Extra Credit 40

Q17.5.8

Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.

Q17.5.8

From the Nernst equation, E°_cell=(\(\dfrac{RT}{nF}\))lnK, we can see that the cell potential (E°_cell) is directly proportional to the temperature in the equation meaning that if the temperature decreases then the cell potential decreases. So, in low temperatures, the E°_cell is low because of the relationship between them resulting in a low cell voltage in the battery of the device. This is equivalent to a battery running out and becoming dead. This observation is also true in reverse: in higher temperatures the cell potential will be higher and because of this battery- powered electronic devices will perform much better than in lower temperatures. In essence, cell potential and temperature determine how effective battery powered devices will perform.

Q12.3.3

Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions:

1. What is the order of the reaction with respect to that reactant?
2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant?

Q12.3.3

1. First, we will write up a equation for the rate law. Rate = k[A]^x where k is the initial rate constant ant x is the order of reaction with respect to the concentration of A.

So, if we tripled the concentration of A,it will increase the rate of reaction by nine so now the equation would look like this. 9x(rate) = [3A]^x. If we assume the concentration was one, the equation will be 9 = 3^x.

Now, we solve for x, the order of reaction, with respect to A and x = 2. Therefore, the order of reaction with respect to that reactant will be 2.

We can use the same procedure of writing out the rate law formula and solving for x do determine the reaction order for almost any reaction. This is especially useful because it allows one to solve for reaction orders which could be fractions and are not easy to determine just by looking at the data and graphs derived from it.

2. We can follow the same procedure in the first part to get the answer of the second part of this question. So, we will start with the general rate law equation again, which is rate = k[A]^x.

If the concentration of A increased by a factor of four and the rate of reaction increases by four. The equation will now be 4x(rate) = [4A]^x. Now, if we assume the concentration was one, the equation will be 4 = 4^x.

Solve for x so x = 1. Therefore, the order of reaction with respect to A, the reactant, will be 1. Once again, this procedure is highly effective in determining rate laws for reactions.

Q12.5.12

In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?

1. the change in free energy per second
2. the change in temperature per second
3. the number of collisions per second
4. the number of product molecules

Q12.5.12

The collision theory is used mostly for gases to predict the rates of chemical reactions. The definition states that in order for a collision to occur, the reacting species (atoms and/or molecules) must collide with one another. However, this collision is only effective in producing a reaction if enough energy is applied (enough to equal the activation energy) and the reactants must be positioned in the ideal manner for a rearrangement of atoms and electrons to occur. In essence, the theory implies that the rate at which a chemical reaction proceeds is equivalent to the frequency of effective collisions. Using this information, we can answer the following question.

In terms of the collision theory, the numbers of collisions per second is proportional to the rate of a chemical reaction. The definition of the collision theory explains why reaction rates increase as concentrations increases. As the concentration increases, the chances for the collisions between molecules is increased because there is more volume per unit of volume assuming the energy of the collision is adequate. Therefore, the rate of a chemical reaction is proportional to the number of collisions per second.

Q21.4.7

Which of the following nuclei is most likely to decay by positron emission? Explain your choice.

1. chromium-53
2. manganese-51
3. iron-59

Q21.4.7

Positron emission is also known as beta plus decay which means that a proton inside the nucleus is converted to a neutron while releasing a positron and an electron neutrino. According to this, when a nuclei undergoes decay via positron emission, the mass number will not change but the number of protons will decrease by 1 since it is being emitted. So, in order to find which nuclei is most likely to decay by positron emission, you will need to find the neutron to proton ratio and positron emission is likely to happen when the ratio is low.

We will first find the neutron to proton ratio and then compare them to each other. Nuclei are determined by the number of protons that they have so this value can always be determined by looking at the periodic table. The number of protons is the atomic number. To find the number of neutrons, you subtract the atomic number(protons) from the mass number.

1. Chromium-53: mass number = 53, atomic number = 24

number of neutrons: 53 - 24 = 29, number of protons: 53 - 29 = 24

neutron to proton ratio: \(\dfrac{29}{24}\) = 1.21

2. Manganese-51: mass number = 51, atomic number = 25

number of neutrons: 51 - 25 = 26, number of protons: 51 - 26 = 25

neutron to proton ratio: \(\dfrac{26}{25}\) = 1.04

3. Iron-59: mass number = 59, atomic number = 26

number of neutrons: 59 - 26 = 33, number of protons: 59 - 33 = 26

neutron to proton ratio: \(\dfrac{33}{26}\) = 1.27

Manganese-51 will most likely decay by positron emission because it does have the lowest neutron to proton ratio and Chromium is a staple isotope and Iron decays by beta emission because it has the highest ratio. Stable isotopes such as Chromium here will not undergo any spontaneous decays because nuclei try to achieve and stay in a state of equilibrium, decay and absorption only occur because nuclei want to reach equilibrium. Likewise Iron here will experience a different kind of beta decay where it will lose an electron and not a positron like Manganese.

