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Problem 57

Find $\Delta S^{\circ}$ for the formation of $\ma…

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Problem 56

Use data from Appendix IIB to calculate $\Delta S_{\mathrm{rxn}}^{\circ}$ for each of the reactions. In each case, try to rationalize the sign of $\Delta S_{\mathrm{rxn}}^{\circ}$
\begin{equation}\begin{array}{l}{\text { a. } 3 \mathrm\ {NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm\ {HNO}_{3}(a q)+\mathrm{NO}(g)} \\ {\text { b. } \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Cr}(s)+3 \mathrm\ {CO}_{2}(g)} \\ {\text { c. } \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g)} \\ {\text { d. } \mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm\ {H}_{2} \mathrm{O}(g)}\end{array}\end{equation}

Answer

PART $A : \Delta S^{\circ}=-288 \mathrm{J} / \mathrm{K}$ and see explanation
$\mathrm{PART} \mathrm{B} : \Delta S^{\circ}=14.7 \mathrm{J} / \mathrm{K}$ and see explanation
$\mathrm{PART} \mathrm{C} : \Delta S^{\circ}=-94.0 \mathrm{J} / \mathrm{K}$ and see explanation
PART $\mathrm{D} : \Delta S^{\circ}=119.6 \mathrm{J} / \mathrm{K}$ and see explanation


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so this question telling us to calculate Don't s of the reaction for each of these reactions and it will try toe rationalize. Why that entropy is the way it is just logically. And so the calculate don't s not. You can do. It's not a products. No, it is. It's not reacting, it's And so you want a son those up from each side. And also you want to include the number of moles for each of your molecules. So, for example, for for a you have three and two gas plus each to a liquid going to too for three. Aye, quieres plus and no gas. And so what you're gonna do is you're gonna look up these tapes these values from where a pendant in your textbook and I've already looked at those values. So no, just plug in so that it's not is equal to two times 1 46 plus 2 10.8 So, as your products Marlys, you're actin ce, so you'll do three times to 42 count for your moles plus 70. And then when you finish your math, you'll get an answer to be minus two a 7.5 jewels for Calvin. And so let's think about why this would be a negative changing entry. And so, if you look at this equation, you're going from 43 moles of a gas, only one more of the gas. And so you're the gas is what is the most disorder has the most entropy in this case. So you're going from more of, Ah, molecular has a lot of in tribute to less of a molecule that has a similar amount of entropy. So your entry will decrease. So do the same thing for B. We have CR 23 as a solid plus carbon monoxide going toe See are solid plus c +02 It was a guess. So we'll look up our values in our textbook, and then we'll get that that that's not equals. So this is 23.8 burning confusion. Uh, Mom is your acting's for just anyone. 0.2 close three. I was 1 97.7 So always make sure your accounting for your moles in your equation, and so this will equal 16.8 jewels for Calvin. And so that's the positive increase in entropy and the way you can rationalize. That is you're going from three miles of gas, the three most of the gas. But instead of seat carbon monoxide, you're having an extra oxygen to his gas molecule. So you're giving it more ways to distribute its energy. So you're basically allowing more macro states to be possible for this energy distribution, so is inherently becoming more and tropic. So that's why the entry it's increasing. All right, quit part. See, s 02 plus oxygen going to s 03 and calculated It s the same way that that's not you get to 56 minus 1/2 times to a 5.2 plus 2 48.2 and so dr s equals minus 94. So that's a negative. Changing entropy in the wake of rash is you're going from 1/2 molds total mo's of gas. Melon kills toe one mole of a gas molecule. So you having two unique gas molecules going to one even if you're counting for the 1/2. So this would be a more and tropic state going to lessen tropics st hiring feel gonna learn to be so That would be why this values naked right for D last one into four because for of hydrogen gas going to natural gas, us water is a gas. So make sure you look up the right phase when you're looking these up in your textbook, so same way that it's not is equal to four times 1 80 8.8 plus when? 90 1.6 minus 2 20 Close four times 1 30.7 So you can look these up in the textbook just tonight. Shanking these numbers correct. But they should come out to be close to two or four jewels for Calvin. And so this is the positive changed entropy in the way you can rationalize that is you're going from. You're still billing for families of gas to five most of the gas. But now you are making age to ER which is Maur you're making for most of age to ah, compared to four moles of age to gas and extra oxygen allows you to distribute energy mawr in a more more micro states. So that increases your entropy is what I'm talking about. The micro states I'm talking about s equals k l n w word W's micro states so the more of these you have, the more is you can distribute the energy, the higher your in tribute can be. So that's why you're increasing in entropy for this reaction.

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