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PT IA Chemistry > Energetics and Equilibria > Flashcards

Energetics and Equilibria Flashcards

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1
Q

what (in informal terms) is a spontaneous process

A
  • one which occurs ‘naturally’ or on its own
  • the reverse of a spontaneous process requires some intervention
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2
Q

explain why exothermicity is not the condition for spontaneous reactions

A
  • there are some endothermic processes which are spontaneous e.g. NH4NO3 dissolving in water
  • or the mixing of inert gases has no energy change
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3
Q

what is the actual definition for what makes a spontaneous process

A
  • In a spontaneous process the entropy of the universe increases
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4
Q

what is entropy? what are the two ways of considering it

A
  • Entropy is a measure of disorder or randomness
  • it can be analysed in a statistical thermodynamics way using energy levels in molecules/atoms relating to translational, vibrational and electronic energies
  • it can also be analysed in a classical way using the heat changes during a reaction
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5
Q

explain how energy levels can be used to help determine entropy (in the molecular interpretation)

A
  • Molecules have quantised energy levels, each molecule has a set of energy levels associated with translation, rotation, vibration and electronic structure
  • in a macroscopic sample (e.g. 10^20 atoms) there are many many ways for the molecules to distribute themselves over the energy levels
  • this can be analysed using statistical thermodynamics
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6
Q

assuming we have x energy levels and N atoms, what is the total number of ways each configuration of energy distributions can be achieved

A

W = number of ways a given energy distribution can be achieved

N = number of atoms
n0 = number of atoms in energy level 0
n1 = number of atoms in energy level 1
nx = number of atoms in energy level x

W = N! / (no! n1! n2! ….. nx!)

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7
Q

what assumption can be made about the likelihood of each configuration and what does this tell us about which configuration we’ll get

A
  • we can assume the system has no preference for one configuration over another, i.e. we can imagine as the molecules collide and move they are constantly changing between configurations
  • this means the most likely configuration we’ll see is the configuration with the greatest W number
  • this configuration is called the ‘most probable distribution’
  • as the number of particles becomes very large, this most probable config. becomes overwhelmingly more probable
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8
Q

How do you find the most probable distribution, what is the outcome of this i.e. what is the Boltzmann Distribution

A
  • to find the most probable distribution you must maximise W for a given total number of particles and energy (beyond scope of IA)
  • the outcome is the Boltzmann distribution
  • in the most probable distribution, the population of level i, ni which has energy Ei, is given by

ni = no e^(-ei/kT)

k = Boltzmann constant
T = temp
R = NA K
NA = Avogadro’s number

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9
Q

what can we say (using the Boltzmann distribution) about populations where the energy of the energy level is much greater than kT or similar/less

A
  • where the energy of the energy level is much greater than kT, the population becomes vanishingly small
  • where the energy of the energy level is comparable (or less than) kT, the population is significant
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10
Q

Give the equation linking entropy and W,

A

S = k ln(W)

i.e. the greater the number of ways a distribution can be achieved, the greater the entropy of the system

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11
Q

what effect does heating the system have on W and S

A
  • When energy as heat is supplied to the system, the energy is accommodated by some of the molecules moving up to higher energy levels
  • this makes the molecules more spread out over the energy levels
  • hence there are more wats of achieving the resultant distribution (W increases)
  • Hence entropy, S, increases
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12
Q

what effect does expanding the system have on W and S

A
  • Quantum mechanics tells us that as the system is expanded, the spacing of the translational energy levels decreases
  • Hence there are more energy levels ‘within reach’ of kT
  • So the molecules are distributed over more levels and W is increased
  • Thus S also increases
  • Increasing vol increases entropy
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13
Q

what effect do physical transformations have on W and S

A
  • a physical transformation such as solid –> liquid/gas is associated with a large increase in number of energy levels available to the system
  • in a liquid/gas the molecules are free to translate so they have access to a larger number of translational energy levels
  • so W is larger
  • so S increases
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14
Q

what effect does changing temperature have on W and S

A
  • supplying energy to a system in the form of heat increases entropy
  • But specifically, for the same amount of ‘heat energy’ added to a system, the entropy increase is greater for a colder system than a hotter one
  • i.e. the increase in entropy resulting from a certain amount of heat energy being added is greater the cooler the system is
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15
Q

what is the classical definition of entropy

A

if an object at a temperature T absorbs a small amount of energy δqrev under reversible conditions then the change in entropy dS is given by

dS = δq(rev) / T

NOTE: we cannot PROVE this is correct but it agrees with all observations (as is the case with all physical theories)

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16
Q

what are some features of the classical definition of entropy which ‘agree’ with the statistical definition

A
  • supplying energy increases entropy: if delta(qrev) > 0 then energy is being supplied, this means for a finite temperature dS > 0 so entropy increases
  • the classical definition also suggests dS is inversely proportional to temperature, i.e. for the same change in energy but at a higher temperature, entropy increases less, as expected from before
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17
Q

how do we split up the universe in terms of the second law of thermodynamics

A
  • The system, this is the thing that we’re interested in
  • the surroundings, this is the rest of the universe
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18
Q

what is the general expression for the entropy change of the universe in terms of system/surroundings

A

deltaSuniv = deltaSsys + deltaSsurr

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19
Q

how can we calculate the entropy change of the system

A

-entropy is a property of matter, hence deltaSsys can be calculated from tabulated values of the entropies of substances

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20
Q

What is meant by a closed system in thermodynamics, what is the result of this

A

A closed system in thermodynamics means matter cannot be exchanged between the system and the surroundings

However, energy as heat can be exchanged

this means deltaSsurr can be calculated directly from

dS = deltaqrev / T

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21
Q

why can we consider any heat exchange as reversible (from the point of view of the surroundings), what does this mean we can use to calculate deltaSsurr

A
  • the surroundings are so large that neither their temperature nor volume are affected by the flow of energy as heat into or out of it

deltaSsurr = qsurr/Tsurr

qsurr = energy as heat absorbed by surroundings
Tsurr = temperature of surroundings

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22
Q

what assumptions/conclusions can we make to simplify the equation for deltaSsurr

A
  • any heat lost by the system is absorbed by the surroundings and vice versa, hence qsurr = -qsys
  • we can assume the system and surroundings are at thermal equilibrium Tsys = Tsurr

hence

deltaSsurr = -qsys / Tsys

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23
Q

using the ‘simplified’ equation for deltaSsurr, how can we calculate the entropy change of the universe

