Download Solubility Solubility is defined as the amount of solute that will

Solubility
Solubility is defined as the amount of solute that will dissolve in a given amount of
solution at a certain temperature. Our main measurement of solubility is molarity.
Molarity- molarity is defined as moles of solute per liter of solution.
Molarity (M) =
moles solute
L solution
Example
What is the molarity of 1 liter of a solution that contains 0.5 moles of NaCl.
Answer
To solve we divide the number of moles by the liters of solution.
Molarity (M) =
0.5 moles
= 0.5 M
1L
Example
What is the molarity of a solution that contains 58.45 grams of NaCl in 2 L of solution.
Answer
The first thing we must do is determine how many moles of solute.
58.4 g
1 mole
= 1 mole
58.4 g
We then divide the number of moles by the volume of the solution.
1 mole
= 0.5 M
2L
Exercise
1 What is the molarity of a solution made by dissolving 4.2 grams of CaCl2 in enough
water to make 125 ml of solution?
0.30 M
2. What is the molarity of the chloride ion in the above solution
0.60 M
72
Equilibrium involving insoluble compounds
When certain salts are placed in water very little of the salt dissolves to come to
equilibrium. A small amont of solid dissovles and then dissociates into ions. The following
is a general equilbrium equation for this process.
Mx A y(s) <-------> x M+ y(aq) + y A-x(aq)
Example
Write the equilbrium equation for the addition of solid AgBr to water.
AgBr(s)
Ag+(aq) + Br-(aq)
Exercise
Write the equilibrium equations for the following compounds.
PbI2
PbI2(s)
Pb2+(aq) + 2I-(aq)
A g3PO4
Ag3PO4 (s)
3Ag+(aq) + PO43- (aq)
PbCO3
PbCO3 (s)
Pb2+(aq) + CO 32-(aq)
Al(OH)3
Al(OH)3 (s)
Al3+(aq) + 3 OH-(aq)
73
Ksp calculations.
Ksp is another example of an equilibrium constant. It can be determined like any other
equilibrium constant if the concentration of the products are known. We only need to know
the product concentrations because the reactants are solids.
Example
AgCl(s) is an insoluble compound and when placed in water the chloride concentration was
found to be 1.26 x 10-4M. What is the Ksp of AgCl(s)?
Answer
AgCl(s)
Ag+(aq) + Cl-(aq)
[Ag+] = [Cl-] = 1.26 x 10-4
Ksp = [Ag+] [Cl-] = (1.26 x 10 -4)2 = 1.58 x 10 -8
Exercise
1. What is the solubilty of a compoumd with MW of 128 if after coming to equilibrium we
find 0.23 g/L in the solution.
1.8 x 10-3M
2. A saturated solution of Al(OH)3 had a pH = 9.0 what is the solubility of the compound at
this temperature.
3.3 x 10-16
3. A solution of PbCl2 has a chloride concentration of 1.6 x 10-2M. What is the Ksp of
Lead(II)chloride?
2.0 x 10-6
74
Determination of solubility given the Ksp
When excess solid is placed in water an equilibrium reaction takes place in which the
solid dissociates into ions.
Solid(s)
Cation(aq) + Anion(aq)
The following equilibrium is what would happen when excess solid AgCl is added to water.
AgCl(s)
Ag+(aq) + Cl-(aq)
The solubility of the silver chloride is equal to the concentration of the silver cation or the
chloride anion. The solubility is defined as the concentration of either the silver cation or the
chloride anion.
Solubility(s) = [Ag+] = [Cl-]
If more than two ions are produced the solubility of the compound must take into account
the multple ions.
CxAy(s)
x C+(aq) + yA -(aq)
Solubility (s) = (1/x[)C+] = (1/y)[A -]
The solubility of lead(II) chloride is defined as the following.
PbCl2(s)
Pb2+(aq) + 2Cl -(aq)
Solubility(s) = [Pb2+] = 1/2 [Cl-]
We calculate the concentration of the aqueous species just as we would with any other
equilibrium using an ICE table and the equilibrium expression.
75
Example
What is the solubility of BaCO3? Ksp = 5.1 x 10-9
Answer
We set up the ice table and plug the values into the equilibrium expression. Note that we
us s instead of x, but everything else is the same as before.
