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

  • Thread starter drtg45
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  • #1
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For a chemical reaction we have, ΔH=-114.1kJ and ΔS=-146.4J/K ---> ΔS=-146.4·10[itex]^{-3}[/itex]kJ/K.

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?:confused:
 

Answers and Replies

  • #2
160
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I prefer to stay away from negative signs when possible.

TΔS>ΔH

ΔS is a negative value, correct? So to quote you, "(the inequality sign changes direction when we multiply or divide both sides by a negative number right?)"

what does that do to your inequality when you divide the negative ΔS back over?
 
  • #3
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Hey thanks for your reply, I just finished posting something about inequalities in the General Math section, it would be great if you could check it out:

https://www.physicsforums.com/showthread.php?t=627753

maybe by doing what I describe in there the problems brought about by the minus sign and the change of direction of the inequality sign can be avoided?
 
  • #4
160
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-ΔH>-TΔS

plug values

(-1)(-114.1kJ)> (-1)T(-146.4·10^−3kJ/K)

multiply. do you have any negative values in your inequality?
 
  • #5
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-ΔH>-TΔS

plug values

(-1)(-114.1kJ)> (-1)T(-146.4·10^−3kJ/K)

multiply. do you have any negative values in your inequality?

That's what I'm talking about, plugging the values before we get to make the mistake and change the direction in the inequality sign.
 
  • #6
160
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(-1)(-114.1kJ)> (-1)T(-146.4·10^−3kJ/K)

Since you have 2 negative numbers being multiplied on either side, they become positive. Now you don't have have negative numbers anymore, so you don't need to flip the inequality sign when you divide.
 

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