CH107 – Physical Chemistry • Atomic and molecular structure • Intermolecular forces & dynamics • Driving forces for equilibrium Instructor (D3): Prof. Arindam Chowdhury, Chemistry, Room 215 Phone: x-7154; 9969437094 Email: arindam.chowdhury@gmail.com arindam@chem.iitb.ac.in
CH107/ D3 Course Information CH107/ D3 Course Information Instructor: Arindam Chowdhury Instructor: Arindam Chowdhury Room No. 215; Department of Chemistry arindam@chem.iitb.ac.in (022)2576 7154 Attendance, marks change and course related issues: Course Secretary: Ms. Charine Astrid Central Facility, Chemistry Department Email: charine@chem.iitb.ac.in Phone: (022)2576 4159
CH-107 / D3 CH-107 / D3 23 lectures and 7 tutorials (Attendance is mandetory Attendance is mandetory) ) 23 lectures and 7 tutorials ( Lectures at Hall 1 Lectures at Hall 1 Mondays 9.30, 9.30, Tuesdays Tuesdays 10.30 and 10.30 and Thursdays Thursdays 11.30 am 11.30 am Mondays 17/09 19/09 19/09 23/09 23/09 24/09 24/09 26/09 26/09 17/09 01/10 03/10 03/10 07/10 07/10 08/10 08/10 10/10 10/10 14/10 14/10 15/10 15/10 01/10 17/10 21/10 21/10 22/10 22/10 24/10 24/10 28/10 28/10 29/10 29/10 31/10 31/10 17/10 04/11 07/11 11/11 11/11 12/11 12/11 04/11 07/11 LCT (31, 32, 33, 22, 23) on Tutorials at LCT (31, 32, 33, 22, 23) on Mondays Mondays 2-3 pm 2-3 pm Tutorials at 23/09 23/09 07/10 14/10 21/10 28/10 07/10 14/10 21/10 28/10 04/11 11/11 04/11 11/11
CH-107/D3 CH-107/D3 Tutorial venue and Teaching Assistants Tutorial venue and Teaching Assistants D3/T1 LCT-31 Shekhar Hansda D3/T1 LCT-31 Shekhar Hansda D3/T2 LCT-32 Arindam Chowdhury Arindam Chowdhury D3/T2 LCT-32 D3/T3 LCT-33 Tuhin Khan Tuhin Khan D3/T3 LCT-33 D3/T4 LCT-22 Avinash Kumar Singh Avinash Kumar Singh D3/T4 LCT-22 D3/T5 LCT-23 Sandip Kar Sandip Kar D3/T5 LCT-23 Emails and phone numbers to be provided Emails and phone numbers to be provided Office hours : 2 hrs/week and mutually convenient time Office hours : 2 hrs/week and mutually convenient time
CH-107 CH-107 Time Table Time Table Duration Half-semester (~8 weeks) Duration Half-semester (~8 weeks) Quiz 19 October 2013 Quiz 19 October 2013 End-Semester Exam Anywhere between 18-29 End-Semester Exam Anywhere between 18-29 November 2013 November 2013 Evaluation Scheme Evaluation Scheme Total 50 Marks Total 50 Marks Quiz 20 Marks Quiz 20 Marks End-Semester Exam 30 Marks End-Semester Exam 30 Marks Passing Marks 15 (To be followed strictly) Passing Marks 15 (To be followed strictly)
GRADING STASTICS Coursewise Statistics Coursewise Statistics Course Code CH 103 – SEM SEM2 Course Code CH 103 – SEM1 SEM1 Course Name Chemistry Course Name Chemistry Total Grades Total Grades 452 416 Given are Given are Out of Which Out of Which AA +AB 22 AA+AB 44 BB 49 BB 96 BC 71 BC 95 CC 105 CC 90 CD 89 CD 49 DD 53 DD 23 FR 63 FR 19
GRADING STASTICS Coursewise Statistics Coursewise Statistics CH 103 – SEM- SEM- CH 103 – SEM- SEM- Course Code Course Code 2 1 2 1 Course Name Chemistry Course Name Chemistry Total Grades Total Grades 462 424 Given are Given are Out of Which Out of Which AA+AB 57 AA + AB 38 AP 1 BB 49 BB 72 BC 69 BC 83 CC 82 CC 64 CD 89 CD 64 DD 48 DD 36 FR 87 FR 47
Why should you study Chemistry? Is there a role of Chemistry in reshaping the modern world? All of Science and Engineering is moving towards interdisciplinary fields of cutting-edge research!!! Knowing only one subject often not good enough! • Nanoelectronics/Nanotechnology: Molecular Electronics • Energy Science – “Solar Energy” conversion • BioTechnology – Disease cure, health, medicine • Atmospheric Science – Go Green - “Save the World”
Plastic Electronics and Displays Conducting-polymers are replacing liquid crystals
Micro & Nano-electronics Intel, 1990s, hundreds of Silicon Transistor, TI 1954 1947, Transistor, Bell Labs Transistors in a single chip Transistors, Intel, 2006, 45 nm separation Next Generation: Molecular Chips
Mechanics of Electrons and Atoms Mult-electron Atoms (Periodic Table) Nano-science And Nanotechnology Electron Atomic/Molecular Microscopy Spectroscopy Multi-atomic Bonding, Molecular Chemical Reactions Structure Molecular Dynamics Biology, Materials Science Intermolecular Forces Condensed Matter And Interactions Physics