Ba2+(aq) +
BaCO3(s)
initial
change
N
CO32+ (aq)
0
.
0
NA
+s
+s
equilibrium NA
s
s
[Ba2 +][CO32-]= s2 = Ksp = 5.1 x 10-9
s = 7.1 x 10-5
Exercise
Determine the solubility of the following compounds using the Ksp values.
CaSO4 Ksp = 9.1 x 10-6
3.0 x 10-3
Al(OH)3 Ksp = 9.1 x 10-6
2.4 x 10-2
HgCl2 Ksp = 9.1 x 10-6
1.3 x 10-2
CaCO3 Ksp = 9.1 x 10-6
3.0 x 10-3
76
Common Ion Effect. (LeChatlier’s)
When a solid is allowed to come to equilibrium in a solution of one of it constituent ions we
call this the common ion effect. As we would expect from LeChatlier’s principle addition of
a product ( the common ion) would shift the equilibrim to the reactant side. An example
would be letting solid silver chloride come to equilibrium in a solution of sodium chloride as
seen in the following example.
AgCl(s)
initial
NA
Ag+(aq) + Cl-(aq)
0
1M
change
NA
+s
+s
equilibrium
NA
s
1M + s
The solution has a chloride ion which is in common with the chloride from the silver chloride.
The net effect is to decrease the solulbility of the silver chloride which is also what we would
predict from LeChatlier’s principle. We solve for the solubility, using the techniques we
learned in the earlier equilibrium section.
Example
What is the solubility of PbCl2 in 0.1M solution of sodium chloride?
PbCl2(s)
Pb2+(aq) +
2Cl-(aq)
NA
0
0.1M
initial
change
NA
+s
+s
equilibrium
NA
s
0.1M + s
We substitute the variables into the equilibrium expression and solve for s.
Ksp = [Pb+2] [Cl-] 2 = [s] [0.1 + s]2 = 1.58 x 10-8
assume s small
-7
s = 1.58 x 10
the asumption that s is small was correct
Precipitation
When two solutions of ions are combined that make an insoluble precipitate it is usefull to
be able to predict if a precipitate will form. To do this we calculate Q as we did earlier and
compare it to Ksp. The following relationships are used to predict precipitation.
77
Q > K precipitate will form
Q = K no precipitate saturated solution
Q < K no precipitate unsaturated solution
Example
25 ml of 2.3 x 10-6 M Pb(NO3)2 was combnined with 25 ml of 1M NaCl. Will a
precipitate form? Ksp =
Answer
Write the solubility equilibrium for the insoluble compund.
PbCl2(s)
Pb2+ (aq) +
2 Cl
-
(aq) .
Ksp = 1.6 x 10-5
Write the Qsp expression.
Qsp = [Pb+2] [Cl -]2
Calculate the concentration of the Ions in the solubility expression using the diultion
equation.
M1V 1 = M2V 2
[Pb2 +] = [(2.3 x10-6M) (25 ml)]/ 50 ml = 1.15 x 10-6
[Cl-] = [(1M) (25 ml)]/ 50 ml = 0.5M
Plug the diluted values into the Q expression.
Qsp = [Pb+2] [Cl -]2 = [(1.5 x 10-6) (0.5) = 7.5 x 10-7
Compare Qsp to Ksp
Qsp < Ksp so a precipitate will not form.
78
Exercise
Determine if a precipitate will form when the following solutions are mixed together.
Assume all concentrations are 1M.
1. 100 ml NaCl and 50 ml Pb(NO3)2 PbCl2 Ksp = 1.6 x 10-5
ppt will form
2. 10 ml NaOH and 5 ml Mg(NO3)2 MgOH2 Ksp = 1.8 x 10-11
ppt will form
3. 20 ml AgNO3 and 20 ml Na2(CO3) Ag2C O3 Ksp = 8.1 x 10-12
ppt will form
4. 15 ml Na2(CO3) and 50 ml PbCO3 PbCl2 Ksp = 1.6 x 10-5
ppt will form
79
Seperation of Ions
We are going to seperate solutions of metal ions based on the differnce in solubility.
