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Problem 35

Find $\Delta S_{\mathrm{rxn}}^{\circ}$ for the co…


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Problem 34

For each reaction, predict the sign and find the value of $\Delta S_{\mathrm{rxn}}^{\circ}$
(a) $3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l)+\mathrm{NO}(g)$
(b) $\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$
(c) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$

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Video Transcript

in each part of this problem were given a chemical reaction. And we want to first form a prediction for what we think the sign will be meaning positive or negative for the change in standard, more entropy of the chemical reaction or delta s of reaction. After that, we want to use this equation in order to calculate a new miracle value for that l tests of reaction. So starting in part, a method that we used to determine whether Delta s of reaction will be positive or negative is to examine the total number of moles of gaseous species on the react inside in the product side, the reason that we focus on Lee on the gashes species is that we know that out of the three states of matter, solid liquid and gas gas particles of any substance move about randomly and are therefore always more disordered and thus have a greater standard molar entropy, then that same substance as it exists as a solid or liquid, since there is more order you a solid or a liquid. And so the effect of non gashes species like the two liquids in this reaction will have a negligible effect since the gas species elicit the greatest amount of of entropy in the system and now are we have to do is count the total number of of moles of the gases species on the reactant and product side, we see that we have three total moles of gas on the reactant. In this warms one mole of of gas on the product side. In the change in entropy of the reaction delta s of reaction, any changes? A final ST minus an initial state. So we produce one mole of gas after we started with three moles of gas and remember that a gaseous species has the greatest amount of entropy or disorder. And so if we have more moles or a greater amount of gas, we have a greater amount of disorder. So as we decrease the number of gas molecules, we increase order and therefore decrease entropy. And so that means we should predict that the change in entropy of the reaction will be negative less than zero. And so that's our prediction for part A. And now we use this equation to find the change in in s of the reaction by first finding the total Moeller entropy of the products and then subtracting the total total Mueller entropy of the reactant. So we start with the product side and we see that we have two moles of H and 03 and one mole of Eno ensue. The designation that I'm using is this term in this equation I'm going to define his pee in this term is our for products in reactant So we're starting with the products and again we have two moles of H and R three and one mole of Eno So to motivation or three and one mole of Eno Then we can look up the standard Moeller entropy values in the appendix for each one of those substances and reason that we have to multiply i n The number of moles in each case is that those standard Miller entropy values in the appendix are given on a personal basis. So when we multiply the total number of moles we get the total amount of entropy for that chemical species in units of jewels per kelvin And when we multiply those values together for each species and then add them Because of this summation, we get this value for the total entropy of the products. And then we use the same steps in order to find the total Mueller entropy of the reactant. Seeing that we have three moles of n 02 and one mole of each 20 we conclude that three moles n 021 more of h 20 and look up their standard more entropy values and multiply them and add them together to get this value. And now that we have r p in our our terms, we just subtract are frumpy and that is how we arrive at this answer part A. We followed the exact same procedure now for Parts B and C, with the only difference being a different given chemical reaction in each case. So, in part B, we again want to start with our prediction. We see that all chemical species are already gas is so we can see that we have a total of one was three so four gas molecules reacting to form to gas molecules su the number of gas molecules decreased again. So therefore the order increase. And so the delta s of reaction, we should again predict to have a negative sign report be then we just use that scene equation again. You calculate out what that delta s of reaction would be So on the B again, start with the products term P. We see that we have two moles of NF three. We look up, it's standard more entropy And that is the only product so that that amount corresponds to the total Mueller entropy of the products And we do that again for the reactant Swe have one mole event too Three moles of F two, one mole of n 23 moles of F two and we multiply each one by the respective standard more entropy is and then add them together Get this value Then after we subtract this value from this value, we get our final answer for Delta s of reaction And just like in part, a notice that the sign matches our prediction. Finally, in part C, we have this reaction we can ignore solid and see that we go from six moles of gas to six plus six or 12 moles of gas. So in this case, the total number of gashes molecules increases And so we increased disorder and so entropy increased as a result of the reactions. So we should expect that to be a positive change. And we used our seem equation again. For the products term, we have six moles of CO two and six moles of H 20 six Moles of co 26 moles of H 20 multiply each one invited by its standard more entropy value and add them together to get the total standard Moeller entropy of the products And then for the reactant, we have one mole glucose and six moles of oxygen one mole of glucose six moles of oxygen area the same calculation after looking up the standard molar entropy values values in the appendix To get the total change the total standard molar entropy of the reactant Then subtract that from the total standard more entropy of the products. You get Delta s reaction to come out to this value and again that is a positive value, which is what we predicted or for that sign in part C

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