Download Using thermodynamics, we can predict whether or not a reaction will

UNIT 15 – KINETICS AND EQUILIBRIUM©
Using thermodynamics, we can predict whether or not a reaction will take place at a
given temperature. If the Grxn is negative, it is spontaneous at that temperature. If the
Grxn is positive, it is not spontaneous at that temperature. HOWEVER, THE Grxn
DOES NOT GIVE US ANY INFORMATION AS TO HOW FAST THE REACTION WILL
OCCUR.
The study of reaction rate is called KINETICS.
For example, given the reaction: hydrogen gas reacts with oxygen gas to form liquid
water, the Grxn at 25oC is -472 kJ, meaning the reaction is spontaneous, but nothing
indicates that it takes approximately 30 years for 1.0 gram of water to form. The
reaction is extremely slow, and the study of kinetics will enable us to predict how fast
reactions will take place.
A reaction rate which is too fast and out of control is called an explosion.
THERE ARE 6 FACTORS WHICH INFLUENCE REACTION RATE:
1. THE TYPE OF REACTION will determine to some extent how fast you might predict
it to be.
(1) Double replacement reactions are very, very fast because the ions are
already formed and simply have to be exchanged which can be done very
rapidly ; nothing changes oxidation number in the equation
(2) On the other hand, synthesis, decomposition and single replacement
reactions are usually slower because at least two elements in each of
these reactions must change oxidation numbers. These types of
reactions are known as REDOX, which is a combination of reduction
(gain of electrons) and oxidation (loss of electrons).
An easy way to remember what oxidation is and what reduction is uses
the following silly sentence "LEO the lion says GER"
LEO (acronym for loss of electrons is oxidation) and GER (acronym for
gain of electrons is reduction). If you don't like this example, consider
using OIL RIG (oxidation is loss, reduction is gain).
Example 15-1 In the synthesis of iron (II) sulfide from iron and sulfur, what is oxidized
and what is reduced? Is there a spectator?
Remember that only the ELEMENT which changes oxidation number is oxidized
or reduced. If something in the equation does not change oxidation number, it is called
a spectator.
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Example 15-2 In the decomposition of mercury (II) oxide into its elements, what is
oxidized and what is reduced? Is there a spectator?
Example 15-3 When a piece of magnesium is dropped into silver nitrate solution, what
is oxidized and what is reduced? Is there a spectator?
2. TEMPERATURE also affects reaction rate. Since temperature is a measure of
kinetic energy, it is reasonable that an increase in temperature causes all
particles to have more kinetic energy and since they are moving around
faster, there is a greater probability of their bumping into each other and
reacting (faster). EACH 10o C RISE IN TEMPERATURE
APPROXIMATELY DOUBLES THE REACTION RATE.
Example 15-4 If the temperature changes from 20oC to 40oC, how much faster does
the reaction go?
If the temperature changes from 15oC to 95oC, what happens to the
reaction rate?
If the temperature changes from 21oC to 58oC, how much faster does the
reaction go?
How to determine the change in reaction rate due to temperature changes:
(1) Find the range of temperature
(2) Divide the range by 10 since each 10 degree increment causes the temp to
double
(3) raise the number 2 to the power of the result you got in #2
Example 15-5 What happens to the reaction rate is the temperature changes from
51.7oC to 212oC?
3. THE CONCENTRATION OF THE REACTANTS, THE RATE LAW, AND THE
RATE CONSTANT (which you will learn to calculate) determine how fast a
reaction will proceed.
Consider the following example:
A +
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B → AB
What do you think would happen if you doubled the concentration of A? There would
be twice as much opportunity for an "A" to find a "B" and react with it to form AB,
If the concentrations of both A and B were doubled, there would be a much greater
opportunity for A to "find" B and react with it.
An increase in concentration of one or both reactants definitely increases the reaction
rate---just exactly how much is determined by the RATE LAW which is
calculated and written from experimental data.
The rate law for the above equation will take the following form
rate of formation of AB =
k (rate constant) [A]x [B]y (square brackets
around a number in
chemistry means
"molarity" or "molar
solution".)
