1 5.1 THE NATURE OF THERMODYNAMICS Thermodynamics is concerned with the relationship between the energy changes that occur in chemical and physical processes. In effect, the laws of thermodynamics are the embodiment of the totality of the experimental facts concerning energy changes that occur in nature. The vast, diverse body of experimental facts concerning these energy changes are summarized in the statements of the three laws of thermodynamics. These can be used to predict the direction in which a process would proceed. Thus this science is of great importance to the chemist, since it can predict whether or not a chemical reaction will proceed spontaneously from a knowledge of the thermodynamic properties of the reactants and products. If thermodynamics tells us that a prospective reaction is favorable, then we are on a safe ground to try to find the proper experimental conditions to make the reaction proceed. For example, when gaseous H 2 and O 2 are mixed, thermodynamics tells us that H 2 O should be formed, since H 2 O is more energetically stable than H 2 and O 2. It is a well-known fact that at room temperature, a mixture of H 2 and O 2 will not produce H 2 O . However, if the mixture is sparked, water is produced with explosive violence, thereby proving the prediction thermodynamics makes concerning this reaction. One limitation of thermodynamics is that it does not give any indication of how fast the reaction will proceed, i.e., in essence, the laws of thermodynamics do not contain time as a variable. Thus thermodynamics is only interested in where the system was initially and where it is after the completion of the process. It does not care how the attainment of the final condition came about or how long it takes to reach this condition. In the case of a chemical reaction, for example, thermodynamics may tell us that the reaction will go spontaneously, but it cannot tell us how long the reaction will take to reach equilibrium or what the path of the reaction will be. The latter two problems are dealt with in the study of chemical kinetics. 5.2 DEFINITION OF THERMODYNAMIC TERMS Before developing the laws of thermodynamics, consideration will be given to the definition of the terms commonly used in this field, since a clear understanding of these definition and terms will aid immensely in the discussions to follow. ENERGY Energy is usually defined as the capacity to do work. Kinetic energy,
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5.1 THE NATURE OF THERMODYNAMICSThermodynamics is concerned with the relationship between the energy changes that occur in chemical and physical processes. In effect, the laws of thermodynamics are the embodiment of the totality of the experimental facts concerning energy changes that occur in nature. The vast, diverse body of experimental facts concerning these energy changes are summarized in the statements of the three laws of thermodynamics. These can be used to predict the direction in which a process would proceed. Thus this science is of great importance to the chemist, since it can predict whether or not a chemical reaction will proceed spontaneously from a knowledge of the thermodynamic properties of the reactants and products. If thermodynamics tells us that a prospective reaction is favorable, then we are on a safe ground to try to find the proper experimental conditions to make the reaction proceed. For example, when gaseous H2 and O2 are mixed, thermodynamics tells us that H2O should be formed, since H2O is more energetically stable than H2 and O2. It is a well-known fact that at room temperature, a mixture of H2 and O2 will not produce H2O. However, if the mixture is sparked, water is produced with explosive violence, thereby proving the prediction thermodynamics makes concerning this reaction. One limitation of thermodynamics is that it does not give any indication of how fast the reaction will proceed, i.e., in essence, the laws of thermodynamics do not contain time as a variable. Thus thermodynamics is only interested in where the system was initially and where it is after the completion of the process. It does not care how the attainment of the final condition came about or how long it takes to reach this condition. In the case of a chemical reaction, for example, thermodynamics may tell us that the reaction will go spontaneously, but it cannot tell us how long the reaction will take to reach equilibrium or what the path of the reaction will be. The latter two problems are dealt with in the study of chemical kinetics.
5.2 DEFINITION OF THERMODYNAMIC TERMSBefore developing the laws of thermodynamics, consideration will be given to the definition of the terms commonly used in this field, since a clear understanding of these definition and terms will aid immensely in the discussions to follow.
ENERGYEnergy is usually defined as the capacity to do work. Kinetic energy, potential energy, chemical energy, and thermal energy are forms of energy particularly useful to the chemist.
Kinetic energy is defined as the energy produced by a moving object. For an object of mass m moving at a velocity v, the kinetic energy is equal to ½ mv2.
Potential energy is energy available by virtue of an object’s position. For instance, because of its altitude, a rock at the top of a cliff has more potential energy and will make a bigger splash if it falls into
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the water below than a similar rock located part way down the cliff.Chemical energy can be considered a form of potential energy
because it is associated with the relative positions and the various attractive and repulsive forces of atoms within a given substance.
Thermal energy is the energy associated with the random motion of atoms and molecules. The thermal energy of a substance is related to its temperature and quantity of matter in it. For example, the thermal energy stored in a bathtub filled with water at 40C is much more than the thermal energy in a cup of coffee at 70C.
HEATHeat can be defined as the transfer of thermal energy between two bodies that are at different temperatures. Thus, we often speak of “heat flow” from a hot object to a cold one, or “heat absorbed” when describing energy changes that occur during a process.
UNITS FOR ENERGYThe SI unit for energy is the Joule, J. The calorie, cal, and the liter atmosphere, L∙atm, are other common units for energy. The relationship between different units of energy is as follows:
1 cal = 4.184 J
1 L∙atm = 101.29 J
SYSTEMS AND SURROUNDINGSThermodynamics is concerned with the results of studies of definite parts of the universe, which are called systems. A thermodynamic system is a part of the universe, which is arbitrarily set off from the rest of the universe by definite boundaries for the purpose of experimental or theoretical study. The remainder of the universe is then, in effect, the surroundings of the system. Usually, however, the surroundings are restricted to a region in the immediate vicinity of the system under study. For example, a cylinder containing a gas is placed in a thermostated water bath. In this case, the gas is the system under study and the thermostat would be the surroundings; the boundaries of the system, of course, are the walls of the cylinder. The boundaries of a system can be fixed, as in the case of a rigid cylinder, or variable, as in the case of a cylinder equipped with a movable piston. Systems can be further subdivided into open and closed systems. An open system is one that can exchange both energy and matter with its surroundings. A closed system can exchange energy only with its surroundings, and the amount of matter in the system is not changed by transfer across the boundaries. For example, a beaker containing an aqueous salt solution (the system) in a thermostated water bath (the surroundings) is an open system, if more salt is added or if water is allowed to evaporate, and if heat energy from the thermostat is allowed to transfer into or out of the solution. However, this system becomes a
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closed system if the beaker is covered and no change in concentration occurs and if heat energy can still pass into or out of the solution. Note that a system is still considered closed if a chemical reaction occurs in the system, thereby changing the amounts of the different constituents, but not the total amount of matter in the system. A liquid in equilibrium with its vapor in a sealed tube is another example of a closed system. Heat energy can be added to cause more liquid to vaporize, but the total amount of matter is not changed, since no vapor can escape. A system in which there is no transfer of either energy or matter across the boundaries is called an isolated system. Thus if the sealed tube containing liquid and vapor were thermally insulated so that heat energy could not enter or leave, the liquid-vapor system would be an isolated system (Figure 5.1).
Figure 5.1 (a) an open system, which allows the exchange of both energy and matter with the surroundings; (b) a closed system, which allows the exchange of energy but not matter; and (c) an isolated system, which allows neither energy nor matter to be exchanged.
THERMODYNAMIC EQUILIBRIUMA system is said to have attained a state of thermodynamic equilibrium when it shows no further tendency to change its properties with time.
The criterion for thermodynamic equilibrium requires that three types of equilibrium exist simultaneously in a system. First of all, the system must be in thermal equilibrium. This means the temperature must be uniform throughout the system, and also, the system must be at the same temperature as its surroundings, if it is not insulated from the surroundings. Any temperature differences that exist will cause heat to flow from a higher temperature to a lower one until the temperature becomes uniform throughout the system. Second, the system must be in mechanical equilibrium . Mechanical equilibrium requires that there be no macroscopic movement within the system itself, or from the system to its surroundings. For example, consider a gas contained in a cylinder equipped with a movable piston. If the piston is pushed in rapidly, the rapid motion will set up pressure and
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temperature differences in the gas, since the pressure and temperature in the region close to the piston will be much greater than in a region away from the piston. Hence, no single pressure measurement will unambiguously describe the system. If enough time is allowed to elapse, these differences will level out until the pressure and temperature throughout the gas will become uniform and constant. The system is then in mechanical equilibrium. In a system where more than one substance is present, thermodynamic equilibrium requires that the composition of the system remains constant with time. For chemical reactions this means no net chemical change can occur, i.e., the system must be at chemical equilibrium. Of course, this equilibrium is a dynamic one in which the forward and reverse reactions are continually occurring at equal rates. Hence there is no net change in the concentration of reactants or products at equilibrium. FUNCTIONS OR VARIABLES OF STATEThermodynamics deals with changes that occur in the properties of sys-tems when the system goes from one equilibrium state to another. Such a change in the values of the properties of the system occurs in what is called a process. In fact, thermodynamics is interested only in the values of the properties, which the system possesses in its initial and final states. It is not concerned with how the system arrived at the final state. Thus as long as the values of property in the initial and final state are the equilibrium values, the quantity of interest in thermodynamics is the change in the value for the process. The change in the value is defined as the difference in the value between the final and initial states. Thus for a compression process, if P2 is the equilibrium pressure of the gas after compression (final state) and P1, is the equilibrium pressure before compression (initial state), then the change in pressure, P, for the compression is
P = P2 – P1 (5-1)
The magnitude of P will be the same, whether the process was carried out in one step or several steps, as long as P2 and P1 are equilibrium values of the pressure for the two states. Not all variables of a system possess this property of being independent of the paths taken between states. The variables that do possess this property are called functions or variables of state. They form an important class of variables that arc very useful in thermodynamics. Pressure, temperature, and volume are examples of state variables. Other very important state variables are obtained from the laws of thermodynamics.
5.3 THE FIRST LAW OF THERMODYNAMICSThe First law is based on the observation that energy can neither be cre-ated nor destroyed in any physical or chemical process. For example, a metal ball of mass m held h m above the ground has potential energy given by mgh. When the ball is dropped it loses potential energy but this
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potential energy is converted into kinetic energy, ½mv2 . The conservation of mechanical energy requires that the total energy of the system, here considered as the ball isolated from its surroundings, to be a constant. Thus one can write for this system that the total energy ET is the sum of the potential and kinetic energy or, mathematically,
ET = mgh + ½mv2 (5-2)
Eventually, of course, the metal ball must strike the ground and come to rest. What happened then to the potential energy that the ball possessed originally? It is well known that this energy is converted into heat energy upon striking the ground. This example illustrates that even though energy cannot be created or destroyed, it can be converted into other forms. In the above example mechanical energy was converted into heat energy. The First law of thermodynamics is not concerned with how much of one form of energy is converted into another. However, it requires that, when all the energy changes have been accounted for, the total energy of the system before and after the change must be the same.
5.4 INTERNAL ENERGY OF A SYSTEMThe internal energy of a substance or system is the grand total of all the different forms of energy that the substance can possess. These include all the kinetic and potential energies that molecules of the substance can possess due to translational, vibrational and rotational motions. It includes potential energies due to the attractive and repulsive forces acting between the molecules and atoms of the substance. Gravitational energy is usually neglected, since it is very small relative to the other forms of energy. It is obvious that the internal energy of a substance depends on its physical state. Thus a substance in the gaseous state will have a greater internal energy than when in the liquid or solid state since the gaseous molecules will have greater translational, vibrational, and rotational motion. The internal energy is an extensive property, since it will depend on the amount of substance present (a property that does not depend on the amount of substance present, like density, is called an intensive property). The large number of energy terms that can contribute to the total internal energy of a substance makes it impossible to determine the absolute value of this quantity for a substance in a given state. However, as was explained above, thermodynamics is interested only in the change in the properties that occur during a process. Thus if the process takes the system from an initial to a final state, where the internal energies are Ei and Ef, respectively, then the change in internal energy E will be given by
E = E f – E i (5-3)
It is this difference in internal energy that can be measured experimentally from the change in properties that occurs in the process, even though the absolute values of Ef and Ei are unknown. For example, consider the process whereby a quantity of heat q is added to the system,
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and at the same time the system performs work equal to w. Both q and w are, of course, measurable quantities, i.e., their magnitudes are known. What is the change in internal energy of the system for this process? The system initially had some indeterminate internal energy Ej and the addition of q joules of heat must increase E j by q. Thus for this step of the process, the final internal energy E' is now
E' = E i + q (5-4)
Now work is performed by the system, and this work must come from the internal energy E' of the system. For instance, for a gas expanding against a piston, the molecules of the gas perform work by pushing against the atmosphere. Because the expansion of a gas results in a decrease in the internal energy, E’, of the system (Figure 5.2), work done by the system is assigned a negative sign. This leads to a final state having E f given by
E f = E' + w (5-5)
FIGURE 5.2 Changes in the internal energy of a system. The system absorbs heat from the surroundings, and then performs expansion work. Note that expansion work, work done by the system is assigned a negative sign. Ef = Ei + q + w and ΔE = Ef Ei = q + w.
