Thermodynamics

Question 1

mole fraction

Answer 1

mole fraction of component J

amount of J (in moles) divided by the total moles, n of molecules in the mixture

Question 2

partial molar volume

Answer 2

partial molar volume of component J in a mixture

change in volume per mole of J added to a large volume of the mixture

V𝐽 is the tangent (derivative) of the curve describing the total
volume, V of the mixture as a function of the amount of J (in moles)

at constant p, T and amount 𝑛𝑖≠𝐽 of any other component in the mixture

Question 3

change in the total volume V of a mixture due to the addition of
small amounts of A and small amounts of B

Answer 3

V is a state function, only depends on the composition of the mixture.

equation can be used independently of how nA and nB are mixed

Question 4

partial molar volume varies with mixture composition

Answer 4

when composition is altered, the environment experienced by each component (and its interaction with the other components) changes

also varies with p and T

Question 5

extensive property

Answer 5

any property (mass, volume) that depends on the absolute amount of matter

Question 6

intensive property

Answer 6

any property (partial molar quantity/volume, density) that depends on the amount of matter normalised to the sample size

Question 7

partial molar Gibbs energy (chemical potential)

Answer 7

the tangent (derivative) of the curve describing the Gibbs energy as a function of the amount of J (in moles)

at constant p, T and amount 𝑛𝑖≠𝐽 of any other component in the mixture.

Question 8

Gibbs energy for binary mixture

Answer 8

Question 9

fundamental equation of chemical thermodynamics (modified to account for changes in composition)

Answer 9

chemical potential changes with changes to p, T and composition

new term accounts for any additional non-expansion work that the system can
perform due to changes of its composition (changes in 𝑛𝐽)

as a consequence of phase changes/chemical reactions

Question 10

modified equation for binary mixtures

Answer 10

Question 11

Gibbs energy of mixing: the case of ideas gases

Answer 11

gas A and gas B initially occupy different compartments of a container - separated by a partition

their chemical potentials is the molar Gibbs energy of the pure substances, μ = Gm.

μ(p) - chemical potential of an ideal gas

Question 12

Gibbs energy before mixing

Answer 12

Question 13

Gibbs energy after mixing

Answer 13

after the partition is removed

p = pA + pB

pA and pB = gases’ partial pressures in the mixture

Question 14

Gibbs energy of mixing

Answer 14

Question 15

Gibbs energy of mixing in terms of mole fraction

Answer 15

𝑥𝐽 is always <1

hence, ln 𝑥𝐽 < 0 so that ∆mixG < 0 (always negative) for all compositions

confirms that mixing ideal gases is always a spontaneous process at any composition

Question 16

entropy of mixing

Answer 16

ln x < 0, ∆mixS > 0 (always positive) for all compositions

mixing gases increases disorder, greater entropy

Question 17

enthalpy of mixing

Answer 17

Using ∆G = ∆H −T∆S,

we obtain that ∆mixH = 0 for all compositions

(i.e. no interaction exist between molecules of ideal gases).

enthalpy in the system constant, therefore the increase in entropy is what drives the mixing of two ideal gases.

Question 18

Gibbs energy (left) and entropy of mixing (right) of two ideal gases

Answer 18

Question 19

When the liquid and gas phases of pure A are at equilibrium, the chemical potentials of the two phases are identical

Answer 19

Question 20

When the liquid phase also contains solute B, the criterion of phase equilibrium still holds for A

Answer 20

Question 21

combine equations for 𝝁𝑨* and 𝝁𝑨

Answer 21

Question 22

Raoult’s law

Answer 22

partial vapour pressure, pA
mole fraction, 𝑥A
vapour pressure of pure A, pA*

Question 23

physical interpretation of Raoult’s law

Answer 23

when there is solute B, the rate at which the solvent A evaporates decreases without altering the rate at which it condenses to the liquid phase from the gas phase

solute molecules (small, black) prevent solvent molecules (large, blue) at the surface of a solution to pass into the gas phase (dashed arrow) but not the opposite exchange (solid arrows).

Question 24

rates of vaporization and
condensation for A

Answer 24

Vapourisation rate = kxA

Condensation rate = k′pA

k and k′ are proportionality constants

Question 25

vapourisation and condensation rates must coincide at equilibrium

Answer 25

for a pure liquid, xA = 1 therefore, k/k′ = pA*, thus giving Raoult’s law.

