Physical chemistry exercise 2 (chemical equilibrium and thermodynamics)

Bài tập lớn Hóa lý (Tiếng Anh), bao gồm câu hỏi liên quan đến Cân bằng hóa học (chemical equilibrium) và Nhiệt động (thermodynamics)

CONTENTS CHAPTER I: CHEMICAL EQUILIBRIUM

Quest 1 1

Solution 1 1

a Using only these data, find ∆rH°, ∆rG°, and ∆rS° at 550 K for synthesis of PCl 5 (g) from PCl 3 (g) and Cl 2 (g) 1

b Plot ∆rG° of decomposition of PCl5 (g) with reaction time at 574 K 5

CHAPTER II: THERMOTHYNAMIC CYCLES Quest 2: 7

Solution 2 7

a Plot the schematic and T-s diagrams 7

b Calculate the work produced by the turbine, the heat supplied in the boiler 8

c Define the thermal efficiency of this cycles 11

REFERENCES 12

EVIDENCE OF TEAM WORKING 13

CHAPTER I: CHEMICAL EQUILIBRIUM

Quest 1 For PCl5(g)  PCl3(g) + Cl2(g), observed equilibrium constants (from measurements on equilibrium mixtures at low pressure) with reaction temperarture (T) the following table:

a Using only these data, find ∆𝑟H°, ∆𝑟G°, and ∆𝑟S° at 550 K for synthesis of PCl 5

(g) from PCl 3 (g) and Cl 2 (g)

b Plot ∆rG° of decomposition of PCl5 (g) with reaction time at 574 K

Solution 1:

a Using only these data, find ∆𝑟H°, ∆𝑟G°, and ∆𝑟S° at 550 K for synthesis of PCl 5 (g) from PCl 3 (g) and Cl 2 (g)

Since the Kp for the reaction PCl5(g)  PCl3(g) + Cl2(g) is equal to [𝐶𝑙2][𝑃𝐶𝑙3]

[𝑃𝐶𝑙5] and the Kp’ for the synthesis of PCl5 (g) from PCl3 (g) and Cl2 (g) is [𝑃𝐶𝑙5]

[𝐶𝑙2][𝑃𝐶𝑙3] , then we can get that the Kp’ = 1/Kp and that why we will make another table for the synthesis of PCl5 to solve the problem easier

49

100 199

25 124

20 187

Based on the table, we do not have enough data to solve the problem with any formula that we have been learnt before to find the equilibrium constants at 550K, so in this case we have to use regression analysis method to find out the path which Kp’ and T are moving on

At first, we start to analyse the data and we get that these data of Kp’ and T make it

to a curve path which have the function approximated with ln(Kp’) = a + b/T But now this function is really hard to solve, we decide to change the problem from solving the logarithmic regression to solving the linear one to make it easier to find out the line

Since ln(Kp’) = a + b/T, then we change it to Y = A + BX by given Y = ln(Kp’), X

= 103/T, A = a , B = b/103 So now we have the other new table to draw the line:

ln(Kp’)

ln (200

49) ln (

100

199) ln (

25

124) ln (

20

187)

103K/T 103/485 103/534 103/556 103/574 The Y axis is stands for ln(Kp’) and X axis is stands for 103K/T

Therefore, we suppose f(No) = No, h(Yi) = Yi, g(Xi) = Xi , by using least square method, we have:

Given A = (

1 X1

1 X2

1 X3

1 X4 ) =

(

1 10

3

485

1 10

3

534

1 10

3

556

1 10

3

574) ,

No = (A

B) , B = (

B1

B2

B3

B4 ) =

(

ln (200

49)

ln (100

199)

ln ( 25

124)

ln ( 20

187)) Now, we find out A and B coefficients through matrix No from this formula:

AT A No = AT B

Then,

No = (AT A)−1 (AT B)

No =

((

1 10

3

485

1 10

3

534

1 10

3

556

1 10

3

574)

T

(

1 10

3

485

1 10

3

534

1 10

3

556

1 10

3

574))

