Δ S f u s and Δ S v a p for methane, ethane and propane has to be calculated. Concept Introduction: Molar entropy of fusion of a substance ( Δ S f u s ) can be determined from its molar enthalpy of fusion ( Δ H f u s ) and melting point (in kelvin) T m using the given formula, Δ S f u s = Δ H f u s T m Molar entropy of vaporization of a substance ( Δ S v a p ) can be determined from its molar enthalpy of vaporization ( Δ H v a p ) and boiling point (in kelvin) T b using the given formula, Δ S v a p = Δ H v a p T b

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Answer 1

Question

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

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Answer 2

Question

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

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Answer 3

Question

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

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Answer 4

Question

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

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Answer 5

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

Answer 6

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

Answer 7

Chapter 23, Problem 23.1P

Interpretation Introduction

Interpretation:

ΔSfus and ΔSvap for methane, ethane and propane has to be calculated.

Concept Introduction:

Molar entropy of fusion of a substance (ΔSfus) can be determined from its molar enthalpy of fusion (ΔHfus) and melting point (in kelvin) T_m using the given formula,

ΔSfus = ΔHfusTm

Molar entropy of vaporization of a substance (ΔSvap) can be determined from its molar enthalpy of vaporization (ΔHvap) and boiling point (in kelvin) T_b using the given formula,

ΔSvap = ΔHvapTb

Answer 8

Expert Solution & Answer

Answer to Problem 23.1P

ΔSfus and ΔSvap for methane, ethane and propane is given below,

Compound ΔSfus(kJ/K⋅mol) ΔSvap(kJ/K⋅mol)CH41.03×10−27.98×10−2C2H63.16×10−28.48×10−2C3H84.12×10−28.71×10−2

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Unit of temperature can be converted to Kelvin using the given conversion,

0°C = 0°C + 273.15=273.15 K

Substitute the melting and boiling points of methane, ethane and propane as follows,

Tm of CH4 = −182.5°C = 90.65 KTm of C2H6 = −182.8°C = 90.35 KTm of C3H8 = −187.6°C = 85.55 KTb of CH4 = −161.5°C = 111.65 KTb of C2H6 = −88.6°C = 184.55 KTb of C3H8 = −42.1°C = 231.05 K

The formula for determining ΔSfus and ΔSvap is given below,

ΔSfus = ΔHfusTmΔSvap = ΔHvapTb

Substitute the values to calculate ΔSfus and ΔSvap for methane, ethane and propane,

Calculate ΔSfus and ΔSvap for methane:

ΔSfus = 0.9370 kJ/mol 90.65 K= 1.03×10−2 kJ/K⋅molΔSvap = 8.907 kJ/mol111.65 K= 7.98×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for ethane:

ΔSfus = 2.859 kJ/mol90.35 K= 3.16×10−2 kJ/K⋅molΔSvap = 15.65 kJ/mol184.55 K= 8.48×10−2 kJ/K⋅mol

Calculate ΔSfus and ΔSvap for propane:

ΔSfus = 3.525 kJ/mol85.55 K= 4.12×10−2 kJ/K⋅molΔSvap = 20.13 kJ/mol231.05 K= 8.71×10−2 kJ/K⋅mol

Answer 12

Ch. 23 - Prob. 23.1P Ch. 23 - Prob. 23.2P Ch. 23 - Prob. 23.3P Ch. 23 - Prob. 23.4P Ch. 23 - Prob. 23.5P Ch. 23 - Prob. 23.6P Ch. 23 - Prob. 23.7P Ch. 23 - Prob. 23.8P Ch. 23 - Prob. 23.9P Ch. 23 - Prob. 23.10P

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Answer to Problem 23.1P

Explanation of Solution

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Chapter 23 Solutions