The law of conservation of energy requires that the internal energy of the final state be equal to the initial energy plus or minus any energy that was gained or lost, respectively, by the system. This result is exactly what is expressed in Eq. (5-5), as can be seen in a clearer fashion when the expression for E' from Eq. (5-4) is used in Eq. (5-5),
E f = E i + q + w (5-6)
and using Eq. (5-3) one has for E,
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E= E f – E i = q + w (5-7)
Equation (5-7) is the mathematical formulation of the First law of thermo-dynamics for energy changes involving heat and work only. q is assigned a positive value when heat is added to the system, thus increasing its internal energy in the final state. If heat were removed from the system, the internal energy of the final state would be less than that of the initial state and therefore q must have a negative sign in Eq. (5-7). Similarly, if work were done on the system by an external source, for example, if a gas in a cylinder were compressed by the application of an external force on the piston, this would result in an increase in the internal energy of the final state and therefore, w would take a positive sign in Eq. (5-7). Table 5.1 summarizes the sign conventions for q and w.
Table 5.1 Sign Convention for Work and HeatProcess SignWork done by the system on the surroundings Work done on the system by the surroundings +Heat absorbed by the surroundings from the system Heat absorbed by the system from the surroundings +
It was mentioned above that the internal energy depends only on the physical state in which the system exists. Thus E in Eq. (5-3) does not depend on the process or path by which the system went from the initial to the final state and only depends on the initial and final states. This means that the internal energy is another state function. This fact leads to a very important conclusion, that it is impossible to construct a machine that will produce work indefinitely without the input of energy, the so-called perpetual motion machine. Consider a process which takes a sys-tem from state A to a higher energy state B by path 1, as is shown in Figure 5.3, and involves an input of heat q1 , and work done by the system equal to w l . Let us imagine another process (path 2 in Figure 5.3), involving an input of heat
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Figure 5.3 Energy changes in a cyclic process.
q2 and work done by the system equal to w2 , which will take us back to state A. The energy changes in these two processes are
E1 = EB – EA = q1 + w1 path I (5-8a)
and
E2 = EA – EB= q2 + w2 path 2 (5-8b)
The change in energy taken around the whole cycle EAA , i.e., from state A through state B and back to state A, is given by the sum of the Eqs. (5-8a) and (5-8b),
EAA = E1 + E2 (5-9)
If the first law of thermodynamics has to be obeyed then E , and E2 should be equal in magnitude but opposite in sign. This means that the net change for the cycle should be
EAA = E1 + (–E2) = 0 (5-10)
Using Eqs. (5-8a) and (5-8b) for E , and E2 in Eq. (5-10), one obtains
q1 + w1 w2 –q2 = 0or
q2 – q1 = w1 – w2
The latter equation states that the net input of heat energy must just equal the net work done by the system in a cyclic process. Only when this is true will the cycle end up at a state with the same internal energy as it had originally. If it were possible to construct an engine that ran in a cycle in which E2 > E1 , then one could continually produce work from the engine without an equivalent expenditure of energy.
WORK AND HEATWhile E is independent of the path, q and w can be different for different paths. For example, if a gas expands against an external force, the work done by the gas will depend on the magnitude of this force. Also, different quantities of heat can be added or removed from the system. Thus in order to reach the same final state, i.e., to expand the gas the same amount, different quantities of work and heat can be involved. These correspond to the different paths taken to reach the final state and the only restriction placed on the different paths is that the sum (q + w) be the same for each path. Therefore, q and w are not thermodynamic state functions, but their sum is a state function which, of course, is E .
Let us look at systems in which the only work done is pressure-volume (PV) work, i.e., mechanical work. From the fundamental definition of work as a force acting through a distance, one has for the differential
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work dw resulting from a force f acting through a differential distance dr,
dw = fdr (5-11)
The work done by the force in moving an object from r1 to r2 is given by the definite integral,
(5-
12)
Consider the case of a gas pushing against a piston, as is shown in Figure 5.4. The gas does work when it pushes against the external pressure, Pext. The external force fext, by definition, equals (Pext) (A), where A is the cross-sectional area of the piston. Using this value for f in Eq. (5-12), one has for the work of expansion
(5-
13)
where dr is the infinitesimal movement of the piston. The minus sign in Equation (5-13) takes care of the convention for work. For gas expansion dr > 0, so PextAdr is a negative quantity. For gas compression (work done on the system), dr < 0, and PextAdr is a positive quantity. The product Adr is equal to the infinitesimal change in the volume dV, and therefore, one has
(5-
14)
where V1 and V2 are the initial and final volume, respectively. When the opposing external pressure is constant, like the
pressure exerted by the atmosphere, Equation (5-14) becomes
w = PextΔV (5-15)
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Figure 5.4 The work done by a gas in a cylinder fitted with a movable weightless piston as it expands against a constant external pressure from r1 to r2 is w = PextΔV.
It is obvious that for an expansion process, dV is positive and for compression dV is negative. Since Pext is always positive, w will be negative when the gas does work against Pext (expansion) and will be positive when work is done on the gas by Pext (compression). That the pressure used in Eq. (5-14) is the external pressure of the surroundings and not the internal pressure of the gas can be shown by the following example.
EXAMPLE 5.1A certain gas aexpands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against vacuum ad (b) against a constant pressure of 1.2 atm.
Solution(a) Since the external pressure is zero, the work w done is
w = PΔV
= 0(6.0 2.0) L
= 0(b) The external, opposing pressure is 1.2 atm, so
w = PΔV
= (1.2 atm)(6.0 2.0) L
= 4.8 L.atm
To convert the answer to joules, we write
= 4.9 × 102 J
WORK DONE BY VAPORIZATION OF LIQUIDS IN OPEN CONTAINERSAn important example of a system involving PV work is the vaporization of a liquid at its normal boiling point. When a liquid vaporizes in an open container, the vapor can be considered to do work against a constant pressure, against the atmospheric pressure in this case. At the normal boiling point vapor of the liquid, the pressure of the vapor can be taken
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equal to the constant external atmospheric pressure. One then has from Eq. (5-8) that the work done in converting a given quantity of liquid, having a volume V l to a vapor, having a volume Vv , against a constant pressure, Patm, which is equal to the vapor pressure Pv, is
w = Pa t mV = Pv(Vv – V1) (5-16)
But V1 << Vg and Vl can be neglected when compared to Vv. In addition, if the vapor is assumed to behave ideally, one has Vv = nRT/Pv, where n is thenumber of moles of liquid that was converted to vapor and Pv is the equilibrium vapor pressure of the liquid at the temperature T. Using these assumptions in Eq. (5-16), one obtains
(5-17)
This equation shows that for the work done in the vaporization of a given amount of liquid depends only on the temperature and is independent of the pressure or volume.
Note that Equation (5-17) also applies to reactions carried out in open containers, which are accompanied by the production of gaseous product. This is illustrated in example 5.3
EXAMPLE 5.2Calculate the work done in vaporizing 1 mole of H2O at 100°C assuming ideality. 1 L∙atm = 101.3 J.
SolutionAt 100°C, the normal boiling point, the vapor pressure of H2O is 1 atm. The molar volume of H2O vapor at this temperature is given by
In this case, Vv = ΔV.The molar volume of the liquid is 0.018 L (1 mol H2O = 18 g = 18
mL = 0.018 L) and is negligible compared to the molar value of the vapor.
The work done in vaporizing 1 mole of water against a constant pressure of 1 atm is given by
or alternately by using w = PvΔV =PvVv = nRT
The energy necessary to perform this PV work of vaporization comes from the heat energy absorbed from the surroundings. However,
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more energy must be absorbed from the surroundings than is necessary to perform this PV work. This additional amount of heat must be supplied to the system in order to separate the molecules from their neighbors in the liquid. The quantity of heat necessary to vaporize 1 mole of H2O at 373 K is obtained from its heat of vaporization, which is equal to 40.69 kJ/mol. The change in the internal energy of a mole of water upon vaporization then is given by Eq. (5-7) as
Thus the internal energy of the vapor is greater than that of the liquid, as has been stated several times in this text, and this is due to the large amounts of heat necessary to vaporize the liquid compared to the PV work done by the vapor.
EXAMPLE 5.3 Calculate the work done when 50 g of iron dissolves in hydrochloric acid in (a) a closed vessel and (b) an open beaker at 25C. Treat the hydrogen gas as perfect and ignore the initial volume of the system.
SolutionThe reaction is
Fe(s) + 2 HCl(aq) FeCl2(aq) + H2(g)
In the course of reaction 1 mole of gas is generated for each mole of iron consumed. The gas drives back the surrounding atmosphere and thereby does the work –PV.
For (a) V = 0 because the vessel is closed and cannot expand. Therefore
w = –PV = 0
For (b) V Vg, and the system expands until its pressure P is 1
atm. Then, we can use PVg = nH2RT.
Note also that the number of moles of hydrogen produced is
equal to the number of moles of iron consumed.
w = –PV –nH2RT = –nFeRT
= 2.2 kJ
5.4 ENTHALPYIn most laboratory work in chemistry, reactions are carried out in open flasks and therefore they occur at the constant pressure of the atmosphere. For constant pressure processes in which pressure-volume work is the only work done, Eq. (5-7) can be written, using Eq. (5-15) for
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w,
qp = E + P V (5-18)
where the subscript on q indicates that the heat change occurs at constant pressure. If, however, the process is carried out at constant volume, V is zero and one obtains
qv = E (5-19)
i.e., the change in internal energy for a constant volume process, in which no PV work is done, is equal to the heat absorbed or evolved by the system. Because of the importance of constant pressure processes, it is convenient to define a new thermodynamic function, the enthalpy, given the symbol H. The enthalpy is defined by the equation
H = E + PV (5-20)
Since E, P, and V are all state functions, it follows that H is also a state function and is therefore, independent of the path taken between initial and final states of the system. One can write, from Eq. (5-18), for the change in enthalpy H,
H = E + (PV) (5-26)
Expanding the quantity (PV) in the usual manner gives
H = E + PV+ VP (5-27)
For a constant pressure process, P = 0, and one has
H = E + PV (5-28)
Substituting E from Eq. (5-23) into Eq. (5-28) gives
H = qp – PV + PV = qp (5-29)
Thus one sees that the change in enthalpy for a constant pressure process, in which the only work done is PV work, is equal to the heat absorbed or evolved in the process. Since H = H f – H i , where Hf
and Hi are the enthalpies of the final and initial states, respectively, then when heat is absorbed by the system, the positive value for H means that the heat content of the final state is greater than that of the initial state. When heat is evolved by the system, H has a negative sign and the heat content of the final state is less than that of the initial state. Using qp and qv for H and E, respectively, in Eq. (5-28), one obtains
qp = q v + PV (5-24)
It is obvious from this equation that the heat absorbed in a
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constant pressure process is greater than the heat absorbed in a constant volume process by an amount equal to the PV work done by the system. This means more heat energy is required to raise a system to a given energy state in a constant pressure process, as compared to a constant volume process, because some of the heat goes into performing PV work.
5.5 MOLAR HEAT CAPACITYThe concepts of enthalpy and internal energy are useful in obtaining an exact definition of heat capacities of gases. Two types of heat capacities could be defined. These are Cv and Cp, the heat capacities at constant volume and pressure, respectively. Using the definition of the molar heat capacity as the heat energy necessary to raise the temperature of one mole of gas through one degree, one has, using qv and qp as the heat absorbed in the constant volume and pressure processes, respectively, needed to give an increase in temperature of T,
(5-25a)
and
(5-25b)
Using Eqs. (5-19) and (5-23) in Eqs. (5-25a) and (5-25b), i.e., using E and H for qv and qp, respectively, gives
(5-26a)
and
(5-26b)
Since qp > qv, as explained above, then Cp > Cv. The exact relationship between Cp and Cv can be obtained as follows. Using Eq. (5-22) in (5-26b) gives
(5-27)
The first term in the expression on the right is equal to Cv by Eq. (5-26a). The second term is simply equal to R for one mole of an ideal gas. (Show this.) Hence, one has
Cp = Cv + R (5-28)
Thus Cp is greater than Cv by R = 8.314 J/mol·K. This quantity is simply the work done by the gas in pushing back the piston against a constant pressure, i.e.
w = PV = RT and for T = 1K, w = R
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Table 5.2 Molar Heat Capacities at 25C and 1 atm
C v ( J /mol ·K) C p ( J /mol ·K)
He , Ne , Ar , Kr , Xe 12 .5 20 .78
H 2 20 .53 28 .84
N 2 21 .06 29 .37
CO 2 28 .81 37 .12
H 2 O(g) 16 .46 24 .77
NH 3 27 .32 35 .63
CH 4 27 .33 35 .64
5.7 THERMOCHEMISTRYUp to now, consideration has been given only to energy changes involved in physical processes. One of the most important applications of the First law to chemistry is in the study of the heat changes that occur in chemical reactions. This study forms the subject of thermochemistry.
