [A]n
This equation is the general rate law. The
exponent n , is called the order with
respect to substance A and must be
determined from experimental data.
Order of a chemical reaction can be said
as the exponent on the concentration for a
specified reactant in a rate law expression.
Determine the rate law equation for the
following reaction , given the experimental
data
3AC
Concentration of A Reaction rate
 0.2M 1.0M/s
 0.4M 4.0M/s
Page 596
Practice problems
The rate law equations you have looked so
far have been for reactions involving only
one reactant.
If more than one reactant is found to
contribute to the rate of the reaction, then
all contributing reactants must appear in
the rate law.
The rate law equation for this will be
rate=k [A] n [B] m
 The value of n is the order with respect to
reactant A. The value of m is the order with
respect to reactant B. The overall reaction
order will be the sum of n and m.
 From the above equation if you double the
concentration of A and the rate doubles then
the reaction is first order with respect to A. If
you double the conc. of B (keeping the conc
of A constant, and the rate quadruples the the
rate of the reaction is second order with
respect to B.
For the reaction A and B for this example
the rate law would be rate=k[A][B]2
Rate laws cannot be derived from a
chemical equation.
2N2O₅↔4NO2+O2
Keq=[NO2] 4 [O2]/[N2O₅]2
rate=k[N2O₅]1
The slowest step in a mechanism, the step
that determines the overall rate of reaction
is the rate determining step.
Mechanism is a proposed sequence of
steps that describes how reactants are
changed into products.
Each step in the mechanism is called as
elementary step.
Page 598
Critical thinking 2,3,5
Practice problems 7and 8 all
 Temperature: An increase in temperature is
accompanied by an increase in the reaction rate.
Temperature is a measure of the kinetic energy of
a system, so higher temperature means higher
average kinetic energy of molecules and more
collisions per unit time.
 For most chemical reactions the rate at which the
reaction proceeds will approximately double for
each 10°C increase in temperature. Once the
temperature reaches a certain point, some of the
chemical species may be altered (e.g., denaturing
of proteins) and the chemical reaction will slow or
stop.
Concentration: A higher concentration of
reactants leads to more effective collisions
per unit time, which leads to an increasing
reaction rate (except for zero order
reactions). Similarly, a higher concentration
of products tends to be associated with a
lower reaction rate.
Medium: The rate of a chemical reaction
depends on the medium in which the
reaction occurs. It sometimes could make
a difference whether a medium is aqueous
or organic; polar or nonpolar; or liquid,
solid, or gaseous.
Surface area: It is easier to dissolve sugar
if it is crushed. Crushing the sugar
increases its surface tension.The larger
surface area allows more sugar molecules
to contact the solution.
 Catalyst: A catalyst is a substance that alters
the rate of a chemical reaction without being
used up or permanently changed chemically.
 A catalyst works by changing the energy
pathway for a chemical reaction. It provides
an alternative route (mechanism) that lowers
the Activation Energy meaning more particles
now have the required energy needed to
undergo a successful collision.
What is activation energy?
The least amount of energy needed to
permit a particular chemical reaction.
 There are 2 types of catalysts:
 Homogeneous catalyst: Homogeneous
catalysts are in the same phase as the
reactants.
 Heterogeneous catalyst: Heterogeneous
catalysts are present in different phases from
the reactants (for example, a solid catalyst in
a liquid reaction mixture), whereas
homogeneous catalysts are in the same
phase (for example, a dissolved catalyst in a
liquid reaction mixture).
Example of Homogeneous catalyst
2H2O2(aq)+ KI(aq)2H2O(l)+O2(g)
Example of Heterogeneous catalyst
Decomposition of H2O2 in presence of MnO2
Hydrogen peroxide is a solution while
manganese dioxide is a solid and can be easily
separated.
Term review all
Page 614
13,22, 23 and 25
Test prep all

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chapter_16.ppt

  • 2.  The time taken for the disappearance of the reactant or the appearance of the product . Rate is a ratio as the amount of reactant disappeared divided by the time.  Average rate: The change in the concentration divided by the total time elapsed.  Rate = amount reacted or produced/ time interval  units: g/s, mol/s, or %/s
  • 3. Instantaneous rate: rate measured between very short interval Initial rate: instantaneous rate at the beginning of an experiment
  • 5. Rate depends on the concentration: In some reactions doubling the concentration doubles the rate of reaction. In some doubling the reaction increases the reaction four folds.This happens in the decomposition of HI to form H 2 and I2.