Contents of Part-I (14 Lectures) • Origin of Quantization: Lecture 1-4 Need of a new theory for electrons, atoms and molecules Postulates of Quantum Mechanics Energy Quantization: Particle in a Potential Well • Electronic Structure in Atoms: Lecture 5-8 Hydrogen Atom and Quantum Numbers Atomic Orbitals and Electron Densities Multi-electronic atoms and the implications of “ Spin ” • Chemical Bonding: Lectures 9-12 Molecular Orbital Theory – Linear Combination of Atomic Orbitals Energetic and electronic structure of diatomic molecules • Molecular-Electronic structure Lecture 13-14 Bonding in polyatomics using hybridization
Contents of Part-II (9 Lectures) • Intermolecular Forces and PE Surfaces Lecture 15-16 • Reaction Dynamics and Kinetic theories Lecture 17-19 • Driving Forces for Chemical reactions Lecture 20-21 • Chemical Potential and Equilibrium Lecture 22-23
Recommended Text • Physical Chemistry – I.N. Levine, 5 th Ed. • Physical Chemistry – P.W. Atkins 2 nd Ed. • Physical Chemistry: Molecular Approach - McQuarrie and Simon Important Websites: CH107 Course Material for 2013 And previous year’s power-point slides: www.chem.iitb.ac.in/academics/menu.php and will be regularly updated in IITB Moodle http://moodle.iitb.ac.in
Classical EM theory can not explain Blackbody Radiation π 8 k T c ρ ν = ν ν ν = 2 ( ) T d b d ; ν λ 3 c All classical theories led to the so called Sun, stars…hot iron rod “Ultraviolet Catastrophe” Theories based on classical physics unable to explain temperature dependence of emitted radiation (radiant energy density)
Max Planck assumed energies of oscillators are discontinuous Assumption: Energy of electronic oscillators were discrete; Assumption: Energy of electronic oscillators were discrete; Proportional to integral multiple of frequencies Proportional to integral multiple of frequencies = ν E nh E = Energy of electronic oscillators Osc v = frequency of electronic oscillators h = Planck’s constant = 6.626 x e -34 joule-sec Note: h came in as a fitting parameter ν 3 a ρ ν = ( ) T d v − bv t / e 1 1858-1947 π ν ν 3 8 h d ρ ν = ( ) T d v 3 c e ν h / k T − 1 B ÷ Planck never believed his theory was right, since he was a classical physicist
Photoelectric Effect Experimental Observations Photodetectors, Photovoltaics, 1. Increasing intensity of light Elevator sensor, smoke detectors increases number of photoelectrons, but not their max. kinetic energy (KE MAX )! 2. Light below a certain wavelength will not cause ejection of electrons, no matter how high it’s intensity! 3. Extremely weak violet light ejects few electrons! But their KE Max >> KE Max of electrons ejected by intense light of longer wavelength = − ω E E Sin kx ( t ) 0 Wave Energy related µ 2 to Intensity E 0 ω Independent of
Einstein: light behaves like particles Borrowing Planck’s idea that ∆ E=hv, Einstein further proposed radiation itself existed as small packets of energy (Quanta), known as PHOTONS µ ν E P = ν E h P Energy of photon is frequency dependent (self contradictory!) 1 = = + φ = + φ 2 E hv KE mv P M 2 1879-1955; Nobel prize φ = Energy to remove e' from surface For explanation of = − φ ≥ KE hv 0 Photoelectric effect M 0
Line Spectra of Atoms 1854-1919 Rydberg’s formula: ν 1 1 1 ν = = = − > R ; n n ÷ λ H 1 2 2 2 c n n 1 2 R H = 109677.57 cm -1
Bohr’s Phenomenological Model (Rutherford-Planck-Einstein-Bohr Model) • Electrons rotate in circular orbits around a central massive nucleus (+ve), and obey laws of classical mechanics. • Allowed orbits are those for which the electron’s angular momentum m e vr = n h/2 π , n=1,2,3,4,…… 1885-1962 • Only certain discrete energy values: “Stationary States” - Atom in such a state does not emit EM radiation (light) • Transition from a stationary state (E a ) to another (E b ), atom emits or absorbs EM radiation (light)
Explanation of atomic spectra nh Quantization of = mvr n=1,2,3,... π 2 Angular momentum π = λ (2 r n ) 4 m e 1 = − E e . n ε 2 2 2 8 h n 0 Spectral Transitions: ∆ E=hc/λ 4 m e 1 1 ∆ = − = ν = E e ÷ h , n n 1,2,3,... ÷ i f ε 2 2 2 2 8 h n n i f Explains Rydberg’s Formula
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