CHEMICAL EQUILIBRIUM
Reversible reactions are those which can proceed either in the
forward or in the backward direction according to the relative
amounts of the substances present.
The two oppositely directed arrows refer to a reversible reaction.
If we mix 1 mole of H2 (2.0 gram) and 1 mole of I2 (254 gram) we
should expect to obtain 2 moles of HI (256 gram). This will be true
if the reaction was irreversible. Actually we obtain about 76% only
of this quantity. This does not at all indicate that the reaction stops
at this limit but that only at this stage the rate of combination of
hydrogen and iodine is equal to the rate of decomposition of
hydrogen iodide and the system is said to be in equilibrium.
Chemical equilibrium is reached when the rate of the forward
reaction equals the rate of the reverse reaction. This does not mean
that the quantities of reactants and products are equal.
LAW OF MASS ACTION:
The influence on concentration of the rate of a chemical reaction is
expressed by the law of mass action. According to this law, the rate
of a chemical reaction is directly proportional to the product of
active masses of the reacting materials. By active mass is meant the
molar concentration of the substance; i.e., the number of moles of
substance per volume (liter) and is designated by two square
brackets.
In general, a reversible reaction may be written as:
where v1 and v2 are the rates (or speed) of the forward and
backward reaction, respectively.
If A and B represents the
concentration of the reactants in moles per liter, then according to
the law of mass action.
where k1: is the specific rate constant for the forward reaction
where k2 : is the specific rate constant for the backward reaction.
At equilibrium, there is no further apparent change and the rate of
the forward reaction becomes equal to that of the backward one,
hence,
and
Where [A], [B], [G],
[H] are the equilibrium concentrations. The
constant Kc is the equilibrium constant with respect to molar
concentration.
At equilibrium we can determine the equilibrium constant, Kc.
For the general reaction:
The equilibrium constant can be expressed as:
in case of ideal diluted solutions.
Uppercase letters represent the various compounds in any reaction
and the lowercase letters represent the coefficients for each
compound.
For example:
To determine Kc, substitute in the concentrations of each compound
at equilibrium.
For example, using
[N2] = 3.0 x 10-2 M
[H2] = 3.7 x 10-2 M
[NH3] = 1.6 x 10-2 M
Calculate the equilibrium constant for the reaction?
You can also calculate the concentration of one compound if you
know the concentrations of the others and Kc
To determine if a reaction is at equilibrium, you use a term called
Qc. It is calculated in the same way as the kc only using the current
concentrations of the compounds. At equilibrium the Qc equals the
Kc. When Qc is larger than Kc the reaction has too many products
and the reaction will move to the left to reach equilibrium.
When
Qc is smaller than Kc the reaction has too many reactants and will
move to the right to reach equilibrium.
For the reaction
Is the reaction at equilibrium, if not in which direction will the
reaction proceed?
Qc is larger than Kc, therefore the reaction will move to the left
(making more reactants).
Relationship between Kc and Kp
Consider the reversible reaction
for which the equilibrium constant Kc is expressed as:
The molar concentration may be expressed as being equal to n/v ,
M = n/V
By applying the ideal gas equation
For each gas it follows that:
Substituting these values of active mass or molar concentration into
the expression
Or
Where: Δn = number of resulting molecules – number of reacting
molecules
Kp = equilibrium constant in case of ideal gases
Kc = equilibrium constant in case of ideal solutions
R = general gas constant (0.082 atm.L. mol-1 .deg-1)
T = Absolute temperature, K.
Three cases may be considered:
i) Δn is 0, e.g.,
H2(g) + I2 (g) = 2 HI(g)
Δn = 2 -1 -1 = 0
Hence,
Kp = Kc
ii) Δn is positive, e.g.,
PCl5(g) = PCl3 (g) + Cl2(g)
Δn = 1 +1 -1 = +1
Hence,
Kp >
Kc
iii) Δn is negative, e.g.,
N2(g) + 3H2 (g) = 2 NH3(g)
Δn = 2 -1 -3 = -2
Hence,
Kp < Kc
Problem:
For the reaction
N2 + 3H2
found to be 0.500
= 2 NH3
at 400°C
Kc was
Calculate Kp value
Solution:
Δn = number of resulting molecules – number of reacting molecules
Δn = 2 – (1+3) = -2
T = 400°C + 273 = 673 k
R = 0.082 atm.L. mol-1 .deg-1
Since
Kp = Kc (RT)Δn
Kp = 0.500 (0.082 x 673)-2 = 1.64 x 10-4
Determination of equilibrium constant:
Many chemical and physical measurements can be used in the
determination of equilibrium constants. The method used depends
on the nature of the reversible reaction under investigation.
