Extra Credit 43

S17.6.2

When solid aluminum is exposed to oxygen, it oxidizes to form solid aluminum oxide. This protective layer helps prevent the oxygen from reaching the aluminum with ease, therefore preventing corrosion and further oxidation. Aluminum oxide is impermeable and and binds strongly to the parent metal. In the case of iron, it is unable to form a solid iron oxide protectant layer, so untreated iron is more prone to corrosion.

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

We know \(rate = k [A]^{a} [B]^{b}\)

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

\(\frac{-[NO]}{dt} = \frac{-[O_3]}{dt} = \frac{[NO_2]}{dt} = \frac{[O_2]}{dt}\)

We need to find the instantaneous rate.

Given in the problem:

• \(k=2.2x10^{7} L/mol/s\)
• \([NO]=3.3x10^{-6}M and [O_3]=5.9x10^{-7}\) (These numbers are the concentrations [A] and [B])
• We are also told that the "reaction is first order with respect to both NO and \(O_3\)". This tells us that a and b are both equal to 1.

\(rate = (2.2 x 10^{7} L/mol/s) [3.3 x 10^{-6} M]^{1} [5.9 x 10^{-7} M]^{1}\)

\(rate = 4.3 x 10^{-5} M/s\)

The overall rate of the reaction is \(4.3 x 10^{-5} M/s\), and the rate of disappearance is the negative equivalent of the formation rate as seen in the balanced reaction.

So, the instantaneous rate of disappearance = \(-4.3 x 10^{-5} M/s\)

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

1. a. Finding the activation energy: (Use a variation of the Arrhenius equation)

\[ln(\frac{k_1}{k_2}) = \frac{-E_a}{R} [\frac{1}{T_2} - \frac{1}{T_1}]\]

Given:

• \(k_1 = 2.1x10^{-11} s^{-1}\) \(T_1 = 27^{o}C\) = 300.15 K
• \(k_2 = 8.5x10^{-11} s^{-1}\) \(T_2 = 37^{o}C\) = 310.15 K

\[ln(\frac{2.1x10^{-11}s^{-1}}{8.5x10^{-11}s^{-1}}) = \frac{-E_a}{8.314 \frac{J}{mol \bullet k}}[\frac{1}{310.15K} - \frac{1}{300.15K}]\]

\[E_a = 108,216.49 J = 108.216 kJ\]

b. Finding the frequency factor:

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

Given:

• \(k=2.1x10^{-11}s^{-1}\)
• T = 27 °C = 300.15 K
• \(E_a = 108,216.49 J\) (found in part a)

\[2.1x10^{-11} s^{-1} = Ae^{\frac{-108,216.49 J}{8.314 \frac{J}{mol \bullet k} (300.15 K)}}\]

\[A = 1.434 x 10^{8} s^{-1}\]

c. Finding the rate constant at 47 °C:

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

Given:

• T = 47 °C = 320.15 K
• \(A = 1.434 x 10^{8} s^{-1}\) (found in part b)
• \(E_a = 108,216.49 J = 108.216 kJ\)
• R = 8.314 \(\frac{J}{mol \bullet K}\)

\[k = (1.4 x 10^{8} s^{-1})e^{\frac{-108,216.49 J}{8.314 \frac{J}{mol \bullet K} (320.15 K)}}\]

\[k = 3.1597 x 10^{-10} s^{-1}\]

2. Finding the time it takes to reach and equilibrium at 27 °C:

\[[A] = [A_o]e^{-kt}\]

Given:

• \([A_o] = 0.150 M\) (initial concentration)
• \(A = 1.65 x 10^{-7} M\)
• \(k = 2.1 x 10^{-11} s^{-1}\) (found in part 1a)

\[1.65 x 10^{-7}M = 0.150e^{-2.1 x 10^{-11}s^{-1}(t)}\]

\[t = 6.533 x 10^{11} seconds\]

3. By assuming that the reaction is irreversible, it simplifies the calculation since we do not have to consider any reactant reacting in the reverse direction, and turning 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. \(^{18}_{}F\)
6. \(^{129}_{}Ba\)
7. \(^{237}_{}Pu\)