Spontaneity, Entropy and Gibbs (Free) Energy

Readings for this section

Petrucci: Chapter 19


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) $\rightarrow$ H2O (l) ΔH = -286 kJ/mol  (Exothermic)
2 NH4NO3(s) $\rightarrow$ NH4+(aq) + NO3-(aq) ΔH = +27.4 kJ/mol  (Endothermic)

Obviously, the evolution of entthalpy 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 alone 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.

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Entropy and the Second law of Thermodynamics

Consider the following diagram:

The diagram on the left depicts an insulated, closed system consisting of two interconnected chambres separated by a valve.  To start our thought experiment, we open the valve.  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. The driving force seems to be an increase in freedom of motion of the individual molecules.  A given instantaneous combination of position and energy of all the molecules taken together is called a microstate.  Clearly, there are more microstates available to the system if the gas expands into both chambres.

Entropy is the scientific name for the macroscopic measure we use to evaluate these microstates of the system and is given the symbol S.

The Second Law of Thermodynamics states:

For any spontaneous process in a closed system the entropy of the system must increase.
Ssys > 0)

or

For any spontaneous process (closed or not) the entropy of the universe must increase.
Suniverse > 0)

The second law, thus, is a statement describing the driving force of all spontaneous process. The tendency towards maximum entropy. We can see an analogy of 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 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 as it merely represents the shifting of energy or matter from one local to another and that heat evolution or absorption must in some way be related to entropy changes.

We have, thus:

ΔS is proportional to heat transferred (q)

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

Define qrev 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):

 
bullet Initially, we are at state A holding the gas in a volume VA in the cylinder with a pressure PA.
  1. 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.
  2. 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.

 

bullet 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. 

 

bullet 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 $\rightarrow$ 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.

\[w_{rev}= -\int_{V_1}^{V_2} P_{ex}dV\] [1]

Since we always have Pex @ Pgas we can use the ideal gas law and write

\[w_{rev} = - \int_{V_1}^{V_2} \frac{nRT}{V} dV = -nRT \int_{V_1}^{V_2} \frac{dV}{V} = -nRT \ln \frac{V_2}{V_1}\] [2]

Recall the relationship P1V1 = P2V2, which leads to

\[\frac{V_2}{V_1}= \frac{P_1}{P_2}\]

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

\[w_{rev} =-nRT \ln \frac{P_1}{P_2} = nRT \ln \frac{P_2}{P_1}\]
[3]

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

\[q_{rev} =w_{rev}\] [4]

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.

\[dS = \frac{d q_{rev}}{T}\] [5]

or more generally, we must integrate between the initial and final state to get the overall change in entropy.

\[dS = \int_{T_1}^{T_2} \frac{dq_{rev}}{T}\] [6]

For the isothermal reversible process we are discussing here, the integral simplifies to

\[\Delta S = \frac{q_{rev}}{T}\] [7]

In order to evaluate this process, we will need to come up with the mathematical relationship for the energy transfer process.  This relationship will depend on the exact situation we encounter.
Also note that this equation does not specify system or surroundings.  If the work is done reversibly then it applies equally well for either.  Since

qrev,system = -qrev, surroundings, [8]

It follows that

ΔSrev,sys = - DSrev, surroundings [9]

Hence,

ΔSrev, universe = 0.  [10]

In words, equation [10] tells us that there is no change in the entropy of the universe for any process done reversibly.

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 qrev so that we can complete the integral.

\[dS = \frac{d q_{rev}}{T} = \frac{C dT}{T}\] [11]

Integrating we get

\[\Delta S(T_1\rightarrow T_2) = \int_{T_1}^{T_2} \frac{C dT}{T} = C \ln \left( \frac{T_2}{T_1} \right)\] [12]

if we assume C is constant.  Equations [11] and [12] use C as an overall heat capacity, referring to our total system.  If we are using specific heat of a substance then we need to multiply by mass of that substance and if we're using molar heat capacities we need to multiply by moles.

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 qrev = DH.  Thus, we can rewrite equation [7] above for any phase transition (fusion, vaporization, sublimation) as follows.

\[\Delta S = \frac{\Delta H_{fusion}}{T_{fusion}}\qquad\qquad\Delta S = \frac{\Delta H_{vap}}{T_{vap}} \qquad\qquad\Delta S = \frac{\Delta H_{sublim}}{T_{sublim}} \] [13]

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.

\[\Delta S(V_1 \rightarrow V_2) = \frac{q_{rev}}{T} = \frac{nRT \ln \left(\tfrac{V_2}{V_1}\right)}{T}=nR\ln\left(\frac{V_2}{V_1}\right)\] [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.

\[\Delta S(P_1 \rightarrow P_2) = \frac{q_{rev}}{T} = \frac{-nRT \ln \left(\tfrac{P_2}{P_1}\right)}{T}=-nR\ln\left(\frac{P_2}{P_1}\right)\] [15]

Note that ΔS 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.

