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Thermodynamics Section - 1
FUNDAMENTALS AND FIRST LAW OF THERMODYNAMICS
MOTIVATION Lets start this chapter with a Question : what will happen if we mix any two reactants of our choice? Does a reaction always take place in such a situation ? We know for sure that some reactions do take place as soon as certain reactants are brought together. But is this the case with any two reactants being brought together ? So that brings us to another question, so as to what are the factors (if there are any) responsible for a reaction to take place ? The study of thermodynamics answers all these questions and a few more. So let us delve into the pages that follow and answer these questions for ourselves. INTRODUCTION Let us consider an example of a chemical reaction that occurs in our daily lives : burning of petrol to drive our vehicles. This is an example of a chemical reaction in which energy is released ( and this energy is used to drive our vehicles.) Another example that we could consider is that of splitting of water into hydrogen and oxygen. This reaction actually requires energy in order to take place. Both these examples illustrate a general point i.e. chemical reactions involve changes in energy. The study of these changes in energy accompanying a chemical reaction is called thermodynamics. Usually these energy changes involve heat-hence the “thermo-” part of the term. However, the study of thermodynamics can be divided into two main aspects. The first is thermochemistry. This practical subject is concerned with how we observe, measure, and predict energy changes for chemical reactions. We will limit ourselves mainly to this part in this chapter. The second part addresses a more fundamental aspect of thermodynamics. There we will learn to use energy changes to tell whether or not a given process can occur under specified conditions, and how to make a process more (or less) favourable. We will study a part of this part of thermodynamics in this unit and another part in the unit on equilibrium. First of all, we will focus mainly on thermochemistry. But before we take up its study in detail, we must refresh our knowledge about energy in brief and study some basic terminology interwined with the topics of this chapter. We will take up explanations of a lot many other terminologies as and when they arise in the course of study of this chapter. Chemistry / Thermodynamics
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THE NATURE OF ENERGY We classify energy into two general types : kinetic and potential. We know from our study of mechanics in physics that the kinetic energy of an object is given by : Ekinetic =
1 2 mv 2
The point to be noted here is that greater the mass and/or velocity of a body the greater will be its kinetic energy and hence, more the work the body will be able to accomplish. For example, the heavier a hammer and the more rapidly it moves, the more work it can accomplish. The work that we do to lift an object is stored in the object as energy. This type of energy that a system possesses by virtue of its position or composition is called potential energy. Energy can take many other forms : electrical energy, radiant energy (light), nuclear energy and chemical energy etc. However, so far as the study of this chapter is concerned we need not worry about them as at the atomic or molecular level, we can think of each of these as either kinetic or potential energy. However, we must use this point to review the fact that energy is interconvertible amongst its various forms. Lets consider an example to review this concept : •A solar powered water-pump has photovolatic cells connected to it. These cells collect energy from sunlight, storing it momentarily in a battery, which later runs an electric motor that pumps water into a storage tank on a hill. What energy conversions are involved in using sunlight to pump water into the storage tank? (Try to answer this Question yourself before you look at the solution given here:) Solution : The photo voltaic cells collect sun’s energy, converting it into electrical energy. This electrical energy is stored in the battery as chemical energy, which is later changed back to electrical energy that runs a motor. As the motor rotates, it changes the electrical energy to kinetic energy of the motor, then of the water, which in turn is changed to potential energy (energy of position) of water as the water moves upward in the gravitation field of earth. Finally, before we end our discussion on energy with the important concept of internal energy lets take a quick glance at the units in which energy is measured. The S.I. unit of energy is kg ⋅ m 2 / s 2 . It has been given the name Joule. The Calorie (cal) is a non-SI unit of energy, originally defined as the amount of energy required to raise the temperature of one gram of water by one degree celsius. Exact definition of calorie is the amount of heat required to raise temperature of one gram of water from 14.5°C to 15.5°C. This amount of heat expressed in terms of Joules is given by:
1 calorie = 4.184 Joules We are now ready to end our discussion on energy with the important concept of internal energy. INTERNAL ENERGY Consider the total energy of a quantity of water as it moves over a dam. This water as a whole has kinetic energy (because of its motion) and potential energy (because of its position above the ground.)
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However, we know that water is made up of molecules, which are made up of smaller particles electrons and nuclei (comprising of neutrons and protons). Each of these particles also possess various kinds of potential and kinetic energies at the molecular / atomic level. All these various kinds of kinetic and potential energies of a substance at the molecular level together make up what is known as the internal energy (denoted by U) of that substance. Some points that should be noted w.r.t. the concept of internal energy are: (1) The kinetic energy and potential energy of the substance as a whole is different from the various energies which make up the internal energy, U. (2)
Therefore, the total energy of a substance can be represented as : Etotal = EK + EP + U (where, EK and EP represent the kinetic energy and potential energy of the substance as a whole and U represents, its internal energy).
(3)
However, normally when we study a substance in a laboratory, the substance is at rest in a vessel. Its kinetic energy as a whole is zero. Moreover, its potential energy as a whole is constant and you can take it to be zero (why ?). In this case the total energy will be equal to the internal energy of the substance.
