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Spontaneity, Entropy and Gibbs (Free) Energy 19/07/2006

Spontaneous Processes

We've seen from experience that some chemical (and other) processes are spontaneous while others are not. What is the driving force which creates spontaneity.

If we look at some processes which we know to be spontaneous, maybe we can find a common trend, or at least eliminate some possibilities.

Consider The following two spontaneous reactions:

(1)  H2(g) + 1/2 O2(g) --->  H2O (l)                 exothermic
                                                     DH = -286 kJ/mol
    
(2)  NH4NO3(s) ----H2O----->  NH4+(aq) + NO3-(aq)    endothermic
                                                     DH = +27.4 kJ/mol

Obviously, the evolution of energy does not drive a reaction as one might first expect. If we look at these two reactions, we have difficulty deciding anything except that Enthalpy change alone is not sufficient to decide if a reaction is spontaneous or not.

Equation (1) is very exothermic and this in itself is a strong tendency towards spontaneity. Equation (2) is endothermic and so there is some other driving force which pushes the reaction.

It turns out that the First Law of Thermodynamics completely eliminates enthalpy in itself as a driving force Although we will see later that it does have a place in the overall equations to determine spontaneity. To understand the driving force, we must introduce a new concept.

Consider the following diagram:

Gas initially in the chamber on the left (shaded) expands adiabatically into the evacuated second chamber (adiabatic means no energy transfer between system and surroundings). The process is obviously spontaneous but energy is not involved. The gas has more room in the increased volume of the double bulb container. This increase in room allows more freedom of motion for the individual gas molecules (randomness). The driving force seems to be an increase in randomness.

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Entropy and the Second law of Thermodynamics

Entropy is the scientific name for 'randomness' and is given the symbol S.

The second law, thus, is a statement describing the driving force of all spontaneous process. The tendency towards maximum Entropy. We can see this in our day to day lives in non scientific ways as well. Our rooms are rarely in perfect order unless we expend energy to keep it so. Without constant input of energy, the room would eventually be quite disorganized (this isn't strictly the same thing but is an interesting parallel).

How do we measure Randomness?

Can we connect it with something we can measure? Say, Entropy?

All process have only two ways to transfer or change internal energy

1. work (organized energy)
2. heat (disorganized energy)

It seems reasonable that we can eliminate work as having anything to do with entropy and that heat evolution or absorption must in some way be related to entropy changes.

We have, thus:

DS is proportional to heat transferred (q)

but q is not a state function whereas we hope to make DS a state function. There are two things we can do to fix this problem.

Define q_rev as the heat transferred for a reversible process (one which is done in infinitely small steps). Reversible simply means that at any step in the process, the system and universe both are essentially at equilibrium and hence the process can go forward or backwards with equal ease.

We can explore the idea of reversible energy transfer by considering the expansion and compression of an ideal gas in a cylinder where Temperature is always constant (isothermal):

	Initially, we are at state A holding the gas in a volume VA in the cylinder with a pressure PA. We first release the pressure to PB. The gas quickly expands to a new volume VB. The PV work done (by the system) is easily measurable as the area under curve 1. To reverse the process, we increase the pressure on the plunger back to PA and the gas then compresses back to VA. The work done (on the system) is the area under curve 2.  Note that the work done to compress the gas was more than the work done by the gas on the surroundings.   Thus, we are loosing energy in this cyclic process.
	Now consider the same process again but where we release the pressure infinitely slowly so that PV is always constant. In this case, we follow path 3 in either direction and we note that the work done on the environment by the system in the expansion phase is exactly the same as the work done by the environment on the system in the compression stage. This is reversible work.
	Note that at all times during this process, the system and the universe are at equilibrium with each other.

Since for any individual infinitesimal step, work dw = –PdV. We must integrate this function over the range V₁ ® V₂ where V₁ is the volume at point A and V₂ is the volume at point B, i.e., we are finding the work done during the reversible expansion phase of the cycle.  Of course, the work for the compression phase would be exactly the same magnitude but opposite in sign.

[1]

Since we always have P_ex @ P_gas we can use the ideal gas law and write

[2]

Recall the relationship P₁V₁ = P₂V₂, which leads to

            .

This relationship allows us to rewrite equation [2] using pressures.

