Question

HELP!!!! 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.

Answer 1

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:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.

For a reaction to be spontaneous, $\Delta G$ must be negative.

$-ve=-ve-[T(-ve)]\\\\-ve=-ve+T$

From above equations, it is visible that $\Delta G$ will be negative only when the temperature will be low.

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