Notice that the rate law which governs the speed of a reaction is based on THREE
things:
(1) the concentration (molarity) of each of the reactants
(2) the power to which each of these reactants is raised
(3) the value of the k or rate constant (which is different for every different
equation.)
The powers to which the concentration of A and the concentration of B are raised are
calculated from experimental data, and the rate constant is also calculated. These
powers are called orders.
For example, if the exponent of A were found to be 2, we would say that the reaction is
2nd order in A; if the exponent of B were found to be 3, we would say that the reaction
is 3rd order in B.
The overall reaction order is the sum of the orders of all of the reactants. If the order of
A is 2 and the order of B is 3, then the overall reaction order is 5.
Example 15-6 Given the following data and experimental data, 2 A + 3 B → 3 C
1.
2.
3.
4.
5.
[A]
[B]
rate at which C is formed (M/sec)
0.050
0.10
0.20
0.050
0.050
0.050
0.050
0.050
0.10
0.20
4.66 X 10-4 M/sec
1.865 X 10-3 M/sec
7.46 X 10-3 M/sec
3.73 X 10-3 M/sec
2.984 X 10-2 M/sec.
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Find the order of A, the order of B, the overall reaction order and the value of k.
How to calculate the orders of each reactant:
(1) To find the order of "A", find two experiments were A changes but B stays the
same.
(2) Divide the molarity of A in one of the experiments by the molarity of A in the
other experiment and raise this value to the "x" power. In the above-given
data, you might choose Experiments 1 and 2 or Experiments 1 and 3 or
Experiments 2 and 3.
(3) Set this factor in the above step equal to the rate of formation of C in the first
experiment you used divided by the rate of formation of C in the second
experiment (the same two experiments you used in A)
(4) Use logarithms to solve for X. This value will usually be a whole number
(integer) or 0 in first year chemistry, but it is possible for it to be a mixed
number or even negative.
(5) Repeat the process again to find the order of B. This time, find two
experiments where the value of B changes, but A remains the same.
This is true in Experiments 1 and 4 or Experiments 1 and 5 or
Experiments 4 and 5. Divide the molarity of B in Experiment 1 by the
molarity of B in Experiment 4 and raise the result to a power "Y".
(6) Now set this factor equal to the rate of formation of C in Experiment 1 divided
by the rate of formation of C in Experiment 4 (the same two experiments
you used in B)
(7) Use logarithms to solve for Y
How to calculate the 'k" or rate constant.
(1) You must have already calculated the orders with respect to both A and B
before you can calculate the "k"
(2) Chose any one of the experiments listed in the data and fill this data into the
rate law you have written raising each to the power (order) you have
calculated for each.. Solve for k and BE CAREFUL OF THE LABEL!
Finding the value of "k" is very valuable because this value of "k" NEVER CHANGES
until you change the reaction as long as the temperature remains constant. Once you
know the rate law orders and the value of "k", you can predict the rate a reaction will
take place at ANY concentrations you choose without having to go to the lab and
experimentally do this (and hope to goodness that you don't get blown up in the
process).
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Example 15-7 Given the reaction in Example 15-6, you have just inherited some
money and have the opportunity to buy lots more A and B. You
know that increasing the concentration of the reactants will
definitely make the reaction speed up, so you decide to increase
the molarity of A to 6.0 M and the molarity of B to 8.0 M. What is
the rate of the reaction at these new concentrations?
Example 15-8 Given the following equation and rate data
1
2
3
4
A + 2B →
[A]
[B]
rate at which C is formed (M/min)
0.250
0.750
0.500
0.250
0.050
0.100
0.100
0.150
1.708 X 10-6 M/min
3.69 X 10-4 M/min
1.094 X 10-4 M/min
4.615 X 10-5 M/min
C
Find the orders of A and B, write the rate law and find the value of k (with label).
(b) What would be the rate of the reaction if the concentration of A were
changed to 1.65 M and the concentration of B changed to 0.33M?
