Physical Chemistry

Physical Chemistry •

Physical Ph sical chemistry chemistr is the branch of chemistry that establishes and develops the principles p p of Chemistry y in terms of the underlying concepts of Physics

Physical Chemistry Main book: Atkins’ Physical y Chemistry y 8th edition, Peter Atkins and Julio de Paula Course web site: http://homes.nano.aau.dk/lg/PhysChem2009.htm

1. 2. 3. 4. 5. 6. 6 7.

Schedule Physical transformation of pure substances. Nucleation. Simple mixtures. Activities. Colligative properties. Two components systems and phase diagrams. Phase diagrams g in the material science. Chemical Equilibrium. Equilibrium electrochemistry. Molecular motion in gases and liquids.Diffusion and Conductivity. The rates of chemical reactions Kinetics of complex reactions.

Oxford On-Line Resources • http://www.oup.com/uk/orc/bin/9780198700722/ • e-book:

Lecture 1 13 10 2008 13-10-2008

• Lecture: – – – – –

thermodynamic functions (recap) single component phase diagrams phase transition and phase bondaries surface tension and nucleation new problems

Thermodynamic Functions 2´ 1 H1, S1, U1 etc

H2´, S2´, U2´ etc

3 H3, S3, U3 etc

2 H2, S2, U2 etc

f

•

•

State functions: properties that depend on the current state only and on the way it was prepared ( (e.g. energy and d enthalpy) th l ) Path functions: relate to the preparation of the state (e.g. work and heat produced)

ΔU = U f − U i

ΔU = ∫ dU i

Exact differential f

q=

∫

i , path

d dq

Enthalpies of transition • For a chemical reaction, if we define ”standard” conditions for reagents g and p products the change g in the thermodynamic (state) function will not depend on the path or intermediate steps. p p • Hess’s law: Standard enthalpy (entropy, Gibbs free energy etc.) of an overall reaction is the sum of the enthalpies (entropies, Gibbs free energies etc.) of the individual reaction it mayy be divided into. • Standard state of a substance at a specified temperature is its pure form at 1 bar CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O(l ) ΔH ø = −890kJ / mol 1mol

2mol

Unmixed reagents in std.form

1mol

2mol

Unmixed products in std.form

Hess’s Law: Example Calculate standard reaction enthalpy of combustion of propene if standard reaction enthalpies p for combustion of p propane p and hydrogenetaion of propene into propane are known.

CH 2 = CHCH 3 ( g ) + H 2 → CH 3CH 2CH 3 ( g )

−124kJ / mol

CH 3CH 2CH 3 ( g ) + 5O2 → 3CO2 ( g ) + 4 H 2O (l )

−2220kJ / moll

1 H 2O(l ) → H 2 ( g ) + O2 ( g ) 2 9 CH 2 = CHCH 3 ( g ) + O2 → 3CO2 ( g ) + 3H 2O (l ) 2

+286kJ / mol −2058kJ / mol

Irreversible processes.Clausius inequality • More energy gy flows as work under reversible condition: − dw ≥ − dw, or dw − dw rev

rev

≥0

• On the other hand as internal energy is a state function

dU = dwrev + dq qrev = dw + dq q ⇒ dqrev − dq = dw − dwrev ≥ 0 ⇒ dqrev / T ≥ dq / T ⇒ dS d ≥ dq d /T • In an isolated system y the entropy py cannot decrease spontaneously dS ≥ 0

Helmholtz and Gibbs energy dq ≥0 T

Clausius inequality:

dS −

At constant volume:

dU dS − ≥0 T

At constant pressure:

dS −

Let’s define: •Helmholtz energy •Gibbs energy

dH ≥0 T

TdS − dU ≥ 0

TdS − dH ≥ 0

A = U − TS

G = H − TS

Helmholtz and Gibbs energy • Helmholtz e o e energy: e gy dAT ,V = 0 – At equilibrium reaction would favorable if dAT ,V < 0 meaning that total entropy of the system increases (dS entropy change of the system system, -dU/T – of the surrounding) – The change in Helmholtz energy is equal to the maximum work accompanying the process

Helmholtz and Gibbs energy • Gibbs energy gy (free ( energy): gy) – At constant T and P chemical reactions are spontaneous p in the direction of decreasing g Gibbs energy – Maximum additional ((non-expansion) p ) work is g given by the change in Gibbs energy. – Standard Gibbs energy gy can be defined as: ΔrG 0 = Δr H 0 − T Δr S 0 Further, standard Gibbs energy of formation can be introduced:

ΔfG = 0

∑

νΔ f G −

Products

0

∑

νΔ f G

Reactants

0

Combining the First and Second Law dU = dq + dw For closed system y in the absence of additional work

