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Chapter 5: Gibbs Free Energy

2012-10-20T20:08:00.000-07:00

Chapter 5-Gibbs Free Energy

Problems: 5.1,3,5,11; concept:5.9,19,21,22,23

1.G-energy to drive chemical reactions

-ΔpG- tabulated (in table)

-ΔrxnG- usually calculated

eq.1: ΔrG- ∑v*Δ_fG_products –∑v*Δ_fG_reactants

-if 0, reaction is at equilibrium

-if negative, reactions proceeds towards right (spontaneous)

-if positive, reaction will not proceed (non-spontaneous)

2.Phase Transitions

-Phase 1 --> Phase 2:

eq.2:ΔrxnG= n* ΔfG_phase
2 – n* ΔfG_phase1 (ΔG is molar)

-if ΔGf phase 2 > phase 1, ΔGrxn is positive (non-spontaneous)

-if ΔGf phase 1 > phase 2, ΔGrxn is negative (spontaneous)

-if ΔGf phase 1 = phase 2, ΔGrxn is 0 (equilbrium)

-stable phase- phase with lowest ΔGf

3.Molar ΔGf vs Pressure- ΔGf,m or ΔGm = Vm*ΔP

-ΔG is directly proportional to changes in pressure

-larger molar volume- ΔG changes more as pressure changes for substances with larger molar volumes

eq.3:Vm=ΔG/ΔP

-solids,liquids: final ΔGm= ΔGm,i + Vm(Pf-Pi), where pressure is in bars

-gases: final ΔGm*(Pf)= ΔGm*(Pi)+RTln(Pf/Pi)

-reactions: ΔGrxn= ∑V*ΔGm,final (products) – ∑V*ΔGm,final (reactants)

4.Molar ΔGf vs Temperature

-for small changes in T, entropy values are nearly constant

eq.4:ΔGm=-Sm*ΔT

5.Phase boundaries- on phase diagrams, show the pressure and temperature combinations at which 2+ phases are stable

-liquid-vapor boundary-liquid is in contact with and in equilbrium with a gas of that composition; pressure of the vapor is its vapor pressure, which substantially increases with temperature

-solid-vapor boundary-solid in contact with and in equilbrium with vapor; sublimination vapor pressure of the solid can be determined in the same way as the liquid vapor pressure

-slope- slope of boundary determined by thermodynamic properties

6. Clapeyron equation- finds slope of boundary (ΔP/ΔT), ONLY fo small changes in P and T

-ΔtrsH- change of heat during transition between phases (where temperature and pressure stay constant)

eq.5: ΔP/ΔT=ΔtrsH/T*ΔtrsV

7.Clausius-Clapeyron equation: Δ(lnP)= ΔvapH/RT² *ΔT

-liquid-gas boundaries

eq.6:lnP’=lnP+ΔvapH/R*(1/T-1/T’), where P’ is final P, and T’ is final T

-Table 5.1- log(P in kPa)=A-(B/T), where A and B are constants (see derivation on p.113)

8.Critical point-terminal point on liquid-gas boundary curve

-highest pressure at which liquid can be condensed

-fluid- above critical point, state of matter is called “fluid”, no boundary

-observed between states of matter as seen between liquid and gas

9.Normal boiling point- boiling temperature at 1 atm

-Standard boiling point- boiling temperature at 1 bar

-Normal boiling point- melting temperature at 1 atm

-Standard melting point- melting temperature at 1 bar

10. Phase Rule- for a system at equilbrium, F=C-P+2

-F=degrees of freedom, C=number of component, P=number of phases, 2=T+P

-components- minimum number of species necessary to define all the phases present in a system

-degrees of freedom- number of intensive variables (P,T,mol fraction, etc.) that can be changed without disturbing equilbrium

-triple point- F=0 (C=1, P=3)

-phase boundary- F=1 (C=1, P=2)

General Biochemistry 2: Lecture 10,11 (Photosynthesis, Photosystems, Cyclic)

2012-10-15T23:53:00.003-07:00

LECTURE 10,11: PHOTOSYNTHESIS, REACTION CENTERS, PHOTOSYSTEMS

A. Photosynthesis- light driven synthesis of CH2O from CO₂; also produce oxygen by oxidating water; overall: CO₂ + 2 H₂O à CH₂O + O₂ + H₂O

Chloroplasts-site of photosynthesis; 1-1k chloroplast per cell, very in size/shape, usualy 5 um long ellipsoids.

-3 membranes: inner, outer, thylakoid membranes (where light driven reactions take place)

-3 regions: intermembrane space, stroma (dark reactions), thylakoid lumen

-thylakoid membrane- light reactions; invaginations of inner membrane cause these, resemble cristae

-thylakoid vesicle- arranged to have disk-like sacs (grana), stacked in piles, connected to each other by stroma lamellae.

-1 chloroplast=10-100 grana

-stroma-concentrated solution w enzymes, dna, rna, ribosomes

B. Reaction Centers

I. Light reactions (thylakoid membrane)- use light to generate nadph, atp and O₂. Resembles ET and OP in mitochondria.

1. Photosynthetic Reaction Center(RC)-primary photochemical rxns take place in RC.

-chlorophyll- green pigments that capture sunlight, located in and around photosystems embedded in thylakoid membrane of chloroplasts. Structure- porphyrin with magnesium at the center.

-Bacteriopheophytin- a bacteriochlorophyll where Mg²⁺ ion is replaced by 2 H⁺; Reaction center from a purple bacteria (rhodopseudomonas viridis) is a transmembrane protein with several chromophores (BchI a and b, nonheme Fe(II), Bpheo b, ubiquinone, menaquinone)

-antenna chlorophyll- chlorophylls not in RC; don’t participate directly in photochemical reaction, but capture photons from sunlight.

2.Light Harvesting Complex B (antenna complex)- has several membrane-bound hydrophobic proteins, each with many chlorophylls and other pigments.

-carotenoid- accesory pigments; non-chlorophyll pigments that absorb wavelengths chlorophylls don’t strongly absorb.

-number of LHCs- many more LHCs than PRCs.

-lower energy state- light captured by antenna complex, transferred to series of antenna pigment molecules, eventually trapped by PRC (10E-10 seconds, >90% efficiency); PRC has lower excited energy state than antenna pigments.

-all O₂ cells- have psysI and II; non-O₂ cells only have psys I.

C.Photosystems

1.Photosystem 1-forms nadph

-max efficency at 700nm; absorbs 4 quanta, makes strong reductant, weak oxidant

-4[ P₇₀₀ + q à P₇₀₀* (strong reductant) à P₇₀₀⁺ (weak oxidizer)+ e]

-4[ P₇₀₀⁺ + e(from P₆₈₀*) à P₇₀₀]

-P700*-strong reductant provides 4e to reduce 2 NADP+ to 2 NADPH

-P700+ - weak oxidant recieves 4 e from weak reductant made by psys II to regenerate ground state psys I P₇₀₀

- ferridoxin-nadp+ reductase-forms nadph in stroma

2.Photosystem II-forms O₂ by oxygen-evolving center (OEC)

-max efficiency drops at >680nm; absorbs 4 q, makes weak reductant and strong oxidant

-4 [P₆₈₀ + q à P₆₈₀* (weak reducer) à P₆₈₀⁺ (strong oxidant) + e]

-4 [P₆₈₀⁺+ e (from water) à P₆₈₀]

-P680+ -strong oxidant takes 4 e from 2water to make 4H+ and O2

-P680*- weak reductant provides 4 e to weak oxidant formed by psysI

3.H⁺ gradient- drives atp synthesis by atp synthase; similar to OP

-thylakoid lumen- 4e goes from weak reducer P680* to weak oxidant P700+, coupled to pumping of 8 H⁺ into thylakoud lumen.