(4)
The exact value of the internal energy of a substance cannot be determined easily. However, the change in the internal energy of a substance as it goes form one state to another can be determined quite easily. This change is represented by ∆U and written as ∆U = U f − U i
(5)
(where U f and U i represent the internal energy of the substance in the final state and initial state, respectively.) We have discussed above in an example that one form of energy can be converted into another form of energy. For example, when water falls over a dam, potential energy is converted into kinetic energy. Some of the kinetic energy of water (as a whole; Ek ) may also be converted into random molecular motion - that is, into internal energy of the water. The total energy, Etotal , of the water, however remains constant, equal to the sum of the kinetic energy, Ek ; the potential energy, E p and the internal energy, U of the water. This result can be stated more generally as the law of conservation of energy : energy may be converted from one form to another, but the total quantity of energy remains constant.
Let us now study some of the basic terminologies associated with the study of this chapter as a whole. (a)
Exothermic Reactions Lets consider the example of burning of petrol considered earlier. We know that when the chemical reaction involving burning of petrol takes place, energy is released that is used to drive our vehicles. Such kind of chemical reactions accompanied by a release of energy are called exothermic reactions. Let us use this point to study such reactions along with some related concepts using a case study. Case study 1
Chemistry / Thermodynamics
Combustion of methane CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O (l ) + 890kJ
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As we can see, this reaction is accompanied by a release of 890 kJ of energy. Hence, this is an example of exothermic reaction. However, things are not as simple as they seem to be. Let us see so as to why. It is known that to start off the combustion of methane we need to provide the reactants a certain amount of energy in form of heat i.e. the reaction between CH 4 and O2 does not come about spontaneously but only after an initial activation (i.e. by supplying energy). But there is an anomaly that arises out of this situation : The initial activation means, the energy supplied to start off the reaction is absorbed by the reactants and hence, their total energy increases. But after the reaction takes place 890 kJ of energy is released. That means the total energy of products is 890 kJ lower than that of the reactants. But wait! which reactants are we talking about here? The reactants in the normal state or in the excited state (i.e. after activation)? The reactant in these different states are at different energy levels and hence, their energy difference with respect to the products will be different. So which of the differences does the 890 kJ of energy release represents? Our common sense would say that since, the reaction started in the excited state, this energy represents the difference between the excited state and the products. But this is not so. This difference is actually between the reactants in their normal state and the products. We explain this as follows : Consider this diagram : total energy of the reactants in excited state
Heat content →
CH3(g)+2O2(g)
amount of energy needed to activate the reaction. (energy difference - 1)
Heat evolved = 890kJ (energy difference -2) CO2(g)+2H2O(l)
reactants → products.
fig. 1
We see that the energy difference -1 between the excited state and the products is indeed much more than energy difference -2 (890 kJ). But if we look closely we realise that the amount by which energy difference 1 is greater than energy difference 2 is same as the activation energy supplied. So, we see that the energy released i.e. 890 kJ is not illustrative of all the energy changes taking place in the reaction but only of the difference in energy in the initial and final states. (We will have a lot to say about the states later in the chapter. But for now the intuitive idea that you have about it shall suffice.) Important points (1) The amount of heat shown in the reaction of the kind illustrated in this case study always refers to the reaction for the number of moles of reactants and products specified by the coefficients in the balanced equation of the reaction. We call this “one mole of reaction”. (2) It is important to specify the physical states of all substances because different physical states have different energy contents.
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(b)
Endothermic Reactions So, we just saw that a chemical reaction which is accompanied by a release of energy is called an exothermic reaction. What about a chemical reaction which is accompanied by an absorption of energy (i.e. the total energy of products is greater than that of reactants. )? Such a reaction is called an endothermic reaction. (We will have more to say about these reactions later.)
(c)
System and Surroundings In the introduction of this chapter, we considered examples of chemical reactions which were accompanied by either absorption or release of energy. The energy released/ absorbed in both these reactions can be given a special name called “heat of reaction”. To understand this concept, we need to know what is meant by a thermodynamic system and its surroundings and to have a precise definition of the term “heat”. Suppose that we are interested in studying the change of a thermodynamic property (such as internal energy) during a chemical change. The substance or mixture of substances under study in which a change occurs is called the thermodynamic system (or simply system). The surroundings are everything in the vicinity of the thermodynamic system. (refer the figure below):
Surroundings
Surroundings
Surroundings
System (under study)
fig. 2 : Illustration of a thermodynamic system: The system consists of the portion of the universe that we choose to study. Everything else constitutes the surroundings.
