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Chapter 7: Thermochemistry
Why do chemical reactions occur?
What is stability?
Thermochemistry
System and the SurroundingsWe call the part of the universe chosen for study a system (the
substance or mixture of substances under study in which a change occur-thermodynamic
system). The surroundings are that part of the universe outside the system in which these
interactions can be detected. Our study will focus on the system’s interactions, that is, on the transfer
of energy and matter between a system and its surroundings.
Open system: is the one that can freely exchange energy and matter with its surroundings.
Closed system: is the one that it can exchange energy with its surroundings, but not matter.
Isolated system: a system that does not interact with its surroundings.
Energy
Energy can be classified as either kinetic or potential.
Kinetic energy, Ek
~ is the energy of a moving object. An object of mass m and speed (velocity) u has
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Ek = 1 mu
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Potential energy, EP
~ is stored energy– perhaps stored in an object because of its height or in a molecule because of
its chemical bonds.
Energy has many forms:
The kinetic energy, which is associated with random molecular motion, is called thermal
energy. In general, thermal energy is proportional to the temperature of a system. And the
thermal energy of a system also depends on the number of particles.
Heat
Chemical energy
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2 2
The SI unit of energy kg·m /s is given the name joule(J). In addition to SI unit, the unit calorie,
cal, is a also commonly used.
E.g. the substances we focus on in an experiment – the starting materials and products are
collectively called the system, while everything else –the reaction flask, the room, the building,
and so on – is called the surroundings. In practice, it is impossible to truly isolate a chemical
reaction from its surroundings. What’s important is not that system be isolated but that we be
able to measure accurately any energy that
enter (leave) the system from (to) surroundings.
One form of energy may be converted from one form to another but the total energy, remain
constant. This result can be state as
Law of conservation of energy:
Consider a water Reservoir
Can you Identify the forms of energy hold by the water 1) in the reservoir, 2) in the water fall
and 3) at the bottom of the dam?
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Internal energy, U,
It includes kinetic energy of molecules, the energy associated with molecular rotations and
vidrations, the energy stored in chemical bonds and intermoleculas attractions and the energy
associated with the electrons in atoms. Internal energy also includes energy associated with the
interaction of protons and neutrons in atomic unclei, although this component is unchanged in
chemical reactionIf the system were isolated from its surroundings so that no energy transfer
could occur between the two, then the total Internal energy (U) of the system would be
conserved and remain constant throughout the reaction. This is a restatement of the
first law of thermodynamic:
Suppose we are interested in studying the change of a thermodynamic property (such as internal
energy) during a physical or chemical change. Now, we need to distinguish between energy
changes produced by the action of force through distance-work-and those involving the
transfer of thermal energy-heat.
Heat :
What is heat?
A system does not contain heat, the energy content of a system is a quantity called internal
energy. Heat is simply a form in which a quantity of energy may be transferred across a
boundary between a system and its surroundings.
When heat, (i. e., energy), goes into a substance one of two things can happen:
1. The substance can experience a raise in temperature.
2. The substance can change state. For example, if the substance is ice, it can melt into water.
Perhaps surprisingly, this change does not cause a raise in temperature.
From kinetic-molecular theory of gases, we know Energy, as heat, passes from a warmer body to
a cooler body. At a molecular level, molecules of warmer body, through collisions, lose kinetic
energy to those of the cooler body thermal energy is transferred-heat flows- until the average
molecular kinetic energies of the two bodies become the same, until the temperatures become
equal.
A process occurring at a constant temperature is said to be isothermal.
It requires heat to raise the temperature of a given amount of substance, and the quantity of heat
depends on the temperature.
The quantity of heat, q, required to change the temperature of a substance depends on:
* how much the temperature is to be changed
* the quantity of substance
•the nature of the substance (type of atoms or molecules)
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Historically, the amount of energy necessary to raise the temperature of 1 g of water by 1°C has
been called one calorie, which is now defined as exactly 4.184 J.
1cal = 4.184 J
Thus, heat lost by a system is gained by its surroundings, and vice versa. Applied Law of
conservation of energy to the exchange of heat, this means that
Heat Capacity and Specific Heat
The quantity of heat required to change the temperature of a system by one degree is called the
heat capacity of the system. If the system is a mole of substance, we can use the term molar
heat capacity. If it is one gram of substance, we call it the specific heat capacity or specific heat.