It was shown in the above discussions that E and H were independent of the path taken in going from the initial to the final state of a system. Thus for reactants going to products in a chemical reaction, the only important consideration is the measurable quantities qv (= E) and qp (= H), the heat changes involved in the reaction occurring at constant volume and constant pressure, respectively. The absolute values of the internal energies and enthalpies of reactants and products are not required, since one is interested only in the change in these values for the reaction. When heat is evolved in a reaction, the final state (products) is lower in energy and by the convention used before, q is negative. For a constant pressure reaction (for instance, a reaction run in an open beaker), qp, and. therefore. H, is negative. This fact is indicated by writing the reaction
Reactants products HT = –q J
where q is the heat evolved in the reaction at the absolute or Kelvin temperature T. For example, the reaction of sulfur and oxygen gas to give sulfur dioxide at 25°C is found to evolve 296.9 kJ for 1 mole of SO2 formed. This is written
S(s) + O2(g) SO2(g) H298 = –296.9 kJ
Note that the physical state of the reactants and products is
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indicated. This is necessary since the internal energy and enthalpy of a substance depends on its physical state, as was explained previously. Thus the heat change for the above reaction would be more if SO2
were formed in the liquid state because the conversion of liquid SO2 to gaseous SO2 requires the input of some energy in the form of heat.
5.8 STANDARD STATESThe specification of the physical state still leaves the state of the system incompletely defined. For instance, sulfur can exist in two solid forms; rhombic and monoclinic. Furthermore, the values of the enthalpy and the internal energy will depend on the temperature and pressure for any given physical state. In order to standardize the data on heats of reactions, it is necessary that some reference point be established for the thermodynamic states of the reactants and the products. The choice of this reference point is purely arbitrary and is called the standard state of the substance. By convention, the following standard states have been adopted. The standard state of a gas is the pure gas at 1 atm; for a liquid it is the pure liquid at 1 atm; and for a solid it is the stable crystalline form existing at 1 atm. For example, graphite and rhombic sulfur are chosen as the standard states of carbon and sulfur, respectively. Although the standard state does not specify a temperature, we will always understand, unless stated otherwise, values measured at 25C. Now that the standard state is defined, one is free to choose any value as a reference point in order to tabulate thermodynamic data. To this end, the most convenient choice is to assign the value of zero to the chemical elements in their standard states. Thus H = 0 for all chemical elements, the superscript zero indicating that the elements are in their standard states. Hence, at 1 atm and 25°C, solid rhombic sulfur, solid copper, liquid bromine, and gaseous oxygen all have H = 0. The enthalpy changes are measured relative to the standard states, as will be discussed in the next section.
5.9 STANDARD ENTHALPY OF REACTIONSThe enthalpy of a reaction is the enthalpy change, Hrxn that occurs during the reaction. When the reaction occurs between the reactants, all in their standard states, and products, also all in their standard states, the enthalpy change is known as the standard enthalpy change of the reactions, Hrxn. For example, when at 1 atm pressure, 2 moles of carbon monoxide are burned in oxygen to produce 2 moles of carbon dioxide, the standard enthalpy change of the reaction is 565.8 kJ
2 CO(g) + O2(g) 2 CO2(g) Hrxn = 565.8 kJ
ENTHALPY OF COMBUSTIONThe heat involved due to the complete combustion of 1 mole of a hydrocarbon with oxygen is known as the enthalpy of combustion. As an example, one mole of propane, C3H8(g), is burned in O2(g) at 1
17
atm and 25°C to give CO2(g) and H2O(l), and the heat evolved is found to be 2220 kJ/mol. Since all substances are at 1 atm and 25°C, they are in their standard states, and the heat evolved is the standard enthalpy of combustion. This change is expressed in an equation as
5.10 HESS’S LAWIn 1840, Hess put forth the following empirical law: the overall heat change at constant pressure or constant volume in a given chemical reaction is the same regardless of whether it takes place in one direct step or in several steps. Thus if compound B could be formed from A directly and it were also possible to go from A to B through intermediates C and D, the enthalpy change for the direct path would be the same as the sum of the enthalpy changes involved in the intermediate reactions. This is shown schematically in Figure 5.5, where the H's are the enthalpy changes for the individual reactions proceeding in the directions shown by the arrows. Hess's law requires that H4 = H1 + H2 + H3. It can be shown that this result can be derived from the First law. Thus since H is a state function, the sum of the H's around the cycle, taken from A through C, D and B and back to A, by the First law must be zero [see Eq. (5-10)]. This means
= H1 + H2 + H3 + (–H4) (5-29)
or
H4 = H1 + H2 + H3
which is the same result predicted by Hess's law. In Eq. (5-29) the sign of H4 was reversed since in going around the cycle in a
18
counterclockwise manner, the reverse of the reaction path A B was traversed. If heat were evolved in going from A B, the First law requires that the same amount of heat be absorbed in the reverse reaction. Thus ΔH is equal in magnitude but opposite in sign for the reverse reaction.
Figure 5.5 Enthalpy changes in a cyclic process
The importance of Hess's law lies in the fact that the enthalpy changes can be calculated for reactions that cannot be performed in the laboratory. In order to obtain reliable H values for reactions, the reactions must proceed rapidly and must go to completion. Many reactions are too slow or involve the formation of side products and therefore do not meet these criteria. Hess's law can be used to obtain H for these reactions. For example, it is not possible to measure directly the heat evolved when graphite is reacted with O2 to give CO, since it is impossible to prevent the formation of CO2. However, the heat evolved in the complete combustion of graphite to CO2 can be measured, as well as the heat evolved in the combustion of CO to CO2
. At 18°C, these reactions are
(1) C(graphite) + O2(g) CO2(g) Hrxn1 = –393.1 kJ
(2) CO(g) + ½O2(g) CO2(g) Hrxn2 = –282.9 kJ
In order to obtain H for the reaction at 18°C
C(graphite) + ½O2(g) CO(g)
from the measured values of H for the above two reactions, reverse reaction (2) and add to reaction (1). The same procedure must be followed with the enthalpy changes for these reactions. Reversing the direction of reaction (2) means the sign of Hrxn2 must be reversed, i.e., the reaction CO2(g) CO(g) + ½O2(g) requires the input of 282.9 kJ. The result then is
C(graphite) + ½O2(g) CO(g) Hrxn = –110.2 kJ
19
ENTHALPY OF FORMATIONWhen a mole of a given compound is formed in its standard state from its elements, all in their standard states, the heat evolved or absorbed is known as the standard enthalpy of formation of the compound, . As an example, the formation of SO2(g) from its elements at 25°C and 1 atm evolves –296.90 kJ/mol of SO2(g) formed, and, therefore,
= –296.90 kJ/mol at 25°C and 1 atm.
S(rhombic) + O2(g) SO2(g) = –296.90 kJ/mol
The heat evolved when 1 mole of methane gas is formed from graphite and hydrogen gas at 25°C is –74.85 kJ/mol. This is repre-sented by
C(graphite) + 2 H2(g) CH4(g) = –74.7kJ/mol
A chemist is interested in the heat change accompanying a chemical reaction, since the value of the equilibrium constant depends on it to a large extent. One would like, therefore, to tabulate the H values for every chemical reaction. However, this is too great an undertaking. The same goal is achieved by determining the heat of formation of some chemical species from its elements in their standard states, . For a general reaction,
aA + bB + . . . cC + dD + . . .
if for products and reactants are known, H for the reaction can be
found from Hess's law in the form
= (products) — (reactants) ( 5-30)
where ni are the coefficients of the products and nj are coefficients of the reactants, in the balanced chemical equation.
Now one can see why it is convenient to choose the enthalpy of the elements in their standard states as zero. Since H for the process is the difference in enthalpies of the product (final state) and reactants (initial state), then for the formation of one mole of methane, CH4, under standard state conditions, the measured value of the enthalpy change for the reaction becomes
= (1 mol) – (1 mol) – (1 mol)
The value obtained is equal to the enthalpy of formation of methane since Δ and Δ are assigned a value of zero.
=
The enthalpy of formation for various compounds are shown in
20
Table 5.4.
TABLE 5.4 Enthalpies of Formation at 25°C
Substance (kJ/mol)
H20(g) –241.8
H20(l) –285.8
HCI(g) –92.3
S02(g) –296.1
S O 3 ( g) –395.2
NH3(g) –46.3
CO(g) –110.5
CO2(g) –393.5
CH4(g) –74.7
C2H2(g) +226.6
NO(g) +90.4NO2(g) +33.85
Many compounds cannot be directly synthesized from their elements. In some cases, the reaction proceeds too slowly, or side reactions produce substances other than the desired compounds. In these cases can be determined by an indirect approach. As an example, consider the formation of benzene, C6H6, from its elements at 25C and 1 atm,
3H2(g) + 6C(graphite) C6 H6 ( l) = ?
The enthalpy change in the written reaction cannot be determined directly since the reaction of carbon and hydrogen results in a mixture of hydrocarbons. However, one can obtain of benzene by noting that benzene, like all organic compounds containing carbon and hydrogen can be burned in oxygen to produce H2O and CO2, whose ’s are known.
EXAMPLE 5.5Pentaborane-9, B5H9, is a colorless, highly reactive liquid that was considered as a potential rocket fuel because it produces a large amount of heat per gram when it explosively burns in oxygen. The idea was abandoned because the solid B2O3 formed by the combustion is abrasive and would quickly destroy the nozzle. The reaction is
2 B5H9(l) + 12 O2(g) 5 B2O3(s) + 9 H2O(l)
The standard heat of formation at 25°C for B5H9, B2O3, and H2O are respectively, 73.2 kJ/mol, 1263.6 kJ/mol, and 285.8 kJ/mol. Calculate the heat liberated per gram of B5H9.
This is the amount of heat released for every two moles of pentaborane-9. The heat released per gram of pentaborane-9 is
Heat released per gram of B5H9 =
22
= 71.58 kJ/g
Note that compounds with positive values release more heat during combustion and are less stable than those with negative values.
5.11 RELATIONSHIP BETWEEN THE HEATS OF REACTION AT CONSTANT PRESSURE AND CONSTANT VOLUMEReactions occurring at constant pressure are the most common reactions in chemical work since they consist of reactions occurring in open vessels in which PV work is done by the system. However, a number of reactions are performed under conditions of constant volume. For instance, the addition of hydrogen to unsaturated hydrocarbons is usually performed in a rigid, closed vessel known as a Paar bomb. Using Eq. (5-22)
H = E + PV
one can see that if there is no net change in volume during a reaction, V = 0 and H = E. Two main types of reactions correspond to this condition. One type includes reactions involving only liquids and solids. In this case, V is negligibly small and PV is usually negligible in comparison to E. The second type includes reactions in which the number of moles of gaseous products equals the number of moles of gaseous reactant, thus resulting in no net increase in volume, and V is essentially zero.
Let us consider the case where there is a change in the number of moles of gaseous substances during the reaction. In this case PV will be a significant quantity and must be taken into account in Eq. (5-22). Assuming the gases to behave ideally, one can write for PV
PV = nRT (5-31)
where n represents the number of moles of gaseous product minus the number of moles of gaseous reactants. Substituting Eq. (5-31) into (5-22) gives
H = E + nRT (5-32)
This equation gives the relationship between the heats of reaction at constant pressure and at constant volume for a reaction involving PV work only.
EXAMPLE 5.6Calculate the heat change for the combustion of n-pentane when the reaction is run in a Paar bomb at 25°C. The reaction is
C 5 H 1 2 ( l ) + 8O 2 (g ) 5 CO2(g) + 6 H2O(l) = –3536.2 kJ
Solution
23
The value of is –3536.2 kJ . The value of n is (5 – 8) = –3 moles, and using Eq. (5-32), one has for the constant volume reaction
E = H –nRT
=
= –3.5362 106 J + 7.43 103 J
= –3.53 J
Notice that the difference between E and H is very small.
5.12 BOND ENERGIESIn many cases it is desired to obtain an estimate of for a reaction in which for a reactant or product is not known and cannot be experimentally determined. Consequently, one cannot use Eq. (5-30) to obtain H for the reaction. A way around this problem can be found in the use of bond energies, which as you will see, yields good estimates of that can be used to calculate . It will be seen that this procedure presupposes a knowledge of the structure of the compound and therefore, does not strictly belong in a discussion of classical thermodynamics.