A

deltaSuniv = deltaSsys - qsys/Tsys

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24
Q

what happens at deltaSuniv = 0 and how can we calculate the temperature at which this occurs, i.e. the critical temperature

A

deltaSuniv = 0
implies
-deltaSsurr = deltaSsys
so
T = qsys/deltaSsys

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25
Q

what the first law of thermodynamics say (in words), what are the forms of energy

A
  • in words, the first law of thermodynamics that energy cannot be created or destroyed but just transformed from one form to another
  • the forms of energy are heat, work and internal energy
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26
Q

what is heat, what do we need to be careful to avoid when talking about heat and what is the SI unit

A
  • heat is the means by which energy is transferred from a hotter body to a cooler one
  • we need to be careful not to think of heat as a substance or fluid
  • its SI unit is Joules (J)
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27
Q

what is work (in relation to work done mechanically)

A
  • a force through a distance
  • if we move a distance x against a force F, we have done work W = fx
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28
Q

what is internal energy, what are the forms of energy that can ‘make up’ internal energy

A
  • it is a property that an object possesses
  • the value of the internal energy depends on the amount of substance, temperature, pressure etc.
  • it is often formed from Ek of particles, energy of interactions of particles and bond energy within chemical bonds
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29
Q

what is the only component of internal energy in an ideal gas

A
  • kinetic energies of particles, i.e. the internal energy is directly proportional to the temperature
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30
Q

what sort of functions are heat, work and internal energy

A
  • heat and work are path functions
  • internal energy is a state function
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31
Q

what is a state function

A
  • a state function is one whose value depends only on the state of the substance under consideration, it has the same value for a given state, no matter how that state was arrived at
  • this is the property that Hess’ Law calculations rely on
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32
Q

What is a path function

A
  • the value which a path function takes depends on the path which the system takes going from A to B
  • e.g. the reaction of hydrogen and oxygen can create different amounts of heat/do different amounts of work depending on the way in which they react.
  • hence heat and work are path functions
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33
Q

express the first law of thermodynamics mathematically and explain the significance

A
  • consider a system changing from state A to state B, the internal energy will change from Ua to Ub through absorbing some amounts of heat, q, and work, w, Hence

deltaU = q + w

  • in going from A to B, q and w can have any values provided that (q+w) is equal to deltaU
  • this is equivalent to saying energy cannot be created or destroyed…
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34
Q

what are the sign conventions that we use for heat and work in relation to internal energy

A
  • the values will be positive when they correspond to increases of internal energy
  • hence when energy is supplied as heat to a system, q is positive
  • when work is done on a system w is positive
  • when work is done by a system, w is negative, it is usually labelled w’ = -w
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35
Q

what are the properties of an ideal gas which specify its state

A
  • pressure, volume, temperature, amount (in mol)
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36
Q

state the ideal gas law and define its quantities

A

pV = nRT
p = pressure (pa)
V = volume (m^3)
n = amount of substance (mol)
R = ideal gas constant
T = temperature (K)

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37
Q

How can we derive the expression for the infinitesimal work done by an expanding gas

A
  • consider a gas within a frictionless cylinder, pressure inside = Pint, pressure outside = Pext
  • the force that the gas is expanding against is Pext so this is what we consider for the work done

deltaw’ = force x distance
= Pext * A * dx
= Pext dV

NOTE: we use w’ as we are considering the work done BY the gas not ON the gas

Hence

deltaw = -Pext dV

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38
Q

what is the work done if a gas expands against 0 pressure

A

0

no force is pushed against or moved through a distance, hence no work is done

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39
Q

what is the expression (and derivation) for the work done by a gas when expanding against constant pressure

A
  • we know deltaw’ = Pext dV
  • so we can integrate it WRT V to get an expression for w’
  • as Pext is a constant, it can be taken out of the integration
  • this leaves us with

w’ = Pext (Vf - Vi)

where Vf is final volume, Vi is initial volume

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40
Q

how can the model be made such that an expanding gas does the maximum amount of work

A
  • to get the maximum amount of work done, we need the maximum force to be moved through the distance
  • in order for the gas to still expand, this means the external pressure must at all times be infinitesimally smaller than the internal pressure
  • this means the external pressure must eb continuously lowered in order to maintain it as slightly lower than the internal temperature
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41
Q

Give the features of a reversible expansion of a gas

A
  • infinitely slow expansion
  • external pressure infinitesimally smaller than internal pressure
  • at equilibrium
  • does maximum work
  • can be transformed from expansion to compression with an infinitesimally small change in external pressure
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42
Q

give the features of an irreversible expansion of a gas

A
  • goes at a finite (not infinitely small) rate
  • not at equilibrium
  • doesn’t do maximum work
  • external pressure must be changed significantly to cause compression to start occuring
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43
Q

what is meant when we use the term maximum work

A
  • the magnitude of the work done is the greatest
  • we avoid saying largest etc. because the sign of the work can change depending if its work done ON or BY
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44
Q

what is the expression (and derivation) for the work done in the reversible isothermal expansion of an ideal gas

A

we know
deltaw’ = Pext dV

from ideal gas equation
Pint = nRT/V

we know Pext is infinitesimally smaller than Pint hence
Pext = Pint = nRT/V

so

w’ = integral(nRT/V)dV

w’ = nRT ln(Vf/Vi)

Vf = final volume
Vi = initial volume

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45
Q

what is an indicator diagram and how do they help us visualise the differences in work done by a gas reversibly and irreversibly

A
  • the work done can be visualised as the area under a curve of Pext against V
  • for a constant external pressure i.e. irreversible expansion, this just gives us a rectangle shape
  • for a reversible expansion, the pressure is a 1/x style line, this gives a greater area for the same Vf and Vi
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46
Q

what can we use (and why) to help us calculate the heat involved in gas expansions

A
  • the first law of thermodynamics
  • this is
    deltaU = q + w = q - w’
  • we know U is a state function but q and w are path functions
  • this means in either the reversible or irreversible expansions the value of U following the expansion will be the same
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47
Q

what can we use (and why) to help us calculate the heat involved in gas expansions

A
  • the first law of thermodynamics
  • this is
    deltaU = q + w = q - w’
  • we know U is a state function but q and w are path functions
  • this means in either the reversible or irreversible expansions the value of U following the expansion will be the same
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48
Q

through the consideration of the first law and the work done in a reversible expansion, what can we say about the heat involved in a reversible expansion

A

we know
DeltaU = q + w = q - w’
and deltaU is constant for reversible or irreversible expansion

  • we know w’irrev < w’rev
  • hence q must also be greater for a reversible expansion

