Extra Credit 43
- Page ID
- 82905
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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.
Solution:
The standard reduction potentials show that aluminum is more spontaneous to oxidize than iron. This proves to be true, because when solid aluminum is exposed to oxygen, it oxidizes to form a solid aluminum oxide (Al2O3). While aluminum is oxidized, it is not corroded. The aluminum oxide layer protects the rest of the aluminum from becoming oxidized because there is no more aluminum exposed to oxygen. This self-passivation gives aluminum metal good corrosion resistance. Iron, however, is unable to self-passivate a solid iron oxide protectant layer, so untreated iron is prone to corrosion. Iron can be coated with aluminum, and subsequently aluminum oxide, as one method to combat corrosion.
Solution 2 (Shriya Ghosh)-
Untreated Aluminum has a very good corrosion resistance in most environments as pure Aluminum spontaneously forms a tough resistant oxide that protects it from further oxidation. This process is known as self-passivation; passivation is the spontaneous formation of a non-reactive fil (usually an oxide) that inhibits further corrosion of a metal. Metals such as Zinc, Titanium and Silicon can also form a protective oxide layer to prevent corrosion through the process of passivation. Pure Iron, on the other hand, is unable to form a protective layer to prevent corrosion and therefore, it corrodes more easily when exposed to oxygen in the environment.
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?
Solution:
To find the instantaneous rate of disappearance of NO, we use the equation \(rate=k[A]^{n}[B]^{m}\) when n and m are experimentally determined rate constants and A and B are the concentration of reactants. The problem tells us that the reaction is first order in respect to each reactant, which means n and m =1. We plug in the given concentrations for the reactants and solve for the rate.
\(rate=2.2\times10^{7}M^{-1}s^{-1}[NO\;M][O_3\;M]=2.2\times10^{7}\not M^{-1}s^{-1}(3.3x10^{-6}\not M)(5.9x10^-7M)=4.3x10^{-5}Ms^{-1}\)
The instantaneous rate of change given these conditions is 4.3x10-5Ms-1.
The rate of disappearance of NO is equal to \(\frac{-1}{1}\frac{d[NO]}{dt}=rate\). d[NO]/dt is the instantaneous rate of change of the concentration of NO in respect to time. Plugging in the rate, we find that \(\frac{d[NO]}{dt}=-4.3x10^{-5}Ms^{-1}\). This means that NO instantaneously disappears at a rate of 4.3x10-5Ms-1.
Solution 2-
The question asks us to find the instantaneous rate of disappearance of NO under given condition. The instantaneous rate of disappearance in [NO] can be described as \(\frac{-d\left [ NO \right ]}{dt}\).
We can use the \(rate = k\left [ A \right ]^{m}\left [ B \right ]^{n}\) formula to find the overall rate of reaction, then evaluate the rate of disappearance of NO.
The problem gives us the following conditions:
- First order reaction in terms of \(\left [ NO \right ]\) and \(\left [ O_{3} \right ]\)
- Rate constant (k) = \(2.2 x 10^{7} M^{-1}s^{-1}\)
- \(\left [ NO \right ]\) = \(3.3 x 10^{-6} M\)
- \(\left [ O_{3} \right ]\) = \(5.9 x 10^{-7} M\)
Step 1: plug in given values into the rate formula, \(rate = k\left [ NO \right ]^{m}\left [ O_{3} \right ]^{n}\), where k = \(2.2 x 10^{7} M^{-1}s^{-1}\), n = 1 and m = 1.
\(rate=2.2\times10^{7}M^{-1}s^{-1}[NO][O_{3}]\)
=\(2.2\times10^{7}\not M^{-1}s^{-1}(3.3x10^{-6}\not M)(5.9x10^-7M)\)
Rate =\(4.3x10^{-5}Ms^{-1}\)
Step 2: solve for rate of disappearance of NO in terms of the overall reaction rate and with respect to time (t)
The rate of disappearance of NO = \(\frac{-d\left [ NO \right ]}{dt}\)
[NO] decreases/disappears at a rate of \(4.3 x 10 ^{-5}Ms^{-1}\)
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)?
Solution:
1. Solve for activation energy:
\(ln(\frac{k_1}{k_2})=\frac{-E_A}{R}(\frac{1}{T_2}-\frac{1}{T_1})\) We use this equation to solve for EA, the activation energy.
