Chemistry Reaction and Synthesis Practice Questions

CHEM20018 Reactions and SynthesisTHE UNIVERSITY OF MELBOURNE
School of Chemistry
2021 Semester 1 Assessment
Subject Number: CHEM20018
Reading Time:
Subject Title: Reactions and Synthesis
15 minutes
Writing Time:
3 hours
This paper has 19 pages.
Authorised materials:
Non-programmable, non-graphical calculators
Unassembled model kits
General Instructions to Students: Students are advised to attempt all questions in each of
the sections in the paper. Note that the sections are of different length. The time (mark)
allocation for each section is indicated at the start of each section.
– Include a scan/photo of your student ID with your upload to Gradescope.
– Your answers can be either handwritten or prepared in an electronic document (or a
combination of both).
– Write at the top of each page your University of Melbourne student ID number and the
question number that you are answering.
– Convert the page/s to a .pdf document and upload to the appropriate question on Gradescope.
Either scan the page using a suitable camera/tablet app or save electronic documents in .pdf
format.
– If you encounter problems during this process email your answers as a .pdf document to the
subject coordinator (pauld@unimelb.edu.au) as close to the exam finishing time as possible. In
a subsequent email explain as clearly as possible the full details of the technical problem.
General Instructions to Students:
Students are advised to attempt questions from the first four sections (A—D) together
with EITHER section E, Chemistry of Materials, or section F, Biological Organic
Chemistry of the paper. Your answers from only one of sections E or F contribute to
your final grade. The marks allocation for each question/section will be in proportion to
the time allocation.
Paper to be held by Baillieu Library – NO
Page 1 of 19
CHEM20018 Reactions and Synthesis
Physical Constants:
Avogadro’s constant, NA = 6.0221367(36) ´ 1023 mol-1
Boltzmann constant, k = 1.380658(19) ´ 10-23 L mol-1
Faraday constant, F = 96 485.309(29) C mol-1
Gas constant, R = 8.314510(70) J K-1 mol-1
Planck constant, h = 6.6260755(40) ´ 10-34 J s
Charge of an electron, e = 1.602 ´ 10-19 C
Vacuum Permittivity, eo = 8.854 ´ 10-12 F m-1
Relative dielectric constant of water = 78.5
Conversion factors:
1 atm = 101 325 N m-2 (note that 1 Pa = 1 N m-2)
RT/F = 0.02569 V and 2.303RT/F = 0.0592 V, at 298.15 K
1 electron volt (eV) = 1.602 ´ 10-19 J
Thermodynamic relations:
dU = dq + dw
dS ≥ dq / T
H = U + PV
G = H – TS
dG = VdP – SdT
DG° = – RT ln(K) = – nFE°
p V = nRT
µ = µ° + RT ln(c/c°)
DrG = DrG° + RT lnQ
0 °C = 273.15 K
! = – ∫ &d(
Madelung constants
Structure type
Sodium chloride (NaCl)
Caesium chloride (CsCl)
Wurtzite (a-ZnS)
Zinc blende (b-ZnS)
A
1.7476
1.7627
1.6413
1.6381
Structure type
Fluorite (CaF2)
Rutile (TiO2)
Cadmium iodide (CdI2)
A
2.5194
2.408
2.355
Page 2 of 19
CHEM20018 Reactions and Synthesis
Section A — Synthesis: Organic Compounds (Prof J. White)
Total time = 45 minutes
Answer ALL questions (1 – 3) in this section.
Question 1
[time = 4 + 6 +5 = 15 minutes]
(a)
Draw all possible enol or equivalent tautomeric forms for the two structures shown
immediately below:
O
O
O
O
(b)
Outline a one-step synthesis for each of the following two compounds from simpler
carbonyl-containing precursors and any other necessary reagents. Mechanistic
detail is not required.
O
OH
O
O
OCH2CH3
CH3O
H
OCH3
CH3
(c)
Using common reagents as needed, show how the following chemical conversion
can be carried out. This will require more than one step and will need to employ
carbon-containing precursors additional to the one shown. Mechanistic detail is not
required.
OCH3
O
O
O
CH2
H 3C
OCH3
H 3C
CH2CH2CH2CH3
Page 3 of 19
CHEM20018 Reactions and Synthesis
Question 2
[time = 2.5 + 2.5 + 2.5 + 2.5 + 2.5 + 2.5 = 15 minutes]
Write a structure for the major product expected from each reaction in the following parts (a)
– (f) below. Stereochemical considerations and mechanistic detail are not required.
O
Cl2 (1 equivalent)
(a)
O
(b)
(i) OH- (catalyst)
(ii) H+/H2O
O
O
O
1) CH3OH/CH3ONa (1 eq.)
(c)
H 3C
OCH3
2) PhCH2CH2Br
3) CH3OH/CH3ONa (1 eq.)
4) CH3CH2Br
O
(i) I2 (excess)
OH- (excess)
CH3
(d)
(ii) H+/ H2O
O
(i) LDA (1 equivalent)
CH3
C
H2
(e)
Ph
(ii)
(iii) H+/ H2O
O
O
(i) CH3O- (1 equivalent)
(f)
OCH3
(ii) H+/H2O
Page 4 of 19
CHEM20018 Reactions and Synthesis
Question 3
[time = 5 + 5 + 5 =15 minutes]
(a) Write structures for compounds A and B involved in the following chemical sequence.
Molecule A has the molecular formula C5H8O4. Mechanistic detail is not required.