“In a reversible process, the magnitude of the heat is a maximum”

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49
Q

what can we say about the heat that is transferred in the isothermal expansion of an ideal gas

explain in words how this is so

A
  • for an ideal gas, internal energy is purely kinetic so is dependent on temp only
  • given the expansion is isothermal, the temperature does not change, thus, deltaU = 0

thus

w’ = qrev = nRT ln(Vf/Vi)

  • the work done by the gas expanding, is exactly compensated for by the heat flowing in
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50
Q

what is an important point to note concerning reversible/irreversible reactions and the definition for entropy

A
  • the definition for an entropy change is

dS = (delta qrev) / T

  • q is a path function, NOT a state function
  • this means even if we want to calculate an entropy change for an irreversible process, we must consider a reversible process to calculate the entropy change
  • NOTE: there are multiple ways of getting around this, but this is the theory
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51
Q

what is the difference/exception in calculating entropy when we are calculating the entropy change of the surroundings

A
  • we know that ‘from the point of view of the surroundings, any heat flow is reversible’. This means the entropy change of the surroundings can be calculated directly from the heat
  • this is because we know that the surroundings are sufficiently large for their volume to be unaffected, hence they do no work
  • hence deltaUsurr = qsurr
  • as the internal energy is a state function this means qsurr is the same for reversible or irreversible processes
  • we also know the temperature of the surroundings will remain unaffected
  • so overall we can say in any case

DeltaSsurr = qsurr / Tsurr

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52
Q

what is the entropy change in the isothermal expansion of an ideal gas

A

we know

dS = deltaqrev / T

we know
w’rev = qrev = nRT ln(Vf/Vi)

so

dS = nR ln(Vf / Vi)

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53
Q

Define internal energy (in terms of heat transfer)

A

the internal energy change, δU, is equal to the energy as heat under constant volume conditions

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54
Q

Define enthalpy

A

the enthalpy change, deltaH, is equal to the energy as heat under constant pressure conditions

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55
Q

what is the equation for infinitesimal changes in internal energy, how can this be changed by considering Pext

A

dU = deltaq + deltaw

we know
deltaw = -PextdV

Hence

dU = deltaq - PextdV

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56
Q

what is the form of the equation for internal energy at a constant volume, explain what it means

A

we know
dU = deltaq - PextdV

hence if volume doesn’t change, dV = 0
so

dU = deltaq (const.vol)

i.e. the heat absorbed under a constant volume is equal to the change in internal energy

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57
Q

what is the general form (normally and infinitesimally) of the relation between heat supplied and temperature change using heat capacities

A

q = c * deltaT
or infinitesimally

deltaq = c dT

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58
Q

what is the molar heat capacity, how can we rewrite the heat capacity equation using it

A

Cm = molar heat capacity = heat capacity per mol = c/n

hence

q = nCmdeltaT

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59
Q

what is the molar heat capacity at a constant volume, how can it be used in an equation/linked to U

A

Cm,v = molar heat capacity at a constant volume = (∂U,m/∂T)v

this works in the same way as normal molar heat capacity but applies only to constant volume conditions

Given we know under constant volumes δq = dU we can write

dU = nCv,m dT
or
dUm = Cv,m dT

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60
Q

give the definition of enthalpy, H, give some of its properties, why is it useful?

A

H = U + pV

  • it is a state function
  • it’s value is equal to the heat transferred at a constant pressure
  • this is useful because most reactions done in labs occur under a constant pressure
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61
Q

give the differential form of enthalpy, H, how can this be used to show H is equal to the heat at a constant pressure

A

dH = dU + pdV + Vdp

substituting
dU = deltaq - pdV
gives

dH = deltaq + Vdp

assuming a constant pressure gives

dH = deltaq(constant pressure)

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62
Q

what is the equation linking heat capacities and enthalpies, what is heat capacity at a constant pressure

A
  • we know q = c dT
  • assuming a constant pressure means

deltaq = Cp dT

but at a constant pressure we know

dH = deltaq
so

dHm = Cp,m dT

or

C,p,m = (delHm/delT)p

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63
Q

how can we calculate the molar enthalpy of a substance at a temperature which it is not tabulated/known at, derive the expression

A

we know

Cp,m = (delHm/delT)p
so
dHm = Cp,m dT (const.pres.)

so integrating both sides gives

Hm(T2) - Hm(T1) = Cp,m[T2-T1]

NOTE: Hm(T1), Hm(T2) are functions of temperature NOT multiplied BUT Cp,m[T2-T1] IS multiplied

so

Hm(T2) = Hm(T1) + Cp,m[T2-T1]

NOTE: we have assumed Cp,m is constant, for small temperature ranges

64
Q

what are the ways that heat capacities are measured/ what are some of their properties over large temperature ranges/how are they usually presented

A
  • Measure a temperature rise for a given heat input to calculate heat capacity
  • Heat capacities do change with temp but usually only small amounts over large ranges
  • they are usually presented in parametrized form
    C(T) = a + bT + c/T^2

a, b, c would be tabulated

65
Q

what is the definition for entropy when considering processes at a constant pressure, and why

A

we know entropy is defined by
dS = deltaq(rev)/T

we also know at a constant pressure
dH = deltaq

so at a constant pressure

dS = dH/T

we do not need to specify dH(rev) because it’s a state function

66
Q

How can we use the equation for entropy in terms of H to derive an expression for absolute entropy at a given temperature

A

we have
dS = dH/T

we also know
dHm = Cp,m(T) dT

so

dSm = Cp,m(T) / T dT

integrating both sides from T = 0 (absolute 0) to a temperature T* gives

Sm(T*) = integral Cp,m(T)/T dT

NOTE: the integral is not trivial as C is a function of T

67
Q

how are absolute entropies usually evaluated practically, how is this complicated if there are phase transitions that occur on this temperature range

A
  • the tricky integral in the expression for absolute entropy must be evaluated graphically
  • we make measurements of Cp,m as a func of temp and plot Cpm(T)/T against T from 0 to T*
  • then find the area underneath the graph
  • if phase changes occur on this interval then we must calculate the entropy change of them too, this is given by
    deltaSm,pc = deltaHm,pc / Tpc

so

S(T) = integral(0 to T)[Cp,m(T)/T dT] + sum(over phase changes)[deltaHm,pc /Tpc]

68
Q

why can we say that absolute entropy at absolute 0 is exactly 0

A
  • at absolute 0, all molecules must be in their lowest energy state
  • hence there’s only one way to arrange them i.e. w = 1
  • so S = kln(w) = 0
69
Q

how can we convert absolute entropies from one temperature to another (constant pressure) (assuming the temperature difference is small)

