Extra Credit 3

Explanation Step by Step:

1. First things first, the correct answer in the text is slightly off, because it should consider the displacement of Sn 2+, not Sn 4+, keep that in mind while calculating.
2. As an logic check, considering the reactivity and the displacement with these other substances, the answer will end up being a range between two different E standards.
3. The next step in terms of logic is that of a rule when considering E standard. If this metal reacts with a substance, then the metal's E standard is less than that substance's E standard that it reacts with.
4. With that in mind, we can substitute the problem's substances into this concept. This gives us the fact that the E standard for the metal M is less than the E standard for HF, but more than the E standard for HCl.
5. When considering displacement, the above statement is true.
6. As a mathematical equation in terms of E standard, we get: HCl < M < HF when considering the E standards of each of those substances.
7. Once more, we will substitute things into this equation to make our estimate for the half reaction of the M, as stated in the question.
8. When factoring in displacement, the E standard for Fe 3+ is .771 Volts, while the E standard for Sn 2+ is -.137 Volts.
9. The above numbers are coming from the Standard Cell Potential Table, which should be accessable during exams and while working on homework problems.
10. While considering the fact that the displacement does not occur for the Sn 2+ substance, we can go ahead and flip the sign on the corresponding E standard.
11. These numbers yield us an equation that looks a little bit like this, which also just so happens to be the solution to the original question:

Explanation Step by Step:

1. First things first, we should consider all of the relevant equations that relate E standard cell, delta G, and K.... Those equations are represented by this big chemistry triangle:
2. 
3. Now, for part a), we will use blue text.
4. When considering the E cell standard, there is another equation that is relevant: E = reduction - oxidation.
5. The reduction half reaction is relative to the oxygen in the system, going from an oxidation state of 0 to -2 within the water in the products. That half reaction and standard cell potential looks like this:
6. The corresponding oxidation half of the reaction is relative to the iodine in the system going from -1 to 0. That half reaction and standard cell potential looks like this:
7. Side Note: this reaction is written as a reduction half reaction, however when plugged into the equation mentioned in part 4 of this explanation, the sign will be flipped, as it is subtracted from the reduction standard cell potential.
8. This equation, after plugging everything in, looks like this:
9. 
10. Now for part b), we will use red text.
11. Now the fun begins, we will now relate the delta G of this reaction to the standard cell potential that we just calculated, using the equation on the top left portion of the triangle above.
12. That equation looks intimidating, however we know all of these variables already: E = .694 V, F = 96485 C (it is a given constant), and n = 4 (number of transferred electrons, shown as "z" in the upcoming picture).
13. When plugging in all of the numbers into this equation, we get this answer:
14. 
15. Now for part c), we will use green text.
16. Part c) is another plug and chug, just like the previous part, however there is a little bit more algebra involved.
17. The equation we will be using is relating E standard cell to K, which is the bottom one in the above triangle.
18. Again, we know all of the variables, which makes the equation less intimidating: E = .694 V, F = 96485 C, n = 4 (again, shown as "z"), R = 8.3145, T = 298 K (we assume, since the question did not specify).
19. After plugging in variables, and simplifying, we end up with this:
20. 
21. Finally, after some algebraic manipulation, we get the answer of:
22. For the last part: part d), we will return to the black text.
23. This question is simply a logic check for this entire problem, which also checks your understanding of equilibrium constants (which should have been mastered in CHE 2B).
24. Since K, the equilibrium constant, is a massive number (10 to the 46th power!!), we can conclude that the reaction favors the products so much that by the end of the reaction, there is a negligible amount of reactants left over, meaning that the reaction moves toward completion when the products and reactants are in their standard states (which is what we based our calculations off of):

This question is relatively easy once you have achieved the proper understanding of the chemical formulas for molecules and what they mean. K can only give up one electron, due to its 4s^1 electron configuration, meaning it cannot possibly be in the middle of this molecule. Due to its proximity to the edge of the molecule, and the fact that it is less electronegative than the Zn in the reactants, the K is what will be removed from the molecule. After the switch of those substances and balancing of the entire reaction, the end product of the balanced chemical equation is:

1. To start things off, the concepts being covered in this question are related to crystal field theory. The term diamagnetic is when the electrons fill up the valence shell, and paramagnetic is when they do not, leaving a spin on the molecule.
2. For part a), we will use blue text.
3. The Cobalt complex in the question has 6 CN molecules attached to it, which all have a -1 charge, and are considered strong field ligands.
4. Since the molecule has a -3 total charge, and there is a total of a -6 charge coming from the CN ligands, that must mean that the Co atom has a +3 charge.
5. With that said, normally Co has 7 d electrons, but when you take away 3 of them, there are 6 d electrons, due to the primary removal of the 2 s electrons in the valence shell.
6. Since the complex has 6 ligands, the complex is an octahedral.
7. With that said, the electron diagram for an octahedral looks like this:
8. 
9. With that said, since the complex is under the influence of strong field ligands, that means the splitting energy (delta o) is very large, putting the molecule under a low spin situation.
10. With a low spin situation, one uses the d electrons (in this case, 6) to fill and pair the bottom row of d orbitals first.
11. In this case, that would leave us with 6 d electrons on the bottom three d orbitals, each in a pair, which yields no spin on the molecule.
12. Considering this situation, the Cobalt complex is considered diamagnetic based on our definition above, leaving the Manganese complex to be considered paramagnetic, which is visible if the question were tackled by analyzing the Manganese first.
13. If we do analyze the Manganese compound, we would find that Mn would have a +2 charge, since I has a -1 charge, and the overall charge is -4. Since there are 6 iodine atoms in the compound, we would need the +2 charge to keep it balanced.
14. With the +2 charge, Mn would lack the s electrons, leaving it with only 5 d electrons.
15. Since I is a weak field ligand, the molecule would instead be under a high spin situation.
16. The electrons would fill every open orbital once before doubling up, leaving one in each orbital. However, since none are filled, the complex is paramagnetic.
17. For part b), we will use red text. (the textbook got this notation incorrect, as the Mn would need a +7 charge to total at a +3 charge for the complex, consider the total charge to be -3 for the molecule, opposed to +3).
18. Here we have a tetrahedral complex, which has a diagram that is similar to an octahedral complex's diagram, except there are three orbitals on top, and two below.
19. There are 4 Br ligands, which are weak field ligands, with a -1 charge on each.
20. To total at a -3 charge, the Manganese metal needs to have a +1 charge.
21. With that said, a Mn +1 metal has 5 d electrons (as the +1 takes away 1 s orbital electron, opposed to a d orbital one).
22. Referring to the previous diagram of electrons, except flipped for the tetrahedral complex, with this weak field ligand, we are looking at a small delta t, yielding a high spin situation.
23. After filling in all orbitals before pairing any of them, we are left with all five of the orbitals with one electron in each, and none of them are paired.
24. This means that we are left with 5 unpaired electrons for this complex, making it diamagnetic.

Step by Step Explanation:

1. To start things off, an important starting point is determining the order of the entire reaction, which is 2, because X is 0 and W is 2.
2. Double checking the units for the rate constant, we confirm that the reaction's order is 2, as the units for that rate constant is 1 over M times min.
3. The equation for a second order rate is: when A and B are reactant concentrations, k is the rate constant, x and y are the corresponding orders for each reactant.
4. Considering that equation and the given information from the question, we can substitute all of those variables with actual numbers: k = .0115, [A] = [W] = 0.095M, x = 2, [B] = [X] = 2.67M, and y = 0.
5. After plugging and chugging, we end with the answer of:

Step by Step Explanation:

1. For part a), we will use blue text.
2. Uranium-235 has 92 protons, Barium-141 has 56, and Krypton-92 has 36.
3. A neutron is symbolized as:
4. In this case, the Uranium is bombarded with a neutron, which means the left side of the equation will look like this:
5. Considering the right side of the equation, we know the notation for Barium and Krypton, however the amount of neutrons is the unknown as of right now.
6. The method for determining how many electrons are emitted as products can be found through analysis of the masses of both sides of the equation, which should equal each other.
7. Considering the left side of the equation has a total of 236 amu, and the Barium and Krypton masses add up to 233 amu, so the missing mass adds up to 3 amu.
8. Considering the mass of one neutron is 1 amu, that would mean 3 neutrons would amount to the missing mass from the right side of the balanced equation.
9. Combining both sides of the equation, the final solution looks like this:
10. For part b), we will use green text.
11. Gallium-70 has 31 protons, Hydrogen-2 has 1, and Germanium-72 has 32.
12. Utilizing the concepts from part a), we see that the masses are equal on both sides of the equation with Gallium and Hydrogen on the left, and Germanium on the right.
13. The amount of protons on both sides are also equivalent, so this problem has a simple notation, which looks like this:
14. For part c), we will use red text.
15. Antimony-121 has 51 protons, alpha particles are symbolized as: , and have 2 protons, Cesium-128 has 55 protons, and a neutron has none.
16. In order to balance the masses on bother sides of the equation, we would need 2 alpha particles, since the right side of the equation has 129 amu, and the left would then total to 129 amu as well.
17. To confirm this, we can also look at the reaction from a proton analysis' perspective.
18. Since Antimony has 51 protons, and Cesium has 55, that means they are 4 protons "away" from each other.
19. Since each alpha particle has 2 protons, there would need to be 2 on the left side of the equation to get Cesium as a product.
20. Using either of these methods works, and would grant you the same equation, which looks like this:
21. The overall solution for each part, without the explanations is imaged below:

Step by Step Explanation:

1. Here we have a question that is much less based on calculations, and more based on conceptual knowledge of Chemistry.
2. Some concepts to keep in mind for this are:
   1. The ratio of neutrons to protons is important, too much of either relative to the other causes instability in the nucleus, so the ratio that is closest to 1:1 is the relatively most stable atom when looking at atoms with the atomic number of 20 and under. When considering atoms with the atomic number between 20 and 83, the necessary amount of neutrons relative to the number of protons shifts up to 1.5:1 due to the amount of repulsion from proton to proton.
   2. The fact of odd and even numbers of neutrons, protons, and mass number also impacts the stability of an atom, if every one of those numbers are even, then the atom is more stable when compared to an atom with those numbers being odd, due to its impact on the nucleus binding energy.
   3. Magic numbers are those in which a specific number of protons and neutrons yield an extremely stable atom, those numbers are 2,8,20,50,82,114, 126 (neutrons), and 184 (neutrons).
3. For part a), we will use blue text.
4. Neither of these atoms have magic numbers, because there are 16 protons in each, and the Sulfur-32 has 16 neutrons, and the Sulfur-34 has 18 neutrons, meaning that will not determine relative stability.
5. Both the number of protons and the number of neutrons are even as well, meaning that will not determine relative stability either.
6. The remaining concept used to determine relative stability: the ratio of neutrons and protons, is the concept we will be using.
7. Since Sulfur has an atomic number below 20, the stability ratio of neutrons to protons is most stable when close to 1:1.
8. As noted above, Sulfur-32 has 16 protons and 16 neutrons, which yields a ratio of 16:16, or 1:1.
9. Sulfur-34 has 16 protons and 18 neutrons, yielding a ratio of 18:16, or 9:8.
10. Utilizing the concept from above, Sulfur-32 is more stable due to the 1:1 ratio of neutrons to protons.
11. For part b), we will use red text.
12. Arsenic-70 has 33 protons, and 37 neutrons, while Arsenic-73 has 33 protons and 40 neutrons.
13. No magic numbers apply in this situation, therefore magic numbers cannot be utilized to determine the relative stability of these two isotopes.
14. Both atoms have odd numbers of protons, and the Arsenic-70 atom has an odd number of neutrons, which leads us to believe that it is less stable than it's Arsenic-73 cousin isotope.
15. On the flip side of that argument, Arsenic-73 has an odd mass number, however that does not necessarily impact the stability of the isotope.
16. With that said, it is more likely to find an odd(p) - even(n) isotope than an odd(p) - odd(n), making Arsenic-73 the more stable isotope.
17. To reinforce that claim, utilizing the ratio of neutrons to protons concept: the ratio for Arsenic-70 is 1.121, while the ratio for Arsenic-73 is 1.212.
18. Considering that, and the fact that the atomic number is between 20 and 83, the desired ratio is 1.5, and Arsenic-73's ratio is closer to that relative to Arsenic-70's ratio. Meaning, again, that Arsenic-73 is the more stable isotope.
19. For part c), we will be using green text.
20. Tungsten-182 has 74 protons and 108 neutrons, while Tungsten-189 has 74 protons, and 115 neutrons.
21. Once again, magic numbers do not apply to these atoms, therefore they cannot be utilized for determining the relative stability of these isotopes.
22. Considering the protons and neutrons, Tungsten-182 has an even number of both, while Tungsten-189 has an even number of protons and an odd number of neutrons.
23. Since it is more likely to find stable isotopes with even(p) - even(n) than it is to find stable isotopes with even(p) - odd(n), Tungsten-182 is considered the relatively stable isotope when compared to Tungsten-189.
24. To reinforce that claim, we can utilize the ratio of neutrons to protons.
25. Tungsten-182 has a ratio of 2.4595, while Tungsten-189 has a ratio of 2.55.
26. Considering the ratio of neutrons to protons, the Tungsten-182 has a ratio closer to the desired ratio of 1.5, due to the atomic number being between 20 and 83, relative to the Tungsten-189 ratio, which means, again, that Tungsten-182 is the more stable isotope when compared to Tungsten-189.
27. Here is a solution for each part, with a different explanation:

Step by Step Explanation:

Here is a few images explaining this process, which will familiarize you with the concepts imbedded in the question before you begin tackling the calculations in the question itself:

1. In order to compute the amount of energy released from the neutron induced fission reaction of Uranium-235 into Krypton-92, Barium-141, and 3 neutrons, we first need to use Einstein's equation:
2. 
3. The method to using this equation will be to find the amount of energy in each of the reactants and each of the products, and then subtract the sum of the product's energy from the sum of the reactant's energy, which will give us the amount of energy released in Joules.
4. Note: the speed of light (c) is 2.9979x10^8, and you will need to convert amu into kg for your calculations (1 amu = 1.6605x10^-27).
5. For the left half of the equation, we have Uranium-235 and one neutron, with masses in kilograms of 3.9029x10^-25 and 1.6749x10^-27, respectively.
6. After plugging each of those numbers into Einstein's equation separately, we see that the energy in Uranium-235 = 3.5077x10^-8 Joules, and the energy in one neutron = 1.5053x10^-10 Joules.
7. After adding the reactants side of the equation, we see that the total energy of the reactants in Joules = 3.5228x10^-8 Joules.
8. Since we already have the mass in kilograms of a neutron: the mass in kilograms of Kypton-92 is 1.5265x10^-25, and the mass for Barium-141 is 2.3414x10^-8.
9. Doing the same process for the right side of the equation, we see that the energy in Krypton-92 = 1.3719x10^-8, the energy in Barium-141 = 2.1043x10^-8 Joules, and the energy in the 3 neutrons = 4.4149x10^-10 Joules.
10. After adding the products side of the equation, we see that the total energy of the products in Joules = 3.5204x10^-8 Joules.
11. Now that we know the total energy of both sides of the equation, subtracting both will grant us the total energy released by the reaction as a whole.
12. After subtracting the products from the reactants, we see that the total released energy of the reaction = 2.4x10^-11 Joules.
13. Since the question asks for the answer in electronvolts per atom, and also in kilojoules per mole, we must convert this number to each of those units, respectively.
14. 1 Joule = 6.242x10^18 electronvolts, to convert that to electronvolts per atom, we don't need to do anything, considering that this reaction is on a 1 atom basis.
15. After the conversion, we get an answer of 1.498x10^8 electronvolts per atom.
16. For the kilojoules per mole unit answer, we will simply divide the number of Joules by 1000 to get kilojoules, and then multiply the number by Avagadro's number (6.022x10^23), which is one mole of the corresponding substance, to then get kilojoules per mole.
17. After this conversion, we see that the answer in kilojoules per mole = 1.4453x10^10 kilojoules per mole.