TdS

− pdV

dU = TdS − pdV

Fundamental equation Can be applied to reversible and irreversible changes

Mathematically:

So,,

⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ ⎟ dS + ⎜ ⎟ dV ⎝ ∂S ⎠V ⎝ ∂V ⎠ S ⎛ ∂U ⎞ ⎜ ⎟ =T ⎝ ∂S ⎠V

⎛ ∂U ⎞ ⎜ ⎟ = −p ⎝ ∂V ⎠ S

Properties of Gibbs energy G = H − TS dG = − SdT + Vdp

⎛ ∂G ⎞ ⎜ ⎟ = −S ⎝ ∂T ⎠ p

⎛ ∂G ⎞ ⎜ ⎟ =V ⎝ ∂p ⎠V

• G decreases with temperature, proportional to entropy

Gibbs-Helmholtz equation

G−H ⎛ ∂G ⎞ = ⎜ ⎟ T ⎝ ∂T ⎠ p

H ⎛ ∂ G⎞ = − ⎜ ⎟ T2 ⎝ ∂T T ⎠ p

Phase diagrams • • • •

Phase – a form of matter that is uniform through in chemical composition and physical state Phase transition – a spontaneous p conversion of one p phase into another Transition temperature – a temperature at which two phases are in equilibrium Metastable phase – thermodynamically unstable phase that persist because the transition is kinetically hindered

Phase boundaries

Phase boundaries

B ili point Boiling i

Boiling temperature – the temperature at which vapour pressure is equal to external pressure

Phase boundaries Supercritical liquid water: 3740C and 218 atm CO2: 310C and 72.9 atm

Water: 273.16K,, 611 Pa

Heating liquid in a closed vessel Applications: Supercritical drying; dry cleaning; supercritical fluid chromatography

Phase boundaries

Water: 273 273.16K 16K and 611 Pa

Melting (freezing) temperature – the temperature at which liquid and solid phases coexist in equilibrium (at given pressure). Melting temperature at 1 atm called normal, normal at 1bar – standard Triple point (T3) - the point at which the three phases coexist.

Phase diagrams: CO2

At atmospheric pressure “dry-ice” forms directly form gas

Phase diagrams: water

Negative slope

Phase diagrams: 4He

Li id li id ttransition Liquid-liquid iti

Stays liquid at 0K

Phase transitions • Chemical potential μ = Gm for a one-component system • At equilibrium chemical potential is the same through the system

Ot e se, if chemical Otherwise, c e ca pote potential t a is s different, we can find a place with higher chemical potential m1 and exchange place at material from there with another p chemical potential m2.

dG = μ2 dn − μ1dn = ( μ 2 − μ1 )dn < 0 Spontaneous process could take place, so the system is out of equilibrium

Phase transitions Temperature dependence

⎛ ∂μ ⎞ ⎜ ⎟ = − Sm ⎝ ∂T ⎠ p

dG = − SdT + Vdp, μ = Gm Melting point vs. applied pressure

⎛ ∂μ ⎞ ⎜ ⎟ = Vm ⎝ ∂p ⎠T

Phase transitions The effect of applied pressure on vapour pressure pressure applied pp to a condensed p phase its vapour p p pressure rises: • when p molecules squeezed and escape as a gas.

p = p* exp(Vm ΔP / RT ) vapour pressure New vapour pressure in the absense of add.pressure

I d d Indeed:

d μ (gas) = d μ (liq) RTdp dp / p = Vm ((liq) q)d dP p

dp RT ∫ = p p*

p* +ΔP

∫

Vm dpp

p*

⎛ p⎞ RT ln ⎜ * ⎟ = Vm ΔP ⎝p ⎠

p = p* exp(Vm ΔP / RT )

p = p* (1 + Vm ΔP / RT )

Location of phase boundaries μα ( p, T ) = μ β ( p, T ) d μ = − S m dT + Vm dp − Sα ,m dT + Vα ,m dp = − S β ,m dT + Vβ ,m dp Cl Clapeyron equation: ti

dp Sα ,m − S β ,m = dT Vα ,m − Vβ ,m

•Exact E t equation ti that th t applies li tto any phase h ttransition iti off pure substance b t

Location of phase boundaries Solid-liquid boundary usually>0

dp Δ fus H = dT T Δ fusV Δ fus H

p

usually>0 and small

T

dT ∫* dp = Δ fusV ∫* T p T

p= p + *

p≈ p + *

Δ fus H

⎛T ⎞ ln ⎜ * ⎟ Δ fusV ⎝ T ⎠

Δ fus H

T −T ) ( V

T Δ fus *

*

T −T* when 1 * T

Location of phase boundaries Liquid-vapour boundary usually>0 and large

dp Δ vap H = dT T Δ vappV

usually>0 and large

If we assume perfect gas and Vm (liq)  Vm (gas)

