CH3F1 ANSWERS TO QUESTIONS FROM PRU.pdf

CH3F1 QUESTIONS FOR WORKSHOP on PRU part of module (surfaces) 1. (i) Use of N() = C-D/2, with a set of values of N() -  allowing determination of D. ln(N) = lnC –D/2 ln() Adsorbate area (σ)/Å2 16 35 53 70 ln σ 2.77 3.55 3.97 4.20 Amount ads/(10-4 mol g-1) 5.00 1.95 1.20 0.98 ln(N) 1.609 0.667 0...

CH3F1 QUESTIONS FOR WORKSHOP on PRU part of module (surfaces) 1. (i) Use of N() = C-D/2, with a set of values of N() -  allowing determination of D. ln(N) = lnC –D/2 ln() Adsorbate area (σ)/Å2 16 35 53 70 ln σ 2.77 3.55 3.97 4.20 Amount ads/(10-4 mol g-1) 5.00 1.95 1.20 0.98 ln(N) 1.609 0.667 0.182 -0.020 Note: You do not need to change any units. ln - ln plot. Slope of plot (next page) gives D = 2.28 (note: answer should be between 2 [flat surface] and 3). Answers between 2.2 and 2.3 expected. (ii) For N2 adsorbate the surface area is 5 x 10-4 mol g-1 x 6.02 x 1023 mol-1 x 16 Å2 = 4.82 x 1021 Å2 g-1. For anthracene, surface area is 0.98 x 10 -4 mol g-1 x 6.02 x 1023 mol-1 x 70 Å2 = 4.13 x 1021 Å2 g-1 The difference in the 2 answers is a consequence of D > 2. Smaller molecule ‘sees more of surface’ and hence the larger surface area of sample. 1 2 2. (a) STM - probe : sharp and hard metal wire (Pt/Ir or W) feedback : electron tunnelling between to tip and substrate (both need to be made of conducting materials) AFM - probe : cantilever with a sharp end (Si or S3Ni4 (silicon nitride)) and known spring constant feedback : elastic force (Hook’s law: F=kx) which is measured by displacement of laser on photodiode. SICM – probe : hollow glass pipette containing electrolyte and a quasi-reference counter electrode feedback : Ion migration limited by the resistance of the tip end and gap. (b) The AFM cantilever is immersed in electrolyte solution. And a 3-electrode electrochemical set up will be used to control the potential at the substrate (WE) measure the WE current, using a bipotentiostat. The potential is applied between the WE and RE and the current measured between the WE and CE. and is Advantages of in situ EC-AFM compared to EC-STM: i) The topographical feedback on AFM is independent of electrochemical current from the substrate. (STM feedback can be convoluted with electrochemical processes at the surface of the substrate and the tip) ii) Study where the electrochemical processes changes the surface with nonconducting material ( e.g. polymerisation and metal passivation with oxide layers) iii) Provides information hydrophobicity etc) of other surface (Exam question from 2012) ANSWER 3. (a) From notes   o  surface tension of pure solvent surface tension of solvent/solute system 3 properties (adhesion, elasticity, (e) From notes. Could supply sketch e.g. The above is an old sketch. These is a more modern sketch in notes. Also – I provided links in lecture notes to some short videos from UCL. The first one is the most relevant. Amphiphile is usually dissolved in a volatile solvent eg, petroleum ether or dichloromethane and then added dropwise to the water surface. After allowing a few minutes for the solvent to evaporate and the amphiplile to spread out across the surface, a film is formed and then slowly compressed by the moving barrier. Measure surface pressure exerted by film on the mica film. In simple form, a force is applied to the torsion wire to keep the mica float in a fixed position. Plot Surface pressure vs. Area to obtain isotherms. The shape of these curves tell us about INTERMOLECULAR INTERACTIONS in the MONOLAYER (from notes) 3. (sorry for the 2 different q 3!) This question highlights how surface tension is related to the degree of molecular interactions in the bulk solvent. Heating the water disrupts hydrogen bonding and hence the surface tension decreases as the temperature increases. 4. This question is largely about solvent effects on adhesion force measurements with AFM.  CH3-CH3 in water is highest, because you bring together 2 hydrophobic surfaces in water. -CH3/water has high surface energy. Some of this area is eliminated when you bring -CH3 on tip into contact with -CH3 on surface. 