Free Energy and Phase & Chemical Equilibria

Question 1

How do you derive the equation where du can be used to determine spontaneity of a reaction? Why is this not the most helpful?

Answer 1

ds>=dq/T
dq<=TdS

du= dq + dw
du <= TdS -Pex dV (from inequality above)
if constant S and V
du<=0 to be spontaneous, or =0 at equilibrium

Hard to maintain a constant entropy in a lab

Question 2

What is the equation for Helmholtz free energy? Derive how the condition for equilibrium can obtained

Answer 2

A= u- TS
dA=du - TdS - SdT
dA<= TdS -P ex dV - TdS -SdT
dA<= -Pex dV - SdT

for a process at constant V and T
dA<= 0, at equilibrium=0

Question 3

What is the equation for Gibbs free energy? Show how it can be used for a reaction at equilibrium?

Answer 3

G= A + pV
dG= dA + pdV + Vdp
dG<= -Pex dV - sdT + pdV + Vdp
at mechanical equilibrium where p=p ex
dG<= Vdp - sdT

for a process at constant p and t, dG<0
and dG=0 at equilibrium

e.g phase transitions, chemical reactions…

Question 4

How can the Gibbs free energy incorporate enthalpy?

Answer 4

G= A + pV
G= U - TS + pV
G= U + pV - TS
G= H - TS

Question 5

How can a systems ability to do non pV work be used within the free energy equations? What is the maximum available work of a system?

Answer 5

dw= -P ex dV + dwe, other work
dG<= dwe or equal at equilibrium
The maximum work from a process at constant T and P is equal to the decrease gibbs free energy

Question 6

What are the terms involved in the fundamental equations?

Answer 6

They show how the 4 energy state functions, U, H, A, G, vary with P, V, S, and T

Can all be derived from definitions of the the energy state functions, and using the first and second law and pv work

Question 7

How can the fundamental equation for H be derived? and A?

Answer 7

H= U + pV
dH= du + pdV + Vdp
dH= TdS - pdV + pdV + Vdp
dH= TdS + Vdp

A= u -TS
dA= du - Tds - SdT
dA= TdS- pdV - TdS - SdT
dA= pdV - SdT

Question 8

Why can we remove the inequality when deriving the fundamental equations but not for spontaneity arguements?

Answer 8

Replacing du rev with du
As valid for changes between equilibrium states

Question 9

How can the maxwell relations be derived?

Answer 9

Start with any fundamental equation

e.g dH= TdS + Vdp
implies H= H(S, p )
dH= (∂H/∂S)p dS + (∂H/∂p)s dP
implies (∂H/∂S)p = T and (∂H/∂p)s = V

and as ∂²H / ∂s∂p = ∂²H / ∂p∂s
so taking the derivative with respect to p for t as already done for s

(∂T/∂P)s = (∂V/∂s)p

Question 10

How do you derive the Gibbs-Helmholtz equation?

Answer 10

G= H -TS
-S= (G-H) / T
From the maxwell equations,
-S=(∂G/∂T)p

(∂G/∂T)p - G/T= -H/T from subbing in and rearranging

but if we take G/T and differentiate implicitly with respect to T, using the product rule

= 1/T (∂G/∂T)p - G/T²
=1/T ((∂G/∂T)p - G/T)

which is 1/T a factor of -H/T

so
∂/∂T (G/T) = - H / T²

Question 11

How do you derive the van’t Hoft equation from the Gibbs Helmholtz?

Answer 11

ΔrG⦵ = -RTlnkp
ΔrG/T⦵=-Rlnkp

∂/∂T (ΔrG/T⦵) = - ΔH / T²

∂/∂T (Rlnkp) = ∂/∂T (ΔrG/T⦵)
∂/∂T (lnkp) = - ΔH / RT²

Question 12

How do you derive an equation to show how G varies with pressure for ideal gases and solids?

Answer 12

dG= Vdp - SdT
using the maxwell relation, (∂G/∂p)T= V

moving the p over and integrate

G2-G1= ∫ v dp
of a perfect gas, using v= nrt/p
G2-G1= nRTln(p2/p1)

using p1=1 bar
G2- G⦵=nRTln(p2/p⦵)
G2= G⦵ + nRTln(p2/p⦵)

for solids
G2-G1=v ∫ dp, as V independent of p
G2= G⦵ + V⦵(p-p⦵)

Question 13

How can the idea of isothermal compressibility used for an equation of how G of solids varies with pressure?

Answer 13

graph lnV against p
slope=-k = (∂lnV/∂p)T= -1/v (∂V/∂p)T
rearrange for V

then same G integral as before, replace V with -1/k (∂V/∂p)T
G2-G1= (v2-v1) / k

-k = (∂lnV/∂p)T, rearrange for v2 in terms of v and e

end up with
G2= G⦵ + V/k(1-e^-k(p2-p⦵)
and as k approaches 0,
G2= G⦵ + v⦵(p2-p⦵)

Question 14

What are the conditions for phase equilibrium and how have these been derived?

Answer 14

Suppose an equilibrium between a and b, where dn moles of a is converted to b, constant p,t
dG= G(b) - G (a) dn
at equilibrium dG=0

so G(b) = G(a)

can be 2 or 3 states. but 3 states have isolated points in the p,t plane

Question 15

What does the p/t graph for water look like?

Answer 15

Y=mx line from origin between s and g
almost backwards line from triple point between solid and liquid
almost quadratic from triple point up ish between liquid and gas

Question 16

How do you calculate the degrees of freedom for a substance?

Answer 16

F= 3 - p, the number of phase in equilibrium

Question 17

Derive the Clapeyron equation

Answer 17

The slope of a pt graph is dp/dT
Consider the points p,t and p+dp, T + dT

at p+dp, T + dT, G(a) + dG(a) = G(b) + dG(b)
at p,T G(a)=G(b)
subtracting the equations

dG(a)= dG(b)

using the fundamental equation dG= Vdp - SdT, which applies to both

V(a)dp - S(a)dT= V(b)dp - S(b)dT
V(b)dp - V(a)dp= S(b)dT- S(a)dT
ΔV(b-a)dp= ΔS(b-a)dT
dp/dT=ΔS/ΔV

where ΔV is the volume change between the phases and ΔS the entropy change between the phases

Question 18

How can the Clapeyron equation be used to incorporate enthalpy?

Answer 18

ΔG=ΔH - TΔS
at equilibrium, ΔG=0
ΔS= ΔH/T

dp/dT=ΔS/ΔV (Clapeyron equation)
dp/dT= ΔH/TΔV

Question 19

How can the Clausius-Clapeyron equation be derived?

Answer 19

Clapeyron= dp/dT= ΔS/ΔT = ΔH/TΔV
For sublimations and vaporisations, we can assume the volume change= volume of the gas, and there is such a large increase in volume, volume solid/liquid negligible
And we can use the ideal gas law for the volume of gas
pV=nrt with 1 mol
V=RT/p

substituting into the Clapeyron equation
dp/dT= ΔH/T/(RT/p) = pΔH/RT²
1/p dp/dT = ΔH/RT²
dlnp/dT=ΔH/RT²

an approximation

Question 20

If given the vapour pressures of a gas at different temperatures, how can the enthalpy changes of vaporisation and sublimation be calculated?

Answer 20

The Clapeyron equation can be written in the form
ln(p)= C -ΔH/RT When integrating out
Sketch a graph of this and it will give two opposing best fits
The first one is vaporisation, with higher temperatures but lower pressures and the second sublimation
Then you can calculate the slopes

Or you can use the equation with suitable data points

Question 21

What is the chemical potential of a pure substance?

Answer 21

The amount of Gibbs energy of one mole of the pure substance

G(m)=μ
G=nμ

Question 22

What is the fundamental equation for dG in an open system?
How does this vary for a mixture?

Answer 22

dG= Vdp - SdT + sum of μdn

for a mixture, μ1dn1 + μ2dn2 …

from exact differential from G=G(P,T,n)

Question 23

What is the Gibbs energy of a system using chemical potential? What is the differential for μ?

Answer 23

G= sum of n x μ for all components of the system

G=nμ
∂G/∂n = μ

Question 24

What is the Gibbs- Duhem equation and how is it derived?

Answer 24

G= sum of nμ
dG= sum of ndμ + sum of μdn

dG=Vdp - SdT + sum of μdn
at constant p and t
dG= sum of μdn

subtracting the equations
0= sum of ndμ

Question 25

How do you calculate the change in Gibbs energy of mixing two gases at constant pressure?

Answer 25

G2= G⦵ + nRTln(p2/p⦵)
and assume 1 molar
μ2= μ⦵ + RTln(p2/p⦵)

mixing n moles of a and n moles of b
Pa= xaP
Pb= xbP
where p is the total pressure and x is the mole fraction
p= external pressure as balanced

before mixing
initial G= na(μa⦵ + RTln(p/p⦵)) + nb(μb⦵ + RTln(p/p⦵))

after mixing
new G= na(μa⦵ + RTln(pa/p⦵)) + nb(μb⦵ + RTln(pb/p⦵))

ΔG=new G - initial G
= naRT(ln(pA/p⦵)-lnp/p⦵)+nbRT(ln(pb/p⦵)-lnp/p⦵)
=naRT(ln(pA/p)+nbRT(ln(pb/p)
=naRTlnxa + naRTlnxb
=nRT(xalnxa + xblnxb)

Question 26

What happens to the chemical potentials upon mixing perfect gases? Comment on the likelihood of this occuring?