-Evolution of one O₂- from 2 waters produces 4H+ in thylakoid lumen.

-redox-active factors- chlorophylls, cytochromes, Q, FeS clusters, plastocyanin help transfer 4e between photosystems.

-plastocyanin- peripheral membrane protein on thylakoid luminal surface, cycles between CuI and CuII, a redox-active factor.

4.cyclic transport- produces atp, but no nadph or O₂

-involves only photosystem I;

-allows cells to adjust ATP:NADPH ratio

-regulation unknown

5.efficiency of non-cyclic ET: 1.25 ATP/quantum (10 atp/8 quanta)

-for every O₂ made or 8 quanta absorbed

-water evolution: 4H+ made in lumen from 2H2Oà4H+ + O₂

-4e from PSII to PSI: 8H+ pumped into lumen

-H⁺ gradient: 12 H⁺/3 H⁺ to power synthase= 4 ATP

-PSI’s p700*: makes 2 NADPH, 3 atp/nadph, equals 6 ATP

6.efficency of cyclic ET: 0.67 ATP/quanta

-for every 4 q absorbed by PSI, 8H+ pumped into lumen, (8H+/3H+ to power synthase)/4 quanta

Dark reactions- use nadph and atp to make ch2o from water and CO₂. in eukaryotes, happens in stroma.

General Biochemistry 2: Lecture 9 (ATP Synthesis)

2012-10-15T23:52:00.004-07:00

LECTURE 9: ATP SYNTHESIS

Energy coupling- free energy released by electron transport must be saved in a form that atp synthase can use.

chemiosmotic theory (1961, peter mitchell)- ETàcoupled to pumping of H⁺ from mitochondria matrix across inner membrane to outside à create H⁺ electrochemical potential gradient across inner membrane à this gradient is energy source which drives atp synthesis

-inner membrane permeability- requires intact inner membrane, impermeable to ions like H⁺, OH-, K+, Cl- (permeable to CO₂, O₂, H₂O)

-electrochemical gradient- ET = H⁺ transport from matrix to outside, creating big gradient across inner membrane

-Potential = 0.168 V (more negative inside);

-pH of matrix = 0.75 units higher than intermembrane space,

-ΔG(transport H⁺ outside) = 21.5kJ/mol

-2,4dinitrophenol (DNP)- compounds that increase permeability of inner membrane to H⁺ allow ET to proceed, but inhibit ATP synthesis. Thus, oxidation (or ET) and phosphorylation can be uncoupled by uncouplers like 2,4-dintrophenol (DNP)

p/o ratio- ratio of atp synthesized to oxygen reduced

p/2e ratio-ratio of atp synthesized to 2-e reduction of some e⁻ acceptor other than o

-FADH₂/O₂(no agent or amytal)=2, (w CN- or DNP)=0

-FADH₂/Fe(w CN-)-3 (no agent or CN-)=1

-NADH/O₂(w CN-)=0

-NADH/Fe(CN6)-3=2

-pH gradient- atp is made by mitochondria/chloroplast when pH gradient artifically generated without ET.

-bacteriorhodopsin-purple membrane protein from halobacteria; pumps H⁺ when illuminated; synthetic lipid vesicles w BR and beef heart mitochondrial ATPase can make ATP when illuminated.

-asymmetric orientation- NADH-Q reductase, Q-Cyt c reductase, Cyt c, cyt c oxidase, ATP synthase all asymmetrically oritented (used mitoc/submitoc particles w inner membrane inside out)

-Efraim Racker- discovered F0 and F1

1.F1- water soluble, peripheral membrane protein; atp hydrolysis in isolated soluble form

-a3b3yE, active in atp synthesis in synthase assembly

2.F0-water insoluble, transmembrane protein; forms proton channel.

-a1b2c9-12 in e. coli, and some addition subunits in mitochondrial F0

3. Binding Change Mechanism- shape change takes place when T has ATP and L has ADP,Pi

-L,O,T(atp)àL(adp,p),O,T(atp) + energyà L,O(ATP),T(adp,p) + ATP,waterà L,O,T(atp)

-shape change takes place when T site has ATP and L site has ADP,Pi; results in conversion of LàT, TàO, OàL;

-shape change requires energy, transmiited to catalytic α3β3 assembly via ye assembly.

-atp synthesize at new t site, original atp released from new o site

General Biochemistry 2: Lecture 8 (Electron Transport Chain, Complexes, Inhibitors)

2012-10-15T23:51:00.000-07:00

LECTURE 8: ELECTRON TRANSPORT CHAIN, COMPLEXES, INHIBITORS

A. NADH oxidation- is highly exergonic- -218 kJ/mol

-NAD⁺ + H⁺ + 2e à NADH (-.315 V)

-oxygen + 2H+ + 2e à water (0.815 V)

Overall: oxygen + NADH + H⁺à water + NAD⁺

ATPàADP: -30.5 kJ/mol

ATPàAMP: -32.2 kJ/mol

2. Efficiency of ATP Synthesis- (3 ATP per NADH) ÷ (total energy per NADH) = 42%

-70% in mitochondria

B. ET Carriers- NADH; FMN; Q; FeS; Cyt a+a3,bl+bh,c,c1; Cua; Cub; O₂

1.Hydrogen carriers- FMN; NAD⁺; Q;

-FMN- add 2H⁺ and 2e, get FMNH2. (1H and 1 e= FMNH*)

-NAD+ - add 2H+ and 2e, get NADH + H⁺.

-Q- add 2H+ and 2e, get QH2.

2. Electron carriers- FeS, cytochromes (ET heme proteins), Cu proteins

-Fe3+ à Fe2+

-Cu2+àCu+

C. Complexes that make up ET chain

1. Complex I- NADH-CoenzQ Oxidoreductase;

-NADH à [[[FMN à2 FeS]]]à Q (passes 2e from NADH to Q)

-inside: 8-9 iron-sulfur clusters and 1 FMN

-4H+ go outside

-possible mechanism: differential H⁺ binding and release caused by protein shape change; H bonded groups in proteins and water are like a proton wire.

AH+BàA+BH. Proton moves from AH to B.

-bacteriorhodopsin-model for proton pump through proton wire

a.Cycle 1- QH₂ + Cyt c1 (Fe3+) à Q*- + Cyt c1(Fe2+) + 2H+ (cystol)

-QH2 from complex 1; pump 2H+ to intermembrane space, then to cytosol;

-donate 1e to ISP, which goes to c1;

-makes 1 Q-*

b.Cycle 2- QH₂ + Q-* + Cytc1(Fe3+) + 2H⁺(mito) à Q + QH₂+Cytc1(Fe2+)

-obtain QH₂ from complex I; pump 2H⁺ to cyto;

-Donate 1e to ISP, goes to c1. Take up 2 H⁺ from mito.