Δ vap H dp = dT T ( RT / p ) Clausius-Clapeyron equation

d ln p Δ vap H = d dT RT 2 * −χ

p= pe

Δ vap H ⎛ 1 1 ⎞ χ= ⎜ − *⎟ R ⎝T T ⎠

Location of phase boundaries solid-vapour boundary

Δ fus H + Δ vap H dp Δ sub H = = dT T Δ subV T Δ subV

≈ ΔVvap

Steeper slope

Ehrenfest classification of phase transitions ⎛ ∂μ β ⎞ ⎛ ∂μα ⎞ ⎜ ⎟ −⎜ ⎟ = Vβ ,m − Vα ,m = Δ trsV , ∂ ∂ p p ⎠T ⎝ ⎠T ⎝ •

⎛ ∂μ β ⎞ ⎛ ∂μα ⎞ Δ trs H − = − + = Δ = S S S ⎜ ⎟ ⎜ trs β ,m α ,m ⎟ Ttrs ⎝ ∂T ⎠ p ⎝ ∂T ⎠ p

First order transition: first derivatives of the chemical potential with respect to temperature and pressure are discontinuous. Second order phase transition: first derivative of the chemical potential is continuous but the second derivative is discontinuous discontinuous. Though the symmetry changes abruptly at the point of transition, the state of the phase changes continuously Lambda-transition – first derivative continuous but heat capacity is infinite at Ttrs.

• •

I

II

Infinite at transition

Phase Transitions First Order Melting; Vaporization; most of solid state transitions (like “tin pest” b (white) ->a (grey))

Second Order Ferromagnetism (H=0); Superconductivity; Superfluidic Liquid crystals

• Phase transition = change g in symmetry y y Abrupt change, two states coexist

Smooth transition Example: XY-ferromagnet

Order parameter: local magnetization

Overcooling/overheating is possible

Surface Tension •

Physical boundary between phases can posses extra energy, therefore work is required do change the area:

dw = γ dσ

Surface tension

•

Minimization of surface energy often results in formation of a bubble or y Let’s consider force balance in a spherical p cavity: y cavity. dσ = 4π (r + dr ) 2 − 4π r 2 = 8π rdr

dw = 8πγ rdr

F = 8πγ r

F = 4π r 2 p

4π r 2 pin = 4π r 2 pout + 8πγ r

pin = pout + 2γ / r Vapour pressure of a droplet (from Kelvin equation) For water:

p = p* exp(2γ Vm / rRT )

p (r = 0.1μ m) = 1.011 p* p (r = 1nm) = 2.95 p*

Surface tension • Nucleation Seed of the new phase will grow only if Gibbs free energy of the system decreases Significant surface energy has to be created to separate phases at the onset of phase transition

ΔG = −

A

4π r Δμ + 4π r 2γ 3Vm 3

∂ 4π r 2 ΔG = − Δμ + 8π rγ ∂r Vm

B

2γ Vm r = Δμ *

As a result at the phase transition only the nucleation center with infinite radius will grow. All finite nucleation center require finite thermodynamic force: Δμ > 0

Problems To solve in class: •

• •

4.7a: An open vessel containing (a) water, (b) benzene, (c) mercury stands in a laboratoryy measuring g 5.0 m × 5.0 m × 3.0 m at 25°C. What mass of each substance will be found in the air if there is no ventilation? (The vapour pressures are (a) 3.2 kPa, (b) 13.1 kPa, (c) 0.23 Pa.) 4.9a Calculate the melting point of ice under a pressure of 50 bar. Assume that the density of ice under these conditions is approximately 0 0.92 92 g cm–3 and that of liquid water is 1.00 g cm–3. 4.10a What fraction of the enthalpy of vaporization of water is spent on expanding water vapour

To solve at home: • P4.5: Calculate the difference in slope of the chemical potential against pressure for water on either side of (a) normal freezing point (b) normal boiling point point. The -3 -3 densities of ice and water at 0ºC are 0.917 g cm and 1.0 g cm and those of water and water vapour at 100ºC are 0.958 g cm-3 and 0.598 g cm-3, respectively. By how much the chemical potential of water vapour exceed that of liquid water at 1 2 atm and 100ºC 1.2 100ºC. • P4.7: 50 dm3 of dry air is slowly bubbled through a thermally insulated beaker containing 250g of water initially at 25ºC. Calculate the final temperature assuming the vapour pressure of water constant at 3.17kPa, its heat capacityy 75.5 JK-1mol-1 and it behaves as a perfect gas. Assume that the temperature of air is constant. If you need any extra data to solve the problems, search for them in the Data section in the end of the Atkins book!