There are two types of seperations that will interest us. They are seperation of an ion to a
certain concentration and maximum seperation of an ion. This is done by adding an anion
that will cause the metal cations to precipitate. The less soluble cation will precipitate first
and the more soluble cation will remain in solution. If the concentration of anion is too large
the more soluble ion will precipitate and we will not be able to seperate the cations. The
following are the rules we use for the seperation of the metal cations.
Ion seperation
1. Determine which ion is the most souble by calculating solubility for each ion.
2. Find the concentration of anion that will make Qsp = Ksp for the most soluble ion.
This will be the maximum amount of anion that can be added to give us a maximum
seperation of the less soluble compound.
Example
What concentration of chloride will give the maximum seperation of a 1M solution of Ag+
and 1M Pb2 + ? AgCl Ksp = 1.8 x 10-10 PbCl2 Ksp = 1.6 x 10-5
Answer
We first find that the AgCl is the least soluble and will be the one we can seperate.
We then find the concentration of Cl- that will give Q=Ksp for the PbCl2.
[Pb+ ] [Cl-]2 = 1.6 x 10-5
[1M] [Cl-]2 = 1.6 x 10-5
[Cl-] = 4.0 x 10-3 this is the maximum amount that can be added before the lead
will precipitate.
We can check and see how much silver will be left after precipitation.
[Ag+ ] [Cl-] = 1.8 x 10-10 we can use the chloride values calculated above and
solve for the silver concentration remaining.
1.8 x 10
Ag =
Cl +
-10
-10
1.8 x 10
-8
=
=
4.5
x
10
4.0 x 10-3
That is the amount of Ag+ that is left along with all the lead cation.
80
Exercise
1. What is the concentration of hydroxide that will give the maximum seperation of 0.5M
Fe3 + and 0.5M Cu2 +? Fe(OH)3 Ks p = 1.0 x 10-38 Cu(OH)2 Ks p = 2.2 x 10-20
[OH-] = 2.1 x 10-10
2. What is the concentration of sulfate that will give the maximum seperation of 0.1M Ba2 +
and 0.1M Ag+ ? BaSO4 Ksp = 1.1 x 10-10 A g2S O4 Ksp = 1.4 x 10-5
[SO42-] = 1.4 x 10-3
Percent Seperation
If we want less than maximum seperation we can determine the amount of the less
soluble cation and then calculate the amount of anion required to do the seperation.
To determine the concetration of anion needed to give a certain seperation we do the
following.
i. Determine the concentration of the less soluble cation that will remain after the
desired precipitation is completed.
81
ii. Determine the concentration of anion that will allow this concentration to remain in
solution. Qsp = Ksp
iii. Make sure that this concentration of anion is less than the concentration of anion that
will give the maximum seperation.
Example
What concentration of chloride will give a 90% 1M solution of Ag+ and 1M Pb2 +? AgCl
Ksp = 1.8 x 10-10 PbCl2 Ksp = 1.6 x 10-5
1. Determine concetration of Ag+ that will remain after the seperation.
0.90 x 1M = 0.9 M Amount remaining = 1- 0.90 = 0.1
2. Determine concentration of chloride that will give a final silver ion concentration of 0.1.
[Ag+ ] [Cl-] = 1.8 x 10-10
[0.1M] [Cl-] = 1.8 x 10-10
[Cl-] = 1.8 x 10-9 this is the maximum amount that can be added to leave the silver at a
concentration of 0.1M
3. We will now check and make sure that the lead will not precipitate. Qsp < Ksp
Q = [Pb+ ] [Cl-]2 = [1M] [1.89 x 10-9]2 = 3.6 x 10-18
3.6 x 10-18 < Ksp = 1.6 x 10-5 so no precipitate
Exercise
1. What concentration of sulfate will give a 80 % seperation of a 1M solution of Ba2 + from
a 1M Solution of Ag+ . Ag2S O4 Ksp = 1.4 x 10-5
BaSO4 Ksp = 1.1 x 10-10
[SO42-] = 5.5 x 10-10
82
2. What concentration of carbonate will give a 99 % seperation of a 1M solution of Mg2 +
from a 1M Solution of Mn2 +. MgCO3 Ksp = 1.4 x 10-5 MnCO3 Ksp = 1.1 x 10-10
[CO32-] =1.4 x 10-3
3. What concentration of sulfate will give a 80 % seperation of a 1M solution of Ba2 + from
a 1M Solution of Ag+ . Ag2S O4 Ksp = 1.4 x 10-5
BaSO4 Ksp = 1.1 x 10-10
[SO42-] = 5.5 x 10-10
83
Kinetics
Kinetics- The study of rates of reactions.
