Respuesta :
Answer:
A and C are true , B and D are false
Explanation:
For A)
from the first law of thermodynamics (in differential form)
dU= δQ - δW = δQ - PdV
from the second law
dS ≥ δQ/T
then
dU ≤ T*dS - p*dV
dU - T*dS + p*dV ≤ 0
from the definition of Gibbs free energy
G=H - TS = U+ PV - TS → dG= dU + p*dV + V*dp - T*dS - S*dT
dG - V*dp + S*dT = dU - T*dS + p*dV ≤ 0
dG ≤ V*dp - S*dT
in equilibrium, pressure and temperature remains constant ( dp=0 and dT=0). Thus
dG ≤ 0
ΔG ≤ 0
therefore the gibbs free energy should decrease in an spontaneous process → A reaction with a negative Gibbs standard free energy is thermodynamically spontaneous under standard conditions
For B) Since the standard reduction potential is related with the Gibbs standard free energy through:
ΔG⁰=-n*F*E⁰
then, when ΔG⁰ is negative , E⁰ is positive and therefore a coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous.
Answer:
Statements A and C are true.
Explanation:
Let's evaluate the given statements:
A) A reaction with negative Gibbs standard free energy (ΔG) will result in a reaction thermodynamically spontaneous, under standard conditions, by the Nernst equation:
[tex] \Delta G = - nFE_{cell} [/tex] (1)
where n: is the number of electrons transferred in the cell reaction, F: is the Faraday constant and E: is the cell potential
For an electrochemical reaction to be spontaneous, that is to say, in a galvanic cell, the change in Gibbs free energy must be negative. From equation (1), we have that a negative ΔG will result in a spontaneous electrochemical reaction.
Hence, statement A is correct, and D is incorrect.
C) A coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous:
On equation (1), for the Gibbs free energy to be negative and, therefore, the reaction to be thermodynamically spontaneous, the standard reduction potential must be positive.
Hence, statement C is correct, and B is incorrect.
I hope it helps you!
