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SupposemL Ilask /5 (led wth0 Il,0. LL mol CO aid mol of H,- Cle) + Hjolg) ~ cole)+ JH,(e)Icjction tcdide pOssiblaComelcte the tablc bclow, hsts the Vutial mclanity ...

Question

SupposemL Ilask /5 (led wth0 Il,0. LL mol CO aid mol of H,- Cle) + Hjolg) ~ cole)+ JH,(e)Icjction tcdide pOssiblaComelcte the tablc bclow, hsts the Vutial mclanity ech compourid Orlan equibnum molanty eJ cblpound aher mcianny Ihe Tex tion n: cach comnoound duc t0 the reaction; tnd &2 cDMI Cuubnum Siano tor the unknown chahqe mclamy & CH You @n leqvc out Ihic smbom mnaune

Suppose mL Ilask /5 (led wth 0 Il,0. LL mol CO aid mol of H,- Cle) + Hjolg) ~ cole)+ JH,(e) Icjction tcdide pOssibla Comelcte the tablc bclow, hsts the Vutial mclanity ech compourid Orlan equibnum molanty eJ cblpound aher mcianny Ihe Tex tion n: cach comnoound duc t0 the reaction; tnd &2 cDMI Cuubnum Siano tor the unknown chahqe mclamy & CH You @n leqvc out Ihic smbom mnaune



Answers

The reaction between $\mathrm{I}^{-}$ and $\left[\mathrm{S}_{2} \mathrm{O}_{8}\right]^{2-}$ occurs as follows: $$ \left[\mathrm{S}_{2} \mathrm{O}_{8}\right]^{2-}+2 \mathrm{I}^{-}--2\left[\mathrm{SO}_{4}\right]^{2-}+\mathrm{I}_{2} $$ $\mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow=\left[\mathrm{I}_{3}\right]^{-}$ The second step occurs so long as there is an excess of $\mathrm{I}^{-}$ present in solution; $\left[\mathrm{I}_{3}\right]^{-}$ absorbs at $353 \mathrm{nm}$ $(\log \varepsilon=4.41) .$ Explain how measurements of the absorbance at $\lambda=353 \mathrm{nm}$ would enable you to measure the change in concentration of $\mathrm{I}_{2}$ during the reaction.

In each part of this problem were given an overall chemical reaction, and we need to determine whether or not this reaction will proceed spontaneously. We know that in order for a reaction to be spontaneous, that value for it's standard cell potential has to be positive in order for that value to be positive. That means that the species that undergoes reduction has to have 1/2 reaction with a greater value of its standard reduction potential than the species that undergoes oxidation. So we can begin by looking at the reaction part and determining which species undergoes oxidation. Which one undergoes reduction. We see that calcium is oxidized from its solid state to its acquiesced to plus state by losing two electrons. And if we look up the value for the standard reduction potential for the half reaction of see a two plus plus two electrons going to see a solid, then we see that it's standing reduction potential comes out to negative 2.87 volts, and now we can see that CD two plus is reduced to CD solid by adding two electrons, and that half reaction has a standard reduction potential of negative 0.40 volts. And now we see that CD, which is reduced, has a larger or less negative value for its standard reduction potential. Then for the reduction of see a two plus, which means that this reaction will proceed spontaneously and again. This is because when we find the overall cell potential, we take the standard reduction potential of the species that is reduced an added TV reverse of the standard reduction potential of the species that is oxidized. And when we do that, that would come out to a positive value. If her positive values for the cell potential, this reaction would be spontaneous. So that is a reasoning that we used to work through the rest of this problem on part B. We see that BR minus is oxidized to be are too, and that has a standard reduction potential of 1.7 volts and s and two plus is reduced to us and solid, and that has a standard reduction potential of negative 0.14 volts. So now we see that the species that undergoes reduction has a smaller value for its standard reduction potential than the species undergoes oxidation. So That means when we go out to you calculate the overall cell potential, it will end up being negative, meaning that this reaction will be non spontaneous. And now, in part C, we can see that silver is oxidized from its solid state to its acquis plus one state, and that has a standard reduction potential of zero 0.80 volts. And we can also see that on I is reduced from two plus two. It's solid state, and that has a student reduction potential of negative 0.25 volts. So, just like in part B, we see that the standard reduction potential for the reaction that undergoes reduction is smaller than for the species that undergoes oxidation. So that means overall cell potential will end up being negative, making this reaction non spontaneous. And now, in party, we see that we have see you plus being oxidized to see you two plus by losing an electron and that has a standard reduction potential of 0.15 volts. We also see that F E is reduced from three plus 22 plus, and that has a standard reduction potential of 0.77 volts and this time we see that the species that undergoes reduction as a larger value for its standard reduction potential than the species that undergoes oxidation. And so when we flip the sign of this value and added to this value, that will come out to a positive overall value for the standard cell potential, making this reaction spontaneous.

