Download A ---> B

Ch 17: Kinetics Pt 1
Dr. Harris
Lecture 18
HW: Ch 17: 5, 11, 18, 23, 41, 50
Reactions Rates
 Chemical kinetics is the area of chemistry that investigates how fast
reactions occur
 Different reactions proceed with different rates
 The rate of a reaction depends on several factors, including:
 reactant concentration
 temperature
 catalysts
 surface area
 Today, we will focus exclusively on the relationship between reaction rates
and reactant concentration
Intro
 Lets take the reaction: A ---> B.
 This reaction tells us that as A is consumed, B is formed at an equal
rate. We can express this mathematically in terms of changing
concentrations by:
−
∆𝐴
∆[𝐵]
=
∆𝑡
∆𝑡
 Imagine we have 10 moles of A in 1 L of solution. If we can freeze time
for an instant, such that the reaction has not yet begin (t=0), the
concentration of A is 10M.
A
[A] = 10 M
[A] = 7 M
[B] = 3 M
B
After 10 seconds, 3 moles
of B have formed.
t=0
= 1 mol of A
= 1 mol of B
t = 10
t = 40
40 more
seconds
t = 80
[A] = 3 M
[B] = 7 M
10 more
seconds
20 more
seconds
[A] = 4 M
[B] = 6 M
t = 20
[A] = 5 M
[B] = 5 M
A ---> B
10
Plotting the data
from previous slide
Concentration (Mol/L)
8
6
A
B
4
2
0
0
10
20
30
40
Time (sec)
50
60
70
80
Reactions Follow a Rate Law
 The graph in the previous slide shows that the disappearance of A
(formation of B) is not linear.
 As [A] decreases, the reaction slows down. This means that reaction
rates depend on reactant concentration.
 This dependence of rate on concentration suggests that reaction rates
follow a rate law, a mathematical expression that ties concentration
and rate together
Instantaneous Rates
• Instantaneous rate at t=0 is the
initial rate
• We can determine the instantaneous
rate by taking the slope of the
tangent at the point of interest
A ---> B
10
Concentration (Mol/L)
• Although the rate of the reaction is
constantly changing with reactant
concentration, we can determine the
instantaneous rate (reaction rate at a
specific time and concentration)
8
6
A
4
2
Tangent at t = 20s, [A] = 5M
0
0
• Note: a tangent line is linear and
ONLY touches the point in question.
It does NOT cross the curve
Instantaneous rate of disappearance of
A at t=20 sec
𝑟𝑎𝑡𝑒20 = −
50
Time (sec)
7.25 − 2.30 𝑀
𝑀
= −.124
0 − 40 𝑠
𝑠
Rates and Stochiometry
 In the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any
given time, the rate of disappearance of A equals the rate of formation of
B. If the stoichiometry is NOT 1:1, we have a much different situation, as
shown below:
2𝐻𝐼 𝑔 → 𝐻2 𝑔 + 𝐼2 (𝑔)
 As you can see, 2 moles of HI are consumed for every 1 mole of H2 and 1
mole of I2 formed. Thus, the disappearance of HI is twice as fast as the
appearance of the products.
Example: N2O5(g) ----> 2NO2(g) + ½ O2(g)
Looking at average rates
average rate of
disappearance
after 10 minutes
.0092 − .0124 𝑀
−
10 𝑚𝑖𝑛
𝑀
−𝟒
= 𝟑. 𝟐 𝒙 𝟏𝟎
𝑚𝑖𝑛
6.4 𝑥 10−4
𝑀
𝑚𝑖𝑛
10−4
𝑀
𝑚𝑖𝑛
1.6 𝑥
average rate of
disappearance after
100 minutes
.0006 − .0124 𝑀
−
100 𝑚𝑖𝑛
𝑀
= 𝟏. 𝟏𝟖 𝒙 𝟏𝟎−𝟒
𝑚𝑖𝑛
2.38 𝑥
10−4
𝑀
𝑚𝑖𝑛
0.59 𝑥
10−4
𝑀
𝑚𝑖𝑛
N2O5(g) ----> 2NO2(g) + ½ O2(g)
slow
fast
Rate Laws
 We see that reducing reactant concentration lowers the reaction rate, but
to what extent? What is the mathematical correlation?
 The equation that relates the concentration of the reactants to the rate of
reaction is called the rate law of the reaction.
 We can derive the rate law of a reaction by seeing HOW THE REACTION
RATE CHANGES WITH REACTANT CONCENTRATION.
