Extra Credit 42.
- Page ID
- 82802
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Q17.6.2
Aluminum (\(E^∘_{Al^{3+} /Al}=−2.07V\)) is more easily oxidized than iron (\(E^∘_{Fe^{3+} /Fe}=−.477V\)) , 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 larger negative on the redox tower is more easily oxidized, if both are exposed to water aluminum is more likely to release its electrons than iron.
When untreated iron and aluminum are placed outside, both of them react with the oxygen present in the air and undergo oxidation forming an oxide layer. There is also a corrosion affect that covers aluminum and protects the lower layer, but oxygen is still able to penetrate the iron corrosion layer. In aluminum case, however, aluminum oxide bonds tightly to the remaining aluminum, but in iron oxide it flakes off exposing remaining iron to react further. For this reason as time passes, most of the iron gets corroded while aluminum remains protected.
Q12.3.5
How will each of the following affect the rate of the reaction: \(CO(g)+ NO_2 (g)⟶ CO_2 (g)+NO(g)\) if the rate law for the reaction is \(rate=k[NO_2][CO]\)?
Solution:
a. Increasing the pressure of \(NO_2\) from 0.1 atm to 0.3 atm
Increasing the pressure in a reaction that involves gases will increases the rate of reaction. Since in this case the pressure is increased from .1 atm to .3 atm, the rate of reaction will be multiplied by 3 because NO2 is a first order reaction being directly proportional to the rate.
b. Increasing the concentration of CO from 0.02 M to 0.06 M.
Same ideas as above for this increase in concentration. Because the rate of reaction is directly proportional to the to the concentrations and is to the power of one, increasing concentration of CO from 0.02 M to 0.06 M will result in a triple of the rate.
S12.5.14
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
| T (K) | k (s−1) |
|---|---|
| 293 | 0.054 |
| 298 |
0.100
|
Solution:
- \(Co (III)+ e^− ⟶ Co (II)\)
- Below is the main equation we will be using:
- \(ln \frac{k_1} {k_2}= \frac{E_a} {R}(\frac{1} {T_2}- \frac{1} {T_1} )\)
Where \(k_1\)is the rate of reaction at \(T_1\) temperature, \(k_2\) is the rate of reaction at \(T_1\) temperature, Ea is the activation energy and R is the gas constant
The values for this reaction at different temperature is as follows:
\(k_1= .054 s^{−1}\) , \(T_1=293K\)
\(k_2= .100 s^{−1}\) , \(T_2=298K\)
\(R= 8.314 J mol^{-1} K^{−1}\)
substituting into the equation:
\(ln \frac{.054s^{-1}} {.100 s^{-1}}= \frac{E_a} {8.314 J mol^{-1}K^{-1}}(\frac{1} {298K}- \frac{1} {293K} )\)
\(-.616= (E_a) (-6.88 x 10^{-6} J mol^{−1})\)
\[E_a= 89.5 kJ mol^{−1}\]
Therefore, the activation energy of the reaction is \(89.5 kJ mol^{−1}\)
Q21.4.9
The following nuclei do not lie in the band of stability. How would they be expected to decay?
a.\(_{15}^{28}P\)
b.\(_{92}^{235} U\)
c. \(_{20}^{37} Ca\)
d. \(_{3}^{9} Li\)
e. \(_{96}^{245} Cm\)
The band of stability shows were an atom is stable because with an atomic number over 20, the atom needs more nuetrons than protons.
The isotope, \(_{15}^{28}P\) has low N/Z ratio, so in order to obtain stability, it undergoes Positron emission.
The isotope, \(_{92}^{235} U\) has greater N/Z ratio. These type of elements undergo either alpha or beta decay in order to attain stability.
The isotope, \(_{37}^{20} Ca\) has low N/Z ratio, so in order to attain stability, it undergoes Positron emission.
The isotope, \(_{3}^{9} Li\) has greater N/Z ratio, therefore in order to attain stability, it undergoes beta decay.
The isotope, \(_{96}^{245} Cm\) has greater N/Z ratio, in order to attain stability, it undergoes alpha decay.
Q20.2.13
Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.