MEASUREMENT OF BOND ENERGIESThe bond energies of simple molecules such as H2, O2, and C12 are usually measured spectroscopically. That is, the light that is emitted by the molecules when they are energized (excited) by a flame or an electric arc is analyzed, and the amount of energy needed to break the bond is computed.For more complex molecules, thermochemical data can be used to calculate bond energies in a Hess's law kind of calculation. We will use the standard heat of formation of methane, CH4, to illustrate how this is accomplished. However, before we can attempt such a calculation, we must take a moment to define a new thermochemical quantity that we will call the atomization energy, symbolized Ha t o m . This is the amount of energy that is needed to rupture all the chemical bonds in one mole of gaseous molecules to give gaseous atoms as products. For example, the change corresponding to the atomization of methane is
CH4(g) C(g) + 4 H(g)
and the enthalpy change for the process is Ha t o m . For this particular molecule, Ha t o m corresponds to the total amount of energy needed to break all the C—H bonds in one mole of CH4; therefore, division of Ha t o m by 4 would give the average C—H bond energy in methane, expressed in kJ/mol (or kcal/mol).
4 H(g) + C(g)
24
2 H2(g) + C(s) CH4(g)
Figure 5.6. Two paths for the formation of methane from its elements in their standard states. Steps 1, 2, and 3 of the upper path involve the formation of gaseous atoms of the elements and the formation of the bonds in CH4.
Figure 5.6 shows how we can use the standard heat of formation, to calculate the atomization energy. Across the bottom we have the
chemical equation for the formation of CH4 from its elements. The enthalpy change for this reaction, of course, is . In this figure we also can see an alternative three-step path that leads to CH4(g). One step is the breaking of H—H bonds in the H2 molecules to give gaseous hydrogen atoms, another is the vaporization of carbon to give gaseous carbon atoms, and the third is the combination of the gaseous atoms to form CH4 molecules. These changes are labeled 1, 2, and 3 in the Figure.
Since H is a state function, the net enthalpy change from one state to another is the same regardless of the path that we follow. This means that the sum of the enthalpy changes along the upper path must be the same as the enthalpy change along the lower path, . Perhaps this can be more easily seen in Hess's-law terms if we write the changes along the upper path in the form of thermochemical equations.
Steps 1 and 2 have enthalpy changes that are called standard heats of formation of gaseous atoms. Values for these quantities have been measured for many of the elements and some of them are given in Table 5.5. Note that for diatomic gaseous molecules these values are exactly half the bond energy in the molecule. Step 3 is the opposite of atomization, and its enthalpy change will therefore be the negative of Ha t o m (recall that if we reverse a reaction, we change the sign of its H).
(Step 1) 2 H2(g) 4 H(g) = 4 [H(g)](Step 2) C(s) C(g) = [C(g)](Step 3) 4H( g ) + C( g ) CH 4 ( g ) = – H atom
2H2(g) + C(s) CH4(g) = [CH4(g)]
Notice that by adding the first three equations, we get the equation for the formation of CH4 from its elements in their standard states. This means that adding the values of the first three equations should give for CH4.
+ + = [CH4(g)]
Let’s substitute for , , and , and then solve for Hatom. First,
Now all we need are values for the ’s on the right side. From Table 5.5 we obtain [H(g)] and [C(g)], and the value of [CH4(g)] as already seen is equal to 74.8 kJ/mol obtained from 5.3.
[H(g)] = 218.0 kJ/mol
[C(g)] = 715.0 kJ/mol
[CH4(g)] = –74.8 kJ/mol
Substituting these values gives
Hatom = 1662 kJ/mol (rounded)
and division by 4 gives an estimate of the average C—H bond energy in this molecule.
Table 5.5 Standard Heats of Formation ofGaseous Atoms from the Elements.
This value is quite close to the value in Table 5.6, which is an average of C—H bond energies in many different compounds. The other bond energies in Table 5.6 are also based on thermochemical data and were obtained by similar calculations.
Table 5.6 Average Bond Energies in Polyatomic MoleculesBond Energy Bond Energy
UESES OF BOND ENERGIESAn amazing thing about covalent bond energies is that they are very nearly the same in many different compounds. This suggests, for example, that a C—H bond is very nearly the same in CH4 as it is in a large number of other compounds that contain this kind of bond. That alone is very useful information, because it explains why molecules having similar bonds behave in similar ways. If the properties of a particular kind of bond varied much from compound to compound, this would not be true.
Because the bond energy doesn't vary much from compound to compound, we can use tabulated bond energies to estimate the heats of formation of substances. This is illustrated in Example 5.7.
EXAMPLE 5.7Use the bond energies in Table 5.5 and 5.6 to estimate the heat of formation of methanol vapor, CH3OH(g).
Solution We solve this problem in much the same way that we calculated the bond energy of methane in the discussion above. We setup two paths from the elements to the compound, as shown in the Figure shown below. The lower path has an enthalpy change corresponding to [CH3OH(g)], while the upper path takes us to the gaseous elements and then through the energy released when the bonds in the molecule are formed. This latter energy can be computed from the bond energies in Table 5.6. As before, the sum of the energy changes along the upper path must be the same as the energy change along the lower path, and this permits us to compute
[CH3OH(g)].
C(g) + 4 H(g) + O(g)
C(s) + 2 H2(g) + 1/2O2(g) CH3OH(g)
1 2 3 4
27
Steps 1, 2, and 3 involve the, formation of the gaseous atoms from the elements, and their enthalpy changes are taken from Table 5.5.
Adding these values gives a total energy input for the first three steps of 1836.2 kJ. In other words, the net for the first three steps is +1836.2 kJ.
The formation of the CH3OH molecule from the gaseous atoms is exothermic—energy is always released when atoms become joined by a covalent bond. In this molecule we can count three C—H bonds, one C —O bond, and one O—H bond. Their formation releases energy equal to their bond energies, which we obtain from Table 5.6.
Bond Energy (kJ)3(C—H) 3 (413 kJ/mol) = 1239
C —O 351O—H 464
Adding these together gives a total of 2054 kJ. for this step is therefore —2054 kJ (because it is exothermic). Now we can compute the total enthalpy change for the upper path.
= (+ 1836 kJ) + (–2054 kJ)
= –218 kJ
The value just calculated must be equal to the for the lower path of the Figure, which is for CH3OH(g). Experimentally, it has been found that for this molecule (in the vapor state) is –201 kJ/mol. At first glance, the agreement doesn't seem very good, but on a relative basis the calculated value (–218 kJ) differs from the experimental one by only about 8%.
Comparisons between measured and calculated bond energies have sometimes helped chemists understand unusual properties of some substances. Consider, for example, the case of benzene, C6H6. This molecule has its six carbon atoms arranged in a hexagonal ring, and bonded to each carbon is one of the hydrogen atoms. One of the Lewis structures that we can draw for the molecule is
Benzene, however, doesn't behave chemically as if this is its
28
structure. It doesn't undergo reactions that are typical of hydrocarbons with carbon -carbon double bonds, and the ring structure has a great tendency to remain intact during chemical reactions. This Lewis structure is inadequate in other ways, too. You learned that the bond length is related to the number of electron pairs that are shared between two bonded atoms. On that basis we would anticipate that some of the carbon - carbon bonds in benzene should be longer than others. Actually, they are all the same, and to account for this we say that the actual structure of benzene is a resonance hybrid. The resonance structures that are drawn to represent the hybrid are
If we used either one of these Lewis structures as if it really existed in an attempt to calculate the heat of formation of C6H6, just as in Example 5.7, the value we would obtain is about +230 kJ/mol. However, the experimentally measured for C6H6 is only about +84 kJ/mol. The difference between the calculated and experimental values is much too large to be just experimental error, and it means that the actual molecule is more stable than we computed it to be by about 146 kJ/mol. In general, it is found that for molecules for which resonance is used to explain equality of bond lengths—as in benzene here and in molecules such as S0 2 and S03, the actual is more exothermic (less endothermic) than the calculated using tabulated bond energies. In other words, resonance leads to an additional stabilization of the molecule, and the energy difference between the calculated and actual is called the resonance energy or stabilization energy.
5.13 DEPENDENCE OF HEAT CAPACITY AND OF ENTHALPY OF REACTION ON TEMPERATURE Experiments with gases have shown that while the heat capacity of a monatomic gas is independent of temperature, except at extremely law temperatures, the heat capacities of polyatomic molecules (gases, liquids, and solids) increases with increasing temperature. Heat capacity equations are often of the form
Cp = a + bT + cT2 + dT3 (5-33a)
or
Cp = a + bT + cT – 2 (5-33b)
29
where a, b, and c are empirically determined constants for a given substance. Using Eq. (5-26b) in the differential form
dH = CpdT (5-34)
and substituting the expression for Cp from Eq. (5-33a and 5-33b) into Eq. (5-34), one obtains an expression for the change in enthalpy, dH, of a substance in some phase for a change in temperature dT
dH = (a + bT + cT 2+ dT3) dT
(5-35a)
or
dH = (a + bT + cT – 2) dT (5-35b)
If one wants to determine the change in enthalpy H for a process in which the temperature changes from T1 to T2 (which is the heat qp
necessary to raise the temperature of a compound in the same phase from T1 to T2 at constant pressure, one integrates Eq. (5-35a and 5-35b) between the temperature T1 and T2. Thus
(5-36)
and
(5-37)
or
(5-38)
and
(5-39)
If either the temperature change or the change in heat capacity over the temperature range is small, i.e., b and c in Eq. (5-33a or 5-33b) are small, then Cp can be taken equal to the constant value a. In this case, Cp is independent of temperature, and integration of Eq. (5-36 or 5-38) gives
H = HT 2 – HT 1 = Cp(T2 – T1) (5-40)
The heat capacities at constant pressure as a function of
30
temperature for various gases are given in Table 5.7.
EXAMPLE 5.8Calculate the heat necessary to raise the temperature of 2 moles of gaseous C12 from 500 K to 1000 K at constant pressure.
SolutionThe expression for Cl2 of Cl2 as a function of temperature is obtained from Table 5.7. Using this in Eq. (5-36), one has for one mole of gas
H = 18515 + 251.25 + 285 = 19051J/mol
Since two moles of Cl2 were involved, H is twice this value or
38102 J. If Cp had been taken as constant and equal to 37.03 J/mol·K over
this temperature range, Eq. (5-50) would give H = 18515 J/mol,
which is much lower than the correct value. Obviously, the vibrational and
rotational moles of motion in C12 are contributing to the heat capacity to
an appreciable extent over this temperature range.
TABLE 5.7 Heat Capacities at Constant Pressure as a Function of T for Some Gases, liquids, and solids. C p ( J/mol·K) = a + bT + cT – 2 *
C(graphite) 16.86 4.77 –8.54Cu 22.64 6.28 0I2 40.12 49.79 0NaCl 45.94 16.32 0*Positive exponent means you use negative exponent in the expression; e.g. if b 103 = 123.0, then b = 123.0 10–3
5.14 ENTHALPY CHANGE OF A REACTION AT ANY TEMPERATUREThe heat capacities of reactants and products can be used to calculate the enthalpy change for a given reaction at some temperature T2 from a knowledge of the enthalpy change for the reaction at some other temperature T1. A calculation of this type is very useful since it eliminates the necessity of experimentally determining H for the reaction at every temperature. Let us see how this calculation is performed. Consider the reaction
aA + bB cC + dD
At T1, the enthalpy change is measured to be H1. One now would like to know the enthalpy change at temperature T2, where T2 > T1. This result can be obtained by considering the cyclic process shown in Figure 5.7.
(aA + bB) at T2 H2 cC + dD) at T2
H’ H”
(aA + bB) at T1 H1 (cC + dD) at T1
Figure 5.7. Alternative paths for converting reactants to products.
1. The reaction aA + bB cC + dD at T1 is performed and the enthalpy change is H1.
2. The reactants aA + bB are now raised from T1 to T2, the enthalpy change for this process is given by Eq. (5-36) as
where Cp(reactants) = aCp(A) + bCp(B)
32
3. Next, the reaction aA + bB cC + dD at T2 is performed. The enthalpy change for this step is H2.
4. Finally, the products are lowered in temperature from T2 to T1.The enthalpy change is
where Cp(products) = cCp(C) + dCp(D)
Referring to Figure 5.7, we see that since H is independent of path
H1 = H’ + H2 + H”
Now H2 is the only unknown, soH2 = H1 – H” – H’
(5-41)
We can change the sign of the second term on the right-hand side of this equation if we reverse the limits of integration. Thus
(5-42)
The expression can be made more compact if we define
Cp = Cp(products) – Cp(reactants)
= cCp(C) + dCp(d) – aCp(A) – bCp(B)
The integrals can be combined to give
(5-43)
(5-
44)
We see now that the difference in H at the two temperatures depends on the difference of the heat capacities of the products and reactants. Often this heat-capacity difference is very small, and H is virtually independent of temperature, particularly over small temperature ranges.