-Generate 1 QH₂, not net gain or loss of QH₂.

Net: QH₂ + 2 Cytc1(Fe3+) + 2H⁺ (mitochondria) à Q + 2 Cytc1(Fe2+) + 4 H⁺ (cystol)

-4 H⁺ pumped out, 2 c1 reduced, use 2 H⁺ from matrix.

2.Complex II- Succinate-CoenzQ Oxidoreductase

-Succinateà[[[FAD (succinate DH, 2H+ released)à3 iron-sulfur clusters à Cyt b560]]]à Q

-I and II not in series

-inside: 3 FeS clusters, succinate dehydrogenase FAD, and cyt b560

-2H+ go outside

3.Complex III- CoenzQ-Cytochrome C Oxidoreductase

-QH2à[[[CytB or FeSàCyt c1]]]à Cyt c

-inside: 2 Cyt b (bL for low and bH for high potential), 1 Cyt c1, and 1 Fe-S cluster.

-4H+ go outside, 2H+ inside

4. Complex IV- Cytochrome c oxidase

-2 Cytc(Fe2+) + 2H⁺ + .5O₂ à[[[CuAàCytaàCyta3-CuB]]]à 2 Cytc(Fe3+) + H₂O

-has cyt a, cyt a3, cua, vub

-high potential- (CuA-Cyta) lower potential than (Cyta3-CuB)

-2H⁺ go to cytosol

- Cyt c- between III and IV. Loosely bound to outer surface of inner membrane. Shuttles e⁻ between Cyt c1 and cyt c oxidase.

D. Donaters/Acceptors

1.Physio e⁻ donating systems:

-β-hydroxybutyrate dehydrogenase- β-HOButyr+NAD⁺àAcetoacetate+NADH

-Succinate DH- Succinate+FADàFumarate+FADH₂

2. Artifical e⁻ donating systems: ascorbateàtetramethyl-p-phenylenediamine (TMPD)àCyt C

3. Physio e⁻ acceptor: O₂ as acceptor of e from Complex IV

4. Artifical e⁻ acceptor: Fe(CN6)3- accepts e from Cyt c_red

E. ΔG=-nFΔE, where F=96494 J/mol*V

1.Site 1(Complex 1)- ΔG=-69.5, ΔE=.36V

-NADH+H⁺+QàNAD⁺+QH₂

-reduce NAD=-.315V; reduce Q=.045V

2.Complex II- no phosphorylation

-FADH₂+QàFAD+QH₂

-ΔE=.085 V, ΔG=-16.4 kJ/mol

3.Site 2 (Complex III)- ΔG=-36.7 kJ/mol, ΔE=0.19V

-QH₂+2Cytc_oxàQ+2Cytc_red
4.Site 3 (Complex IV)- ΔG=-112kJ; ΔE=.58V;

-2CytRED+ 2H⁺+ .5O₂à2CytOX+ H₂O

F.Inhibtors

Rotenone and Amytal- block complex I

Antimycin A- blocks complex III

CN^- -blocks complex IV

G.P/O value- ratio of ATP made over oxygens used

- β-HOButyr/NADH and O₂= 3

- Succinate/FADH₂ and O₂= 2

- β-HOButyr/NADH and Fe(CN6)-3= 2

- Succinate/FADH₂ and Fe(CN6)-3= 1

- ascorbate and O₂= 1

-phosphorylation and oxidation tightly coupled, NADH+P+O₂ wont happen until ADP is added.

-Reconsitution- isolated components and phospholipid vesicles remake active sites.

General Biochemistry 2- Lecture 7 (oxidative phosphorylation, mitochondria)

2012-10-15T23:50:00.003-07:00

LECTURE 7: OXIDATIVE PHOSPHORYLATION, MITOCHONDRIA

A. Oxidation of glucose: glucose + 6 oxygen à 6 CO₂ + 6 water; (ΔG=-2823 kJ/mol=-680 kcal/mol)

1.1st half: glucose + 6waterà 6 CO₂ + 24 H⁺ + 24e;

-12 e pairs -from glycolysis and tca cycle transferred to 10 NAD⁺ and 2 FAD

2.2nd half: 24 H⁺ + 24 e + 6 oxygen à 12 water

-oxidative phosphorylation- process of 2^nd half, makes ATP

3. substrate level phosphorylation- unrelated to mitochondrial OP, part of glycolysis makes ATP from ADP; substrate level atp synthases do this.

-(1,3bisPGa)à3Pga+ATPà2PGaàPEP+H₂OàPyruvate+ATP; kinase, mutase, enolase, kinase.

B.Oxidative phosphorylation- atp is made as e- are transferred from NADH/FADH2 to O2 by series of e- carriers (over 10 redox centers in 4 enzyme complexes)

-respiratory assemblies- takes place in inner membrane of mitochondria

-ratio of NADH/FADH₂- oxidation of NADH makes 3 ATP, oxidation of FADH2 makes 2 ATP.

-H⁺ gradient- coupled to pumping H+ out of mito matrix across inner membrane; ATP IS MADE WHEN H⁺ FLOWS BACK IN.

1. Mitochondrion-site of euk oxidative metabolism

-1948: Eugene Kennedy and Albert Lehniger learned that redox proteins in mitochondria (pyruvate dehydrogenase, TCA enzymes, FA oxidation enzymes, and enzymes in redox proteins for ET and OP)

-ellipsoid- (1 um long, .5 um wide), change size/shape; 1 cell=2k mitochondria

-volume- takes up lots of space (liver cells=20% of volume, heart muscle=more than 50%)

2.Outer Membrane- smooth, porous

-porins that allow medium-sized molecules to go through (<10kda p="p">

3.Inner Membrane- cristae(inward folds); contains 75% proteins.;

-freely permeable only to oxygen, CO2, and water. Can create and maintain ionic gradient w inner membrane.

4..Cristae-compartments made by folds in inner membrane. Vary in number/structure

-surface area- increases inner membrane SA in relation to mitochondrial volume

-blowfly flight muscle mito=400m² of inner membrane surface/g of mito protein

5.Matrix-space inside inner membrane; gel like, contains 50% of proteins

-has DNA, RNA, ribosomes

-dramatically changes volume and organization during changes in resp activity.

C. Transporting NADH into mitochondria- NADH molecules made by glycolysis are in cytosol (most NADH made by TCA cycle in mitochondria)

-mito inner membrane permeability- can’t transport NADH

-needs shuttle systems for transporting reducing agents of cystol nadh into mito

1.malate-aspartate shuttle- 1 cystol NADH à 1 mitochondria NADH à 3 ATP.

-in reverse: cystol OAA+NADHàmalate + NAD⁺ à malate into mitoà(mito malate + NAD+àOAA+NADH);

2. glycerophosphate shuttle- 1 cystol NADHà1 mitochondria FADH₂à2 ATP

-flavoprotein dehydrogenase- outer surface of inner membrane; can accept electrons from 3-phospholgycerol and give electrons to ET chain. Similar to succinate dehydrogenase.

