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

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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.

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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.

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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.

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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.

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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

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Chapter 19 Solutions

Interpretation:

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