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Problem 1: Given at temperature 298 K, the equilibrium constant for the dissociation of acetic acid in water is KD = 1.75 x 105. 1.1. At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x105 mol.dm3 CH3COOH only. 1.2. At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x105 mol.dm3 CH3COOH and 101 mol.dm3 NaOH. Calculate the ionic strength of this solution. Then use the DebyeHūckel 1st limiting law to calculate the mean activity coefficients of electrolytes inside this solution.1.3. At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x105 mol.dm3 CH3COOH and 102 mol.dm3 HCl. Calculate the ionic strength of this solution. Then use the DebyeHūckel 1st limiting law to calculate the mean activity coefficients of electrolytes inside this solution.
Project 2: Problem 1: Given at temperature 298 K, the equilibrium constant for the dissociation of acetic acid in water is KD = 1.75 x 10-5 1.1 At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x10-5 mol.dm-3 CH3COOH only CH3COO H CH3COOH 10 Initial Final 10 x x x x 1.75 10 x 2.2097 10 5 10 x 2.2097 10 100 44.1948 10 1.2 At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x10-5 mol.dm-3 CH3COOH and 10-1 mol.dm-3 NaOH Calculate the ionic strength of this solution Then use the Debye-Hūckel 1st limiting law to calculate the mean activity coefficients of electrolytes inside this solution CH3COOH NaOH CH3COO Na H2O KD 10 Initial Final 0.1 0.09995 10 Then our solution we have [CH3COO ] 5 10 5M ;[Na ] 10 1M ;[OH ] 0.09995M I 1 z i2Ci [CH3COO ]( 1)2 [Na ](1)2 [OH ]( 1)2 10 0.1 0.09995 0.1 2 exp A z z I CH3COONa exp 0.509 1( 1) 0.1 0.8513 1.3 At this temperature, calculate the degree of dissociation of acetic acid in its aqueous solution containing 5x10-5 mol.dm-3 CH3COOH and 10-2 mol.dm-3 HCl Calculate the ionic strength of this solution Then use the Debye-Hūckel 1st limiting law to calculate the mean activity coefficients of electrolytes inside this solution CH3COOH NaOH CH3COO Na H2O 10 Initial Final 0.01 9.95 10 10 Then our solution we have [CH3COO ] 5 10 5M ;[Na ] 0.01M ;[OH ] 9.95 10 3M I 1 z i2Ci [CH3COO ]( 1)2 [Na ](1)2 [OH ]( 1)2 10 0.01 9.95 10 0.01 2 exp A z z I CH3COONa exp 0.509 1( 1) 0.01 0.9512 1.4 Try to qualitatively compare the conductivities of the solutions mentioned above The solution has the most conductivity and the first solution hs the least conductivity o o Zn Ag 2 Zn Ag Problem 2: Consider the electrochemical cell with = – 0.76 V = 0.7996 V Zn ZnSO4 ( aZn2 0.3 ) AgNO3 ( aAg 0.3 ) Ag 2.1 Write the electrode half-reactions and the overall reaction occurring in this cell The electrode half reactions : Anod ( ) : Zn Zn 2 2e Cathod () : Ag e Ag The overall reaction : Zn 2Ag Zn 2 2Ag o 2.2 Calculate the standard emf E298 and the actual emf E298 of this cell at 298 K Try to comment about this value 0.0592 0.0592 o E E Ag logaAg 0.7996 log0.3 0.8306V Ag n 0.0592 0.0592 o E E Zn logaZn2 0.76 log0.3 0.744V 2 Zn n E E E 0.8306 ( 0.744) 1.5746V Thus , the EMF of the cell is E 1.5746V o o o Erxn E Ag E Zn 0.7996 ( 0.76) 1.5596V 2 Ag Zn o 2.3 Calculate the standard Gibbs energy G298 and the actual Gibbs energy G298 of the overall reaction occurring in this cell at 298 K Try to comment about this value o o Gcathod nE Ag F 10.7996 96500 77.161103 J / mol Ag o Ganod nE Zn F ( 0.6) 96500 115.8 103 J / mol 2 Zn o o Grxn nErxn F 1.5596 96500 301103 J / mol So the value of Gibbs energy is smaller than O so the reaction is theorically thermodynamically flavored Gcathod nE F 10.8306 96500 80.153 103 J / mol Ganod nE F ( 0.744) 96500 143.592 103 J / mol Grxn nErxnF 1.5746 96500 303.898 103 J / mol So the value of Gibbs energy is smaller than O so the reaction is actually thermodynamically flavored 2.4 Try to propose solutions to increase the actual emf E298 of this cell at 298 K - Decrease the concentration of Ag+ solution - Increase the concentration of Zn2+ solution Problem 3: An electrolytic cell might be considered a counter-part of an electrochemical cell However, both electrodes in an electrolytic cell are immersed in the same electrolyte solution, while each electrode in an electrochemical cell has its own electrode solution (environments) Try to discuss the reasons, advantages and limitations of this construction difference An electrochemical cell generates electrical energy through a spontaneous chemical reaction between two separate electrode solutions The electrodes are connected by an external circuit, allowing electrons to flow from the anode to the cathode The flow of electrons is accompanied by the flow of ions in the solution, which maintains the electrical neutrality of the system On the other hand, an electrolytic cell is an electrochemical cell in which electrical energy is used to drive a non-spontaneous chemical reaction In an electrolytic cell, the electrodes are immersed in the same electrolyte solution, and an external power source is used to supply energy to the system The construction difference between the two types of cells is due to their different functions In an electrochemical cell, the