As we have pointed out, a system consists of a substance or a mixture of substances (or simply, matter). And when it undergoes a chemical change it is (most definitely) accompanied by a change in energy. Depending upon how this matter and energy are shared by the system with the surroundings we categorise systems into 3 types:
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(i) open system A system is said to be open if it can exchange matter as well as energy with the surroundings. For example, in the figure 2 above if the vessel remains uncovered, we could term it as an open system. (ii) closed system A system is said to be closed if it can exchange only energy with the surroundings but not matter. For example, if we cover the vessel in figure 2 the system will not be able to exchange matter with its surroundings. However if the vessel is made up of a conducting material like copper or steel it can exchange energy with the surroundings. (iii) isolated system A system is said to be isolated if it can exchange neither matter nor energy with the surroundings. As an example we could again consider the same vessel in figure 2 but this time covering it with an insulating cover and considering it to be made up of an insulating material as well. It would then be able to exchange neither matter nor energy with the surroundings. (However, in reality, no known system is isolated. A thermos flask comes closest to being an isolated system) NOTE : In the example of figure 2 above, the surroundings and the system are separated by the walls of the vessel. Such a wall that separates a system from its surroundings is called boundary. The importance of boundary lies in the fact that it allows us to control and keep track of all movements of matter and energy in or out of the system.
Chemistry / Thermodynamics
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Heat Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings. As long as a system and its surroundings are in thermal contact (i.e. they are not thermally insulated from one another), energy (heat) flows between them to establish temperature equality, or thermal equilibrium. Heat flows from a region of higher temperature to one of lower temperature. Once the temperatures become equal, the heat flow stops. (Note: Once heat flows into a system, it appears in the system as an increase in its internal energy. It is incorrect to say that the system has heat, because heat is only an energy flow.) Heat and temperature are sometimes confused. The distinction is clear in the kinetic picture of a gas. According to the kinetic theory, the absolute temperature of a gas is directly proportional to the average kinetic energy of the molecules. When we add heat to a gas, we increase its internal energy and therefore its total kinetic energy. This increase in kinetic energy will be distributed over the molecules in the sample. Therefore, the increase in average kinetic energy per molecule(and therefore the increase in temperature) depends on the size of the gas sample. A given quantity of heat will raise the temperature of a sample more if the sample is small. Heat is denoted by the symbol q. The algebraic sign of q is chosen to be positive if heat is absorbed by the system and negative if heat is evolved. The sign of q can be remembered this way : when heat is absorbed by a system, energy is added to it; q is assigned a positive quantity. On the other hand, when heat is evolved by a system, energy is subtracted from it; q is assigned a negative quantity. A still easier way to remember it could be that gain of heat is taken as positive and loss of heat is taken as negative (as in our everyday life.)
(e)
The thermodynamic state of a system It is defined by a set of conditions that completely specifies all the properties of the system. This set commonly includes the temperature, pressure, composition (identity and number of moles of each component), and physical state (gas, liquid or solid) of each part of the system. Once the state has been specified, all other properties-both physical and chemical are fixed.
(f)
State Functions Lets define a thermodynamic system as 50 grams of water at 25°C at 1 atm as in the figure below : 50 g H2O (l) 25°C
fig. 3 Chemistry / Thermodynamics
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Our system could have arrived at that state by our cooling 50 grams of water from 100°C or by our melting 50 grams of ice and subsequently warming the water to 25°C (as shown in the figure below) 50 g H2O(l) 100°C
50 g H2O (s) 0°C
50 g H2 O (l) 25°C
Cooling
Heating
fig. 4 No matter which way we arrived at our system we will find that the internal energy of the water after getting to 25°C and 1 atm. is same in both the cases i.e. the internal energy of a system depends only on the present state of the system and not on how that state was attained. Such a property of a system that depends only on its present state(which is determined by variables such as temperature and pressure) and is independent of any previous history of the system is known as state function. So, we can say that internal energy is a state function i.e. a change in the internal energy ∆U does not depend upon how the change was made, but only on initial and final states of the system. Another simple analogy that we could consider to further clarify the definition of state function is that of displacement and distance. Consider the following figure : Y
Path - A X
Path - B
Path - C
fig. 5
!!!" Here XY is the displacement vector between X and Y. But we could go from X to Y in infinite number of ways, three of which (path -A, B and C) are shown in the figure. If we consider the distance between X and Y we find that it depends upon the path chosen to cover the distance between X and Y. But no matter which path is chosen to cover the distance between X and Y, the displacement in all these cases is one and the same. So, displacement can be termed as a state function ( but not distance.) Some examples of state functions are listed here. You are requested to convince yourself that each of these is indeed a state function. (a) Temperature (e) Enthalpy (b) Pressure (f) Entropy (c) Density (g) Heat capacity (d) Refractive index (h) Gibbs free energy ((e) – (h) will be discussed later in this chapter). Chemistry / Thermodynamics
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(g)
Intensive and Extensive Properties An intensive property of a substance is a property which does not depend upon the amount of the substance. For example, temperature, melting point, density, pressure, boiling point etc. are all intensive properties. The intensive properties are particularly useful because many of them can be used to identify substances. On the other hand, an extensive property of a substance is a property that depends upon the amount of substance. For example, mass, volume, length, heat, enthalpy ( to be discussed later) are all extensive properties.
(h)
Work Earlier we saw that one way in which energy flows in or out of a system is in form of heat. Another form in which energy flows into or out of a system is work. Let us consider an example to understand this concept. Suppose we introduce two gases in an insulated cylindrical vessel with a movable piston (with weight, k) as shown in the figure.