Solutions:
This equation will be composed of what we have, zinc and water. The cell potential of zinc is above that of water, so zinc will give its electrons to water. This redox reaction will oxidize zinc and reduce a water molecule. The water molecule will split into H2 and 2OH creating the basic pH.
This is the full equation:
\(Zn(s) + 2H_2O(l) \rightarrow Zn^{2+}(aq) + H_2 (g) + 2OH^- (aq)\)
Q20.5.8
Blood analyzers, which measure pH, \(PCO_2\), and \(PO_2\), are frequently used in clinical emergencies. For example, blood \(PCO_2\), is measured with a pH electrode covered with a plastic membrane that is permeable to CO2. Based on your knowledge of how electrodes function, explain how such an electrode might work.
Hint: \(CO_2(g)+ H_2O (l) \rightarrow HCO^{3-} (aq) + H^+ (aq)\)
Solution:
- The pH in the body is directly proportional to \(PCO_2\) in the blood. Therefore, the amount of how much \(CO_2\) gets reduced to \(HCO^{3-}\)
is what determines the pH. Because in the question it is given as permeable to \(CO_2\), only \(CO_2\) will get reduced - The electrode accounts for the pressur of CO2 which gives a proportional amount of its concentration. Furthermore, the pH is determined by H+. These two amounts give the rate of the reaction and if the blood is too acidic or basic
Q24.6.4
For an octahedral complex of a metal ion with a \(d^6\) configuration, what factors favor a high-spin configuration versus a low-spin configuration?
Solution:
- In electron configuration high splitting energy will have lower spin and low splitting energy will have high spin. For \(d^6\) if it has low splitting energy, there will be total of four unpaired electrons resulting in many/ more spins. On the other hand, however, if it is spread in a high splitting energy it will be diamagnetic, meaning all the electrons will be pared up and there will not be any spins. In addition, splitting energy does depend on the ligands that are bounded.
- High spin: weak field ligands, paramagnetism, and small delta o
- Low spin: strong field ligands, diamagnetism and large delta o
Q14.7.10
At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (ES), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations:
enzyme (E) + substrate (S)⇌enzyme-substrate complex (ES)⇌enzyme (E) + product (P)enzyme (E) + substrate (S)⇌enzyme-substrate complex (ES)⇌enzyme (E) + product (P)enzyme (E) + substrate (S)⇌enzyme-substrate complex (ES)⇌enzyme (E) + product (P)enzyme (E) + substrate (S)⇌enzyme-substrate complex (ES)⇌enzyme (E) + product (P)enzyme (E) + substrate (S)⇌enzyme-substrate complex (ES)⇌enzyme (E) + product (P)
This can also be shown as follows: \[{E+S}\mathrel{\mathop{\rightleftarrows}^{\mathrm{k_1}}_{\mathrm{k_{-1}}}}ES\mathrel{\mathop{\rightleftarrows}^{\mathrm{k_2}}_{\mathrm{k_{-2}}}}{E+P}\]
Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?
Determining the rate constants individually:
- \(k_1[E][S]=[ES]\)
- The above equation being the first forward reaction
- \(k_{-1} [ES]=[E][S]\)
- The above eqaution being the reverse equation
- \(k_2[ES]=[E][P]\)
- The above equation would be the second reaction
- \(k_{-2} [E][P]=[ES]\)
- The above equation would be the reverse second reaction
k1 and k-2 are appearance because those reactions form/ create ES; whereas k-1 and k2 are disappearance due to their reactions getting rid of ES.
If a reaction is showing the appearance of ES it will have a positive sign in the equation but if the rate law is disappearance it will be shown with a negative sign.
Therefore the complete, combined equation of the reaction will be :
\(\frac {Δ[ES]}{Δt}=−(k_2+k_{-1})[ES]+k_1[E][S]+k_{-2}[E][P]\)
This would be the increase of [ES] because each reaction is going towards the concentration of ES.
Δ[ES]this equation will be approximately equal to zero and show a zero order because the rate of appearance of ES equals rate of disappearance of ES. The other intermediates being constants also contribute to the reason why its about proportional to zero. The complete equation will then look like,
\(\frac {Δ[ES]}{Δt}=−(k_2+k_{-1})[ES]+k_1[E][S]+k_{-2}[E][P]\approx 0\)