In this question, we want to fill out the given table for a first order decomposition reaction. So we know that's his first order read. She equals K times A. So for experiment one rate should equal 7.5 Times 0.1, Which is 0.05. For experiment too, there is no change in the temperature, only eight and there is no change in catalyst, there's only a change in the concentration of A. So K should be the same for experiment to as as it is for experiment one than to find the rate we get, that weight should equal 0.5 times 0.2. So the rate should equal zero point one. Yeah. Mhm. Finally, for the activation energy, there is no change to the state of the catalyst. So it should Be the same as an experiment one. For experiment three, the weight should be equal two K, which is one times 0.1. So it should equal to BB equal to 0.1. And given that there is a catalyst, or given no other information, we cannot determine the activation energy for Experiment three, for experiment, for we get that the rate should be equal 2 0.6 times 0.1, So it should be 0.6 And because there is no, we're going back to the case where there is no catalyst. The activation energy should be the same as in one and two, so it should be 32.

For the reaction that we are given in the problem statement, we went to calculate the values of the overall self potential of standard conditions the standard changing Gibbs free energy and also the equilibrium constant K at 25 degrees Celsius. If we examine the overall reaction that were given and we noticed that see you pluses both oxidized and reduced, we're given the value of the standard reduction potential for the reduction of see you plus and for the oxidation of C. U Plus and to see you two plus, we look up the standard reduction potential for the reduction of see you two plus two C u plus and we reverse it. And when we reverse its sign from positive to negative, that corresponds to a potential of negative 0.15 volts. So now we see that when we combine these 2/2 reactions together and cancel off one mole of electrons on either side that we have to see you plus ions combining for the reactant to produce. See you two plus and see you. And this is the overall reaction that were given. So to find the standard cell potential, which is the first value that we need to calculate. We just add these two potentials together so that 0.5 to minus 0.15 and that comes out to 0.37 volts. And the next value we need to find is the change in Gibbs Free Energy. It's standard conditions. We know that Delta G is equal to negative N f E and is a number of moles of electrons that we transferred and we saw that that was one mole of electrons. F is Faraday's constant, which is 96,000 485. Cool ums. Her mole of electrons and E is the standard cell potential, which we just calculated the 0.37 volts or jewels per Coolum. So now we see that most of electrons and cool ums cancel out. So we're left with energy units of jewels. If we divide the overall result by 1000 we can get Delta G in units of killing jewels. When we do that, we should find that the changing gives free energy. It's standard conditions where this reaction is about negative 36 killer jewels. Now we use the nursed equation to sell for the equilibrium constant at 25 degrees Celsius. We know that at equilibrium the cell potential is equal to zero volts. And based on the nursed equation, this is equal to the standard cell potential, which we calculated to be 0.37 volts, minus 0.591 divided by N, she said, was one mole of electrons times the log of K instead of queues. Since we're at equilibrium, we re arrange this now to solve four K should find that it comes out to about two times 10 to the power of six.

Here were given information for a specific chemical reaction. We have phosphorus Penta oxide plus water yields essentially phosphoric acid and we're essentially given a table for the different quantities and we're asked to find essentially the delta the the standard free energy of the reaction. So in this case we have to remember that delta jeannot equals delta H. Now of the reaction minus T delta S not of the reaction. And were given that this reaction occurs At 298 Kelvin. And we have a long list of a data table. So r N trapeze and jewels per mole kelvin. So we have to remember to convert it to kill jules since our and therapies are given in kilograms promote And were given that the entropy for the first component 2984, -285.8 -1281 Or entropy is 228.969.91. 110.5. So essentially in this case we basically for delta H not of reaction, we take the summation of the basically heats of formation for the the entropy of formation of the products minus delta H, nods half of the heat of formation of the reactions. So here we're seeing that we're making four moles of phosphoric acid, so that's our only product. So this would be four times negative 1,281. While for the reactant we essentially see that we're making reactant, we have basically six miles. We're consuming six bowls of water, So six times negative 285.8. and we're essentially consuming one mole of phosphorus oxide. And essentially for the first term We have negative 5,124, I'll add units later -2698. And basically adding everything together. We can find that essentially the Uh the entropy of the reaction is -425. About negative 425 killer goals per mole. And for the uh the entropy of reaction is the same process. We're going to summit the entropy of the products and minus the entropy of the reactant. So once again we have to look at their data table again, we're making four moles of essentially phosphoric acid. Then we're losing essentially six moles of water and we're losing one mole of phosphorus Penta oxide And you see that's 440- -648.4. When we find that the entropy of the reaction is negative 206 jaws prime or kelvin. However, this has to be intelligible since we have to agree, we have need units agreement with a century and kill people. So this is negative 0.206 killer jewels promote Calvin. So now let's plug everything back into our equation And remember that we're running this at 298 Kelvin and essentially we can find by plugging our values in Delta G. Not for this reaction. At 298 Kelvin is negative. 364 killer goals promote and this is our final answer.


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