Third Law of Thermodynamics states:

The entropy of a pure crystalline material at absolute zero (0 K) is zero.

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 (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.

First, we define the system to be the iron and the water inside the insulated container. We recognise that the insulation means no heat is lost or gained (qtotal = 0).  We also recognise that heat was moved within the system and because none was lost, we can state the following:

Heat transferred out of metal = heat transferred into water
                     qFe = -qH2O
             mFe CFe ΔTFe = -mwater Cwater ΔTwater
mFe CFe {TFe,final -TFe,init}= -mwater Cwater {Twater,final -Twater,init}

We will need to look up the heat capacities of the iron and of water.  Since we are using gram amounts of iron and water, we look up the specific heats of the two.

CFe = 0.45 J K-1g-1;          CH2O = 4.18 J K-1g-1

Note that the final temperature of the iron and the water will be the same so TFe,final = Twater,final ; let's call it Tf. so now we substitute these values to get

100.0 g 0.45 J K-1g-1 {Tf - 100°C} = - 100.0 g 4.18 J K-1g-1 {Tf - 0°C}

Tf = 9.719

Since we know the initial and the final temperature, we can use equation [12], modified for specific heats and mass measurements to determine the change in entropy.

ΔS = mC ln(T2/T1)

For iron:

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

      = -12.5 J/K

For water:

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

      = 14.6 J/K

ΔST = ΔSiron + ΔSwater = -12.5 + 14.6 = +2.1 J K-1

Since the overall ΔS is positive the process is spontaneous.

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

\[\Delta S=\frac{\Delta H_{vap}}{T}=\frac{40650\; \mathrm{J}\;\mathrm{mol}^{-1}}{373.15\;\mathrm{K}} = 108.9\; \tfrac{\mathrm{J}}{\mathrm{K\; mol}}\]


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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 (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:

ΔS = S(final) - S(initial) [16]

 

For a chemical reaction at standard conditions, we can write

Δ = (products) - (reactants).

[17]

For example,

4 Fe(s) + 3 O2(g) $\rightarrow$ 2 Fe2O3(s)

ΔS° = 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

NOTE 1: Large decrease in entropy is due to the fact that oxygen gas (reactant) becomes part of the solid iron oxide (product).
NOTE 2: This large decrease in entropy must be offset by a large release of enthalpy (exothermic) so as to increase the entropy of the surroundings to compensate for the loss of entropy in the system.

.


Another example

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

Δ =  S°(CaO) + (CO2) - (CaCO3)
=  (38.1 J/mol K) + (213.7 J/mol K) - (92.2 J/mol K)
=  158.9 J/mol K             DS is positive just like we predicted

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) $\rightarrow$ 2 Fe2O3(s)                                 Δ = -1648.4 kJ/mol

We can thus calculate

Δsurroundings =  -Dsystem/T = -(-1648.4 kJ/mol)/298.15K) = + 5529 J/mol K

Now, combining with the Δsystem we just calculated, we get:

Δuniverse = Dsystem + Δsurroundings
                   = -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 Δuniverse 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.

Δuniverse  = Dsystem + Δsurroundings

[18]
 

Δuniverse  = Dsystem +  -Dsystem/T

[19]
 

TΔuniverse  = TΔsystem +  -Δsystem

[20]

for sign consistency with enthalpy, we will multiply both sides by -1

-TΔuniverse  = -TΔsystem +  Δsystem

[21]

now, replace the left-hand side with a new parameter called the change in Gibbs Energy DG° and rearrange the right-hand side.  You will sometimes see this called "the change in Free energy".  Free energy is an old term that is falling out of favour slowly  It implies that the energy is somehow free.

ΔG° = D - TΔ  
[22]

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

The actual defining function for Gibbs Energy is

G = H - TS

[23]

 

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
universe positive) hence, Δ (-TΔuniverse) is negative.

We can now summarize:

ΔG Negative $\Rightarrow$ spontaneous

ΔG zero $\Rightarrow$ equilibrium

ΔG Positive $\Rightarrow$ 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.

case ΔH ΔS ΔG Spontaneous?
1 - + - yes
2 - - - @ low T
+ @ high T
yes
no
3 + - + no
4 + + + @ low T
- @ high T
no
yes

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

ΔG = G(final) - G(initial)

[24]

For a chemical reaction at standard conditions, we can write

Δ = ΣΔGf°(products) - ΣΔGf°(reactants).