What we are going to study is the rate of chemical reactions and things that are
experimentally and theoretically related to rates.
Rate- a rate is defined as an experimentally determined positive quantity, that relates the
change in concentration, of a reactant, or product as a function of time. This is represented
mathematically below.
Rate of change of a reactant
1 change in concentration of reactant
1 final concentration - initial concentration
rate = - (
)=- (
)
n
change in time
n
final time - initial time
Example
In the following chemical reaction the initial concentration of CH3Br was found to be 1M if
after 1.0 min the concentration of CH3Br is 0.45 M what is the rate.
C H3Br + CH3OH -------> CH3OCH3 + HBr
(0.45M - 1.0M) 
1  final concentration - initial concentration 
0.55 M
= - 
 =
rate = - 

 1 min - 0 min 
n
final time - initial time
min
This will give us a average rate because we will find that often the rate will change as the
reaction proceeds. Also you might observe that the negative sign in front of the equation
makes the value positive, as required by the definitition of rate. The rate is the
experimentally determined change in concentration versus the change in time.
Rate of change of a product
rate =
1 change in concentration of product
1 final concentration - initial concentration
(
)= (
)
n
change in time
n
final time - initial time
Notice that it is not necessary to put a negative sign in front, because the quantity will be
positive.
Example
In the reaction above after 2 minutes the final concentration of HBr was found to be 0.7M
what is the rate.
rate =
1 final concentration - initial concentration
0.7M - OM
M
(
) =
= 0.35
n
final time - initial time
2 min - 0 min
min
84
As was stated before the rates determined by these methods are the average rates. We
are often interested in the instantaneous rate which can be determined from a graph or by a
computer program. The graphical method is shown below.
Procedure for determining instantaneous rates.
Step 1
An experiment is done in which the change in concentration of the species of interest is
followed as a function of time. We look at the concentration as time increases.
Data table from a concentration vs time experiment for the decomposition of N2O5.
N2O 5(aq) ------> 4NO2(aq) + O2(aq)
concentration N2O5 (M)
time(s)
1.00
0
0.88
200
0.78
400
0.69
600
0.61
800
0.54
1000
0.48
1200
0.38
1600
0.30
2000
Step 2
The data is graphed as illustrated on the next page.
85
1.0
0.80
conc.
N2O5 0.60
(M) 0.40
.. .
Conc. N2O5 vs Time
....
. .
0.20
400
800
1200
1600
2000
Time (s)
We can find the instantaneous rate by taking a tangent to the point at the desired
concentration, as is listed below for two points.
concentration N2O 5
Rate (M/s)
0.90M
5.4 x 10-4
0.45M
2.7 x 10-4
Excercises
Using the data for the decomposition of N2O 5 answer the following questions.
1 What is the average rate between 800-1200 seconds?
3.3 x 10-4 M/s
2. What is the average rate between 0-400 seconds?
3.25 x 10-4 M/s
3. From examination of the instantaneous rate how does rate change with concentration?
Decreases
86
Rates and rate laws
We are interested in the experimental results but what is of greater interest is how does
the rate vary as a function of concentration. We can graph our experimental data as
instantaneous rate vs concentration and come up with the following sort of graph. I have not
used numbers as I am discussing this in a qualitative manner.
Rate vs concentration
Rate (M/min)
Concentration of reactant (M)
This is a line and we know that the general equation of a line is the following.
y = mx + b
y= vertical axis
x= horizontal axis
m= slope of line ( change in x / change in y)
b= y intercept ( where the line crosses the y axis)
With this in hand we can write an equation that relates the rate of a reaction with the reactant
concentration. This equation that goes with the graph above takes the following form.
rate = k[reactant]
[ ] = concentration brackets
k= rate constant (slope of the line)
87
General Rate Law
Not all rate vs concentration graphs are linear, and to account for this we can write a more
general rate law that has the following form.
rate = k [ ] m
m = order of the reaction.