O
CH3
A +
O
O
(i) CH3O- (1 equivalent)
H3CO
(ii) H+ H2O
CH3
O
HO
(ii) B
CH3
(iii) H+ heat
O
(b)
O
(i) CH3O- (1 equivalent)
OCH3
Write structures for compounds C, D and E in the following chemical conversion.
Hint: Compounds C and D are b-keto esters
C
1) CH3ONa (1 eq.)
in CH3OH/
2) CH3Br
O
D
O
1) CH3ONa (1 eq.)
in CH3OH/
Ph
2) Compound E
3) OH- excess
CH3
4) H+ heat
(c)
Devise a plausible synthesis of the diketone F below (hint it involves two carbonyl
containing precursors).
O
O
Ph
F
End of Section A
Page 5 of 19
CHEM20018 Reactions and Synthesis
Section B — Thermodynamics (Prof T. Smith)
Total time = 30 minutes
Answer ALL questions (4 – 5) in this section.
Question 4
[time = 4 + 3 +3 = 10 minutes]
a) Draw an appropriate diagram and explain why the change in internal energy, △U, for
an adiabatic expansion/compression (changing volume) of a gas can be expressed in
terms of the specific heat capacity at constant volume, Cv.
b) What are the three processes by which energy can be transferred in a heat exchange
system?
c) The reaction:
N2(g) + 3H2 (g) → 2NH3 (g)
is spontaneous at room temperature but becomes nonspontaneous at much higher
temperatures. From this fact alone, obtain the signs of ∆Ho and ∆So, assuming that ∆Ho
and ∆So do not change with temperature. Briefly explain your reasoning.
Question 5
[time = 3 + 3 + 3 + 3 + 3 + 5 = 20 minutes]
a) The thermodynamic Carnot cycle consists of 4 pressure-volume steps for an ideal gas
describing the processes occurring in a heat engine. Are the values for work, heat and
the change in entropy, positive, negative or equal to zero for the total Carnot cycle?
b) 3 moles of an ideal gas expand at 298K against an external pressure of 1 atm. from an
initial volume of 120 mL to a final volume of 570 mL. In one experiment the expansion
is carried out irreversibly and in a second measurement the expansion is performed as a
reversible process. Calculate the difference in the work done between the reversible
and irreversible expansions.
c) Calculate the final temperature, the work and the change in internal energy when 0.6
moles of neon, initially at 273 K, undergoes a reversible adiabatic expansion from 0.3
dm3 to 0.75 dm3.
d) Calculate ΔS for the system 10g of ice initially at 0 oC heated to 40 oC at 1 atm. (given
ΔfusHɵ(H2O) = 6.0 kJmol-1 and Cp,m (H2O(l)) = 75.24 JK-1mol-1)
e) Calculate the efficiency of a heat engine having a hot reservoir at 90 oC and a cold
reservoir of 5 oC. How can the efficiency of a heat engine be improved?
Continued….
Page 6 of 19
nonspontaneous at higher temperatures means that ∆S Eg.
α ~0,E < Eg. Absorbance Band gap VB Wavelength /nm A Typical Insulator-ZnO Consider ZnO: Its band gap is 3.7eV, corresponding to light with a wavelength of ~380nm. This is in the UV. Hence no visible light can excite the crystal. All visible light passes through and the material is colourless if the crystal is pure. A good crystal will allow complete transmission. White means there is surface roughness that the incident light is being reflected or scattered incoherently. Q. Are the band gaps of glass and chalk 400nm? ZnO is used in suncreams. In older formulations suncream is white, but modern formulations use much smaller particles and the suncream can be colourless. End of Lecture 6 • Summary • There are three basic types of materials – insulator, semiconductor, metal. • They arise from bands being filled, half filled or empty. (dband metals are more complex). • The bandgap determines the colour of the material. Materials with bands tend to have very high absorption coefficients for all energies above the band gap and to be more or less transparent below the bandgap. Please read up on band structure!! Egg Carton Analogy Conduction band Band gap Valence band Lecture 7 – More on Band Structure, Energy Levels and Doping Goals (Lecture 7): • • • • To learn about doping of crystals. To link different forms of spectroscopy to the band structure. To understand how we establish energy levels in solids. Recognise different recombination pathways for charge carriers. Semiconductor, Metal and Insulator Colours (Rehash) In molecules, the electronic structure is broken up into molecular orbitals. There are two electrons in each orbital, and the basic rule for absorption is E1- E2 = hv The same basic rule applies to crystals but now, the absorption is between bands. Eg Q. What colour is this material? Yellow! Your eye perceives a mixture of red and green light as yellow. The blue is absorbed so red and green are reflected. Molecular vs Band Absorption Spectra • A simple molecule has a HOMO and LUMO. The difference in energy is E=hv = ELUMO-EHOMO.
• Light is only absorbed if it matches this energy. The spectrum
is a single sharp line. (A real molecule has many such sharp
lines. They are also broadened by vibrations.)
LUMO
HOMO
Absorbance
E=hv = ELUMO-EHOMO
Wavelength /nm
Band Absorption Spectra
• The density of states (filled and empty orbitals is high and
increases above the band gap. There are many ways for
transitions to occur between a filled and empty state.
• Absorption is very high at ALL wavelengths above the band
gap.
• Below the band gap, absorption falls close to zero very quickly.
Higher
bands
CB
α = A(E − Eg)3/2 ,E > Eg.