Movable piston of weight, k Gaseous reaction mixture
fig. 6 Let us suppose that the reaction between the gases is exothermic and the no. of moles of gases before and after the reaction is same. Now, since, the reaction is exothermic, the temperature of the system will increase. According to the ideal gas law, if volume is kept constant too, pressure will increase and will move the piston upwards. In moving the piston upwards the system does ‘work’ on the piston (which can be considered a part of surroundings) and hence, expends energy which appears as a decrease in the internal energy of the system. So, we conclude that if a system does work, w on its surroundings, its internal energy reduces by the same amount. We can sum this up as : Work is done by the system on the surroundings w is negative. Work is done on the system by the surroundings
w is positive.
We are now ready to enter the realm of thermodynamics. We start with the study of “first law of thermodynamics”. FIRST LAW OF THERMODYNAMICS We know that Internal energy, U of a system is a state function, i.e. it is a property of a system that depends only on its present state, which is completely determined by variables such as temperature and pressure. Thus, 1 mol of water at 0°C and 1 atm pressure has a definite quantity of energy. So, when a system changes from one state to another, its internal energy changes from one definite value to another We can calculate the change in internal energy, ∆U , from the initial value of the internal energy, U i and the final value of the internal energy, U f . ∆U =U f –Ui Chemistry / Thermodynamics
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But, normally, we are interested in changes in internal energy rather than in absolute values. We measure these changes for a thermodynamic system by noting the exchanges of energy between the system and its surroundings. These exchanges of energy are of two kinds : heat and work (both have been discussed earlier.) The distinction between heat and work is illustrated in the system shown in the figure 7 below: Work Weight Piston Gas Heat
fig.7 : Exchanges of heat and work with the surroundings. A gas is enclosed in a vessel with a piston. Heat flows from the surroundings, which are at a higher temperature. As the temperature of the gas increases, the gas expands, lifting the weight (doing work).
This system consists of a gas in a vessel equipped with a movable piston. On top of the piston is a weight, which we can consider as part of the surroundings. We momentarily fix the position of the piston so it does not move. Suppose the temperature of the surroundings is raised, and as a result heat passes from the surroundings to the vessel. If we find that the energy of the surroundings decreases by 215 J in this way, we know from the law of conservation of energy that the internal energy of the system must have increased by just this quantity. We write q = +215 J , using the sign convention that heat (denoted by the symbol q) absorbed by the system is positive (energy is added to the system) and that heat evolved by the system is negative (energy is subtracted from the system). As the temperature of the gas in the vessel increases, the gas pressure increases at fixed volume. We now allow the piston to move, so the gas expands and lifts the piston and the weight on top of it. In lifting the weight, the system does work. The energy gained by the weight equals the force of gravity on the weight times the height to which the weight was raised (the distance it was moved). Suppose this energy is 105 J. Because the surroundings, which include the weight, have gained 105 J of energy, the system must have lost 105 J of energy (energy cannot be created or destroyed). We write w = –105 J, adhering to the sign convention that work done on the system is positive (energy is added to the system) and work done by the system is negative (energy is subtracted from the system.) The system in the figure above gains internal energy from the heat absorbed and loses internal energy via the work done. This observation is actually a consequence of the general law :
∆U =q +w Chemistry / Thermodynamics
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In writing this equation, we have actually written the first law of thermodynamics, mathematically. As a statement this would read as : The first law of thermodynamics states that the change in internal energy of a system, ∆U , equals q + w. (So, for the system shown in figure above, ∆U =(+215 J )+(–105 J )=110 J ). (Important note : We should note that the first law of thermodynamics is actually the law of conservation of energy applied to thermodynamic systems.) HEAT OF REACTION Consider a system in which a chemical reaction occurs. Before the reaction, the system and surroundings are at the same temperature, say 25°C. When the reaction starts, however, the temperature changes. Suppose the temperature of the system falls. In that case, heat flows from the surroundings into the system. When the reaction stops, heat continues to flow until the system returns to the temperature of its surroundings (at 25°C). Heat has flowed into the system; that is, the system has absorbed heat and q is positive. (This is an endothermic process). Suppose on the other hand, that the temperature of the system rises. In this case, heat flows from the system to the surroundings. When the reaction stops, heat continues to flow until the system returns to the temperature of its surroundings (at 25°C). In this case, heat has flowed out of the system; that is the system has evolved heat and q is negative. This is an exothermic process. We are now ready to define the heat of reaction : The heat of reaction (at a given temperature) is the value of q required to return a system to the given temperature at the completion of the reaction. We summarise the above discussion as follows: Type of reaction Endothermic
Result on system Energy added
Sign of q +
Exothermic
Energy substracted
–
Please try to answer the following problem by yourself before seeing its solution. •
The equation for the combustion of butane, C4 H10 , is C4 H10 ( g ) + (13 2)O2 ( g ) → 4CO2 ( g ) + 5 H 2O( g ) Which one of the following generates the least heat ? Why ?
(a)
Burning one mole of butane
(b)
Reacting one mole of oxygen with excess butane.
(c)
Producing one mole of CO2 by burning butane.
