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

20,0 Complete the electron pushing mechanism to y drawing the necomery unicaciones and carved on for Step 1: Add curved arms for the tint step, traiment with NalilĻ. The Nation 458 Step 2: Added for the second step, inalment with), how the "counterion bar Step 3: Daw the products of the last simplom organic and one incoganic spacient, including all nonbonding

please provide the structure for this problem, thank you!

Draw the Fischer projection from the skeletal structure shown below. HO OH OH OH OH H Q Drawing Atoms, Bonds and Rings Charges I ☐ T HO H H OH HO I CH2OH H OH Drag H OH -CH2OH CHO -COOH Undo Reset Remove Done

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.