It is important that you understand what these orders mean. If you have a
reactant which is 1st order and you double the concentration, 1st order
means that it would also cause the reaction rate to double. If you have a
reactant which is 2nd order and you double the concentration, 2nd order
means that it would cause the reaction rate to be 22 faster or 4 times faster.
If you have a reactant which is 3rd order and you double the
concentration, 3rd order means that it would cause the reaction rate to be
23 or 8 times faster.
Since the reactants are multiplied together in the rate law, if you have two
reactants, both of which are 1st order and their concentrations are
doubled, it would cause the reaction rate to be [2]1 times [2]1 or 4 times
faster. If you have two reactants, one of which is 1st order and the
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other is 2nd order and you double the concentration of both, it would cause
the reaction rate to be [2]1 times [2]2 or 8 times faster. And if you have two
reactants, both of which are 2nd order and each of their concentrations are
doubled, it would cause the reaction rate to be [2]2 times [2]2 or 16 times
faster.
WHAT DO YOU DO IF THERE ARE 3 REACTANTS?
Example 15-9 Given the following equation and experimental data: A + B + C → Prod.
1`
2
3
4
[A]
[B]
[C]
rate products form
0.50
0.50`
0.75
0.50
0.50
0.25
0.25
0.50
0.50
0.50
0.50
0.25
0.6875 M/sec
0.085938 M/sec
0.193359 M/sec
0.34375
Write the rate law and find the value of "k".
WHAT IF ONE OF THE ORDERS IS "O"?
Example 15-10 Given the following equation and exp data: A + B + C → Prod.
1
2
3
4
[A]
0.15
0.25
0.35
0.25
[B]
0.25
0.25
0.35
0.25
[C]
0.45
0.35
0.35
0.45
rate products form (M/min)
1.055 X 10-3 M/min
8.2 X 10-4 M/min
2.251 X 10-3 M/min
1.055 X 10-3 M/min
Write the rate law and find the value of "k"
(1) You can find the order of "C" using Experiments 2 and 4. Then you can find the
order of "A" by using Experiments 1 and 4, and it turns out that the order of "A" is
0--that is, increasing the concentration of "A" does not change the rate at all.
(2) There is not a place where both "A" and "C" stay the same while "B" changes,
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BUT since "A" is zero order and does not affect the reaction rate at all when its
concentration changes, you may ignore it and simply find a place where "B"
changes and "C" stays the same (and it doesn't matter what "A" does.
(3) When you write the rate law, I prefer that you completely leave "A" out of the
rate law since anything raised to the 0 power is 1. If you must put it in, be sure
that you show clearly that the order of "A" is 0.
WHAT DO YOU DO IF THERE IS NOT INFORMATION GIVEN IN THE DATA TABLE
TO FIND THE ORDERS OF ALL OF THE REACTANTS AND ONE OF THE
REACTANTS IS NOT ZERO?
Example 15-11 Given the following kinetics data, find the orders of each of the
reactants: A + B → Products
1.
2.
3.
[A]
[B]
rate products form
0.25
0.35
0.45
0.25
0.25
0.35
1.13 X 10-4,M/min
2.21 X 10-4 M/min
5.10 X 10-4 M/min
You can use experiments 1 and 2 to find the order of "A", but there is no way to
find "B" and "A" is NOT 0.
(1) First, find the order of "A". Then write the mathematical equation to solve for
"B" just as you normally would and then multiply it by the same two
experiments of "A" raised to the order you have determined for "A".
Example 15-12 This same type of reasoning can work if there are 3 reactants and you
can only find the orders for two of them. When you start to find the
last order, you must include the factors for the reactants you know
raised to their appropriate powers and solve mathematically.
A + B + C → Products
1.
2.
3.
4.
[A]
[B]
[C]
rate products form
0.15
0.25
0.25
0.55
0.30
0.30
0.40
0.45
0.40
0.40
0.40
0.50
3.159 X 10--5 M/sec
1.4625 X 10-4 M/sec
1.95 X 10-4 M/sec
3.64599 X 10-3 M/sec
Find the orders with respect to each reactant and the value of k.
