Extra Credit 32
- Page ID
- 83003
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q 12.1.5: A study of the rate of the reaction represented as 2A⟶B gave the following data:
a. Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
average rate = (Δ[A])/ Δt
Use the table to plug in these values into the equation.
0-10: -[(.625-1)/ (10-0)] = .0375 M/s
10-20: -[(.370 - .625)/(20-10)] = .0255 M/s
b. Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?
The instantaneous rate of disappearance is about .05 molarity per second (M/s). This is because the slope of the tangent line of the graph at t=15s is about -.05 M/s.
c. Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.
rate = .5(Δ[A])/ Δt = (Δ[B])/ Δt
(Δ[A])/ Δt = 2(Δ[B])/Δt
(Δ[B])/Δt = .5(-[(.625-1)/ (10-0)]) = .5(.0375) M/s = .0375 M/s
instantaneous rate of formation for B at 15 s = .5(.05) = .025 M/s
Q 12.5.3: What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
The activation energy is the minimum amount of energy needed for a reaction to take place. The activated complex refers to the molecular compounds existing at the highest energy state in the reaction path of a chemical reaction. The activation energy of a chemical reaction is measured by finding the change in energies between the reactants and the activated complex.
Q 14.6.5: Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health?
Radical = highly reactive species
James bond needed to purge his body of radicals in order to stabilize it. Having highly reactive species inside the body would cause instability and potential harmful/unwanted reactions may occur, as the radicals would be looking for species to bond to and react with.
Q 17.4.5: Use the data in Table P1 to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.
NOTE: For each solution, the Nernst equation will be used, however Ecell=0 since there is no net transfer of electrons during equilibrium.
a. AgCl(s)⇌Ag+(aq)+Cl−(aq)
oxidation half reaction: AgCl (s) + e- ⇌ Ag (s) + Cl- E˚ = .2223
reduction half reaction: Ag+(aq) + e- ⇌ Ag (s) E˚ = .7996
E˚cell = E˚reduction - E˚oxidation = .2223 - .7996 = -.5773
E˚cell = [(RT)/nF] ln(K) = (.0591v/n) log(K)
-.5773 v = (.0591v/1) log(K)
k = 1.7 x 10-10
b. CdS(s)⇌Cd2+(aq)+S2−(aq) at 377 K
Cd(s) ⇌ Cd2+(aq) + 2e- E˚ = -.4030
CdS(s) + 2e- ⇌ Cd(s) + S2-(aq) E˚ = -1.17
E˚cell = E˚reduction - E˚oxidation = -1.17 - (-.4030) = -.767
E˚cell = [(RT)/nF] ln(K) = [(8.3145 J/mol K)(377K)/(2)(96485)] ln(K)
K = 2.6 x 10-21
c. Hg2+(aq)+4Br−(aq)⇌[HgBr4]2−(aq)
this is not a redox reaction
Hg: +2 -> +2
Br: -1 -> -1
d. H2O(l)⇌H+(aq)+OH−(aq) at 25°C
H2(g) ⇌ 2H+(aq) + 2e- E˚ = 0.0v
H2O(l) + e- ⇌ 1/2*H2(g) + OH-(aq) E˚ = -.8277v
The K value corresponds to the Kwater, which has a value of 1 x 10-14.
Q: 20.4.22: Calculate E°cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?
E°cell = E°cathode - E°anode
Cathode (reduction) = Zn2+ (aq) + 2e- ⇌ Zn (s) E˚ = -.7618 v
Anode (oxidation) = Cl2(g) + 2e− ⇌ 2Cl− E˚ = 1.396 v
E°cell = E°cathode - E°anode = -.7618 v - 1.396 v = -2.1578 v
To calculate ΔG°, use the Nernst equation.
ΔG° = -nF(E°cell) = (-2)(96485 C/mol)(-2.1578 v) = 416390.7 J
Because the ΔG° is positive (and E°cell is negative), the reaction will not occur spontaneously.
Q 20.2.3: In each redox reaction, determine which species is oxidized and which is reduced.
a. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
Oxidation: Zn(s) → ZnSO4(aq)
Oxidation state of Zn(s) = 0 and the oxidation state of Zn in ZnSO4(aq) = +2, so the zinc loses two electrons and is oxidized.
Reduction: H2SO4(aq) → H2(g)
Oxidation state of H in H2SO4 (aq) = +1 and the oxidation state of H2(g) = 0, so the hydrogen gains an electron and is reduced.
b. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
Oxidation: Cu(s) → Cu(NO3)2(aq)
Oxidation state of Cu(s) = 0 and the oxidation state of Cu in Cu(NO3)2(aq)= +2, so the copper loses two electrons and is oxidized.
Reduction: 4HNO3(aq) → 2NO2(g)
Oxidation state of N in 4HNO3(aq) = +5 and the oxidation state of 2NO2(g) = +4, so the nitrogen gains an electron and is reduced
c. BrO3−(aq) + 2MnO2(s) + H2O(l) → Br−(aq) + 2MnO4−(aq) + 2H+(aq)
Oxidation: 2MnO2(s) → 2MnO4−(aq)
Oxidation state of Mn in 2MnO2 = +4 and the oxidation state of Mn in 2MnO4−= +7, so the manganese loses three electrons and is oxidized.
Reduction: BrO3−(aq) → Br−(aq)
Oxidation state of Br in BrO3- = +5 and the oxidation state of Br-= -1, so the bromine gains six electrons and is reduced.
Q 21.3.7: The mass of the atom 199F is 18.99840 amu.
number of protons = 9
number of neutrons = 19-9 = 10
mass of protons=1.0078 amu
mass of neutrons=1.0087 amu
1 amu = 931.5 MeV
Δm = (mass of protons +mass of neutrons) - mass of the nucleus (given)
a) Calculate its binding energy per atom in millions of electron volts.
Δm = [(9)(1.0078)+(10)(1.0087)] - (18.9984) = .1588 amu
--> convert into MeV
(.1588 amu) x (931.5 MeV/1 amu) = 147.9 MeV
b) Calculate its binding energy per nucleon.
--> divide by the number of nucleons (protons+neutrons)
(147.9 MeV)/(55 nucleons) = 7.79 MeV
Q20.9.7 (DONE BY MARLEE NUNEZ, NOT PHASE 1 STUDENT)
What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl2 is electrolyzed using a current of 12.4 A for 1.0 h?
Use equation:
n = It/F
to get the number of moles of electrons.
Convert hours to seconds -> (1 hr)(60 min/1 hr)(60 sec/1 hr)= 3600 sec
n = (12.4 A)(3600 s)/(96485 C*mol-1) = .462663 mol e-
Next find out how many moles of electrons are transferred in the reaction MgCl2 -> Mg(s) + Cl2(g).
Mg2+(aq) + 2e- -> Mg(s)
2Cl-(aq) -> Cl2(g) + 2e-
Therefore 2 moles of electrons are transferred. Now use stoichiometry to find the volume of chlorine gas.
(.462663 mol e-)*(1 mol MgCl2/2 mol e-)*(1 mol Cl2/1 mol MgCl2)*(22.4 L Cl2/1 mol Cl2) = 5.18 L Cl2