A
  • if the temperature difference is small then we can assume Cp,m is a constant
  • so we simply evaluate the integral for the absolute entropy but now without Cp,m(T)
  • this gives

Sm(T2) = Sm(T1) + Cp,m ln(T2/T1)

70
Q

what is an alternative (and usually easier way) to deal with the second law, what is the definition of the new quantity

A
  • we can introduce a quantity called Gibbs Free Energy
  • it is defined by

G = H - TS

  • on a spontaneous process, the change in G is negative
71
Q

How can we derive the condition that in a spontaneous process, DeltaG is -ve, link DeltaG to DeltaSuniv

A

G = H -TS
dG = dH - TdS - SdT
-dG/T = -dH/T + dS
(assuming constant temperature)

also
DeltaSuniv = deltaSsurr + DeltaSsys
so
dSuniv= -deltaqsys/ Tsurr + dSsys

if we assume the process is taking place at a constant pressure so dH = deltaqsys, and that the system, surroundings are the same temp then we obtain

dSuniv = -dH/T + dSsys

  • hence dSuniv = -dG/T
  • so given in any spontaneous process Suniv > 0, in any spontaneous process, G<0
72
Q

Give the two main points about Gibbs Free Energy and spontaneity of reactions

A

G decreases or deltaG is negative in a spontaneous reaction

At equilibrium, G reaches a minimum i.e. dG = 0

73
Q

give the finite changes form of G

A

deltaG = deltaH - TdeltaS

(assuming no change in temperature)

74
Q

what is the master form of the equation for internal energy, how do we form it

A

dU = deltaq + deltaw
if we assume the only work done is from pV then
dU = deltaq - pVext
we know deltaqrev = TdS from the entropy definition, and because it’s reversible pint = pext

dU = TdS - pdV

75
Q

what is the master equation for H, how is it derived

A

dH = TdS + Vdp

H = U + PV
dH = dU + pdV + Vdp
dH = TdS - pdV + pdV + Vdp
then cancel

76
Q

what is the master equation for G, how is it derived

A

dG = Vdp - SdT

G = H - TS
dG = dH -TdS - SdT
dG = TdS + Vdp - TdS - SdT
then cancel

77
Q

What is the expression for the variation of the Gibbs energy with pressure for a constant temperature, how is it derived

A

our expression for dG is
dG = Vdp - SdT
at a constant temperature dT = 0 so

dG = VdP
so
(delG/delp)T = V
and for an ideal gas
V = nRT/p
so

integral(dG) = integral( nRT/p dp)
as this is at a constant T this gives

G(p2) - G(p1) = nRT ln(p2/p1)

78
Q

what is the general way that the variation of free energy with pressure at a const. temp. (slightly different to derived version)

A
  • we take p1 to be standard pressure (10^5 pa)
  • G(p1) = G”
  • we also divide by n to give a molar quantity
    so

Gm(p) = G”m + RT ln(p/p”)

79
Q

what is the expression for the variation of Gibbs energy with temperature at a constant pressure (Gibbs-Helmholtz)

A

start with master eq.
dG = Vdp - SdT
impose constant pressure i.e. dp = 0
so

dG = -SdT
so
(delG/delT)p = -S

use
d/dT (GT^-1) = T^-1(dG/dT) - T^-2 G

substituting from above for G and dG/dT and simplifying yields

d/dT ( G/T ) = -H/T^2

This is the Gibbs Helmholtz equation

80
Q

what is the partial pressure of a gas i in a mixture

A

pi = xi ptot

pi = partial pressure
ptot = total pressure
xi = mole fraction of i

xi = ni/ntot

81
Q

How can we calculate the variation of the gibbs energy with pressure (const temp) of a gas component of a mixture

A
  • we know
    Gm(p) = G”m + RT ln(p/p”)

we can apply the same concepts to a component of an ideal gas because in an ideal gas the components don’t interact i.e. they act the same as if the other components are not there, so

Gm,i(pi) = Gm,i” + RT ln(pi/p”)

82
Q

how can we calculate the Gibbs energy of a mixture of ideal gases (A,B,C,D….)

A

simply multiply each gas’ free energy by its molar proportion

G = nA Gm,A(pA) + nB Gm,B(pB) ……

83
Q

what can we introduce to extend the ideas about free energy in a mixture beyond ideal gases

A
  • chemical potential, given by μ
  • this is a property of a species which depends on its partial pressure or concentration
84
Q

what is the expression for the gibbs free energy of a general mixture in terms of its chemical potentials

A

G = naμa + nbμb …..

85
Q

Give the equation for the chemical potential of a species in terms of its partial pressure

A

μi(pi) = μi” + RT ln(pi / p”)

μi” is the chemical potential of i under standard conditions

p” is standard pressure (10^5)

86
Q

Give the equation for the chemical potential of a species in terms of its concentration

A

μi(ci) = μi” + RT ln(ci / c”)

μi” is the chemical potential of i under standard conditions

c” is the standard concentration (1moldm^-3)

87
Q

what is the chemical potential of solids or pure liquids

A

its chemical potential is always equal to its standard chemical potential μi”

88
Q

Give the definition of standard state, what is the important thing we should remember about it

A

The standard state of a substance is the pure form at a pressure of one bar and at the specified temperature

NOTE: the specified temperature NOT necessarily 298K

89
Q

define the standard enthalpy of formation

A

the standard enthalpy of formation deltaH” is the standard enthalpy change for a reaction in which one mole of the compound is formed from its constituent elements each in their reference states (the most stable state at the stated temperature and 1 bar)

90
Q

How can we use the standard enthalpy changes of formation to calculate the standard enthalpy change of reaction

A

for a reaction

vaA + vbB —> vpP + vqQ

delta(r)H” =
vpdelta(f)H”(P) + vqdelta(f)H”(Q)
-
vadelta(f)H”(A) + vbdelta(f)H”(B)

i.e. enthalpies of formation for products (in stoichiometric proportions) - enthalpies of formations of reactants (in stoichiometric porportions)

91
Q

How can we use standard entropy changes of formation to calculate the standard changes in entropy for a reaction

A
  • the same way which we did for enthalpies

for a reaction

vaA + vbB —> vpP + vqQ

delta(r)S” =
vpdelta(f)S”(P) + vqdelta(f)S”(Q)
-
vadelta(f)S”(A) + vbdelta(f)S”(B)

92
Q

How can we calculate change in standard Gibbs Free energy of reaction from standard enthalpy/entropy changes of formation

A
  • calculate standard enthalpy change of reaction using method previously described
  • calculate standard entropy change of reaction using method previously described
  • combine as
    Delta(r)G” = Delta(r)H” - T*Delta(r)S”
93
Q

Does Delta(r)H” change with temperature?
What can we use to help calculate this?

