Helmholtz and Gibbs Energy Examples

1 INITIAL DEMONSTRATION 1

These are the examples to be used along with the powerpoint lecture slides. The problems are numbered to match the tags in the the lower left hand corner of the powerpoint slides. The numbers of the examples are # the in the HG EX# tags on the slides.

1 Initial Demonstration

This ﬁrst example takes liquid sodium acetate at room temp and with a small perturbation it crystallizes with a release of heat. Here is a nice demonstration of a clear increase in order (∆S < 0) but the reaction clearly proceeds spontaneously. Have a student make note of the temperature, which has increased. The example here is the building of a mountain of a solid from a stream of liquid and the hand warmers. You can also use 5 minute epoxy for an example, where you take a number of smaller molecules and make one very big molecule (seems like a loss of dissorder?), again the student should notice that the mixture warms up. Less order (entropy), but still goes spontaneously and there seems to be a release of heat to the surroundings... Perhaps there is a balance between the need to maximize disorder and minimize the energy.

2 Calculating the entropy of the universe? That’s a lot of work.

• The First Law gave us energy balance.

• The Second Law gave us a criterion for a spontaneous process, ∆S ≥ 0. This is great, but the change in entropy here is the TOTAL change and this means we must determine the change for both the system and the surroundings.

Calculating the change in a state function for the Universe is not always an easy task, so we would like to consider the criteria for a spontaneous process from only the properties of the system. In fact, we would like to deﬁne a thermodynamic state function in terms of the system alone that would provide this information. The total change in entropy is simply the sum of the entropy change for the system and the entropy change of the surroundings (system + surrounding = everything),

dStot = dSsys + dSsur 2 CALCULATING THE ENTROPY OF THE UNIVERSE? THAT’S A LOT OF WORK.2

Consider a system at constant T . Since it is at a constant temperature it must be interacting with the environment via heat exchange to maintain the same T , ie. it is not isolated. Now consider a process that occurs in the system that results in the transfer of an inﬁnitesimal amount of heat, δq, from the system to the surrounding. So the change in the entropy for δqsur the surrounding is deﬁned as dSsur = T , and the total entropy change is now, δq dS = dS + sur tot sys T Our convention is that heat transferred from the system to the surrounding is negative, and we note that the amount of heat transferred to the surrounding must be equal and opposite to the amount of heat transferred from the system (it is the same heat),

−δqsys = δqsur

Using this relationship, δq dS = dS − sys tot sys T This is now an expression for the total change in the entropy (change in the Universe) in terms of only the system! Just what we wanted.

Look at a constant pressure process,

δqsys = dHsys

From the Second law we know that,

dStot ≥ 0 Substitute our result in terms of the system for the total entropy change, dH dS = dS − sys ≥ 0 tot sys T Multiply by T on both sides,

T dSsys − dHsys ≥ 0 Move things to the other side (which means ﬂip the ≥ and change the signs),

dHsys − T dSsys ≤ 0

INTERESTING, we have a criterion for spontaneity in terms of the system alone.

Deﬁne a new state function, G = H − TS 3 MELTING AND BOILING OF WATER 3

Note that this is a state function since it is a combination of other state functions. For our system at constant temperature,

dG = dH − T dS − SdT = dH − T dS (constant temperature)

And this is exactly what we have above (dHsys − T dSsys ≤ 0),

dG ≤ 0 (<)spontaneous, (=)equilibrium

Consider ∆G for a general isothermal process, A → B,

∆G = ∆H − T ∆S ∆G ∆H = − ∆S T T where the ﬁrst term on the rhs is the entropy from exchange with the surroundings (increase thermal disorder in the surroundings), and the second term on the rhs is the entropy produced in the system.

3 Melting and boiling of water

(Taken from McQuarrie and Simon pgs. 305-306 Gold or 885-886 Red) Consider a process where the sign of ∆G changes sign over a very small change in temperature. Consider the vaporization of liquid water,

H2O(l) −→ H2O(g)

The molar Gibbs energy of vaporization is expressed,

¯ ¯ ¯ ∆vapG = G[H2O(g)] − G[H2O(l)] ¯ ¯ = ∆vapH − T ∆vapS

Near 100◦C and at one atm,

¯ −1 ∆vapH = 40.65 kJ · mol

¯ −1 −1 ∆vapS = 108.9 J · K · mol So we have, ¯ −1 −1 −1 ∆vapG = 40.65 kJ · mol − T (108.9 J · K · mol ) 4 NON-PV WORK: ELECTROLYSIS OF WATER 4

And at 373.15 K we have,

¯ −1 −1 −1 ∆vapG = 40.65 kJ · mol − (373.15 K)(108.9 J · K · mol ) = 0

¯ Notice that ∆vapG = 0 shows us that liquid and vapor are in equilibrium, as expected at 373.15 K. At 373.15 K and 1 atm the Gibbs energy of the liquid and vapor are equal and the transfer from liquid to vapor is a reversible process.