-FAD à dihydroxyacetone, NAD à3-phosphoglycerol

D. ATP in cytosol produces ADP and Pi. Making ATP in mitochondria needs them as substrates.

1. ADP-ATP Translocator- transmembrane protein dimer, electrogenic antiport.

-electrogenic antiport- ADP (3-) transported from cytosol to mitochondria in exchange of the transport of ATP (4-) out.

-membrane potential- regulates net transport of ADP in and ATP out (more negative inside mitochondria)

2. Phosphate Transport- Pi in cytosol transported into mitochondria by P carrier

-phosphate carrier- Pi-H⁺ symport, driven by pH gradient (more acidic outside mitochondria)

Solution Manual Elements of Physical Chemistry 5e

2012-10-10T02:22:00.001-07:00

MIRROR (compressed version): https://www.dropbox.com/s/i5zs0ejvto81apw/Solution%20Manual%20Elements%20of%20Physical%20Chemistry%20Atkins%205e.rar
MIRROR 2(compressed, also need to create an account): http://www.4shared.com/rar/DT87L-2g/Solution_Manual_Elements_of_Ph.html

DOWNLOAD ONLY THE SOLUTIONS FOR CH1-CH5: https://www.dropbox.com/s/9jt4xrsus22q018/KIC%20Document.pdf?dl=0

Elements of Physical Chemistry Solutions Manual, by Peter Atkins (Author), Julio dePaula (Author), Charles Trapp (Author), Marshall Cady (Author)

Publication Date: September 25, 2003 | ISBN-10: 1429224002 | ISBN-13: 978-1429224000 | Edition: 5

"Solution manual to textbook Elements of Physical Chemistry by Atkins, 5th edition (5e)."

Buy it on Amazon

2012-10-01T16:14:00.000-07:00

3.2,3,5,7,10,14,15,16,19,20,22,24,26,30,32
4.1,2,3, etc., skip 10,13,20, 22, 24, 26, 30, 31, 32, 33

Physical Chemistry Chapter 2 Questions/Answers

2012-09-30T19:35:00.000-07:00

2.) Calculate the work done by 2.0 mol of a gas when it expands reversibly and isothermally from 1.0 dm³ at 300 K to 3.0 dm³

W=-nRTln(V_f/V_i)

=-(2.0mol)(8.31)(300)ln(3.0dm³/1.0dm³)

=-5.5kJ

3.) A sample of methane of mass 4.50 g occupies 12.7 dm³ at 310 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 30.0 kPa until its volume has increased by 3.3 dm³. (b) Calculate the work that would be done if the same expansion occurred isothermally and reversibly.

a.)W=-P_ext ΔV

=-(30000 Pa)*(3.3dm³)*(1m³/1000dm³)

=-99 J

b.)W=-nRTln(V_f/V_i)

=-(4.5g*(16.04 g/mol))*(8.31)*(310K)*ln(16/12.7))

=-167 J

4.) In the isothermal reversible compression of 52.0 mmol of a perfect gas at 260 K, the volume of the gas is reduced from 300 cm3 to 100 cm3. Calculate w for this process.

W=-nRTln(V_f/V_i)

=-(.052mol)*(8.31)*(260K)*ln(100cm³/300cm³)

=123.4 J

6.) A strip of magnesium metal of mass 12.5 g is dropped into a beaker of dilute hydrochloric acid. Given that the magnesium is the limiting reactant, calculate the work done by the system as a result of the reaction. The atmospheric pressure is 1.00 atm and the temperature is 20.2°C.

Mg_(s) + HCl _(l) --> MgCl_{2 (aq)} + H₂ _(g)

Moles of H₂= (12.5 g magnesium)*(1mol/24.305g)*(1mol H₂/1mol Mg)=0.514 mol

Volume of H₂= nRT/P= (0.514 mol)*(8.31)*(293.2 K)/(1 atm)=12.38 L

1 atm=101325 Pa; 1m³=1000L;

W=P_ext* ΔV= (101325 Pa)*(.01238 m³)

=-1.25 kJ

9.) What is the heat capacity of a sample of liquid that rose in temperature by 5.23°C when supplied with 124 J of energy as heat?

C=q/ ΔT=124 J/ 5.23 K

=23.7 J/K

10.) A cube of iron was heated to 70°C and transferred to a beaker containing 100 g of water at 20°C. The final temperature of the water and the iron was 23°C. What is (a) the heat capacity, (b) the specific heat capacity, and (c) the molar heat capacity of the iron cube? Ignore heat losses from the assembly

heat lost by iron=heat gained by water

mcΔT_iron= mcΔT_water

m*c*(70-23 K)=(100 g)*(4.18 J/g*C)*(23-20)

heat capacity = 26.68 J/K

molar heat capacity= heat capacity/molar mass=26.68/55.84= 0.478 J/mol*K

13.) When 229 J of energy is supplied as heat to 3.00 mol Ar(g), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas.

C_P=q/ (n*ΔT)=229/(3mol*2.55K)=29.9 J/mol*K

C_p=C_v+R; C_v=C_P-R=29.9-8.31=21.6 J/mol*K

2012-09-12T21:18:00.000-07:00

Internal Energy: U
-the total energy of a system
-4 components: vibration, rotations, translation of atoms/molecules, entropy
-depends on T and P (higher T = higher U)
-extensive property- depends on mass
-Um (molar internal energy) = U/n
-not measured absolutely
-ΔU=ΔProducts - ΔReactants
=W+q (W=energy transferred by work, if P is constant, W=0), (q=energy transferred by heat)
For a perfect gas, isothermal expansion results in the energy of the system being constant so, Δ=0 and q=-W.

Internal energy is a "state function", it depends only on the final and initial states, not on the pathway
Conclusion: ΔU, ΔG (Gibb's free energy), ΔH (enthalpy), and S (entropy) are all independent of the pathway, "state functions"
ΔU, ΔG, ΔH, and S are thermodynamic properties

1st Law of Thermodynamics
Isolated systems:
1.) cannot do work
2.)U is constant (1st law)

How do you measure change in U (ΔU)?
-conduct experiments at fixed V, thus V=0, and W=0, so ΔU=q at constant V (use bomb calorimeter)

Bomb Calorimeter
-used to measure C, C=q/ΔT, at constant V, Cv=ΔU/ΔT

enthalpy
-in the real world, most reactions take place in ways that are not at constant V
-very hard to work with ΔU in many circumstances, thus we use enthalpy (H)
H=U+PV
-extensive property (depends on mass);
Hm=H/n= Um*PVm -> ΔH=Qp @constant P
ΔH=ΔU+Δ(PV) *can measure ΔH but not H*

Chapter 2 Notes (unfinished)

2012-09-10T21:55:00.001-07:00

Chapter 2-Thermodynamics

thermodynamics- the study of the transformation of energy, particularly the relationship b/w heat and work
chemistry-revolves around the study of energy
-energy- the capacity to do work
-work- done to achieve motion against an opposing force
system-the part of the universe in which we are interested
surroundings-the region outside the system , from where we observe the system
3 types open-exchange matter/energy w/surroundings

closed-exchange energy w/surroundings

isolated-no exchange of energy or matter
1.) Energy can be exchanged b/w closed system and surroundings
-doing work (PV); expansion work; adiabatic (Q=0, no heat change);
-heating (energy is transferred because of temp differences); diathermic
2.) Both are modes of energy transfer, not forms of energy
-heat vs temp: T is an intensive property, and it determines the direction of energy flow
-heat, “q”, is energy in transit
diathermic-walls that permit heating (or cooling) as a transfer of energy, ex. a metal container
adiabatic-walls that do not permit energy transfer
-expansion means system does work, pushes against an external force