[3]

Since the expansions and compressions were done isothermally, the internal energy E of the gas has not changed (DE = 0 = q + w ) thus, we can write

This function, although interesting is not a state function since the exact position of the reversible isothermal curve depends on the absolute temperature. Thus, to finally make the connection between S and q, we must factor out T, thereby creating a state function.

Let's use this new function under a variety of energy-transfer processes:

Temperature change

We start with the derivative equation [5] we introduced above and we need to develop the relationship for q_rev so that we can complete the integral.

dS =	d qrev	=	CdT	[11]
	T		T

Integrating we get

[12]

if we assume C is constant.

In general, C is a function of temperature [= C(T)] and this equation would need to be evaluated numerically.  for our purposes, however, we'll not bother with this complication.  The assumption that C is constant for relatively small temperature changes will give us satisfactory results.

Phase Transformation

A system undergoing a phase transformation is at equilibrium (the two phases are in equilibrium with each other).  This is the same as saying that the energy is transferred reversibly between the system and the surroundings.  Thus, the amount of energy transferred in a constant pressure situation, and thereby, also constant temperature is simply q_rev = DH.  Thus, we can use the equation relating q_rev to DS which we saw above as follows.

Isothermal Expansion/Compression of an Ideal Gas

For an Ideal gas isothermal process, we need to further develop the ideas we left.  We combine equations [2], [4] and [7] to arrive at the following.

[14]

or if we're using the pressure (in this same gas PV change), we can combine equations [3], [4] and [7] to arrive at the following.

[15]

Note that DS is often expressed as a molar quantity.  If we are working with one mole quantities, then n = 1 and the equations [14] and [15] simplify slightly.  Now we have all the tools necessary to calculate the entropy change for any non-chemical energy transfer process. With one more crucial piece of information, we can do more.

Now we have all the tools necessary to calculate the absolute entropy for any pure material at any temperature (say, Standard Conditions). Hence, we can measure a standard entropy of a material S° (unlike the situation we saw earlier where we cannot define the absolute zero of enthalpy).

Thus, it is possible to tabulate lists of standard enthalpies for materials.

Examples:

A block of iron weighing 100.0 g is taken from a boiling water bath (NBP) and placed in an insulated container containing 100.0 g water at 0°C. Calculate The entropy change for the iron, the water and the whole process.

C_Fe = 0.45 J K^-1g^-1;          C_H2O = 4.18 J K^-1g^-1

Note that T_Fe,final = T_water,final ; let's call it T_f. so now we substitute these values to get

100.0 g × 0.45 J K^-1g^-1 ×{T_f - 100°C} = - 100.0 g × 4.18 J K^-1g^-1 {T_f - 0°C}

T_f = 9.7₁₉

DS = nC ln(T₂/T₁) or using specific heat capacities, DS = mC ln(T₂/T₁)

For iron:

DS = 100.0 g × 0.45 J K^-1g^-1 ln[(273.15+9.719)/373.15]

      = -12.5 J/K

For water:

DS = 100.0 g × 4.18 J K^-1g^-1 ln[(273.15+9.719)/273.15]

      = 14.6 J/K (we see that although, we've thrown sig figs to the wind in this calculation, we cannot really make the assumption above.)

DS_T = DS_iron + DS_water = -12.5 + 14.6 = +2.1 J K^-1

Since the overall DS is positive the process is spontaneous.

Example: What is the entropy change for the vaporization of one mole of water at the NBP?

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Standard Molar Entropies

The standard molar entropy of a substance is the entropy of one mole of that substance at Thermodynamic standard conditions (25°C, 1 bar, 1M conc,...)

We can always determine the absolute value for S° (J K^-1mol^-1) since we know the absolute zero of entropy (third law).

The Appendix lists S° values for many chemicals.

Since S is a state function, we can write the following:

For a chemical reaction at standard conditions, we can write

For example,

4 Fe(s) + 3 O₂(g) ® 2 Fe₂O₃(s)

DS° = 2 S°(Fe₂O₃) - {4 S°(Fe) + 3 S°(O₂)}

     = 2(87.4 J/mol K) - 4(27.3 J/mol K) - 3(205.0 J/mol K)

       = -549.4 J/mol K

.

Another example

CaCO₃(s) ® CaO(s) + CO₂(g)                
Here we see that a gas is evolved and hence we expect that DS is positive.

We see that as number of moles of gas increases, S increases and vise versa.