A

YES

use the heat capacity

94
Q

what is delta(r)Cp”

A

the standard change in heat capacity of a reaction under const. pressure

for a reaction

vaA + vbB —> vpP + vqQ

delta(r)Cp” =
vpdelta(f)Cp,m”(P) + vqdelta(f)Cp,m”(Q)
-
vadelta(f)Cp,m”(A) + vbdelta(f)Cp,m”(B)

95
Q

what is the expression for how Delta(r)H” changes with temperature, how can we derive it

A

we know
dHm / dT = Cp,m (molar and const. pressure)
so
dDelta(r)H” / dT = delta(r)Cp”

integrating both sides wrt T (and assuming Cp,m is a constant) gives

Hm(T2) - Hm(T1) = Cp,m*[T2-T1]

so

Delta(r)H”(T2) = Delta(r)H”(T1) + delta(r)Cp” * [T2-T1]

delta

96
Q

what is the expression for how Delta(r)S” changes with temperature, how can we derive it

A
  • we already know the equation for how S changes with temperature

Sm(T2) = Sm(T1) + Cp,m,*ln(T2/T1)

so applying this to changes in entropy gives

Delta(r)S”(T2) = Delta(r)S”(T1) + Delta(r)Cp”*ln(T2/T1)

97
Q

How is the equilibrium constant Kc or Kp often written, why is this wrong how should it be written

A

for a reaction
vaA + vbB –rev.—> vpP + vqQ

usually Kc is written as
Kc = [P]^vp [Q]^vq / [A]^va [B]^vb

and Kp as
Kp = (pp)^vp (pQ)^vq / (pa)^va (pb)^vb

this is only wrong because equilibrium constants should technically be dimensionless, to obtain this we divide each conc. in Kc by the standard conc. 1moldm^-3 and each partial pres. in Kp by 1 bar

this gives

Kc = (cP/c”)^vp (cQ/c”)^vq …

and

Kp = (pp/p”)^vp (pQ/p”)^vq …

in reality we can write it either way as long as we know we mean the second way

98
Q

What is the expression delta(r)G of a general reaction in a large reaction mixture in terms of chemical potential, why do we use a large reaction mixture

A

consider
vaA + vbB —rev.–> vpP + vqQ

  • we have a large amount of this mixture such that one mole of reaction has no significant change on the conc’s

when one mole of reaction occurs we have

delta(r)G = vpμp + vqμq - vaμa - vbμb

99
Q

how can our expression for delta(r)G for a general reaction help us deduce the conditions for when an equil moves either way and what the equil point is

A

delta(r)G = vpμp + vqμq - vaμa - vbμb

  • the values of the chemical potentials are dependent on the species’ partial pressure or conc.
  • Hence the sign of delta(r)G depends on the composition of the mixture
  • where they are such that delta(r)G < 0, the forwards reaction is favoured
  • where they are such that delta(r)G>0 the backwards reaction is favoured
  • over time the concentrations/pressures of the reaction mixture will change such that the Gibbs energy becomes a minimum
  • at the minimum/at equil. delta(r)G = 0
  • thus
    (vpμp + vqμq - vaμa - vbμb)eq. = 0
100
Q

For a reaction mixture which is sufficiently large that the concentrations do not change, what does delta(r)G represent in relation to G and composition

A
  • it is the gradient of a G against composition graph
101
Q

How can we rewrite delta(r)G for a reversible reaction such that it only relies on partial pressures (or conc’s)

A
  • we know
    delta(r)G = vpμp + vqμq - vaμa + vbμb

we can define each μ in terms of partial pressures

μi(pi) = μi” + RTln(pi/p”)

then substitute this into the delta(r)G equation for each component and rearrange to give

delta(r)G = [vpμp” + vqμq” - vaμa” - vbμb”] + RT(vpln(pp/p”) + vqln(pq/p”) - valn(pa/p”) - vbln(pb/p”))

or equiv. for concs

102
Q

how can we use our expression for delta(r)G in terms of chemical potentials as funcs of partial pressures/conc.’s to derive the important expression for delta(r)G” in terms of K

A

we know
delta(r)G = [vpμp” + vqμq” - vaμa” - vbμb”] + RT(vpln(pp/p”) + vqln(pq/p”) - valn(pa/p”) - vbln(pb/p”))

we can recognise that the first term is the same as delta(r)G”

we can also clean up the logs to give

delta(r)G = delta(r)G” + RT ln[ (pp/p”)^vp (pq/p”)^vq / (pa/p”)^va (pb/p”)^vb ]

we know that at equil delta(r)G = 0 and we can recognise the term inside the ln is K
Hence

delta(r)G” = -RT ln[K]

all of the same concepts apply if deriving using concs

103
Q

explain why solids are not included in equilibrium constant equation (using the delta(r)G” definition)

A
  • we know that the chemical potentials of solids remain the same as their standard value regardless of pressures etc.
  • hence we have no ln(..) term in their expression so they only contribute to the delta(r)G” first term in the derivation and aren’t included in the K expression in the ln
104
Q

what is the equation for K in terms of delta(r)G”, what is significant about the sign of delta(r)G”, what else can we note about the range of values that K can take

A

K = exp(-delta(r)G” / RT)

  • if delta(r)G” < 0 then the argument of exp > 0, so K > 1 so products are favoured
  • if delta(r)G” > 0 then the argument of exp < 0, so K < 1 so reactants are favoured
  • we can see that due to the exponential relationship, a small change in delta(r)G” will give a large change in K so K can cover a wide range of values
105
Q

what does a plot of ln(K) against delta(r)G” give

A
  • a straight line of negative gradient RT
106
Q

state le chatalier’s principle

A

‘When a system in equilibrium is subjected to a change, the composition of the equilibrium mixture will alter in such a way as to counteract that change’

107
Q

explain, using free energy, why increasing the concentration of the reactants causes equilbrium to move towards the products

A
  • we can imagine a system at equilibrium, this would be the lowest point on a G against composition graph
  • from this we can consider adding some reactant/ increasing reactant concentration
  • this will move us to the left on the composition against G graph
  • as a result G will increase
  • this means moving back towards equilibrium will become a spontaneous process

the same will occur (but reversed) if the concentration of the the products increases etc.

