Change in Free Energy when Heating a Substance

  • #1
Summary:
Heating anything implies a change in enthalpy and entropy which should produce a change in free energy (Gibbs). However, free energy is defined at constant temperature and temperature was increased from A to B. What can I do?
Hello everyone!

I'm new to the forum, and I've been trying to solve a problem that seems farily but I can't still convince myself of having the right (or wrong) answer.

Imagine you warm up one mol of for example potassium hydride from standard temperature 298.15K to 673.15K (400 deg C). It is clear that there is going to be a change in both enthalpy and entropy which luckly can be calculated from NIST's Chembook [1]. The enthalpy change involved in heating this substance can obtained from the Shomate equation which evaluates to +20.77 kJ which makes sense since energy needs to be absorbed by the substance in order to heat it up. Using the Shomate equation again the enthalpy at 400 deg C can be calculated. By subtracting it from its enthalpy at 298.15K we get the enthalpy change which is equal to 100.99 J/K. This also makes sense since warming up things should increase their degree of "disorder".

So far so good! Now the tricky part. The change in Gibbs free energy is defined as: dG = dH - T*dS. This equation involves only one temperature since it is supposed to be defined for constant temperatures. I came up with different ideas on what temperature to use including the average between the starting and finishing temperatures, the final temperature, and even the change in temperature dT.

I think this last one (dT) could actually be correct since the Gibbs free energy is defined as G = H - T*S, and thus a differential version of it where temperature is not constant would look like this dG = dH - d(T*S) [2]. Nevertheless, I get negative values for the dG meaning that warming up this chemical is exergonic or "spontaneous". My intuition tells me that this can't be true. Although "spontaneous" is a very vague concept (or misused word), I belive nature should opose to warming up things. If you heat a metal with a blow torch until it is glowing red, and then you stop you don't see the metal's red color increase in intensity for a while before going off. As soon as you cut the heat source it starts cooling off which is a more "natural" thing to do.

I would appreciate if you could give me any insights on what would be the right approach for calculating this dG, specially on which concepts I am missing.

Thank you very much for your help :)


Refs
[1] https://webbook.nist.gov/cgi/cbook.cgi?ID=C7693267&Units=SI&Mask=2#Thermo-Condensed
[2] https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php
 

Answers and Replies

  • #2
DrDu
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You should simply integrate:
## G_2-G_1=\int_{T_1}^{T_2} dG/dT \; dT##.
 
  • #3
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You are correct to say that the correct equation for the free energy change is $$\Delta G=\Delta H-\Delta (TS)$$ But that doesn't mean that the process occurs spontaneously. That is only implied if the system is an open (flow) system in contact with a constant temperature reservoir .
 
  • #4
DrDu
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You should simply integrate:
## G_2-G_1=\int_{T_1}^{T_2} dG/dT \; dT##.
More specifically, from ##dU=TdS-pdV## and ##G=U+pV-TS## you get ## dG=-SdT-Vdp##, so that at constant pressure ## G_2-G_1=-\int_{T_1}^{T_2} S(T) dT##.
 
  • #5
More specifically, from ##dU=TdS-pdV## and ##G=U+pV-TS## you get ## dG=-SdT-Vdp##, so that at constant pressure ## G_2-G_1=-\int_{T_1}^{T_2} S(T) dT##.
Hello DrDu, thank you very much for your explaination. Si I just take the the Shomte equation for enthalpy S(T), integrate it from T1 to T2, and that should be it?
Thanks!
 
  • #6
You are correct to say that the correct equation for the free energy change is $$\Delta G=\Delta H-\Delta (TS)$$ But that doesn't mean that the process occurs spontaneously. That is only implied if the system is an open (flow) system in contact with a constant temperature reservoir .
Hello @Chestermiller , thank you very much! So is it correct to simply use the temperature difference to calculate the change in free energy?
 
  • #7
Lord Jestocost
Gold Member
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In practice, one proceeds in the following way to calculate the Gibbs energy values for chemical species (see, for example, the Outotec HSC Chemistry Software). The enthalpy, ##H##, and entropy, ##S##, can be calculated by means of the heat capacity, ##c_p##, at constant pressure:
$$H(T)=H_f(298.15)+ \int_{298.15}^T c_p\, dT+\Sigma H_{tr}$$
Here, ##H_f(298.15)## is the enthalpy of formation at 298.15##K## and ##H_{tr}## is the enthalpy of transformation of the substance when there are transformations within the considered temperature interval.
For the entropy one has:
$$S(T)=S(298.15) + \int_{298.15}^T (c_p/T)\, dT+\Sigma H_{tr}/T_{tr}$$
Here, ##S(298.15)## is the standard entropy of the substance.
The Gibbs energy, ##G(T)##, can then be written:
$$G(T)=H(T)-TS(T)$$
 
  • #8
DrDu
Science Advisor
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Hello DrDu, thank you very much for your explaination. Si I just take the the Shomte equation for enthalpy S(T), integrate it from T1 to T2, and that should be it?
Thanks!
Yes, integrate the Shomate equation for entropy (not enthalpy) S(T). The equations given for H(T) and S(T) by Lord Jestocost are a more simple but less precise approximation (at least as long as cp is assumed to be constant and not temperature dependent, otherwise they are exact). As Chestermiller pointed out, you could alternatively take the differences of H and TS at the two temperatures. All results should coincide.
 
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  • #10
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Usually the change is Gibbs free energy with temperature is calculated from: $$\frac{d(G/T)}{d(1/T)}=H$$
 

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