Answer: The correct statement is low temperature only, because entropy decreases during freezing.
Explanation:
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
Where,
[tex]\Delta G[/tex] = change in Gibb's free energy
[tex]\Delta H[/tex] = change in enthalpy
T = temperature
[tex]\Delta S[/tex] = change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and [tex]Delta S[/tex] is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the [tex]\Delta H[/tex] is also negative.
For a reaction to be spontaneous, [tex]\Delta G[/tex] must be negative.
[tex]-ve=-ve-[T(-ve)]\\\\-ve=-ve+T[/tex]
From above equations, it is visible that [tex]\Delta G[/tex] will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.
Answer:
Low temperature only, because entropy decreases during freezing.
Explanation:
