Bài tập lớn Hóa lý 2

Khảo sát hấp phụ của nitrogen trên bề mặt chất bằng phương trình, khảo sát tốc độ phản ứng bằng phần mềm matlab

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF CHEMICAL ENGINEERING

GROUP ASSIGNMENT PROJECT

Course: Physical Chemistry 2 (CH2109) – 222

Group No: 222 – A01 – 06

Lecturer:

Students:

1

Ho Chi Minh City – 04/2023

I Content of Questions and Solutions

Question 1: Using the following data of the adsorption of N2 on solid X at -196oC, in which: x= P/P0: relative pressure, V: volume adsorbed cc (STP)/g:

- Plot the adsorption isotherm

- Calculate the specific surface area of the solid X

Solution 1: The diagram is plotted basing on the Brunauer-Emmett-Teller equation (BET)

which is chosen thanks to be suitable for given information about P/P0 and V BET

equation: P/ P0

V(1- P/ P0)=

1

vmC+

C-1

vmC .( P/ P0) This equation can be simplified as linear

equation y = ax+b, P/ P0

V(1- P/ P0) is y, ( P/ P0) is x, C-1v

mC is a and

1

vmC is b Through linear regression method, we can identify:

{ C-1vmC = a ≈ 0.07712

1

vmC= b ≈ 4.176× 10

-3

⟹{C ≈ 185.63vm ≈ 12.9 cc = 12.9ml The molar volume of N2 can identified by ideal gas law PV= nRT So:

1×0.082×(0+273.15)

3ml/ mol

The specific surface area between solid X and gas N2 is identified by equation

´

S0=vm×N×A

V×m

Having:

vm=12.3ml ; N=6.022×1023; A=16.2×10-20m

V=22.4L/ mol =22.4×103 ml/ mol ; m=1g (assume)

⇒ ´S0=vm×N×A

12.9×6.022× 1023×16.2× 10-20

2/g

3

Certified reference materials for porosity properties

First we assume v m is unchanged even and calculate new molar colume of N2 at 77K

V77K=nRT

1×0.082×77

⇒ ´S0=vm×N×A

12.9×6.022× 1023×16.2× 10-20

2

/g

As can be seen on the given tables, SSA of solid X is surpassed when compares to other porous materials yet it is not one of the highest SSA materials

Question 2: Search data from the Internet for the above battery and:

- Describe the composition of the battery

- Write half-reaction at the negative electrode of a cell in the battery

- Write half-reaction at the positive electrode of a cell in the battery

- Calculate the standard emf of a cell in the battery

Solution 2:

5

Question 3: Your company has two liquid streams available containing solutes that are

not profitably marketable at the present time One stream contains an aqueous solution of

A The second stream contains an aqueous solution of species B Species A can react either with species B or with itself according to the following stoichiometric equations and rate expressions:

(a) At a specific temperature (T): k1= 0.01 (M-1.min-1), k2= 0.01 (M-1.min-1), k3= 0.002 (M-0.5)

When a 1L aqueous solution containing 1M of A mixed quickly with a 1L aqueous

solution containing 1M of B:

- Plot the concentration of V as a function of reaction time (up to 90% A reacted)

- Plot the concentration of W as a function of reaction time (up to 90% A reacted)

(b) You have two beakers containing samples of the two streams and desire to carry out a

small-scale laboratory experiment in which you maximize the formation of species V In what manner should you carry out this experiment; that is, in what order and at what rate would you add each beaker of the reactants to an empty container?

(c) If the activation energies for the rate constants k1, k2, and k3 are 90, 40, and 50 kJ/mol,

respectively, what additional statements can you make regarding the operating temperature recommended for maximizing production of species V?

Solution 3:

a We have:

d CA

dt = - k1CACB- 2 k2CA2

1+ k3CV0.5

d CB

dt = -k1CACB= - d CV

d CW

2 k2CA2

1+ k3CV0.5

Using matlab to calculate the change in the concentration of 4 substances by time, with the value shown below

7

The change in the concentration of A, B, V and W by time

After 352.2987 seconds, 90% of A reacted

b To maximize the formation of species V, we need to favor the reaction between species A and B over the reaction of A with itself This is because the formation of V is directly proportional to the rate of the reaction between A and B, and inversely proportional to the rate of the reaction of A with itself

The stoichiometric equations and rate expressions given are:

From these equations, we can see that the formation of V depends on the concentration of both A and B, as given by the rate expression rV = k1CACB The rate of the reaction between A and B can be increased by increasing the concentrations of A and

B, and by increasing the rate constant k1

On the other hand, the formation of W depends only on the concentration of A, as given by the rate expression rW =2 k2CA

2

1+ k3C0.5V The rate of the reaction of A with itself can be increased by increasing the concentration of A, the rate constant k2, and the square root

of the concentration of V, which is given by CV0.5

To maximize the formation of V, we need to control the addition of A so that the rate

of the reaction between A and B is faster than the rate of the reaction of A with itself This can be achieved by limiting the concentration of A in the reaction mixture so that the second reaction does not occur to a significant extent

Here's how we can carry out the experiment in a step-wise manner to maximize the formation of V:

1 Begin by adding the aqueous solution of species B to an empty container

2 Start stirring the solution and maintain a constant stirring rate throughout the experiment

3 Slowly add the aqueous solution of species A to the container, while monitoring the concentration of V

4 Control the addition of A so that the rate of the reaction between A and B is faster than the rate of the reaction of A with itself This can be achieved by adjusting the rate of addition of A to keep the concentration of A low compared to the concentration of B

5 Continue adding A until the maximum concentration of V is reached

It's important to note that the rate of the reaction between A and B is given by

rV = k1CACB, which means that the concentration of both A and B should be kept as high

as possible to maximize the rate of the reaction However, we need to be careful not to

9

add too much A, as it could result in the second reaction taking place and reducing the yield of V

c If the activation energies for the rate constants k1, k2, and k3 are 90, 40, and 50 kJ/mol, respectively

Applying Arrhenius equation:

k2 = -

Ea

R (1T2 -

1

T1)

1 Choose k1’ = 2k1 = 0.02 M-1min-1

⟹(1T2

-1

T1)= -6.403× 10-5⟹ k2'

=0.0136 M-1min-1; k3'=0.00294 M-0.5

CV' =0,286948, CW' =0,106518

⇒Selectivity(2)=C'V

C'W+ C'V =

0,286948 0,106518+0,286948=0,72928

2 Choose k1’ = 3k1 = 0.03 M-1min-1

11

⟹(1T2

-1

T1)= -1.015×10-4⟹ k2'=0.0163 M-1min-1; k3'=0.00368 M-0.5

CV' =0,308202 M , CW' =0,095899 M

⇒Selectivity(2)=CV

'

C'W+ C'V =

0,308202 0,095899+0,308202=0,762656

So temperature increase  Selectivy increases