The following are some representative examples.
1) Equilibrium constant for the homogeneous liquid reaction of
acetic acid and ethyl alcohol,
CH3COOH + C2H5OH
= CH3COOC2H5
+ H2O
To determine Kc for this reaction, different quantities from the acid
and alcohol are placed in sealed glass tubes then left in a thermostat
at a certain fixed temperature. After reasonable sufficient time, the
remaining amount of acetic acid is quantitatively determined by
titration with sodium hydroxide solution of appropriate known
concentration. From the acid and alcohol it is possible to calculate
the concentration of the reaction constituents at equilibrium, hence
Kc can be calculated.
2) Equilibrium constant for the gaseous reaction
N2(g) + 3H2(g)
=
2NH3(g)
The equilibrium constant for this reaction can be determined by
passing a mixture of N2 gases and hydrogen at a 1:3 ratio in an iron
coil immersed in a thermostat at a certain temperature. In order to
accelerate the reaction, the inside of the coil may be covered with a
catalyst.
The outlet gases from the coil are then rapidly cooled,
collected and then analysed quantitatively for nitrogen, hydrogen
and ammonia, hence the value of Kp can be determined.
Le Chatelier Principle:
This principle describes the effect of temperature, volume, pressure,
concentration, etc. on the position of equilibrium in a system. It
stated that the equilibrium is shifted in the direction that opposes
such effect.
endothermic
Accordingly, increase of temperature will favor
reactions,
increase
of
pressure
will
shift
the
equilibrium in the direction accompanied with a fewer number of
molecules.
As an example for chemical equilibrium,
N2 + O2
= 2 NO
- 43,200 Cal
The combination of N2 and O2 to form NO is an endothermic process.
According to Le Chatelier principle, increase of temperature will
favour the formation of NO. The reaction on the other hand, is not
accompanied with a change in volume, pressure, temperature,
therefore, has no effect on the position of equilibrium. The excess
amounts of reactants added consumed in the formation of more
nitric oxide to maintain the position of equilibrium.
As an example of a physical equilibrium,
Ice = liquid water -heat
By raising the temperature the equilibrium will shift in the direction
of ice to melt. By increasing the pressure the equilibrium will shift
in the direction accompanied with a decrease in volume (ice
converts to liquid water).
Application of the law of mass action
A) Homogeneous Gaseous reaction
1. Reactions occurring without change in the number of molecules:
Consider the formation of hydrogen iodide from hydrogen and iodine
Δn = number of resulting molecules – number of reacting molecules
Δn for this reaction = zero
Therefore,
Kp = Kc
Suppose we start with (a) moles of H2 and (b) moles of I2. When
equilibrium is established, let (x) mole be the quantity reacted of
hydrogen and iodine to form (2x) moles of HI. The remaining
quantity of H2 is then (a-x) mole, and of iodine (b-x) mole.
If V is the volume of the system, it follows that:
By applying the law of mass action :
Or
It will be noted that the volume V of the system does not appear in
the last equation.
Therefore, the volume and consequently the
pressure should not alter the position of equilibrium in such system.
The above conclusion may be arrived at by considering Kp.
According to Dalton's law of partial pressure,
In the above example
By applying the law of mass action :
It follows that
Which is the same value for Kc previously derived. So, the pressure
has no effect on the position of equilibrium.
Problem:
A mixture of 7.9 cc H2 and 33.1 cc I2 vapors was heated at 444 °C
until equilibrium was reached. If the equilibrium constant for the
reaction is 36.68, Calculate the number of cc of H2 present when
equilibrium is attained.
Solution:
H2 + I2
= 2 HI
Let X cc of H2 and I2 be changed to HI
Then the volume of H2 remaining at equilibrium = (a-x) = (7.9 – X)
cc
The volume of I2 remaining at equilibrium = ( b-x) = (33.1 – X) cc
The volume of HI formed at equilibrium = (2 X) cc
For the reaction
4x2
Kp = ----------(a-x) (b-x)
4x2
36.68 = --------------(7.9-x) (33.1-x)
4x2 = 36.68 (7.9-x)(33.1-x)
X = 7.8 or 38.2
38.2 can not be the solution of the problem, therefore, x = 7.8
Hence, the number of cc of H2 present at equilibrium = (7.9 -7.8) =
0.1 cc
2. Reactions occurring with a change in the number of molecules:
Consider the dissociation of nitrogen tetra oxide to nitrogen dioxide
If x represents the degree of dissociation (i.e., fraction of mole that
dissociates at equilibrium).