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4. Pressure - changes in pressure affect the reaction rate ONLY IF THE REACTANTS
ARE GASES! An INCREASE in pressure DECREASES the volume occupied
and therefore, INCREASES the concentration (i.e. when the pressure is
doubled, the volume occupied is halved and therefore the concentration is also
doubled)
5. Surface Area - an increase in surface area increases the reaction rate. This applies
to reactants which are solids only; the surface area can be increased by
grinding the reactants into smaller pieces (the smaller the pieces, the greater the
surface area and therefore the greater the reaction rate)
Zn(solid) + HCl (aq) → ZnCl2(aq + H2(g)
Zn(powder)
(bubbles as hydrogen gas escapes;
gets hot because this is an exothermic
reaction, but very controllable)
+ HCl(aq) → BOOM!!!!
(reaction rate is too fast and out of
control due to the tremendous amount
of surface area of the solid reactant).
To give you an idea of how much you
have increased the surface area, if you
began with a cube of zinc 1"on the
edge (which is 6 in2 of surface area)
and you began halving the cube and
halving it and halving it until the particle
was the size of a talcum powder
particle, you would have over 2 acres of
surface area!!
6. Catalyst - a very specific substance which increases the reaction rate.
There are two types of catalysts:
(1) Heterogeneous - a catalyst which is in a different phase than the reaction it
is catalyzing; also called a contact catalyst, these catalysts do not actually
enter into the reaction--they only provide a surface for the reaction to take
place on and serves to weaken the bonds of one of the reactants which
adsorbs to it. This then makes it easier for the other reactant to react with
the one with weak bonds and thus the reaction rate is faster.
The catalytic converter in your car is a heterogeneous catalyst which
provides a surface for the reaction of carbon monoxide plus oxygen
yields carbon dioxide to take place quicker. Common specific substances
which are used in catalytic converters are platinum, V2O5 and MnO2.
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(2) Homogeneous - a catalyst which is in the same phase as the reaction it is
catalyzing; common examples are enzymes and acids and usually liquid
reactions are being catalyzed. These catalysts actually enter into the
reaction and then are returned unchanged at the end of the reaction to be
used again.
In the following example, X is the catalyst (catalyst is usually written over the yield
arrow)
A + B → AB
Rather than A simply bumping into B and reacting, we think it happens like this:
A + X → AX
(A reacts with the catalyst to form a compound AX)
Then AX + B → AB + X (returning X unchanged to be used again)
Up to now we have concerned ourselves mainly with chemical reactions which proceed
in the forward direction only. However, MOST chemical reactions are known to reverse
themselves at some point and tend to go forward at the same rate as they go in the
reverse direction. To your eye, it appears that the reaction has "stopped", but it hasn't!
It is simply going in the reverse direction at the same rate that it is going in the forward
direction. We call this condition EQUILIBRIUM. Thermodynamically, equilibrium is the
point when Grxn = 0. We show that a reaction has reached equilibrium by drawing not
only a forward pointing arrow, but another arrow as well (either above or below the first)
pointing backward. You can also indicate equilibrium by using a forward pointing halfarrow and a reverse pointing backward half-arrow written over it.
Equilibrium can occur at any point in a reaction, AND IT IS UNIQUE TO THAT
PARTICULAR REACTION WHERE IT DOES REACH EQUILIBRIUM.
Only in reversible equations are you able to predict the orders of the reactants in the
FORWARD reaction and the orders of the reactants in the REVERSE reaction
by looking at the balancing coefficients in the equation and using the balancing
coefficients as the orders!
Two men. Guldberg and Waage discovered this relationship and combined both the rate
constants for the forward and reverse reactions into one chemical equation They called
it the Law of Mass Action.
Given the following chemical reaction at equilibrium:
2 A (g) + 3 B (g)  2 C(g) + 4 D(g)
Keq
=
[C]2 [D]4
[A]2 [B] 3
This is called the equilibrium expression. You must
remember that only reactants and products which are
gases or aqueous solutions can be included in this
expression (i.e. only substances whose
concentrations can change).