The specific heat capacity (sp ht) is the quantity of heat required to raise the temperature of one
gram of a substance by one degree Celsius. The specific heat of water depends somewhat on
temperature, but over the range from 0 –100 °C, its value is about
1.00 cal/g °C =4.184 J/g °C
To find the heat required raising the temperature of a sample
E.g. Calculate the heat absorbed by 15.0g of water to raise its temperature from 20.0°C to 50.0°C
at constant pressure. The specific heat of water is 4.184 J/(g °C)
___1 Choose the incorrect statement
(a)The surroundings are the part of the universe that is studied
(b)Thermal energy is energy associated with radom molecular motion
(c) Chemical energy is associated with chemical bonds and intermolecy=ular forces
(d) Energy is the capacity to do work
(e) Work is done when a force acts through a distance
___2 The Metric system unit of energy that was originally designed to measure heat is the
(a) joule
(b) Btu
(c) Horse power
(d) Calorie
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Experimental Determination of Specific Heat
E.g. A 99.8g of Aluminum is heated to the temperature of boiling water (100°C). The hot Al is
then dumped to the 22°C of 50.0g of water, and the temperature of the final Al-water mixture is
45.5°C. What is the specific heat of Al?
Heat of Reaction and CalorimetryChemical energy is associated with chemical bonds and
intermolecular attractions.
A heat of reaction, qrxn, is the quantity of heat exchanged between a system and its surroundings
when a chemical reaction occurs within the system, at constant temperature.
An exothermic reaction is one that produces a temperature increase in an isolated system or, in
a nonisolated system, gives off heat to the surroundings.
An endothermic reaction, the corresponding situation is a temperature decrease in an isolated
system or a gain of heat from the surroundings.
Heats of reaction are experimentally determined in a calorimeter, a device for measuring
quantities of heat absorbed or evolved during a physical or chemical change.
i) Bomb Calorimetry- is a constant volume calorimeter and is ideal for measuring heat evolved in
a combustion reaction.
The system is everything within the double-walled outer jacket of the calorimeter. This includes
the bomb and its contents, the water in which the bomb is immersed, the thermometer, the stirrer,
and so on. The system is isolated from its surroundings.
ii) The coffee-cup calorimeter
is a constant pressure calorimeter. Since this is a constant pressure process, the heat can be
directly related to enthalpy change, H. It is used whenever gases are not involved.
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E.g. Two solutions, 100.0mL of 1.00M AgNO3(aq) and 100.0mL of 1.00 M NaCl(aq), both
initially at 22.4°C, are mixed in a Styrofoam cup calorimeter and allow to react. The
temperature rises to 30.2°C. Determine qrxn per mole of AgCl(s) in the reaction.
Work
E.g. A gas is compressed from 30.0 L to 20.0 L against a constant opposing pressure of 0.500
atm. What is the magnitude of the work involved? Is work done by or on the gas?
V The first Law of Thermodynamics
A system contain only internal energy and does not contain energy in the form of heat(q) or
work(w). Heat or work exists only during a change in the system.
If we consider that an isolated system is unable to exchange either heat or work with its
surrounding, them Uisolated sys. = 0, and we can say
The energy of an isolated system is constant.
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* Any energy entering the system carries a positive sign. q>0 heat is absorbed by the sys. w>0 if
work is done on the sys.
* Any energy leaving the system carries a negative sign.
* In general, the internal energy of a system changes as a result of energy entering or leaving the
sys. as heat or work. U>0 if more energy enters the sys than leave.
The internal energy of a system is a function of state.
To describe a system complete, we must indicate the state of the system -its temperature,
pressure and the kinds and amount of substances present.
E.g. a sample of pure of water at 20°C and under one atm is in a specific state. The density of
water in this state is 0.99820 g/mL which is a unique value-a function of state-in this way.
Obtain three different sample of water, the density at the state we specified for all the sample is
0.9820 g/mL.
The value of a function of state depends on the state of the system, and not how that state was
established.