(c)
Producing one mole of H 2O by burning butane.
Chemistry / Thermodynamics
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Solution: The equation says that 1 mole of butane reacts with (13/2) moles of oxygen to yield 4 moles CO2 and 5 moles of H 2O . The reaction yields a certain amount of heat, which we can symbolise as q. So, (a) yields heat q. On the other hand, (b) involves only 1 mole oxygen (and not (13/2) moles). So, the heat involved in (b) will be equal to (2/13) q. In (c), only 1 mole of CO2 is involved which is 1/4th of that given in the equation. This means that (c) yields (1/4) q. Similarly, (d) yields (1/5)q. Therefore, (b) yields the least heat. PRESSURE–VOLUME WORK AND ∆U According to the first law of thermodynamics, ∆U = q + w
i.e. the change in internal energy of a system is sum of the heat changes accompanying the reaction and the work done by/on the system. We wish to focus upon the latter part of this sum i.e. work done on/by the system. But the only type of work involved in most chemical and physical changes is pressure-volume work. Let us see so as to what it means by the following case study : Case study 2
Consider the reaction Zn( s ) + 2 HCl (aq ) → ZnCl2 (aq ) + H 2 ( g ).
When this reaction is carried out in beaker open to the atmosphere, the reaction is exothermic, evolving 152.4 kJ of heat per mole of zinc consumed. Since the reaction occurs open to atmosphere, we can say that it occurs at atmospheric pressure. Hence, we may conclude that reaction occurs at constant pressure. The heat changes accompanying a reaction taking place at constant pressure, can be given a special notation : q p . So, in the reaction above, we may write q p = −152.4 kJ . Now, note that before the reaction takes place, no gases are involved. So the hydrogen gas that is produced increases the volume of the system. As hydrogen is evolved, work must be done by the system to push back the atmosphere. How can we calculate this work? Imagine for the moment that the atmosphere is replaced by a piston and weights, whose downward force from gravity F creates a pressure on the gas equivalent to that of the atmosphere. The pressure P equals F divided by the cross sectional area of the piston, A. (see the figure :) H2
HCl
Zn (Before reaction) The beginning of the reaction. The constant pressure of the atmosphere has been replaced by a piston and weight to give an equivalent pressure.
Chemistry / Thermodynamics
fig. 8
(After reaction) The reaction produces hydrogen gas. This increases the volume of the system, so that the piston and weight are lifted upward. Work is done by the system on the piston and weight.
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Suppose the increase in volume of the system due to the production of hydrogen is ∆V . Because the volume of a cylinder equals its height, h times its cross-sectional area, A, the change in volume is ∆V = A × h (where h is the change in height of the piston). ∴ h = ∆V A . Now, the work done by the system in expanding equals the force of gravity times the distance the piston moves. So, we have, w = −F × h = −F ×
∆V F = − × ∆V A A
(The negative sign is given because w is work done by the system and represents energy lost by it.) Now, note that F A is actually pressure (atmospheric pressure in our case). Therefore we can write,
w = − P∆V This formula tells us that we can calculate the work done by a chemical reaction carried out in an open vessel (i.e. at constant atmospheric pressure) by multiplying the atmospheric pressure, P by the change in volume of the chemical system, ∆V . In order to illustrate what we have just discussed, suppose that in the above reaction when 1 mole of Zn reacts with excess HCl, 1 mole of H 2 is produced. At 25°C and 1atm (= 1.01×105 Pa), this amount of H 2 occupies 24.5 L(= 24.5 ×10−3 m3 ). The work done by the chemical system in pushing back the atmosphere is W = − P∆V = −(1.01× 105 Pa) × (24.5 × 10−3 m3 ) = −24.7 × 102 J = −2.47 kJ
If we apply the first law of thermodynamics to this chemical system, we can relate the change in internal energy of the system to the heat of reaction. So, we have, ∆U = q p + w = q p − P ∆V = −152.4 kJ − 2.47 kJ
= −154.9kJ We can sum this up as : When the reaction occurs, the internal energy changes in going from reactants to products. This energy change, ∆U , equals −154.9kJ . Energy leaves the system mostly as heat ( q p = − 152.4 kJ ) but partly as expansive work ( w = −2.47 kJ ).
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HEAT CHANGE AT CONSTANT VOLUME, qvV From the discussion above ∆U = q − P∆V (assuming that P – V work is the only work that can be done.)
Now, suppose we carry out a reaction in a constant volume container. Then since the volume is constant, ∆V = 0. Then we can write the above equation as ∆U = qv
(the subscript v indicates that the volume is constant.)
i.e. if the volume is constant, the heat of reaction is same as change in internal energy. So far, so good. But the experimental condition of constant volume is difficult to guarantee for many reactions. In fact, constant pressure, considering that many experiments occur exposed to atmosphere, is often an easier experimental parameter. But if the pressure is constant and volume is not then we know that ∆U = q p − P ∆V . ⇒
q p = ∆U + P ∆ V .