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Gibbs Energy

Look at the rusting iron situation, we know it is spontaneous yet there is a decrease in system entropy.

We must consider the entropy change of the whole universe before we can explain for sure why the system will be spontaneous.

The measured value of enthalpy for the reaction

4 Fe(s) + 3 O₂(g) ® 2 Fe₂O₃(s)                                 DH° = -1648.4 kJ/mol

We can thus calculate

DS°_surroundings =  -DH°_sys/T = -(-1648.4 kJ/mol)/298.15K) = + 5529 J/mol K

Now, combining with the DS°_system we just calculated, we get:

DS°_universe = DS°_system + DS°_surroundings
                   = -549.4 J/mol K + 5529 J/mol K = 4980 J/mol K

Very large and positive. This is a spontaneous reaction according to second law

So, we were able to calculate a parameter DS°_universe that tells us definitively whether or not a system is at equilibrium.  However, we would prefer to not be forced to deal with universe parameters if at all possible.   Let's rework the previous equation somewhat to try to incorporate only system thermodynamic quantities.

now, replace the left-hand side with a new parameter called DG° and rearrange the right-hand side.

Since these are all system parameters, we will drop the subscripts from now on.

The actual defining function for Gibbs Energy is

but since we can't actually measure the absolute enthalpy of a substance, neither can we measure the absolute Gibbs energy of a substance. Hence, this remains a defining equation only. It is not useful for any actual calculations or measurements we may wish to do.

We know that the second law states that the entropy of the universe must increase for any spontaneous process (DS°_universe positive) hence, DG° (° -TDS°_universe) is negative.

We can now summarize:

• DG Negative Þ spontaneous
• DG zero Þ equilibrium
• DG Positive Þ non-spontaneous (spontaneous in reverse direction)

Let's set up a table of the thermodynamic measurable for a reaction system and see what we can learn regarding system spontaneity.

Since G is a state function (made up of H, T, and S which are each state functions) we can write 

For a chemical reaction at standard conditions, we can write

We must use Standard Gibbs Energy of Formation (DG_f°) since we cannot define the absolute zero of Gibbs energy. This is exactly analogous with the case of Standard Enthalpy of Formation (DH_f°).

Note that we now have two ways to calculate DG°.  Equations [22] and [25] both allow us to determine a value for DG°.  The difference comes in the temperature factor that is explicit in [22].  Using [22], we can calculate Gibbs energy for any temperature (assuming  DH° and DS° don't change with temperature).   However, equation [25] does not have temperature explicitly and so the value of DG° we calculate using it will depend on the temperature at which the DG_f° values were tabulate.  Typically, that is 25ºC.We will now do a few examples to familiarize ourselves with the calculations. For each of the following, calculate DG°, DH° and DS°.  The temperature in each of these examples is standard temperature (25ºC).

a) CaCO₃(s) ® CaO(s) + CO₂(g)

Gibbs-energy change can be calculated in one of two ways, we'll do both here.

b) N₂(g) + 3H₂(g) ® 2NH₃(g)

Gibbs-energy change can be calculated in one of two ways, we'll do both here.

c) H₂(g) + Cl₂(g) ® 2 HCl(g)

Gibbs-energy change can be calculated in one of two ways, we'll do both here.

In these cases, the two methods of determining DG° are equivalent. This is only true at standard temperature (298.15 K). In this case, you can choose the equation which best suits the data you have. If the temperature is not standard temperature then you have no choice but to use the second equation and to assume that DH° and DS° do not change with temperature. This assumption is not strictly true but is not a bad approximation for many cases.

We've seen that we can predict spontaneity of reactions using thermodynamic function DG. However, Thermodynamics gives us no information regarding the actual reaction path, specifically, how fast the reaction will occur. Hence, although Thermodynamics may predict that a reaction is spontaneous, it does not tell if the reaction will occur with any appreciable speed, if at all.

Temperature dependence of Gibbs Energy

Let's explore the effect temperature has on DG. We've seen that equation [22]

DG° = DH° - T DS°

describes a temperature effect on G. This can be manifest in several ways.

At room temperature (25°C) the following is true (calculated previously)

CaCO₃(s) ® CaO(s) + CO₂(g)                        

DS = 158.9 J/mol K
DH = 178.3 kJ/mol
DG = 130.9 (non spontaneous)

Now, let's try 1000 K

DG° = 178.3 kJ/mol - 1000 K×158.9 J/molK = 19 kJ/mol

Now, let's try 2000 K

DG° = 178.3 kJ/mol - 2000 K×158.9 J/molK = -13 kJ/mol (it's spontaneous at this temperature)

At what T would DG = 0?

