Respuesta :
Answer:
c. I₂ (s)
Explanation:
Gibbs free energy -
It is a thermodynamic quantity , which is given by the change in enthalpy minus the product of the change in entropy and absolute temperature.
i.e.,
ΔG is given as the change in gibbs free energy ( KJ )
ΔS is given as the change in entropy ( KJ /K )
ΔH is given as the change in enthalpy ( KJ )
T = temperature ( Kelvin ( K ))
ΔG = ΔH - TΔS
The sign of ΔG determines the reaction spontaneity , as
ΔG = negative , the reaction is spontaneous and
If ΔG = positive , the reaction is non spontaneous .
The standard free energy of formation will be zero , only for the compounds that are in their pure form ,
Hence, from the given option,
I₂ (s) is naturally found in solid state , therefore , ΔG = 0
The standard Gibbs free energy of formation of [tex]\rm I_2[/tex] has been equal to zero. Thus option c is correct.
Standard Gibbs free energy of formation has been the amount of change in energy that is required for the formation of 1 mole of a substance. The value of Gibbs free energy has been zero for the naturally occurring elements as they are not required to be formed and occur naturally.
(a) [tex]\rm H_2O\\[/tex]
It has been a compound made by the combination of naturally occurring [tex]\rm O_2[/tex] and [tex]\rm H_2[/tex]. Since it has been formed, the standard Gibbs free energy has not been equal to zero.
(b) O
The naturally occurring form of oxygen has been [tex]\rm O_2[/tex], the O state of the element has to be formed. Thus the standard Gibbs free energy has not been equal to zero.
(c) [tex]\rm I_2[/tex]
The di-iodide form of the iodine has been the naturally occurring state of iodine in nature. Thus, the standard Gibbs free energy of formation for [tex]\rm I_2[/tex] has been equal to zero.
The standard Gibbs free energy of formation of [tex]\rm I_2[/tex] has been equal to zero. Thus option c is correct.
For more information about the Gibbs free energy of formation, refer to the link:
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