two electrodes are separated by a salt bridge or a porous barrier to prevent the mixing of the electrode solutions, which would neutralize the charge and halt the reaction In contrast, an electrolytic cell requires the mixing of the reactants to occur at the electrodes for the reaction to take place One advantage of using separate electrode solutions in an electrochemical cell is that it allows for greater control over the reaction conditions, such as the pH, temperature, and concentration of the reactants This is because each electrode solution can be independently adjusted to optimize the reaction In contrast, in an electrolytic cell, the conditions at both electrodes must be the same, which can limit the range of reactions that can occur Another advantage of separate electrode solutions is that it allows for the use of different electrode materials for the anode and cathode This can be important for certain reactions, where specific electrode materials are required for optimal performance In an electrolytic cell, the same material must be used for both electrodes, which can limit the range of reactions that can occur One limitation of using separate electrode solutions is that it can be more difficult to maintain a constant potential difference between the electrodes, as the potential difference depends on the concentration of the reactants in the electrode solutions In an electrolytic cell, the potential difference can be precisely controlled by the external power source, which can be advantageous for certain reactions In summary, the construction difference between an electrochemical cell and an electrolytic cell is due to their different functions While separate electrode solutions offer greater control over the reaction conditions and electrode materials, the use of the same electrolyte solution in an electrolytic cell allows for the driving of nonspontaneous reactions and precise control of the potential difference Project 3: Problem 1: Consider 0.1 kg liquid water at temperature 298 K (specific weight kg/L) filled in a closed glass vessel Given the interfacial tension glass / water at this temperature is 0.07275 J/m2 Calculate the interfacial energy in case the glass vessel is 1.1 A cube 1.2 A sphere 1.3 A circular cylinder cm length V m rwater 0.1(kg ) 0.1L 10 4m3 1(kg / L) If the glass vessel is Tube Vtube a and Atube a 3V 10 2.154 10 3m2 Winterfacial Atube 0.07275 2.154 10 1.567 10 J Sphere Vsphere 3V r and Asphere 4 r 4 4 10 3 4 0.0104m2 Winterfacial Asphere 0.07275 0.0104 7.5797 10 J Tube Vtube r 2h and Acylinder 0.36448m2 Winterfacial A 0.07275 0.36448 0.026516 J Compare the results obtained and formulate a general statement about effect of the interfacial shape on its energy So the sharp has more area the more the interfacial energy Problem 2: The following data refer to the adsorption of the red–mauve dye from beetroot juice on porcelain at 25o C Equil Conc C, mmol.dm-3 0.012 0.026 0.047 0.101 0.126 Adsorb Amount n, nmol 2.94 4.98 7.94 9.00 9.59 2.1 Show that the data obey the Langmuir adsorption isotherm Equil Conc C, mmol.dm-3 0.012 0.026 0.047 0.101 0.126 Adsorb Amount n, nmol 2.94 4.98 7.94 9.59 C/n 0.00408 0.00522 0.00591 0.01122 0.01 f(x) = 0.08 x + R² = 0.99 0.01 0.01 0.01 0.01 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 12.33nmol 0.0811 0.0811 K 27.9655 0.0029 2.2 Demonstrate that 1.2×10−8 mol of dye adsorb to form a monolayer nmax 0.01313 2.3 Estimate the area of a single dye molecule if the radius of a plate was 17.8 cm (we assume the formation of a complete monolayer) S 0.0995 So Ao nmaxN Ao o 1.3398 10 17m2 / molecular 9 23 nmaxN 12.33 10 6.023 10 So R 0.1782 0.0995m2 Problem 3: Given the saturated vapor pressure of N2 gas over pure liquid N2 at temperature 77 K is 76 kPa Based on the experimental data obtained from sorption of N2 at this temperature onto a solid adsorbent in the following table, equil PN2 , kPa 12.46 15.57 19.97 24.64 30.43 Adsorb V, mm3 970 1020 1100 1200 1350 The values for adsorbed volumes are already corrected to 273 K, 1.0 atm, and referred to 1.0 g adsorbent 3.1 Construct the corresponding BET equation 0.03 0.02 f(x) = 0.03 x + 0.01 R² = 0.99 0.02 0.01 0.01 0.15 0.2 0.25 0.3 0.35 0.4 0.45 P Po K 1P K P P VmK V m o V 1 A B X P o Y 3.2 Find the specific surface area of this of N2 molecule is 0.16 nm2 equil PN2 , kPa 12.46 15.57 Adsorb V, mm3 970 1020 P/Po (PN2/76) 0.164 0.205 1-P/Po 0.836 0.795 Y ((P/Po)/V*(1- P/Po)) 0.016 0.017 X (P/Po) 0.164 0.205 adsorbent, given the effective cross section 19.97 1100 0.263 0.737 0.018 0.263 24.64 1200 0.324 0.676 0.019 0.324 30.43 1350 0.4 0.6 0.022 0.4 V K 0.0261 m C 1.429 Vm 26.812(mm / g ) K 0.0112 VmK So we can find the value of V m 26.812 mm3/ g Because T 0o C (273K ) and P 1atm So V 22.4L 22.4 103mm3 VmN A A 26.812 6.023 1023 0.16 10 19 So 11.535m2 Vm 22.4 10 1 Problem 4: Write a short essay to express your opinions about the applicability of all formulas (equations) in colloid chemistry, e.g formulas relating stationary sedimentation, sedimentation equilibrium, optical properties, etc Try to explore and briefly