4  Adhesion force decreases dramatically CH3-CH3 in ethanol, because –CH3/ethanol surface energy is smaller.  Note that COOH-COOH is relatively small, because although there is strong hydrogen bonding when these groups contact, there is already strong H-bonding of the surfaces with water. 5. (SIMILAR TYPE OF PROBLEM TO Q1) (i) Material area is: Adsorbate cross-sectional area x NA x Amount adsorbed N2 16 Å2 x 20.0 x 10-4 x 6.02 x 1023 g-1 = 192 m2 g-1 164 m2 g-1 Benzene as above (35 x 7.8) 153 m2 g-1 Naphthalene (53 x 4.8) Apparent area of sample decreases as the molecular probe measuring the surface goes up – a consequence of the fractal dimension of the sample (ii) ln [N()] = ln C – (D/2) ln  ln (20) = ln C – (D/2) ln (16) ln (4.8) = ln C – (D/2) ln (53) ln(20/4.8) = – (D/2)ln (16/53) (1.427/-1.198) = -D/2 Hence D = 2.38 6. For part (a) see question 3. (b) Data and images taken from Biomacromolecules2007,8, 8, 2611-2617 (“Chitosan as a Lipid Binder: A Langmuir Monolayer Study of Chitosan−Lipid Interactions”). There is evidently association of chitosan with -linolenoic acid, causing the molecular area to increase for a given surface pressure (1 mark). The paper suggests that (a) molecules of fatty acids may form electrostatic complexes with chitosan through interactions of their carboxylic groups with −NH 3+ groups of chitosan [note the pH of the solution], (b) chitosan molecules may get accommodated in the monolayers, possibly through hydrophobic interactions. (1 mark for either of these suggestions or any other relevant ideas). 7. 5 6 7 8 8. Read off Fi = 4.4; F0 = 1.9; F = 3.08 D = r2/(4tD); tD is time for fluorescence to attain value (F  - Fo)/2 + F0 = 0.7 +1.9 = 2.6 giving tD = 35 s so D = (12m)2/(4 x 35 s) = 1 m2 s-1 Mobile fraction = (F - Fo)/(Fi - Fo) = (3.08 - 1.9)/((4.4 - 1.9) = 0.47 9. (i) Contact mode or tapping mode AFM. The output would be a sequence of topographical images. The best answers need to give some thought about measuring the velocity as stipulated in the question (not covered in lectures). (If students have done very extensive additional reading, they might have come across disengaging one scan direction and scanning back and forth perpendicularly over a step). (Answers also might mention mass transport). Other microscopes covered, e.g. SICM, would not have the resolution. [l-cystine dissolution and growth covered as example in lecture material]. (ii) In-situ electrochemical STM. Expect diagram with 4 electrode set up, explaining role of each electrode, including insulated STM tip. [Copper electrodissolution covered as an example in lecture material]. (iii) Best would be SECCM. Will also accept SECM-AFM as a technique, but would need good description of the tip and set up (not covered for application to an electrochemical problem). SECM could also be a possibility, but would expect some discussion of issue from lack of positional feedback and need to deploy small-scale tip, and – again – the set up. Expect to see diagrams and description of the electrochemical set ups. Knowledge of activity at edge > terrace sites will be credited. [note that we did not cover this in details this year and I also indicated that SECCM is not needed this year]. 10. The tunnelling current has an exponential dependence, with a negative sign, on the tip-sample separation. As such, the tunnelling current will exponentially decay from the initial value of 3.5 nA. Even without having all the data necessary to evaluate this decay quantitatively, it should be remembered that for variations of about 1 Å in the tip-sample separation, the tunnelling current varies by roughly one order of magnitude. As a consequence, the schematics sketch should indicate that after the tip-sample separation has increased to 1.0 nm – i.e. it has undergone a variation of 6 Å from its initial value – the current has decreased by several orders of magnitude and thus it will be essentially zero. The following could be an example of the schematic plot requested: 9 10 11. 11