Extra Credit 43

All Phase II corrections, made by Isabella Schrammel, are indicated in red ink.

Q17.6.2 (Part II)

Aluminum (E°_Al³⁺/Al = −2.07 V) is more easily oxidized than iron (E°_Fe³⁺/Fe = −0.477 V), and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation.

S17.6.2 (Part II)

Before looking at anything else, we must first know the meaning of "corrosion." Corrosion is the natural (spontaneous) oxidation of a metal, with the effect of converting a refined metal into a more chemically stable (lower potential energy) form.

We can see aluminum is more easily oxidized than iron from their given reduction potentials. Aluminum has a more negative reduction potential (-2.07 V) as opposed to iron (-0.477 V), and a smaller reduction potential indicates a smaller tendency to be reduced, and rather a greater tendency to be oxidized.

Knowing that corrosion involves oxidation, we might first expect that aluminum (as the more easily oxidized metal) would be corroded before iron. In actuality, most of the aluminum is protected through a small amount of proceeding corrosion; this process is called self-passivation. self-passivation. Phase II addition: Self-passivation can be defined as the spontaneous formation of a non-reactive "shield" on the surface of the metal (in this case Aluminum) that acts as a barrier for further corrosion. Aluminum reacts with oxygen in the air to form Al₂O₃ on the surface of the metal, and this coating protects the rest of the metal. This process could be compared to a vaccine - a patient being exposed to a small amount of a virus protects them from worse effects later on!

Iron, however, does not act in self-passivation, so it is not as protected from corrosion as aluminum. self-passivation, so it is not as protected from corrosion as aluminum.

Q12.3.6

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction NO + O₃ → NO₂ + O₂ is first order with respect to both NO and O₃ with a rate constant of 2.20 × 10⁷ L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 × 10⁻⁶ M and [O₃] = 5.9 × 10⁻⁷ M?

S12.3.6

A good first step for questions dealing with kinetics is to write out a rate law. Rate laws are expressions expressing the rate of the reaction in terms of the concentrations of the reactants in the equation raised to some experimentally determined order as an exponent, all multiplied by an experimentally determined rate constant k. In this case, we are told that both of the reactants (NO and O_3, on the left-hand side of the equation) are first order, so they will both be raised to the power of 1. Thus, we can write the rate law expression for this reaction as:

rate = k[NO][O₃]

The instantaneous rate of the reaction is the rate at which reactants are disappearing and products are appearing at any instant in time. In this question, we are not given any time in units of time, but rather the concentrations at the time we are interested in. Using this information, as well as the given k, we have all the values we need to plug into the rate law we found! In doing so, we find the answer to be:

rate = (2.20x10⁷ L/mol/s)(3.3x10^-6 M)(5.9x10^-7 M) = 4.3x10^-5 mol/L/s.

Phase II addition: In this problem, it is important to note that the rate law given (rate = k[NO][O₃]) implies that the reaction NO + O₃ → NO₂ + O₂ is an elementary reaction. The order of a rate law does not always correspond to the stoichiometric coefficients of a reaction, unless the reaction is an elementary reaction.

Q12.5.15

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C₁₂H₂₂O₁₁ + H₂O ⟶ C₆H₁₂O₆ + C₆H₁₂O₆ follows a first-order rate equation for the disappearance of sucrose: Rate = k[C₁₂H₂₂O₁₁] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

1. In neutral solution, k = 2.1 × 10⁻¹¹ s⁻¹ at 27 °C and 8.5 × 10⁻¹¹ s⁻¹ at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10⁻⁷ M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

S12.5.15

1. This is a kinetics question. We are already told this is a first-order reaction and given the corresponding rate law. Now to get started on the subproblems:

With the parentheses at the end, we are told this question involves the Arrhenius question! However, even without that hint, we know we must use the Arrhenius equation since the terms "activation energy" (E_a) and "frequency factor" came up. The Arrhenius equation is

[Equation 1] k = Ae^-E_a/RT

Where k is our rate constant, A is a constant known as the frequency factor, E_a is the activation energy, R is the gas constant in terms of 8.3145 J/molK, and T is temperature (K). We can manipulate this equation into a form more usable to us. By taking the natural log of both sides, we arrive at the equation:

[Equation 2] ln k = ln A (-E_a/RT)

This equation can also be expressed in ratio form:

[Equation 3] ln (k₂/k₁) = -E_a/R x (1/T₂ - 1/T₁)

We can use this Equation 3 to first find the value of the activation energy using the information given to us. We know that at 27°C (in 27+273 = 300 K, let's call this T₁), k = 2.1 × 10⁻¹¹ s⁻¹ (let's call this k₁). At 37 °C (in 37+273 = 310 K, let's call this T₂), k = 8.5 × 10⁻¹¹ s⁻¹ (let's call this k₂). Plugging in these values into the equation, and using our constant R, we can find E_a:

ln (8.5x10^-11 s^-1/2.1x10^-11 s^-1) = -E_a/8.3145 J/molK x (1/310 - 1/300)

1.398 = -E_a/8.3145 x (1/310 - 1/300)

-13002.6 = -E_a/8.3145

E_a = 108000 J = 108 kJ (there are 1000 J in 1 kJ).

The activation energy of the reaction will be the same no matter what temperature, so this is one part of our necessary answer. The frequency factor is a constant, so this will also be the same at any temperature. Thus, we can plug in some of the given data for either one of the temperatures into Equation 1, as well as the activation energy we just found. Let's use the data for 27 °C, where k = 2.1 × 10⁻¹¹ s⁻¹. Plugging values into Equation 1 to solve for A, we get:

2.1 × 10⁻¹¹ s⁻¹ = Ae^{-108 kJ/(0.0083145 kJ/molK)(300 K)}

A = 1.3x10⁸ s^-1.

Finally, we can use our activation energy and our frequency factor to find the rate constant k at the desired temperature of 47°C (320 K) using Equation 1 again:

k = (1.3x10⁸ s^-1)e^{-108 kJ/(0.0083145 kJ/molK)(320 K)} = 3.0 x 10^-26 s^-1

2. We know this is a first order reaction. We can use the integrated form of the first order rate law: ln([A_t]/[A₀])= -kt, where [A_t] is the concentration of a reactant at a certain time t and [A₀] is the initial concentration. We are given the k value for the temperature concerned in Part 1: k = 2.1 × 10⁻¹¹ s⁻¹ at 27 °C . We are told the initial concentration is 0.150 M, and the equilibrium concentration (wherein equilibrium occurs at the time t we are trying to solve for) is 1.65 × 10⁻⁷ M.

We can plug in values into our equation to find the time required for the sucrose solution to reach equilibrium without a catalyst:

ln([A_t]/[A₀])= -kt

ln ([1.65 × 10⁻⁷ M]/[0.150 M]) = -(2.1 × 10⁻¹¹ s⁻¹)t

-13.720 = -(2.1 × 10⁻¹¹ s⁻¹)t

t = 6.53x10¹¹ s x (1 hr/60 s) = 1.09x10¹⁰ hours.

Phase II Correction: I believe k sound equal 3.0 x 10^-10. I agree with all the steps you took, however, I think there might have been a calculation error in multiplying everything out.

Part 2 (done by Isabella Schrammel in Phase II):

Because this reaction is a first order reaction, we can use the following equation to determine the concentration of C₁₂H₂₂O₁₁:

ln[A] = ln[A₀] - kt

We are given [A] = 1.65 x 10^-7 , [A₀] = .150 M , and k = 2.1 x 10^-11

Thus we can solve for t

ln(1.65 x 10^-7 /.150) / -2.1 x 10^-11 = t = 6.5 x 10¹¹ s = 1.8 x 10⁸ hours

3. Assuming the reaction is irreversible, or one-way, simplifies the calculations because we don't have to consider the concentration of sucrose changing if it converted back into a reactant.

Q21.4.10

Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:

1. 6-2 He
2. 60-30 Zn
3. 235-91 Pa
4. 241-94 Np
5. ¹⁸F
6. ¹²⁹Ba
7. ²³⁷Pu

S21.4.10

Atomic number is equal to the number of protons in an atom. The number of neutrons can be found by subtracting the number of protons (Z) from the mass number (A). The belt of stability is a figure that can help us predict what kind of radioactive decay an isotope will undergo.

1. This isotope falls above the belt of stability. This isotope has 2 protons, and 4 neutrons. The belt of stability has the ideal neutron/proton ratio of 1. Correction (Phase II): The belt of stability has an ideal neutron: proton ratio of 1 for atomic numbers up to 20. After atomic number 20, the neutron: proton ration deviates from 1. This isotope has n/p = 2. Isotopes above the belt of stability are prone to beta emission.
2. This isotope falls on the belt of stability - its n/p ratio = 1. Therefore, this isotope is probably stable and will not undergo decay. Correction (Phase II): I believe Zinc-60 is actually below the belt of stability (see graph), meaning it would undergo positron emission or electron capture. As mentioned above, the n/p ratio = 1 only for elements with atomic numbers < 20.
3. This isotope is most likely to undergo alpha decay. Alpha decay is most likely in heavier elements on the periodic table - those with atomic numbers above 83. This isotope has an atomic number of 91. Gamma radiation may also occur, as gamma radiation often accompanies alpha or beta decay.
4. This isotope is most likely to undergo alpha decay. Alpha decay is most likely in heavier elements on the periodic table - those with atomic numbers above 83. This isotope has an atomic number of 94. Gamma radiation may also occur, as gamma radiation often accompanies alpha or beta decay.
5. This isotope falls on the belt of stability - its n/p ratio = 1 and the element has an atomic number less than 20 (correction, Phase II). Therefore, this isotope is probably stable and will not undergo decay.
6. This isotope falls above the belt of stability. This isotope has 56 protons (since that is the atomic number of Ba on the periodic table), and 73 neutrons. The belt of stability has the ideal neutron/proton ratio of 1. This isotope has n/p = 1.3 Isotopes above the belt of stability are prone to beta emission. Correction, Phase II: This isotope actually appears to fall on the belt of stability, which is, in the graph provided, the area within the yellow zone.
7. This isotope is most likely to undergo alpha decay. Alpha decay is most likely in heavier elements on the periodic table - those with atomic numbers above 83. This isotope has an atomic number of 94. Gamma radiation may also occur, as gamma radiation often accompanies alpha or beta decay.