[25]

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

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


a) CaCO3(s) $\rightarrow$ CaO(s) + CO2(g)

Δ = S°(products) -S°(reactants)
= (CaO) + (CO2) -(CaCO3)
= 38.1 J/mol K + 213.7 J/molK -92.4 J/molK
= 158.9 J/mol K
Δ = ΣΔHf°(products) -ΣΔHf°(reactants)
= ΔHf°(CaO) + DHf°(CO2) -ΔHf°(CaCO3)
= -635.1 kJ/mol + (-393.5 kJ/mol) -(-1206.9 kJ/mol)
= 178.3 kJ/mol

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

ΔG° = Δ-  TΔ
= 178.3 kJ/mol - 298.15 K 0.1589 kJ/molK
= 130.9 kJ/mol
or
ΔG° = ΣΔGf°(products) -  ΣΔGf°(reactants)
= ΔGf°(CaO) + DGf°(CO2) -  DGf°(CaCO3)
= -603.5 kJ/mol + (-394.4 kJ/mol) - (-1128.8 kJ/mol)
= 130.9 kJ/mol

b) N2(g) + 3H2(g) $\rightarrow$ 2NH3(g)

Δ = S°(products) -S°(reactants)
= 2S°(NH3) -(N2) -3S°(H2)
= 2(192.7 J/molK) -191.5 J/molK -3(130.6 J/molK)
= -197.9 J/mol K
Δ = ΣΔHf°(products) -ΣΔHf°(reactants)
= 2DHf°(NH3) -ΔHf°(N2) -3ΔHf°(H2)
= 2(-45.9 kJ/mol) -(0) -3(0)
= -91.8 kJ/mol

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

ΔG° = Δ-   TΔ
= -91.8 kJ/mol - 298.15 K -0.1979 kJ/molK
= -32.8 kJ/mol
or
ΔG° = ΣΔGf°(products) -   DGf°(reactants)
= 2DGf°(NH3) -ΔGf°(N2) -3ΔGf°(H2)
= 2(-16.4 kJ/mol) -(0) -3(0)
= -32.8 kJ/mol

c) H2(g) + Cl2(g) $\rightarrow$ 2 HCl(g)

Δ = S°(products) -S°(reactants)
= 2S°(HCl) -(H2) -(Cl2)
= 2(186.8 J/molK) -130.6 J/molK -(233.0 J/molK)
= 20.0 J/mol K
Δ = ΣΔHf°(products) -ΣΔHf°(reactants)
= 2DHf°(HCl) -ΔHf°(H2) -ΔHf°(Cl2)
= 2(-92.3 kJ/mol) -(0) -(0)
= -184.6 kJ/mol

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

ΔG° = Δ-  TΔ
= -184.6 kJ/mol - 298.15 K -0.0200 kJ/molK
= -190.6 kJ/mol
or
ΔG° = ΣΔGf°(products) - ΣΔGf°(reactants)
= Gf°(HCl) -ΔGf°(H2) -ΔGf°(Cl2)
= 2(-95.3 kJ/mol) -(0) -(0)
= -190.6 kJ/mol

In these cases, the two methods of determining Δ 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 Δ and D 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 ΔG. 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 ΔG. We've seen that equation [22]

Δ = Δ - T D

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) $\rightarrow$ CaO(s) + CO2(g)                        

ΔS = 158.9 J/mol K
ΔH = 178.3 kJ/mol
ΔG = 130.9 (non spontaneous)

Now, let's try 1000 K

Δ = 178.3 kJ/mol - 1000 K158.9 J/molK = 19 kJ/mol

Now, let's try 2000 K

Δ = 178.3 kJ/mol - 2000 K158.9 J/molK = -13 kJ/mol (it's spontaneous at this temperature)

At what T would ΔG = 0?

0 kJ/mol = 178.3 kJ/mol - T158.9 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 25C 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

ΔGphase-change = 0 = Δphase-change - TDphase-change

\ Tphase-change = D / ΔS°.

[26]

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

Tb(H2O) = ΔHv°(H2O) / ΔSv°(H2O)

Tb(H2O) = { ΔHf°(gas) - ΔHf°(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 Δ 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 ΔG gives us information about the spontaneity of the reaction system (no superscript ° so not standard).

positive $\Rightarrow$ spontaneous to left

zero $\Rightarrow$ equilibrium

negative $\Rightarrow$ 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 $\Rightarrow$ spontaneous to left

Q/K = 1 $\Rightarrow$ equilibrium

Q/K < 1 $\Rightarrow$ 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.

G-G° = (H-H°) - T(S-S°)

[28]

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

[29]

for one mole of gas or

[30]
NOTE 1: Many Texts write equation [30] as

G = G° + RT ln (P),

which is only correct if P is in units of bars (100 kPa) since P° = 1 bar by definition. Any other units of pressure will result in the incorrect answer for G.

