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Individual and Overall Reaction Orders
For the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g):
The rate law is rate = k[NO]2[H2]
The reaction is second order with respect to NO, first
order with respect to H2 and third order overall.
Note that the reaction is first order with respect to H2
even though the coefficient for H2 in the balanced
equation is 2.
Reaction orders must be determined from experimental
data and cannot be deduced from the balanced
equation. Dr. N. Singh

PLAN: We inspect the exponents in the rate law, not the coefficients
of the balanced equation, to find the individual orders. We add
the individual orders to get the overall reaction order.
SOLUTION:
(a) The exponent of [NO] is 2 and the exponent of [O2] is 1, so the
reaction is second order with respect to NO, first order
with respect to O2 and third order overall.
Determining Reaction Orders from Rate Laws
PROBLEM2: For each of the following reactions, use the give rate law
to determine the reaction order with respect to each
reactant and the overall order.
(a) 2NO(g) + O2(g) → 2NO2(g); rate = k[NO]2[O2]
(b) CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2
(c) 3
H2O2(aq) + 3I-(aq) + 2H+(aq) →I -(aq) + 2H2O(l); rate = k[H2O2][I-]
Dr. N. Singh

Problem 2
(b) The reaction is 3 order in CH3CHO and 3 order overall.
2 2
(c) The reaction is first order in H2O2, first order in I-, and
second order overall. The reactant H+ does not appear in
the rate law, so the reaction is zero order with respect to H+.
Dr. N. Singh

Determining Reaction Orders
For the general reaction A + 2B → C +D,
the rate law will have the form
Rate = k[A]m[B]n
Todetermine the values of m and n, we run a series of
experiments in which one reactant concentration
changes while the other is kept constant, and we
measure the effect on the initial rate in each case.
Dr. N. Singh

Table 2 Initial Rates for the Reaction between A and B
Initial Rate Initial [A] Initial [B]
Experiment (mol/L·s) (mol/L) (mol/L)
1 1.75x10-3 2.50x10-2 3.00x10-2
2 3.50x10-3 5.00x10-2 3.00x10-2
3 3.50x10-3 2.50x10-2 6.00x10-2
4 7.00x10-3 5.00x10-2 6.00x10-2
[B] is kept constant for experiments 1 and 2, while [A] is doubled.
Then [A] is kept constant while [B] is doubled.
Dr. N. Singh

Rate 2
Rate 1
=
k[A] m [B]n
2 2
1
k[A] m
[B]n
1
Finding m, the order with respect to A:
We compare experiments 1 and 2, where [B] is kept
constant but [A] doubles:
=
[A]
m
2
m
[A]1
=
[A]2
m
3.50x10-3 mol/L·s
1.75x10-3mol/L·s
=
5.00x10-2 mol/L
2.50x10-2 mol/L
[A]1
m
Dividing, we get 2.00 = (2.00)m so m = 1
Dr. N. Singh

Rate 3
Rate 1
=
k[A] m [B]n
3 3
1
k[A] m
[B]n
1
Finding n, the order with respect to B:
We compare experiments 3 and 1, where [A] is kept
constant but [B] doubles:
=
[B]
n
3
n
[B]1
=
[B]3
n
3.50x10-3 mol/L·s
1.75x10-3mol/L·s
=
6.00x10-2 mol/L
3.00x10-2 mol/L
[B]1
m
Dividing, we get 2.00 = (2.00)n so n = 1
Dr. N. Singh

Table 3 Initial Rates for the Reaction between O2 and NO
O2(g) + 2NO(g) → 2NO2(g) Rate = k[O2]m[NO]n
Initial Rate
Initial Reactant
Concentrations (mol/L)
Experiment (mol/L·s) [O2] [NO]
1 3.21x10-3 1.10x10-2 1.30x10-2
2 6.40x10-3 2.20x10-2 1.30x10-2
3 12.48x10-3 1.10x10-2 2.60x10-2
4 9.60x10-3 3.30x10-2 1.30x10-2
5 28.8x10-3 1.10x10-2 3.90x10-2
Dr. N. Singh