α ~0,E < Eg. Absorbance Band gap, Eg VB Wavelength /nm Egg Carton Analogy Conduction band Photoexcitation creates Both an electron and a hole. Band gap Surface trap Valence band A Typical Insulator-ZnO Consider ZnO: Its band gap is 3.7eV, corresponding to light with a wavelength of ~380nm. This is in the UV. Hence no visible light can excite the crystal. All visible light passes through and the material is colourless if the crystal is pure. A good crystal will allow complete transmission. White means there is surface roughness that the incident light is being reflected or scattered incoherently. Q. Are the band gaps of glass and chalk 400nm? ZnO is used in suncreams. In older formulations suncream is white, but modern formulations use much smaller particles and the suncream can be colourless. Semiconductors – Excess Energy is Lost • Many semiconductors have very high absorption coefficients. • Note: excess energy above the band gap is lost as heat (fs-ps). CB VB For every photon, all excess energy is lost above the bandgap. So for example if the band gap is 1.7eV (700nm), then a photon with energy of 3eV, (400nm) is losing 1.3eV or 42%. Semiconductors – Recombination To get back to the ground state, the electrons and holes can recombine through several different processes. Right: Electrons at the bottom of the CB lose energy by non-radiative pathways (surface states, defects, recombination centres). A few materials will lose the energy as fluorescence. recombination Centre (donor or acceptor) Surface states Dangling bonds Band edge PL Crimson indicates photon released Red indicates non-radiative (heat, vibrational loss). Energy Levels in Solids cf Vacuum Physicists usually use vacuum as the energy reference. An isolated electron at infinity is defined to have zero energy. Chemists use the hydrogen electrode as a common reference. E∞vac IP = Φ m E.A E.A Ec Ef The Fermi energy is IP Ef Hatched means An empty band. i.e. allowed but empty Levels. White means nonallowed energy levels. Φ sc Metal Semiconductor Solid orange means filled. Hashed line is unfilled energy levels equivalent to the “redox potential” in solution phase and defines the mean electron energy. It is the energy in the solid at which Ev the probability of occupation is ½. Energy Level Spectroscopy (a) Near Edge X-ray Absorption Fine Structure (b) X-Ray Photoelectron Spectroscopy (c )X-Ray Emission Spectroscopy (d) Ultraviolet Photoelectron Spectroscopy p-type and n-type doping Band Edges for Common Semiconductors E (ev) vs Vacuum Courtesy Lu et al. Adv. Mater. 2015, DOI: 10.1002/adma.201503270 Doping Silicon with Phosphorus Left Hand Side: Photoconductivity In pure silicon Right Hand Side: Electrical conductivity In P-doped silicon Common Dopants in Silicon Gp 5 elements lead to n-type donor levels; Gp 3 dopants produce p-type acceptor levels End of Lecture 7 Summary • In an intrinsic or pure semiconductor, electrons and holes are created in equal numbers, either thermally or by light. • Electrons and holes recombine mostly non-radiatively (otherwise the world around us would fluoresce!). • In most materials, there are bulk traps and defects or surface traps and defects that make recombination occur in ps-ns. • We can measure the energy levels of a crystal on the vacuum scale to do quantitative measurements and for device fabrication. • The Fermi level plays a similar role to chemical potential or Eo in solution chemistry. Lecture 8 – Conductivity in Semiconductors Goals • Quantify the links between doping and electrical conductivity. • Quantify the effects of temperature on conductivity. Lecture 8 – Conductivity Measurements 1. When an electric field is applied to a macroscopic material, a current can flow due to both ions (ionic conduction) or the motion of electrons and holes (electronic conduction). 2. From Ohm’s Law (V=IR), we can readily derive an equation for the RA resistivity: ρ= l 1 3.Sometimes we use the conductivity σ = ρ 4. An alternative way to express Ohm’s Law is J = σ E 5. When a field is applied, there is a steady state current J=nev. Defining the mobility as µ = v / E yields an expression for the conductivity σ = n e µ Check this using : J=I/A; E = V/l, sigma= l/RA RT Metal Electrical Conductivities Metal s 1/(Ohm-m) Silver 6.8x107 Copper 6.0x107 Gold 4.3x107 Aluminium 3.8x107 Iron 1.0x107 Brass 1.6x107 Platinum 0.94x107 Carbon Steel 0.6x107 Stainless Steel 0.2x107 What happens at contact? • When two materials touch, the electrons try to move from the phase with higher Fermi energy to the one with lower Fermi energy. For metals, a voltage called the contact potential difference forms at the surface. CPD for Metals Symbol Metal Work Function (eV) Ag silver 4.7 Au gold 4.8 Ca calcium 3.2 Cu copper 4.1 K potassium 2.1 Mn manganese 3.8 Na sodium 2.3 Ni nickel 5.2 Q: The threshold wavelength for the photoelectric effect is 271 nm for tungsten and 262 nm for silver. What is the contact potential developed when silver and tungsten are placed in contact? A. The photon wavelengths correspond to work functions of 4.58eV and 4.71eV, so the CPD = 0.13eV/e = 0.13V Aside: The Boltzmann Distribution In science, a Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. The distribution is expressed in the form: !! ∝ # "#! /%& where