If the heat capacities reactants and products do not change appreciably over the temperature range T1 to T2, then we can write
33
H2 = H1 + CpT (5-45)
On the other hand, if the heat capacities change appreciably over the temperature range T1 to T2, then the empirical expressions for Cp as a function of T for each substance must be used in Eq. (5-44) in a manner exactly analogous to that used in Example 5-8.
EXAMPLE 5.9The hydrogenation of methyl chloride was studied at 248C (521 K). It was found that at that temperature H521 = –82.32 kJ. Calculate by using Cp values reported at 400 K. Cp(CH4(g)) = 41.28 J/mol·K, Cp(HCl(g) = 29.11 J/mol·K, Cp(CH3Cl(g) = 48.46 J/mol·K, and Cp(H2(g) = 28.56 J/mol·K.
where a = (19.9 – 3.95 – 32.22 = –16.29, b = (20.95 10 – 2)T –
(15.63 10 – 2)T – 0.192 10 – 2)T = 15.13 10 – 2T and so on.
= 4039.92 – 2211,54 + 271.11 – 11.78
= 2087.71
= + 2087.71
= – 45.756 kJ + 2.088 kJ = – 43.668 kJ
Notice that, like in the previous Exercise, at approximately 250C lower termperature, the reaction is only slightly less exothermic, and for practical purposes H can be considered insensitive to temperature over a few hundred degrees temperature change.
5.15 THE SECOND LAW OF THERMODYNAMICS: FEASIBILITY OF ENERGY CHANGEThe Second law is formulated in order to complete the generalization of the experimental facts concerned with the way in which energy forms can be interconverted. The First law is only concerned with energy balance, i.e., if one form of energy is created during a process, an equal quantity of energy, either in the same form or in a different form, must be used up. For example, when two metal bars at different temperatures are brought together, q calories of heat will flow from the bar at the higher temperature to the one at the lower temperature until thermal equilibrium is established. The reverse process can be considered to occur also, i.e., –q calories of heat can be lost by one bar and gained by the second, thus restoring the system (the two bars) to its original state. Neither process
35
violates the First law since all energy changes are accounted for. However, the second process violates our knowledge, based on numerous observations, that a bar of metal at a uniform temperature can never, without some external help, become hotter at one end and colder at the other. Hence our experience places restrictions on how processes involving energy changes can occur that are not considered in the First law. These restrictions are formulated into expressions of the Second law.
As can be seen from the example just given, the Second law is concerned with the feasibility of an energy change occurring. Thus the Second law states that heat can never flow from a colder to a hotter body without some outside aid.
5.16 SPONTANEOUS PROCESSES AND ENTROPYOne of the main objectives in studying thermodynamics, as far as chemists are concerned, is to he able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration). This knowledge is important whether one is synthesizing compounds in a research laboratory, manufacturing chemicals on an industrial scale, or trying to understand the intricate biological processes in a cell. A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous. We observe spontaneous physical and chemical processes every day, including many of the following examples:
· A waterfall runs downhill, but never up, spontaneously.· A lump of sugar spontaneously dissolves in a cup of coffee,
but dissolved sugar does not spontaneously reappear in its original form.
· Water freezes spontaneously below 0°C, and ice melts spontaneously above 0°C (at 1 atm).
· Heat flows from a hotter object to a colder one, but the reverse never happens spontaneously.
· The expansion of a gas in an evacuated bulb is a spontaneous process. The reverse process, that is, the gathering of all the molecules into one bulb, is not spontaneous.
· A piece of sodium metal reacts violently with water to form sodium hydroxide and hydrogen gas. However, hydrogen gas does not react with sodium hydroxide to form water and sodium.
· Iron exposed to water and oxygen forms rust, but rust does not spontaneously change back to iron.
These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction. If we assume that spontaneous processes occur so as to decrease the energy of a system, we can explain why a ball rolls downhill. Similarly, a large number of
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exothermic reactions are spontaneous. An example is the combustion of methane:
CH4( g ) + 2 O2 (g) CO2( g ) + 2 H2O(l) = –890.4 kJ
Another example is the acid-base neutralization reaction:
H+(aq) + OH–(aq) H2O(l) = –56.2 kJ
But consider a solid-to-liquid phase transition such as
H20(s) H2O(l) = 6.01 kJ
In this case, the assumption that spontaneous processes always decrease a system's energy fails. Experience tells us that ice melts spontaneously above 0C even though the process is endothermic. Another example that contradicts our assumption is the dissolution of ammonium nitrate in water:
NH4NO3(s) (aq) + (aq) = 25 kJ
This process is spontaneous, and yet it is also endothermic. The decomposition of mercury(II) oxide is an endothermic reaction that is nonspontaneous at room temperature, but it becomes spontaneous when the temperature is raised:
2HgO(s) 2Hg(l) + O2(g) = 90.7 kJ
From a study of the examples mentioned and many more cases, we come to the following conclusion: Exothermicity favors the spontaneity of a reaction but does not guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is possible for an exothermic reaction to be nonspontaneous. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy.
5.17 ENTROPY In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, which is nearly equivalent to E for most processes. The other is entropy (S), which is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system, the smaller its entropy. One way to illustrate order and disorder is with a deck of playing cards. A new deck of cards is arranged in an ordered fashion (the cards are from ace to king and the suits are in the order of spades to hearts to diamonds to clubs). Once the deck has been shuffled, the cards are no longer in sequence by number or by suit. It is possible, but extremely unlikely, that by reshuffling the cards we can restore the original order. There are many ways for the cards to be Out of sequence but only one way for them to
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he ordered according to our definition.For any substance, the particles in the solid state are more
ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state. So for the same molar amount of a substance, we can write
Ssolid < Sliquid << Sgas
In other words, entropy describes the extent to which atoms, molecules, or ions are distributed in a disorderly fashion in a given region in space.
One way to conceptualize order and disorder is in terms of probability. A probable event is one that can happen in many ways, and an improbable event is one that can happen in only one or a few ways. An ordered state has a low probability of occurring and a small entropy, while a disordered state has a high probability of occurring and a large entropy.
As we will see, it is possible to determine the absolute entropy of a substance, something we cannot do for energy or enthalpy. Standard entropy is the absolute entropy of a substance at 1 atm and 25°C. It is this value that is generally used in calculations. (Recall that the standard state refers only to 1 atm. The reason for specifying 25°C is that many processes are carried out at room temperature.) Table 5.7 lists standard entropies of a few elements and compounds; A more extensive listing can be found in Appendices of General Chemistry books. The units of entropy are J/K or J/K∙mol for 1 mole of the substance. We use joules rather than kilojoules because entropy values are typically quite small. Entropies of elements and compounds are all positive (that is, S° > 0). By contrast, the standard enthalpy of formation ( ) for elements in their stable form is zero, and for compounds it may be positive or negative.
Referring to Table 5.8, we can make the following observations. The entropy of water vapor is greater than the entropy of liquid water, because a mole of water has a much smaller volume in the liquid state than it does in the gaseous state. In other words, water molecules are more ordered in the liquid state because there is less space for them to occupy. Similarly, bromine vapor has a higher entropy value than liquid bromine, and iodine vapor has a greater entropy value than solid iodine. For different substances in the same phase, the molecular complexity and molar mass determine which ones have higher entropy values. Both diamond and graphite are solids, for example, but diamond has the more ordered structure. Therefore, diamond has a smaller entropy value than graphite. Both neon and helium are monatomic gases, but neon has a greater entropy value than helium because its molar mass is Greater.
Table 5.8 Standard Entropy Values (S) for Selected Substances.Substance S(J/K·mol)
Like energy and enthalpy, entropy is a state function. Consider a certain process in which a system changes from some initial state to some final state. The entropy change for the process, S, is
S = Sf – Si (5-46)
where Sf and Si are the entropies of the system in the final and initial states, respectively. If the change results in an increase in randomness, or disorder, then Sf > Si or S > 0. Thus, both melting and vaporization processes have S > 0. The solution process usually leads to an increase in entropy. The following example deals with the entropy changes of a system resulting from physical changes.
EXAMPLE 5.11
Predict whether the entropy change is greater than or less than zero for each of the following processes: (a) freezing ethanol, (b) evaporating a beaker of liquid bromine at room temperature, (c) dissolving sucrose in water, (d) cooling nitrogen gas from 80°C to 20°C.Answer
(a) This is a liquid-to-solid phase transition. The system becomes more ordered, so that S < 0.
(b) This is a liquid-to-vapor phase transition. The system becomes more disordered, and S > 0.
(c) A solution is invariably more disordered than its components (the solute and solvent). Therefore, S > 0.
(d) Cooling decreases molecular motion; therefore, S < 0.
PRACTICE EXERCISEHow does the entropy of a system change for each of the following processes? (a) condensing water vapor, (b) forming sucrose crystals from a supersaturated solution, (c) heating hydrogen gas from 60°C to 80°C, (d) subliming dry ice
5.18 THE SECOND LAW OF THERMODYNAMICS
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STATED IN TERMS OF ENTROPYThe connect ion between entropy and the spontane i ty o f a react ion i s expressed by the second law of thermodynamics: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Since the universe is made up of the system and the surroundings, the entropy change in the universe (Suniv) for any process is the sum of the entropy changes in the system (Ssys) and in the surroundings (Ssurr). Mathematically, we can express the second law of thermodynam-ics as follows:
For a spontaneous process: Suniv = Ssys + Ssurr > 0 (5-47)
For an equilibrium process: Suniv = Ssys + Ssurr = 0 (5-48)
For a spontaneous process, the second law says that Suniv must be greater than zero, but it does not place a restriction on either Ssys
or Ssurr. Thus it is possible for either Ssys or Ssurr to be negative, as long as the sum of these two quantities is greater than zero. For an equilibrium process, Suniv is zero. In this case Ssys and Ssurr must be equal in magnitude, but opposite in sign. What if for some process we find that Suniv is negative'? What this means is that the process is not spontaneous in the direction described. Rather, it is spontaneous in the opposite direction.
ENTROPY CHANGES IN THE SYSTEMTo calculate Suniv we need to know both Ssys and Ssurr. Let us focus first on Ssys Suppose that the system is represented by the following reaction:
aA + bB cC + dD
As is the case for the enthalpy of a reaction, the standard entropy of reaction is given by
= [cS(C) + dS(D)] – [aS(A) + bS(B) (5-49)
or, in general, using to represent summation and m and n for the stoichiometric coefficients in the reaction,
= n S(productd) – mS(reactants) (5-50)
The standard entropy values of a large number of compounds have been measured in J/K∙mol. To calculate (which is Ssys), we look up their values in an Appendix and proceed according to the following example.
EXAMPLE 5.12
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From the absolute entropy values the chemicals taking part in the reactions given below, calculate the standard entropy changes for the following reactions at 25°C. S° (CaO) = 39.8 J/K∙mol, S°(C02) = 213.6J/K∙mol, S°(CaCO3) = 92.9 J/K∙mol, S° (NH3) =193 J/K∙mol, S°(N2) =192 J/K∙mol, S°(H2) = 131 J/K∙mol, S°(HCl) = 187 J/K∙mol, S°(Cl2) = 223 J/K∙mol.
This result shows that when I mole of gaseous nitrogen reacts with 3 moles of gaseous hydrogen to form 2 moles of gaseous ammonia, there is a decrease in entropy equal to –199 J/K.
Thus the formation of 2 moles of gaseous HCl from 1 mole of gaseous H2 and 1 mole of gaseous C12 results in a small increase in entropy equal to 20 J/K.
The results of Example 12 are consistent with results observed for many other reactions. Taken together, they support the following general rules:
· If a reaction produces more gas molecules than it consumes [Example 12(a)], S° is positive.
· If the total number of gas molecules diminishes ]Example 12(b)], S° is negative.
· If there is no net change in the total number of gas molecules ]Example 12(c)], then S° may be positive or negative, but will he relatively small numerically.
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These conclusions make sense, given that gases invariably have greater entropy than liquids and solids. For reactions involving only liquids and solids, predicting the sign of S° is more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase in entropy.
The following example shows how knowing the nature of reactants and products makes it possible to predict entropy changes.
EXAMPLE 5.14Predict whether the entropy change of the system in each of the following reactions is positive or negative.
Answer (a) Two gases combine to form a liquid. Therefore, S is negative.(b) Since the solid is converted to two gaseous products, S is positive.(c) We see that the same number of moles of gas is involved in the
reactants as in the product. Therefore we cannot predict the sign of S, but we know the change must be quite small.