Measure work
-when a system does work, W is negative
-when work is done on the system, W is positive (unit of measure is Joules)
Expansion Work-for reactions that produce gas
W=-Pexternal*deltaV
=measure of work done by the system
-if no Pexternal, no work is done
exothermic-releases energy as heat (feels warm)
endothermic-absorbs energy (feels cool)
-

Chapter 1 Notes/Questions/Answers (Properties of Gases)

2012-09-08T00:57:00.000-07:00

Chapter 1 Notes/Questions/Answers

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Equation of State (EoS)- any gas can have its state specified by the following properties: Volume, Pressure, Temperature, N (number of moles);
Very few EoS are known, only:

Low pressure gases
Water and dilute aqueous solutions

Perfect Gas EoS: PV=nRT; P=f(n,V,T); R= gas constant=8.314 J/mol*K=1.99 cal/mol*K
Perfect- individual particles don’t interact at all, no chemical or physical interactions; most likely to be approached at low pressures.
Ideal Gas Law- ideal mixtures are mixtures in which all the interactions (particle-to-particle) are equal, but not necessarily zero. Perfect gases have 0 interactions, ideal gases have equal interactions.

Boyle’s Law- Pressure and volume are inversely proportional (at constant T, P ∝1/V).
Charle’s Law- Volume and temperature are directly proportional (V=A+BT, where A, B are constants).
Avogadro’s Principle- Volume and moles are directly related. All gases contain an equal number of molecules at a given volume.

-Vm(molar volume)=Vgas/number of moles; Vm=24.8 L at 1 bar.

Uses of Perfect Gas Law

PV=nRT-used to calculate P,V,n, or T when other 3 are given
P1V1/T1=P2V2/T2- combined gas equation; used when you have a known set of conditions that change
Vm=RT/P- used to calculate molar volume of a perfect gas at any P,T

Standard Ambient T and P (SATP)-298.15 K, 1 bar (1.013 atm); molar volume=24.8 L/mol
Partial Pressures of Gas Mixtures

PT=PA+PB...Pn; Total Pressure= Sum of Partial Presures;

PA=XA*PTotal, XA is the mole fraction (moles of A/total moles);

-what influences pressure of a certain gas is not its molecular weight (SF6 and H2 contribute equally), but moles of the gas.

Kinetic Model of Gases

Gas consists of molecules in ceaseless random motion
Size of molecules is negligible compared to distance traveled
Molecules don’t interact except during collision

Gas Pressure-each molecule’s impact on the container walls gives a brief force on the wall;

-infinite collisions;

-F=ma, thus heavier mass = higher concentration or higher speed of molecules (related to temperature), all causing more gas pressure

Maxwell Distribution of Speeds of Molecules
-at lower temperatures, the average speed is low and the range is small
-at higher temperatures, the average speed is higher and the range is broad
-there’s a threshold of energy level for chemical reactions, thus, at higher temperatures, molecules are more likely to reach that threshold and more likely to participate in that chemical reaction
-heavy molecules move slower and have smaller range, lighter molecules move faster and have a broad range

Diffusion-process by which molecules of different substances mingle with each other
-rates in solids are very low, liquids intermediate, and gases quick
Effusion-escape of gas through small opening
-both processes are temperature dependant (as temperature rises, rate rises)
Separation-distance between molecules; Forces almost exclusively electrostatic; Too close, they push. Too far, pull.
-Equilibrium exists between 2 molecules when minimum amount of energy consumed in keeping molecule stabilized via push and pull forces.

Molecular interactions affect bulk properties of gases, modify EoS
-isotherms (temp stays the same along line)are similar to ideal gases at high temps (ex.CO2 @ 50°C), very diff at low temps.
From A→B→C: CO2 is gas, as piston is pushed down, P increases and the gas compresses (Boyle’s Law);
From C→D→E: CO2 condenses. @ E it has condensed to a compact liquid, and piston rests on surface of liquid. No P change since it’s condensing, and the molecules are close enough that they attract each other and cohere into a liquid.
From E→F: CO2 is a already a very compact liquid, need a lot of P to further compress it (trying to force the molecules closer together when they’re already in contact/trying to overcome strong repulsive interactions

Gas can’t be condensed to liquid by P unless under the critical temperature.
-when you heat a liquid in a container, the pressure of the container increases, and the liquid becomes less dense and the gas becomes more dense. After a certain temperature, Tc, the densities of the liquid and gas are the same, and a single state of matter (super-critical fluid) exists (for all real gases)

Virial EoS: P=(nRT/V) (1+nB/V+nC2/V2...nXn/Vn), where B, C are empirical coefficients that fit experimental data
Van der Waals EoS: P=(nRT/(V-nb))-a(n/V)2

Questions/Answers
1.1) What pressure is exerted by a sample of nitrogen gas of mass 3.055 g in a container of volume 3.00 dm3 at 32°C?
V=(3.00 dm3)((10cm/1dm)^3)=(3000cm3)(1mL/cm3)=(3000 mL)(1L/1000mL)=3L;
n=(3.055 g)(1mol/28.01g)=.109 mol;
T=32°C, C+273=K→ 32+273= 305 K;
R=8.314 Jmol*K
PV=nRT→P(3L)=(.109mol)(8.314J/mol*K)(305 K)=92.1 kPa

1.3) Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect, a sample was collected in a container of volume 300.0 cm3. At 14.5°C its pressure is found to be 34.5 kPa. What amount (in moles) of NO has been collected?
P=34.5 kPa;
V=300cm3(1mL/cm3)=300mL(1L/1000mL)=.3 L
R=8.314 J/mol*K
T=14.5°C, C+273=K→14.5+273= 287.5 K;
pV=nRT→(34.5kPa)(.3L)=n(8.314 J/mol*K)(287.5K)= 0.00433 mol

1.5) The effect of high pressure on organisms, including humans, is studied to gain information about deep-sea diving and anaesthesia. A sample of air occupies 1.00 dm3 at 25°C and 1.00 atm. What pressure is needed to compress it to 100 cm3 at this temperature?
P1V1/T1=P2V2/T2→1.00atm*1.00dm3/25°C=P2*(100cm3(1dm3/1000cm3)/25°C=10 atm

1.7) Until we find an economical way of extracting oxygen from sea-water or lunar rocks, we have to carry it with us to inhospitable places, and do so in compressed form in tanks. A sample of oxygen at 101 kPa is compressed at constant temperature from 7.20 dm3 to 4.21 dm3. Calculate the final pressure of the gas.
P1V1/T1=P2V2/T2→101kPa*7.20dm3/T1=P2*4.21dm3/T2;T1=T2 ; P2= 173 kPa

1.9) Hot-air balloons gain their lift from the lowering of density of air that occurs when the air in the envelope is heated. To what temperature should you heat a sample of air, initially at 315 K, to increase its volume by 25 per cent?
P1V1/T1=P2V2/T2 ;T1=315 K; P1=P2; V2=125% of V1; V1/315 K=1.25*V1/T2=393.75 K