 For any reaction aA + bB ----> cC + dD
𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]𝒎 [𝑩]𝒏
 In this expression, k is the rate constant, m and n are reaction orders.
Reaction Orders and the Method of Initial Rates
 Lets go back to the previous reaction: N2O5(g)
2NO2(g) + ½ O2(g)
 Below is a table of data, showing the initial reaction rate as a function of the
starting concentration of N2O5 (g). We perform multiple experiments to collect
enough data to determine our rate law.
Experiment
[N2O5]o (M)
Rate, M/s
1
0.010
.018
2
0.020
.036
3
0.040
.072
 We see that when we double [N2O5]o, the rate also doubles. When we quadruple
[N2O5]o, the rate quadruples. Thus, the rate is directly proportional to [N2O5]o by
the rate constant, k.
 This means that the reaction is FIRST ORDER WITH RESPECT TO [N2O5] (m=1).
We can write the rate law as: 𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝟐𝑶𝟓]𝟏 = 𝒌[𝐍𝟐𝐎𝟓]
Reaction Orders
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝟐𝑶𝟓]𝟏 = 𝒌[𝐍𝟐𝐎𝟓]
Run
[N2O5]o (M)
Rate, M/s
1
0.010
.018
2
0.020
.036
3
0.040
.072
 The overall reaction order is the sum of the individual reaction orders. In
our previous example, there was only one reactant, so the overall order is 1
(1st order reaction).
 We can easily solve for k by plugging in any corresponding rate and
concentration. Lets plug in the values from run # 1
.018
𝑀
= 𝑘(.010 𝑀)
𝑠
𝑘 = 1.8 𝑠 −1
Rate Laws/Reaction Orders
𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]𝒎 [𝑩]𝒏
• Reaction orders must be determined experimentally. You can not
assume based on the stoichiometry.
• When you have multiple reactants, you must determine the reaction
order of each one. To do this, you must vary the concentration of
only one reactant at a time while holding the others fixed.
• Let’s attempt to determine the rate law for the reaction below:
2NO(g) + O2(g) ---> 2NO2
Example: 2NO(g) + O2(g) ---> 2NO2
 Using the data below, determine the rate law of this reaction in the form:
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝒎 [𝑶𝟐 ]𝒏
Experiment
[NO]o (M)
[O2]o (M)
Rate (M/s)
1
.0126
.0125
2.82 x 10-2
2
.0252
.0250
1.13 x 10-1
3
.0252
.0125
5.64 x 10-2
• This time, we have two reactants. Lets start by determining the value of
‘m’. To do so, we hold [O2]o fixed and vary [NO]o. This will show how the
rate depends on [NO]o.
• In experiments #1 and #3, [O2]o is fixed, so we will use these to find ‘m’.
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝒎 [𝑶𝟐 ]𝒏
Experiment
[NO]o (M)
[O2]o (M)
Rate (M/s)
1
.0126
.0125
2.82 x 10-2
2
.0252
.0250
1.13 x 10-1
3
.0252
.0125
5.64 x 10-2
• Remember, rate is proportional to [NO] by the power m. The factor of
change in the rate is equal to the factor of change of [NO] to the mth
power:
order
𝑚
5.64 𝑥 10−2
.0252
=
2.82 𝑥 10−2
.0126
factor of
rate change
m=1
• The reaction is 1st order with respect to [NO]
factor of change
in [NO]
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝒎 [𝑶𝟐 ]𝒏
Run
[NO]o (M)
[O2]o (M)
Rate (M/s)
1
.0126
.0125
2.82 x 10-2
2
.0252
.0250
1.13 x 10-1
3
.0252
.0125
5.64 x 10-2
• Now we can find ‘n’ by varying [O2]o and holding [NO]o fixed. We can use
experiments #2 and #3 for this. This will show how the rate depends on [O2]o.
factor of
rate change
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝟏 [𝑶𝟐 ]𝟏
5.64 𝑥 10−2
.0125
=
1.13 𝑥 10−1
.025
n=1
𝑛
order
factor of change in [O2]
The reaction is 1st order with respect
to [O2] and 2nd order overall.
Pay Attention to the Units of k, As They Change with
Overall Reaction Order
 The rate constant, k, is the constant of proportionality between rate and
concentration.
 Higher values of k = faster reactions
 It is important to note that the units of k depend on the overall reaction
order.