PRACTICE EXERCISEDiscuss qualitatively the sign of the entropy change expected for each of the following processes:(a) I2(g) 2 I(g)(b) 2 Zn( s ) + 02(g) 2 ZnO(s)(c) N2(g) + O2(g) 2 NO(g)
ENTROPY CHANGES IN THE SURROUNDINGSNext we see how the Ssurr is calculated. When an exothermic process takes place in the system, the heat transferred to the surroundings enhances motion of the molecules in the surroundings. Consequently, there is an increase in disorder at the molecular level, and the entropy of the surroundings increases. Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the surroundings because molecular motion decreases. For constant-pressure processes the heat change is equal to the enthalpy change of the system, Ssys. Therefore, the change in entropy of the surroundings, Ssurr is proportional to
Ssurr –Ssys
The minus sign is used because if the process is exothermic, Ssys
is negative and Ssurr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, Ssys is
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positive and the negative sign ensures that the entropy of the surroundings decreases.
The change in entropy for a given amount of heat also depends on the temperature. If the temperature of the surroundings is high, the molecules are already quite energetic. Therefore, the absorption of heat from an exothermic process in the system will have relatively little impact on molecular motion and the resulting increase in entropy will be small. However, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a more drastic increase in molecular motion and hence a larger increase in entropy. By analogy, someone coughing in a crowded restaurant will not disturb too many people, but someone coughing in a library definitely will. From the inverse relationship between Ssurr and temperature T (in Kelvins)—that is, the higher the temperature, the smaller the Ssurr and vice versa—we can rewrite the above relationship as
(5-61)
Let us now apply the procedure for calculating Ssys and Ssurr
to the synthesis of ammonia and ask whether the reaction is spontaneous at 25°C:
N2(g) + 3 H2(g) 2 NH3(g) H = –92.6 kJ
From Example 12(b) we have Ssys = –199 J/K, and substituting Hsys (–
92.6 kJ) in Equation (5-51), we obtain
The change in entropy of the universe is
Suniv = Ssys + Ssurr
= –199 J/K + 311 J/K = 112 J/K
Because Suniv is positive, we predict that the reaction is spontaneous at 25°C. It is important to keep in mind that just because a reaction is spontaneous does not mean that it will occur at an observable rate. The synthesis of ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur.
5.19 PHASE TRANSITIONSSince during a phase transition there is a change in molecular order, there should be a change in entropy. At the temperature at which a phase transition occurs (the melting point or boiling point) two phases exist at equilibrium. If a system is at equilibrium, then there is no tendency for spontaneous change in either direction. Under constant pressure conditions Equation (5-48) becomes
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Ssys = –Ssurr
(5-52)
The change in entropy as a result of vaporization, fusion, and sublimation can be written as
(5-53)
where Hfus, Hvap, and Hsub are the heats of fusion, vaporization, and sublimation, respectively, and Tfus, Tboil, and Tsub are the fusion, boiling, and sublimation temperatures, respectively.
As an example, let us first consider the ice-water equilibrium. For the ice water transition, Hsys is the molar heat of fusion, equal to 6010 J/mol and T is the normal melting point. The entropy change is therefore
Thus when 1 mole of ice melts at 0°C, there is an increase in entropy of 22.0 J/K. The increase in entropy is consistent with the increase in disorder from solid to liquid. Conversely, for the water ice transition, the decrease in entropy is given by
In the laboratory we normally carry out unidirectional phase changes, that is, either ice to water or water to ice. We can calculate entropy change in each case using the equation S = H/T as long as the temperature remains at 0°C. The same procedure can be applied to the water-steam transition. In this case H is the heat of vaporization and T is the boiling point of water. Example 5.14 examines the phase transitions in benzene.
EXAMPLE 5.14The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid liquid and liquid vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
SolutionAt the melting point, the system is at equilibrium. Therefore S, the entropy of fusion is given by
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= 39.1 J/K · mol
Similarly, at the boiling point we have
= 87.8 J/K · mol
Comment Since vaporization creates more disorder than the melting process, Svap > Sfus.
5.20 ENTROPY CHANGES WHEN A SUBSTANCE IS HEATED IRREVERSIBLY AT CONSTANT PRESSURE.To obtain the change of entropy as a substance is heated irreversibly, it is necessary to discover a reversible path between the same initial and final temperatures. Since S is a State function, S is the same for the irreversible as for the reversible process. Therefore, the equations at constant pressure are
(5-54)
Assuming Cp is independent of temperature, we can write
If the system is heated very slowly the change in the surroundings is equal and opposite in sign to that for the system, and
Ssys + Ssurr = Suniv = 0
The entropy changes for temperature changes at constant volume are analogous to those at constant pressure except that Cv
replaces Cp.
(5-
65)
Again for a reversible change, the entropy change for the system plus surroundings is zero.
5.21 THE THIRD LAW OF THERMODYNAMICS AND ABSOLUTE ENTROPY
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Finally it is appropriate to consider the third law of thermodynamics briefly in connection with the determination of entropy values. So far we have related entropy to molecular disorder—the greater the disorder or freedom of motion of the atoms or molecules in a system, the greater the entropy of the system. The most ordered arrangement of any substance with the least freedom of atomic or molecular motion is a perfect crystalline substance at absolute zero (0 K). It follows, therefore, that the lowest entropy any substance can attain is that of a perfect crystal at absolute zero. According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at the absolute zero of temperature. As the temperature increases, the freedom of motion also increases. Thus the entropy of any substance at a temperature above 0 K is greater than zero. Note also that if the crystal is impure or if it has defects, then its entropy is greater than zero even at 0 K because it would not be perfectly ordered.
The important point about the third law of thermodynamics is that it allows us to determine the absolute entropies of substances. Starting with the knowledge that the entropy of a pure crystalline substance is zero at 0 K, we can measure the increase in entropy of the substance when it is heated to, say, 298 K. The change in entropy, S, is given by
S = Sf – S i
= Sf
since S i is zero. The entropy of the substance at 298 K, then, is given by S or Sf, which is called the absolute entropy because this is the true value and not a value derived using some arbitrary reference. Thus the entropy values quoted so far are all absolute entropies. In contrast, we cannot have the absolute energy or enthalpy of a substance because the zero of energy or enthalpy is undefined.
For a constant pressure process, provided there is no change in the phase of the substance, the entropy change of a chemical substance is given by
(5-56)
Hence, to determine the absolute entropy value of some chemical substance at temperature T one must express the heat capacity of the substance as a function of T (i.e., Cp = a + bT + cT 2 + dT3) and then determine the entropy value either numerically by the integration of Eq (5-56), or graphically by determining the area under the curve for the graph of Cp/T vs T (Figure 5.8).
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Firure 5.8
It should be realized that if the crystalline substance melts and then turns into a gas before it reaches 298°K, the calculation of the standard entropy given by Eq. (5-56) should be modified so as to include the entropy change on fusion and vaporization given by Eq. (5-53), and the entropy changes due to heating to 298°K. Thus Eq. (5-56) becomes
(5-
57)
If we try to calculate the entropy from 0 to 298 using this equation, with Cp dependent on temperature, we get the following for the first, third and the fifth term
This obviously does not work for a change in temperature from 0 K to Tf because at 0 K, we have lnT term which goes to infinity. One approach to overcome this complication is to measure values of Cp down to a few Kelvins, and then use Debye theory of heat capacity of solids at low temperatures,
(5-58)
Notice that Equation (5-58) contains only one constant, k, which can be determined from a value of Cp in the region below 20 K. The integral for the entropy then becomes
EXAMPLE 5.15The molar Cp of a solid at 10 K is 0.43 J/K∙mol. What is the entropy of the solid at that temperature?
Solution
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Cp at 10 K is described by the Debye Equation
The value of k in the above Equation is
The entropy of the solid at 10 K is
Substituting
S = 0.14 J/K∙mol
Figure 5.9 shows the change (increase) in entropy of a substance with temperature. At absolute zero, it has a zero entropy value (assuming that it is a perfect crystalline substance). As it is heated, its entropy increases gradually because of greater molecular motion (described by the first term of Eq. (5-57). At the melting point, there is a sizable increase in entropy as the more random liquid state is formed. Further heating increases the entropy of the liquid again due to enhanced molecular motion (third term of Eq. 5-57)). At the boiling point there is a large increase in entropy as a result of the liquid to gas transition. Beyond that temperature, the entropy of the gas continues to rise with increasing temperature (Fifth term of Eq. (5-57)). The entropy change of the overall process from 0 K to T2 is the sum of the five terms in Eq. (5-57).
As another example, consider the calculation of absolute entropy of methylammonium chloride, CH3NH3Cl, which undergoes phase transitions involving three allotropic forms between 0 K and 298 K. Heat capacities for this solid in its various allotropic forms have been determined very precisely down to 12 K.
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Figure 5.9 Entropy increase of a substance as the temperature rises from absolute zero
Solid methylammonium chloride exist in the form from 0 K up to 220.4 K, in the form from 220.4 to 264.5 K, and in the form from 264.5 K to 298 K. In the following equations, Cp is the heat capacity of the form, Cp’ of the form, and Cp” of the form. One can calculate the entropy by integrating Equation (5-64) for each allotrope in the temperature region in which it is most stable and then adding the two entropies of transition to the integral thus obtained. The details at a pressure of 1 atm are as follows:
a) CH3NH3Cl(s, , 0 K) CH3NH3Cl(s, , 12.04 K)
b) CH3NH3Cl(s, , 12.04 K) CH3NH3Cl(s, , 220.4 K)
c) CH3NH3Cl(s, , 220.4 K) CH3NH3Cl(s, , 220.4 K)
d) CH3NH3Cl(s, , 220.4 K) CH3NH3Cl(s, , 264.5 K)
e) CH3NH3Cl(s, , 264.5 K) CH3NH3Cl(s, , 264.5 K)
f) CH3NH3Cl(s, , 264.5 K) CH3NH3Cl(s, , 298 K)
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Addition of steps (a) through (f) gives
g) CH3NH3Cl(s, , 0 K) CH3NH3Cl(s, , 298 K)
S = S = S1 + S1 + S2 + S3 + S4 + S5 + S6 = 138.6
Thus for methylammonium chloride, CH3NH3Cl, the absolute entropy, S, =138.6 .
5.22 VARIATION OF THE ENTROPY CHANGE OF A REACTION WITH TEMPERATURERecall that entropy, like enthalpy, is a State function and arguments used to develop Hrxn at any temperature apply to develop an equation similar to Equation (5-43) shown below
The equation for calculating Srxn at a temperature T2 from a knowledge of Srxn at T1 is
(5-59)
5.23 GIBBS FREE ENERGYThe second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the Universe; that is, Suniv > 0. In order to determine the sign of Suniv, for a reaction, however, we would need to calculate both Ssys and Ssurr. However, we are usually concerned only with what happens in a particular system, and the calculation of Ssurr can be quite difficult. Therefore, we generally rely on another thermodynamic function to help us determine whether a reaction will occur spontaneously if we consider only the system itself.
From Equation (5-47), we know that for a spontaneous process, we have
Suniv = Ssys + Ssurr > 0
Substituting –Hsys /T for Ssurr, we write
Suniv = Ssys –
Multiplying both sides of the equation by T gives
TSuniv = –Hsys + TSsys > 0
Now we have a criterion for a spontaneous reaction that is expressed only in terms of the properties of the system (Hsys and Ssys) and we
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can ignore the surroundings. For convenience, we can change the above equation, multiplying it throughout by –1 and replacing the > sign with <:
–TSuniv = Hsys – TSsys < 0
This equation says that for a process carried out at temperature T, if the changes in enthalpy and entropy of the system are such that Hsys – TSsys is less than zero, the process must be spontaneous.
In order to express the spontaneity of a reaction more directly, we can use another thermodynamic function called Gibbs free energy (G), or simply free energy:
G = H – TS (5-60)
All quantities in Equation (5-60) pertain to the system, and T is the temperature of the system. You can see that G has units of energy (both H and TS are in energy units). Like H and S, G is a state function.
The change in free energy (G) of a system for a constant-temperature process is
G = H – TS (5-61)
In this context free energy is the energy available to do work. Thus, if a particular reaction is accompanied by a release of usable energy (that is, if G is negative), this fact alone guarantees that it is spontaneous, and there is no need to worry about what happens to the rest of the universe.
Note that we have merely organized the expression for the entropy change of the universe, eliminating Suniv, and equating the free-energy change of the system (G) with –TSuniv, so that we can focus on changes in the system. We can now summarize the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of G as follows:
G < 0 The reaction is spontaneous in the forward direction.G > 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction.G = 0 The system is at equilibrium. There is no net change.
STANDARD FREE-ENERGY CHANGESThe standard free-energy of reaction ( ) is the free-energy change for a reaction when it occurs under standard-state conditions, when reactants in their standard states are converted to products in their standard states. Table 5.9 summarizes the conventions used by chemists to define the standard states of pure substances as well as solutions.