We will only investigate reactions in which m is equal to 0 , 1 or 2; which are refered to as
zero , first or second order repectively. We can also write a general rate law for reactions
that have fractional or negative orders, but that is a topic that would be dealt with in more
advanced classes. The following graphs show how rate changes with concentration for
each of the orders.
0 order
1st order
Rate
2nd order
Rate
Rate
Concentration
Concentration
Concentation
Experimental Determination of rate laws.
The rate law must be determined experimentally from data collected in the lab. Using
the rate law we can theoretically determine what happen to the rate when the concentration
of reactant is doubled. We can then do experiments in which the concentration of a reactant
is doubled and then determine the order from the change in rate as compared to the
theoretical expectation. The following is a summary of how rate change versus a doubling
in concentration.
Order
Concentration
Change in rate
0
x2
no change
1
x2
x2
2
x2
x2
88
Example
Using the relationships above and the experimental data, determine the order of the
reaction. Use the data to determine the rate constant for the reaction.
N2O 4(g) --------> 2NO2(g)
Experiment
Conc. N2O 4(M)
rate(M/s)
1
0.12
0.34
2
0.24
0.68
3
0.48
1.36
The concentration was doubled in experiment 2 compared to experiment 1. The data
shows that the rate doubled in experiment 2. This shows that the reaction is 1st order with
respect to N2O 4. The rate law can be written as the following.
rate = k[N2O 4]
To determine the rate constant we can use the data from one of the experiments and plug
the values into the rate law and solve for k.
k = rate / [N2O 4] = 0.34 / 0.12 = 2.8 s-1
Exercise
1. Using the experimental data in the following table write the rate law for the
decomposition of Ozone.
2O3(g) -------> 3O2(g)
Experiment
Conc. O3(M)
rate(M/s)
1
0.10
1.10
2
0.20
4.40
3
0.40
17.60
Rate = k[O3]2
k = 110 1/M s
89
2. For the reaction in which A decomposes to B the following data was collected.
A (aq) ------> B(aq)
Experiment
Conc. A (M)
rate(M/s)
1
0.36
0.94
2
0.48
0.94
3
0.72
0.94
Determine the order, rate constant and write the rate law for the reaction.
rate = k
k = 0.94 M/s
Determination of rate laws for reactions with 2 reactants
Like before the rate law of reactions with 2 reactants must be determined by
experiment. To do this we will double the concentration of one reactant and hold the other
reactant constant and determine the order by the change in rate.
Example
The following experimental data was collected for the reaction of hydrogen and nitrogen
monoxide. Determine the order for each reactant and write the rate law. Use the data to
determine the rate constant.
2H2(g) + 2NO(g) ------> N2(g) + 2H2O (g)
Experiment
Conc. H2 (M)
Conc. NO (M)
rate(M/s)
1
0.10
0.10
0.10
2
0.20
0.10
0.20
3
0.10
0.10
0.10
4
0.10
0.20
0.40
90
Answer
We will determine the order of each reactant in the manner described above. If we
compare experiment 1 to experiment 2 we held the concentration of NO constant and
doubled the concentration of H2, and we saw a doubling in the rate. This would indicate that
the reaction is 1st order with respect to H2. .
rate = k [H2] [NO]2
We can use the data from experiment 1 to determine the rate constant for the rate law.
k=
rate
0.10 M/S
=
= 100M-2S-1
[H2 ][NO] [0.1M][0.1M]2
Exercise
1. In a solution at constant H+ concentration, I- reacts with H2O 2 to produce I2 and water.
2H+ (aq) + 2I-(aq) + H2O 2(aq) -----> I2(aq) + 2H2O (l)
Experiment
Conc. I- (M)
Conc. H2O 2 (M)
rate(M/s)
1
0.10
0.10
0.10
2
0.20
0.10
0.20
3
0.10
0.10
0.10
4
0.10
0.20
0.40
Write out the rate law and solve for the rate constant.
rate = k[I-][H2O 2]2 k = 100 1/M2 s
2. The data below are for the reaction of NO with Cl2 to form NOCl.
Experiment
Conc. NO (M)
Conc. Cl2 (M)
rate(M/s)
1
0.050
0.050
1.0 x 10-3
2
0.10
0.050
2.0 x 10-3
3
0.10
0.10
Write out the rate law and solve for the rate constant.