Rate 2
Rate 1
= 2 2
2 1
k[O ] m
[NO]n
1
Finding m, the order with respect to O2:
We compare experiments 1 and 2, where [NO] is kept
constant but [O2] doubles:
=
k[O2] m [NO]n [O2] m
[O ]m
2 1
2 =
[O ]
2 2
[O ]
2 1
m
6.40x10-3 mol/L·s
=
2.20x10-2 mol/L
3.21x10-3mol/L·s 1.10x10-2 mol/L
Dividing, we get 1.99 = (2.00)m or 2 = 2m, so m = 1
The reaction is first order with respect to O2.
m
Dr. N. Singh

Sometimes the exponent is not easy to find by inspection.
In those cases, we solve for m with an equation of the form
a = bm:
log a
=
log 1.99
log b log 2.00
m = = 0.993
This confirms that the reaction is first order with respect to O2.
Reaction orders may be positive integers, zero, negative integers, or
fractions.
Dr. N. Singh

Finding n, the order with respect to NO:
We compare experiments 1 and 3, where [O2] is kept
constant but [NO] doubles:
Rate 3
=
[NO]3
Rate 1 [NO]1
n
=
12.8x10-3 mol/L·s 2.60x10-2 mol/L
3.21x10-3mol/L·s 1.30x10-2 mol/L
n
Dividing, we get 3.99 = (2.00)n or 4 = 2n, so n =2.
The reaction is second order with respect to NO.
The rate law is given by: rate = k[O2][NO]2
log a
=
log 3.99
log b log 2.00
n = = 2.00
Alternatively:
Dr. N. Singh

Sample Problem 3 Determining Reaction Orders from Rate Data
PROBLEM: Many gaseous reactions occur in a car engine and exhaust
system. One of these reactions is
NO2(g) + CO(g) → NO(g) + CO2(g) rate = k[NO2]m[CO]n
Use the following data to determine the individual and
overall reaction orders:
Experiment
Initial Rate
(mol/L·s)
Initial [NO2]
(mol/L)
Initial [CO]
(mol/L)
1 0.0050 0.10 0.10
2 0.080 0.40 0.10
3 0.0050 0.10 0.20
Dr. N. Singh

Sample Problem 3
PLAN: We need to solve the general rate law for m and for n and
then add those orders to get the overall order. We proceed by
taking the ratio of the rate laws for two experiments in which
only the reactant in question changes concentration.
rate 2
rate 1 [NO ]
2 1
k [NO ]m [CO]n [NO ]
2 2 2
= 2 2
1
k [NO2]m [CO]n
1
=
0.080 0.40
0.0050 0.10
=
m
SOLUTION:
Tocalculate m, the order with respect to NO2, we compare
experiments 1 and 2:
m
16 = (4.0)m so m = 2
The reaction is second order in NO2.
Dr. N. Singh

k [NO ]m [CO]n [CO] 3
[CO]1
2 3 3
=
k [NO2]m [CO]n
1 1
n
rate 3
rate 1
= =
0.0050 0.20
0.0050 0.10
n
1.0 = (2.0)n so n = 0
The reaction is zero order in CO.
rate = k[NO2]2[CO]0 or rate = k[NO2]2
Sample Problem 3
Tocalculate n, the order with respect to CO, we compare experiments
1 and 3:
Dr. N. Singh

Sample Problem 4 Determining Reaction Orders from Molecular
Scenes
PROBLEM: At a particular temperature and volume, two gases, A (red)
and B (blue), react. The following molecular scenes
represent starting mixtures for four experiments:
(a) What is the reaction order with respect to A? With respect to B?
The overall order?
(b) Write the rate law for the reaction.
(c) Predict the initial rate of experiment 4.
PLAN: We find the individual reaction orders by seeing how a change
in each reactant changes the rate. Instead of using
concentrations we count the number of particles.
Expt no:
Initial rate (mol/L·s)
1
0.50x10-4
2
1.0x10-4
3
2.0x10-4
4
?
Dr. N. Singh

Sample Problem 4
SOLUTION:
(a) For reactant A(red):
Experiments 1 and 2 have the same number of particles of B, but
the number of particles of A doubles. The rate doubles. Thus the
order with respect to A is 1.
For reactant B (blue):
Experiments 1 and 3 show that when the number of particles of B
doubles (while A remains constant), the rate quadruples. The
order with respect to B is 2.
The overall order is 1 + 2 = 3.
(b) Rate = k[A][B]2
(c) Between experiments 3 and 4, the number of particles of A
doubles while the number of particles of B does not change. The
rate should double, so rate = 2 x 2.0x10-4 = 4.0x10-4mol/L·s
Dr. N. Singh

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