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Exp.2 Time-resolved Thermal Lens Calorimetry

Exp.2 Time-resolved Thermal Lens CalorimetryAffiliation이화여자대학교 화학나노학과AbstractIn this experiment, we use an optical system of He-Ne laser beam and a computer program to find the heat capacity and molar absorptivity of solutions. Since the solution contains solute, a thermal lens effect occurs, and the oscilloscope displays a decreasing light intensity over time. The heat capacity for methanol is 203.60J/molK in both solutions. The heat capacity of acetone is 142.51J/molK and 238.68J/molK with malachite green oxalate and azulene respectively. The molar absorptivity of azulene in methanol is 0.547cm-1M-1 while the molar absorptivity of azulene in acetone is 0.383cm-1M-1.IntroductionWhen studying the reaction between matter and the energy radiated by the reaction, we rely on spectroscopy. It is effective in analyzing and investigating molecular structures and dynamics since each atom or molecule displays a distinctive spectrum. There are five major types of spectroscopy including Infrared different medium. When the light source is directed to the solution, the light hits the solution and is absorbed by the solute. In this state, we assume the solvent has no absorption mechanism and that the light will just transmit through the solvent. The solute that absorbed the light then sends the electron to the excited state and the electron wants to come back to the ground state. Ways to do it are by emitting heat or by emitting fluorescence.When the solution is composed of pure solvent, absorption does not occur. The convergence angle should be the same as the divergence angle since the refraction index is consistent over medium. However, when the solution consists of solute molecules, absorption occurs. As the heat spreads, the spreading of the transmitted beam occurs. This heat causes the decrease of the refractive index, and therefore following Snell’s law increases the refractive angle. Note that the temperature of the sample is the highest at the center of the beam where tfitting the data to the equations above. As a result, we can find the heat capacity of the solvent and molar absorptivity of the dye material can be calculated using Beer’s law.(3)InstrumentsOptical systemVolumetric flasks; 5mL, 50mL10mL pipetteCuvetMethanolAcetoneMalachite green oxalateAzuleneExperimental methodsPrepare 5mL of azulene solution with methanol and acetone as a solvent respectively. Prepare 50mL of malachite green oxalate with methanol and acetone as a solvent respectively. Be sure to use the right volumetric flask for the solution being made. Never tilt the pipette after it touched the chemicals.MethanolAcetoneMalachite green oxalate0.004635g0.004635gAzulene0.032g0.032gTurn on the He-Ne laser instrument and let it warm up for 10 minutes before getting into the further experiment.Adjust the reflective mirrors, chopper, and focusing lens in a way that allows the laser light to pass through the lens and be directed into the oscilloscope.Prepare a sample in a cell or a cuveten we compare the two equations I(t)= and the above, we know the value of I(0) and I(. I(0) – I(=0.69253, I()=2.28658, tc=0.00336msSince , is calculated as follows.=With this data, heat capacity is calculated as follows.203.60J/molKThe molar absorptivity of dye can be obtained from Beer’s law and it requires the absorbance value that can be obtained from equation 3. Converting equation 3 in a form of Beer’s law, the absorptivity of dye is obtained.=263.69cm-1M-1Repeat the same process with methanol solution containing azulene.I(0) – I(=0.76815, I()=2.44422, tc=0.00336msThe value of the heat capacity is then203.60J/molK=0.547cm-1M-1(0) – I(=0.96759, I()=2.53151, tc=0.00163msThe value of the heat capacity is then142.51J/molK=204.90cm-1M-1I(0) – I(=0.78531, I()=2.20044, tc=0.00273msThe value of the heat capacity is then238.68J/molK=0.383cm-1M-1ConclusionSince we used He-Ne laser beam in this experiment, we could observe the red light traveling in the optical system. He-Ne laser beam has 6nsmitted beam spreading occurs and refractive angle decreases following Snell’s law. As the laser light enters the sample solution, the heating causes the spreading of the beam and the power per unit area detected decreases.We would have detected the fixed power over fixed area if it were not for the absorption of the solute. However, because of the heating, the same amount of light gets spread into larger area, showing the decreased power density. The thermal lens effect grows stronger as time goes by due to the heating. We obtained the electrical signal of the laser through conversion by photodiode and an exponentially decreasing shaped graph is displayed. Using the computer program, we could obtain the values necessary to deduce the heat capacity and molar absorptivity of dye.Although we put different dye materials in solvent, theoretically we should have obtained the same data for the heat capacity once the solvent is the same. While the value of heat capacity for methanol was 203.th