Question 26

combine Raoult's law for A and B with Dalton’s law of partial pressures to construct vapour-pressure diagram for the mixture

Answer 26

Question 27

ideal solutions (or ideal mixtures)

Answer 27

mixtures that follow Raoult’s law at all compositions substituting Raoult’s law in the expression of the liquid chemical potential: expression shows that: in a solution, the solvent (A) chemical potential is always reduced by the solute B.

Question 28

Gibbs energy before mixing of two liquids

Answer 28

Question 29

Gibbs energy after mixing of two liquids

Answer 29

Question 30

Gibbs energy of mixing of two liquids

Answer 30

Question 31

entropy of mixing of two liquids

Answer 31

Question 32

enthalpy of mixing of the two liquids

Answer 32

the rise in the system’s entropy is the reason for the mixing, as ∆mixH = 0

Question 33

volume variation due to mixing is null for ideal solutions

Answer 33

Question 34

ideal gas

Answer 34

molecules do not interact

Question 35

ideal solution

Answer 35

molecules interact A-B, A-A and B-B interactions are all equivalent energetically

Question 36

phase-diagrams of binary mixtures

Answer 36

a: single phase (liquid) b: bubble-point line c: dew-point line, single phase (vapour of same composition as the liquid in a)

Question 37

vapour pressure (fixed temperature)

Answer 37

Using Raoult’s law, express the total vapour pressure p of an ideal binary mixture as: Hence, at a given temperature, p depends linearly on composition.

Question 38

vapour pressure diagrams (fixed temperature)

Answer 38

zA - overall composition xA - mole fraction in liquid yA - mole fraction in vapour total vapour pressure, p of an ideal binary mixture as a function of its overall composition, zA on LHS, points of compositions a and b have the same vapour pressure - represent a mixture with liquid of composition a=xA in equilibrium with gas of composition b=yA all points above the upper line (i.e. high pressure) correspond to points where only the liquid phase is stable (zA = xA) all points below the curve (i.e. low pressure) correspond to points where only the vapour phase is stable (zA = yA) for the points contained between the two curves, the two phases coexist and zA quantifies how much A is present in total in the two phases

Question 39

isopleth

Answer 39

vertical line on vapour pressure diagram - it is a line of constant composition

Question 40

tie line

Answer 40

horizontal line on vapour pressure diagram - connects the compositions of the 2 phases in equilibrium

Question 41

lever rule: focusing on the points in the area between the curves where we have the coexistence of liquid and vapour - a point in this region (green point) gives quantitative information about their relative amounts

Answer 41

α and β are the two phases - nα and nβ are their amounts (in moles)

Question 42

temperature-composition diagrams (fixed pressure)

Answer 42

important for distillation where pressure rather than temperature is kept constant

Question 43

colligative properties (physical properties)

Answer 43

due to presence of solute B, dilute solutions show colligative properties - these are only determined by how much B there is rather than what B is

Question 44

colligative properties examples

Answer 44

- elevation of the boiling point - depression of the freezing point - osmotic pressure these properties are discussed assuming that solute B is: * not volatile - it is not present in the vapour phase * not soluble in the solid solvent this means that the solute only influences the chemical potential of the liquid phase (and not the solid or vapour phases)

Question 45

origin of all colligative properties

Answer 45

attributed to the fact that, in a solution, a solute B always reduces the chemical potential of the liquid solvent A both freezing point and boiling point shift.

Question 46

elevation of boiling point - consider eqm between solvent’s liquid and vapour phases

Answer 46

A = solvent B = solute - only contributes to the chemical potential of the liquid phase as it is non-volatile --- T = the mixture’s boiling temperature (increases) Tb = boiling temperature of pure A ∆vapH = vaporisation enthalpy of pure A. equation shows that solute B induces an increase in the boiling point from Tb to Tb + ∆T, proportional to its mole fraction xB

Question 47

depression of freezing point - consider eqm between solvent's liquid and solid phases

Answer 47

A = solvent B = solute - only contributes to the chemical potential of the liquid phase due to its insolubility in the solid phase --- T = the mixtures freezing temperature (decreases) Tf = freezing temperature of pure A ∆fusH = fusion enthalpy of pure A