−1

((

1 10

3

485

1 10

3

534

1 10

3

556

1 10

3

574)

T

(

ln (200

49)

ln (100

199)

ln ( 25

124)

ln ( 20

187))

) Thus, we have No ≈ ( −22.095

11.406 )

So A = a = −22.095 and B = 𝑏

103 = 11.406 ⇒ 𝑏 = 11.406 ∗ 103 = 11406 Hence, the function we have just solved is 𝑌 = −22.095 + 11.406 𝑋 and the original function actually is:

ln(𝐾𝑝′) = −22.095 +11406

𝑇

So now we can easily find the ln(Kp’) at 550K from that function:

𝐥𝐧 (𝑲𝒑′)@𝑻=𝟓𝟓𝟎𝑲 = −𝟐𝟐 𝟎𝟗𝟓 +𝟏𝟏𝟒𝟎𝟔

𝑻 = −𝟐𝟐 𝟎𝟗𝟓 +

𝟏𝟏𝟒𝟎𝟔

𝟓𝟓𝟎 = −𝟏 𝟑𝟓𝟔𝟖 For more information about the linear regression that we have just found out, by some basic statistical methods, we got that the path which ln(Kp’) and 1000/T are moving

on have the correlation coefficient rxy which is 0.9998, very close to 1 (or R2 = 0.9996 from the chart below) that means X and Y have the super close relation to each other and that’s why it is such an extremely suitable line for us to use to calculate and solve the problem

we got

Picture 1: Linear regression line for synthesis of PCl 5 (g) from PCl 3 (g) and Cl 2 (g)

Then, the Gibbs energy of the reaction at 550K is:

∆𝒓𝑮 = −𝑹𝑻 𝐥𝐧(𝑲𝒑′)@𝑻=𝟓𝟓𝟎𝑲 = −𝟖 𝟑𝟏𝟒 ∗ 𝟓𝟓𝟎 ∗ (−𝟏 𝟑𝟓𝟔𝟖)

= 𝟔𝟐𝟎𝟒 𝟐𝟑𝟗𝟒 (𝑱𝒎𝒐𝒍−𝟏)

So the reaction has such a nonspontaneous forward direction

From the above result, we’ve already got the function

𝑙𝑛(𝐾𝑝′) = −22.095 +11406

𝑇 (∗) Then, we use the Gibb-Helmholtz equation and Van’t Hoff equation to calculate the enthalpy change for a reaction of synthesis of PCl5(g) from PCl3(g) and Cl2(g) at 550K:

𝒅𝒍𝒏(𝑲𝒑′)

∆𝒓𝑯

𝑹𝑻𝟐

⇔ 𝑑(−22.095 +

11406

𝑇 )

∆𝑟𝐻

𝑅𝑇2

Therefore, we can easily prove that the function (*) above is differentiable and continuous at T = 550K so that why it has a derivative at T = 550K, now we got:

y = 11.406x - 22.095 R² = 0.9996

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

(K P

1000/T

Plot of Ln(Kp') against 1000/T

∆𝒓𝑯 = −𝟏𝟏𝟒𝟎𝟔

𝑻𝟐 𝑹𝑻𝟐 = −𝟏𝟏𝟒𝟎𝟔 ∗ 𝟖 𝟑𝟏𝟒 = −𝟗𝟒𝟖𝟐𝟗 𝟒𝟖𝟒 (𝑱𝒎𝒐𝒍−𝟏)

Next step, by using ∆𝑟𝐺 = ∆𝑟𝐻 − 𝑇∆𝑟𝑆, we can calculate the entropy change for a reaction of PCl5’s decomposition at 550K:

∆𝒓𝑺 =∆𝒓𝑯 − ∆𝒓𝑮

−𝟗𝟒𝟖𝟐𝟗 𝟒𝟖𝟒 − 𝟔𝟐𝟎𝟒 𝟐𝟑𝟗𝟒

𝟓𝟓𝟎 = −𝟏𝟖𝟑 𝟔𝟗𝟕𝟕 (𝑱𝑲

−𝟏𝒎𝒐𝒍−𝟏)