1-x will represents the fraction of undissociated N2O4 and 2x the
number of moles of NO2 formed. If the total volume is V and the
total pressure is P, therefore
By applying the law of mass action for the equation
N2O4(g) = 2 NO2(g)
Total number of moles present at equilibrium = 1-x + 2x = 1+x
The partial pressures will be
PNO2
= P. 2x / (1+x)
PN2O4
= P. (1-x) / (1+x)
Substituting in the expression
(PNO2)2 / (PN2O4)
= Kp
(P. 2x / 1+x)2 / (P. 1-x / 1+x)
It follows that
= Kp
P2. 4x2 . [1/ (1+x)2 ]/ (P. 1-x / 1+x)
= Kp
4x2 P . [1/ (1+x)2 ]. [1/1-x] . [1+x]
= Kp
The last
p
equation shows that the pressure
should affect the
position of equilibrium but not the equilibrium constant Kp. Thus if
P increases, x should decrease so that
Kp remains constant at
constant temperature.
B) Heterogeneous reaction
1) Equilibrium involving gases and solids
When the reaction involves one or more solids in equilibrium with a
gas or a liquid, the reaction is a heterogeneous one.
As an example consider the dissociation of calcium carbonate
CaCO3 = CaO(s) + CO2 (g)
By applying the law of mass action, it follows that
Kp = PCaO . PCO2 / PCaCO3
But since PCaO and PCaCO3 are constant at any one temperature
constant Kp is used where
Kp = PCO2
So at any particular temperature the equilibrium constant is
determined solely by the equilibrium pressure of carbon dioxide.
This equilibrium pressure of carbon dioxide is generally calculated x
from the dissociation pressure of calcium carbonate at that
particular temperature.
As in the case of homogeneous equilibrium, the influence of
pressure on heterogeneous equilibrium can be predicted by means
of Le Chatelier principle.
Ionic equilibrium
and applications of law of mass action
Electrolytic solutions are divided into two main groups:
a) Strong electrolytes:
This group includes strong acid like (HNO3,
HCl, H2SO4).
Strong
alkalies like (KOH, NaOH) and salt solutions. By the word (strong)
we mean that the solutions of these substances are completely
ionized.
Thus if we have a 0.1N solution of HCl the whole
concentration of the acid will be present as H+ and
Cl- each of
which will be 0.1 g ion. Such state is represented as:
b) Weak electrolytes:
Which includes weak acids like acetic acid
CH3COOH, oxalic acid
H2C2O4, weak alkalies like NH4OH. By the word (weak) we mean
that such solutions is not completely ionized. For example, if we
have a 0.1N solution of acetic acid, then only part of this
concentration will ionize to give H+ and CH3COO- while the rest will
remain in the undissociated from CH3COOH. A state of equilibrium
will then be established between the formed ions and the
undissociated acid as follows:
In this case it is clear that the concentration of each of CH3COOand H+ will not be equal 0.1
Ostwald law of dilution:
Consider 1 g equivalent of a weak acid HA dissolved in V liters of
solution. If α is the degree of dissociation of the acid. We have at
equilibrium.
If in general the acid concentration is C g. mole. Then we have
By applying the law of mass action we obtain
Where ka is the equilibrium constant, which in this case is known as
the and [A-] we obtain
The above relation is known as Ostwald law of dilution.
It indicates that if C increases, α should decrease.
Since Ka is a
constant. For weak electrolytes the degree of dissociation is usually
very small compared to unity and the Ostwald law may be simpled
as
A relation similar to that given above can be obtained by applying
the law of mass action to the ionization of a weak base BOH, i.e.
Where kb is the ionization constant for the weak base BOH.
Neglecting α with respect to unity.
The following table shows the ionization constant for some weak
acids (ka) and weak bases (kb).