\(ln(\frac{8.5x10^{-11}\not s^{-1}}{2.1x10^{-11}\not s^{-1}})=\frac{-E_A}{8.3145\frac{J}{mol\not K}}(\frac{1}{300\not K}-\frac{1}{310\not K})\)
\(1.398=E_A\frac{J}{mol}(1.293x10^{-5})\)
\(E_A=108100\frac{J}{mol}=108.1\frac{kJ}{mol}\)
Solve for frequency factor:
\(k=Ae^{\frac{-E_A}{RT}}\) We use this equation to solve for A, the frequency factor.
\(2.11x10^{-11}s^{-1}=Ae^{\frac{-108100\frac{\not J}{\not mol}}{8.3145\frac{\not J}{\not mol\not K}300\not K}}\)
\(2.11x10^{-11}s^{-1}=A(1.509x10^{-19})\)
\(A=1.4x10^{8}s^{-1}\)
Solve for the rate constant at 47°C/320K:
\(k=Ae^{\frac{-E_A}{RT}}\) We use this equation to solve for k, the rate constant.
\(k=1.4x10^{8}s^{-1}(e^{\frac{-108100\frac{\not J}{\not mol}}{8.3145\frac{\not J}{\not mol \not K}320\not K}})\)
\(k=3.17x10^{-10}s^{-1}\)
2. Solve for time to reach equilibrium:
\([A]=[A_0]e^{-kt}\) We use this equation to solve for t, the time.
\(1.65x10^{-7}\not M=(0.15\not M)e^{-2.11x10^{-11}s^{-1}(t)}\)
\(1.1x10^{-6}=e^{-2.11x10^{-11}s^{-1}(t)}\)
\(ln(1.1x10^{-6})=\not ln(\not e^{-2.11x10^{-11}s^{-1}(t)})\)
\(-13.72=-2.11x10^{-11}s^{-1}(t)\)
\(t=6.5x10^{11}\not s(\frac{1 \not {min}}{60 \not s})(\frac{1\not {hour}}{60\not {min}})(\frac{1day}{24\not {hour}}\))
\(t=7.53x10^{6}days\)
3. Assuming that the reaction is irreversible means that we do not have to deal with glucose or fructose back-reacting into sucrose. We only care about the concentration of sucrose. Also, having this reaction be reversible would change the reaction order and the equation used to answer the previous question. With a different equation, we would find a different value of t.
Solution 2-
1. To solve for the activation energy (Ea), we will the formula: \(ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\)
with the following givens:
- \(k_{1} = 2.1x10^{-11} s^{-1}\)
- \(k_{2} = 8.5x10^{-11} s^{-1}\)
- \(R = 0.0083145 \frac{kJ}{mol*K}\)
- \(T_{1} = 300 K\)
- \(T_{2} = 310 K\)
Step 1: Solve for Ea
\(ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\)
\(ln(\frac{8.5x10^{-11} s^{-1}}{2.1x10^{-11} s^{-1}})=\frac{E_a}{0.0083145\frac{kJ}{mol K}}(\frac{1}{300K}-\frac{1}{310K})\)
\(1.398=E_a\frac{kJ}{mol}(1.293x10^{-5})\)
\(E_a = 108.1\frac{kJ}{mol}\)
Step 2: Solve for frequency factor A
\(k=Ae^{\frac{-E_a}{RT}}\)
\(2.11x10^{-11}s^{-1}=Ae^{\frac{-108.1\frac{kJ}{mol}}{0.0083145\frac{kJ}{mol*K}300 K}}\)
\(A = \frac{2.1x10^{-11}s^{-1}}{e^{\frac{-108.1 \frac{kJ}{mol}}{0.0083145*300K}}}\)
\(A=1.4x10^{8}s^{-1}\)
Step 3: Solve for rate constant, k, at 47°C (320K)
\(k=Ae^{\frac{-E_a}{RT}}\)
\(k=1.4x10^{8}s^{-1}(e^{\frac{-108.1\frac{kJ}{mol}}{0.0083145\frac{kJ}{mol*K}320 K}})\)
\(k=3.17x10^{-10}s^{-1}\)
2. Solve for time to reach equilibrium:
\([A]=[A_0]e^{-kt}\)
\(1.65x10^{-7} M=(0.15 M)e^{-2.1x10^{-11}s^{-1}(t)}\)
\(1.1x10^{-6}=e^{-2.1x10^{-11}s^{-1}(t)}\)
\(ln(1.1x10^{-6})= ln(e^{-2.1x10^{-11}s^{-1}(t)})\)
\(-13.72=-2.1x10^{-11}s^{-1}(t)\)
\(t=6.5x10^{11}\not s(\frac{1 \not {min}}{60 \not s})(\frac{1\not {hour}}{60\not {min}})(\frac{1day}{24\not {hour}}\))
\(t=7.53x10^{6}days\)
3. By assuming that the reaction is irreversible, we do not have to account for glucose and fructose reacting together to produce sucrose. In addition to this, assuming that the reaction is irreversible, we base our calculations only on the forward reaction order; the reverse reaction would have a different reaction order and the formula used in part (b) would be different, therefore the result would be a different value of t.