  • 6. An expression for the rate of a reaction as a function of the concentration of one or more of the reactants. Rate=k [A]n This equation is the general rate law. The exponent n , is called the order with respect to substance A and must be determined from experimental data.
  • 7. Order of a chemical reaction can be said as the exponent on the concentration for a specified reactant in a rate law expression.
  • 8. Determine the rate law equation for the following reaction , given the experimental data 3AC Concentration of A Reaction rate  0.2M 1.0M/s  0.4M 4.0M/s
  • 10. The rate law equations you have looked so far have been for reactions involving only one reactant. If more than one reactant is found to contribute to the rate of the reaction, then all contributing reactants must appear in the rate law. The rate law equation for this will be rate=k [A] n [B] m
  • 11.  The value of n is the order with respect to reactant A. The value of m is the order with respect to reactant B. The overall reaction order will be the sum of n and m.  From the above equation if you double the concentration of A and the rate doubles then the reaction is first order with respect to A. If you double the conc. of B (keeping the conc of A constant, and the rate quadruples the the rate of the reaction is second order with respect to B.
  • 12. For the reaction A and B for this example the rate law would be rate=k[A][B]2 Rate laws cannot be derived from a chemical equation. 2N2O₅↔4NO2+O2 Keq=[NO2] 4 [O2]/[N2O₅]2 rate=k[N2O₅]1
  • 13. The slowest step in a mechanism, the step that determines the overall rate of reaction is the rate determining step. Mechanism is a proposed sequence of steps that describes how reactants are changed into products. Each step in the mechanism is called as elementary step.
  • 14. Page 598 Critical thinking 2,3,5 Practice problems 7and 8 all
  • 15.  Temperature: An increase in temperature is accompanied by an increase in the reaction rate. Temperature is a measure of the kinetic energy of a system, so higher temperature means higher average kinetic energy of molecules and more collisions per unit time.  For most chemical reactions the rate at which the reaction proceeds will approximately double for each 10°C increase in temperature. Once the temperature reaches a certain point, some of the chemical species may be altered (e.g., denaturing of proteins) and the chemical reaction will slow or stop.
  • 16. Concentration: A higher concentration of reactants leads to more effective collisions per unit time, which leads to an increasing reaction rate (except for zero order reactions). Similarly, a higher concentration of products tends to be associated with a lower reaction rate.
  • 17. Medium: The rate of a chemical reaction depends on the medium in which the reaction occurs. It sometimes could make a difference whether a medium is aqueous or organic; polar or nonpolar; or liquid, solid, or gaseous.
  • 18. Surface area: It is easier to dissolve sugar if it is crushed. Crushing the sugar increases its surface tension.The larger surface area allows more sugar molecules to contact the solution.
  • 19.  Catalyst: A catalyst is a substance that alters the rate of a chemical reaction without being used up or permanently changed chemically.  A catalyst works by changing the energy pathway for a chemical reaction. It provides an alternative route (mechanism) that lowers the Activation Energy meaning more particles now have the required energy needed to undergo a successful collision.
  • 20. What is activation energy? The least amount of energy needed to permit a particular chemical reaction.
  • 21.  There are 2 types of catalysts:  Homogeneous catalyst: Homogeneous catalysts are in the same phase as the reactants.  Heterogeneous catalyst: Heterogeneous catalysts are present in different phases from the reactants (for example, a solid catalyst in a liquid reaction mixture), whereas homogeneous catalysts are in the same phase (for example, a dissolved catalyst in a liquid reaction mixture).
  • 22. Example of Homogeneous catalyst 2H2O2(aq)+ KI(aq)2H2O(l)+O2(g) Example of Heterogeneous catalyst Decomposition of H2O2 in presence of MnO2 Hydrogen peroxide is a solution while manganese dioxide is a solid and can be easily separated.
  • 23. Term review all Page 614 13,22, 23 and 25 Test prep all