Combustion of ethane gas is an exothermic reaction. Which of the following best describes the temperature conditions that are likely to make the combustion of
ethane gas a spontaneous change?
Any temperature, because combustion of ethane leads to an increase in entropy.
Any temperature, because combustion of ethane leads to a decrease in entropy.
O Low temperature only, because combustion of ethane leads to an increase in entropy.
High temperature only, because combustion of ethane leads to a decrease in entropy

An spontaneous change ocurrs when ΔS>0. The boiling point of ethane is −88.5 °C = −127.4 °F = 184.6 K a very low temperature. In the day to day work at much higher temperatures. So, any temperature, because combustion of ethane leads to an increase in entropy

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OlaMacgregor OlaMacgregor · Answer 2 · 2019-08-26T11:31:35+02:00

Answer: Option (A) is the correct answer.

Explanation:

A combustion reaction is defined as the reaction in which hydrocarbons react with oxygen and results in the formation of carbon dioxide and water.

Combustion reactions are always exothermic in nature. It means that enthalpy of reaction is negative.

Also, it is known that [tex]\Delta G = \Delta H - T \Delta S[/tex]

As it is given that [tex]\Delta H[/tex] is negative so, if entropy is also negative then value of [tex]\Delta G[/tex] will be as follows.

[tex]\Delta G = \Delta H - T \Delta S[/tex]

= [tex](-ve) - (-ve)[/tex]

= -ve

If entropy is +ve then value of [tex]\Delta G[/tex] will be as follows.

[tex]\Delta G = \Delta H - T \Delta S[/tex]

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

= -ve

This shows that there will be increase in entropy.

Thus, we can conclude that the statement any temperature, because combustion of ethane leads to an increase in entropy, best describes the temperature conditions that are likely to make the combustion of ethane gas a spontaneous change.