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Ch.19 - Free Energy & Thermodynamics
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All textbooksTro 6th EditionCh.19 - Free Energy & ThermodynamicsProblem 57b

Chapter 19, Problem 57b

Use data from Appendix IIB to calculate ΔS°rxn for each of the reactions. In each case, try to rationalize the sign of ΔS°rxn . b. C(s) + H2O(g) → CO(g) + H2(g)

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Hello everyone today we are being asked to calculate the delta S. Of the reaction or the total entropy of the reaction. And then we have the following reaction listed out here. So the first thing we want to do is write this reaction as if it were to dissolve in a solution. So we have the initial calcium which is a solid plus two moles of liquid water. We're gonna say that's gonna dissolve into one calcium two plus ion Aquarius and two hydroxide groups. Also Aquarius. And then we have hydrogen gas here We then must look at the different values that we have for our total entropy of each compound. So for calcium or see a we have a value of 41.6 jewels per mole kelvin for water Which is a liquid. Of course we have 70 jewels per mole kelvin For our calcium two plus ion which is the highest form of it. We have negative 53. jewels per mole kelvin For our hydroxide concentration. Which is also in the acquis form we have negative 10.9 jewels per mole kelvin. And then finally for our hydrogen gas we have a total entropy of 130.7 jewels per mole kelvin. And so now to find the total delta S. Or the total entropy of the reaction, we're going to take the amount of molds that we have for each compound and we're going to multiply those by the total entropy of each compound itself. So for example we're going to start with the products and so we're going to start with our calcium R. C. A. Two plus and we only have one of those. So we're going to take that negative Or 53.1. We're going to add that to our second product which is hydroxide and we have two of those. So we're going to do two times negative 10.9. And we're going to finally add that to our last product which is hydrogen gas. And we only have one of those. So we have 103.7 And we're gonna subtract that by doing the same thing but with the reactant. And so our first reaction is calcium solid. Since I only have one of those, we have only 41.6 here we're gonna add that to our two moles of water which gives us an entropy of 70. And then we're gonna lastly Close that off with a bracket and that's going to give us a total entropy of the reaction as negative 125.8 jewels per mole kelvin. And now we have reached our final answer. I hope this helped. And until next time
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