Download Kinetics Intro. Worksheet

Chang, 7th Edition, Chapter 13, Worksheet #1
S. B. Piepho, Fall 2002
Chemical Kinetics: Introductory Concepts
Chemical kinetics is the study of the rates of chemical reactions and the mechanisms by
which reactions occur. A rate is the change of a property (in this case, concentration) per unit
time. The rate of a chemical reaction is found by following the rate of disappearance (or
decomposition) of one of the reactants or the rate of appearance (or formation) of one of the
products.
Suppose we consider the reaction
N2(g) + 3 H2(g)  2 NH3(g)
(1)
Since three H2 molecules react with one N2 molecule to produce two NH3 molecules, the rate of
disappearance of H2 will be three times the rate of disappearance of N2, and the rate of
appearance of NH3 will be twice the rate of disappearance of H2. Thus,
rate
=
1/ the rate of
rate of
3
disappearance of N2 = disappearance of H2 =
rate
=
 [N2 ]

 t 
rate
=
vN2
=
=
1  [H 2 ] 

3  t 
1
vH
3 2
=
=
1/
times the rate of
appearance of NH3
2
1 [NH3 ]
2  t 
1
vNH 3
2
(2)
Note that rates are positive numbers (time doesn’t go backwards!). That is the reason for the
negative sign in the expressions for the rate of disappearance of N2 and H2 in the Eq (2); for
example, since [N2 ] is a negative number, we need to multiply [N2 ] t  by (-1) to obtain a
positive rate.
In general, for the chemical equation
aA + bB  cC + dD
the rate is given by
rate 
1  A  1  B 
1 C 
1 D 





a  t  b  t 
c  t 
d  t 
rate  v 
1
1
1
1
vA  vB  v C  v D
a
b
c
d
(3)
Since the rates of appearance and disappearance of all reactants and products are related by
the equation stoichiometry, it doesn’t matter which rate we actually measure – experimental
convenience governs our choice. However, since the rates differ by stoichiometric ratios, we
must specify the substance for which our rate is defined.
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Chang, 7th Edition, Chapter 13, Worksheet #1
S. B. Piepho, Fall 2002
Rate Laws
The rate of most reactions changes with time. Initially, when concentrations of reagents are
highest, the rate is fastest. As reactants are consumed and products form, the forward reaction
slows down and the backward reaction speeds up. Eventually, either the reaction reaches
equilibrium, or it goes to completion and the rate goes to zero. At equilibrium, the forward
reaction rate equals the backwards reaction rate, and there is no further net change in
concentration (see Chapter 14).
In many cases the rate of reaction obeys a simple equation known as the rate law. A rate law
is an equation expressing the reaction rate as the product of a rate constant k and the
concentrations of species involved in the reaction raised to various powers:
rate = v = k [A] m [B] n ....
(4)
A fast reaction is characterized by a large value for k. As a reaction proceeds, its rate changes
with concentration according to the rate law, but k remains constant. The rate constant k depends
on temperature, but is independent of concentration.
The powers to which the concentrations are raised (m, n, ...) are typically positive integers
(but may be fractions or negative numbers). They may also be zero, but in that case the factor is
usually omitted from the rate law, since, for example, [A]o = 1.
Reaction Order
The overall order of a reaction is the sum of the exponents in the rate law (Eq (4)). The
order of reaction in A is the power to which [A] is raised in the rate law, and so on for B, C, etc.
If the concentration of a substance does not appear in the rate law, the reaction is zeroth order in
that species. For example, if the reaction
2 A + B + C  D + 2E
has the rate law
rate = v = k [A] [B] 2
the reaction is first order in [A], second order in [B], zeroth order in [C], and third order
overall.
The rate law must be determined experimentally. The reaction orders have no necessary
relation to the stoichiometric coefficients in the equation for the reaction.
Units
The units of a rate are always M s-1 = mol L-1 s-1. Since in the rate law equation (Eq (4)), the
units on the left must equal the units on the right, the units of k depend on the overall order. For
example, if a reaction is first-order overall, k will have units of s-1; if it is second-order overall, k
will have units of L mol-1 s-1; and so on.
______________________________________________________________________________
1. For the reaction, 2 N2O5(g)  4 NO2(g) + O2(g), the rate of formation of NO2(g) is
4.0 x 10-3 mol L-1s-1.
(a) Calculate the rate of disappearance of N2O5(g)
(b) Calculate the rate of appearance of O2(g).
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Chang, 7th Edition, Chapter 13, Worksheet #1
S. B. Piepho, Fall 2002
Answers: (a) 2.0 x 10-3 mol L-1 s-1; (b) 1.0 x 10-3 mol L-1 s-1
Hint: Eq(3) gives rate  v  1 v N2 O 5  1 v NO 2  v O2 so here
2
4
1
-3
-1
1
rate  v 
v
= ( /4)( 4.0 x 10 mol L s-1) = 1.0 x 10-3 mol L-1s-1.
4 NO2
 
 
 
______________________________________________________________________________
2. The reaction 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) is found experimentally to be second
order in NO(g) and first-order in H2(g).
(a) Write the rate law for the reaction.
(b) What is the overall order of the reaction?
(c) What are the units for the rate constant k?
(d) If [NO] is doubled (while keeping [H2] constant), by what factor will the reaction rate
increase?
(e) If [H2] is doubled (while keeping [NO] constant), by what factor will the reaction rate
increase?
Answers: (a) rate = v = k [NO]2 [H2] ; (b) third order overall; (c) L2 mol-2 s-1; (d) 4-fold; (e) rate will double.
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