CHEM Revision Lectures PDF

Title CHEM Revision Lectures
Course The Chemical Basis of Biology and Human Health
Institution University of Otago
Pages 3
File Size 95.9 KB
File Type PDF
Total Downloads 104
Total Views 161

Summary

Notes from the revision lectures given ...


Description

CHEM Revision Lectures Module 1 – Chemical bonding, chemical equilibria and chemical energetics L1-17  No multi choice on topics in Terms test (no module 1, so from module 2-5)  Timing: written questions 27 mins each, multi choice 72 mins New Guest lecturers – 2 multi choice questions (look at past papers)  Bonding theory, equilibrium, acid and base, thermodynamics (7-9 marks spread across Q1-2) Bonding revision (3-4 marks) -

Lewis structures (probs only MCQs) VSEPR (not likely) Valence bond theory  Multiple bonds: sigma and pi bonding (e.g. ethene 2p orbitals overlapping to form pi bond)

Stoichiometry revision -

n=m/M and c=n/v equilibrium constant

Thermodynamics revision -

enthalpy change = sum of products – sum of reactants (taking into account stoichiometry) G = H – TS (S needs to be changed to kJK-1mol-1) Equilibrium theory Kc = [C]c[D]d/[A]a[B]b G = -RTlnK Le chatelier’s principle Q and K Ksp and Qsp

Acid-base revision -

pH = -log[H3O+] Ka = {H3O+}[A-]/[HA] Kw = [H3O+][OH-] pH + pOH = pKw pH = pKa + log [A-]/[HA] Henderson Hasselbalch equation only for buffers

Module 2 – Chemical kinetics, rates of equation  Kinetics (7-9 marks) + 7MCQs Kinetics revision  



Rate always conc/time usually molL-1s-1 Rate = -1/x d[X]/d[t] negative for reactants, positive for products Initial rates method Integrated rate law for first order reaction -> spell out answer clearly, comparing experiments keeping one concentration constant Rate constant using rate law, units! ln[A]t = ln[A]o – kt half life for 1st order reaction = ln2/k Arrhenius equation ln(k1/k2) = -Ea/R (1/T1 – 1/T2)



Energy profile diagram A+B reactants –Ea-> transition state -> C+D products (total energy difference is deltaG) Mechanisms 2 types only first step slow – for elementary step can write down rate law directly from stoichiometry of this step (State this) first step always a fast equilibrium – focus on slow step, write down rate law directly from stoichiometry of this elementary step, use fast equilibrium constant K to remove reaction intermediate from rate law, biological stuff

Module 3 – Electron transfer reactions  







 





Definitions: oxidation, reduction, oxidant, reductant, oxidation numbers (increase in oxidation number = oxidation) Balancing redox reactions – in acidic solution (most common) adding water to balance oxygens and H+ to balance hydrogens, then balance for electrons In basic solutions, add the same number of OH- ions to each side as there are H+ ions, then balance the charge Galvanic cells provide a way to measure the potential difference V between two redox couples – how much one oxidant wants to get electrons compared to the other anode-salt bridge-cathode electron flow from anode (oxidation) to cathode (reduction) Cell diagram Zn / Zn2+ (aq, c) // Cu2+ (aq, c) / Cu (standard cell potential) Reading on potentiometer tells us which way the electrons are going, positive value means reduction on the right (electrons from left to right) Ecell = ERHS - ELHS To determine actual values of individual E value, build a galvanic cell with the standard hydrogen electrode on the LHS By considering standard reduction potentials we can determine the direction of spontaneous electron transfer Ereaction = Ereduction process + Eoxidation process (Ecell will always be +ve) G = -nFEcell (positive value of E will give negative value for G -> spontaneous) relate E to K: Ecell = RT/nF ln K (n is number of electrons transferred in the reaction) Non-standard conditions: Nernst equation E = Eo - RT/nF ln Q (reaction quotient actual concentrations) Transition metal complexes: (more common in multi-choice) Compound formed from transition metal cation and some number of ligands bonded to it Ligands – contain one or more donor atoms (normally O or N) which have a lone pair of electrons, with which a coordinate covalent bond can be formed to the cation Denticity of ligand = number of donor atoms so number of bonds that can form to cation (monodentate = 1, bidentate = 2) Ligands which can form two or more bonds to a cation also called chelating ligands Donor atoms found in the side chains of amino acids residues in proteins are important in holding in place transition metal cations in the active site of metalloenzymes (pH dependant) Electronic configuration of transition metal cation Coordination number of cation is number of bonds which ligand’s donor atoms make to it Different coordination numbers have different characteristic coordination geometries: 6= octahedral, 5= square pyramidal or trigonal bipyramidal, 4= square planar or tetrahedral Influence of transition metal on the ligand: ligand activation – pull of positively charged cation on electrons from water ligand makes its bonds easier to break and lose H+ -> acts as a weak acid (important part of how many enzymes catalyse reactions)



Influence of ligand on transition metal ion: change the reduction potential, if ligand more electronegative/electron-withdrawing, then reduction potential increases/more positive - Adding fluorine atoms to ligands pulls electron density away from metal ion, increasing its desire to an electron and so increasing its reduction potential - Adding CH3 to ligands pushes electron density towards to the metal ion, stabilising it and making it less reactive towards an electron/nucleophile

Module 4 – Reactions to carbon compounds     

C-OH hydroxyl is a poor leaving group as it is a short, strong bond, hard to break due to attraction between C and O (also similar size) SN1 reaction carbocation formed is planar, so nucleophile can attack on either side, giving a racemic mixture, 0 in plane polarised light For rate law and Sn1/Sn2 use equilibrium step to substitute in K In nucleophilic acyl substitution, the C=O breaks first because it is a weak pi bond, giving a tetrahedral intermediate In acetone vs water as a solvent, solubility of the nucleophile can determine if reaction is an equilibrium and therefore what the end products are...


Similar Free PDFs