G. H. Lovett Monsanto Company TexasCtty.Texas77590
An Elementary View of Thermodynamics for Calculating Chemical Equilibria A review for teachers
When thermochemical calculations are used to analyze a chemical reaction, two enerrv changes in the reacting system are or prime importance. 0"; energy rhange is designa;rd by the symhd A H and precisely, it'n(9t informatively. called the enthalpy change of the system. In a constant pressure reaction involving no gases, AH equals the total energy which must be interchaneed between the svstem and the surroundines t o maintain t11esystem at a amstant temperature. Sinre heat is usually theonly form of energy interrhangrd, 111 is frequently rererred to as the heat 01' reaction. AH is calculated by suhtrarrine the enthal~vof the reactants from the e n t h d ~ v othe f produ&. If AH isnegative, heat will he evolved froKthe reacting svstem into the surroundinps. AH is imoortant in making an energy balance around a chemical reaction, and it is also a factor in calculating the other important energy change AG. AG represents the change in knergy that-is available to do useful work. I t is usually referred to as the change in free energy. If AG is negative, work can be performed by the reacting system under ideal conditions. AG can he measured as the eledropotential of a galvanic cell for a few reactions, but i t is also important in other reactions because it is the dominant quantity in calculating the equilibrium constant. Much of the confusion surrounding thermochemical calculations results from the apparently inconsistent relationship of A H and AG. Although AH is one quantity in the equation for calculating AG, AH can he much larger than AG in one reaction and much smaller in another. I t is the purpose of this article to clarify the relationship between AHand AG by examining a simple, hypothetical reaction in which the inequality of A H and AG is caused by a single factor: the he& of fusion of the reactant. If the effect of the heat of fusion in this simple reaction can be understood, the more complicated reactions can be faced with the confidence that the inequality of AH and AG is nothing more than the same principle at work many times. Before looking at the hypothetical reaction, it is necessary to discuss the nature of the energy involved in a chemical reaction. The second law of thermodynamics restricts the conversion of thermal energy to useful work according to the equation for the ideal heat engine = QI(TI- T P ) (1) T. where W = work performed by heat engine, Q1 =quantity of thermal energy absorbed from reservoir a t TI, TI = temperature of reservoir supplying thermal energy, OK, and Tp = temperature of reservoir into which thermal energy is rejected, n A>. u
Equation (1)still applies if Tz is larger than TI.In this case the heat engine becomes a heat pump, and work must be supplied to raise the thermal energy from TI to T2. Energy can he classified according to whether or not its wes conversion to useful work is restricted hv .eon. . (1). . . Both t ". of energy are involved in a chemical reaction. The energy renresented hv the heat of reaction at 0°K is not restricted. ~ work. It is which means that it is totally convertible t c useful R simoliatir statement of the third law of thermod\.narnirs that at 0°K 626 / Journal of Chemical Education
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Figure 1. AHand A G a s functionof absolute temperature for reaction A B. Heat capacnies of A and B are equal, AHo = -2000 callmole. A melts at 15O0K with heal af fusion of 1000 cal/male. No other phase changes in A or B from
0-600°K.
The energy stored in a chemical compound above absolute zero is thermal energy, and it is restricted by eqn. (1)in its ability to be converted to useful work. The stored thermal energy of a compound can be calculated from its heat capacity and the various latent heats associated with phase changes that occur in the compound. A compound may he referred to as a heat reservoir because of its ability to store thermal energy. If the third law says that AH and AG are equal a t O"K, their inequality a t elevated temperatures must be caused by the thermal energy stored in the reactants and products. Consider the simole chemical reaction in which one mole of reactant A is converted to one mole of product B. Prior to the reaction, A has a certain auantitv of thermal enerw stored in it between O°K and the reaction temperature. As x i s converted to B, A disa~oearsas a heat reservoir and B anoears as a new heat rese&ir. The thermal e n e r a that had heen stored in A must either hc stored in 13 or it must flow into rhe surroundinas. If A and B are heat reservoirs of the same size and shape, ail the thermal enernv ... stored in A can he stored in B at the same timperature lwei. I n this case A l l and AG =,ill remain equal LO each u t h ~ and r also equal to A11 at 0°K. In the combuition of carbon and oxygen td carbon dioxide there is such a good balance of stored thermal energy in the reactants and product that even at 1500°K AH and AG are almost equal to AH at n " w
A&.