Acid
Ka
Base
Kb
HCOOH
1.8 x 10 – 4
NH4OH
1.7 x 10 - 5
CH3COOH
1.8 x 10 – 5
CH3NH2
5.0 x 10 - 4
C6H5.COOH
6.3 x 10 – 5
C2H5NH2
4.1 x 10 - 10
Example:
Calculate the degree of ionization and hydrogen ion concentration
for 0.005 M solution of acetic acid knowing that its ionization
constant is 1.8 x 10-5
Solution:
Ka = α2 C
α2 = Ka / C
α2 = 1.8 x 10 -5 / 0.005
α = 6 x 10-2
Since
[H +] = C α = 6 x 10-2 x 0.005 = 3.0 x 10-4
Example:
Calculate the percent ionization of a 1.00 M solution of HCN acid,
knowing that ka is 4.8 x 10-10
Solution:
Ka = α2 C
α2 = Ka / C
α2 = 4.8 x 10 -10 / 1.00
α = 2.2 x 10-5
Ionization of polybasic acids
When a polybasic acid is dissolved in water, the various hydrogen
atoms undergo ionization to different extents. For a dibasic acid
H2A, the primary and secondary ionization can be represented as
follows:
If the dibasic acid is a weak electrolyte the law of mass action may
be applied and therefore:
[H+] [ HA- ] / [H2A] = k1
[H+] [A- -] / [HA-] = k2
K1 and k2 are known as the primary and secondary dissociation
constants respectively. Each stage of the dissociation process has
its ionization constant, and the magnitude of these constants give a
measure of the extent to which each ionization has proceeded at
any given concentration.
A tribasic acid (e.g. orthophosphoric acid) will similarly give three
dissociation constants K1, K2, and K3 according to:
Common ion effect
Consider a weak electrolyte such as acetic acid.
In solution the
following equilibrium exists:
Consider the addition of a strong electrolyte having ion in common
to this solution, e.g. sodium acetate.
The new equilibrium will be
The acetate ion is now called the common ion. The result is that the
equilibrium of acetic acid is shifted to the left, i.e. the degree of
ionization is decreased.
Another examples:
Rule:
Addition of a strong electrolyte
to a weak electrolyte
containing a common ion, results in decrease of concentration of the
common and uncommon ion for the latter.
Electrolyte: The common ion effect provides a valuable method for
controlling the concentration of the ions furnished by a weak
electrolyte.
Solubility Product Constant
In general, when ionic compounds dissolve in water, they go into
solution as ions. When the solution becomes saturated with ions,
that is, unable to hold any more, the excess solid settles to the
bottom of the container and an equilibrium is established between
the undissolved solid and the dissolved ions.
Consider a sparingly (slightly) soluble electrolyte e.g. silver chloride.
When it is shaken up with water, some of it (an exceedingly small
quantity) passes in solution to form a saturated solution of the salt.
The following equilibrium exists between the insoluble AgCl and
ions in solution
Applying law of mass action (or Ostwald law of dilution) to such an
equilibrium then
Since the silver chloride is present in the solid state its active mass
may be taken as unity. Therefore, the above equation reduces to:
This new constant S is known as the "solubility product" of silver
chloride and can be written as Ksp .
Like all equilibrium constants, the Ksp is temperature dependant, but
at a given temperature it remains relatively constant.
It is also like any equilibrium expression, each ion concentration in
the expression is raised to the power of its coefficient in the
solubility equation.
In general, the solubility product (Ksp), is the equilibrium constant
for the solubility equilibrium of a slightly soluble ionic compounds.
Slightly soluble substances such as AgCl, PbSO4, etc. posses small
solubility product and can therefore be easily precipitated in
solution. On the other hand, substances possessing higher solubility
have high solubility products and cannot be precipitated easily from
solution.
In general any substances cannot be precipitated except if the
product of its ions in solution exceeds the solubility product of its
sparingly soluble compound.
For any electrolyte having the general formula AxBy in contact with
its saturated solution, the equilibrium between the solid and ions in
solution is
and the solubility products is given by
The solubility product of a sparingly soluble substance may be
derived from a knowledge of the solubility of the substance in pure
water. For example if the solubility of silver chloride in water is "S"
gram mole per liter, then the concentration of both silver and
chloride ions is "s" gram ion per liter, therefore,
Solubility product = [Ag+] [Cl-] = s X s = solubility (s)2
Or
solubility (s) = solubility product (S)½
In case of a salt giving more than two ions in solution, e.g. Ag2CrO4
S = [Ag+]2 [CrO4--] = (2s)2 (s) = 4s3
Solubility (s) = (s/4)1/3
Example:
The solubility of silver chloride is 0.0015 gram per liter, calculate the
solubility product knowing that the MW of AgCl is 143.5
Solution:
Solubility in gram mole/ Liter = 0.0015 / 143.5 = 1.05 x 10-5.
