Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①

Do the Lone Pairs get added bc its valence e's are a total of 6 for oxygen and that completes it or due to other reasons. How do we know the particular indication of such.

NGLISH b) Identify the bonds present in the molecule drawn (s) above. (break) State the function of the following equipments found in laboratory. Omka) a) Gas mask b) Fire extinguisher c) Safety glasses 4. 60cm³ of oxygen gas diffused through a porous hole in 50 seconds. How long w 80cm³ of sulphur(IV) oxide to diffuse through the same hole under the same conditions (S-32.0.0-16.0) (3 m 5. In an experiment, a piece of magnesium ribbon was cleaned with steel w clean magnesium ribbon was placed in a crucible and completely burnt in oxy cooling the product weighed 4.0g a) Explain why it is necessary to clean magnesium ribbon. Masterclass Holiday assignmen PB 2

Answer 2

Question

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Answer 6

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Answer 7

Chapter 19, Problem 84QRT

Interpretation Introduction

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Answer 8

Expert Solution & Answer

Explanation of Solution

Clausius-Clapeyron equation can be given as,

ln P = −ΔvapHRT + C −−−−−−−−−−−−−−−−−−−(1)

Where,

−ΔvapH is molar vaporization enthalpy.

C is the constant that is characteristic of liquid.

If there are two sets of pressure and temperature, then the Clausius-Clayperon equation can be recast as shown below,

ln (P2P1) = −ΔvapHR[1T2 − 1T1] −−−−−−−−−−−−−−−−−−−(2)

From the phase diagram, two pressure-temperature points can be estimated roughly. The obtained pressure and temperature is given below,

P1 = 10−4 atm T1 = 393 KP2 = 10−5 atm T2 = 368 K

Sublimation enthalpy for monoclinic sulfur can be calculated by substituting the values in equation (2).

ln (P2P1) = −ΔvapHR[1T2 − 1T1]ln (10−510−4) = −ΔvapH8.314 J K−1 mol−1[1368 K − 1393 K]ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)

Rearranging the above equation, the sublimation enthalpy can be calculated,

ln (10−1) = −ΔvapH8.314 J K−1 mol−1(0.00017 K−1)−ΔvapH = ln (10−1) × 8.314 J K−1 mol−1 × 10.00017 K−1 = −2.302 × 48905.882 J mol−1 = 1.12 × 105 J/mol

Sublimation enthalpy of monoclinic sulfur is 1.12 × 105 J/mol.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 19.1 - Prob. 19.1E Ch. 19.1 - Prob. 19.2E Ch. 19.2 - Prob. 19.3CE Ch. 19.2 - Prob. 19.4CE Ch. 19.3 - Prob. 19.5E Ch. 19.4 - Prob. 19.1PSP Ch. 19.4 - Prob. 19.6E Ch. 19.4 - Prob. 19.7E Ch. 19.4 - Prob. 19.2PSP Ch. 19.5 - Prob. 19.8CE

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Interpretation: Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions

Interpretation:

Sublimation enthalpy of monoclinic sulfur has to be calculated using Clausius-Clapeyron equation.