When there is an imbalance between the thermal energy stored in A and B, any movement of thermal energy from one temperature level to another as A is converted to B must be in agreement with the second law (eqn. (1)). AG of the reaction must he corrected for anv useful work that is either nroduced or consumed by this movement. The movement of thermal enerev can be from one tem~eraturelevel in A to another in B, fro& A to the surroundin&, or from the surroundings to B.
The quantitative change in AG due to an imbalance of stored thermal energy can he best illustrated by a simple numerical example. Assume the following thermal data for the reaction A B. The heat of reaction at O°K is -2000 callmole. A and B have equal heat capacities from 0-600°K. A melts at 150°K with a heat of fusion of 1000 callmole. There are no other phase changes in A or B between 0 and 600°K. How do AH and AG vary from 0-600°K? AH and AG are plotted versus temperature on Figure 1. The following should be noted. AH remains constant at -2000 cal from 0-150°K. The melting of A at 150°K produces a step change in AH equal to the heat of fusion. AH remains constant at -3000 callmole from 150-600°K. AG is also equal to -2000 callmole from 0-15O0K. There is no step change in AG a t 150°K, but AG steadily increases as the reaction temperature rises from 150-600°K. The changes in AH and AG between 150-600°K are due entirely to the heat of fusion of A. The step change in AH a t 150°K takes place because A gains 1000 cal of thermal energy as it melts. When liquid A is converted to solid B, this extra 1000 cal must flow into the surroundings since it cannot he stored in B. AG is only influenced by how much useful work is required to move the extra 1000 cal from A into the surroundings. The surroundings are at the reaction temperature, and the heat of fusion is stored in A at 150°K independent of the reaction temperature. Equation ( I )can he used to calculate the work needed for this movement Q1(Tl- Tz) - 1000(150 - T) W= T.L 15n Although no work is required to move the 1000 cal from A at 150°K to the surroundings at 15O0K,work input is required as the reaction temperature Trises above 150°K. This work must be deducted from the work which could otherwise have been produced by converting A to B. At 150°K, 2000 cal of work could be produced by the reaction. At 450°K 2000 cal of work are required to move the excess thermal energy from A to the surroundings leaving AG = 0. At 600°K 3000 cal are required for the movement, and a net 1000 cal of work must he supplied to the reacting system if the conversion is to take place (AG = 1000). I t should he apparent from Figure 1 that the energy which makes AH negative (an exothermic reaction) a t O0K also makes AG negative. The energy which makes AH more negative above 0°K makes AG more positive. Unfortunately when a chemist observes calories being evolved by an exothermic reaction, the calorie making AG more negative looks just like the calorie which is making AG more positive. The chemical equilibrium, however, can sense the difference between the two. An equation for AG as a function of T between 150 and 600°K can be written AG= AHo- Qta(Tra- T) (2) +
A
senarates the contribution of thermal enerwto AG from the cokribution of the unrestricted energy. ~ q & i o n (3)is a little more convenient than e m . ( 2 ) since AH at the reaction temperature is easier to measure than M l at V K . Bothequations ha\,e disad\,anmres for calculnting l(: of real reactions. In eqn. (3) t h e factor Q r . ~ I ~ r i c abe n regarded as a correction factor that must be applied to AH at T in order to calculate AG at T. I t corrects for Qfs cal of thermal energy stored in the reactant a t Tr.. Had B melted a t Tm with a heat of fusion of Qm, a second correction factor with an opposite sign would have been required. The equation for AG would have been (4)
Eauation (4) can handle the calculation of AG for a svstem wheri unr qt;nntity uf thermal energy in the reacrant and one auantit\.in the uruduct are causing - l(: and 1H u)he uneaual. In a real chemical reaction with multiple reactants, multiple products, unequal heat capacities, and many phase changes eqn. (4) will be unwieldy. One solution to this problem would be to apply the correction factor to every quantity of thermal energy stored in both the reactants and products. Similar quantities stored in the products and reactants would cancel each other during the calculation. A new number is required for each component involved in the reaction. The new number will consist of each increment of thermal energy stored in the component from O°K up to the reaction temperature divided by the absolute temperature at which the energy is stored. The number will he a function of the temperature, pressure, and state of the component, but once calculated it can be tabulated and used in other reactions involving that component. For comnonent A this number would be and for B