We see that the heat of reaction is not same as the change in internal energy if the volume of the system changes after reaction takes place. i.e. the concept of internal energy is not sufficient to study the heat changes in a reaction taking place at constant pressure. In order to study the heat changes in a reaction at constant pressure, we define a new thermodynamic property of the system, called Enthalpy.
ENTHALPY The enthalpy is defined as H = U + PV. U, P and V are all state functions, so H is also a state function. This means that for a given temperature and pressure, a given amount of a substance has a definite enthalpy. Let us now show the relationship of ∆H to the heat of reaction, q p . Note that ∆H is the change in enthalpy which is equal to the final enthalpy H f minus the initial enthalpy H i i.e. ∆H = H f – H i
Now, let us substitute defining expression for H f and H i , using the subscripts f and i to indicate final and initial, respectively to get: ∆H = (U f + Pf V f ) – (U i + PV i i) ⇒
⇒
∆H = (U f – U i ) + ( Pf V f – PV i i)
∆H = ∆U + ( Pf V f – PV i i)
This expression gives us the change in enthalpy of a system. However if the reaction occurs at constant pressure, we can rewrite the above equation as : Chemistry / Thermodynamics
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∆H = ∆U + PV f – PVi ⇒
(where P is the constant pressure)
∆H = ∆U + P (V f – Vi )
⇒
∆H = ∆U + P ⋅ ∆V
But we know that for a reaction occurring at constant pressure, ∆U = q p – P∆V . Substituting this in the above equation, we get : ∆H = ( q p – P ∆V ) + P ∆V
⇒
∆H = q p
So, we see that the heat change taking place in a system at constant pressure is same as the change in enthalpy of the system. Important Points 1. Enthalpy is an extensive property. 2. The change in enthalpy for a reaction at a given temperature and pressure is called the enthalpy of reaction. It is given by: ∆H = q p , i.e. the enthalpy of reaction is same as heat of reaction at constant
3.
4.
pressure. The enthalpy of reaction must not be confused with enthalpy of reactants and products. The former denotes a change in enthalpy whereas the latter denotes absolute enthalpies. The difference between ∆U and ∆H is the amount of P - V work done by the system, – P∆V . The volume change accompanying many reactions is close to zero, which makes P∆V , and therefore the difference between ∆U and ∆H small. It is generally satisfactory to use ∆H as the measure of energy changes during most chemical processes. Solids and liquids do not expand or contract significantly as pressure changes ( ∆V ≈ 0). In reactions in which equal no. of moles of gases are produced and consumed at constant temperature and pressure, essentially no work is done. By the ideal gas law, P∆V = ( ∆n) RT and ∆n = 0, where ∆n equals the no. of moles of gaseous products minus the no. of moles of gaseous reactants. Thus, the work term w has a significant value only when there are different numbers of moles of gaseous products and reactants so that the volume of the system changes. In such a case, P ∆V = ( ∆n) RT can be substituted in ∆H = ∆U + P∆V to obtain : ∆H = ∆U + (∆n) RT
5.
In order to measure ∆U , experimentally, we need a calorimeter whose volume is constant. For this a bomb calorimeter is used. ∆H is measured experimentally in an ordinary calorimeter which is open to atmosphere.(Calorimetry will be covered in detail in the LOCUS study material of Physics for Heat)
6.
Because the energy changes of so many processes are measured under conditions of constant pressure, the change in enthalpy for a process is usually easier to measure than the change in internal energy. As such, although the internal energy is the more fundamental quantity, the enthalpy is the more common.
7.
Because ∆H is a common state function, we base the definitions of some terms on enthalpy, not internal energy. For example, the heat changes accompanying an exothermic/ endothermic reaction are written in terms of ∆H .
Chemistry / Thermodynamics
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16
THERMO CHEMICAL EQUATIONS Consider the reaction of hydrogen and oxygen to produce water. Is the ∆H accompanying this reaction always the same? The answer is NO. Let us see so as to why. If the product is water vapour, 2 moles of H 2 burn to release 483.7 kJ of heat. i.e.
2 H 2 ( g ) + O2 ( g ) → 2 H 2O( g ); ∆H = −483.7 kJ On the other hand, if the product is liquid water, the heat released is 571.7 kJ. i.e.
2 H 2 ( g ) + O2 ( g ) → 2 H 2O(l ); ∆H = −571.7 kJ (In this case, additional heat is released when water vapour condenses to liquid.) From the above example it is clear that when we are studying heat changes accompanying chemical reactions we must specify the states of reactants and products along with the heat changes accompanying the reaction. This concept is so important that we give this type of specifying a reaction a new name; called as Thermochemical Equation. We define a thermochemical equation as the chemical equation for a reaction (including state labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. We further elucidate the above definition in the following example : Consider the thermochemical equation for the combustion of ethyl alcohol : C2 H 5OH (l ) + 3O2 ( g ) → 2CO2 ( g ) + 3H 2O(l ); ∆H = −1367kJ The state labels are mentioned and so is the ∆H for the reaction. The only thing left to be explained is “molar interpretation”. In giving a molar interpretation to the above thermochemical equation, we can say that for every 3 moles of H 2 O produced 1367 kJ of heat is released in the above reaction (or) for every 1 mole of C 2 H 5OH consumed the heat released is still 1367 kJ and so on. The following are important rules for manipulating thermochemical equations: 1.