4.0 x 10-3
rate = [NO][Cl2] k = 0.4 1/M s
91
Integrated rate law
The rate law is used to determine the rate of a reaction, for a concentration of a reactant in
the rate law. As we saw earlier rate is defined as a change in concentration divided by a
change in time. We can use some principles of calculus to determine a concentration at
some time, given the initial concentration and the rate law. We will not show how to derive
the integrated rate laws but have summerized the results in the following table.
Order
Rate Law
Integrated rate law
0
rate = K
[ ]t - [ ]o = -kt
1
rate = k[ ]
ln
2
rate = k[ ]2
1
1
= kt
[ ]t
[ ]o
[ ]t
= - kt
[ ]0
Example
A first order reaction had a rate constant of 0.12 s-1. If the initial concentration was 0.25M
how long will it take to reach 0.10M.
Answer
To solve this reaction we will use the first order integrated rate equation, which is
rearanged to solve for time with the given conditionsl.
t = ln
[ ]o
[ ]t
k = ln
0.25
0.12 s-1 =
0.10
We can solve for any variable we choose as long as all the other parameters are known.
Exercise
1. A first order reaction required 25 s to go from an initial concentration of 1M to 0.65M.
What is the rate constant for this reaction.
k = 1.7 x 10-2 s-1
2. The decomposition of HI in the gas phase is a second order reaction. The rate constant
for the reaction was found to be 1.2 x 10-3 M-1s-1. If the initital concentration of HI is
0.36M how much will remain after 45 min.
0.17M
92
3. A first order reaction had a rate constant of 3.45 x 10-6 s-1. How long will it take for a
3.2 x 10-2M solution to go to 1.0 x 10-4 M?
1.7 x 106 s
Half life
The time required for a reaction to reach half of its original concentration is called the half life
of a reaction. If we started with a 1M concentration the time it takes to get to 0.5M would be
called the half life. We can determine the half life for a reaction using the integrated rate law
and the definition of the half life. The following example shows the derivation of the first
order half life.
Example
To derive the first order half life we start with the integrated rate law and use the
relationship 1/2[ ]o = [ ]t at t1/2.
ln
[ ]o
= kt 1/2
[ ]t 1/2
ln
t 1/2 =
[ ]0
1/2[ ] 0 ln 2
0.693
=
k
k
k
Notice that the first order half life is independent of concentration.
Summary of half life equations
Order
Equation
0
t 1/2 =
[ ]0
2k
1
t 1/2 =
0.693
k
2
t 1/2 =
1
k[ ]0
93
Example
A first order reaction had a k = 4.6 x 10-2 s-1 what is the half life for this reaction.
Answer
t 1/2 =
0.693
0.693
=
= 15 s
k
4.6 x10-2 s-1
Exercise
1. If a 2nd order reaction had a half life of 265 s what is k for this reaction.
k = 3.74 1 / s [ ]o
2. The rate constant for the 1st order decomposition of N2O 5 was found to be 0.35 min-1.
How long will it take for a 0.80M concentration of N2O 5 to decompose to 0.20M?
4.0 min
3 A 0 order reaction had a half life of 3.7 hours using the integrated rate law, how long it will
take 0.30 atm to react to leave 0.2 atm of the reactant left. ( calculate k from the half life)
94
Linear plots
The integrated rate laws can be used to determine the order of a reaction graphically by
writting the rate law in a linear form. When the data is graphed the linearity of the graph will
tell us the order. The following is the linear form for the 0 order reaction.
[ ]0 - [ ]t = kt
Rearranging the equation into standard y = mx +b linear form gives the following equation.
[ ]t = - kt + [ ]0
y is the concentration at some time t and x is time t. The slope is the rate constant, and the
intercept is the initial concentration. This means that if we graphed our data with the
concentration vs time and the graph was liner it would be indicative of a 0 order reaction.
[ ]
If the data gave us this graph it would be a 0 order
reaction,
time
The following is a summary of facts relating to rate laws, including what type of graph will
give a linear plot.