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 $\rightarrow$ yC + zD

As a special note, at this point, we should realize that Q is in fact more general than we've discussed here. we are using a = P/P° as the definition of activity. In fact, we can also define activities of solutes in solution in terms of concentrations as a = C/C°. This extension is perfectly analogous to the pressure case and the equation so derived is the same. Thus, we can write the general formula valid for all reactions as

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

ΔG = 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)
=
= ΔG° + RT ln Q

ΔG = D + RT  ln Q

[30]

Note that the value for ΔG 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, ΔG is not a constant of reaction. In fact, we could think of it as a slope of the G versus reaction coordinate curve.

Note that when the slope of the curve is down (negative) the reaction proceeds forward to equilibrium and when it is up (positive) the reaction proceeds backwards towards equilibrium. At equilibrium, the slope is zero.

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

0 = DG° + RT ln K

[31]

or

ΔG° = -RT ln K

or

ln K = - ΔG°/RT

or

K = exp(- ΔG°/RT)

[32]

SOmetimes, we wish to calculate ΔG (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 DG by subtracting equations [30] - [31]

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

to get

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

ΔG = RT ln (Q/K)
[33]
We now have a relation that can be used to determine Gibbs energy from equilibrium-type information and vice versa.
NOTE 1: The derivation shown above is only valid if we use relative activities to calculate our Q and K values. If we use concentrations in molarity and/or pressures in bars then we can safely use the equations in the boxes here since the standard values are 1 M and 1 bar. Thus, to convert from concentrations to activities by dividing by the standard leaves us with the same numbers if and only if we use these units.

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

N2(g) + O2(g) reaction arrows 2 NO(g)

Δ = 2 ΔGf°(NO) - ΔGf°(N2) - ΔGf°(O2)
        = 2 ( 86.6 kJ/mol)
        = 173.2 kJ/mol


Example:

Calculate ΔG for the following reaction at room temperature

N2(g) + 3 H2(g) reaction arrows 2 NH3(g)

NOTE: 1 bar = 100 kPa .

To convert from pressures in bars to activities, we simply drop the units since standard pressure is 1 bar (100 kPa).

if PN2 = 10. bar, PH2 = 10. bar and PNH3 = 1.0 bar

K = 5.68105 (calculated earlier in notes)

at standard conditions we calculate

Δ = -32.8 kJ/mol

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

This difference in ΔG can be explained using Le Chtelier'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) reaction arrows 2 CO(g)

From thermodynamic data tables, we find the following:

compound ΔHf°
(kJ/mol)
ΔGf°
(kJ/mol)

(J/mol K)
C(s) 0 0 6
CO2(g) -393.5 -394 214
CO(g) -110.5 -137 198

Δ = 2(-110.5) - 0 - (-393.5) = 172.5 kJ mol-1

Δ = 2(198) - 6 - (214) = 176 J mol-1 K-1

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

Δ = Δ- TΔ
        = 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 {-D/ RT) = exp {-49.3 kJ mol-1 / (8.3145 J mol-1 K-1 700 K)}
    = 2.1 10-4

Remember that ΔH° and ΔS° are assumed to be constant over the temperature range (298 K - 700 K). This is not strictly true. DG° has been somewhat 'corrected' for temperature. In this case, the '°' symbol on the ΔG° implies that all other conditions but temperature are standard state (poor symbology).

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 D and Δ 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 Δ is constant with temperature.  We can use this equation to quickly tell the effect of T on K.

Consider an exothermic reaction (Δ 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

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

wmax = -DG

Consider the oxidation of glucose:

C6H12O6(s) + 6 O2(g) reaction arrows 6 CO2(g) + 6 H2O(l)

Δ = -2808 kJ/mol        Δ = -2870 kJ/mol

Now consider a person 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

This means the person needs 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.

Actually, The only way one can achieve 100% efficiency is if the waste heat from the system is absorbed by a heat sink at absolute zero (0 K). (cf. Murphy's version of the laws of thermodynamics).

Coupled Reactions

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

Consider the following reaction

Cu2S(s) reaction arrows 2 Cu(s) + S(s) Δ = + 79.5 kJ/mol
Δ = - 22.4 J/mol K
i.e., ΔG is positive at all temperatures:
Δ = + 86.2 kJ/mol

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

S(s) + O2(g) reaction arrows SO2(s) Δ = -296.8 kJ/mol Δ = -300.1 kJ/mol
Cu2S(s) reaction arrows 2 Cu(s) + S(s) Δ = + 79.5 kJ/mol Δ = + 86.2 kJ/mol
add these up to get
Cu2S(s) + O2(g) reaction arrows 2 Cu(s) + SO2(g) Δ = -217.3 kJ/mol Δ = -213.9 kJ/mol

Now, the overall reaction is both exothermic and spontaneous.

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

ATP3- + H2O reaction arrows ADP2- + H2PO4-

Δ = -30 kJ/mol,      Δ = -20 kJ/mol,      ΔS° = +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: September 05, 2012.