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bekas [8.4K]
2 years ago
15

Anyone who wanna be my frien this is my sn id hassaan_navee21I am b​

Chemistry
1 answer:
Marysya12 [62]2 years ago
4 0

Answer:

sure

Explanation:

ill add u later

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Read 2 more answers
HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make
vazorg [7]

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

<u>Explanation:</u>

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.

8 0
3 years ago
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