Path-Dependent Function
Unlike internal energy and changes in internal energy, heat and work are not functions of state
but path-dependent function. Their values depend on the path followed when a system undergoes
a change.
E.g. How much work, in joules, is involved when (a) 0.225 mol N2(g) at a constant temperature
of 23°C is allowed to expand from 1.00L to 2.50L against an external pressure of 2.19atm? (b)
How much work done if the N2(g) first expand to 1.50 L at 3.64atm than to final 2.50L with the
final pressure 2.19 atm?
E.g. Travel trip from LA to Boston.
You are the system and your position is a state function because how you got to wherever you
are is irrelevant. Because your position is a state function, the change in your position when you
travel from LA to Boston (You can fly non-stop from LA to Boston or go from LA-ChicagoBoston) is independent of the path you take. However, any work you do in making the trip, for
example any money or time you spend are not state function. The money or time you spend
depends on how you make the trip.
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Heat of Reaction: U and H
Consider the reactants in a chemical reaction as the initial state of a system and the products as
the final state. Then,
At a constant volume: Such as a combustion reaction in a bomb calorimeter. The original
reactants and products are confined within the bomb, and that the reaction occurs at constant
volume. V=0, and no work is done. That is w = –PV=0. Denoting the heat of reaction for a
constant-volume as qv,
At a constant pressure: In general, we do not carry out chemical reaction in bomb calorimeters
but in beaker or flask under constant-pressure. In these cases, the heat of reaction, qp, is
different from qv. If the process is carry out at constant temperature and pressure (Pf =Pi) and
with work limited to pressure-volume work,
Summary:
The heat of reaction at constant pressure relate to H = U+PV
The heat of reaction at constant volume, U, does not in general equal H. So, to obtain H for
reaction in constant volume (E.g. bomb calorimeter), we must added the correction PV.
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Enthalpy (H) and Internal Energy (U) Change in Chemical Reaction
The heat of reaction at constant pressure, H, and the heat of reaction at constant volume, U,
are related by the expression
U = H –PV
└ the energy associated with the change in volume of the system under a constant external
pressure.
Consider the following reaction
2CO(g) + O2(g) 2CO2(g)
If assume the heat of this reaction under constant-pressure at a constant temperature of 298K, we
get –566.0KJ, indicating that 566.0KJ of energy has left the system as heat:
qp = H = -566.0 KJ.
To evaluate the pressure-volume work, we begin
PV = P (Vf-Vi)
use ideal gas law
PV = RT (nf-ni)
Here nf is the # of moles of gas in the products and ni is the # of moles of gas in the reactant
(2mol CO + 1 mol O2)
PV = 0.0083145 KJ/mol K x 298 K x [2-(2+1)]mol = -2.5 kJ
The change in internal energy
U = H-PV
= -556.0KJ –(-2.5kJ)
= -563.5 KJ
at constant volume qv = U = -563.5 KJ
This calculation show that PV term is quite small compared to H and that U and H are
almost the same. An additional interesting fact here is that the volume of the system decreases as
a consequence of work done by the surroundings on the system.
In general, the PV term is much smaller than the U term so that the total internal energy
change of a reaction is approximately equal to H, also called the heat of reaction.
The heat required to vaporize a fixed quantity of liquid is called the enthalpy (or heat) of
vaporization. When the fixed quantity is one mole, we called this quantity the molar enthalpy of
vaporization.
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E.g. What is the enthalpy change when 7.00g of ice is brought form –10.0C to 30°C?
Specific heat of ice = 2.01 J/g °C and an enthalpy of fusion = 6.01 KJ/mol. Specific heat of
water = 4.18 J/g °C
Enthalpy Diagrams
H<0 means that the enthalpy of the products is lower than that of reactants. This decrease in
enthalpy appears as heat evolved to the surroundings. The reaction is exothermic reaction.
In reaction
N2(g) + O2(g)2NO(g) H = +180.50 KJ
H >0 the products have a higher enthalpy than reactants. To produce this increase in enthalpy,
heat is absorbed from the surroundings. The reaction is endothermic reaction.
Enthalpy
Products
H >0
Products
H<0
reactants reactants
An enthalpy diagram is a digrammatic representation of enthalpy changes in a process.