describe methods to fractionate colloid systems In colloid chemistry, formulas and equations are essential for understanding and predicting the behavior of colloidal systems However, not all formulas are applicable in every situation, as the complexity and variability of colloid systems can make it difficult to apply simplified equations It is important to understand the limitations and assumptions behind each equation and to use them appropriately One example of an equation used in colloid chemistry is the Stokes-Einstein equation, which relates the diffusion coefficient of a particle to its size and the viscosity of the medium This equation assumes that the particle is spherical and that the medium is homogeneous, which may not be the case in all colloid systems Other factors, such as surface charge and interactions with other particles, can also affect the diffusion coefficient Another example is the Debye-Hückel equation, which describes the effect of ionic strength on the electrostatic interactions between charged particles This equation assumes that the ions in the solution are well-mixed and that the charges are evenly distributed on the particles, which may not be true in all colloid systems In addition, the equation does not take into account the effects of other factors, such as surface charge density and particle size Despite these limitations, formulas and equations are still valuable tools in colloid chemistry They provide a framework for understanding the behavior of colloidal systems and can help guide experimental design and interpretation It is important, however, to use them with caution and to consider the limitations and assumptions behind each equation One method for fractionating colloid systems is centrifugation, which separates particles based on their sedimentation rate By subjecting a colloid solution to centrifugal force, larger and denser particles will sediment more quickly than smaller and less dense particles This allows for the separation of particles based on their size and density Another method is field-flow fractionation, which separates particles based on their size and shape using a combination of a flow field and a perpendicular field The flow field drives particles through a narrow channel, while the perpendicular field selectively deflects particles based on their size and shape This allows for the separation of particles based on their size and shape, which can provide valuable information about the structure and properties of the colloidal system In conclusion, formulas and equations are valuable tools in colloid chemistry, but their applicability depends on the specific system and the limitations and assumptions of each equation Fractionation methods, such as centrifugation and field-flow fractionation, can provide valuable information about the structure and properties of colloidal systems, but careful interpretation and consideration of the limitations of each method is necessary Problem 5: Write a short essay to express your opinions about the viscosity of liquid phases Viscosity is a measure of a fluid's resistance to flow In liquid phases, viscosity plays an important role in determining the behavior and properties of the liquid Viscosity can vary widely among liquids, and can be affected by factors such as temperature, pressure, and the presence of solutes One key property of high-viscosity liquids is their resistance to deformation This can be seen in the behavior of liquids like honey or molasses, which flow slowly and resist being poured The high viscosity of these liquids is due to their high molecular weight and the presence of long, tangled chains of molecules In contrast, lowviscosity liquids like water flow easily and have less resistance to deformation Viscosity can also be affected by temperature, with most liquids becoming less viscous as temperature increases This is due to the increased thermal energy of the molecules, which allows them to move more easily and reduces the resistance to flow However, some liquids, such as glass-forming liquids, can exhibit unusual temperature-dependent viscosity behavior, with viscosity increasing rather than decreasing as temperature increases This is thought to be due to the formation of dense clusters of molecules that hinder molecular motion The viscosity of liquid phases is also important in industrial and engineering applications, where it can affect the performance and efficiency of processes such as mixing, pumping, and coating Understanding and controlling the viscosity of liquids in these applications is essential for achieving the desired results In conclusion, viscosity is an important property of liquid phases that can vary widely among different liquids and can be affected by temperature, pressure, and other factors The behavior of high-viscosity liquids is characterized by resistance to deformation, while low-viscosity liquids flow easily Understanding and controlling viscosity is essential in both scientific and industrial settings, and can impact the performance and efficiency of a wide range of processes