Physical Chemistry - Coursework Example

CHEMISTRY WORK: Q1.)Cyclic Voltammogram. Cyclic voltammogram makes use of 6m M Ascorbic acid in a 1M KNO3 matrix, the reference electrode is Saturated Calomel Electrode(SCE) , in this experiment the scan starts 0.00V(red) and Working Voltage is +o.80 V. A strong peak is reached at 0.35V relative to SCE indicating oxidation of dehydroascorbic acid. The Red Curve shows the first half of the scan 0.00V to 0.80V and back to 0.50V. Chemical Reaction: 2 electron Oxidation of Ascorbic Acid. The process is irreversible because the oxidation of ascorbic acid produces Dehydroascorbic Acid which is very unstable and it decomposes quickly before the reduction cycle can convert it back, thus due to Dehydroascorbic instability the process is irreversible (Naylor, 1995) Question 2 Polyaniline, the polymer resulting from oxidative polymerization of aniline, is built up from reduced (B – NH – B – NH) and oxidized (B – N = Q = N - ) repeat units, where B denotes benzenoid and Q denotes quinoid ring, we shall denote polyaniline as PA and ammonia as NH3 With NH3 gas the reaction mechanism is a deprotonation-reprotonation one, in that reaction leads to disappearance of charge carriers(polarons) and increase in electrical resistance Reaction mechanism with NH3: PAH++NH3 PA+NH4 + Reaction with Carbon monoxide: Polyaniline showed no sensitivity to carbon monoxide, the electro catalytic oxidation of Carbon Monoxide is weak. Mechanism; PAH++CO PAH++CO 3.) E CELL We need to apply Nernst equation, we also need EOCell, n and Q. Fe3+ + e- Fe2+ EO Fe3+ / Fe2+ = +0.77V Cl2 +2e- 2Cl - EO CL2 / 2CL -=+1.36V E0 cell= E0 cathode-E0 Anode= +1.36-0.77v=0.59v Overall reaction: Cathode: Cl2 +2e- 2Cl – *1 Anode: Fe2+ Fe3+ + e- *2 Cl2 + 2 Fe2+ 2Cl- + 2 Fe 3+ Q = (Cl -) 2 (Fe 3+ )2 (Fe 2+) 2 N=2 Fit in Nernst equation; E CELL =E0 CELL 0.059 log Q N ECELL= O.59V - 0.059 log (1.0M)2 ( 0.3M)2 2 (0.1M)2 E CELL=0.562V B. Applied Kinetics: Green, Atmospheric and Catalyst Chemistry 1. Ԏ for CH4= TCH= 1 = 1 =8.13yrs ∑ K1 N1 3x10-15 molecules-1 cm3 s-1×4.1x1013 molecules cm-3 Average Lifetime for CH4 is 8.13 years, in this case it’s a first order chemical loss for X(CH4) With Rate Constant KC=3×10-15, The chemical loss is L=Kcm so that Tc is simply the inverse of the rate constant (Naylor, 1995) It’s a methane oxidation mechanism that involves one step; i.e CH4 + OH CH3 + H20 Question 2 (a) The most abundant oxidants in the Earths atmosphere are O2(oxygen), O3(ozone) and OH(hydroxyl ) Radical group. These oxidants have large bond energies and are hence relatively unreactive except toward radicals (O2 only toward highly unstable radicals). With a few exceptions, oxidation of non-radical atmospheric species by O2 or O3 is negligibly slow. Of the three, OH radical is identified as a strong oxidant in the stratosphere. OH reacts rapidly with most reduced non-radical species, and is particularly reactive toward H-containing molecules due to H-abstraction reactions converting OH to H2O. Oxygen and Ozone:O2 and O3-oxygen is a principal constituent of dry air, accounting for 21% of the atmospheric volume, atmospheric oxygen is regulated by a slow atmospheric lithosphere cycle involving conversion of O2 to Carbon dioxide. Elements that combine with oxygen remove it from atmosphere; these are described as oxygen sinks, oxygen also combines with other elements to make oxides, the process of chemical change involving oxygen molecules and electrons is both oxidation and reduction, thus its relative activity in the atmosphere (Naylor, 1995) Ozone is generated by the energetic action of solar ultraviolet radiation (UV) on Ordinary Oxygen in the presence of stabilizing agent like Nitrogen. The reaction is reversible and O3 reverts to diatomic oxygen. It’s a strong oxidizing agent readily reacting with other chemical compounds to make toxic oxides, it’s also responsible in providing the strong oxidant OH and its also an important greenhouse gas, however its being depleted at high rate due to emissions due to human activities and natural systems, the proportion of remaining after specified is the Airborne Fraction (Naylor, 1995) OH: it’s a radical and a strong oxidant in the stratosphere, it reacts rapidly with most reduced non-radicals species, and is very reactive towards H-containing molecules due to H-abstraction reactions converting OH to H2O, it quickly oxidizes species like CO and CH4 within