Table 5.9 Conventions for Standard StatesState of Matter Standard StateGas 1 atm pressure
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Liquid Pure liquidSolid Pure solidElements* (element) = 0Solution 1 molar concentration*The most stable allotropic form at 25C and 1 atm.
To calculate we start with the equation
aA + bB cC + dD
The standard free-energy change for this reaction is given by
= [c (c) + d (D)] – [a (A) + b (B)]
or, in general,
= n (products) – n (reactants) (5-62)
where n and m are stoichiometric coefficients. The term is the standard free energy of formation of a compound, that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states. For the combustion of graphite:
C(graphite) + O2(g) CO2(g)
the standard free-energy change [from Equation (5-62)] is
As in the case of the standard enthalpy of formation, we define the standard free energy of formation of any element in its stable form as zero. Thus
(C, graphite) = 0 and (O2) = 0
Therefore the standard free-energy change for the reaction in this case is numerically equal to the standard free energy of formation of CO2:
= (CO2)
Note that is in kJ, but is in kJ/mol. for a number of compounds can be found in the Appendices of most General Chemistry textbooks. Calculations of standard free-energy changes are handled as shown in the following example.
EXAMPLE 5.16Calculate the standard free-energy changes for the following reactions at 25°C.
In the preceding example the large negative value of for the combustion of methane in (a) means that the reaction is a spontaneous process under standard-state conditions, whereas the decomposition of MgO in (b) is nonspontaneous because is a large, positive quantity. Remember, however, that a large, negative does not tell us anything about the actual rate of the spontaneous process; a mixture of CH4 and O2 at 25°C could sit unchanged for quite some time in the absence of a spark or flame.
APPLICATIONS OF EQUATION (5-61)In order to predict the sign of G, according to Equation (5-61) we need to know both H and S. A negative H (an exothermic reaction) and a positive S (a reaction that results in an increase in disorder of the system) tend to make G negative, although temperature may influence the direction of a spontaneous reaction. The four possible outcomes of this relationship are:· If both H and S are positive, then G will be negative only when
the TS term is greater in magnitude than H. This condition is met when T is large.
· If H is positive and S is negative, G will always be positive, regardless of temperature.
· If H is negative and S is positive, then G will always be negative regardless of temperature.
· If H is negative and S is negative, then G will be negative only when TS is smaller in magnitude than H. This condition is met when T is small.
The temperatures that will cause G to be negative for the first and last cases depend on the actual values of H and S of the system. Table 5.10 summarizes the effects of the possibilities just described.
We will now consider two specific applications of Equation (5.61).
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TABLE 5.10 Factors Affecting the Sign of G in the Relationship G = H – TS
H S G Example
+ +Reaction proceeds spontaneously at high
temperatures. At low temperatures, reaction is
spontaneous in the reverse direction.
H2(g) + I2(g)2HI(g)
+ –G is always positive. Reaction is spontaneous in the
reverse direction at all temperatures.3O2(g)2O3(g)
– +G is always negative. Reaction proceeds
spontaneously at all tempe-ratures.2H2O2(l)2H2O(l)+O2(g)
– –Reaction proceeds spontaneously at low
temperatures. At high temperatures, the reverse
reaction becomes spontaneous.
NH3(g)+HCl(g)NH4Cl(s)
TEMPERATURE AND CHEMICAL REACTIONCalcium oxide (CaO), also called quicklime, is an extremely valuable inorganic substance used in steelmaking, production of calcium metal, the paper industry, water treatment, and pollution control. It is prepared by decomposing limestone (CaCO 3) in a kiln at a high temperature:
CaCO3(s) CaO(s) + CO2(g)
The reaction is reversible, and CaO readily combines with CO 2 to form CaCO3. The pressure of CO2 in equilibrium with CaCO3 and CaO increases with temperature. In the industrial preparation of quicklime, the system is never maintained at equilibrium; rather, CO2 is constantly removed from the kiln to shift the equilibrium from left to right, promoting the formation of calcium oxide.
The important information for the practical chemist is the temperature at which the decomposition of CaCO3 becomes appreciable (that is, the temperature at which the reaction becomes spontaneous). We can make a reliable estimate of that temperature as follows. First we calculate H and S for the reaction at 25°C, using the data available in an Appendix. To determine H we apply Equation (5.30):
For reactions carried out under standard-state conditions, Equation (5-71) takes the form
G = H – TSso we obtain
G = 177.8 kJ — (298 K)(160.5 J/K) (1 kJ/1000J)
= 130.0 kJ
Since G is a large positive quantity, we conclude that the reaction is not favored at 25°C (or 298 K). In order to make G negative, we first have to find the temperature at which G is zero; that is,
0 = H – TS
= 1108 K or 835C
At a temperature higher than 835°C, G becomes negative, indicating that the decomposition is spontaneous. For example, at 840°C, or 1113 K,
G = H – TS
= 177.8 kJ – (1113 K)(160.5 J/K)(1kJ/1000 J)
= –0.8 kJ
Two points are worth making about such a calculation. First, we used the H and S values at 25°C to calculate changes that occur at a much higher temperature. Since both H and S change with temperature, this approach will not give us an accurate value of G, but it is good enough for rough estimates. Second, the standard state does not specify temperature, therefore we can use the symbol G. Third, we should not be misled into thinking that nothing happens below 835°C and that at 835°C CaCO3 suddenly begins to decompose. The fact that G is a positive value at some temperature below 835°C does not mean that no CO2 is produced, but rather that the pressure of the CO2
gas formed at that temperature will be below 1 atm (its standard-state value). In fact, the pressure of CO2 at first increases very slowly with temperature; however, it becomes easily measurable above 700°C. The significance of 835°C is that this is the temperature at which the equilibrium pressure of CO2 reaches 1 atm. Above 835°C, the equilibrium pressure of CO 2 exceeds 1
55
atm. If a system is at equilibrium, then there is no tendency for
spontaneous change in either direction. The condition G = 0 applies to any phase transition.
PHASE TRANSITIONSAt the temperature at which a phase transition occurs (the melting point or boiling point), G = 0 so Equation (5-61) becomes
0 = H – TS
where H is Hfus, Hvap, or Hsub; the heat of fusion, vaporization, or sublimation, and T is Tfus, Tboil, or Tsub; the fusion, boiling, or sublimation temperature.
Notice that the above equation is the same as Equation (5-53) and can be applied to calculate entropy changes during phase transitions. (See Example 5.14).
5.24 FREE ENERGY AND CHEMICAL EQUILIBRIUMSuppose we start a reaction in solution with all the reactants in their standard states (that is, all at 1 M concentration). As soon as the reaction starts, the standard-state condition no longer exists for the reactants or the products since their concentrations are different from 1 M. Under conditions that are not standard state, we must use G rather than G to predict the direction of the reaction. The relationship between G and G is
G = G + RT ln Q (5-63)
where R is the gas constant (8.314 J/K∙mol), T is the absolute temperature of the reaction, and Q is the reaction quotient, that is, a number equal to the ratio of product concentrations, each raised to the power of its stoichioinctric coefficient at some point other than equilibrium. We see that G depends on two quantities: G and RT ln Q. For a given reaction at temperature T the value of G is fixed but that of RT ln Q is not, because Q varies according to the composition of the reacting mixture. Let us consider two special cases:
Case 1: If G is a large negative value, the RT ln Q term will not become positive enough to match the G term until a significant amount of product has formed.
Case 2: If G is a large positive value, the RT ln Q term will be more negative than G is positive only as long as very little product formation has occurred and the concentration of the reactant is high relative to that of the product.
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At equilibrium, by definition, G = 0 and Q = K, where K is the equilibrium constant. Thus
0 = G + RT ln K
or
G = –RT ln K (5-64)
In this equation, Kp is used for gases and Kc for reactions in solution. Note that the larger the K is, the more negative G is. For chemists, Equation (5-74) is one of the most important equations in thermodynamics because it allows us to find the equilibrium constant of a reaction if we know the change in standard free energy and vice versa.
It is significant that Equation (5-64) relates the equilibrium constant to the standard free energy change G rather than to the actual free energy change G. The actual free energy of the system changes as the reaction proceeds and becomes zero at equilibrium. On the other hand, both G and K are constants for a particular reaction at a given temperature. In fact, if G < 0. the products are favored over reactants at equilibrium. Conversely, if G > 0, there will be more reactants than products at equilibrium. Table 5.11 summarizes the three possible relations between G and K, as predicted by Equation (5-64).
For reactions having very large or very small equilibrium constants, it is generally very difficult, if not impossible, to measue the K values by monitoring the concentrations of all the reacting species. Consider, for example, the formation of nitric oxide fro molecular nitrogen and molecular oxygen:
N2(g) + O2(g) ⇄ 2 NO(g)
At 25C, the equilibrium constant Kc is
The very small value of Kc means that the concentration of NO at equilibrium will be exceedingly low. In such a case the equilibrium constant is more conveniently obtained from G. (as we have seen, G can be calculated from H and S.) On the other hand, the equilibrium constant for the formation of hydrogen iodide from molecular hydrogen and molecular iodine is near unity at room temperature:
H2(g) + I2(g) ⇄ 2 HI(g)
For this reaction it is easier to measure K and then calculate G using Equation (5-64) than to measure H and S.
Table 5.11 Relation between G and K as Predicted by the Equation G = –RT ln KK ln K G Comments
> 1 Positive Negative Products are favored over reactants at euilibrium.= 1 0 0 Products and reactants are equally favored at equilibrium.
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< 1 Negative Positive Reactants are favored over products at equilibrium.
It is important to remember when solving Equation (5-64) that the unit of G is J or kJ and those of RT ln K are J/mol. The reason for the difference is that in deriving Equation (5-64), the unit "mol" was absorbed into the logarithmic term (ln K). Thus in using this equation, we will add (1 mol) to the RT ln K term to make the units consistent on both sides of the equation. The following examples illustrate the use of Equations (5-63) and (5-64).
EXAMPLE 5.17
Using data listed in an Appendix, calculate the equilibrium constant (Kp) for the following reaction at 25°C:
Comment This extremely small equilibrium constant is consistent with the fact that water does not decompose into hydrogen and oxygen gases at 25°C. Thus a large positive G favors reactants over products at equilibrium. Note that we have added the (1 mol) term on the right side of Equation (5-64) here to make the units consistent.
EXAMPLE 5.18
Using the solubility product of silver chloride at 25°C (1.6 10–
10), calculate G for the process
AgCl(s) ⇄ Ag+(aq) + Cl(aq)
SolutionBecause this is a heterogeneous equilibrium, the solubility product is the equilibrium constant.
Ksp = [Ag+][CI–] = 1.6 10–
58
10
Using Equation (5-64) we obtain
G = –(1 mol)(8.314 J/K∙mol)(298 K) ln 1.6 10–10
= 5.6 104 J = 56 kJ
Comment The large, positive G indicates that AgCI is slightly soluble and that the equilibrium lies mostly to the left.
As another example, let us consider the decomposition of isopropyl alcohol into acetone and hydrogen gas in the presence of a suitable catalyst:
(CH3)2CHOH(g) ⇄ (CH3)2CO(g) + H2(g)
At 452.2 K and a total pressue, P, of 0.564 atm , the equilibrium constant of this reaction, K, is equal to 0.444. Equation (5-64) can be used to calculate the standard free energy change whatever the total pressure is, as long as the system is at equilibrium.
The positive value of ΔG does not imply that the reaction under consideration may not proceed spontaneously under any conditions. ΔG refers to the reaction
(CH3)2CHOH(g. P = 1 atm) ⇄ (CH3)2CO(g, P = 1 atm) + H2(g, P = 1 atm)
where each substance is in its standard state; that is, at a partial pressure of 1 atm. The positive value of ΔG only means that the reaction will not proceed spontaneously under these conditions. However, if we were to start with isopropyl alcohol at a partial pressure of 1 atm and no acetone or hydrogen, the alcohol would decompose spontaneously at 452.2 K and more than 50% dissociation could occur. Yields can be made even better is one of the products is removed continuously (Le Chatelier’s principle)
We also might calculate ΔG for one set of conditions with the substances not all in their standard states; for example,
(CH3)2CHOH(g, P = 1 atm) ⇄(CH3)2CO(g, P = 0.100 atm) + H2(g, P = 0.100
atm)
For this calculation we refer to Equation (5-63), which relates ΔG to ΔG and the non-equilibrium partial pressures
G = G + RT ln Q
Applied to our reaction, we get
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= 3053 J + (1 mol)(8.314 J/K∙mol)(452.2 K) ×
= 15.5 × 103 J
Thus, if one is considering a given reaction in connection with the preparation of some substance, it is important not to be misled by positive value of ΔG, because ΔG refers to the reaction under standard conditions, when the pressures of each of the component gases is 1 atm and solutions have a concentration of 1 M. It is quite possible that appreciable yields can be obtained even though a reaction will not go to completion. Such was the decomposition of isopropyl alcohol. Only when ΔG has a very large positive value, perhaps greater than 50 kJ, one can be assured without calculations of the equilibrium constant that no significant transformation would occur.