pi is the probability of the system being in state i, Ei is the energy of that state, and a constant kT of the distribution is the product of Boltzmann's constant k and thermodynamic temperature T. The term system here has a very wide meaning; it can range from a single atom to a macroscopic system such as a natural gas storage tank! Because of this the Boltzmann distribution can be used to solve a very wide variety of problems. The distribution shows that states with lower energy will always have a higher probability of being occupied. The ratio of probabilities of two states is known as the Boltzmann factor and characteristically only depends on the states' energy difference: !! ∝ # #(%! #%" )/() !" Courtesy - wikipedia n.p = Constant! • Using the Boltzmann distribution it can be shown that ⎛ −E g ⎞ n = p = (N v Nc ) exp ⎜ ⎟ = ni ⎝ 2kT ⎠ 1/2 The product of the concentration of electrons and holes is a constant, irrespective of doping., Nv and Nc correspond to roughly one orbital per atom so the density or concentration is similar to the number of atoms per cm3. cf H2O: H2O + H2O ! H3O + + OH − ⎡ H3O + ⎤ ⎡OH − ⎤ ⎦⎣ ⎦ = exp(−ΔG / RT ) Kw = ⎣ ⎡⎣ H2O ⎤⎦ ⎡⎣ H2O ⎤⎦ ( [H ]= [OH ]= ⎡⎣ H2O ⎤⎦ ⎡⎣ H2O ⎤⎦ + − ) exp(−ΔG /2RT ) 1/2 Problems Question: How many electrons are in the conduction band of pure Si at 400K if Eg = 1.1eV, Nc = Nv = 5x1019 cm-3? Answer : Eg = 1.1eV = 1.1 x 1.6 x 10-19 J = 1.76 x 10-19 J. kT = 1.38 x 10-23 x 400 = 5.5 x 10-21 J. −1.76x10 −19 12 −3 n = 5x10 x exp[ ] = 6x10 cm 2 * 400 *1.38x10 −23 19 Question: What is the conductivity of pure Si at 400K if the mobility of the electron and hole is 1500 and 450 cm2 V -1 s-1? Answer: σ = neµe + peµh = 6x1012 cm −3 *1.6 *10 −19 C *(1500 + 450)cm 2V −1s −1 = 0.00187CV −1s −1cm −1 Question: What current density flows if a field of 10 V/cm is applied to such a sample? Answer: j(Ccm −2 s −1 ) = σ E = 0.00187CV −1s −1cm −1 *10Vcm −1 = 0.0187Ccm −2 s −1 = 0.0187Acm −2 . Semiconductor Conductivities Conductivity of semiconductors is due to both types of charge carriers. The relative importance of each depends on the doping. σ = n e µe + p e µh Question: For intrinsic silicon, Room Temp conductivity is 4 x 10-4 (Ohm-m)-1, and the electron and hole mobilities are 0.14 and 0.048 m2/Vs. What are the electron and hole concentrations at room temperature? Answer: Intrinsic materials have n=p. σ 4 × 10 −4 (Ω − m)−1 16 −3 n= p= = = 1.33 × 10 m e ( µe + µh ) 1.6 × 10 −19 C × (0.14 + 0.048)m 2 / Vs Doping silicon with phosphorus Question: Phosphorus is used to dope silicon with 1023 m-3 P atoms. Calculate the conductivity of the sample. Answer: The intrinsic doping level of Si is ~1018m-3. So the dopant dominates the conductivity. Use: σ = n e µe = 10 23 × 1.6 × 10 −19 C × 0.14m 2 / Vs = 2240(Ω − m)−1 Although the mobility can drop with increasing temperature, increasing temperature boosts the thermal population of intrinsic carriers and also helps to ionise deep donors /acceptors. ln σ = C − Eg 2kT Temperature Dependence of Ge • Question: If the room temperature (298K) conductivity of intrinsic Ge is 2.2 (Ohm-m)-1, estimate its conductivity at 423K. • Answer: First we need to find C. We need the band gap for Ge (0.67eV). Eg 0.67 × 1.602 × 10 −19 J C = ln σ + = ln(2.2) + = 13.83 −23 2kT 2 *1.38 × 10 × 298 • Now at 423K Eg 0.67 × 1.602 × 10 −19 J ln σ = C − = 13.83 − = 4.64 −23 2kT 2 *1.38 × 10 × 423 • Hence conductivity is 104 (Ohm-m)-1. Questions to think about • Q1. Is a silver atom metallic? • Q2. What colour do you think silver atoms might be? • Q3. Is a heavily doped semiconductor a metal? How to make transparent metals? CB Sn(IV) 3.1eV ITO : 50% of all the world’s indium goes Into laptops and LCD screens. The price of Indium has gone up a factor of 10 in the last decades as supplies have dwindled. VB * In2O3 - Eg >> kT (0.025eV) so low intrinsic conductivity.
* It is doped with Sn(IV), a shallow donor.
* But due to high band gap is transparent to visible light.
* ITO: 8% Sn(IV) in Indium(III) Oxide
* Sputter Coated, CVD or by Sol-gel dip-coating from InCl3 / SnCl4 mixtures.
* Some absorbance by free electrons.
* Slight metallic shine, ! > 0.1S/m.
What might be some other options??
20018 Lecture 9:
Solar Cells and New Materials
Some Current Research Directions
Outline
Goals- To understand the basic way that a p-n junction works.
– To understand the basic way a solar cell works and the limits
placed on them by band structure.
– To exemplify the increasing role of hybrid materials with
electrochromic cells.
– Review the learning goals of Lectures 1-9.
Formation of a Depletion Layer
(a) Two metals prior to electrical contact.
(b) After contact, here through an external wire, electrons flow from the low work
function to high work function metal. The build up of charge changes the work function
until the two metals are in equilibrium.
(c) One can apply a voltage to remove the electric field or alter the Fermi level
difference between the metals.
Note: The long line is the positive terminal of a battery.
Formation of a Depletion Layer or pn Junction
Unlike metals, the doping density in a semiconductor is quite small. So to build up the contact
potential difference (VB), the charges come from a finite distance inside the semiconductors,
sometimes 100nm in depth.