the troposphere. The lifetime of OH in air is given by TOH = 1 ∑I KI NI Where ni is the number density of species i reacting with OH, ki is the corresponding rate constant, and the sum is over all reactants in the air parcel. One finds that CO is the dominant sink of OH in most of the troposphere, and that CH4 is next in importance. The resulting OH lifetime is of the order of one second. Because of this short lifetime, atmospheric concentrations of OH are highly variable; they respond rapidly to changes in the sources or sinks. Thus a calculation of OH requires knowledge of its known concentrations averaged over time and space. In conclusion the average lifetimes of the oxidants or any other material is noted as the removal of a chemical compound as burden divide by the loss denoted as: T= BURDEN LOSS RATE Q2b.) the lifetime of Z is a first order reaction, Z-hex-3-en-1-ol(leaf alcohol) is calculated as; T Z HEX-3-EN-1-OL = 1 ∑K(OH+LA+NO3+LAl+O3+LAl)×∑n(OH) +(NO3)+(O3) LA denotes leaf alcohol; T Z-hex-3-en-1-ol = 1 ∑(1.1×10 -10)+(2.7×10-13)+(8.5×10-17)×∑(7.5×10 5)+(5.8×10 8)+(7×10 11) =0.014 of a day 0.014×24×6o=21seconds Question 3a Albedo is the fraction of solar energy (shortwave radiation) reflected from the Earth back into space. It is a measure of the reflectivity of the earths surface. Example Ice, especially with snow on top of it, has a high albedo: most sunlight hitting the surface bounces back towards space. Water is much more ××orbent and less reflective. So, if there is a lot of water, more solar radiation is absorbed by the ocean than when ice dominates. Question 3b E(in)= Fs x (1-A) x πr2 Fs = Solar Flux = 1368 W m-2 Short wave radiation A=Albedo=0.29 E in = 1368 ×( 1-0.29) × πr2 Eout= s x (T earth)4 x 4πr2 s = Stefan-Boltzmann constant Long radiation: = 5.67 x 10-8 W m-2 K-4 We will estimate that the rate of (light) energy coming to earth and rate of (heat0 emitted back into space and this has to be in balance so that; E IN = E OUT FS ×(1-A)× πr2 = S ×(T EARTH)4 ×4 πr2 SOLVING FOR TEMPERATURE; T EARTH= FS (1-A) 1/4 4S TEARTH = (1368 W/M2)(1-0.29) 1/4 (4)(5.67 × 10 -8 W/M 2 K4) = 256K = -170 C LETS INCREASE THE ALBEDO AND SEE HOW THIS AFFECTS TEMPERATURE: EG. PUT ALBEDO TO BE 0.31 TEARTH = (1368 W/M2)(1-0.31) 1/4 (4)(5.67 × 10 -8 W/M 2 K4) = 254K = -190 C THUS THE ORIGINAL ALBEDO WAS 0.29 WITH A TEMPERATURE OF 256K ( -170 C) WHEN THE ALBEDO WAS INCREASED TO 0.31 THE TEMPERATURE DECREASED TO 254K ( -190 C) Question 3C Albedo=0.29, decrease by 20%= 80×0.29 = 0.232 100 Substitute in the equation the new albedo: TEARTH = (1368 W/M2)(1-0.232) 1/4 (4)(5.67 × 10 -8 W/M 2 K4) = 260.8K approximately 261K= -120C The New average temperature is TEARTH =261 K = -12O C C MOLECULAR INTERACTIONS Question 1a Leonnard jones interactions: We set the minimum to be 0 and set dv/DR=O AND SOLE FOR r we therefore have; that is set at minimum dv =0 and solve for r dr dv = 4ϵ -12 б12 + 6 б6 =O dr r13 r7 12б1 2 = 6 б6 r 13 r 7 2б6 = r min 6 r MIN = 2 1/6 б. Q1b Set net force to be at 2 (i.e the minimum of the two molecules) dv =2 and solve for r (1) dr 24 б1 2 = 6 б6 r 26 r 7 4б2 = r min 27 R = 26 1/6 б 7 Question 2a If the energy for 0.3nm is -20.0 kJ mol-1. The energy for 0.5n will be weaker it will be -33.3 kj mol-1, in that Typically the strength of dipole-induced dipole interaction is not enough to mutually orient the molecules. The interaction needs to be angle-averaged. The interaction (Debye interaction or inductive interaction) pictorial showing the interactions; There is ease of distortion of the electron cloud of a molecular entity by an electric field (such as that due to the proximity of a charged reagent). It is experimentally measured as the ratio of induced dipole moment (uind) to the field E which induces it: Question 2b Van der Waals forces is dependent on Pressure, volume and Temperature, if the temperature is increased by 20%, , at high temperature van der Waals break apart and in some molecules it’s the melting point of the molecules and causes rearrangement of the molecules since this are weak bonds held by weak inter molecular forces, other parameters like pressure and volume could also change drastically.in large molecules like the extra energy is produced that is exothermic reaction like dimers (Naylor, 1995) Bibliography T.De V. Naylor. Polymer Membranes: Materials, Structures and Separation Performance - Review Report (Rapra Review Reports)1995 Read More