EXAMPLE 5.19
The standard free-energy change for the reaction
N2(g) + 3 H2(g) ⇄ 2 NH3(g)
is –33.2 kJ and the equilibrium constant Kp is 6.59 105 at 25°C. In a
certain experiment, the initial pressures are PH2 = 0.250 atm, PN2 =
0.870 atm, and PNH3 = 12.9 atm. Calculate G for the reaction at these
pressures, and predict the direction of reaction.
SolutionEquation (7-73) can be written as
G = G + RT ln Qp
=
=
= –33.2 103 J + 23.3 103 J
= –9.9 103 J = –9.9 kJ
Since G is negative, the net reaction proceeds from left to right.
Comment Notice that Qp is smaller than it with Kp. This is to be expected considering the very high pressure of ammonia relative to the reacting gases (Le Chatelier’s principle).
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TEMPERATURE DEPENDENCE OF EQUILIBRIAWhile Le Chatelier’s principle provides a qualitative guide for predicting how equilibria are affected by changes in temperature, we can obtain a quantitative relation between K and T by using the thermodynamic concepts available to us. To derive an expression, we combine two fundamental relations: Eqs. (5.61) and (5-64)
ΔG = ΔH TΔS
ΔG = RT ln K
(5-
65)
This equation says that if we accept the approximation that ΔH and ΔS are constants independent of temperature, then lnK becomes linearly dependent on 1/T. According to Equation (5-65), for an exothermic reaction K increases as T increases. These facts are consistent with the qualitative conclusions based on Le Chatelier’s principle.
Not only does the sign of ΔH indicate the direction in which K changes; for a given temperature variation, the magnitude of ΔH determines how rapidly K changes as a function of temperature. According to Equation (5-65), plotting lnK as a function of 1/T should give a straight line whose slope is ΔH/R, Thus the more negative ΔH is, the faster lnK should decrease as T increases, and vice versa (Figure 5.9).
Figure 5.9 lnK versus 1/T for (a) exothermic reaction with ΔH = 40.0 kJ; (b) endothermic reaction with ΔH = 85 kJ.
Another form of Equation (5-65) that is particularly useful for numerical calculations can be obtained by writing
,
61
and then subtracting the first of these expressions from the second. The result is
(5-
66)
Equation (5-66) shows that if we know the value of ΔH and of the equilibrium constant at one temperature, we can calculate K at any other temperature. Also, if we measure K at two different temperatures, we can calculate ΔH by Equation (5-66). Thus, it is possible to obtain ΔH for a reaction without ever doing a calorimetric experiment.
EXAMPLE 5.20For the reaction
NO(g) + 1/2O2(g) ⇄ NO2(g)
ΔG = 34.9 kJ and ΔH = 56.5 kJ at 298 K. Calculate the equilibrium constant at 298 K and ay 598 K.
Solution
K = 1.28 × 106
To find the equilibrium constant at 598 K, we cannot use Equation (5-64) since we do not know ΔG at that temperature. However, we can use Equation (5-66):
K2 = 14.4
The equilibrium constant of this exothermic reaction is smaller at the higher temperatue, which is consistent with Le Chatelier’s principle.
COMMON UNITS AND CONVERSION FACTORS FOR THERMODYNAMICS
1 cal = 4.184 J
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1 L∙atm = 24.21 cal
1 L∙atm = 101.29 J
1 atm = 1.01325 105 N∙m–2 (Newton meter–2)
1 atm = 760 mm Hg
1 atm = 101,325 Pa (Pascal)
1 torr = 1.000 mm Hg
1 torr = 133.32 Pa
1 J = 1 kg∙m2∙s–2
1 N (Newton) = kg∙m∙s–2
1 Pa = N∙m–2
1 Pa = kg∙m–1∙s–2
1 J = kg∙m2∙s–2
0C = 273.15 K (Kelvin)
R (gas constant) = 8.31433 J∙K–1∙mol–1
R = 1.9872 cal∙K–1∙mol–1
R = 0.082053 L∙K–1∙mol–1
F (Faraday constant) = 9.64870 104 C∙mol–1 (coulombs mole–1)
F = 2.3061 104 cal∙V–1∙mol–1
1 eV (electron volt) = 1.601 10–19 J
PROBLEMS
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I. CaCO3(s) decomposes according to
CaCO3(s) CaO(s) + CO2(g) H = 42.4 kcal/mol
Calculate E at 950C by assuming that the change in volume of the solid is negligible. R = 1.987 cal/mol.K.
II. Using the chemical equations given below, calculate the standard enthalpy, H, of the following reaction
HCOOH(l) CO(g) + H2O(l)
C(s) + ½O2(g) CO(g) Hf = –26.4 kcal/mol
H2(g) + ½O2(g) H2O(l) Hf = –68.3 kcal/mol
H2(g) + O2(g) + C(s) HCOOH(l) Hf = –97.5 kcal/mol
III. Predict whether the following reactions take place spontaneously: a) at 298 K and 1 atm, b) at 800 K and 1 atm
SO3(g) + NO(g) SO2(g) + NO2(g)
H = 9.98 kcal/mol; S = 0.01369 kcal/mol. Assume H and S do not vary with temperature.
IV. Using the following equations, calculate the standard free energy of formation, Gf for ozone at 298 K.
3/2 O2(g) O3(g)
Given:Hf (O3) = 34.0 kcal/mol
S(O3) = 56.8 cal/mol.K
S(O2) = 49.0 cal/mol.K
V. Calculate G and Kp for the reaction
NO(g) + O3(g) ⇄ NO2(g) + O2(g)
Given:Gf (NO2 gas) = 12.39 kcal/mol at 298 K
Gf (O2 gas) = 0
Gf (NO gas) = 20.72 kcal/mol at 298 K
Gf (O3 gas) = 39.0 kcal/mol at 298 K
R = 1.987 cal/mol.K
VI. Calculate Kp and Kc at 25C for the following equilibrium:
NO(g) + ½O2(g) ⇄ NO2(g)
Given:Gf (NO2 gas) = 12.39 kcal/mol at 298 K
Gf (NO gas) = 20.72 kcal/mol
R = 1.987 cal/mol.KR = 0.0821 L.atm/mol.K
VII. Calculate G and H for the following reaction
SO2(g) + ½O2(g) SO3(g)
Assuming that H and S are independent of temperature, calculate the equilibrium constant at 298 K and 600 K.
Given:
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Gf (SO2 gas) = –71.79 kcal/mol
Hf (SO2 gas) = –70.96 kcal/mol
Gf (SO3 gas) = –88.52 kcal/mol
Hf (SO3 gas) = –94.45 kcal/mol
VIII. The enthalpies of the following reactions are for 0C and 1 atm:
N2(g) + 3H2(g) 2NH3(g) H1 = –24.4 kcal
N2(g) + O2(g) 2NO(g) H2 = +43.14 kcal
H2(g) + ½O2(g) H2O(g) H3 = –57.9 kcal
The molar heat capacities at constant pressure for N2, H2, and NH3 respectively are:
Cp(N2) = 6.94 cal/mol.K
Cp(H2) = 6.90 cal/mol.K
Cp(NH3) = 8.63 cal/mol.K
a. Calculate the standard enthalpy of formation for NH3 at 298 K.
b. Calculate for 0C and 1 atm, the enthalpy for the oxidation reaction of NH3(g):
2NH3(g) + 5/2 O2(g) 2NO(g) + 3H2O(g)
IX. Calculate the bond energy C—H in CH4 using the following data:
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = –890.5 kJ
CO2(g) C(graphite) + O2(g) H = +392.0 kJ
2H2O(l) 2H2(g) + O2(g) H = +570.1 kJ
H2(g) 2H(g) H = +435.75 kJ
C(graphite) C(g) H = +715.5 kJ
X. Consider the following equilibrium in aqueous solution:
A ⇄ B + H2O H = 22134 J
with [A] = 4[B] at 25C.a. Calculate the equilibrium constant of the above reaction at 50C (assume
that H and S are independent of temperature).b. Calculate E, G, and S.
XI. Calculate the standard enthalpy of formation of gaseous benzene,
, by the following two methods and comment on the significant difference in the values obtained ascribing the difference to the correct energy factor.
Method a: Use a Hess diagram and the following data:
[H(g)] = 218.0 kJ/mol or H—H bond energy = 2 218.0 kJ/mol,
[C(g)] = 715.0 kJ/molAverage CC bond energy = 615 kJ/molAverage C—C bond energy = 348 kJ/molAverage C—H bond energy = 413 kJ/mol
Method b: Use = –393.5 kJ/mol, = –285.8 kJ/mol, and the enthalpy of combustion of benzene
XII. For which of the following processes would S have a negative value?
a) 2 Fe2O3(s) 4 Fe(s) + 3 O2(g)
b) Mg2+(aq) + 2 OH–(aq) Mg(OH)2(s)
c) 3 H2(g) + 3 C2H4(g) 3 C2H6(g)
XIII. The standard molar entropy of NH3(g) is 192.45 J /mol·K at 298 K and its heat
capacity is given by c = a + bT + cT 2 where a = 29.75 J /mol·K b = 25.1 10 –
3 J /mol·K and c = –1.55 10 5 J /mol·K.
d) Calculate the change in the s tandard molar entropy, S, from
25ºC at 500ºC.
e) Calculate the standard molar entropy of NH3(g) at 500ºC.
XIV. Make use of the following data on methanol, CH3OH, and a) calculate the entropy change when 1 mole of liquid methanol at 25C is heated to its normal boiling point and then completely evaporated at that temperature. b) Calculate the entropy of 1 mole of gaseous methanol at its boiling point.Given: S for liquid methanol at 25C = 126.8 J/K∙molBoiling point of methanol = 64.6CΔHvap =35.25 J/molCp (J/K∙mol) = 15.15 + 0.10 × 103T 29.50 × 106T2
XV. Ethyl hydroperoxide (CH3CH2-O-O-H) in the gaseous state is rumored to have a value of = 140.0 kJ/mol. Use the data given below to prove that the value is feasible.
[C(g)] = 715.0 kJ/mol
[H(g)] = 218.0 kJ/mol
[O(g)] = 249.2 kJ/mol
Average bond energies in polyatomic molecules
Bond type C-H C-C C-O O-O O-HKJ/mol 413 348 351 146 464
XVI. Using a flow calorimeter, a chemist studied the hydrogenation of methyl chloride
CH3Cl(g) + H2(g) CH4(g) + HCl(g)
and found that at 248C (541 K), = –82.32 kJ/mol. Make the adjustment to the standard temperature (298 k) so that this result could be incorporated into tables of standard data. Cp values at 400 K for this calculation are as follows:
Cp(CH4(g)) = 41,28 J/mol K,
Cp(HCl(g)) = 29.11 J/mol K,
Cp(CH3Cl(g)) = 48.46 J/mol K,
Cp(H2(g)) = 28.56 J/mol K,
XVII. Given the following Hrxn values, expressed as per mol of reactant:
Calculate Hrxn at 298 K for the following reaction
N2O4(g) + N2H4(l) 2N2(g) + 2H2O2(l)
XVIII. Given the thermodynamics data listed in the table below for the following reaction:4 HCl(g) + O2(g) 2 H2O(g) + 2 Cl2(g)
O2(g) Cl2(g) H2O(g) HCl(g)
H (kJ/mol) 0 0 –241.81 –92.31
So (J/K · mol) 205 223 189 187
Cp (J/ K · mol) 30.0+1.110–3 T 36.9+3.010–3 T 30.0++2.110–3 T 26.5+1.010–3 T
a. Calculate at T1 = 298 K the standard enthalpy change, , and the
change in the standard entropy, .
b. Calculate at T1 = 298 K the internal energy change, . R = 8.314 J/ K · mol
c. Calculate at T1 = 298 K the standard free energy change, , and comment on the significance of the value obtained. Calculate the equilibrium constant (Kp) for the reaction at 298 K.
d. Determine the value of at T2 = 500 K. Comment briefly on the significance of the value obtained in relation to the value at T1 (i.e. if it is consistent with any principal in chemistry.
e. Determine the value of at T2 = 500 K. Comment briefly on the significance of the value obtained in relation to the value at T1.
XIX. Using data given below, calculate the standard enthalpy of formation of gaseous acetaldehyde, CH3COH.
Standard enthalpy of sublimation for carbon is
C(s) C(g) = 718 kJ/mol
Bond Dissociation Energies of Diatomic Molecules and Average Bond Energies for Bonds in Polyatomic Molecules in kJ/mol