Therefore in a liter of solution
the concentration of silver ions = 1.05 x 10-5 gram ions
and since
S = [ Ag+] [ Cl-]
S = (1.05 x 10-5 ) (1.05 x 10-5) = 1.1 x 10-10
Problem:
What is the molar solubility of barium fluoride in a solution that is
0.15 M NaF at 25 °C
Solution:
Since the solution is already 0.15 M in F-1 ions, we must make an
addition to our equilibrium concentrations.
BaF2(s)
Ba++(aq)
"x"
+
2F-(aq)
"2x +0.15"
(at equilibrium)
Ksp = [Ba++] [ F-]2
Because BaF2 is only slightly soluble, you might expect "2x" to be
negligible compared to 0.15. In that case
(2x +0.15 ) ≈ (0.15) and substituting into the Ksp expression, we
get
1.0 x 10-6 = (x)(0.15)2
solving for x, we get
x= 4.44 x 10-5 M
7. THERMOCHEMISTRY
Introduction:
This chapter deals with energy and heat, two terms used widely by
both the general public and Scientists. Scientifically, these terms
have quite different meanings.
capacity to do work.
Energy can be defined as the
Heat is a particular form of energy that is
transferred from a body at a high temperature to one at a lower
temperature when they are brought into contact with each other.
Two centuries ago, heat was believed to be a material fluid
("caloric"); we still use the phrase "heat flow" to refer to heat
transfer or to heat effects in general.
Thermochemistry refers to the study of the heat flow that
accompanies chemical reactions. Our discussion of this subject will
focus upon

The basic principles of heat flow.

The experimental measurement of the magnitude and
direction of heat flow, known as calorimetry.

The concept of enthalpy (heat content) and enthalpy
change, ΔH.

The
calculation
of
ΔH
for
reactions,
using
thermochemical equations and enthalpies of formation.

Heat effects in the breaking and formation of covalent
bonds.

The relation between heat and other forms of energy, as
expressed by the first law of thermodynamics.
Principles of heat flow:
In any discussion of heat flow, it is important to distinguish between
system and surroundings. The system is that part of the universe
upon which attention is focused. In a very simple case it might be a
sample of water in contact with a hot plate.
The surroundings,
which exchange energy with the system, comprise in principle the
rest of the universe.
For all practical purposes, however, they
include only those materials in close contact with the system. If we
heat some of water in a beaker, the surroundings would consist of
the hot plate, the beaker holding the water sample, and the air
around it.
When a chemical reaction takes place, we consider the substances
involved,
reactants
and
products,
to
be
the
system.
The
surroundings include the vessel in which the reaction takes place
(test tube, beaker, and so on) and the air or other material in
thermal contact with the reaction system.
State properties:
The state of a system is described by giving its composition,
temperature, and pressure.
For example, if the system which
consists of 50.0 g of H2O(L) at 50.0 °C and at 1 atm is heated, its
state changes, perhaps to one described as 50.0 g of H2O(L) at 80.0
°C and 1 atm.
Certain quantities, called state properties, depend only upon the
state of the system, not upon the way the system reached that
state. Put it another way, if X is a state property, then
ΔX = X final – X initial
That is, the change in X is the difference between its values in final
and initial states. Most of the quantities that you are familiar with
are state properties; volume is a common example.
You may be
surprised to learn, however, that heat flow is not a state property;
its magnitude depends upon how a process is carried out.
Direction and sign of heat flow:
If the hot plate is turned on, there is a flow of heat from the
surroundings into the system, 50.0 g of water.
This situation is
described by stating that the heat flow, q , for the system is a
positive quantity.
The (q) value is (+) when heat flows into the system from the
surroundings.
Usually, when heat flows into a system, its temperature rises. In
this case, the temperature of the 50.0 g water sample might
increase from 50.0 to
80.0 °C. When the hot plate is shut off, the
hot water gives off heat to the surrounding air. In this case, q for
the system is a negative quantity.
The (q) value is (-) when heat flows out of the system into the
surroundings.
Here, as in usually the case, the temperature of the system drops
when heat flows out of it into the surroundings. The 50.0 g water
sample might cool from 80.0°C back to 50.0°C.