Then Since it is conventional to calculate AH by subtracting the enerw -. content of the reactants from the energy content of the products, the same convention will be used in calculating AS
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1 fa
where AHo = heat of reaction a t O°K in callmole, Q,, = heat of fusion of A in callmole, TI,= melting point of A in OK, and T = reaction temperature OK. Equation ( 2 ) can be rearranged AG= AHo-Qr,+- Qf.T T, AH = AH0 - Qta from 150-600°K and A G = M + QrnT , 1 fa
from 150-600°K. (3) Equation (2) calculates AG a t T by applying a correction factor to the value of AH at absolute zero. Equation (2) can be rearranged into eqn. ( 3 )where AG is calculated by applying a different correction factor to AH at T. Equation ( 2 )clearly
Figure 2. Plot of log Kversus 1000ITfor
reaction A
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B. Thermal data same
as Figure 1. - - - - - - corresponds to slope of indicated AH. -- gives path 01 reaction.
Volume 53, Number 10, October 1976 / 627
AS=SB-SA ..~
and
While the number S for a component is called its entropy, there is no need to treat this number with awe. Entropy is used as a means of keeping track of the temperature level at which thermal energy is stored in a component of a chemical reaction. All that is needed to calculate the entropy of the component is heat capacity data from O°K up to the reaction temperature, the latent heats of phase changes that occur in this temperature range, and the temperatures at which the phase changes occur. A common statement of the third law of thermodynamics is that the entropy of a pure crystalline solid at absolute zero mav he taken as zero. All that is needed to predict the equilibrium constant of a chemical reaction is the entrow of each comnonent involved and rn at the reaction tem&rature. A chkmical reaction behaves as a perfect thermodynamic machine in sensing its equilibrium. The ability to predict the equilibrium is limited only by the accuracy of the available thermodynamic data. Since a chemist is really interested in the equilibrium constant K rather than AG, the log of K is plotted versus 1000IT on Figure 2 for the reaction of A B. K is related to AG by the equation
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AG=-RTlnK=AH-TAS
and
628 / Journal of Chemical Education
(6)
~~
If the AH of a reaction is constant. a olot of in K versus 11T will give a straight line with a slopebf--AHl~and a y interceDt of ASIR. The dashed lines on Figure 2 show the slone corresponding to the indicated valuesof rn written on the lines. The reaction of A B follows the AH = -2000 line from k150"K. When A melts at 150DKthe slope changes to -3000. I t follows this slope to 600°K. While it will not be apparent from one example, the examination of other combinations of heats of reaction and stored thermal energy imbalance will reveal a pattern. The heat of reaction at absolute zero will have the controlling influence on K a t very low temperatures. The stored thermal energy imbalance will controlK at very hieh temneratures. If a reaction with A S = 0 (a balance of thermal energy stored in the products and reactants) is followed from a low temDerature to a hieh temverature on Fieure 2, a constant A H line will be tracedhith a.value of K ="1at 10001T = 0. Anv excess thermal enerevstored in the ~ r o d u c t s sr i l l bend the line up [make the rea&n more endoihermicl so that K > 1 at IOU0 T = 0.Anv excess thermal enerr\ stored in the reactants will bend the line down (make thereaction more exothermic) and makeK < 1at 1000lT = 0. In most real reactions the inequality of A?l and AG will he due to many separate increments of thermal energy in both reactants and products. Each increment will have an effect on AG according to eqn. (I), and the net change in AG from its value at O°K will be the algebraic sum of all these effects.
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