Answer: Only using these data, for synthesis of PCl5 (g) from PCl3 (g) and Cl2 (g)

at 550K, we got:

∆𝑟𝐺 = 6204.2394 (𝐽𝑚𝑜𝑙−1)

∆𝑟𝐻 = −94829.484 (𝐽𝑚𝑜𝑙−1)

∆𝑟𝑆 = −183.6977 (𝐽𝐾−1𝑚𝑜𝑙−1)

b Plot ∆rG° of decomposition of PCl5 (g) with reaction time at 574 K

Follow the idea that the reaction ends at equilibrium and ∆rG equals 0 at this point,

we can depict how ∆rG change over the time of reaction

Based on the data we have, we cannot form a direct relation between ∆rG and reaction time So we can find another way to present ∆rG by a factor that depends on the reaction time In this situation, we will use Πp as that factor

We can define the ∆rG by using this formula:

∆rG = RTln Πp

Kp (I) Because Kp depends only on the temperature so in this situation it is constant Πp

will change with the reaction and defined by:

So we consider Πp as a variable to draw the plot: Πp/Kp

Picture 2: The change of ∆ r G with Π p /K p

The plot is based on the formula (I) At the first point, we don’t have any product

so Πp ≈ 0, we have ∆rG at a very low position When the products are produced more and more, the Πp is increasing over the time and to the point at equilibrium, the reaction ends,

Πp = Kp = 9.35 so we have ∆rG equals 0 at this point

So we can draw the another plot ∆rG0 of decomposition of PCl5 (g) with reaction time because Πp depends on reaction time:

-70 -60 -50 -40 -30 -20 -10 0 10 20

∆ r

Πp/Kp

Picture 3: ∆rGo of decomposition of PCl5 (g)

with reaction time

CHAPTER II: THERMOTHYNAMIC CYCLES

Quest 2: A simple ideal Rankine cycle which uses water as the working fluid operates its

condenser at 30 ℃ and it boiler at 350℃ Assuming that steam enters the turbine at

saturated vapor condition

a Plot the schematic and T-s diagrams

b Calculate the work produced by the turbine, the heat supplied in the boiler

c Define the thermal efficiency of this cycles

Solution 2:

a Plot the schematic and T-s diagrams

Solution: A steam power plant operating on the simple ideal Rankine cycle is considered The work produced by the turbine, the heat supplied in the boiler and the thermal efficiency of the cycle is to be determined

Assumptions: 1 Steady operating conditions exist

2 Kinetic and potential energy changes are negligible

Analysis: The schematic of the power plant and the T-s diagram of the cycle are

shown in picture 4 and 5 We note that the power plant operates on the ideal Rankine cycle

Therefore, the pump and the turbine are isentropic, there are no pressure drops in the boiler and condenser, steam leaves the con- denser and enters the pump as saturated liquid at the condenser pressure and assuming that steam enters the turbine at saturated vapor condition First we determine the enthalpies at various points in the cycle, using data from steam tables (Tables A-4, A-5, and A-6)

b Calculate the work produced by the turbine, the heat supplied in the boiler

State 1: saturated liquid

State 2: compressed (subcooled) liquid

State 3: saturated vapor

State 4: saturated vapor and saturated liquid mixture

Picture 4: The schematic of the power plant

Picture 5: The T-s diagram

The cycle consists of four processes:

1  2: Isentropic compression in a pump (Wpump,in)

2  3: Constant pressure heat addition in a boiler (Qin)

3  4: Isentropic expansion in a turbine (Wturb,out)

4  1: Constant pressure heat rejection in a condenser (Qout)

Some equations applied in the cycle:

Pump (Q=0): Wpump in = h 2 – h 1 = v(P 2 – P 1 )

h1 = hf@P1; v ≈ v1 = vf@P1

Boiler (W=0): Qin = h 3 – h 2

Turbine (Q=0): Wturb,out = h 3 – h 4

Condenser (W=0): Qout = h 4 – h 1

Thermal efficiency of the Rankine cycle:

Ŋth = 𝑊𝑛𝑒𝑡

𝑄𝑖𝑛 = 1 -

𝑄𝑜𝑢𝑡 𝑄𝑖𝑛 ; Wnet = Qin – Qout = Wturb,out – Wpump, in

Point 1 (Condenser outlet or Pump inlet):

Since point 1 is on the saturated liquid line:

h1 = hf@30⁰C = 125,74 𝑘𝐽

𝑘𝑔 P1 = P@30⁰C = 4,2469 kPa

v1 = vf@30⁰C = 0,001004 𝑚3

𝑘𝑔.𝐾

sfg@30⁰C = 8.0152 𝑘𝐽

𝑘𝑔.𝐾

Point 2 (Pump outlet or Boiler inlet):

s2 = s1 = 0.4368 𝑘𝐽

𝑘𝑔.𝐾 (isentropic process) P2 = Psat liq at 350⁰C = 16529 kPa

h2 = h1 + Wpump in = h1 + vf @30⁰C(P3 – P1)

= 125,74 𝑘𝐽

𝑘𝑔 + 0,001004

𝑚3

𝑘𝑔 (16529 kPa – 4,2469 kPa) = 142,33

𝑘𝐽 𝑘𝑔

Point 3 (Boiler outlet or Turbine inlet):

The water enters the turbine at saturated vapor condition:

h3 = hg@350⁰C = 2563,9 𝑘𝐽

𝑘𝑔 P3 = P sat.water @ 350⁰C = 16529 kPa

s3 = sg@350⁰C = 5,2114 𝑘𝐽

𝑘𝑔.𝐾

Point 4 (Turbine outlet or Condenser inlet):

The entropy of the process from point 3 to point 4 doesn’t change

s4 = s3 = 5,2114 𝑘𝐽

𝑘𝑔.𝐾

Since at point 4 is the mixture of saturated liquid and saturated vapor so we have to identify the quality of the mixture before calculating h4

Mixture’s quality x4 = 𝑆4−𝑆𝑓4

5,2114−0,4368

Then, we can calculate the enthalpy at point 4 through mixture’s quality x4

h4 = hf4 + x4hfg4 = hf@30⁰C + x4.hfg@30⁰C

= 125,74 𝑘𝐽

𝑘𝑔 + 0,5957.2429,8 𝑘𝐽

𝑘𝑔 = 1573,16

𝑘𝐽 𝑘𝑔

Calculate the heat supplied in the boiler

Qin = h3 – h2 = 2563.9𝑘𝐽

𝑘𝑔 - 142,33 𝑘𝐽

𝑘𝑔 = 2421.57 𝑘𝐽

𝑘𝑔

Calculate the work produced by the turbine

Wout = h3 – h4 = 2563.9 𝑘𝐽

𝑘𝑔 – 1573.16 𝑘𝐽

𝑘𝑔 = 990.74 𝑘𝐽

𝑘𝑔

Calculate the net power

Wnet = Wout – Wpump in = 990.74 𝑘𝐽

𝑘𝑔 – 16.59 𝑘𝐽

𝑘𝑔 = 974.15 𝑘𝐽

𝑘𝑔

c Define the thermal efficiency of this cycles

𝜂𝑡ℎ = 𝑊𝑛𝑒𝑡

974.15

Thus, the thermal efficiency is 40,23%

REFERENCES

[1] Robert G.Mortimer (2008) Physical Chemistry 3rd Edition Elsevier publisher,

Canada

[2] Yunus A Cengel, Michael A Boles (2011) Thermodynamics: An Engineering

Approach 7th Edition McGraw-Hill pulisher, USA

[3] Ph.D Nguyen Van Dung Chapter 4: Thermodynamic cycles

[4] P Atkins and J De Paula (2006) Physical Chemistry 8th Edition Oxford University

Press, New York, 2006