Basic Si Bulk Junction Solar Cell
thickness
x1
x2
The free electrons in the n-type region neutralise holes in the positive region. The number
must be equal, so Q= Nd A x1 = NA A x2 where A is the cross-sectional area.
Challenge – Can we do better than silicon?
6
Optimal Band Gap for Photovoltaics
The ideal bandgap for a
single junction PV device
operating at AM 1.5 and
around 1- 10 suns is 1.4eV.
The materials should be
highly absorbing and ideally
should be able to be doped
over a wide concentration
range.
Current leaders are Si, CdTe,
CuInS2 while the best multijunction cells use
GaAs/GaInP/Ge.
See Boeing Spectrolab – efficiency is 41% but cost is C. US$70,000 /m2, compared with ~$500
for a 22% commercial Si panel.
The Future – Flexible Electronics
• Need materials that can handle
bending without cracking.
• Need to process < 180oC • Need to be able to process through printing and solution based processes such as dipcoating, spin coating etc. CSIRO-UM-Monash working on organic conducting polymers for use in flexible solar cells. (David Jones and Wallace Wong) Image Courtesy of CSIRO Power windows – To Absorb or Not to Absorb Perovskite nanocrystals eg CsPbI3 offer photostable PV cells which can be printed onto ITO glass. The concentration can be altered to harvest different amounts of light. These appear to be better than OPVs at present. Image Courtesy of Monash University Photochemical water splitting Australia wants to position itself as an energy superpower, converting solar energy into green hydrogen for export. Question: What is the minimum energy needed to split water into H2 and O2? Answer: Alternative reaction 2H+ + 2 e- è H2 Eo=0.0V vs NHE 2H2O è O2 + 4H+ + 4e- Eo=1.23V vs NHE 4 h+ (vb) + 2H2O è O2 + 4H+ Comparing Vacuum and Solution Energy Levels Tamirat et al.,”Using hematite for photoelectrochemical water splitting: a review of current progress and challenges”, Nanoscale Horizons 2016. Challenge – Harnessing the NIR Spectrum 12 Upconversion Courtesy Timothy Schmidt UNSW The “Grätzel Cell”- Dye Sensitized Titania The “Artificial Leaf” is the motivation for many chemical systems for harnessing solar energy. You will make these in 30015 Chem Prac! How Does it Work? S + hv → S* S* + TiO2 → S+ + e _ (cb) S+ + 3I_ → S + I3_ e _ (cb) → e _ (Pt) + Work I3_ + e _ (Pt) → 3I_ Nanocrystals of Titania Make it Possible The titania particles are about 10nm in size and crystalline. The nanoporous electrode is ~ 1micron thick. The roughness factor is about 800. i.e. the actual surface area of a 1cm2 piece of glass coated with this particle film is 800cm2. Sensitizers These must absorb as much solar radiation as possible,but E(S+/S*) must remain about 200mV above Ecb(TiO2) for efficient injection. Electrochromic Windows • Normally based on WO3, MoO3 or IrO2. These metals all change colour when reduced or oxidized. Protons or cations must diffuse in to neutralize the electronic charge WO 3 + nH + + ne - ! WO 3H n • Made by coating ITO with WO3 colloids using sol-gel processing, but also by electrochemical deposition, thermal evaporation, or CVD. A molecular based electrochromic window Absorption of V+ in solution (black), film (grey) and background spectra with no V+ Boeing Dreamliners use WO3 as the colour changing element in their windows. Learning Outcomes – Bonding • Be able to identify the 3 primary types of chemical bonding. • Be able to identify the 3 secondary types of chemical bonding. • Be able to write down the Lennard-Jones equation. • Be able to differentiate the LJ equation. • Be able to write and use Coulomb’s equation and calculate force from energy and energy from force. • Understand the difference between electric potential and electric force. • Be able to interconvert units. • Reading: Atkins, most Phys Chem Texts. Learning Outcomes – Bonding • Be able to understand the Born-Haber cycle – also done in main part of course! • Be able to calculate d spacing for a simple cubic lattice. • Be able to calculate the first 3 terms of the Madelung constant for a simple cubic lattice. • Be able to calculate the binding energy and bond energy for a Coulomb lattice. • Reading: Callister, "Inorganic Chemistry" (6th Ed.), notes from Brendan Abrahams lectures! Learning Goals – Colour and Band Structure * Be able to explain the origin of band formations in crystals using molecular orbital concepts. • Be able to estimate the band gap of a material from its colour, and decide whether it is likely to be semiconducting, insulating or metallic. Reading: Callister – very good. Philip Hoffman – Solid State Physics Kittel- Introduction to Solid State Physics. Learning Goals – Electrical Properties * Be able to explain how dopants affect the conductivity of a solid, how one can make a material n-type or p-type and which dopants are likely to render a material p-type or n-type. • Be able to estimate the intrinsic conductivity of a material from the density of states and the band gap at temperature T. * Be able to use Ohm’s Law in its different forms, e.g. to be able to calculate conductivities from mobilities (and vice versa) for intrinsic and doped materials.e.g. σ = n e µe + p e µh * Be able to calculate the conductivity of a material at a second temperature given data at the first temperature using: ln σ = C − Reading: Callister Eg 2kT The End! The End! CHEM20018 Weeks 4 and 5 Thermodynamics Prof. TREVOR SMITH Room 269, Chemistry West trevoras@unimelb.edu.au CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 1 Housekeeping • Lecture content outline and (sometimes abbreviated) lecture slides available on LMS before lecture. • Discussion board through Canvas page • Tutorials start this week (week 4) – Face-to-face and Zoom • Problems posted on Canvas before tutorials • Have a go at these yourselves - probably best after lecture 3 • Worked solutions provided after tutorials • Submit queries through discussion board rather than Zoom chat or email CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 3 Library resources • Note: – Library resources – Library guide for students in chemistry is available at: • https://unimelb.libguides.com/chemistry – Science/chemistry librarian: ERC Library • https://chemistry.unimelb.edu.au/people/professional-staff • Recommended Reference Text: “‘Atkins’ Physical Chemistry” – Atkins, de Paula, Keeler, 11th Ed., Ch. 2 & 3 (some copies are available in the Reserve Collection in the ERC Library. “Student's Solution Manual” 11th Ed. (odd numbered questions); 10th Ed solutions also available. Call no. 541.3 ATKI CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 4 Lecture outline CHEM20018 Reactions and Synthesis SEMESTER 1, 2022 Note: Atkins, de Paula, Keeler “Atkins’ Physical Chemistry” 11th edition is the recommended reference text for this subject. 