Q21.4.10
Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:
- \(^{6}_2He\)
N=4, P=2. Beta emission to \(^{6}_3Li\)
2. \(^{60}_{30}Zn\)
N=30, P=30. Positron emission to \(^{60}_{29}Cu\), positron emission to \(^{60}_{28}Ni\).
3. \(^{235}_{91}Pa\)
N=144, P=91. Beta emission to \(^{235}_{92}U\), alpha decay to \(^{231}_{90}Th\), beta emission to \(^{231}_{91}Pa\), alpha decay to \(^{227}_{89}Ac\), beta emission to \(^{227}_{90}Th\), alpha decay to\(^{223}_{88}Ra\), alpha decay to \(^{219}_{86}Rn\), alpha decay to \(^{215}_{84}Po\), alpha decay to \(^{211}_{82}Pb\), beta emission to \(^{211}_{83}Bi\), alpha decay to \(^{207}_{81}Tl\), beta emission to \(^{207}_{82}Pb\).
4. \(^{241}_{94}Np\)
N=147, P=94. Beta emission to \(^{241}_{95}Am\), alpha decay to \(^{237}_{93}Np\), alpha decay to \(^{233}_{91}Pa\), beta emission to \(^{233}_{92}U\), alpha decay to \(^{229}_{90}Th\), alpha decay to \(^{225}_{88}Ra\), beta emission \(^{225}_{89}Ac\), alpha decay to \(^{221}_{87}Fr\), alpha emission to \(^{217}_{85}At\), alpha decay to \(^{213}_{83}Bi\), beta emission to \(^{213}_{84}Po\), alpha decay to \(^{209}_{82}Pb\), beta decay to \(^{209}_{83}Bi\), alpha decay to \(^{205}_{81}Tl\).
5. 18F
N=9. P=9. Positron emission to \(^{18}_8O\).
6. 129Ba
N=73. P=56. Positron emission to \(^{129}_{55}Cs\), positron emission to \(^{129}_{54}Xe\).
7. 237Pu
N=143. P=94. Positron emission to \(^{237}_{93}Np\), alpha decay to \(^{233}_{91}Pa\), beta emission to \(^{233}_{92}U\), alpha decay to \(^{229}_{90}Th\), alpha decay to \(^{223}_{88}Ra\), beta emission to \(^{225}_{89}Ac\), alpha decay \(^{221}_{87}Fr\), alpha decay to \(^{217}_{85}At\), alpha decay to \(^{213}_{83}Bi\), beta emission to \(^{213}_{84}Po\), alpha decay to \(^{209}_{82}Pb\), beta decay to \(^{209}_{83}Bi\), alpha decay to \(^{205}_{81}Tl\).
The spontaneous mode of decay is predicted by looking at a graph of number of neutrons plotted against number of protons. Within this graph there is a band of stability, in which stable nuclei are found. Surrounding the band in the graph on this page (https://upload.wikimedia.org/wikiped...sotopes_en.svg) are color-coded plots that tell the spontaneous form of decay by the neutron-to-proton ratio. By following the decay and writing the reaction for each step of decay, one is able to finally reach a stable nucleus.
Solution 2-
To determine which type of decay has taken place, we can formulate general decay equations that can help us to figure out changes in mass number (A) and proton number (Z).
Alpha Decay
\(_{Z}^{A}\textrm{X}\rightarrow _{2}^{4}\textrm{He}+_{Z-2}^{A-4}\textrm{Y}\)
Beta Decay
\(_{Z}^{A}\textrm{X}\rightarrow _{-1}^{0}\textrm{e}+_{Z+1}^{A}\textrm{Y} + _{0}^{0}\textrm{v}\)
Beta (positron) Decay
\(_{Z}^{A}\textrm{X}\rightarrow _{+1}^{0}\textrm{e}+_{Z-1}^{A}\textrm{Y} + _{0}^{0}\textrm{v}\)
Gamma Ray
\(_{Z}^{A}\textrm{X}\rightarrow _{0}^{0}\textrm{gamma}+_{Z}^{A}\textrm{Y}\)
1. \(^{6}_2He\)
N=4, P=2. Beta emission to \(^{6}_3Li\)
2. \(^{60}_{30}Zn\)
N=30, P=30. Positron emission to \(^{60}_{29}Cu\), positron emission to \(^{60}_{28}Ni\).