0 kJ/mol = 178.3 kJ/mol - T×158.9 J/molK = 19 kJ/mol
T = 1121 K (848°C) the system is at equilibrium         (if all other conditions are held to standard conditions)

Note that all these values were considered to be Standard values of Gibbs Energy.  Recall that 25ºC is not required for standard state, it is merely the temperature at which we normally tabulate these standard values.

Phase-Change Temperatures

We can also use equation [22] to calculate the  the phase-change temperature for any phase change since by definition, at that temperature, the two phases are in equilibrium with each other.

Thus, we can write

DG_phase-change = 0 = DH°_phase-change - TDS°_phase-change

For example,  Let's find the boiling point of water.

T_b(H₂O) = DH_v°(H₂O) / DS_v°(H₂O)

T_b(H₂O) = { DH_f°(gas) - DH_f°(liq)} / {S°(gas) - S°(liq)}
               = {–242 -(–286)}×10³ / {189 -70} = 370 K ( c.f. 373 K)

Other dependencies of Gibbs Energy

We've seen that calculations of DG° are useful for predicting spontaneity of systems where all conditions are standard. Further, we explored the situation where the temperature is varied away from standard conditions. Now we will look into the more general case where other conditions are non-standard.

It is quite uncommon to deal with a reaction mixture, for example, where all concentrations are exactly 1 M. or with a gas-phase reaction where all partial pressures are 1 bar (100 kPa). In reality, we mostly deal with non-standard conditions of our reaction mixtures. How do we handle these?

We know that the sign of DG gives us information about the spontaneity of the reaction system (no superscript ° so not standard).

• positive Þ spontaneous to left
• zero Þ equilibrium
• negative Þ spontaneous to right

We also know that by comparing Q with K we can determine whether a reaction will be spontaneous or not

• Q/K > 1 Þ spontaneous to left
• Q/K = 1 Þ equilibrium
• Q/K < 1 Þ spontaneous to right

It turns out that these two functions are related (need the same kind of logic we saw in the earlier derivations at the start of this chapter)

Let's begin our exploration using gas-phase reactions. We will quickly see that we can extend the results to cover all types of reactions.

Enthalpy changes in most system processes are relatively pressure independent. Therefore, we need only consider the effect pressure has on the entropy changes in a process to be able to understand how pressure affects the Gibbs-energy changes in that process.

S is relatively independent of pressure for solids and liquids but not for gases. For an ideal gas, what is the change in the value of G for as we move from standard pressure to non-standard?  We will use equation [23] for the two states and subtract them.

but since there is no concentration or pressure dependence for enthalpy, we know that
(H-H°) = 0.   Additionally, we can replace the (S - Sº) term using equation [15] where
P₁ = Pº and P₂ = P.

for one mole of gas or

We recall that we already have a function which is called activity

(a = P/P°)

which we can use here. Thus,

G = G° + RT ln(a)

We recall as well that since a = 1 for pure liquids and solids, this formula is not exclusive to gases.

Now consider a general chemical reaction

wA + xB ® yC + zD

The general (not necessarily standard-conditions) Gibbs energy change for this reaction is

DG =	yGC+ zGD- wGA- xGB
=	yG°C+ zG°D– wG°A–xG°B + RT (y ln aC+z ln aD–w ln aA–x ln aB)
=	yG°C + zG°D–wG°A–xG°B           + RT (ln aCy+ln aDz–ln aAw–ln aBx)
=
=	DG° + RT ln Q

Note that the value for DG depends on the reaction conditions at the moment of measurement. The instant we allow the reaction proceed to any extent, Q will change and hence, so too will DG. Thus, DG is not a constant of reaction. In fact, we could think of it as a slope of the G versus reaction coordinate curve.