자연과학| 2022.06.02| 11페이지| 3,500원| 조회(310)

열량계, 화학실험, 화학실험기법

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Exp 3. Computational Chemistry with Electronic Structure Methods

Exp 3. Computational Chemistry with Electronic Structure Methods(Quantum Chemistry Calculation: SN2 Reaction)화학실험기법1Affiliation이화여자대학교 화학나노학과AbstractIn this experiment, we use GaussianView in computational chemistry to conduct three experiments. In the first part, we calculate the energy of 1,2-Dichloro-1,2-Difluoroethane stereoisomers. The energy of RR form is 749692 kcal mol-1. It has a dipole moment of 2.7567 Debye. The energy of meso form -749699 kcal mol-1. And the dipole moment is 0.0184 Debye. In the second part, we calculate and visualize molecular orbitals of ethylene and formaldehyde. The total energy of formaldehyde is -113.86 a.u. and the total energy of ethylene is -78.032 a.u. Ethylene’s HOMO depicts the ∏ bonding orbital while LUMO depicts the ∏ * antibonding orbital. Formaldehyde has two plane of the node with HOMO in the bonding orbital and LUMO in the ∏ * antibonding orbital. In the last part, we construct an optimized transition state of a simple SN2 reaction and detn redistribution of atoms or molecules in certain reactions.Though this technique provides a precise interpretation and prediction of many chemical reactions, there are some limitations such as the time required to conduct the calculation and the size limit of molecules that can be studied. To conduct this experiment, we need The Gaussian09, a program for quantum chemical calculation. However, since it is not user-friendly, we use GaussianView which provides better instructions for first-time users.This experiment is divided into three parts. In the first part, we calculate the energy of 1,2-Dichloro-1,2-Difluoroethane stereoisomers. There are three possible structures – RR form, SS form, and meso form. RR form and SS form might seem as if they are the same structures, however, they rotate in different directions and are enantiomers. Unlike these two, meso form is achiral.In the second part, we calculate and visualize molecular orbitals of ethylene and formaldehyde. Ethylene and formalangle by following.Bond length: 1.316 (C-C), 1.073 (H-C-C-H)Bond angle: 122.0 (C-C-H)Dihedral angle: 0 or 180.0 (H-C-C-H)Open Gaussian Calculation Setup and calculate the energy and dipole moment with RHF/6-31G(d).Record the obtained dipole moment after calculation and compare HOMO with LUMO.Create a new molecular group to make formaldehyde. Adjust the bond length and angle by following.Bond length: 1.26 (C=O), 1.07 (C-H)Bond angle: 120.0 (O-C-H)Open Gaussian Calculation Setup and calculate the energy and dipole moment with RHF/6-31G(d).Record the obtained dipole moment after calculation and compare HOMO with LUMO.Create a new molecular group to make F-CH3-Cl. Adjust the bond length and angle by following.Bond length: 2.20 (C-F), 2.20 (C-H)Bond angle: 120.0 (O-C-H)Open Gaussian Calculation Setup and find the structure of the optimized transition state using RHF/6-31G(d) and Transition State Optimization. In this step, verify that there’s only one negative value of vibrational frequency032 a.u.When we look at the dipole moment of two molecules, formaldehyde has 3.0406 Debye while ethylene has zero dipole moment. Since ethylene has a symmetrical molecular structure, the dipole moment cancels out each other. Unlike ethylene, formaldehyde contains an oxygen atom that has high electronegativity consequently creating a dipole moment towards oxygen.The total number of electrons in both formaldehyde and ethylene is sixteen. When we fill up the orbitals from the lowest energy level, the highest occupied molecular orbital will be the 8th molecular orbital and thereby the lowest unoccupied molecular orbital will be the 9th.Ethylene has a C=C double bond that consists of a pi bond and a sigma bond. As a weak nucleophile, the pi bond between carbon atoms provides electron pairs to other molecules that lack electrons. This pi bond is formed by the overlap of two lobes in a 2p atomic orbital. In the case of a HOMO, the coefficient of both orbitals is positive and vice versa, when the fluorine atom and chlorine atom going over a transitional state by breaking and creating a sigma bond. The reason for the negative charge in a vibrational frequency is that the molecule has maximum energy in that certain state.Total energyReactant: -598.545927 hartreeTransition state: -598.540393 hartreeProduct: -598.599033 hartreeWith this data, we can calculate ΔH˚and activation energy.ΔH˚= -598.599033 hartree - -598.545927 hartree = -0.053106 hartree = -33.325 kcal mol-1Ea = -598.540393 hartree - -598.545927 hartree = 5.534*10-3 hartree = 3.473 kcal mol-1Through the IRC, we now know that the energy state of a single transition state resembles the reactant rather than the product. Since the energy of the product is lower than that of the reactant, the reaction enthalpy value is negative and this reaction is exothermic. This observation suits well with the Hammond postulation that states that the transition state of the reaction resembles either reactant or the product that is en 5

