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

  1. Aug 12, 2012 #1
    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:
     
  2. jcsd
  3. Aug 12, 2012 #2
    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?
     
  4. Aug 12, 2012 #3
    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?
     
  5. Aug 12, 2012 #4
    -ΔH>-TΔS

    plug values

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

    multiply. do you have any negative values in your inequality?
     
  6. Aug 12, 2012 #5
    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.
     
  7. Aug 12, 2012 #6
    (-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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