9th, 10th edition chapter references are also indicated. THERMODYNAMICS (weeks 4 and 5) Prof. T. Smith LECTURE CONTENT Lectures 1 & 2: heat, work, first law of thermodynamics, reversible and irreversible expansion/compression of gases. Adiabatic processes, thermochemistry ATKINS’ CHAPTER REFERENCE 2A, 2B (11th, 10th Ed) 2.1, 2.2, 2.3 (9th Ed) 2C, 2E (11th, 10th Ed) 2.6, 2.7, 2.8, 2.9, 2.11 (9th Ed) Lectures 3 & 4: Adiabatic processes, thermochemistry Entropy and the directions of spontaneous change, the second law of thermodynamics, statistical view of entropy, entropy as a state function 2C, 2E (11th,10th Ed) 2.6, 2.7, 2.8, 2.9, 2.11 (9th Ed) 3A (11th, 10th Ed) 3.1, 3.2 (9th Ed) Lectures 5 & 6: the Carnot cycle and state functions, the third law of thermodynamics, 3A, 3B, 3C, 3D (11th, 10th Ed) absolute entropies, free energy 3.2, 3.3, 3.4, 3.5 (9th Ed) Applications of thermodynamics CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 5 Thermodynamics (6 lectures): • Heat, work first law of thermodynamics, reversible versus irreversible processes in terms of expansion/compression of ideal gases, free expansion. • Molecular basis of enthalpy and its temperature dependence, thermochemistry & enthalpy calculations, adiabatic processes, adiabatic expansion/compression of ideal gases Temperature dependence of reaction enthalpies, Entropy and the direction of spontaneous change, statistical mechanical (molecular) view of entropy, entropy and the Second Law of thermodynamics, Entropy as a state function • Entropy change with temperature, entropy change and phase transitions, reaction entropy changes, Third Law of thermodynamics, absolute entropies and absolute zero. Thermodynamic cycles and entropy changes, Carnot heat engines, Gibbs Free energy and spontaneity, standard free energies, reaction free energies CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 6 Lecture 1 • Introduction and review – Heat, work, internal energy – The First Law of Thermodynamics – Irreversible and reversible expansion of gases – cf. Atkins 2A CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 12 Thermodynamics • Thermodynamics means “heat” and “movement” • It is about the interconversion of different forms of energy – Thermodynamics is the quintessential Victorian science – Began and grew with the development of the steam engine concerned with the conversion of heat energy into mechanical energy • In chemistry can deal with the conversion of chemical energy into electrical energy (redox reactions) or into heat (thermochemistry) CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 13 Arnold Sommerfeld • “Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two points.The third time you go through it, you know you don't understand it, but by that time you are so used to the subject, it doesn't bother you anymore..” 1868 – 1951 CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 14 History/Timeline • Industrial Revolution ~1780s → • The nature of heat ????? – “Caloric”, a substance – Movement of molecules/atoms. • The more heat the greater the kinetic energy of the particles The thermodynamicists representative of the original eight founding schools of thermodynamics. CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 15 History/Timeline • Joseph Black ~ 1760 discovery of latent heat – “I have ... put a lump of ice into an equal quantity of water, heated to the temperature 176[°F] (80°C), and the result was, that the fluid was no hotter than water just ready to freeze. Nay, if a little sea salt be added to the water, and it be heated only to 166 (74.4°C) or 170 (76.6°C), we shall produce a fluid sensibly colder than the ice was in the beginning, which has appeared a curious and puzzling thing to those unacquainted with the general fact.” Heating curve for water S.S. Zumdahl, “Chemistry” 2nd Ed. CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 16 Steam engines • Newcomen “atmospheric” steam engine (1712) – operated by condensing steam drawn into the cylinder - creating a partial vacuum which allowed the atmospheric pressure to push the piston into the cylinder. – first practical device to harness steam to produce mechanical work https://en.wikipedia.org/wiki/ Newcomen_atmospheric_engine CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 17 • James Watt ~ 1770 improved on the Newcomen steam engine • Had a more efficient condenser or cold sink – liquid to vapour energy (expansion) CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 18 Thermodynamics and Chemistry • Conversion of chemical energy into – Electrical energy – redox reactions – Heat – thermochemistry • The ‘Thermodynamic ‘Laws’ developed through 19th C by physicists and engineers relating conversion of heat → mechanical energy are universal – they apply to the inter-conversion of all types of energy CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 19 1st Law of Thermodynamics (law of energy conservation) ΔU = q + W • U is the total ‘internal energy’ of a system – the sum of the kinetic and interaction energies of particles in a system – it can be changed by the exchange of heat (q) or work (w) with the surroundings. • ΔU = Uf - Ui – the difference between the final and initial internal energy of a system. • The ‘system’ and its ‘surroundings’ make up the ‘Universe’. CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 20 q: the energy transferred into the system through heating – +ve for heat transferred into the system, – -ve for heat transferred out of the system to the surroundings. w: Work – work