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

Chapter 18, Problem 133

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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Well everyone's in this video talking about this reaction right over here and we're being asked to use the given thermodynamic data to calculate for my entropy for br two at 25 degrees Celsius. So my delta G. Of reaction. This formula is equal to delta H. Of reaction minus T. S. Over temperature multiplied by delta S. Of reaction. And my K. P. Value is equal to the pressure of 02. So auction gas. So for my delta H of reaction, the sequel to while we have zero plus zero minus two times negative. 100.4 killer jewels Permal. Once you put that into a calculator we get that delta H of reaction is equal to 200.8 killer jules Permal. Again we're using the formula for delta G. Here. So delta G of reaction. I'm just rewriting this equal to delta H. Of reaction minus T. Delta S of reaction. Alright so we're already given the delta G value that is 3.8 kg joules per mole is equal to our delta H value. Just solve for that's 200.8 kg jewels Permal. We're subtracting this with T. Which is temperature. That's a 25°C. We want the C to be in Kelvin's. So we're adding 273.15. Multiplying this by delta s of reaction. It's something we are solving for. Just leave it as is. Alright let's go ahead and isolate our delta s of reaction If we do so we get - 3.8. Kill Jules Remote minus 200.8 kg joules per mole divided by 298.15. Cape. For kelvin's, this gives me the delta s of reaction Is equal to 0.0234781. Killer joules per kelvin to convert the killer jewels unit into jewels. So we're doing a direct conversion here. So for every 1000 jewels that we have, we have one killer jewel Then we can see from my unit cancelation that the killer jewels will cancel. So putting these values into the calculator, I get the numerical value to be 23.4781 units being jewels per kelvin. I wanna go and scroll down for more space here. So for my deltas of reaction we have the entropy of my products minus the entropy of my reactant. So we have two delta acid reaction, that is 23.4781 jules per kelvin equaling two. We have two moles Times The 42.6 Jewels Per Mole Times Kelvin plus the one more, Multiplied by the entropy of R. b. R. two - moles, multiplied by 107.1 jewels per more times Calvin. Alright, we're gonna go ahead and isolate the term that we need to solve for. So that is one mold multiplying by the entropy of B. R. To this equals to 23.4781 jules per kelvin plus 214.2 jewels per kelvin minus 85.2 jewels per kelvin. So basically just simplifying here And once we get this entropy of BR two value, this should give us 152.5 units being jewels per mole times Kelvin. All right, so this right here is going to be my final answer for this problem.
Related Practice
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
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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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
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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Textbook Question
A mixture of 14.0 g of N2 and 3.024 g of H2 in a 5.00 L container is heated to 400 °C. Use the data in Appendix B to calculate the molar concentrations of N2, H2, and NH3 at equilibrium. Assume that ∆H° and ∆S° are independent of temperature, and remember that the standard state of a gas is defined in terms of pressure.

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Textbook Question

The lead storage battery uses the reaction: (b) Calculate ∆G for this reaction on a cold winter's day (10 °F) in a battery that has run down to the point where the sulfuric acid concentration is only 0.100 M.

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