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The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make this a spontaneous change? Any temperature, because entropy increases during freezing. Any temperature, because entropy decreases during freezing. Low temperature only, because entropy decreases during freezing. High temperature only, because entropy increases during freezing.

Respuesta :

Answer: The correct statement is low temperature only, because entropy decreases during freezing.

Explanation:

It is given that freezing of methane is an exothermic reaction.

The equation for Gibb's free energy is given by the relation:

[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

As, methane is freezing, this means that [tex]Delta S[/tex] is decreasing and its sign is negative. This reaction is an exothermic reaction, which means that the [tex]\Delta H[/tex] is also negative.

[tex]-ve=-ve-[T(-ve)]\\\\-ve=-ve+T[/tex]

So, for the reaction to be spontaneous [tex](\Delta G=-ve)[/tex], the temperature must be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

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