For a chemical reaction we have, ΔH=-114.1kJ and ΔS=-146.4J/K ---> ΔS=-146.4·10[itex]^{-3}[/itex]kJ/K.(adsbygoogle = window.adsbygoogle || []).push({});

Question:Determine the temperature range at which the reaction is spontaneous.

A reaction is spontaneous when ΔG is negative in the equation ΔG=ΔH-TΔS, so I do:

ΔG=ΔH-TΔS

0>ΔH-TΔS

-ΔH>-TΔS

[itex]\frac{-ΔH}{-ΔS}[/itex]<T (the inequality sign changes direction when we multiply or divide both sides by a negative number right?)

[itex]\frac{ΔH}{ΔS}[/itex]<T (signs cancel each other)

[itex]\frac{-114.1kJ}{-146.4·10^-3kJ/K}[/itex]<T (we input the values)

780K<T , so this tells us that the T must be greater than 780K in order for the reaction to be spontaneous, but then when I check my answer in the original equation:

ΔG=ΔH-TΔS

ΔG=-114.1kJ-(900K)(-146.4·10[itex]^{-3}[/itex]kJ/K) ---> ΔG=18kJ , My inequality answer I suppose is wrong because if T is greater than 780K (in this example I used 900K) then ΔG is positive and the reaction is non spontaneous.

ΔG=-114.1kJ-(200K)(-146.4·10[itex]^{-3}[/itex]kJ/K) ---> ΔG=-85kJ , In here if we use a T lower than 780K the reaction is spontaneous.

I know I made a mistake somewhere because these results are not in accordance with each other, I'm guessing I made the mistake in the inequality while solving for T but I'm honestly not seeing it!, Where did I mess up?

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# Temperature range at which a reaction is spontaneous, Where is the mistake?

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