done on a system, +ve – work by the system on the surroundings, -ve H: Enthalpy – Heat at constant pressure CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 21 State functions • A property with a value that depends only on the current state of the system & independent of how that state has been prepared. – e.g. internal energy • a function of the variables that determine the current state of the system – e.g. a change in pressure may result in a ΔU CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 22 Extensive properties • A property that depends on the amount of the substance present – e.g. internal energy: U (1J = 1 kg m2 s-2) • Compare with “intensive” properties – e.g. kJ mol-1 CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 23 U, q and w • The internal energy (U) of a system is the sum of the kinetic energy (e.g. particle translational, vibrational, rotational motion) and the potential energy (e.g. particle attraction/repulsion) of the components: • Utotal = Ukinetic + Upotential CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 24 Some definitions: • Isolated system – no exchange of material, heat or work with surroundings • First law: “internal energy of an isolated system is constant” – The energy in a closed system is conserved – neither created or destroyed • e.g. in an isolated, insulated box of fixed volume, q=0, w=0, ΔU = 0. – Convert heat into work & vice versa (doesn’t say anything about efficiency of conversion – 2nd Law) CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 25 Some definitions: • Isothermal process – Temperature remains constant (ΔT = 0). – For an isothermal process ΔU = 0, and thus q = -w. • Adiabatic process – No heat exchange between the system and the surroundings. q = 0 and thus ΔU = w. • (e.g. process in an insulated container). CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 26 Work (w) • Mechanical means Work (w)of energy transfer • Joule’s experiment • Mechanical means of energy transfer • Joule’s experiment 1844-1854 Mathematica animation https://demonstrations.wolfram.com/JoulesExperiment/ (for= an process) • dU = dw ∝ dT • dU dwadiabatic ∝ dT (i.e. temp. is a measure of internal energy). – i.e. temp. is a measure of internal energy CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 27 Crookes Radiometer • “Light mill” • Invented 1873 by the chemist Sir William Crookes • Several explanations put forward Tutorial discussion: what is the accepted explanation for the rotation? CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 28 Lecture 2 • Review of 1st Law of Thermodynamics – reversible expansion – heat and heat capacity – enthalpy and enthalpy change – adiabatic processes cf. Atkins’ 11th Ed. 2A, 2B, 2C CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 29 Work (w) • In chemical reactions we’re often concerned with the generation of gases – these will lead to volume changes of our system (e.g. work will be done by the system on the surroundings when gases are generated). dw = − PexdV w = dw = − PexdV ∫ ∫ • Total work is the integral of the work done when a system expands through a volume dV against an external pressure Pex CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 30 e.g. Types of engines • Two-stroke engine: – http://www.animatedengines.com/twostroke.html • Stirling engine: – http://www.animatedengines.com/vstirling.html – efficient and promising for current-day applications CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 31 Work (w) w = dw = − PexdV ∫ ∫ • The volume of a gas can change either: – Irreversibly (volume change occurs spontaneously) • OR – Reversibly (equilibrium is maintained during the process) CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 32 Irreversible expansion of a gas – expansion against a constant external pressure, work done is pexΔV Irreversible expansion of a gas • Expansion against a constant external pressure, work done is pexΔV Figure 2A.6 Atkins “Physical Chemistry CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne Physical ChemistrFigure 2A.6 Atkins “Physical Chemistry 33 Work (w) • Expansion of a gas against a constant external pressure (IRREVERSIBLE CHANGE) – e.g. Calculate the work done on a system of 2 moles of an ideal gas if it expands irreversibly against an external pressure of 2 atm. Initial Vol = 500mL, final Vol = 1000mL. • (but Pex is a constant with value 2 atm) 1atm = 101325 Pa 1 Pa = 1 kg m-1s-2 1 J = 1 kg m2 s-2 • Note: be consistent with units – pascals and m3 CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 34 Reversible expansion of a gas – pressure and volume changes occur in small increments so they kept in equilibrium. Reversible expansion of are a gas Obtain the maximum amount of work this way • Pressure and volume changes occur in small increments so they are kept in equilibrium. – Obtain the maximum amount of work this way Figure 2A.7 Atkins “Physical Chemistry CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne Physical ChemiFigure 2A.7 Atkins Physical Chemistry 35 Work (w) • Consider the reversible, isothermal expansion of 2 moles of an ideal gas from a volume of 500 mL to 1000 mL (T=298K). • In this case P is not constant but P = nRT/V from ideal gas equation. w = dw = − PdV ∫ ∫ CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 36 Special case: • The work associated with the free expansion of a gas (against zero pressure) is zero. case: i.e. pSpecial ex = 0 dw=-pexdV w=0 • The work associated with the free expansion of a gas (against zero pressure) is zero. i.e. pex = 0 ⇒ dw=-pexdV ⇒ w = 0 CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne 37 Should now know: • First law of thermodynamics - “internal energy of an isolated system is constant” • Internal energy of a system can be altered by the exchange of heat and/or work with the surroundings • Work associated with reversible expansion of a gas > irreversible expansion work
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1st Law of Thermodynamics (law of
energy conservation)
ΔU = q + w
• U is the total ‘internal energy’ of a system
– it can be changed by the exchange of heat (q) or work
(w) with the surroundings.