108
Q

how can we derive (quick approach) the relationship (equation) between temperature and position of equilibrium, what assumption do we have to make for this equation to be useful

A

we know
delta(r)G” = -RTln(k)
and
delta(r)G” = delta(r)H” - Tdelta(r)S”

setting them equal and rearranging gives

ln(k) = -delta(r)H”/R (1/T) + delta(r)S”/R

  • delta(r)H” and delta(r)S” do change slightly with temperature but over a modest temperature range we can assume they’re constant
  • what this tells us is that how the position of equilibrium changes with temp depends on the sign of delta(r)H”
109
Q

how can we use our quick approach derivation of the effect of temperature on K to determine how temperature affects the position of equilibrium/link to le chatalier’s

A

we know (from the quick approach)

ln(k) = -delta(r)H”/R (1/T) + delta(r)S”/R

this means:
- if delta(r)H” > 0 then the reaction is endothermic, so increasing T makes -delta(r)H”/RT<0 less -ve so ln(k) and thus K increase with increasing T

  • if delta(r)H” < 0 then the reaction is exothermic so increasing T makes -delta(r)H”/RT>0 less +ve so ln(k) and thus k decrease with increasing T

this is all as expected given what we know about le chatalier’s

110
Q

why can we not use heat measurements to calculate delta(r)H” for reactions which don’t completely go to products

A
  • deltaH is equal to the heat absorbed at constant pressure
  • so delta(r)H” is the heat absorbed when the reactants, in their standard states, are completely converted to the reactants, in their standard states
  • so we can’t calculate it for equilibrium reactions as they never COMPELTELY go to their products
111
Q

using the quick derivation of how ln(k) varies with temp, what would a graph of ln(k) against 1/T show?

A

we know
ln(k) = -delta(r)H”/R (1/T) + delta(r)S”/R

thus, for a ln(k) (on Y) against (1/T) (on x) graph:
– grad = -delta(r)H”/R
– Y-int = delta(r)S”/R (watch for this, it would be very extrapolated so probably not very accurate)

112
Q

how can we use the Gibbs-Helmholtz equation to derive the van’t Hoff equation, link it to the quick derivation of variation of ln(k) with temp.

A

the Gibbs-Helmholtz equation states:

d/dT (G/T) = -H/T^2
d/dT (delta(r)G”/T) = -delta(r)H”/T^2
d/dT(-Rln(k)) = -delta(r)H”/ T^2
dln(K)/dT = delta(r)H”/RT^2

this is the van’t Hoff eq. :
- it says that if delta(r)H”>0 then dln(k)/dT > 0 so ln(k) (so k also) will increase with temperature (as expected from previously)
- and vice versa for delta(r)H”<0

113
Q

how can we use the van’t Hoff eq. to derive an explicit equation for ln(k) at one temperature in terms of delta(r)H” and ln(k) at another temperature

A

dln(k)/dT = delta(r)H”/RT^2
dln(K) = delta(r)H”/RT^2 dT
ln(k) = delta(r)H”/R -1/T

ln(k(T2)) = ln(k(T1)) - delta(r)H”/R [1/T2 - 1/T1]

NOTE: we have assumed delta(r)H” to be constant over the temp range

114
Q

what other expression can be derived from the van’t Hoff equation that confirms the graph of ln(k) against 1/T

A

the van’t Hoff eq. is
dln(k)/dT = delta(r)H”/RT^2

integrating both sides wrt T gives

ln(k) = -delta(r)H”/R (1/T) + c

(c is some constant of integration)

this confirms that the graph has a gradient of
-delta(r)H”/R

115
Q

what occurs to the position of an equilibrium when pressure changes?
what happens to k?

A
  • the position of equilibrium will move to the side which has fewer moles of gas
  • k, however does not change, in fact the position of equilibrium changes in order to keep k constant
116
Q

how do we know that k is unaffected by pressure

A

delta(r)G” = -RTln(k)

none of these quantities depend on pressure, p

117
Q

what is the degree of dissociation, what does it measure

A
  • the extent to which a reaction has gone to products can be characterised by a parameter α, this is the degree of dissociation
  • where α = 0, there’s no dissociation, where α = 1, there’s full dissociation
  • α can be described as the fraction of A2 (or other) molecules that have dissociated/reacted
118
Q

do a derivation explaining how K does not change with pressure but how α does for the reaction

A2(g) —reversible—> 2A(g)

A

we start with n0 moles of A2, the system is at an equilibrium pressure ptot
amount in moles of A is
2αn0
the amount in moles of A2 is
n0(1-α)
total number of moles is
n0(1+α)

pA2 = n0(1-α)/n0(1+α) ptot = (1-α)/(1+α) ptot
pA = 2n0α / n0(α+1) ptot = 2α/(α+1) ptot

Hence,
k = (pA/p”)^2 / (pA2 / p”)
k = (4α^2 / (1-α^2)) (ptot/p”)

as k doesn’t change we know α must change as ptot is varied
assuming α«1
k = 4α^2 ptot/p”
α = sqrt(kp”/4ptot)

119
Q

using our expression for α in terms of ptot, explain the observations of how reactions change with pressure

A

α = sqrt(kp”/4ptot)
Hence, assuming the reaction is a dissociation of one mol of gas to two we know that when ptot increases, α will decrease

120
Q

how can we think of redox reactions occurring, what can we do to make this energy useful

A
  • we can think of redox reactions as involving the transfer of electrons
    e.g.
    Cu(2+)(aq) + Zn(m) —> Cu(m) + Zn(2+)(aq)
  • we tend to think of this as two electrons moving from the zinc to the copper, just by mixing the components we cannot obtain anything useful so we must set up an electrochemical cell
121
Q

how is an electrochemical cell formed

A
  • an electrochemical cell is formed from two separate half cells connected by a wire
  • at one half cell a species is oxidised, their ions go into solution and the electrons pass round the wire to the other half cell where another species’ ions are taken out of solution and reduced
122
Q

what is the form of energy we can obtain from an electrochemical cell, how can we measure it

A
  • it is electrical energy, it can either occur in the form of electrical current or we can measure the electrical potential that the cell develops between its terminals under the conditions of zero current flow
123
Q

what do we call the reactions that might occur in a half cell/ how would we write the equation for it

A
  • it’s a half equation between the reduced and oxidised form at each half cell
  • the reduced and oxidised form are said to make a couple

e.g.

Cu(2+)(aq) + 2e- —> Cu(m)

124
Q

give the 4 main cell conventions

A

1) the cell is written on paper clearly identifying the LHS and RHS

2) each half cell reaction is written as a reduction

3) having written the LHS and RHS half cell reactions as reductions involving the SAME NUMBER OF ELECTRONS the conventional cell reaction is found by taking
RHS Half eq. - LHS half eq.

4) the cell potential is that of the RHS measured relative to the LHS

125
Q

what is the shorthand way that cells are written as, explain what each part means

A

standard notation uses vertical lines to represent the boundaries between phases with a double line where there is a liquid junction, a comma is used where two species are in the same phase

e.g.

Cu(m) | Cu(2+)(aq) || Zn(2+)(aq) | Zn(m)

126
Q

what is the equation linking the cell potential and delta(r)Gcell for a cell

A

delta(r)Gcell = -nFE

n = number of electrons involved in the balanced chemical equation which represents the cell reaction

F = Faraday’s constant = 96485 Cmol^-1

E = cell potential

127
Q

how can we derive an expression for delta(r)Scell given our expression for delta(r)Gcell, in practice how can we use this to determine delta(r)Scell

A

delta(r)Gcell = -nFE

(delG/delT)p = -S
delta(r)Scell = - (del delta(r)Gcell / delT)p

delta(r)Scell = - (del(-nFE)/delT)p

delta(r)Scell = nF(delE/delT)p

  • measure E over a range of temperatures
  • find gradient
  • substitute with other numbers
128
Q

give the equations for delta(r)Gcell and delta(r)Scell, how can these be used to determine delta(r)Hcell

A

delta(r)Gcell = -nFE
delta(r)Scell = nF(delE/delT)p

we can use
delta(r)Gcell = delta(r)Hcell - Tdelta(r)Scell

129
Q

Why must we use chemical potentials when considering how cell potential changes with conc.’s

A
  • Cell potential changes as the conc.’s of species changes because delta(r)Gcell depends on the chemical potentials of the species present
  • chemicals potentials change with conc.’s
130
Q

give the equations for how the chemical potentials of gases and ideal solutions change with conc. (or partial pressure)

explain why this doesn’t work for ionic solutions, what do we use instead

A

The chemical potential of a gas varies with partial pressure according to
μi(pi) = μ”i + RTln(pi/p”)

The chemical potential of an ideal solution according to
μi(ci) = μ”i + RTln(ci/c”)

The problem is that ionic solutions are almost never ideal solutions so this expression doesn’t work

Instead we define a new property called activity, a, such that
μi(ai) = μ”i + RTln(ai)

131
Q

what is the definition of activity, why is it not as weird as it seems

A
  • activity is actually defined as the number such that

μi(ai) = μ”i + RTln(ai)

  • this seems weird but the chemical potential can be directly measured so it still works
132
Q

what happens to the activity as the concentration tends to 0

A

as conc. tends to 0

ai —> (ci/c”)

i.e. it starts to act as an ideal solution

133
Q

what is the nernst equation, derive it

A
  • the nernst equation links the chemical potential at one conc./partial pressure to another

consider a cell with reaction

vaA + vbB —> vpP + vqQ

we know from before

delta(r)G = vpμp + vqμq - vaμa - vbμb

we also know how delta(r)G changes with partial pressures

delta(r)G = delta(r)G” + RT ln[ (pp/p”)^vp (pq/p”)^vq / (pa/p”)^va (pb/p”)^vb ]

Hence, replacing partial pressures with activities and using delta(r)G”cell = -nFE”
gives

-nFE = -nFE” + RT ln[ (ap)^vp (aq)^vq / (aa)^va (ab)^vb ]

so

E = E” - RT/nF ln[ (ap)^vp (aq)^vq / (aa)^va (ab)^vb ]

134
Q

what should we do in our questions in terms of activities and solids in the nernst equation

A
  • although not fully accurate, for now we should just replace activities with partial pressures/conc.’s
  • if a solid or pure liquid is present then we omit their chemical potentials
135
Q

how can we do nernst equations for half cell reactions, how can this be used to find the overall potential for the cell

A
  • in exactly the same way as before, write out the equation, substitute into the nernst equation
  • now we have E1/2 as it is the potential of the half cell
    E1/2(RHS) and E1/2(LHS)

E = E1/2(RHS) - E1/2(LHS)

136
Q

what are standard half cell potentials, how are they formed, what is the reference point

A
  • we cannot directly measure half cell potentials so an arbitrary zero point is chosen and everything is referenced against that
  • the arbitrary zero point is chosen to be a hydrogen half cell
  • this consists of H2(g) at a pressure of 1bar in contact with an aqueous solution of H+ ions at unit activity, an inert Pt electrode is used to make the electrical contact
137
Q

where can we find standard half cell potentials

A
  • they are tabulated (at 298K)
138
Q

what is the difference between a spontaneous cell reaction and the conventional cell reaction

A
  • the conventional cell reaction is the reaction as the cell is written down on paper i.e. half reaction(RHS) - half reaction(LHS)
  • the spontaneous reaction is the reaction that would actually occur under the conditions if current were allowed to flow
  • NOTE: as potentials are affected by the conc.’s of involved species, the direction of the spontaneous reaction can be made to change
139
Q

how can we determine if the spontaneous cell reaction is the same or opposite to the conventional cell reaction

A
  • If E >0, delta(r)Gcell < 0 so reaction is spontaneous in same direction as the conventional cell reaction
  • If E<0, delta(r)Gcell > 0 so reaction is spontaneous in opposite direction to conventional cell reaction
140
Q

what is a metal/metal ion half cell, give an example of the reaction and nernst equation

A
  • consists of a metal in contact with a solution of its ions

Ag(+)(aq) + e(-) —–> Ag(m)

E = E”(Ag+, Ag) - RT/F ln(1/a(Ag+))

141
Q

what is a gas/ion half cell, give an example of the reaction and nernst equation

A

These consist of a gas in contact with a solution containing related ions, an inert Pt electrode provides the electrical contact

e.g.
Cl2(g) + 2e(-) —-> 2Cl(-)(aq)
or
2H2O(l) + 2e- —-> 2OH(-)(aq) + H2(g)

E = E”(Cl2, Cl(-)) - RT/2F ln( a(Cl-)^2 / (p(Cl2)/p”))

or

E = E”(H2O, OH-, H+) - RT/2F ln ((a(OH-)^2)( p(H2)/p”))

142
Q

Can solvent molecules/ions be used to balance half equations, how are they considered in the nernst eq.