1.11) A diving bell has an air space of 3.0 m3 when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g cm−3 and assume that the temperature is the same as on the surface.
P2=ρ*g*h=1.025gcm31kg1000g100cm1m3*9.8ms2*50m=502.25 kPa
T1=T2; V1=S.T.P.=100kPa; P1V1/T1=P2V2/T2→100kPa*3.0m3/T1=502.25kPa*V2/T1=0.597m3

1.13) Atmospheric pollution is a problem that has received much attention. Not all pollution, however, is from industrial sources. Volcanic eruptions can be a significant source of air pollution. The Kilauea volcano in Hawaii emits 200–300 t of SO2 per day. If this gas is emitted at 800°C and 1.0 atm, what volume of gas is emitted?
P=1.0atm*101.3kPa1atm=101.3kPa;
T=800°C; C+273=K→800°C+273=1073 K;
n=200tons-300tons907185g1ton1mol64.064g=2.8326*106 to 4.248*106mol
R=8.314Jmol*K
PV=nRT→101.3kPa*V=2.8326*106 to 4.248*106mol*8.314Jmol*K*1073K=
=2.49*108 to 3.74*108L

1.15 A gas mixture being used to simulate the atmosphere of another planet consists of 320 mg of methane, 175 mg of argon, and 225 mg of nitrogen. The partial pressure of nitrogen at 300 K is 15.2 kPa. Calculate (a) the volume and (b) the total pressure of the mixture.
moles methane= .32 g*1mol16.04g= .02 mol
moles argon= .175 g*1mol39.948g= .00438 mol
moles nitrogen=.225g*1mol28.014g= .01606 mol
Pnitrogen=Xnitrogen*Ptotal→15.2kPa=.01606mol.02+.00438+.00803mol*Ptotal=61.246kPa
PV=nRT→61.246kPa*V=.02+.00438+.00803mol*8.314Jmol*K*300K=1.32L

1.17 A determination of the density of a gas or vapour can provide a quick estimate of its molar mass even though for practical work mass spectrometry is far more precise. The density of a gaseous compound was found to be 1.23 g dm−3 at 330 K and 25.5 kPa. What is the molar mass of the compound?
ρ=1.23 gdm3*1kg1000g10dm1m3=1.23 kgm3
PV=nRT→Pmρ=mMRT→MP=ρRT→M*25.5kPa=8.314Jmol*K*1.23kgm3*330K=132g/mol

1.19 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K. Calculate (a) their partial pressures and (b) the total pressure.
PV=nRT→P*22.4dm31m10dm3=2.0+1.0mol*273.15K*8.314Jmol*K=304kPa
Phydrogen=Xhydrogen*P=2/3*304kPa=202.6kPa
Pnitrogen=Xnitrogen*P=1/3*304kPa=101.3kPa

Chapter 0 Notes/Questions/Answers (Introduction)

2012-09-07T10:09:00.000-07:00

Chapter 0 (Introduction) Notes/Questions/Answers

DOWNLOAD for better formatting: https://www.dropbox.com/sh/5xa5upls8rhev73/S6a4qKWz9c

Notes

Know the difference between solids, liquids, gases;

F=m*a; W=F*d.

Energy- capacity to do work;

-kinetic energy- energy due to motion; Ek=1/2 mv2

-potential energy- energy due to position; Ep=mgh

Etotal= Ek+Ep; Etotal is constant if forces act on the body, can convert Ek to Ep and vice versa

Pressure-measured in pascals (Pa); F/A= 1 N/1 m²=1 kg/m*s²

1 atm = 1.013 bar; 1 bar = 10⁵ Pa;

Questions/Answers

0.3) Calculate the work that a person of mass 65 kg must do to climb between two floors of a building separated by 4.0 m.

W=F*d=(65 kg* 9.8 m/s²) * (4.0 m) = 2600 J

0.5) A car of mass 1.5 t (1 t = 103 kg) travelling at 50 km h−1 must be brought to a stop. How much kinetic energy must be

dissipated?

KE= ½mv2= ½(1.5 t)(103 kg/t)(50km/hr)(1 hr/3600 s)(1000 m/1km)= 14.9 kJ

0.8) Calculate the minimum energy that a bird of mass 25 g must expend in order to reach a height of 50 m.

PE=mgh=(.025 kg)(9.8 m/s²)(50m)=12 J

0.12) Express (a) 108 kPa in torr, (b) 0.975 bar in atmospheres, (c) 22.5 kPa in atmospheres, (d) 770 Torr in pascals.

a.) (108 kPa)(.007501 torr/1 Pa)(1000 Pa/1 kPa)=810 torr

b.)(.975 bar)(1 atm/1.013 bar)= 0.962 atm

c.) (22.5kPa)(1000Pa/1 kPa)(1 bar/10⁵Pa)(1 atm/1.013 bar)= 0.222 atm

d.) (770 torr)(1 Pa/0.007501 torr)=1.03*10⁵Pa

0.13) Calculate the pressure in the Mindañao trench, near the Philippines, the deepest region of the oceans. Take the depth there as 11.5 km and for the average mass density of sea water use 1.10 g cm−3.

(11.5 km)(1000 m/km)(1.1g/cm³)(1kg/1000g)(100 cm/1m)^3(9.8m/s²)= 1.24*10⁸Pa

0.18) Given that the Celsius and Fahrenheit temperature scales are related by qCelsius/°C = (qFahrenheit/°F − 32), what is the temperature of absolute zero (T = 0) on the Fahrenheit scale?

At absolute 0, K=0; C=K-273=0-273=-273; C=(f-32)/1.8; -273=(f-32)/1.8= -459.67 °F

0.24) The molar mass of the oxygen-storage protein myoglobin is 16.1 kg mol−1. How many myoglobin molecules are present in 1.0 g of the compound?

(1.0 g)(1 mole/16100 g)(6.02*10²³ molecules/mole)= 3.74*10¹⁹ molecules

0.28) What mass of sodium chloride should be dissolved in enough water to make 300 cm3 of 1.00 M NaCl(aq)?

Molarity= moles/Liters; 1.00 M= moles/((300 cm³)(1 mL/1 cm³)(1 L/1000 mL))=0.3 moles;

0.3 moles=grams/molar mass; 0.3 moles= grams/(58.443 g/mol)= 17.5 g

0.30) Calculate the mole fractions of the molecules of a mixture that contains 56 g of benzene and 120 g of methylbenzene (toluene).

Moles of benzene= (56 g)(1 mole/78.118 g)=.717 mol

Moles of toluene= (120 g)(1 mole/92.138 g)= 1.3 mol

X_benzene=moles benzene/moles total=.717/(1.3+.717)= .35

X_toluene=moles toluene/moles total=1.3/(1.3+.717)= .64

Essential Cell Biology 2nd Edition Alberts pdf

2012-09-05T19:11:00.001-07:00

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MIRROR 2: torrent on tpb

Essential Cell Biology, by Bruce Alberts (Author), Dennis Bray (Author), Karen Hopkin (Author)

Publication Date: September 25, 2003 | ISBN-10: 081533480X | ISBN-13: 978-0815334804 | Edition: 2

"Essential Cell Biology, Second Edition contains basic, core knowledge about how cells work. It has a proven track record in providing students with a conceptual and accessible grounding in cell biology. The text and figures have been prepared to be easy-to-follow, accurate, clear and engaging for the introductory student. Each section follows logically from the previous one, telling a story, rather than being a collection of facts. Questions integrated throughout each chapter encourage the reader to pause, think about what they have read, and attempt to apply the new knowledge in ways that test their understanding. Based on user feedback, the Second Edition now offers increased coverage of genetics and more experimental background. It is completely up-to-date."

