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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium
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All textbooksMcMurry 8th EditionCh.18 - Thermodynamics: Entropy, Free Energy & EquilibriumProblem 79

Chapter 18, Problem 79

For the melting point of sodium chloride, ΔHfusion = 28.16 kJ/mol and ΔSfusion = 26.22 J/(K·mol). Does NaCl melt at 1100 K? Calculate the melting point of NaCl.

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everyone in this video, we're giving the delta H. Of fusion and the delta S. A. Fusion of glucose. So these values here and at 100 degrees Celsius we're being asked if glucose melts and if it does not melt then what is the melting point for glucose? So we were given here from this problem or what we know we know that glucose melts at when our doctor us of total. So the total delta s. is equal to zero. We can calculate R. T. So temperature when they dealt us total is equal to zero. We know that the equation for adults as total. If you go to the delta S for the system plus the delta S. Of our surroundings. So this basically means since our delta S of total is zero that the delta S of our system is going to be equal to my delta S of fusion and we know delta us of surroundings is equal to the negative delta H. Of fusion over R. T. And temperature. So we're gonna go ahead and do some math now. So again our delta S total is equal to zero. So first we have our 47.5 jewels per kelvin times more. We're gonna go ahead and add this. Well we have a negative 19. killer jewels per mole. We want to convert this killer jewels unit into jewels. So we have 1000 jewels for every one kg jule. It's all divided by T. So once I kind of simplify this we get The 47.5 jewels, her Kelvin times more. You going to 1. times 10 to the positive four value. All over. Are units on top being jewels time small. And it's all over our T. Value. We're gonna go ahead and isolate for T. And we get one we get 1.993 times 10 to the positive four jewels per molt over 47.5 jewels per kelvin times month. So once we saw fly that we get to T. So temperature is equal to 4 19. Calvin's. So we want to convert this back into Celsius to match it with the melting point. So how to do this is by subtracting 273.15 and we get our T. Value to be 1 46 degrees Celsius. So then to answer this question, Go ahead. First scroll down a little. All right, so we can say that at 100°C glucose does not melt. But instead the multi point is actually 146 degrees Celsius. So the reason why glucose does not melt at 100 degrees Celsius is because its melting point is actually at 100 and 46. So this highlight is going to be my final answer for this problem
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