3. \(^{235}_{91}Pa\)
N=144, P=91. Beta emission to \(^{235}_{92}U\), alpha decay to \(^{231}_{90}Th\), beta emission to \(^{231}_{91}Pa\), alpha decay to \(^{227}_{89}Ac\), beta emission to \(^{227}_{90}Th\), alpha decay to\(^{223}_{88}Ra\), alpha decay to \(^{219}_{86}Rn\), alpha decay to \(^{215}_{84}Po\), alpha decay to \(^{211}_{82}Pb\), beta emission to \(^{211}_{83}Bi\), alpha decay to \(^{207}_{81}Tl\), beta emission to \(^{207}_{82}Pb\).
4. \(^{241}_{94}Np\)
N=147, P=94. Beta emission to \(^{241}_{95}Am\), alpha decay to \(^{237}_{93}Np\), alpha decay to \(^{233}_{91}Pa\), beta emission to \(^{233}_{92}U\), alpha decay to \(^{229}_{90}Th\), alpha decay to \(^{225}_{88}Ra\), beta emission \(^{225}_{89}Ac\), alpha decay to \(^{221}_{87}Fr\), alpha emission to \(^{217}_{85}At\), alpha decay to \(^{213}_{83}Bi\), beta emission to \(^{213}_{84}Po\), alpha decay to \(^{209}_{82}Pb\), beta decay to \(^{209}_{83}Bi\), alpha decay to \(^{205}_{81}Tl\).
5. 18F
N=9. P=9. Positron emission to \(^{18}_8O\).
6. 129Ba
N=73. P=56. Positron emission to \(^{129}_{55}Cs\), positron emission to \(^{129}_{54}Xe\).
7. 237Pu
N=143. P=94. Positron emission to \(^{237}_{93}Np\), alpha decay to \(^{233}_{91}Pa\), beta emission to \(^{233}_{92}U\), alpha decay to \(^{229}_{90}Th\), alpha decay to \(^{223}_{88}Ra\), beta emission to \(^{225}_{89}Ac\), alpha decay \(^{221}_{87}Fr\), alpha decay to \(^{217}_{85}At\), alpha decay to \(^{213}_{83}Bi\), beta emission to \(^{213}_{84}Po\), alpha decay to \(^{209}_{82}Pb\), beta decay to \(^{209}_{83}Bi\), alpha decay to \(^{205}_{81}Tl\).
Q20.2.14
Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.
What has been oxidized? What has been reduced?
Balance the chemical equation.
Solution:
For the copper half reaction:
Cu(s) → Cu2+(aq)
To balance the charge on both sides we add 2 electrons to the products.
Cu(s) → Cu2+(aq) + 2e-
The reactant is losing electrons so this reaction is oxidation. Copper has been oxidized.
For the nitric acid half reaction:
HNO3(aq) → NO(g)
To balance the oxygens, we add two oxygens via water molecules to the products.
HNO3(aq) → NO(g) + 2H2O(l)
To balance the hydrogens we add three hydrogen ions to the reactants.
HNO3(aq) + 3H+(aq) → NO(g) + 2H2O(l)
To balance the charge, we add three electrons to the reactants side.
HNO3(aq) + 3H+(aq) + 3e- → NO(g) + 2H2O(l)
The reactant is gaining gaining electrons so this reaction is reduction. Nitrogen has been reduced.
To balance the half reactions together, we multiply the copper half reaction by three and the nitric acid half reaction by two.
3Cu(s) + 6e-+ 6H+(aq) + 2HNO3(aq) → 3Cu2+(aq) + 6e- + 2NO(g) + 4H2O(l)
3Cu(s) + 6H+(aq) + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)
Solution 2-
| Rules for Balancing Redox Reactions |
| 1. Balance atoms that are not Oxygen or Hydrogen |
| 2. Balance number of Oxygen atoms on both sides of equation |
| 3. Balance number of Hydrogen atoms on both sides of equation by adding H+ ions |
| 4. Balance charge on both sides of equation by adding electrons |
| 5. Balance equation in terms of acidic or basic solution |
Step 1: find the two half reactions involved in this redox reaction then determine which element was oxidized and which was reduced
Copper half-reaction:
\(Cu_{s} \rightarrow Cu^{+2}_{aq} + 2e^{-}\)
Nitric Acid half-reaction:
HNO3 (aq) + 3H+(aq) + 3e --> NO(g) + 2H2O(l)
From the half reactions, we can see that Cu(s) lost electrons, so it was oxidized; HNO3 (aq) gained electrons, so it was reduced.