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Elements of Physical Chemistry Atkins 5th Edition pdf

2012-09-05T18:32:00.000-07:00

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Title: Elements of Physical Chemistry, by Peter Atkins (Author), Julio de Paula (Author)

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TECH 0705 Post-Lab Questions: Seperating Acids and Neutral Compunds by Solvent Extraction

2011-09-28T22:11:00.000-07:00

1. Calculate percent recovery... not hard
2. When the first washing was done with NaOH, the strong attraction of the OH- pulled both compounds into the aqueous solution, leaving nothing to react with the acidification of the NaHCO3 layer.
OR
Appartently, you had benzoic acid, naphthalene, and 2-naphthol dissolved in ether. If you mistakenly extracted that solution with aqueous NaOH, then:

C6H5COOH + NaOH ===> C6H5COO-Na+ + H2O

C10H7OH + NaOH ===> C10H7O-Na+ + H2O

C10H8 + NaOH ===> No reraction

So both the benzoate and the 2-naphtholate would have gone into the water solution and the naphthalene remained in the ether solution. If you had extracted the ether solution with NaHCO3 solution next, only the naphthalene would be left in the ether solution, nothing would go into the NaHCO3 solution, and no crystals would be formed on acidification of the NaHCO3 solution.

3. p-tert-butylphenoxide anion (The sodium salt of p-tert-butylphenol)

4.Dichloromethane would sepaprate into the lower layer of the separatory funnel instead of the top, making extraction more difficult. Instead of using the funnel's stopcock, one would need to extract the top layer from above, leaving the bottom layer intact for further washes. (Check density)
The only change you would make is you would save the opposite layer. For example, if you were extracting an organic into the organic layer (tbme), you would save the top layer and discard the bottom (aqueous). If you used dichloromethane instead, you would save the bottom layer (MeCl2) and discard the top (aqueous).

5. Make a flowchart (http://i.imgur.com/gji6F.gif) also refer to (http://www.sas.upenn.edu/~zhall/flowchart.html) for details and cool stuff and unicorns and cute puppies

6. Lower/wider melting point range than expected implies less purtiy, nuff said

TECH 0705 Results: Seperating Acids and Neutral Compunds by Solvent Extraction

2011-09-28T21:27:00.000-07:00

Results
SET 1:
Weight Before Extraction
p-Toluic Acid-0.513g
Acetanilde-0.256g
p-Tertbutylphenol-0.517g

Weight After Extraction
p-Toluic Acid-0.760g
p-Tertbutyl phenol-0.149g

Expected Melting Point Experimental Melting Point
p-Toluic Acid 180-182 174-180

p-Tertbutylphenol 98-101 91-98

SET 2:

Weight of initial acetanilide: 0.25606 g
Weight of initial p-toluic acid: 0.6079 g
Weight of initial p-tert-butylphenol: 0.5547 g
Mass of filter paper to filter out p-toluic acid: 0.013 g
Mass of test tube used to collect p-toluic acid: 7.934 g
Mass of filter paper to filter out p-tert-butylphenol: 0.006 g
Mass of test tube used to collect p-tert-butylphenol: 7.986 g
Mass of large test tube used to collect acetanilide: 38.7405 g
Weight of p-toluic recovery and test tube: 8.507 g
Weight of p-tert-butylphenol recovery and test tube: 8.172 g
Weight of acetanilide recovery and test tube: 39.141 g
Melting range of p-tert-buyltphenol: 82-89 C.
Melting range of p-toluic acid: 179-185 C.

SET 3:
Mass of the acetanilide: 0.31 g.
Mass of the p-toluic acid: 0.57 g.
Mass of the p=tert-butylphenol: 0.45 g.

Isolating P-toluic Acid:
Mass of the filter paper: 0.0061 g.
Mass of dry p-toluic acid, test tube and filter paper: 5.476 g.
Mass of recovered p toluic acid: 0.084 g.
Measured melting point of p- toluic acid: 179.3-180.2 degrees Celsius
Actual melting point of p-toulic acid: 180-182 degrees Celsius

Isolating p-tert-butylphenol:
Mass of the filter paper: 0.0066 g.
Mass of the empty test tube: 5.324 g.
Mass of the dry p-tert-butylphenol, test tube, and filter paper: 5.64 g.
Mass of collected p-tert butylphenol: 0.3094 g.
Measured melting point of p-tert-butylphenol: 98.1 degrees Celsius
Actual melting point of p-tert-butylphenol: 98-101 degrees Celsius

Isolating Acetanilide
Mass of anhydrous Na₂SO_{4: 1.003 g

Mass of 100 mL test tube: 5.288g.

Mass of dry acetinilide and test tube: 5.43 g.

Mass of recovered acetinilide: 0.142 g.

Measured melting point of acetanilide: 88.9 degrees Celsius

Actual melting point of acetinilide: 113-115 degrees Celsius}

TECH 0707: TLC Plates Post-Lab

2011-09-15T02:56:00.000-07:00

Post-Lab

1a) The eluents in order of increasing polarity are hexane, toluene, dichloromethane, ethyl acetate and methanol.
1b) Hexane is very non polar so it causes the Rf the stock solution to stay low, and doen't allow for seperation of the compounds in the solution. Toulene and dichloromethane allows the compounds in the solution to seperate because it has an intermediate polarity and would thus give a range of Rf values based on the polarity of the individual compounds. Ethyl acetate and methanol are both very polar so they don't allow the compounds that are more non-polar to seperate out and would carry them all together towards the eluent front. This would then give a higher Rf value.
1c) Compound polarity and the polarity of the eluent determine how big of a Rf value you will get, if the compound is non-polar but is put into a very polar eluent, your Rf value will be higher because the stronger attraction from the eluent.

2) We chose to use toluene as the eluent because it seperated the compounds well. Hexane, ethyl acetate and methanol didn't allow the compounds to seperate at all so it would have been foolish to chose them as an eluent. The only other option was dichloromethane, which also seperated the compounds from the solution but we didn't choose it for arbitrary reasons.

3a) If a sample that is too large is applied while spotting the plate it may lead to poor seperation .
3b) If a spotted plate is put into a developing chamber that has an eluent level above the origin the compounds that were spotted on the plate may be dissolved in the eluent, with none left on the TLC plate.
3c) If you allow the plate to remain in the developing chamber until the eluent reaches the top of the plate you will not be able to have a eluent front to compare to the spots to give Rf values.
3d) If you remove the plate from the chamber when the eluent front has moved only halfway up the plate , you will not allow the plate to fully develop and the compounds to reach their polarities.
3e) If you develop the plate in an uncovered chamber you could allow the eluent to evaporate, and not travel up the TLC plate.