This same reasoning can be applied to a reaction where the system
consists of the reaction mixture (products and reactants). We can
distinguish between:
-
an endothermic process (q>0), in which heat flows from the
surroundings into the reaction system.
An example is the melting of ice:
The melting of ice absorbs heat from the surroundings, which might
be the water in a glass of iced tea. The temperature of the
surroundings drops, perhaps from 25 to 0 °C, as they give up heat to
the system.
-
an exothermic process (q<0), in which heat flows from the the
reaction system into the surroundings.
An example is the combustion of methane gas:
This reaction evolves heat to the surroundings, which might be the
air around a Bunsen burner in the laboratory or a potato being
baked in a gas oven. In either case, the effect of the heat transfer is
to raise the temperature of the surroundings.
Magnitude of heat flow:
In any process, we are interested not only in the direction of heat
flow but also in its magnitude. The magnitude of q is ordinarily cited
in joules (J) or kilo-joules (kJ).
1 kJ = 103 J
The SI unit of heat, the joule, is named for James Joule, who carried
out very precise thermometric measurements that established the
first law of thermodynamics.
Most of the remainder of this chapter is devoted to a discussion of
the magnitude of the heat flow in chemical reactions or phase
changes. Here, however, we will focus upon a simpler process in
which the only effect of the heat flow is to change the temperature
of a system. In general, the relationship between the magnitude of
the heat flow, q , and the temperature change, Δt , is given by the
equation:
The quantity C appearing in this equation is known as
the heat capacity of the system. It represents the amount of heat
required to raise the temperature of the system 1°C and has the
units J/°C.
For a pure substance of mass m, the expression for q can be written
as
The quantity , C, is known as the specific heat; it is defined as the
amount of heat required to raise the temperature of one gram of a
substance 1°C.
Specific heat, like density or melting point, is an intensive property
which can be used to identify a substance. Water has an unusually
large specific heat, 4.18 J/g.°C. This explains why swimming is not
a popular pastime in northern Minnesota in May.
Even the air
temperature rises to 90°F, the water temperature will remain below
60°F. Metals have a relatively low specific heat. When you warm
water in a stainless steel saucepan, nearly all of the heat is absorbed
by the water, very little by the steel.
Example:
How much heat is given off by a 50.0 g sample of copper when it
cools from 80.0 to 50.0°C , Knowing that the specific heat of copper
is 0.382 J/g.°C?
Solution:
The temperature change is
Δt = t final – t initial
= 50.0°C - 80.0°C
= - 30.0°C
q = m x C x Δt
= 50.0 g x 0.382 J/g.°C x (-30.0 °C)
= - 573 J
The negative sign indicates that heat flows from copper to the
surroundings.
Specific heats of a few common substances
Substance
C (J/g.°C)
Substance
C (J/g.°C)
Br2(L)
0.474
Cu(s)
0.382
Cl2(g)
0.478
Fe(s)
0.446
C2H5OH(L)
2.43
H2O(g)
1.87
C6H6(L)
1.72
H2O(L)
4.18
CO2(g)
0.843
NaCl(s)
0.866
Heat flow in an reaction is measured using a calorimeter device.
Enthalpy:
We have referred several times to " the heat flow for the reaction
system, " symbolized as q
reaction.
At this point, you may well fined
this concept a bit nebulous and wonder if it could be made more
concrete by relating q
products.
reaction
to some property of reactants and
This can indeed be done; the situation is particularly
simple for reactions taking place at constant pressure. Under that
condition, the heat flow for the reaction system is equal to the
difference in enthalpy (H) between products and reactants. That is,
Enthalpy is a type of chemical energy, sometimes referred to as "
heat content. " Reactions that occur in the laboratory in an open
container or in the world around us take place at a constant
pressure, that of the atmosphere. For such reactions, the equation
just written is valid, making enthalpy a very useful quantity.
Reactions which accomplished by the release of energy in the form
of heat are called Exothermic Reactions, whereas reactions that
absorbs heat from the surroundings are called
Endothermic
Reactions.
If you wish to show that a reaction is exothermic, there is a simple
way of doing it.
For example,
2 Na(s) + Cl2(g)
→
2 NaCl(s)
+ heat
However, sometimes we wish to be more specific and say exactly
how much heat is released.
