[Thermodynamics] Calculate change in entropy of closed reversible system

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Homework Help Overview

The problem involves calculating the change in entropy for mercury as it freezes under specific conditions, utilizing concepts from thermodynamics. The original poster presents the enthalpy change associated with the phase transition and attempts to apply relevant equations to find the entropy change.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the change in entropy using the enthalpy change and the mass of mercury. Participants question the calculations of moles and the application of the equations.

Discussion Status

Participants are actively engaging in clarifying the calculations and definitions involved in the equations. Some guidance has been provided regarding the correct interpretation of variables and units, but there is no explicit consensus on the final reasoning yet.

Contextual Notes

There is a noted confusion regarding the conversion between grams and moles, as well as the proper application of the equations for enthalpy and entropy. The discussion reflects the challenges of ensuring accurate unit conversions and understanding the relationships between the variables involved.

Ortix
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Homework Statement


Mercury is a silvery liquid at room temperature. The freezing point is -38.9 degrees celsius at atmospheric pressure and the enthalpy change when the mercury metls is 2.29 kJ/mol. Wat is the entropy change of the mercury if 50.0 g of mercury freezes at these conditions? The molarmass of mercury is 200.59 g/mol. Assume the process is reversible

Homework Equations


Q=m(h_2-h_1) (enthalpy equation)
Q=mT(s_2-h_1) (entropy equation)

The Attempt at a Solution



first I calculated what the energy change is per kg with what the enthalpy change is. h2-h1 is 2.29 kJ/mol and since there are 4.0118 mols I got 9.187 J/kg.

So then I used Q/m = T(delta S)

I rewrote Q/m to 459.35 J (since there are 50 grams of the substance) and divided that by T which is 234.1. The answer I get is 1.96 J/K whereas the answer is -2.44 J/K.

Not sure what I'm doing wrong.
 
Last edited:
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Hi Ortix! :smile:

Perhaps you calculated the number of moles the wrong way around?
If 1 mol is 200.59 g, how many moles is 50.0 g?
 
Hey Serena,

Very stupid mistake indeed. It would be 50/200.59=0.249264669 but this would result in a much smaller solution and still not negative (however, it would be negative entropy since the mercury is being frozed, in other words, energy is being taken out)

I quickly worked it out on my laptop calculator (in bed on my laptop) and my answer is: 0.0488 J/K which is obviously wrong.
 
Yup, that's why it's negative.

Did you take the 2.29 kJ/mol into account?
Or else what did you calculate?
 
I got the answer! Pretty simple calculation, but could you perhaps tell me what "m" exactly is in the equations and what the units are?
Q=m(h_2-h_1) (enthalpy equation)
Q=mT(s_2-h_1) (entropy equation)

Because I ended up with the right answer being in the form Q/mT when it should be Q/T. I think I'm doing something wrong in the conversion of mol to gram.

EDIT:
I think I got it. I equated both equations and let the m's drop out. Since enthalpy is given as specific enthalpy per mol, i just multiplied it by the amount of moles present in the substance and got the total enthalpy change. Divide that by T and that is the answer!

Can anyone tell me if my reasoning is correct? :)
 
Last edited:
Yep. That's it. :)

m is the mass in kilograms.

And for the record, your 2nd equation should read: Q=mT(s_2-s_1) (entropy equation)
 
Oh yeah, that was a typo :) Thanks for putting me on the right track! :D
 

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