4) You could use a TLC plate to tell when the reaction is complete when the solution shows a compound that has a low polarity as indicated by the TLC plate.

1. a) The five eluents could be arranged as follows in order of increasing polarity.

Hexane < Toluene < Dichloromethane < Ethyl acetate < Methanol

b) The effects of eluent polarity on the Rf value of each stock solution mixture compound.

Based on the experiment I could say that two of the compounds were either too polar or too non polar which means that either did not move from the origin at all (hexane- too non polar) or they went very close to the top of the plate (methanol – too polar).

The rest three compounds (dichloromethane, ethyl acetate and toluene) gave us pretty good separation by leaving two to three spots on the plate.

c) As stated in general chemistry II “Like dissolves like” means that polar compounds will interact will polar and non polar with non polar. Therefore, the Rf value will depend on plarity.

We selected Toulene as our eluent because it gave us the best separation by leaving three spots on the plate. Dichloromethane gave us also three spots but in the middle of the experiment something did not work out very well and we decided to take it out of the list. The other compounds either did not separate at all or gave us one or two spots.
a.) If we apply too much of the sample on the plate, we are risking of getting not very good results because we might not be able to distinguish the spots and they might get mixed up if the distance between them is very small. Moreover, we will get poor separation.

b.) If we put the plate in the chamber with eluent level above the level of the plate, we risk the sample to dissolve from the plate into the eluent layer.

c.) If we forget and leave the plate in the chamber until the eluent front reaches the top of the plate, then the mixture compound may keep moving along the plate which will lead us to not valid Rf measurements.

d.) If we remove the plate from the chamber before it reaches the eluent front, then we might not be able to see the full process of separation. This will also lead us to not valid Rf values.

e.) If we leave the chamber open and keep developing the plate, then the eluent will evaporate in the air and will not allow proper separation of the plate which again will lead to wrong Rf values if any at all.

4. If you prepared benzhydrol by the reduction of benzophenone with sodium borohydride, we could use TLC to decide when the reaction was completed by performing the same experiment with compounds that will give separation and mark the spots. Then do the experiment again with sodium borohydride and spot it. The result will be on the same spot as benzhydrol.

TECH 0707: TLC Plates Results

2011-09-15T02:53:00.000-07:00

Results

I was assigned to do the Thin-Layer Chromatography on dichloromethane as a reagent for the first part of the lab. The middle set of dots on my TLC plate was due to error, while spotting the plate my capillary tube must have touched the TLC plate intentionally.

The distance between the origin and eluent front of dichloromethane was 6.0 cm.
From other team members TLC Plates: Hexane didn't move the eluent very far from the origin and didn't seperate so we decieded it is non-polar. Ethyl acetate and Methanol did move the eluent far up closer to the eluent front, thus showing that it is more polar, but also didn't seperate the compounds within the mixture. Toluene was the only other eluent that seperated the mixture of Biphenyl, Benzhydrol, and Benzophenone into it's individual compounds. So as a group we decided to use Toulene as being thMy assigned compound for this section of the lab was biphenyl. I spotted the stock solution on the left side of the TLC plate and biphenyl on the right side. After the plate had been developed, it appears that toluene takes much longer to evaporate than dicholoromethane. The distance between the origin and the topmost spot was 3.8 cm and the distance between the origin and the eluent front was 5.4 cm.
e best for seperating the mixtures later in this lab since my TLC plate had an error (the middle dots).

My unknown solution was #13 , and Toluene was used as the eluent. The left-most set of spots is the stock solution and the right-most set of spots is the unknown. Based on the results of the TLC plate, my unknown solution was comprised of Biphenyl and Benzhydrol. The distance between the origin and the eluent front is 5.8 cm, the distance between the origin and Biphenyl is 4.5 cm, and the distance between th origin and Benzhydrol is .5 cm.

TECH 0708: Adsorption Column Chromatography Post-Lab

2011-09-15T02:49:00.000-07:00

Post-Lab

2. As hexane was added to the column, Acetylferrocene did progress down the column although very slowly. It wasnt until TBME was added that acetylferrocene went down the column at a more rapid pace.
3. Had hexane been the only eluent added to the column, the ferrocene would have come out of the column but since the acetylferrocene is more polar, it has a stronger attractive force to the alumna and wouldn't proceed down the column bed.
4. If you had used TBME as the only eluent added, you wouldn't achieve seperation, both the ferrocene and acetylferrocene would proceed down the column bed together since ferrocene is more non-polar and holds less attraction to the absorbant,alumna.

TECH 0708: Adsorption Column Chromatography Results

2011-09-15T02:47:00.000-07:00

Results

    After the hexane was added to the pipet, yellow was the first color to separate out of the solution, as was expected since ferrocene is less polar. It seperated out fairly quickly once the hexane had reached the 50/50 ferrocene/ acetylferrocene mixture. Acetylferrocene gradually does seperate out with hexane and moves down the column bed much more slowly than ferrocene and is dark orange in color. As the ferrocene moves down the column, the yellow and dark orange colors do seperate. Air was used to force the hexane down the pipet at a faster rate, it worked well to move the hexane as well as ferrocene down the column.
    Once the ferrocene was through the column, the acetylferrocene moved more slowly when TBME was added to the column, than the speed at which the ferrocene moved down the column with hexane. Some of the Acetylferrocene / TBME that was collected accidently spilled. We were also told to stop collecting the acetylferrocene eluent when the drops were clearer although not completly colorless.
    After spotting TLC plates at different ratios of eluents, as a group we determined that the best separation of the ferrocene and acetylferrocene came from a 25% Ethyl Acetate/ 75% Toluene mix.

TECH 0701: Melting Points

2011-09-15T02:42:00.001-07:00

Conclusion

In this experiment, we need to be careful when measuring the melting point of compounds because if we put too much of the sample into the capillary tube there would be poor heat transfer and if the compound is not a very fine powder, we would not get very good heat transfer. Mistakes can occur when measuring the melting point because there might be some king of changes in the compound such as sagging or shrinking and one can take it as the start of melting which actually is not. Another process that we must be aware of when taking the melting point of a compound is called sweating. Sweating is not melting. Sweating refers to process when we might have traces of moist due to not very well drying of the tubes.
All these possible errors can occur and this would lead us to wrong results.

OR

Within this lab we observed the characteristics of melting points and how they could be used to identify an unknown substance.

When we melted benzoic acid the melting point we had found was between 123-125 degrees celcius which is higher than what the literature had stated and when we had melted mandelic acid, we had observed it starting to melt earlier than the literature had stated although it did complete melting at the accepted temperature.

It was determined that for the unknown substance #14 the melting point had started at 153 degrees celcius and had completely melted at 155 degrees celcius. Given this range, we know that the unknown substance could be benzilic acid which has a melting point between 150-154 degrees celcius or adipic acid which has a melting point between 152-154 degrees celcius. Based on my results of melting the unknown compound with benzilic acid and then with adipic acid I am led to believe that unknown compound #14 was adipic acid. When melted together with adipic acid, the unknown had a narrower range of when it began to melt and when it completed melting as well as it's melting point had not been lowered due to an impurity as was the case of when the unknown had been mixed together with benzilic acid and had melted.