VII Indirect determination of H: Hess’s Law
1 H is an extensive property. Enthalpy change is directly proportional to the amounts of
substances in a system. (When the thermochemical equation is multiplied by any factor, the
value of H for the new equation is obtain by multiplying the H in original equation by the
same factor) Consider the standard enthalpy change in the formation of NO from its elements at
25°C.
2 H change sign when a process is reversed. If reverse a process, the change in a function of
state reverse sign. Thus, H for the decomposition of one mole of NO(g) is –H of the formation
of one mole of NO(g)
3 Hess’s law of constant heat summation. When two reactions are added together, the
individual reaction enthalpies are summed.
To describe the standard enthalpy change for the formation of NO2(g) from N2(g) and O2(g)
1/2N2(g) + O2(g)  NO2(g) H = ?
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Since enthalpy is a state function that means the enthalpy change for a reaction is independent of
the details of how we do the reaction. The enthalpy change depends only on the initial state and
final state. The reaction can be accomplished in one step with excess amount of oxygen or two
of steps: First, 1/2N2(g) + 1/2O2(g)  NO (g) then NO(g) + 1/2O2(g)  NO2(g), each of
which has a known enthalpy change.
Hess’s Law states: if a process occurs in stages or steps (even if only hypothetically), the
enthalpy changes for the overall process is the sum of the enthalpy changes for the individual
steps.
VIII Standard Enthalpy of Formation
The standard state (standard thermodynamic condition) of a solid or liquid substance is the pure
elements or compound at a pressure of 1 bar (105 Pa) and at the temperature of interest. The
temperature given in this text is 298.15 (25°C) unless otherwise stated.
The standard enthalpy of formation (H°f) of a substance is the enthalpy change that occurs in
the formation of one mole of the substance in the standard state from the reference forms of the
elements in their standard state.
The reference forms of the elements in all but few cases are the most stable forms of the
elements under standard state. Because the formation of the most stable form of an element from
itself is no change at all, the standard enthalpy of formation of a pure element in its reference
form is 0.
Listed here are the most stable forms of several elements at 298K,
Na(s) H2(g) N2(g) O2(g) C(graphite) Br2(l)
Note that te standard enthalpy of formation of an element will depend on the form of the element
E.g. C(graphite)  C(diamond)
H°f = 1.9 KJ
One of the few cases in which the reference form is not the most stable form with the element is
phosphous. Although overtime solid white phosphous converts to solid red phosphous, white
phosphous has been chosen as the reference form.
E.g. P(s, white) P(s, red) H°f = -17.6 KJ
the standard enthalpy of formation are H°f [P(s, white)] =0 and H°f [P(s, red)] =-17.6
Standard Enthalpies of Reaction
The measured enthalpy change for a reaction has a unique value only if the initial state(reactants)
and final state(products) are precisely described. If we define a particular state as standard for
the reactants and products, the standard enthalpy change, H° or H°rxn, is the enthalpy
change in a reaction in which the reactants and products are in their standard states
E.g. Consider the reaction CH4(g) + 4Cl2(g) CCl4(l) + 4HCl(g) H° = ?
C(graphite) +2H2(g)  CH4(g)
H°f = -74.81 KJ/mol
C(graphite) +2Cl2(g)  CCl4(l)
H°f = -135.4 KJ/mol
1/2H2(g) +1/2Cl2(g)  HCl(g)
H°f = -92.31KJ/mol
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Use Hess’s Law
From here we see that
The equation above is a specific application of the following more general relationship for a
standard enthalpy of reation
The standard enthalpy change for a reaction is
E.g. The overall reaction that occurs in photosynthesis in plants is
6CO2(g) + 6H2O(l) C6H12O6(s)+ 6O2(g) H° = 2803 KJ
Determine the standard enthalpy of formation of glucose, C6H12O6(s), at 298K.
Ionic Reaction in Solution
The reference ion assigned is H+(aq) and its enthalpy of formation is assigned as zero in its
aqueous solution. Then, the enthalpy of formation of other ions is compared to this reference
ion.
E.g. Given that H°f [AgI(s)]= -61.84 KJ/mol, what is the standard enthalpy of change for the
precipitation of silver iodide?
Ag+(aq) + I-(aq) AgI(s)