Now consider a temperature just below the boiling point. At 363.15 K,

¯ −1 −1 −1 ∆vapG = 40.65 kJ · mol − (363.15 K)(108.9 J · K · mol ) = 1.10 kJ · mol−1

¯ The positive value of ∆vapG indicates that the formation of water vapor from liquid water is not spontaneous at this temperature and 1 atm.

Now consider a temperature just above the boiling point. At 383.15 K,

¯ −1 −1 −1 ∆vapG = 40.65 kJ · mol − (383.15 K)(108.9 J · K · mol ) = −1.10 kJ · mol−1

¯ The negative value of ∆vapG indicates that the formation of water vapor from liquid water is a spontaneous process at this temperature and 1 atm.

*You can check to see that the same thing happens for the melting of water given that at 273.15 K (example 8-1 on pg 306 Gold or 22-1 on pg 886 Red in McQuarrie and Simon),

¯ −1 ∆fusH = 6.01 kJ · mol

¯ −1 −1 ∆fusS = 22.0 J · K · mol

4 Non-PV work: Electrolysis of water

(This is from example 8-2 on pg 307 Gold or 22-2 on pg 887 Red of McQuarrie and Simon.) You may have heard a lot of talk about the hydrogen economy. The idea is to burn hydrogen for energy, which is nice and clean since it leaves you with water, 1 H (g) + O (g) −→ H O(l). 2 2 2 2 4 NON-PV WORK: ELECTROLYSIS OF WATER 5

The next question should be, “Where do I get the hydrogen?” The answer is usually that you can use electrical power (hopefully produced in a clean way, waves, sun, ...) to split water. This means that we turn the equation around and provide electrical work to run the reaction, electrolysis. 1 H O(l) −→ H (g) + O (g) 2 2 2 2 So, your boss at the water splitting plant asks you, “What is the minimum voltage required to split water into hydrogen and oxygen?” (Note at constant T and P , such as 298.15 K and 1 bar) First, you remember the deﬁnition of electrical work,

work = charge × voltage

and then you remember that you already know how to calculate the electrical (non-PV ) work for a reaction, it’s just the Gibbs energy by deﬁnition. So you use the tables to calculate

∆rH and ∆rS, and using these you calculate,

¯ ∆G = wnonPV (reversible, constant T and P )

∆G¯ = 237.1 kJ · mol−1 (reversible, T = 298.15 K and P = 1 bar) Now to get the voltage all you need is the charge. The oxidation state of hydrogen goes from +1 to 0 and that of oxygen goes from −2 to 0. This means two electrons are transferred for every H2O(l) molecule. That is two time Avagadro’s number of electrons per mole. The total charge of two moles of electrons is,

total charge per mol = 2(1.602 × 10−19 C · e−1)(6.022 × 1023 e) = 1.929 × 105 C

So we can now calculate the minimum (reversible) voltage required,

∆G¯ ∆G¯ = w = charge × volts −→ volts = non-PV charge Use the fact that 1 J = 1 CV And we have, 237.1 × 103 J · mol−1 volts = = 1.23 volts 1.929 × 105 C · mol−1 5 ENTROPY FROM PVT EQUATION OF STATE 6

5 Entropy from PVT equation of state

(This is example 8-3 on pgs 309-310 Gold or 22-3 on 889-890 Red of McQuarrie and Simon) ¯ ¯ ¯ Calculate ∆S for an isothermal expansion from V1 to V2 for a gas that obeys the equation of state, P (V¯ − b) = RT Using our derived expression we have,

¯ Z V2 ∂P ∆S¯ = dV¯ V¯1 ∂T V¯ ¯ Z V2 dV¯ = R ¯ V¯1 V − b ¯ V2 − b = R ln ¯ V1 − b Notice that we got this result back in chapter 6, but we had to know that dU = 0 for an isothermal process that followed the given equation of state. That information was not required here.

