ANSWERS WORKSHOP 2 2019-2020 PDF
Document Details
Uploaded by CheaperBlueLaceAgate
Warwick
2019
Tags
Summary
This document includes questions and answers from a chemistry workshop 2 exam paper from 2019. The document covers topics like adsorption, surface areas, and different types of microscopy. It includes calculations related to surface tension and concepts from physical chemistry.
Full Transcript
CH3A9 QUESTIONS FOR WORKSHOP 2 (on PRU part of module) 1. Adsorption studies were carried out on a copper catalyst and the maximum amount of adsorption with various adsorbates was determined, as summarised in the table below. Adsorbate Adsorbate cross-sectional area/Å2 Amount adsorbed/10-4 mol g-1)...
CH3A9 QUESTIONS FOR WORKSHOP 2 (on PRU part of module) 1. Adsorption studies were carried out on a copper catalyst and the maximum amount of adsorption with various adsorbates was determined, as summarised in the table below. Adsorbate Adsorbate cross-sectional area/Å2 Amount adsorbed/10-4 mol g-1) N2 Benzene Naphthalene Anthracene 16 35 53 70 5.00 1.95 1.20 0.98 (i) Using a graphical method determine the fractal dimension of the catalyst. [25%] (ii) What are the surface areas determined from the measurement with N2 and anthracene? Comment on any differences. [15%] (part exam question from 2012) ANSWER (i) Use of N() = C-D/2, with a set of values of N() - allowing determination of D. See Topic 4 (MK & PRU) if you need more information 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 Slope gives D = 2.25 (note: answer should be between 2 [flat surface] and 3) For N2 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. 2. (a) The figure below illustrates a basic configuration of scanning probe microscopy (SPM). Describe type of probe and feedback utilised in scanning tunnelling microscopy (STM), atomic force microscopy (AFM) and scanning ion conductance microscopy (SICM) [10%] ANSWER: STM - probe : sharp and hard metal wire (Pt/Ir or W) In STM, the measured quantity is the tunnelling current, originating from 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 In AFM, the measured quantity is the displacement of a laser on a photodiode detector, originating from forces between the tip and the sample causing a deflection of the cantilever according to Hooke's law. SICM – probe : hollow glass pipette In SICM, the measured quantity is the ion current, originating from charge transfer due to migration of charged ions in solution. The measured quantity is fed into a control system, which tries to keep the quantity constantly equal to the reference value. (b) How can a setup be designed for in situ electrochemical (EC)-AFM and what are the advantages of it compared to in situ EC-STM? [10%] ANSWER: the AFM cantilever is immersed in electrolyte solution. And 3-electrode electrochemical set up will be added to independently control potential at the substrate. Advantages over in situ EC AFM: 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) ii) Study where the electrochemical processes changes the surface with non-conducting material (Polymerisation and metal passivation with oxide layers) iii) Provides information of other surface properties (adhesion, elasticity, hydrophobicity etc) 3. (a) The Gibbs adsorption isotherm at constant temperature may be written as: d RTs dcs cs (1) Define the terms , s and cs [10%] The surface tension of aqueous solutions of a non-ionic surfactant was measured as a function of concentration at 20 oC with the following results: surfactant concentration/ µM 10 20 50 100 Surface tension/ mN m-1 65.0 63.3 61.1 59.4 (b) Explain the trend in the data. [10%] (c) Assuming the surface excess is constant over this range of data, derive a form of equation (1) from which you can construct a linear plot to determine the surface excess. Produce the plot and use it to determine both the surface excess and the average area occupied by a surfactant molecule at the solution/air interface. [40%] (d) Explain how surface pressure and surface tension are related. [10%] (e) Describe how the Langmuir film balance may be used to measure the surface pressurearea characteristics of surfactant films. [15%] (Question from 2012) ANSWER 3. (a) From notes: - surface tension, s - surface excess of solute, cs - concentration of solute (b) Tests understanding of notes. Decrease in surface tension with increasing solution concentration largely due to increased surface concentration at the water/air interface. (c) Unseen problem not covered in lectures. Integrate eq (1) to: RT ln c const Plot vs ln c and slope is -RT Slope = -2.4 mN m-1 = -RT . T = 293K, So = 1.0 x 10-6 mol m-2 This gives area per molecule as 1/(1 x 10-6 x 6.02 x 1023) = 166 Å2 (1.66 nm2) (d) From notes (e) Tests understanding of notes. Could supply sketch, e.g. 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 4. (a) What is atomic force microscopy (AFM) and how may be used to measure interaction forces between two surfaces? [20%] (b) The AFM force curves ((a)-(e)) between a Silica (SiO2) microsphere and a Titania (TiO2) crystal were obtained at pH values of 8.8, 7.2, 6.3, 5.8, and 3.2, with 1 mM KNO3 background electrolyte. Explain which curve corresponds to each pH value in the figure based on the following information. [20%] Isoelectric point of SiO2 = pH 3.0 Isoelectric point of TiO2 = pH 5.6 (Isoelectric point: pH where the net charge = 0) (a) (b) (c) (d) (e) ANSWER (a) Tip or modified tip translated to and from a surface and the deflection monitored (via laser from back of tip). Piezos used for tip movement. Deflection converted to force using Hooke’s Law. Example below of force curves for a general surface. Can also measure colloidal forces and binding force, etc. (b) ae : pH 8.8 pH 3.2 The surface charges of both materials are mainly determined by the pH. Silica has an isoelectric point around pH 3.0, while the isoelectric point of titania is pH 5.6. As a consequence at high pH, where both materials are negatively charged, an electrostatic repulsion is observed. The repulsion decreases as the pH decreases, and at pH 3.2, i.e. below the isoelectric point of titania, there is an electrostatic attraction as well as a van der Waals force resulting in an overall attraction between the two surfaces. 5. The plot below shows the surface tension of the air-water interface as a function of temperature. Explain the trend in the data. (Part question from 2016) [10%] ANSWER 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.