The standard free energies of formation (AGF) of PCI,(g) and PCI,(g) are -286 and -325 kJ/mol, respectively. The equilibrium constant for the following reaction at 25°C equals (R8314 J/molK) OA 15X107 OB 27X10-¹ OC 1.7 X 10 OD 8.1X10 PC, (g) PCI, (g) + Cl₂(g)
The standard free energies of formation (AGF) of PCI,(g) and PCI,(g) are -286 and -325 kJ/mol, respectively. The equilibrium constant for the following reaction at 25°C equals (R8314 J/molK) OA 15X107 OB 27X10-¹ OC 1.7 X 10 OD 8.1X10 PC, (g) PCI, (g) + Cl₂(g)
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![**Understanding the Relationship Between Standard Free Energies and Equilibrium Constants**
The standard free energies of formation (ΔGf°) of PCl3(g) and PCl5(g) are –286 kJ/mol and –325 kJ/mol, respectively. The equilibrium constant for the following reaction at 25°C is to be determined:
Given the reaction:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
The universal gas constant (\( R \)) is 8.314 J/mol·K.
**Options:**
a) \( 1.5 \times 10^{-7} \)
b) \( 2.7 \times 10^{-3} \)
c) \( 1.7 \times 10^{-4} \)
d) \( 8.1 \times 10^{-5} \)
This problem involves calculating the equilibrium constant (K) using the Gibbs free energy values of the participating compounds. The relationship between the Gibbs free energy change of the reaction (ΔGr°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G_r° = -RT \ln K \]
Where:
- \( \Delta G_r° \) is the standard Gibbs free energy change for the reaction.
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin.
- \( K \) is the equilibrium constant.
For more detailed explanation and calculations, students are encouraged to refer to specific thermochemistry sections of their textbooks or consult chemical thermodynamics resources available in the educational website’s repository.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18d7602f-9f25-445c-bda8-4d38e99e2d82%2Fccff083f-8bc5-4dba-ade4-65caf69a79c8%2Ffiq1th8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Relationship Between Standard Free Energies and Equilibrium Constants**
The standard free energies of formation (ΔGf°) of PCl3(g) and PCl5(g) are –286 kJ/mol and –325 kJ/mol, respectively. The equilibrium constant for the following reaction at 25°C is to be determined:
Given the reaction:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
The universal gas constant (\( R \)) is 8.314 J/mol·K.
**Options:**
a) \( 1.5 \times 10^{-7} \)
b) \( 2.7 \times 10^{-3} \)
c) \( 1.7 \times 10^{-4} \)
d) \( 8.1 \times 10^{-5} \)
This problem involves calculating the equilibrium constant (K) using the Gibbs free energy values of the participating compounds. The relationship between the Gibbs free energy change of the reaction (ΔGr°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G_r° = -RT \ln K \]
Where:
- \( \Delta G_r° \) is the standard Gibbs free energy change for the reaction.
- \( R \) is the gas constant (8.314 J/mol·K).
- \( T \) is the temperature in Kelvin.
- \( K \) is the equilibrium constant.
For more detailed explanation and calculations, students are encouraged to refer to specific thermochemistry sections of their textbooks or consult chemical thermodynamics resources available in the educational website’s repository.
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