# chapter 16

```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&deg;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
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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