Order
Rate law
Integrated rate law
0
rate = k
[ ]o - [ ]t = kt
1
rate = k[ ]
ln
2
rate = k[ ]2
t1/2
linear graph
t 1/2 =
[ ]0
2k
[ ] vs time
[ ]o
= kt
[ ]t
t 1/2 =
0.693
k
ln [ ] vs time
1
1
= kt
[ ]t
[ ]o
t 1/2 =
1
k[ ]0
1
vs time
[]
95
Exercise
Using the data from the following table. graph the data and determine the order from the
linear graph. Use the slope of the linear graph to detemine the rate constant and then write
the complete rate law for the decomposition of N2O 5.
Data table from a concentration vs time experiment for the decomposition of N2O 5.
N2O 5(aq) ------> 4NO2(aq) + O2(aq)
concentration N2O5 (M)
time(s)
1.00
0
0.88
200
0.78
400
0.69
600
0.61
800
0.54
1000
0.48
1200
0.38
1600
0.30
2000
1 st order
Activation energy
Reactions involve the breaking and making of chemical bonds. The breaking of bonds
require the input of energy which is provided by the collision of molecules. For the reaction
to happen, a certain minimum amount of collision energy is required for the reaction to
proceed. This minimum energy is called the activation energy (Ea). The following are
some characteristics of activation energy.
1. Ea is a positive quantity.
2. Ea is independent of temperature and concentration.
3. Ea is dependent on the type of reaction. For any given reaction we find that activation
energy will always be the same and will be greater for a slow reaction and smaller for a fast
reaction.
96
Reaction Rate and Temperature
Experimentally we find that reaction rate increases with temperature, and as a general
and approximate rule for every 10o C increase in temperature the rate will double. The
following equation relates k and temperature.
-E a
k = Ae RT
R = 8.314 J/ Mole K
T in K
A = pre exponential factor
For this equation to be useful we must know Ea and A, which must be determined by
experiment. If we rearange the equation into a linear form this will allow us to use graphical
methods along with experiment to determine A and Ea. The following is the the equation
rearranged in linear form.
ln k =
-E a 1
+ ln A
R T
If we graph the ln of k versus I/ T we can determine activation energy and the
preexponential factor A. Ea will be the negative slope of the line multiplied by R, and the
ln of A will be the y intercept.
ln A
slope =
-Ea
R
ln k
1/T
97
Exercise
1 The following data are for the gas-phase decomposition of acetaldehyde.
 1 
k

 M s
T (K)
0.0105
0.101
0.60
2.92
700
750
800
850
Graph the data and determine the activation energy and the preexponential factor.
2. The following data were obtained for the reaction
SiH4(g) -----> Si(s) + 2H2(g)
k (s-1)
0.048
T (K)
500
2.3
49
590
600
700
800
Graph the data and determine the activation energy and the preexponential factor.
We can also use the linear relationship to determine how the rate constant changes with
temperature. We can with a little mathematics write the equation in the following form.
k2 Ea  1
1 
ln
=
−
k1 R  T1 T2 
Example
A certain reaction had an activation energy of 9.3 x 104 J. At 27oC, k = 1.25 x 10-2 1/ M s.
Calculate k at 127oC.
Answer
Using the equation above we can write the following.
ln
k2 9.32 x 104  1
1 
=

−
 = 4.06
k1
8.314  300 400 
98
k2
= e 4.06 = 1.15 x 10 4
k1
k2 = 1.15x10 4 k1 = (1.15 x10 4 )(1.25 x10 -4 ) = 1.44 x 10 2
1
Ms
1. A reaction had an activation energy of 6.2 x 103J. k was measured to be 1.5 x 10 -1s-1
at 100C. The temperature was increased by 10oC to 110oC. What is the value of the
rate constant at the higher temperature.
k2
6.2 x 10 3  1
1 
ln
=

−

.15
8.314  383 373
k2 = 0.14
2. At 200oC a reaction had a k= 1.2 x 10-1 s-1, and at 250 oC the k was found to be
3.8 s-1. What is the activation energy for this reaction.
ln
3.8
Ea  1
1 
=

−
 Ea = 1.4 x 10-5
.12
8.314  473 523
99