Question 48

osmosis

Answer 48

the net unprompted movement of solvent molecules through a semipermeable membrane into a region of higher solute concentration

Question 49

semipermeable membrane

Answer 49

a membrane that allows only solvent and not solute molecules to pass

Question 50

osmotic pressure Π

Answer 50

the pressure that, if applied to region of higher solute concentration, will oppose osmosis Π stems from the equilibrium between pure solvent A (at pressure p) and A in an ideal solution (at higher pressure p + Π) at both sides of semipermeable membrane

Question 51

importance of osmosis

Answer 51

* measurements based on osmosis (osmometry) - used to measure the molar masses of macromolecules. * osmosis helps cells keep their shape. * osmosis is used in dialysis - the purification of solutions. * reverse osmosis is applied to desalinate seawater.

Question 52

osmosis figure

Answer 52

solvent A diffuses from the left of the semipermeable membrane (low solute concentration) to its right (high solute concentration) to dilute the solution and establish equilibrium. this movement can be halted by a pressure Π being applied to the compartment on the right ∴ at equilibrium, there is no net mass transport - the concentrations on both sides of the membrane do not change

Question 53

chemical potential of A at equilibrium

Answer 53

must coincide at both sides of the membrane - although presence of B would lower chemical potential of A, this effect is counterbalanced by the increased pressure Vm = the molar volume of A

Question 54

assume solution is dilute - amount of solute B is very small compared to solvent A

Answer 54

for a dilute solution, since ln xA = ln(1 − xB) ≈ −xB and assuming Vm to be constant in the range of integration (which holds for small values of Π), we obtain

Question 55

van ’t Hoff equation

Answer 55

𝜫 = [B]RT [B] = nB/V B molar concentration.

Question 56

osmotic pressure can be measured from hydrostatic pressure

Answer 56

Π = ρgh ρ = density of solution at equilibrium g = gravitational acceleration (9.81 m s-2) h = height of solution in column at equilibrium the spontaneous passage of A from the bottom container to the solution in the upper container generates a difference in hydrostatic pressure. - the osmotic pressure can then be measured from this hydrostatic pressure

Question 57

criterion of spontaneous change

Answer 57

at constant T and p, a system naturally evolves towards a minimum of Gibbs free energy G - this also applies to chemical reactions

Question 58

extent of reaction, ξ - measures by how much the reaction proceeds (units: moles)

Answer 58

equilibrium A ⇌ B If a very small amount ∆ξ of A becomes B, the amount of A will change by ∆nA = −∆ξ the amount of B will change by ∆nB = +∆ξ ___ A variation of ξ by ∆ξ will cause the amount of any reactant/product J to vary by ∆nJ = νJ∆ξ

Question 59

reaction Gibbs energy ∆rG

Answer 59

Since chemical potentials depend on composition, ∆rG also varies with the stage of the reaction (as the composition changes)

Question 60

Gibbs energy vs Extent of reaction

Answer 60

(1) Spontaneous reaction in the forward case for ∆rG < 0 (2) Spontaneous reaction in the the reverse case for ∆rG > 0 (3) equilibrium is reached at the point of zero slope (when ∆rG = 0), where the Gibbs energy has its minimum

Question 61

perfect gas (ideal gas law)

Answer 61

𝑝𝑉 = 𝑛𝑅𝑇 with R = 8.314 J K-1 mol-1 law represents an approximate equation of state for all gases (real gases) - works progressively better to describe a real gas behaviour the lower the pressure (p → 0), i.e. when gas molecules are far apart departures from this law become significant when gas molecules are packed together - high pressure and low temperature * V → 0 when p → ∞ or T → 0; * molecules do not interact; * a gas does not become a liquid or a solid upon cooling.

