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What is the relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction?
- a)?G is unrelated to Kc.
- b)?G > 0 when Qc > Kc, indicating a spontaneous reaction.
- c)?G < 0="" when="" qc="" />< kc,="" indicating="" a="" non-spontaneous="" reaction.="" kc,="" indicating="" a="" non-spontaneous="" />
- d)?G = 0 at all equilibrium states.
Correct answer is option 'B'. Can you explain this answer?
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What is the relationship between the equilibrium constant Kc and the G...
Relationship between Kc and ?G:
The relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction is described by the equation ?G = -RTln(Kc), where R is the gas constant and T is the temperature in Kelvin.
?G > 0 when Qc > Kc:
- When the reaction quotient Qc is greater than the equilibrium constant Kc, the reaction is not at equilibrium.
- In this case, the Gibbs free energy (?G) is greater than zero, indicating that the reaction is non-spontaneous in the forward direction.
Spontaneity of the reaction:
- If ?G is negative, the reaction is spontaneous in the forward direction.
- If ?G is positive, the reaction is non-spontaneous in the forward direction.
- If ?G is zero, the reaction is at equilibrium.
Therefore, when Qc is greater than Kc, the reaction will proceed in the reverse direction to reach equilibrium, and the value of ?G will be greater than zero. This relationship helps us understand the spontaneity and direction of a reaction based on the values of Kc and Qc.
The relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction is described by the equation ?G = -RTln(Kc), where R is the gas constant and T is the temperature in Kelvin.
?G > 0 when Qc > Kc:
- When the reaction quotient Qc is greater than the equilibrium constant Kc, the reaction is not at equilibrium.
- In this case, the Gibbs free energy (?G) is greater than zero, indicating that the reaction is non-spontaneous in the forward direction.
Spontaneity of the reaction:
- If ?G is negative, the reaction is spontaneous in the forward direction.
- If ?G is positive, the reaction is non-spontaneous in the forward direction.
- If ?G is zero, the reaction is at equilibrium.
Therefore, when Qc is greater than Kc, the reaction will proceed in the reverse direction to reach equilibrium, and the value of ?G will be greater than zero. This relationship helps us understand the spontaneity and direction of a reaction based on the values of Kc and Qc.
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What is the relationship between the equilibrium constant Kc and the G...
When ?G is greater than 0 and Qc is greater than Kc, it indicates a spontaneous reaction in the forward direction.
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What is the relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction?a)?G is unrelated to Kc.b)?G > 0 when Qc > Kc, indicating a spontaneous reaction.c)?G d)?G = 0 at all equilibrium states.Correct answer is option 'B'. Can you explain this answer?
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What is the relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction?a)?G is unrelated to Kc.b)?G > 0 when Qc > Kc, indicating a spontaneous reaction.c)?G d)?G = 0 at all equilibrium states.Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about What is the relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction?a)?G is unrelated to Kc.b)?G > 0 when Qc > Kc, indicating a spontaneous reaction.c)?G d)?G = 0 at all equilibrium states.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the relationship between the equilibrium constant Kc and the Gibbs free energy (?G) of a reaction?a)?G is unrelated to Kc.b)?G > 0 when Qc > Kc, indicating a spontaneous reaction.c)?G d)?G = 0 at all equilibrium states.Correct answer is option 'B'. Can you explain this answer?.
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