Step 2: After balancing number of electrons on both sides of equations, we get
3Cu(s) + 6e- + 6H+(aq) + 2HNO3 (aq) --> 3Cu+2(aq) + 6e- + 2NO(g) + 4H2O(l)
Step 3: Overall reaction after canceling number of electrons on both sides
3Cu(s) + 6H+(aq) + 2HNO3 (aq) --> 3Cu+2(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?
Solution:
The voltage in a concentration cell is created by the concentration difference in the cell. In the Nerst equation, \(E_{cell}=E^{0}_{cell}-\frac{RT}{nF}lnQ\), if the lnQ is negative, then Ecell is positive and the reaction progresses until the concentrations of ions are at equilibrium. The cell will spontaneously go to equilibrium as long as Ecell is positive and the lnQ is negative. The E0cell is equal to 0 volts because E0cell is calculated by E0cathode-E0anode. Because it is the same half-reaction at the cathode and the anode, E0cell is equal to a value minus itself, which is 0. While there is a voltage value before the concentration cell reaches equilibrium, after the equilibrium is reached the voltage will remain at 0 volts. When the voltage of a battery reaches 0, the battery is dead because the electrons are equally flowing between the half-cells. This does not create a potential and so it cannot be used. This same principle applies to biology, as the cell is always avoiding equilibrium so the necessary reactions will occur in the intended direction.
Solution 2-
A concentration cell is an electrolytic cell that is made up of two half-cells with the same electrodes which differ in their concentrations. Concentration cells work by diluting the higher concentration compartment solution and concentrating the dilute compartment solution. This process creates a voltage as the concentration cells tries to establish equilibrium with transfer of electrons from the lower concentration cell to the higher concentration cell.
At equilibrium, the voltage drops to zero volts and there is no current flow (the battery dies). If changes are made to the cell that push it away from equilibrium, the cell potential will go up; if we make changes that push it toward equilibrium, cell potential will go down. The standard electrode potential (E°cell) of a concentration cell, when its electrodes are the same, is zero volts. The cell potential value is dependent on concentration according to the following formula:
\(E_{cell}=E^{0}_{cell}-\frac{RT}{nF}lnQ\)
When Q = 1, the concentration of products and reactants are equal, so the Ecell and E°cell are equal to each other.
Q24.6.5
How can CFT explain the color of a transition-metal complex?
Solution:
Crystal field splitting explains the difference in energy when transition metal complexes have split energy d-orbitals. The magnitude of crystal field splitting is proportional to the energy necessary to excite a low-energy electron up to a higher orbital. This energy comes in the form of visible light and the human eye sees the corresponding complementary color. For example, a strong field ligand would create a large ΔO that takes high-energy light to excite up an electron. This high energy light may be violet in color, which is absorbed by the complex. Yellow light, the low energy wavelength complementary to violet, is not absorbed so we see that color. On the other hand, a weak field ligand will make a small ΔO that needs very little energy to excite up an electron. This low energy light may be orange, which is absorbed by the compound and excites electrons. Blue light, the complement to orange light, is not absorbed by the compound, so it is reflected into the human eye. We see this compound as blue. Clear compounds, however, have no d-electrons to excite or no space to fit electrons. They do not need to absorb light, so we see them as clear or colorless.
Solution 2-
The electromagnetic spectrum is made up of photons of different wavelengths. The amount of energy needed to split d-orbitals is described by the crystal field theory. If ΔO is large, then much energy is required to promote electrons into the higher energy orbitals, the electrons will instead prefer to pair in the lower energy orbitals which causes a low spin complex. Vice versa, if ΔO is small, then it takes little energy to occupy higher orbitals and the electrons will remain unpaired which causes a high spin complex.
A photon equal to the energy difference ΔO can be absorbed, and electrons can get promoted to the higher energy level. As certain wavelengths are absorbed, color mixing occurs which causes coordination complexes to be of different colors. In general, a larger ΔO indicates that higher energy photons are absorbed and the complex will be of a color on the right end of the electromagnetic spectrum; this relationship can be explained by the formula: ∆=hc/λ.
If a complex absorbs a color of a particular wavelength, then it will reflect the color opposite of the absorbed color on the color wheel.