At equilibrium, Q = K and DG = 0, so we have

or

SOmetimes, we wish to calculate DG (non-standard) in circumstances where we do not know the thermodynamic data necessary to use equation [30] .  In this case, we can get rid of DGº by subtracting equations [30] - [31]

DG = DG° + RT ln Q
   0 = DG° + RT ln K

to get

DG = RT ln Q - RT ln K
      = RT ln (Q/K)
 

We can use the above equations to calculate K if we know DG°, or vice versa,

N₂(g) + O₂(g) ® 2 NO(g)

DG° = 2 DG_f°(NO) - DG_f°(N₂) - DG_f°(O₂)
        = 2 ( 86.6 kJ/mol)
        = 173.2 kJ/mol

Example:

Calculate DG for the following reaction at room temperature

N₂(g) + 3 H₂(g) « 2 NH₃(g)

if P_N2 = 10. bar, P_H2 = 10. bar and P_NH3 = 1.0 bar

K = 5.68×10⁵ (calculated earlier in notes)

at standard conditions we calculate

DG° = -32.8 kJ/mol

Hence, the stated conditions have a more negative DG than standard conditions, i.e., more spontaneous.

This difference in DG can be explained using Le Châtelier's principle. We added more reactants to a system which tends to drive the reaction further to the right than standard conditions do. (more spontaneous)

Example:

Calculate K for the following reaction at 700K:

C(s) + CO₂(g) « 2 CO(g)

From thermodynamic data tables, we find the following:

DH° = 2(-110.5) - 0 - (-393.5) = 172.5 kJ mol^-1

DS° = 2(198) - 6 - (214) = 176 J mol^-1 K^-1

DG° = 2(-137) - 0 - (-394) = 120 kJ mol^-1 (only at 25°K so not useful for this question)

DG° = DH° - TDS°
        = 172.5 kJ mol^-1 - 700K × 176 J mol^-1 K^-1
        = 49.3 kJ mol^-1                 (valid for 700 K with all other conditions standard)

K = exp {-DG° / RT) = exp {-49.3 kJ mol^-1 / (8.3145 J mol^-1 K^-1 × 700 K)}
    = 2.1 × 10^-4

We can also use these ideas to be able to develop an expression which can give us an equilibrium constant at any temperature if we know DH° and DS° and K at some other temperature.

We start from equation [32] and then expand it using equation [22].

We know K₁ at T₁ and wish to find out K₂ at T₂. 

We can write for temperatures T₁ and T₂ the following two cases:

                                    (2)

                                    (1)

now subtract case (1) from (2) to get the following

[34]

This is the van't Hoff equation, which we have already seen earlier in this course.   It assumes again that DH° is constant with temperature.  We can use this equation to quickly tell the effect of T on K.

Consider an exothermic reaction (DH° is negative). What will happen to K as the temperature is increased?

Looking at the right-hand side of the equation, if T₂ > T₁ the RHS will be negative.

This means that (on the LHS) the argument to the natural logarithm function must be less than one, i.e., K₂ < K₁.

Hence, as T ­, K ¯ for an exothermic reaction.

Examples

DG can be thought of as the 'maximum' amount of work which can be obtained from a spontaneous process at constant T and P.

w_max = -DG

Consider the oxidation of glucose:

C₆H₁₂O₆(s) + 6 O₂(g) ® 6 CO₂(g) + 6 H₂O(l)

DH° = -2808 kJ/mol        DG° = -2870 kJ/mol

Now consider a woman of 60 kg climbing a 100 m high hill. How much glucose will she need to consume to obtain sufficient energy for the climb.

w = mgh = 60 kg × 9.8 ms^-2 × 100 m = 60 kJ

need 3.8g of sucrose if 100% of energy released from oxidation of sucrose is used in climbing. Assuming, of course that 100% of the food energy is converted into climbing work.

Coupled Reactions

Sometimes, we can use spontaneous reactions to drive non-spontaneous ones in the direction we desire.

Consider the following reaction

If we couple this reaction with a highly spontaneous one, we can make the overall process proceed spontaneously.

Now, the overall reaction is both exothermic and spontaneous.

In biological reactions, the ADP « ATP reversible reaction is coupled with any process which requires driving

ATP^3- + H₂O ® ADP^2- + H₂PO₄^-

      DG° = -30 kJ/mol,      DH° = –20 kJ/mol,      DS° = +34 J/mol K

Thus, in biological systems no compound can be created which requires more than 30 kJ of energy for one mole of any given reaction step in the synthesis process. This is one of the biggest single limiting factor in the diversity of life on the planet.

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Michael J. Mombourquette.
Copyright © 1997
Revised: July 19, 2006.

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