𝟏 𝑵𝑶− 𝟏
 Ex: 𝑹𝒂𝒕𝒆 = 𝒌 𝐍𝑯+
𝟒
𝟐
 Rate is always in molarity per unit time (sec, hr, etc). Concentration is
always M (mol/L). Thus, we have:
𝑀
1
=𝑘 𝑀 𝑀 →𝑘=
= 𝑀 −1 𝑠 −1
𝑠
𝑀𝑠
 Recall for a
1st
order reaction: 𝑘 =
𝑠 −1
Units of k for a 2nd
order reaction
Units of k for a 1st
order reaction
Example
𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]𝒎 [𝑩]𝒏
 Determine the relative (m & n) and overall (m+n) reaction order of the
reaction below. Then, derive the rate law and determine the value of k
(with correct units)
N𝑂2 + 𝐶𝑂 → 𝑁𝑂 + 𝐶𝑂2
Experiment
[NO2]o (M)
[CO]o (M)
Rate (M/s)
1
.0300
.200
1 x 105
2
.0900
.200
9 x 105
3
.300
.0400
1 x 107
4
.300
.0800
1 x 107
• Tripling [NO2] causes the rate to increase nine-fold. This means that the
rate is proportional to the square of [NO2], so the reaction is second order
with respect to NO2 (n=2). Doubling [CO] does nothing. Thus, the rate does
not depend on [CO], and is zero order with respect to CO (m=0). Overall 2nd
order.
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶𝟐 ]𝟐 [𝑪𝑶]𝟎 → 𝒌 [𝑵𝑶𝟐 ]𝟐
k = 1.11 x 108 M-1s-1
Example: 2NO(g) + Br2(L) ---> 2NOBr (g)
• Using the information below, determine the rate law of this reaction in the form:
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝒎 [𝑩𝒓𝟐 ]𝒏
Experiment
[NO]o
[Br2]o
Rate (M/s)
1
0.10
0.20
24
2
0.25
0.20
150
3
0.10
0.50
60
1.) Find m. We can use runs 1 & 2:
150
.25
=
24
.10
2.) Find n. We can use runs 1 & 3
60
.50
=
24
.20
𝑚
→ 6.25 = 2. 5𝑚
m=2
𝑚
→ 2.5 = 2.5𝑛
n=1
𝑹𝒂𝒕𝒆 = 𝒌 [𝑵𝑶]𝟐 [𝑩𝒓𝟐 ]𝟏
• The reaction is 2nd order with respect to [NO], 1st order with respect to
Br2, and the reaction is overall 3rd order.
Determining the Overall Rate Order of A Reaction
Graphically
 As we have shown, a first-order reaction depends on the concentration of
a single reactant to the 1st power. For the reaction: A----> products
𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]
 Using calculus, we can convert this to:
𝐥𝐧[𝑨]𝒕 = −𝒌𝒕 + 𝐥𝐧[𝑨]𝒐
natural log of
concentration at
time t
y axis
rate
constant
m (slope)
time
x axis
natural log of starting
concentration
b
 This equation is in y = mx + b form. Therefore, for any 1st order reaction,
the plot of the natural log of [A]t vs time will be linear. The slope of the
line will be –k.
Plotting 1st Order Reactions
𝐥𝐧[𝑨]𝒕 = −𝒌𝒕 + 𝐥𝐧[𝑨]𝒐
b
slope = -k
units: s-1
time values on x-axis
natural log of [A]t on y-axis
Determining the Overall Rate Order of A Reaction
graphically
 A second-order reaction depends on the concentration of [A] to the 2nd
power. For the reaction: A ----> B
𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]𝟐
y
𝟏
𝟏
= 𝒌𝒕 +
[𝑨]𝒕
[𝑨]𝒐
m•x
b
 Therefore, for any 2nd order reaction, the plot of the inverse of [A]t vs time
will be linear. The slope of the line will be k.
Plotting a 2nd Order Reaction
𝟏
𝟏
= 𝒌𝒕 +
[𝑨]𝒕
[𝑨]𝒐
b
slope = k
units = M-1 s-1
time values on x-axis
1/[A]t on y-axis
Determining Overall Rate Order From Plotting TimeDependent Data
𝐥𝐧[𝑨]𝒕 = −𝒌𝒕 + 𝐥𝐧[𝑨]𝒐
not linear:
NOT 1st order
𝟏
𝟏
= 𝒌𝒕 +
[𝑨]𝒕
[𝑨]𝒐
linear!
2nd order
• We can determine if a process is first or second order by plotting the
data against both equations. Which ever fitting method yields a linear
plot gives the overall order.