A

YES, generally they are in their pure form so do not contribute to the nernst eq. e.g. Water

143
Q

What is a Redox half cell, give an example of the reaction and nernst eq.

A

This is where both the reduced and oxidised forms are present in solution

e.g.
MnO4(-)(aq) + 8H(+)(aq) + 5e(-) —> Mn(2+)(aq) + 4H2O(l)

We know we can exclude water from the nernst eq. so we obtain

E = E”(MnO4(-), Mn(2+)) - RT/5F ln(a(Mn2+) / a(MnO4(-)) (a(H+))^8))

144
Q

What is a metal/insoluble salt/anion half cell, give an example of the reaction and nernst eq.

A

These consist of a metal coated with a layer of the insoluble salt formed from the metal and the anion in solution

e.g.

AgCl(s) + e(-) —-> Ag(m) + Cl(-)(aq)

E = E”(AgCl, Ag, Cl(-)) - RT/F ln(aCl-)

145
Q

what is the significance/advantage of metal/insoluble salt/anion half cells

A
  • their potential depends on the conc. of the anions, this is much more convenient than a gas half cell
146
Q

what is a liquid junction, why are they used, what is a common example

A
  • If the two half cell solutions are different then there’s an issue
  • if we want the cell to produce a potential then the two solutions must be in contact but they can’t be mixed or the electrons would not flow round the external circuit
  • so we use a porous material between them which allows contact but prevents rapid mixing, the problem with this is that liquid junction potential can form over it and reduce the potential of the full cell
  • so we actually use a salt bridge, this is a tube containing a conc. solution of a salt solution (usually KCl or KNO3), the ends of the tubes dip into the solutions and this is a better way to reduce liquid junction potential
147
Q

What is the condition for Aox to reduce Bred and why

A

Aox will oxidise Bred if

E”(Aox, Ared) > E”(Box, Bred)

this is because the overall reaction would be

nbAox + naBred —-> nbAred + naBox

hence

E = E”(Aox, Ared) - E”(Box, Bred)

so if our condition is satisfied then E>0 so delta(r)Gcell < 0 and the reaction is spontaneous

148
Q

What is the better way, using electrochemical potentials to work out what will oxidize/reduce what

A

MPBOA - More positive, better oxidising agent

i.e. the greater the half cell potential the better oxidising the species is. This means it is more likely to be reduced.

out of two given equations with two different potentials, the feasible reaction is the one where the higher potential one is reduced and the lower potential one is oxidised

149
Q

why do we not need to worry about how many electrons are involved in a redox when comparing E values

A

consider

delta(r)Gcell = -E/nF
so
E = -delta(r)Gcell/nF

i.e. we can think about potential as a sort of free energy PER ELECTRON

150
Q

does changing the conditions away from standard conditions affect the value of the cell potentials

A
  • yes but not significantly

e.g. in the Zn(2+), Zn system, reducing the conc. of Zn(2+) by a factor of 10 changes the cell potential by -0.03V, so unless two potentials are very similar, we don’t need to worry

151
Q

how can we use the idea of cell potentials to calculate the solubility product of a reaction

A

1) determine your dissociation that the calculation is for, e.g.
AgI(s) —REV.—> Ag(+)(aq) + I(-)(aq)

2) determine Ksp (the solubility product)
Ksp = a(Ag+) * a(I-)

3) write out two half cell reactions which we can use to form our dissociation and find their half cell potentials e.g.
AgI(s) + e(-) —> Ag(m) + I(-)(aq), E” = -0.15V
and
Ag(+)(aq) + e(-) —> Ag(m), E” = +0.8V

4) manipulate these to get the desired dissociation reaction e.g.
1-2 in our case gives
AgI(s) - Ag(+)(aq) —> I(-)(aq)
AgI(s) —-> Ag(+)(aq) + I(-)(aq)

5) determine E” for this reaction using the two from the seperate reactions
e.g. E(1) - E(2) gives
-0.95V

6) use this along with
delta(r)G”cell = -nFE”
and
delta(r)G”cell = -RTln(Ksp)
to calculate anything needed

152
Q

how can we use the idea of cell potentials to calculate the enthalpy of formation of ions

A

1) consider two half cell reactions, one of which must contain the ion of interest
e.g.
H(+)(aq) + e(-) —> 1/2 H2(g)
Ag(+)(aq) + e(-) —-> Ag(m)
we are interested in Ag(+)

2) combining them in the way that it is feasible and calculating the overall potential from their half cell potentials gives a suitable overall eq. e.g.
Ag(m) + H(+)(aq) —-> Ag(+)(aq) + 1/2 H2(g)
E” = - 0.8V

3) calculate delta(r)G for this reaction in the usual way using
delta(r)G”cell = -nFE”

4) also calculate delta(r)G”cell using the enthalpies of formations of the species present, noting that many cancel to 0 such as delta(f)G(H+) = 0 from the half cell

5) from this a comparison can be done between the calculated value using -nFE” and the enthalpies of formation to obtain the necessary enthalpy

153
Q

what is a concentration cell, what can we say about their standard cell potentials

A

a concentration cell has the same electrode on the RHS and LHS but with different concentrations or pressures of the species involved

we know the standard cell potential will always be zero because in the standard potential you consider the potential between the two half cells when they’re both at standard conc., this would be zero as the two sides would be identical

154
Q

knowing what we do about the standard potential of a conc. cell, what can we say about the reaction and hence deduce what the cell potential is

A
  • we can calculate the cell potential at non-standard conditions using the nernst eq.

this is

E = E” - RT/nF ln[ (ap)^vp (aq)^vq / (aa)^va (ab)^vb ]

but we know E” = 0 and we know all reactants will cancel apart from our two different conc solutions, giving

A(high.conc.) —> A(low.conc.) overall

so we get

E = - RT/nF ln[(a(low conc. sol)) / (a(high conc.sol))]

155
Q

when considering changes in deltarH” by plotting a graph using eq. const. what method is best to use

A
  • use the van’t Hoff eq.
  • DON’T just go directly from deltarG” = -RTln(k)
  • it is best to do it this way because it avoids assumptions about deltar(s) being independent of temp and linked to Y-int
  • instead calc. deltarS” by findign deltarG” at a certain temp. by using deltarG” = -RTln(k) and then using the enthalpy value calculated
156
Q

why might an acid dissociating be associated with a DECREASE in entropy

A
  • the charged ions which result tend to decrease disorder/ increase order as they interact with each other in favourable ways

PT IA Chemistry (9 decks)

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