자연과학| 2022.06.02| 11페이지| 3,500원| 조회(1,595)

화학실험, 화학실험기법, SN2반응

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Exp 1. Quenching Study with Absorption and Fluorescence Spectroscopy

Exp 1. Quenching Study with Absorption and Fluorescence Spectroscopy화학실험기법1Affiliation이화여자대학교 자연과학대학 화학나노학과AbstractLuminescence quenching occurs when the excited molecule or atom releases energy in a form of light as the relaxation process takes place. In this experiment, we used UV/vis spectrometer and fluorescence spectrometry to understand the mechanism of fluorescence quenching. Since the quencher reduces the fluorescence, we could observe decreasing fluorescence intensity depending on the concentration of the quencher. The fluorescence yield for coumarin 1 was 4.56373 and the second-order quenching constant was 3.32*108 M-1s-1. Comparing this value with the diffusion rate constant 2.777*108 M-1s-1, there was a 19.55% errorIntroductionFluorescence is the luminescence caused by the absorption of radiation at a certain wavelength. When the fluorescence intensity is decreased by adding a quencher, it is called fluorescence quenching. This can happen by excited-state reactions, moleculm of light. In this experiment, we will study the mechanism of fluorescence decay. The equation for fluorescence yield is as follows.refers to the fluorescence yield, A stands for the absorbance of the sample solution, n refers to the refractive index, S stands for the area of the fluorescence spectrum, while ref and x stand for reference and unknown respectively. We can simplify the equation by assuming the value for nref is equal to the value of nx given the same solvents are used at low concentrations for each reference and the unknown.Moving on to the quenching part, there may be many reasons to explain quenching. However, we used the mechanism of fluorescence quenching by adding a chemical quencher in this experiment. This quenching mechanism can be further explained through the Stern-Volmer equation.refers to the fluorescence yield when the quencher is absent, while refers to the fluorescence yield when the quencher is present. is the fluorescence lifetime of a molecule when the ceneBenzene0.857.3MaterialsQuinine sulfate, H2SO4, Coumarin 1, NaCl, volumetric flasks: 500mL 1ea, 10mL 6ea, beaker: 250mL 2ea, fluorescence spectrometer, UV-vis spectrometer, and sonicatorMethods(Experiment 1) Finding the Fluorescence Yield of Coumarin 1Prepare 250mL of 1M H2SO4. To prevent burns, pour distilled water up to half of the 250mL flask before putting the acid. Then, pour 95% H2SO4 into the flask and fill the rest with distilled water.To make 1M with 95% H2SO4, we need 13.33mL of 