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

In summary: You can just solve for T as usual.In summary, to determine the temperature range at which the reaction is spontaneous, we use the equation ΔG=ΔH-TΔS and solve for T by dividing both sides by -ΔS. However, it is important to be careful with the direction of the inequality sign when plugging in values, as changing the sign of any negative values will affect the final answer.
  • #1
drtg45
5
0
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:
 
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  • #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?
 
  • #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?
 
  • #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?
 
  • #5
ChiralWaltz said:
-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
(-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.
 

1. What is the temperature range at which a reaction is spontaneous?

The temperature range at which a reaction is spontaneous can vary depending on the specific reaction and its reactants. However, in general, reactions tend to be spontaneous at higher temperatures and less spontaneous (or non-spontaneous) at lower temperatures.

2. Can a reaction be spontaneous at any temperature?

No, a reaction cannot be spontaneous at any temperature. As mentioned before, the temperature range at which a reaction is spontaneous depends on the specific reaction and its reactants. Some reactions may not be spontaneous at any temperature, while others may have a wider range of temperatures at which they are spontaneous.

3. What is the mistake in determining the temperature range for a spontaneous reaction?

The mistake in determining the temperature range for a spontaneous reaction is often assuming that all reactions will be spontaneous at the same temperature range. As mentioned before, the temperature range for a spontaneous reaction varies and is dependent on the specific reaction and its reactants.

4. Is the temperature range for a spontaneous reaction always the same?

No, the temperature range for a spontaneous reaction is not always the same. As mentioned before, it varies depending on the specific reaction and its reactants. Additionally, external factors such as pressure and concentration can also affect the temperature range for a spontaneous reaction.

5. How can the temperature range for a spontaneous reaction be determined?

The temperature range for a spontaneous reaction can be determined through experimentation and observation. By manipulating the temperature and measuring the rate of the reaction, scientists can determine the temperature range at which the reaction is spontaneous. Additionally, thermodynamic calculations can also be used to predict the temperature range for a spontaneous reaction.

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