Gibbs Free EnergyThird Law of ThermodynamicsDirections: Answer the following items1. Calculate the stande25°C.2. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol,respectively. Calculate the entropy changes for the solid → liquid and liquid vaportransitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.3. Using data listed in Table I below, calculate the equilibrium constant (K) for the followingreaction at 25°C:2H2O1) → 2Hzg) = O2g)%3DInorganic Substances書AH°; (kJ/mol)AG°r (kJ/mol)-137.3S° (J/K'mol)SubstanceCOCO22)H22)H2OaOzg)-110.5197.9-393.5-394.4213.6131.0-285.8-237.269.9205.0Organic Substances-74.85CH4)Mg)MgOs)Table 1. Table 1. Selected Thermodynamic Data at 1 atm and 25°C*-50.8186.232.5-601.8569.626.78*The AH°s, AG, and S° values for ions are based on the reference states AH (H.) = 0, AGr (H.)3 0, and S (H.)%3D0.

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2. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid liquid and liquid vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.

2. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid liquid and liquid vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Gibbs Free Energy
Third Law of Thermodynamics
Directions: Answer the following items
1. Calculate the stande
25°C.
2. The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol,
respectively. Calculate the entropy changes for the solid → liquid and liquid vapor
transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
3. Using data listed in Table I below, calculate the equilibrium constant (K) for the following
reaction at 25°C:
2H2O1) → 2Hzg) = O2g)
%3D
Inorganic Substances
書
AH°; (kJ/mol)
AG°r (kJ/mol)
-137.3
S° (J/K'mol)
Substance
CO
CO22)
H22)
H2Oa
Ozg)
-110.5
197.9
-393.5
-394.4
213.6
131.0
-285.8
-237.2
69.9
205.0
Organic Substances
-74.85
CH4)
Mg)
MgOs)
Table 1. Table 1. Selected Thermodynamic Data at 1 atm and 25°C*
-50.8
186.2
32.5
-601.8
569.6
26.78
*The AH°s, AG, and S° values for ions are based on the reference states AH (H.) = 0, AGr (H.)3 0, and S (H.)%3D0.

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