95% H2SO4.Create a solution containing 1mg of quinine sulfate using 50mL of the solution made in step 1. Prepare the other solution with 3mg coumarin 1 dissolved in 100mL of ethanol. Use sonicator to help dissolution of solutes.Turn on the spectrometer and let it warm up for 10 minutes.Put the reference in and click ‘blank’ to start the measurement. In this case, the reference should be the pure solvent of the solution we’re about to measure.Click ‘sample’ and start measuring.Obtain the absorbance and Discussion(Experiment 1) Fluorescence Yield of Coumarin 1We obtained the absorbance graph of Coumarin 1 as above. Through this data, we figured out that the absorbance of Coumarin 1 at wavelength 350nm is 0.14103776. When extracting the absorbance of coumarin 1 at a wavelength of 350nm, it is ideal to let the absorbance be lower than 0.1. However, even after diluting the coumarin 1 prepared in the lab several times, we obtained a value above 0.1. Using this data might have caused a subtle error in succeeding procedures.And for quinine sulfate, the absorbance at a wavelength of 350nm is 0.0334.Using the fluorescence spectrometer, we obtained the graph of Coumarin 1 as a function of wavelength. The second peak has an area of 2.1581e+05 and the apex wavelength is 447.2nm.The second peak area of quinine sulfate is 1.4145e+05 and the apex is 455.4nm.Since the refractive indexes of the solutions are not known, we regard each value is equivalent. Then the equation simplifies as follows. tion of quencher increases, the fluorescence decreases. Since NaCl is used as a quencher in this case, the energy of quinine sulfate is transferred to NaCl, decreasing the fluorescence.The ratio of the fluorescence yield between quinine sulfate with and without NaCl can be obtained using the equation0M0.1M0.2M0.3M0.4M0.5MA (350nm)0.124180.1309690.1248610.1251450.1215070.1232150.248690.2603420.2498660.2503560.2440500.*************.7113.571.4751.2341.327330797.8454.2285.5209.9167.319.18816.1425.6734.9243.81The x-axis is the concentration of the quencher, and the y-axis is the ratio of fluorescence yield with and without a quencher. According to the Stern-Volmer equation, the y-intercept is 1 and the slope of the graph is K, the Stern-Volmer constant. Therefore, K is equal to 6.4498. Since K is equal to , 2nd order quenching rate constant is 3.32*108 M-1s-1.Using the Debye-Smoluchowski equation, we can derive the diffusion rate constant.Since the viscosity of sulfuric acid at room tempera

자연과학| 2022.06.02| 10페이지| 3,500원| 조회(1,470)

화학실험, 화학실험기법, 형광 분광법

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Exp 4. Phase Diagram by Differential Scanning Calorimetry (DSC) 평가A좋아요

