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Ch.19 - Chemical Thermodynamics

Chapter 19, Problem 23b

The normal boiling point of Br2(𝑙) is 58.8β€Š Β°C, and its molar enthalpy of vaporization is Δ𝐻vap=29.6β€ŠkJ/mol. (b) Calculate the value of Δ𝑆 when 1.00 mol of Br2(𝑙) is vaporized at 58.8β€Š Β°C.

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Hello everyone. Today we are being asked to consider a system with 3.62 moles of acetone. Were then asked to calculate the change in entropy that occurs in a system when it condenses at its boiling point of 56.1Β°C. And the heat of vaporization is 29.1 kg per mole. So the first thing I wanna do is you want to recall our equation for entropy which is entropy or delta S. Is equal to Q rev. And rev is just the heat that's transferred in a reversible reaction. And then our tea is going to of course stand for temperature. And so we're gonna just plug in our variables that we have and see what we get. So four hour Q rev or the heat transferred at a reversible process is going to be this 29.1 kg jewels per mole. And we also have to consider that our answer has to be in units of jewels per mall kelvin. So this killer jewels must be converted into regular jewels. And to do that we can just use the conversion factor That there's 10 to the third jewels per one killer jewel. And lastly we must divide by our temperature. And we said as before we need to end units of Kelvin. So we must take that 56.1Β°. Associates and add to 73.15 to it. And we calculate that we get a final answer of 88. jewels per mole kelvin. We must then take into account the molds that we have and so we have to take this value this 88. jewels per mole kelvin. We must multiply by that 3.62 moles of acetone that we have. Our units are gonna cancel that. We're gonna be left with the final value of about roughly 320 jewels per kelvin overall. I hope this helped, and until next time.