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.
Trang 1Project 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
3
5
3
5
5
2
5
5
5 10
5 10 x
2.2097
5 10
44.1948 5
10 10
D
Initial
x
x
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
3
2
3
5
5
0 0.09995
5
09995
10
5 1 2
[ 2
0
i i
Initial
Final
3
2
5 1
2 exp
5 10
0.851
CH COONa
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
2
5
5
3
5 3
3
0 9.95
[
1
i i
CH COOH NaOH CH COO Na H O
Initial
Final
Then our solution we have CH COO M Na M OH M
3
2
3
3
2
1
2 exp
exp 0.509 1 ( 1) 0.01 0.951
CH COONa
A z z I
1.4 Try to qualitatively compare the conductivities of the 3 solutions mentioned above
The solution 3 has the most conductivity and the first solution hs the least
conductivity
Problem 2:Consider the electrochemical cell with 2
o
Zn Zn= – 0.76 V
o
Ag Ag= 0.7996 V
Zn ZnSO4 (a Zn2 0.3) AgNO3 (a Ag 0.3
) Ag
Trang 22.1 Write the electrode half-reactions and the overall reaction occurring in this cell
2
2
:
: 2
The overall reaction
2.2 Calculate the standard emf 298
o
E and the actual emf E298 of this cell at 298 K Try
to comment about this value
2
0.3 0.8306
0.8306 ( 0.744) 1.5746 ,
1
5
2
o
o
o
Ag Ag Zn Zn
o
rxn
V
E
Thus the EMF of the ce
E
n E
ll is E
E
a n
E
E
V
0.7996 ( 0.76) 1.5596
2.3 Calculate the standard Gibbs energy 298
o
G and the actual Gibbs energy G298of the
overall reaction occurring in this cell at 298 K Try to comment about this value
2
3
3
So the value of Gibbs energy is smaller than O so th
1.5596
cathod Ag Ag
o anod Zn Zn
rxn rxn
3
3
3
e reaction is theorically thermodynamically flavored
So the
cathod
anod
rxn rxn
value of Gibbs energy is smaller than O so the reaction is actually thermodynamically flavored
2.4 Try to propose 2 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
Trang 3The 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 non-spontaneous reactions and precise control of the potential difference
Project 3:
Problem 1: Consider 0.1 kg liquid water at temperature 298 K (specific weight 1 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 1 cm length
Trang 4
2 2
3
4
0.07275
4
water
tube
n
er
c
e
i terfa ial
If the glass vessel is
Tube
Sphere
V
2 2
4
2 3
3
4
3 10
4 0.0104 7.5797 1
h
0.07275
0
0
0.36448 0.36448 0.026 6
interfacial
cylinder int i
sphere
tu e
erfac al
b
m
Tube
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
2 0.005221 0.005919 0.011222 0.013139
0
0
0
0.01
0.01
0.01
0.01
0.01
f(x) = 0.08 x + 0
R² = 0.99
max
0.0811
0.0029
K
2.2 Demonstrate that 1.2×10−8 mol of dye adsorb to form a monolayer
Trang 52.3 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)
17 2
max
o
o
S
n N
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
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
0.01
0.01
0.02
0.02
0.03
f(x) = 0.03 x + 0.01
R² = 0.99
1
o
o
Y
P
P
V
P
3.2 Find the specific surface area of this adsorbent, given the effective cross section
of N2 molecule is 0.16 nm2
equil PN2 , kPa 12.46 15.57 19.97 24.64 30.43
Adsorb V, mm 3 970 1020 1100 1200 1350
Trang 6
3
3
1.429
22.4
m
m m
m o
m A
o
C
V K
V K
V N A
S
2
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 2 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
Trang 7Problem 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,
low-viscosity 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