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Ch.19 - Chemical Thermodynamics
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All textbooksBrown 14th EditionCh.19 - Chemical ThermodynamicsProblem 51d

Chapter 19, Problem 51d

Using S° values from Appendix C, calculate ΔS° values for the following reactions. In each case, account for the sign of ΔS°.

d. 2  CH3OH(𝑔) + 3  O2(𝑔) ⟶ 2  CO2(𝑔) + 4  H2O(𝑔)

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Hey everyone, we're given the following entropy values and were asked to calculate and provide the correct sign of our change of entropy for the following reaction. Now to calculate our change in entropy we know that this is going to be the entropy values for our products minus the entropy values of our reactant. So plugging in the compounds into this formula, we would have to take n times the entropy value of carbon dioxide and n is going to represent the number of moles we have. And we're also going to add n times the entropy value of water. And we're going to subtract n times the entropy value of isopropyl alcohol. And we're going to add n times the entropy value of our oxygen gas. So let's go ahead and plug in these values. Starting off with the entropy value of carbon dioxide, we can see that we have six mole of carbon dioxide and we're going to multiply six times 213. joules per mole times kelvin. Next we're going to add eight mole times The entropy value of water, which is 188.83 jewels per mole kelvin. Next we're going to subtract the entropy values of our reactant. So that's going to be two more from our isopropyl alcohol and we're going to multiply two mole by 322.6 joules per mole times kelvin. Next we're going to add nine more from our oxygen gas And we're going to multiply nine x 205.0 jewels per mole kelvin. Now when we calculate this out and cancel out our units, we end up With a change of entropy of plus 302.0 jewels over Kelvin, which is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.
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