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DO NOT EVER INCLUDE PURE LIQUIDS NOR SOLIDS IN THIS EXPRESSION
SINCE THEIR CONCENTRATIONS CANNOT CHANGE--JUST LEAVE THEM OUT!
Do not include the following types of substances in a Keq expression
(1) any substance which has (l) after it such as Br2, Hg, water, CH3OH(l)
(2) any substance which is a solid such as Zn, CaCO3(s), ice, etc.
ALWAYS include substances which can CHANGE concentrations such as gases
and solutions.
The forward rate constant and the reverse rate constant have been combined into an
equilibrium constant Keq . By convention Keq is a pure number (no label) --it is a
comparison of the concentrations of PRODUCTS to REACTANTS present in the system
at equilibrium.
If the Keq > 1, this means that there are more products than reactants present at
equilibrium
If the Keq < 1, this means that there are more reactants than products present at
equilibrium
THE VALUE OF THE Keq NEVER CHANGES EXCEPT WITH TEMPERATURE AND A
DIFFERENT REACTION. Changes in concentrations of products or reactants
will cause a SHIFT in the direction of the reaction to re-establish equilibrium,. but
it will not change the value of the Keq
Le Chatelier's Principle states when a stress is applied to a reaction at
equilibrium, the reaction will shift in a direction to relieve the stress.
Stresses which we will consider are the following
(1) addition of more reactants (increasing the concentration of reactants)
causes a shift AWAY from the stress or to the right -------->
(2) addition of products (increasing the concentration of products) causes a
shift AWAY from the stress or to the left <---------(3) removal of products causes a shift to “fill the hole” which is left by the
removal or a shift to the right ------->
(4) removal of reactants will not allow the forward reaction to begin and
causes a shift to fill the hole left by the removal or a shift to the left <---(5) Pressure changes affect gases in the equation only.
(a) an increase in pressure causes a shift to the side of the equation
with the LEAST number of moles of gas.
(b) a decrease in pressure causes a shift to the side of the equation
with the MOST number of moles of gas
(5) Temperature changes cause shift based on whether the reaction is
exothermic or endothermic.
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Just remember to write the thermochemical equation and put the heat
on the correct side of the equation and then simply treat it as a product or
a reactant, depending on which side of the equation the heat factor is on.
(a) raising the temperature (adding heat) to an exothermic reaction
causes a shift to the left <--------- since heat is a product
(b) raising the temperature (adding heat) to an endothermic reaction
causes a shift to the right --------> since heat is a reactant
Example 15-13 Given the following reversible equation at equilbrium
3 A(g) + 2 B(g)  4 C(g) + 2 D(g)
If, at equilibrium, it is found that the concentration of A = 0.50 M, the concentration of
B = 0.060 M, the concentration of C = 1.2 M and the concentration of D = 0.095 M, find
the value of the Keq
Example 15-14. Given the Haber process (synthesis of ammonia) and  Hrxn = -96 kJ
(a) Which way would the reaction shift to re-establish equilibrium if more
Hydrogen gas were added?
(b) Which way would the reaction shift to re-establish equilibrium if the
ammonia were removed as soon as it was formed?
(c) Which way would the reaction shift if it were heated?
(d) Which way would the reaction shift if the size of the reaction container
Were doubled?
(e) Which way would the reaction shift if the amount of nitrogen available
were reduced
(f) if the pressure inside the container were doubled?
(g) if the system were cooled?
(h) if a catalyst were added?
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Example 15-15 Given the synthesis of gaseous hydrogen chloride from its elements
and Hrxn = -536 kJ
(a) Which way would the reaction shift if the pressure were increased?
(b) Which way would the reaction shift if the container is heated?
(c) Which way would the reaction shift if the size of the container is
halved?
(d) Which way would the reaction shift if the HCl is removed as soon as it
is formed
(e) Which way would the reaction shift if a contact catalyst were added?
(f) Which way would the reaction shift if the concentration of chlorine were
doubled?
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