Question 62

gas expansion

Answer 62

driven by repulsive interactions between molecules

Question 63

gas compression

Answer 63

driven by attractive forces between molecules

Question 64

real gas (virial equation of state)

Answer 64

temperature-dependent coefficients B, C, etc. are known as the second, third, etc. virial coefficients - account for non-ideal behaviour. - the first virial coefficient is always 1

Question 65

real liquid mixtures - positive deviation from Raoult's Law

Answer 65

A-B interactions less favourable than A-A and B-B interactions - causing an increased vapour pressure compared to ideal case A and B do not like each other - go from liquid to vapour phase earlier as they prefer to be further apart in vapour phase. in liquid phase A and B would be forced to be closer together

Question 66

real liquid mixtures - negative deviation from Raoult's Law

Answer 66

A-B interactions more favourable than A-A and B-B interactions - causing a decreased vapour pressure compared to ideal case A and B like each other - go from liquid to vapour phase later as they prefer to be closer together in liquid phase in vapour phase A and B would be forced to be further apart

Question 67

using Raoult's Law for real solutions

Answer 67

can still use Raoult's Law to describe behaviour of solvent A in real solutions (when xA is close to 1)

Question 68

Henry's Law

Answer 68

describes the behaviour of the solute B at low concentrations (i.e. when xB is close to 0) KB, is the slope of the tangent to the experimental vapour pressure curve as a function of mole fraction of B at xB = 0. KB is experimentally determined and has the dimensions of pressure

Question 69

ideal-dilute solutions

Answer 69

solutions where the solvent and the solution follow Raoult’s law and Henry’s law, respectively

Question 70

ideal-dilute solution diagram

Answer 70

small amount of solute (B) - B is mainly surrounded by solvent (A) solvent molecules are mainly surrounded by other molecules of the same nature, whilst solute molecules are not - small amount of B in a lot of A Henry Law: the vapour pressure of an ideal-dilute solution is proportional to the mole fraction xB

Question 71

azeotropes

Answer 71

a max/min can occur in temperature-composition diagram max can appear in presence of favourable A-B interactions (HCl/H2O mixture) - due to favourable interactions, molecules prefer to stay in liquid phase rather than escaping to the gas phase more than for an ideal mixture min can appear in presence of unfavourable A-B interactions (ethanol/water mixture) - due to unfavourable interactions, molecules escape to the gas phase faster than in the ideal case azeotropes form at these points - vapour and liquid have the same composition

Question 72

negative azeotrope (e.g. hydrochloric acid and water)

Answer 72

a high boiling point c denotes azeotropic composition ~ vapour and liquid have the same composition

Question 73

implication of azeotropes on distillation

Answer 73

the composition at max/min is reached during distillation - evaporation of an azeotrope takes place without a change in composition - therefore, distillation can no longer separate the two components beyond this composition (point c)

Question 74

positive azeotrope (e.g. ethanol and water)

Answer 74

a low boiling point c denotes azeotropic composition ~ vapour and liquid have the same composition

Question 75

activity - effective mole fraction

Answer 75

contains the information of how the liquid deviates from ideal behaviour (assume ideality of liquid phase) The activity of a liquid in a mixture can be obtained in experiments by measuring its vapour pressure at all compositions

Question 76

activity coefficient, γ

Answer 76

γ<1 for favourable A-B interactions γ>1 for unfavourable A-B interactions activity coefficient changes with T, p and mole fraction (tending to 1 when 𝑥𝐴 tends to 1). only for ideal mixtures, 𝛾𝐴 = 1 at all temperatures, pressures and compositions

Question 77

equation relating acitivity, α to activity coefficient, γ

Answer 77

Question 78

colligative properties with non-ideal solutions: osmosis

Answer 78

assume that the solute B is not volatile or soluble in the solid phase so that it only influences the liquid chemical potential experimentally determined constant B = the osmotic virial coefficient osmosis is often used to measure molar masses of macromolecules. - solutions of these big molecules are hardly ideal so use 𝛱 = 𝜌𝑔ℎ

Question 79

excess functions, X^E

Answer 79

for real mixtures quantify the deviation of the measured value of a thermodynamic function of mixing from that of the corresponding function expected assuming ideal behaviour.

Question 80

excess Gibbs free energy

Answer 80

Question 81

excess Gibbs free energy of mixing

Answer 81

Question 82

excess entropy

Answer 82

Question 83

excess enthalpy and excess volume

Answer 83

directly defined by the measured enthalpy and volume of mixing

Question 84

excess free Gibbs energy can be measured by:

Answer 84

1. Measuring the vapour pressures of A/B mixtures, e.g. for chloroform/acetone 2. Calculating their activities 3. Calculating the activity coefficients 4. Calculating G^E In this case, for example, mixing acetone and chloroform is more favourable than mixing an ideal mixture.