6 Correction for non-ideality

This is adapted from pgs 899-910 in Red McQuarrie and Simon

Remember when we calculated S for diatomic nitrogen there was a “correction for non- ideality” added to the total. Now we can see how to get this correction from the equation of state that describes the real gas. Start from the general equation we derived from the Maxwell relation for a change in the entropy expressed in terms of P , V , and T ,

Z P2 ¯ ¯ ¯ ¯ ∂V ∆S = S(T,P2) − S(T,P1) = − dP (constant T ) P1 ∂T P

Z P2 ¯ ¯ ¯ ∂V S(T,P2) = S(T,P1) − dP (constant T ) P1 ∂T P

We choose a reference point for P1. One choice is the limit of low pressure since all gases

will be ideal in this limit, P1 → 0,

Z P2 ¯ ¯ ¯ ∂V S(T,P2) = S(T,P → 0) − dP (constant T ) (P →0) ∂T P 6 CORRECTION FOR NON-IDEALITY 7

We now choose an EOS to describe the gas and put it into the integral. If we assume the gas is ideal all the way up to the pressure of interest, P2, then we can use the ideal EOS to

solve the integral on the right, and we get the entropy for an ideal gas at P2,

Z P2 ¯ RT ¯id ¯ R V = , S (T,P2) = S(T,P → 0) − dP (constant T ) P (P →0) P

Now compare this to the general case where we would use the EOS for the real gas at P2, and with the same reference of the ideal gas at low P ,

Z P2 ¯ ¯real ¯ ∂V S (T,P2) = S(T,P → 0) − dP (constant T ) (P →0) ∂T P

The diﬀerence between the entropy for an ideal gas at P2 and the real gas at P2 is the correction for non-ideality,

Z P2 ¯ ¯id ¯real ∂V R S (T,P2) − S (T,P2) = − dP (constant T ) (P →0) ∂T P P Note that the reference entropy, S¯(T,P → 0), dropped out when we took the diﬀerence ¯ideal ¯real between S and S at P2.

If the pressure P2 is one bar, then for the gas treated as ideal you have the standard molar ¯id ◦ entropy and we denote this with the superscript ◦, S (T,P2 = 1 bar) = S (T ), Z 1 bar ∂V¯ R S◦(T ) − S¯real(T,P = 1 bar) = − dP (constant T ) (P →0) ∂T P P For the real gas around 1 bar we could use the virial expansion to second order for the EOS, ¯ ¯ P V P ∂V R dB2V = 1 + B2V (T ) + ··· = + + ··· RT RT ∂T P P dT Putting this into the integral the correction for non-ideality becomes simply the derivative of the second virial coeﬃcient, Z 1 bar R dB R S◦(T ) − S¯real(T,P = 1 bar) = + 2V + ··· − dP (P →0) P dT P Z 1 bar dB = 2V dP (P →0) dT

dB 1 bar = 2V P dT (P →0 bar) dB = 2V (1 bar) dT 7 THE RUBBER BAND PROBLEM 8

This diﬀerence is typically small. For nitrogen at 298 K, dB 2V = 0.192 cm3 · mol−1 · K−1 dT Multiply by one bar, convert cm3·bar to J, and the correction for nitrogen at 298 K is,

S◦(T = 298 K) − S¯real(T = 298 K,P = 1 bar) = 0.02 J · K−1 · mol−1

7 The rubber band problem

I. (This is Marc’s shortened and improved version. Changes made by Dave 2006 ) The Question: When I heat this rubber band will the restoring force be higher or lower?

pull

Let’s see if we can derive a relationship between the restoring force, l, and the temperature. When a rubber band is stretched, the differential work done on the system is given by, dw = fdl − P dV where f is the restoring force and l is the length, and fdl is the non-PV work. Note that the first derm fdl is positive since a positive dl represents work done on the system (analagous to our sign definition for P dV work, but in that case positive dV is defined as work done by the system). We can assume the change in the volume of the rubber band when stretched is very small. At constant V and T , the non-PV work is equal to the change in the Helmholtz energy - let’s derive that, A = U − TS We choose a reversible path, and this will give us the limit of maximum work that the rubber band can do when contracting, or the minimum work required to stretch the 7 THE RUBBER BAND PROBLEM 9

rubber band.

dA = dU − T dS − SdT

= δqrev + δwrev − T dS − SdT = T dS + fdl − P dV − T dS − SdT = fdl − P dV − SdT

And for constant pressure and temperature,

dA = fdl (= dwnon-PV )

Now we can write the restoring force in terms of the change in the Helmholtz energy,

