Extra Credit 40
- Page ID
- 83012
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q17.5.8
Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.
Answer:
The Nernst Equation, which is
\[E= E^{\circ}-\frac{RT}{nF}lnQ\displaystyle \]
T is directly proportional to E. This means that as T increases, E will also increase and if T decreases, then E will also decrease. Since E is the cell voltage measured from the battery to the device, a decreased E value means less voltage from the battery to the device. Therefore, if temperatures are low and E is directly proportional to temperature then E is also low and this means worse performance.
Q12.3.3
Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions:
- What is the order of the reaction with respect to that reactant?
- Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant?
Answer:
- The order of the reaction in respect to this particular reactant is 2. Since tripling the concentration of the reactant increases the rate of the reaction by nine, which is three squared, then you know that the rate law for the reaction in respect to this reactant is
\[Rate\propto \left [ A \right ]^{2\displaystyle }\]
The order of the reaction is the sum of the exponents. Since there is only one exponent, then the order of the reaction in respect to this reactant is 2.
2. The order of the reaction in respect to that reactant is 1, You know this since changing the reactant concentration changes the rate of the reaction by the same factor, so the rate of the reaction is directly proportional to the change in concentration of the reactant. Therefore the rate of the reaction is
\[Rate\propto \left [ A \right ]\displaystyle\]
Since the order of the reaction is equal to the sum of all the exponents, and there is only one exponent, one, then the order of the reaction is one.
Q12.5.12
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
- the change in free energy per second
- the change in temperature per second
- the number of collisions per second
- the number of product molecules
Answer:
In terms of collision theory, rate of the chemical reaction is proportional to the change in temperature per second as well as the number of collisions per second. The collision theory states that in order for two species to react, three conditions must be met. First, the reactants must come into physical contact with each other. Second, both reactants must come in contact with each other in the correct orientation. Finally, they must collide with enough energy to break the chemical bonds in the reactants. The first and final criteria are temperature dependent in that a change in temperature would change how often the two reactants collide since increased temperature means increased kinetic energy. A temperature change would also change the amount of energy each reactant has since increased temperature means increased kinetic energy.
The answer is not the change in free energy per second since by the collision theory, there is no relation between the free energy of the reaction and the rate of the reaction. The answer is also not the number of product molecules since the reaction rate is not influenced by the amount of product molecules present.
Q21.4.7
Which of the following nuclei is most likely to decay by positron emission? Explain your choice.
- chromium-53
- manganese-51
- iron-59
Answer:
Manganese-51 is the most likely to decay by positron emission. The element with the neutron to proton ratio closest to one is the one most likely to decay by positron emission since it is most likely to be neutron poor. The number of neutrons and protons can be calculated using the mass number and the atomic number (which represents the number of protons the element has.) For chromium-53. The 53 represents that there are 53 protons and neutrons in the chromium isotope. By looking at the the periodic table, you can see that chromium has 24 protons. By subtracting the atomic number (number of protons) from the mass number (number of protons and neutrons,) you find the number of neutrons. After finding this number, for chromium, 29 neutrons, you divide the number of neutrons by the number of protons.
\[{29\over 24} \displaystyle = 1.21\]
You do this calculation again for the other two elements and get 25 protons and 26 neutrons for Manganese.
\[{26\over 25} \displaystyle = 1.04\]
You do this again for Iron and get 26 protons and 33 neutrons.
\[{33\over 26} \displaystyle = 1.26\]
After comparing all of the values, you see that 1.04 is closest to one out of all the choices, so Manganese-51 is the most likely to decay by positron emission.
Q20.2.11
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.
- Platinum wire is dipped in hydrochloric acid.
- Manganese metal is added to a solution of iron(II) chloride.
- Tin is heated with steam.
- Hydrogen gas is bubbled through a solution of lead(II) nitrate.
Answer:
The activity series is a list of elements in the order of reactivity. The elements at the top of the series are most reactive since they are most easily oxidized. Conversely, the elements at the bottom are least reactive.
- No reaction occurs since Platinum is really low on the activity series and therefore does not react with aqueous Hydrogen ions (since HCl is a strong acid and it will dissociate into its corresponding ions) to release bimolecular Hydrogen.
- A reaction occurs since FeCl2 dissociates in solution and MnCl2 forms.
- Net Ionic Equation: \[Mn(s)+2Cl(g)\rightarrow MnCl_{2}(s)\displaystyle\]
- Complete Ionic Equation: \[Fe^{2+}(aq)+Mn(s)+2Cl(g)\rightarrow MnCl_{2}(s)+Fe^{2+}(aq)\displaystyle\]
- No reaction occurs since based on the activity series, Tin only reacts with simple acids to make bimolecular Hydrogen.
- A reaction occurs since lead (II) nitrate is soluble in water and will form nitric acid.
- Net Ionic Equation: \[H_{2}(g)+NO_{3}(aq)\rightarrow 2HNO_{3}(aq)\displaystyle\]
- Complete Ionic Equation:\[Pb^{2+}(aq)+H_{2}(g)+NO_{3}(aq)\rightarrow 2HNO_{3}(aq)+Pb^{2+}(aq)\displaystyle\]
Q20.5.6
Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?
Answer:
The cell potentials of two half reactions cannot just be simply added to find the cell potential of the whole reaction since cell potentials are not a state function. A state function is when the value depends on the state of the substance and not how the substance reached that state. Cell potentials are not state functions since how the substance reached its current state can be different from reaction to reaction. The sum of the two half reactions equals the cell potential of the overall reaction when the substances are in standard state conditions. Standard state conditions are when molarity is 1M for all solutions present, partial pressure is equal to 1atm for all gases present and temperature is 25°C.
Q24.6.2
In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?
Answer:
Crystal Field Theory is based off the interactions of the electrons in the d orbital of transition metals. Once degenerate sets of electrons split into different energy levels because of the electron-electron repulsion since they are so close together. Placing them at the different points of an octahedral removes their degeneracy but does not change the inherent energy of each of the electrons. The splitting is called D-Orbital Splitting since it is the splitting of the electrons in the d orbital. The splitting magnitude depends on the VESPR geometry of the molecule since different geometry (octahedral compared to square planar and tetrahedral) have different electron distributions. Also, the spin of the electrons depends on the ligand to which the metal is bound and whether it is strong field or weak field.