Extra Credit 43
- Page ID
- 82854
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q17.6.2
Aluminum (EoAl3+/Al = -2.07 V) is more easily oxidized than iron (EoFe3+/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
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.)
- 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).
- 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.
- 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
- \(^{6}_{2}He\)
- \(^{60}_{30}Zn\)
- \(^{235}_{91}Pa\)
- \(^{241}_{94}Np\)
- \(^{18}_{}F\)
- \(^{129}_{}Ba\)
- \(^{237}_{}Pu\)
A21.4.10
To determine the modes of spontaneous radioactive decay that each unstable isotope might proceed, we need to look at the number of neutrons and protons in each isotope, and then refer to the belt of stability (pictured below) to determine which types of decay will move the isotope to be more stable and lie on the belt.
1. \(^{6}_{2}He\) ; \(\beta\) emission
4 neutrons, 2 protons places this point slightly above the belt of stability. Looking at the graph, \(\beta\) emission is preferred.
2. \(^{60}_{30}Zn\) ; positron (\(\beta^+\)) emission
30 neutrons, 30 protons places this point slightly below the belt of stability towards the bottom of the graph so positron (\(\beta^+\)) emission is preferred.
3. \(^{235}_{91}Pa\) ; \(\beta\) emission
144 neutrons, 91 protons places this point slightly above the belt of stability. One thing one should also consider is that Z>83 which would indicate decay via alpha emission, however, due to the large number of protons in relation to the number of neutrons, this point lands slightly to the left of the belt, making it prefer \(\beta\) emission.
4. \(^{241}_{94}Np\) ; \(\beta\) emission
147 neutrons, 94 protons places this point slightly above the belt of stability. But once again, with the same reasoning as the last example, we must consider that Z>83 which would indicate decay via alpha emission, however, due to the large number of protons in relation to the number of neutrons, this point lands slightly to the left of the belt, making it prefer \(\beta\) emission.
5. \(^{18}_{}F\) ; positron (\(\beta^+\)) emission
9 neutrons, 9 protons places this point slightly below the belt of stability towards the bottom of the graph so positron (\(\beta^+\)) emission is preferred.
6. \(^{129}_{}Ba\) ; positron (\(\beta^+\)) emission
73 neutrons, 56 protons places this point really close to the belt, which makes it hard to see. But approximations put the ratio just below the belt, making it prefer positron (\(\beta^+\)) emission.
7. \(^{237}_{}Pu\) ; positron (\(\beta^+\)) emission
143 neutrons, 94 protons places the point right below the belt of stability, near the red are in the graph shown above which should indicate \(\alpha\) decay but because its low ratio of neutrons to protons, positron (\(\beta^+\)) emission is actually preferred.