CH4(g) + 2 O2(g)
→
CO2(g)
+ 2 H2O(g) + 890 KJ
Each of the four substances involved in this reaction contains a
certain quantity of energy and this energy content is called the
Enthalpy (H) of the Compound. In this example, 890 KJ of energy
was released as heat. This means that one mole of carbon dioxide
gas plus two moles of water vapor must contain 890 KJ of energy
less than one mole of methane gas plus two mole of oxygen gas.
Because the products themselves posses 890 KJ of energy less than
the reactants, we say that the change in Enthalpy (∆H) of the
reaction is -890 KJ, (∆H = -890 KJ).
Our equation can be alternatively written as:
CH4(g) + 2 O2(g)
→
CO2(g)
+ 2 H2O(g)
∆H = - 890 KJ
If heat is absorbed during the reaction, the enthalpy of the products
will be greater than that of the reactants. The change in Enthalpy
(∆H) will, therefore, have a positive sign.
An example of such a
reaction is as follows:
2 HgO(s)
→
2 Hg(ℓ) +
O2(g)
∆H = + 181.7 KJ
The study of the energy changes that take place during chemical
reactions is called thermochemistry.
The following figure shows the enthalpy relations for an exothermic
reaction such as
CH4 (g) + 2 O2 (g)
→
CO2 (g) + 2 H2 (ℓ)
ΔH < 0
Here, the products, 1 mol of CO2 (g) and 2 mol of H2O(ℓ), have a lower
enthalpy than the reactants, 1 mol of CH4 (g) and 2 mol of O2(g). The
decrease in enthalpy is the source of the heat evolved to the
surroundings.
In an exothermic reaction, the products have a lower enthalpy than
the reactants; thus, ΔH is negative, and heat is given off to the
surroundings.
The following figure shows the situation for an endothermic process
such as
H2O (s)
H2O (ℓ)
ΔH > 0
Since liquid water, H2O(ℓ), has a higher enthalpy than ice, heat must
be transferred from the surroundings to melt the ice.
In an endothermic reaction, the products have a higher enthalpy
than the reactants so ΔH is positive, and heat is absorbed from the
surroundings.
In general, the following relations apply for reactions taking place at
constant pressure.
The enthalpy of a substance, like its volume, is a state property. A
sample of one gram of liquid water at 25.0 °C and 1 atm has a fixed
enthalpy, H. In practice, no attempt is made to determine absolute
values of enthalpy.
Instead, scientists deal with changes in
enthalpy, which are readily determined. For the process
1.00 g H2O (ℓ, 25.0°C, 1atm)
1.00 g H2O (ℓ, 26.0°C, 1atm)
ΔH is 4.18 J because the specific heat of water is 4.18 J/g.°C.
THERMOCHEMICAL EQUATIONS:
A chemical equation that shows the enthalpy relation between
products and reactants is called a thermochemical equation. This
type of equation contains, at the right of the balanced chemical
equation, the appropriate value and sign for ΔH.
The thermochemical equation for the formation of HCl from the
elements is found to be
In other words, 185 KJ of heat is evolved when two moles of HCl are
formed from H2 and Cl2.
Rules of thermochemistry:
To make effective use of thermochemical equations, three basic
rules of thermochemistry are applied.
Rule 1. The magnitude of ΔH is directly proportional to the amount
of reactant or product. The amount of heat that must be absorbed
to boil a sample of water is directly proportional to its mass.
In
another case, the more gasoline you burn in your car's engine, the
more energy you produce.
This rule allows you to find ΔH corresponding to any desired amount
of reactant or product.
Example:
Consider the thermochemical equation for the formation of two
moles of HCl from the elements.
Calculate ΔH when
a) 1.00 mol of HCl is formed
b) 1.00 g of Cl2 reacts.
Solution:
(a)
2 mol HCl
1 mol HCl …………….
…………… ΔH – 185 KJ
?
1 mol x -185 KJ / 2
ΔH = – 92.5 KJ
(b) 1 mol Cl2 ………………. ΔH – 185 KJ
71 g Cl2 ………………. ΔH – 185 KJ
1 g Cl2 ………………..
?
ΔH = 1 g X -185 KJ / 71 g
= -2,6 KJ
This relation between ΔH and amounts of substances is equally
useful in dealing with chemical reactions or phase changes.
Example:
Using a coffee-cup calorimeter, it is found that when an ice cube
weighing 24.6 g melts, it absorbs 8.19 KJ of heat. Calculate ΔH for
the phase change represented by the thermochemical equation
H2O (s)
H2O (L)
ΔH = ?