When a thermochemical equation is multiplied by any factor, the value of ∆H for the new equation is obtained by multiplying the value of ∆H in the original equation by that same factor. This is because enthalpy of a system is an__________property. (please fill in the blanks yourself).
2.
When a chemical equation is reversed, the value of ∆H is reversed in sign.
3.
The coefficients in a balanced thermochemical equation refer to the numbers of moles of reactants and products involved. In the thermodynamic interpretation of equations we never interpret the coefficients as number of molecules. Thus it is acceptable to write coefficients as fractions rather than as integers, when necessary.
As discussed previously, for most of the solved problems, a critical thinking box is included before solution to every problem. You are requested to think about the approach you would take to solve the problem. Only then study the critical thinking box and compare your approach to the approach given. This kind of study will ensure that any errors in your approach are rectified and also help you to develop your own thinking process. You should then try to attempt the problem yourself and only then, see its detailed solution. Chemistry / Thermodynamics
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Example – 1 Suppose we use two identical pencil batteries under same thermodynamic conditions in a torch and a toy respectively. In a torch, the internal energy of a battery is converted to heat( and radiant energy), whereas in a toy it is converted to heat and mainly, work. So, q and w for both the systems are different. Does this violate the first law of thermodynamics ? Critical thinking The first law of thermodynamics says, ∆U = q + w. But it does not mention how these changes of q and w come about. So, the relative contribution of q and w to ∆U may vary but their sum will be constant for a ∆U between two thermodynamic conditions. Solution: If the battery is discharged in a torch, no mechanical work is accomplished. All the energy lost from the battery appears as radiant energy and heat. However, if the battery is used in a toy, the same change in the state of the battery produces mechanical work and heat. However, for a given thermodynamic condition, the change in the state of the system (i.e. battery) and thus the change in ∆U are the same in both cases. This is illustrated in the following diagram. (a)
(b)
Charged battery Heat Heat + radiant energy
∆U
Energy lost by battery
Work
Discharged battery
When a battery is discharged in lighting a torch, all the energy of the battery appears as radiant energy and heat; no work is done. When the battery is used in the toy car, work is done in moving the car from place to place. Thus, the work done by the system (the battery) is not a state function because its magnitude depends on the particular path by which the system gets from its initial state to its final state.
fig. 9 So, we see that the amount of work done in the two cases is different, as is the amount of heat released. So, we may conclude that although the ∆U is always same for a given change in the system, the way in which the change is performed will determine the relative contributions of q and w, the means by which the energy is transferred. So, we see that the first law of thermodynamics is not violated. Example – 2 How much heat is evolved when 907 kg of ammonia is produced according to the following thermochemical equation ? N 2 ( g ) + 3H 2 ( g ) → 2 NH 3 ( g ); ∆H = −91.8kJ Chemistry / Thermodynamics
LOCUS
18
Critical thinking The calculation involves converting grams of NH 3 to moles of NH 3 and then to kilojoules of heat. Grams of NH 3 → moles of NH 3 → kilo joules of heat We can obtain the conversion factor for the second step from the thermochemical equation, which says that the production of 2 mol NH3 is accompanied by qp = -91.8 kJ.
Solution: 907 kg NH 3 = 9.07 × 105 g NH 3 Hence, 9.07 × 105 grams of NH 3 ×
1 mol NH 3 17.0 grams NH 3
×
−91.8kJ = −2.45 × 106 kJ . 2 mol NH 3
Thus, 2.45 × 106 kJ of heat evolves. Example – 3 For each of the following chemical reactions carried out at constant temperature and constant pressure, predict the sign of w and tell whether work is done on or by the system. (a)
Ammonium nitrate, commonly used as a fertilizer, decomposes explosively.
2 NH 4 NO3 ( s ) → 2 N 2 ( g ) + 4 H 2O ( g ) + O2 ( g ) (b)
H 2 ( g ) + Cl2 ( g ) → 2 HCl ( g )
(c)
2SO2 ( g ) + O2 ( g ) → 2SO3 ( g )
Critical thinking For a process at constant temperature, w = − P∆V = − ( ∆n ) RT . For each reaction, we evaluate ∆n, the change in number of moles of gaseous substances in the reaction. ∆n = (no. of moles of gaseous products - no. of moles of gaseous reactants.) Because both R and T
(on the kelvin scale) are positive quantities, the sign of w is opposite from that of ∆n; it tells us whether the work is done on ( w = + ) or by ( w = − ) the system. Solution: (a) Here there are no gaseous reactants. ∆n = 2 mol N 2 ( g ) + 4 mol H 2O ( g ) + 1mol O2 ( g ) − 0 mol = 7 mol − 0 mol = +7mol.
i.e ∆n is positive, so w is negative. This tells us that work is done by the system. The large amount of gas formed by the reaction pushes against the surroundings. Chemistry / Thermodynamics
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19
(b) ∆n = 2 mol HCl ( g ) − 1 mol H 2 ( g ) + 1mol Cl2 ( g ) = 2 mol − 2mol = 0 mol.