∂A f = ∂l T But what do you do with this?! We know that, A = U − TS so we know, ∂A ∂U ∂S = − T ∂l T ∂l T ∂l T Now we assume this is a perfect elastomer, which means,

∂U = 0 ∂l T and we get, ∂A ∂S = −T = f ∂l T ∂l T ∂S f = −T ∂l T Now the question is: ∂S Is positive or negative? ∂l T (Have a class discussion and refer to the ﬁgure.) The answer is, ∂S < 0 ∂l T 7 THE RUBBER BAND PROBLEM 10

So looking at our expression for f we see that,

f ↑ as T ↑

Additional problem: What happens if you stretch a rubberband adiabatically? If you stretch a rubberband quickly this is eﬀectively adiabatic (no time for heat transfer). So what happens to the temperature of the rubberband?

dU = ∂q + ∂w adiabatic−→ dU = dw = fdl

Now deﬁne a constant l heat capacity,

∂U Cl = ∂T l note that U only depends on T ,

Z ∂U U = dT ∂T l ∂U dU = dT ∂T l and we have,

dU = CldT = fdl If you suddenly stretch a rubberband you apply a force (is this a constant force?) over a distance that the ruberband is stretched. Z Z fdl = CldT

f∆l = Cl∆T

dU Assuming dT is positive then if ∆l is positive, ∆T is positive. As a result the rubberband should heat up when you stretch is quickly (give it a try!).

II. (This is homework problem 8-54 Gold or 22-54 Red in McQuarrie and Simon) This problem can help to put together many of the concepts from the Helmholtz/Gibbs section of the class. This example is only to be done if extra time is left at the end of the last Helmholtz/Gibbs lecture. Warning: This is a long example. 7 THE RUBBER BAND PROBLEM 11

When a rubber band is stretched, it exerts a restoring force, f, which is a function of its length L and its temperature T . The work involved is given by, Z w = f(L, T )dL

First note that the work is positive (no negative sign such as we have in the deﬁnition of PV work). This is because the force on the rubber band is a restoring force acting to contract the rubber band. Note that L seems to be playing the role of V for the expansion of a gas. Next we derive the diﬀerential expression for the energy dU,

dU = δq + δw

δq = T dS dU = T dS + fdL Next we can derive the dependence of the energy on length, L. First take the partial derivative of both sides with respect to L at constant T ,

∂U ∂S = T + f ∂L T ∂L T We would like to express this in terms of more easily determined quantities T , L, and ∂S f, so we will derive a Maxwell’s relation for ∂L T . Use the deﬁnition of the Helmholtz energy,

A = U − TS −→ U = A + TS −→ dU = dA + T dS + SdT

dU = T dS + fdL dA + T dS + SdT = T dS + fdL dA = −SdT + fdL

Compare this with the total derivative of A(T,L),

∂A ∂A dA = dT + dL ∂T L ∂L T and we ﬁnd, ∂A ∂A = −S = f ∂T L ∂L T 7 THE RUBBER BAND PROBLEM 12

Now equating the cross derivatives,

∂2A ∂2A = ∂L∂T ∂T ∂L ∂2A ∂S = − ∂L∂T ∂L T ∂2A ∂f = ∂T ∂L ∂T L

∂S ∂f − = ∂L T ∂T L We can plug this result back into our original expression for the change in the energy with length, ∂U ∂S = T + f ∂L T ∂L T ∂U and we get ∂L T in terms of the quantities T , L, and f, ∂U ∂f = −T + f ∂L T ∂T L Now consider an ideal rubber band. For an ideal rubber band the force is related to the temperature by a function of the length,

∂f f = T φ(L) giving us = φL ∂T L and plugging this in, ∂U = −T φ + T φ = 0 ∂L T This says that the energy of an ideal rubber band is independent of the length at constant temperature. This is the same as having the energy independent of volume for an ideal gas at constant temperature. Next consider what happens if you stretch a rubber band quickly, which would be adiabatically. In this case, dU = dw = fdL We can deﬁne a constant length heat capacity,

∂U CL = ∂T L 7 THE RUBBER BAND PROBLEM 13

For an ideal rubber band U only depends on T , so we have,

Z ∂U U = dT ∂T L ∂U dU = dT = CLdT ∂T L Using dU = fdL,

CLdT = fdL If you suddenly stretch a rubber band you apply a force f over the distance the the rubber band is stretched. Z Z fdL = CL dT

and this is approximately,

f∆L = CL∆T So if ∆L is positive (stretch) then ∆T is positive. If you stretch the rubber band quickly it heats up (give it a try!).