Solution:
24.6 g ice ……………………..
1 mol (H2O) = 18 g ……………
ΔH = + 8.19 KJ
?
ΔH = 18 X + 8.19 / 24.6
= + 6.0 KJ
H2O (s)
H2O (L)
ΔH = + 6.0 KJ
This calculation shows that 6.00 KJ of heat must be absorbed to
melt one mole of ice:
H2O (s)
H2O (L)
ΔH = + 6.00 KJ
ΔH per one mole of ice = 6.00 KJ
ΔH per gram of ice = 6.00 KJ / 18.02 g
= 0.333 KJ/g
Rule 2. ΔH for a reaction is equal in magnitude but opposite in sign
to ΔH for the reverse reaction. Another way to state this rule is to
say that the amount of heat evolved in a reaction is exactly equal to
the amount of heat absorbed in the reverse reaction. This again is a
common-sense rule. If 6.00 KJ of heat is absorbed when a mole of
ice melts,
H2O(s)
H2O(L)
ΔH = + 6.00 KJ
Then 6.00 KJ of heat should be evolved when a mole of liquid water
freezes.
H2O(L)
H2O(s)
ΔH = - 6.00 KJ
Example:
Given
H2(g) + ½O2(g)
Calculate ΔH for the equation:
2 H2O(L)
2 H2(g) + O2(g)
H2O(L)
ΔH = - 285.8 KJ
Solution:
Applying rule 1:
H2(g) + ½O2(g)
2 H2(g) + O2(g)
H2O(L)
ΔH = - 285.8 KJ
2 H2O(L)
ΔH = 2 (- 285.8 KJ) = -571.6
2 H2O(L)
ΔH = -571.6 KJ
KJ
2 H2(g) + O2(g)
Applying rule 2:
2 H2O(L)
2 H2(g) + O2(g)
ΔH = +571.6 KJ
Rule 3. The value of ΔH for a reaction is the same whether it
occurs in one step or in a series of steps.
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ………
This relationship is referred to as Hess' law
Example:
Given
Calculate ΔH for the reaction
Solution:
To get one mole of CO on the right side, reverse Equation (2) and
divide the coefficients by two.
succession.
Applying Rule 1 and rule 2 in
Now add Equation (1) and simplify:
Summarizing the rules of thermochemistry:
1. ΔH is directly proportional to the amount of reactant or product.
2. ΔH changes sign when a reaction is reversed.
3. ΔH for a reaction has the same value regardless of the number of
steps.
Enthalpies of formation:
Meaning of ΔHf°:
The standard molar enthalpy of formation of a compound, ΔHf° , is
equal to the enthalpy change when one mole of the compound is
formed at a constant pressure of 1 atm and a fixed temperature,
ordinarily 25°C, from the elements in their stable states at that
pressure and temperature.
From the equations
It follows that:
ΔHf° AgCl (s) = -127.1 KJ/mol
ΔHf° NO2 (s) = +33.2 KJ/mol
Enthalpies of formation, ΔHf° (KJ/mol), of compounds at 25°C, 1atm
AgBr(s)
-100.4
BaCl2 (s)
-858.6
AgCl(s)
-127.1
BaCO3(s)
-1216.3
AgI(s)
-61.8
C2H2(g)
+226.7
AgNO3(s)
-124.4
C2H6(g)
-84.7
Ag2O(s)
-31.0
C2H4(g)
+52.3
Al2O3 (s)
-1675.7
ZnS(s)
-206.0
Notice that, with a few exceptions, enthalpies of formation are
negative quantities. This means that the formation of a compound
from the elements is ordinarily exothermic.
Conversely, when a
compound decomposes to the elements, heat usually must be
absorbed.
The enthalpy of formation of an element in its stable state at 25°C
and 1atm is taken to be zero.
That is,
Example:
ΔHf° Br2(L) = ΔHf° O2(g) = zero
Calculate ΔH° for the combustion of one mole of propane, C3H8,
according to the equation
Solution:
ΔH° = heat of formation of products – heat of formation of reactants
Heat of formation of an element (at 25 °C, 1 atm) is equal zero
H°f, Br2 = 0
H°f, O2 = 0
Expressing ΔH° in terms of enthalpies of formation,
Taking the enthalpy of formation of O2(g) to be zero and substituting
values for the other substances tables.