Exp 4. Phase Diagram by Differential Scanning Calorimetry (DSC)Affiliation이화여자대학교 자연과학대학 화학나노학과AbstractIn this experiment, we use Differential Scanning Calorimetry of Naphthalene and p-dichlorobenzene to figure out the liquid-solid phase diagram in a binary system. Since substances undergo physical or chemical changes through releasing or absorbing heat, we can observe the phase transition by calculating their heat capacity. Before we begin the further experiment, we use indium to calibrate the DSC instrument. Then, make different samples with varying mole fractions of Naphthalene and p-dichlorobenzene and see if there is a change in the heat flux if the mixture has varying proportions of the substances. Lastly, calculate the melting point, molar freezing-point-depression constant, activity coefficient, and eutectic point of the mixture based on the data obtained from the DSC diagram. We can calculate the total heat of fusion by multiplying the molar mass of the substance by the enthal, heat is inputted, while during exothermic transition heat is extracted. Since DSC records the heat flux as a function of time or temperature, a dramatic change is a spot where a phase transition takes place. Since this experiment is about a binary system, heat is continuously supplied until the two substances completely melt. While cooling down, the temperature drops until the substances turn back into a solid form.In this experiment, we use the heat flux DSC which consists of a measuring cell with a furnace and integrated sensors. Heat flux DSC has a single heat source and is resistant to contaminations like emitted gas. Also, it can run throughout a broad range of temperatures and is long-lived. However, it takes a very long time to measure, and the voltage change in response to the temperature change is small – resulting in a high possibility for errors. Before proceeding with the experiment with Naphthalene and p-DCB, we use known information about Indium to compare the calibratiof fusion asRlnyAXA=ΔHfus(Tm)(1/ Tm-1/T)ΔHfus is the heat of fusion of substance A at its normal melting point (Tm). Using this equation, derive the heat of fusion and activity coefficients (kf) of each substance. Once you obtained the value for ΔHfus, calculate the molar freezing-point-depression constant for the substance where MA stands for the molecular weight of A.kf=RTm2MA/ΔHfusExperimental methodsDifferential Scanning Calorimetry, sample pans, Indium standard, Naphthalene, p-dichlorobenzeneCut the pellets into small pieces to get ~10mg of Indium, and weigh it precisely in the sample pan. Then, insert it into the DSC for calibration.Read the manual and turn the DSC on to let it warm up before proceeding with any measurement. Be sure not to touch the furnace with bare hands since it is very sensitive and there is a possibility of contamination.Under a rate of 10oC/min and in the range of 100~180oC obtain the thermogram of the Indium standard. Let DSC cool down once done.While keep, which means 11mg of Indium would have amount of the heat of fusion. The theoretical value of heat of fusion is 0.314J while the experimental value indicates 0.284J with the percent error (=)9.55%.The thermogram shows a sharp peak at 80.76oC which is the melting point of pure Naphthalene. The known melting point of pure Naphthalene is 80.3oC. In this experiment, we used 10mg of Naphthalene and obtained the heat of fusion at 175.16J/g. Therefore, the total heat of fusion can be calculated using the equation. . Since 10mg of Naphthalene was used, considering the known heat of fusion of Naphthalene is 18.98kJ/mol, 1mol of Naphthalene is equal to 128g, which means 10mg of Naphthalene would have amount of the heat of fusion. The theoretical value of heat of fusion is 1.483J while the experimental value indicates 1.752Jwith the percent error (=)18.14%.The known molar weight of p-DCB is 147.00g/mol and it has a melting point of 53.5oC with the heat of fusion of 4.54kJ/mol. We used 0.28mg of phthalene: 95.29J/g128g/mol=12197.12J/molUsing this data, we can derive the value for and activity coefficient.p-DCB :Naphthalene :Using RlnyAXA=ΔHfus(1-Tm/T) equation, we can obtain the activity coefficient.p-DCB :Naphthalene : .In this case, we obtained a thermogram that might be seen as one peak or the intersection of two different peaks. Since it was hard to measure the separate area of two possible peaks, let’s assume that this is a single peak. This thermogram shows a moderate peak at 30℃. We can calculate the heat of fusion per 1mol of each substances. The heat of fusion for p-DCB is (141.73J/g*147.00g/mol=)20834J/mol, and for Naphthalene, it is (141.73J/g*128g/mol=)18141J/mol. Using these values, we can calculate and the activity coefficient.p-DCB :Naphthalene :Using RlnyAXA=ΔHfus(1-Tm/T) equation, we can obtain the activity coefficient.p-DCB :Naphthalene :When we measured pure p-DCB, it showed a sharp peak at 53.07℃ with the heat of fusion 132.44J/g. Using this information, we

자연과학| 2022.06.02| 10페이지| 3,500원| 조회(237)

DSC, 화학실험기법, 화실기

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