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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

Chapter 18, Problem 128

The normal boiling point of bromine is 58.8 °C, and the standard entropies of the liquid and vapor are S°[Br2(l) = 152.2 J/(K*mol); S°[Br2(g) = 245.4 J/(K*mol). At what temperature does bromine have a vapor pressure of 227 mmHg?

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Hello. Everyone in this video, we're being told that the molecular iodine supplies at 100 and 84 degrees Celsius were given the standard molar entropy is for solid and gaseous iodine. So this question was being asked to determine the temperature where iodine exhibits a vapor pressure of exactly 433 or 430 millimeters of mercury. The direction that we have going on is R. I two. And it's solid state being an equilibrium of our eye to in its gaseous state. So for my delta s of reaction it's going to be for the entropy of my products minus the entropy of my starting materials. In this case we only have one of each. So it does make the calculation a little bit easier. So starting off with my products, we have one more of this. I'm gonna go ahead and multiply this by the given entropy. So this is given to us as to 60. jules per mole times kelvin's. I'm gonna go ahead subtract this from the start materials. Again, we just have one bowl of this And they give an entropy for this is going to be 116.1 joules, per mole times Kelvin's. So you can see here that the moles will cancel each other out. All right, So once I put everything into vocabulary I get the value of 1 44. jewels per kelvin actually want to convert those jewels into killer jewels. So again we have 144.6 jewels per kelvin covering this into killer jewels. So we have 1000 jewels for everyone kill a jewel Then you can see that my jewels unit will cancel. Once I put the values into my calculator I guess the final delta as a reaction to be 0.1446. Kill Jules for Calvin. Alright so for our face change this is an equilibrium process. So my delta G of reaction and will equal to zero. Let's put this in writing as well. So face change means that my delta G of the reaction is going to equal to zero. Now for my formula for this my delta G of reaction is equal to my delta age of reaction minus T. So temperature multiplied by delta S. Of my reaction. So we see here that we are no temperature. We already have the delta S. Let's go ahead and calculate for delta H. So delta H of reaction is then equal to T delta S. Of my reaction. And this is only possible because our delta G of reaction were said to be equal to zero. So we're basically isolating our delta age of reaction. So again we have all these values already. So let's go ahead and plug those in. So starting off with the temperature that's given to us at 1 84 degrees Celsius, we want to convert this new kelvin's so we can add 273.15 to this. And then for my delta S of reaction that's already said to be 0.1446 kg jewels per kelvin's. Alright so once I put this into my calculator I get that my delta H. Of reaction is equal to 66.10389 killer jewels. Alright so scrolling down for more space over here. So for my K. P. This is equal to the pressure of my eye to gas and this is equal to 430 millimeters of mercury. Let's convert this into A. T. M. We do a direct conversion. So for every one A. T. M. We have 760 millimeters of mercury. Of course that unit will cancel leaving us with just A. T. M. So once I put that into my calculator we get the value of 0.56579 A. T. M. And let's go ahead and recall from a delta G reaction to be equal to this. I'm actually gonna go ahead and add on is also equal to we have another equation that the delta G. Of reaction is also equal to negative R. Times T. Multiplied by the natural log of K. P. Alright so solving for R. T. Value here again the doctor H. Of reaction is equal to negative T. Times delta S. Of reaction and this equal to negative R. Times T. Times natural log of K. P. Again we're solving for T. Here. So for our tea natural log of K. P. Minus T. Delta s of reaction is equal to negative delta H. Of the reaction isolating T. Here we get T. It was just basically factoring out T. So our natural log of K. P minus T. Delta S. Of reaction. It's still equal to the negative delta H. Of reaction. So T. Is then is equal to negative adults. Age of reaction over. Are times the natural log of K. P minus the delta S. Of reaction is all under one denominator. So not calculating finally for tea and putting in some actual values here, go ahead scroll down for that. So T. Is equal to we have negative 66.10389 killed jules per mole. Go ahead and divide this. We have our our this is our gas law constant. So 8.314 times 10 to the negative three kill jules. Permal times kelvin's Go ahead and multiply this. But the natural log of K. P. to that. 0. -0.1446 killer jewels per Kelvin. Alright, so once I put everything into my calculator again we get the numerical value to be 442. 477 kelvin's gonna convert this back into degrees Celsius. So we're subtracting this time to 73.15. Once I put that into my calculator I get my temperature to equal to 169.5 degrees Celsius. So this right here is going to be my final answer for this problem. Thank you all so much for watching.
Related Practice
Textbook Question
Is it possible for a reaction to be nonspontaneous yet exo-thermic? Explain.
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Textbook Question

Trouton's rule says that the ratio of the molar heat of vaporization of a liquid to its normal boiling point (in kelvin) is approximately the same for all liquids: ∆Hvap/Tbp ≈ 88 J/(K*mol) (a) Check the reliability of Trouton's rule for the liquids listed in the following table.

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Textbook Question

Trouton's rule says that the ratio of the molar heat of vaporization of a liquid to its normal boiling point (in kelvin) is approximately the same for all liquids: ∆Hvap/Tbp ≈ 88 J/(K*mol) (b) Explain why liquids tend to have the same value of ∆Hvap/Tbp.

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Textbook Question
Tell whether reactions with the following values of ΔH and ΔS are spontaneous or nonspontaneous and whether they are exothermic or endothermic. (a) ΔH = - 48 kJ; ΔS = + 135 J>K at 400 K (b) ΔH = - 48 kJ; ΔS = - 135 J>K at 400 K (c) ΔH = + 48 kJ; ΔS = + 135 J>K at 400 K (d) ΔH = + 48 kJ; ΔS = - 135 J>K at 400 K
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Textbook Question
The following reaction, sometimes used in the laboratory to generate small quantities of oxygen gas, has ∆G° = -224.4 kJ/mol at 25°C:

Use the following additional data at 25 °C to calculate the standard molar entropy S° of O2 at 25°C: ∆H°f(KClO3) = -397.7 kJ/mol, ∆H°f(KCl) = -436.5 kJ/mol, S°(KClO3) = 143.1 J/(K*mol), and S°(KCl) = 82.6 J/(K*mol).
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Textbook Question
Suppose that a reaction has ΔH = - 33 kJ and ΔS = - 58 J>K. At what temperature will it change from spontaneous to nonspontaneous?
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