Extra Credit 43

Q17.6.2

Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel.

A17.6.2

In order for a metal to be used as a sacrificial anode in cathodic protection, the metal should be more readily oxidized that the parent metal it is protecting. To determine this, we look at a standard reduction potential table and find the values for each metal.

Ag = 0.80 V

Au = 1.50 V

Mg = -2.37 V

Ni = -0.25 V

Zn = -0.76 V

As the standard reduction potential becomes more negative, the strength as a reducing agent increases. Because the standard reduction potential of Fe is -0.447 V, we would need to use a metal or metals whose reduction potential is greater than that value, in the negative direction - those metals would lie lower on the standard reduction potential table than Fe.

From using the table and the values we found, we can see that the two metals that satisfy the need are Mg and Zn, therefore those two metals could be used a sacrificial anodes in the cathodic protection of an underground steel storage tank.

Q17.6.2

Aluminum (E∘Al3+/Al=−2.07V)] is more easily oxidized than iron (E∘Fe3+/Fe=−0.477V), 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.

A17.6.2

Untreated aluminum has a better corrosion resistance than untreated iron because aluminum will spontaneously form a thin oxide layer that protects itself from further oxidation. Aluminum, which is more easily oxidized, will oxidize more quickly than iron, but after forming the initial first layer of aluminum oxide, further corrosion is blocked by that layer. Aluminum oxide is impermeable and varies from oxides of other metals in that in binds very strongly to the parent metal. When iron oxidizes, it does not form a protective layer, thus the metal will continue to corrode.

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?

A12.3.6

NO + O₃ → NO₂ + O₂

rate = R = k[NO][O₃]

k is the rate constant for the reaction, and is given to be 2.20x10⁷ M/s, so the rate equation would be,

R = (2.20x10⁷ M/s)[NO][O₃]

The rate can also be defined in terms of the individual species of the reaction.

-[NO]/dt = -[O₃]/dt= [NO2]/dt= [O2]/dt

To find the rate of disappearance of NO, you would use the first equation. You could then set this equation equal to the overall rate reaction to get the following,

-[NO]/dt = (2.20x10⁷ M/s)[NO][O₃]

To find the instantaneous rate of disappearance of NO, you would plug in the given values for the concentrations of NO and O3.

-[NO]/dt= (2.20x10⁷ M/s)(3.3x10^-6 M)(5.9x10^-7 M)

= - 4.284x10^-5 M/s

We have found a negative value because the NO is being used up, so its concentration is decreasing over time as the reaction proceeds. We would adjust this to find that the rate is equal to 4.284x10^-5 M/s.

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)?

A12.5.1

To find the activation energy:

\[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}}\]

k₁ = 2.1 × 10⁻¹¹ s⁻¹ T₁ = 27 °C = 300.15 K

k₂ = 8.5 × 10⁻¹¹ s⁻¹ T₂ = 37 °C = 310.15 K

\[E_a = \dfrac{\left( 8.314 J/mol*K\right) \ln \dfrac{8.5*10^{-11}s^{-1}}{2.1*10^{-11}s^{-1}}}{\dfrac{1}{300K}-\dfrac{1}{310K}}\]

E_a = 108,216.49 J = 108.216 kJ

To find the frequency factor:

\[k=Ae^{\frac{-E_a}{RT}}\]

\[2.1*10^{-11}s^{-1}=Ae^{\dfrac{-108216.48J}{\left(8.314J/mol*K\right)\left(300K\right)}}\]

2.1 × 10⁻¹¹ s⁻¹ = Ae^{-108,216.49 J/[(8.314 J/mol K)(300.15 K)]}

A = 1.434 x 10⁸ s^-1

To find the rate constant at 47°C:

T = 47°C = 320.15 K

\[k=Ae^{\frac{-E_a}{RT}}\]

\[k=\left(1.434*10^8s^{-1}\right)e^{\dfrac{-108216.48J}{\left(8.314J/mol*K\right)\left(320K\right)}}\]

_k = Ae^{-108,216.49 J/[(8.314 J/mol K)(320.15 K)]}

k = 3.1675 x 10^-10

To find time it takes to reach equilibrium at 27°C:

[A] = [A_o]e^-kt

1.65 × 10⁻⁷ M = (0.150M)e^{(2.1 × 10^-11)t}

1.1 × 10^{−6 =} e^{(2.1 × 10^-11)t}

t = 6.533 x 10¹¹ seconds --> 7.56 x 10⁶ days

By assuming that the reaction is irreversible in part (b), we are able to simplify our calculation because we do not have to consider any reactant that turns into product and then back into reactant, which in this case would be sucrose turning into glucose and fructose, and either or turning back into sucrose.

Q21.4.10

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

1. \(^{6}_2He\)
2. \(^{60}_{30}Zn\)
3. \(^{235}_{91}Pa\)
4. \(^{241}_{94}Np\)
5. \(^{18}_{}F\)
6. \(^{129}_{}Ba\)
7. \(^{237}_{}Pu\)