Q20.2.11

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

Q20.2.11

For each reaction, we will use the activity series to determine if a reaction occurs. The activity series is a table that has elements in decreasing order of reactiveness from top to bottom. First, we need to write the overall equation for the situation. Then, we will know which two elements we need to compare them to each other. Using the activity series, if the first element is more reactive than the second, then a reaction will occur. If the first element is less reactive than the second, then a reaction will not occur. An example of an activity series table is given below. As you can see, Lithium is the most reactive metal.

1. The overall equation for this problem is \[Pt(s)+HCl(aq)\rightarrow PtCl_{2}(aq)+H_{2}(g)\]

From the equation, we can now tell that we will compare platinum to hydrogen. On the activity series, we see that platinum is below hydrogen which means that platinum is less reactive than hydrogen and we needed the platinum to be more reactive than the hydrogen for a reaction to occur. Therefore, a reaction will not occur.

2. The overall equation for this problem not knowing if it a reaction will occur is \[Mn(s)+FeCl_{2}(aq)\rightarrow MnCl_{2}(aq)+Fe(s)\]

Now we will compare the two elements (Manganese and Iron) to each other. On the activity series table, Manganese is higher than Iron so we can predict that a reaction will occur because manganese is a reactant which is more reactive than Iron, the product, which is less reactive.

3. The overall equation for Tin being heated with steam is \[Sn(s)+2H_{2}O(g)\rightarrow SnO_{2}(s)+2H_{2}(g)\]

We are going to compare Tin to hydrogen on the activity series. Tin is above hydrogen on the table meaning that tin is more reactive than hydrogen. Therefore, a reaction will occur because the first element, the reactant is higher and more reactive then the second, the product.

4. The overall equation for this situation is \[H_{2}(g)+Pb(NO_{3})_{2}(aq)\rightarrow Pb(s)+2H^{_{+}}(aq)+2NO_{3}^{_{-}}\]

In this equation, we will compare hydrogen (H_{2}) to solid lead. On the activity series, it shows that hydrogen is right below lead. With hydrogen gas being on the reactant side, it means that hydrogen is less reactive then lead in this situation. Therefore, a reaction will not occur.

Q20.5.6

Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?

Q20.5.6

Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reactions because the cell potential (E°) is not a state function meaning that it does not depend on the path taken to reach a specific value. Cell potential is an intensive property, which means that it does not depend on how many times the reaction occurs. Because Gibbs Free Energy (G°) is a state function, the sum of G° values for the individual reactions gives us G° for the overall reaction and is proportional to both the potential and number of electrons transferred. The sum of two half-reactions corresponds to the overall reaction when you first, find the values of G° = (-nFE°) for each individual half-reaction then you can use that to find the cell potential and now can add them up for the overall reaction. This is because G° is proportional to the cell potential which then can you sum the potentials of the two half-reactions to get the overall reaction.

Q24.6.2

In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?

Q24.6.2

The Crystal Field Theory describes the breaking of orbital degeneracy in transition metal complexes due to the presence of different ligands and qualitatively describes the strength of the metal-ligand bonds. Ligands are ions or neutral molecules that bond to a central metal atom or ion. Ligand field complexes such as octahedrals, tetrahedrals and square planars causes degenerate sets of d orbitals to split into different energy levels. The energy levels are usually split into e_g and t_2g for both octahedral and tetrahedral complexes and for a square planar complex, the d orbital is split into four different levels. The splitting of d orbitals into different energy levels is known as crystal field splitting (_o), where the subscript o stands for octahedral. The magnitude of the splitting depends on the charge of the metal ion, the position of the metal on the periodic table and the nature of the ligand meaning what kind of complex it is (ie. tetrahedral, octahedral,and square planar). Lastly, the crystal field splitting does not change the total energy of the d orbitals. The splitting of octahedral, tetrahedral, and square planar orbitals are given below. Notice how e_g and t_2g are similar in octahedral/tetrahedral (although switched) but are different for square planar.