Answer 26

Chemical potentials of the gases are lower in the mixture than pure A and B
Spontaneous and entropy driven

Question 27

Following on from the equation for the the change in Gibbs energy for mixing perfect gases, show by calculation the process is entropy driven. Also calculate the maximum change in Gibbs energy?

Answer 27

ΔG=nRT(xalnxa + xblnxb)
from the Maxwell relation,
(∂G/∂T)p=-S
-S= nR(xalnxa + xblnxb)

ΔG=ΔH-TΔS
ΔH=ΔG + TΔS = 0, so entropy driven

X= total Xa=1-Xb

ΔG=nRT(1-xb)lnx(1-xb) + xblnxb)
differentiating with respect to Xb
=nRT(lnxb - ln(1-xb))
maximum when xb= 1/2

ΔG=nRT(1/2ln1/2 + 1/2ln1/2)
= -nRTln2

Question 28

How has gibbs energy been defined in terms of extent of reaction?

Answer 28

dn= component x d extent (dE)
dG= Vdp - SdT + sum of μ dn
at constant p/t

dG= sum of μ x component dE
dG= ΔG dE

Question 29

What is the condition for equilibrium?

Answer 29

ΔG=0

Question 30

How has Kp been derived, and the equation associated with it?

Answer 30

μj = μj⦵ + RTln(Pj/P⦵)
summing each of the components
ΔG= ∑ μj
= ∑μj⦵ + ∑j RTln(Pj/P⦵)
=ΔG⦵ + ∑j RTln(Pj/P⦵)
ΔG=ΔG ⦵+ RT ∑ln(Pj/P⦵) ^ j
ΔG=ΔG ⦵+ RT lnKp

when at equilibrium, ΔG=0
ΔG⦵=-RT lnKp

Question 31

What is the equation for Kp? What are the units and what factors does Kp depend on?

Answer 31

Kp= e^-ΔG⦵/RT
dimensionless, as only depends on ΔG⦵ so not total pressure
It is dependent on temperature

Question 32

Why does temperature affect Kp? Why is there an approximation? How can accuracy be increased for this approximation?

Answer 32

From the van’t Hoft equation:
∂/∂T (lnkp) = - ΔH / RT²
and after integrating everything out

lnKp(T2) approximately = lnKp(T1) + ΔH⦵/R (1/T1 - 1/T2)
so dependent on temperature
Assumes enthalpy change independent on temperature so not exact

Replace enthalpy with Cp values and equations…

Question 33

How does pressure affect the position of equilibrium and how can these calculations be carried out?

Answer 33

Increased pressure, push to the side with fewer gaseous moles

The products will have a degree of disassociation between 0 and 1
Use algebra to start with mole changes and stoichiometry, think A level table
Put into partial pressures and the Kp expression with standard pressure

Question 34

What is the general equation for chemical potential e.g for a solid, real gas…

Answer 34

μj = μj⦵ + RTlnaj
where aj is the activity of a component

Question 35

What happens to substances in terms of state at a high enough temperature?

Answer 35

The the liquid/gas phase boundary disappears and results in a fluid state

Question 36

What do the PV graphs for a substance look like before, after, and at the critical temperature? Explain

Answer 36

Above, follows inverse proportionality of a gas, increase pressure, lower volume
Below, from the right, curve for a gas, then a straight line across and then straight up
This is because gases follow inverse proportionality, liquids and gases have a range of volume for that phase, and liquids can have higher pressures but similar volumes so straight up
At Tc, negative cubic shape with point of inflection, in between two temperatures

Question 37

What are the conditions for the critical point? How can the critical point be found for a substance?

Answer 37

The point where P=Pc, T=Tc, and V=Vc, above this acts as fluid
Singular

∂P/∂V= 0 and also the second derivative

Question 38

Why does a critical point occur? Why isn’t it seen in perfect gases?

Answer 38

Arises from intermolecular interactions, perfect gases ignore these
From derivative of ideal gas equation, so way for it to equal 0 at any pressure

Question 39

What happens to the isothermal compressibility at the critical point? And Cp-Cv?

Answer 39

Compressibility tends to infinity as dividing by ∂P/∂V which is 0
As does Cp-Cv