Q20.2.14

Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu²⁺(aq) and nitric oxide gas.

1. What has been oxidized? What has been reduced?
2. Balance the chemical equation.

S20.2.14

1. Copper has been oxidized, since it goes from having an oxidation state of 0 in Cu(s) to 2+ in Cu²⁺ (aq). Oxidation is the loss of electrons, and copper is shown to lose electrons by its charge getting more positive. Therefore, nitrogen must be reduced, since redox reactions always involve one oxidation reaction being coupled with one reduction reaction. Phase II addition: Nitrogen transitions from an oxidation state of +5 to +2. Because the oxidation number decreases, further evidence that nitrogen is reduced is provided.

2. The unbalanced equation is as follows: Cu(s) + HNO₃ (aq) → Cu²⁺ (aq) + NO (g)

There are several steps in balancing a redox equation. We apply these steps to each half reaction. Our oxidation half reaction is Cu (s) → Cu²⁺ (aq), and our reduction half reaction is HNO₃ (aq) → NO (g). The steps are as follows:

1. Balance all elements besides hydrogen and oxygen:
   1. Cu (s) → Cu²⁺ (aq)
   2. HNO₃ (aq) → NO (g)
2. Balance oxygen using water molecules:
   1. Cu(s) → Cu²⁺ (aq)
   2. HNO₃ (aq) → NO(g) +2H₂O (l)
3. Balance hydrogen using H⁺ ions:
   1. Cu(s) → Cu²⁺ (aq)
   2. 3 H⁺ (aq) + HNO₃ (aq) → NO(g) +2H₂O (l)
4. Balance charge using electrons:
   1. Cu(s) → Cu²⁺ (aq) + 2e^-
   2. 3e^- + 3 H⁺ (aq) + HNO₃ (aq) → NO(g) +2H₂O (l)
5. Multiply each reaction by whatever integer necessary so that electrons will cancel out:
   1. 3(Cu(s) → Cu²⁺ (aq) + 2e^-) = 3Cu(s) → 3Cu²⁺ (aq) + 6e^-
   2. 2(3e^- + 3 H⁺ (aq) + HNO₃ (aq) → NO(g) +2H₂O (l)) = 6e^- + 6 H⁺(aq) + 2HNO₃ (aq) → 2NO(g) +4H₂O (l)
6. Add both half reactions and cancel out species that appear on both sides:
   1. 6 H⁺ (aq) + 3 Cu(s) + 2HNO₃ (aq) → 3Cu²⁺ (aq) + 2NO(g) +4H₂O (l)

Q20.5.9

Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?

S20.5.9

The same species may be present in the compartments of a concentration cell, but the different compartments house different concentrations. A voltage is generated as an equilibrium is being reached between the compartments as the cell with the higher concentration becomes more dilute, and the cell with the lower concentration becomes more concentrated. Voltage is generated by the flow of ions between the cells of differing concentrations. The voltage is 0 when the cell is at equilibrium.

Q24.6.5

How can CFT explain the color of a transition-metal complex?

S24.6.5

CFT stands for Crystal Field Theory, a model describing how electrons fit in d orbitals in ligands. Some ligand field d orbitals are higher energy and others are lower. The absorption of visible light promotes low energy d orbital electrons to higher energy orbitals, and this transition gives off colors. Solutions that are colorless therefore either have no d electrons to be promoted, or they will have completely full d orbitals, where there's no space to which an electron can be promoted.

Phase II addition: Different ligands produce different field splitting patterns in transition-metal complexes. High spin complexes, produced by weak field ligands, are less likely to absorb high energy light due to the likely presence of electrons in the upper e_g level. Low spin complexes, however, produced by strong field ligands, are more likely to absorb high energy light to due to the likely absence of electrons occupying e_g orbitals (and therefore, the potential for t₂g electrons to be excited).