Thus, w = 0, and no work is done as the reaction proceeds. We can see from the balanced equation that for every two moles (total) of gas that react, two moles of gas are formed, so the volume neither expands nor contracts as the reaction occurs. (c) ∆n = 2 mol SO3 ( g ) − 2 mol SO2 ( g ) + 1 mol O2 ( g ) = 2 mol − 3 mol = −1mol.
i.e. ∆n is negative, so w is positive. This tells us that work is done on the system as the reaction proceeds The surroundings push against the diminishing volume of the gas. Example – 4 Calculate the difference between heats of reaction at constant pressure and constant volume for the reaction
2C6 H 6 (l ) + 15 O2 ( g ) → 12 CO2 ( g ) + 6 H 2O (l ) at 25º C in kJ/mol. Critical thinking Note that we have to find the difference between ∆H and ∆U . Is it necessary to know exact values of ∆H and ∆U to calculate this difference? If it is, then we cannot calculate them from the given data. Think about the relation that connects ∆H and ∆U . Can it be used to calculate the required difference with a slight manipulation?
∆H = ∆U + P ∆V
Solution:
⇒
∆H = ∆U + ( ∆n ) RT
⇒
∆H − ∆U = ( ∆n ) RT
Here, ∆n = 12 – 15 = – 3. ∴
∆H − ∆U = ( −3) (8.314 ×10 −3 kJ K −1mol −1 ) ( 298 K )
= – 7.43 kJ / mol.
Example – 5 An insulated container contains 1 mol of a liquid. Molar volume of this liquid is 100 ml at 1 bar. When liquid is steeply pressed to 50 bar, volume decreases to 30 ml. ∆U for the process is found to be 600 bar.ml. Find ∆H for the process. Chemistry / Thermodynamics
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Critical thinking We see that pressure is not constant. So, there is no q p for this process. So, how do we calculate the value of ∆H ? The key to finding this lies in going back to the basic definition of H i.e. H = U + PV . Solution: Here, Pi = 1 bar
Pf = 50 bar
Vi = 100 ml.
V f = 30 ml.
∆U = 600 bar − ml
Now, by basic definition of Enthalpy; H = U + PV . ∴
and
H f = U f + Pf V f
...(I)
H i = U i + PV i i
...(II)
Subtracting (II) form (I), we get, H f − H i = U f − U i + Pf V f − PV i i ⇒
∆H = ∆U + Pf V f − PV i i.
Substituting the values we get, ∆H = 600 bar − ml + (50 × 30 − 100 × 1)bar − ml = (600 + 1500 − 100)bar − ml
= 2000 bar ml
Chemistry / Thermodynamics
LOCUS
21
TRY YOURSELF - I 1.
Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce hydrogen gas in (a) a closed vessel of fixed volume.
2.
(b) an open beaker at 25º C.
From the conditions and the given definitions of the system, determine whether there is work done by the system, work done on the system, or no work done. (a)
A balloon expands as a small piece of dry ice (solids CO2) inside the balloon sublimes (balloon = system)
(b) The space shuttle’s cargo bay doors are opened to space, releasing a little bit of residual atmosphere. (cargo bay = system). (c)
Gaseous CHF2Cl, a refrigerant is compressed in the compressor of an air conditioner, to try to
liquefy it. (CHF2Cl = system ) . 3.
Indicate which state function change is equal to heat of the reaction for each process described: (a)
The ignition of a sample in a bomb calorimeter.
(b) The melting of an ice cube in a cup. (c)
The cooling down of the inside of a refrigerator.
(d) A fire in a fire place. 4.
A 1.00 L sample of a gas at 1.00 atm pressure and 298 K expands isothermally and reversibly to 10.0 L. It is then heated to 500 K, compressed to 1.00 L, and then cooled to 25º C. What is ∆U for the overall process?
5.
Under what conditions will ∆U be exactly zero for a process whose initial conditions are not the same as its final conditions?
6.
What is ∆H of the reaction
PbO ( s ) + C ( s ) → Pb ( s ) + CO ( g ) if 23.8 kJ must be supplied to convert 49.7 g lead (II) oxide to lead? 7.
Which is more exothermic, the combustion of one mole of methane to form CO2(g) and liquid water or the combustion of one mole of methane to form CO2 ( g ) and steam? Why? (No calculations are necessary).
Chemistry / Thermodynamics
LOCUS
8.
9.
22
For each of the following reactions carried out at constant pressure, state whether work is done by the system on the surroundings or by the surroundings on the system, or whether the amount of work is negligible. (a)
C6 H 6 (l ) → C6 H 6 ( g )
(b)
1 N ( g ) + 3 H ( g ) → NH ( g ) 3 2 2 2 2
(c)
SiO2 ( s ) + 3C ( s ) → SiC ( s ) + 2 CO ( g )
How much heat is evolved from 10 grams of methane in the following reaction assuming you have an excess of oxygen?
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O (l ) ∆H = −890.3 kJ
Chemistry / Thermodynamics