• From previously:
w = dw = − PdV
∫
∫
= − Pex(Vf − Vi) = − PexΔV (for the irreversible expansion of a gas)
w = − nRT ln
Vf
( Vi )
(for the reversible expansion of a gas)
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Heat (q)
• Supplying heat energy to a system causes a temperature rise
(e.g. for a system of constant volume, where w=0)
dU = dq α dT
dU = CvdT
• The integrated result is:
ΔU = CvΔT
• Cv is the constant volume heat capacity (J K-1)
• ΔT = Tf – Ti (the difference between final and initial temp.)
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Constant volume heat capacity Cv
Physical Chemistry Fundamentals: Figure 2.10
• Assume constant (over a
small T range)
Cv =
ΔU
ΔT
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Enthalpy (H)
Physical Chemistry Fundamentals: Figure 2.12
• However ΔU does not equal the energy
transferred as heat when there is a
change in volume
– some of the energy associated with the
heating is transferred to the surroundings as
work.
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Enthalpy (H)
• Define enthalpy as:
H = U + PV
since
dH = dU +
d(PV)
• For a process where P is constant:
dU = dq + dw
and dw = -PdV
dH = (dq − PdV) + PdV
dH = dq
ΔH = q (if P is constant)
• Enthalpy change is heat exchanged at constant
pressure.
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Enthalpy (H)
• Definition:
ΔH is qp (heat exchanged at constant pressure)
• ΔH is related to ΔU by:
ΔH = ΔU + PΔV
• For an ideal gas PΔV = ΔngRT
– (where Δng is the change in the number of moles of
gas during a reaction) – thus:
ΔH=ΔU+ ΔngRT
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ΔH and ΔU
Consider reaction
• Zn(s) + 2H+(aq) ➛ Zn2+(aq) + H2(g)
ΔH = -154 kJ mol-1 at const. T of 298K.
• What is ΔU for the reaction?
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Temperature Dependence of Enthalpy
• Note that since ΔH = qp (at constant pressure) and qp = CpΔT (if Cp
is constant with T):
dH = CpdT
– where Cp is the constant pressure heat capacity (can show Cp = Cv + nR for
an ideal gas)
• In general, change in enthalpy of a substance with temperature can
be determined using:
H(T2) − H(T1) =
T2
∫T
C dT
1
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Adiabatic processes (q=0)
• For an adiabatic expansion/compression of a gas,
(thermally insulated container):
dU = dw = CvdT
• Energy changes only occur via work, and the temperature of
the system must change.
• Expansion – temperature of the system will decrease
• Compression – temperature of the system will increase
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
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Adiabatic expansion
Physical Chemistry Fundamentals: Figure 2.17
• Can be divided into two steps – consider the work component of
each step:
Isothermal expansion
Temperature change
at constant volume
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Adiabatic processes
• Consider reversible adiabatic expansion/compression of an ideal
gas.
dU = dw = CvdT = -PdV
(in a reversible process P is not constant, P changes as V changes,
P = nRT/V)
−nRT
CV dT =
dV
V
CV
−nR
dT =
dV
( T )
( V )
• Integration gives:
CV ln
Tf
( Ti )
= − nR ln
Vf
( Vi )
(Vf, Vi and Tf, Ti are the final and initial vol. and temp.)
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Adiabatic processes
• Rearrangement (see Atkins 11th Ed. Topic 2E, p67; or 10th
Ed., Ch. 2E, p101) gives:
1/c
Vi
Tf = Ti
( Vf )
where c =
Cv,m
R
• and Cv,m is the molar heat capacity (units of JK-1mol-1)
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Example
• Calculate the final temperature, w and ΔU when 0.020
mol of argon, initially at 298K, undergoes a reversible
adiabatic expansion from 0.5 dm3 to 1.00 dm3
(Cv,m for argon is 12.48 J K-1 mol-1)
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Should now know:
• Definition of enthalpy – heat transferred at constant
pressure
• Relationship between ΔH and ΔU for a chemical
reaction
• Dependence of ΔH on temperature
• Energy changes in adiabatic processes
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Lecture 3
• Thermochemistry
– standard enthalpy changes
– enthalpies of chemical reactions
– temperature dependence of reaction enthalpy
• cf. Atkins 10th Ed. 2C, 2E
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Enthalpy (H)
• Definition: ΔH = qp
(heat exchanged at constant pressure)
• ΔH is related to ΔU by:
ΔH = ΔU + Δ(PV)
for an ideal gas PΔV = ΔngRT
(where Δng is the change in the number of moles of gas during a
reaction) – thus:
ΔH = ΔU + ΔngRT
CHEM20018 Thermodynamics, 2022 ©Trevor Smith, University of Melbourne
Physical Chemistry Fundamentals: Table 2.4
54
Standard Enthalpy Change (∆Hɵ)
• Standard enthalpy change, ∆Hɵ, refers to a process where the initial
and final substances are in their ‘standard states’ (state at 1 bar
pressure and a specified temperature, usually 298K).
H2O(l) ➛ H2O(g)
ΔvapHɵ (373 K) = +40.66 kJmol-1
(Vaporisation)
Endothermic process (∆Hɵ>0)
H2(g) + ½O2(g) ➛ H2O(l)
(Combustion)
ΔcHɵ (298K) = -286 kJmol-1
Exothermic reaction (∆Hɵ