Extra Credit 39

Q17.5.7

Explain what happens to battery voltage as a battery is used, in terms of the Nernst equation.

A17.5.7

\[E=E^o - \frac{0.025693}{n}lnQ\]

\[E=\frac{0.025693}{n}lnK - \frac{0.025693}{n}{n}lnQ\]

\(E\) is positive as long as \(K>Q\), which is the case when there are more reactants than products relative to the equilibrium concentrations. When the system is unbalanced like this, the reaction spontaneously shifts towards products, which corresponds to a \(+E\). This value is highest at the begnning and as \(K\) shifts towards \(Q\) as the system equillbrates, \(E\) decreases. Once \(K=Q, E=0\) and the battery is "dead."

Q12.3.2

Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

a.) What is the order of the reaction with respect to that reactant?

b.) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

A12.3.2

\([R_1]\) = Reactant 1 concentration \([R_2]\) = Reactant 2 concentration

a.) To solve for the reactant order, set up the rate law in the form \(r=k[A]^m [B]^n\) for the initial and doubled conditions.

\[r=k[R_1]^m [R_2]^n\]

\[4r=k[2R_1]^m [R_2]^n\]

\[\frac{4r}{r} = \frac{k[2R_1]^m [R_2]^n}{k[R_1]^m [R_2]^n}\]

\[4=2^m\]

\[m=2\]

b.) To solve for the reactant order, set up the rate law in the form \(r=k[A]^m [B]^n\) for the initial and tripled conditions.

\[r=k[R_1]^m [R_2]^n\]

\[3r=k[R_1]^m [3R_2]^n\]

\[\frac{3r}{r} = \frac{k[R_1]^m [3R_2]^n}{k[R_1]^m [R_2]^n}\]

\[3=3^n\]

\[n=1\]

Q12.5.11

An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?

A12.5.11

The given information is \(T_1=30+273.15=303.15K\), \(T_2=37+273.15=310.15K\), \(R=8.314 \frac{J}{mol \cdot K}\), and \(k_2=1.47 k_1\). The rate constant \(k_2\) is equal to \(1.47 k_1\) because \(k\) is a temperature dependent constant.

\[ln(\frac{k_2}{k_1})=\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})\]

\[ln(\frac{1.47 k_1}{k_1})=\frac{E_a}{8.314 \frac{J}{mol \cdot K}}(\frac{1}{303.15K} - \frac{1}{310.15K})\]

\[ln(1.47)\frac{J}{mol}=E_a(\frac{1}{8.314})(7.4451 \times 10^{-5})\]

\[E_a=43000 \frac{J}{mol} = 43.0 \frac{kJ}{mol}\]

Q21.4.6

Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability a.) if they are below the band of stability and b.) if they are above the band of stability.

A21.4.6

a.) If the heavy nuclides are below the band of stability, that means they are neutron poor. To stabilize, they need to decay in a way to increase their neutron-to-proton ratio. Two ways they can do this is through \( _{-1}^{0}\textrm{e}\) capture:

\[_{1}^{1}\textrm{p} + _{-1}^{0}\textrm{e} \to _{0}^{1}\textrm{n}\]

or \(_{+1}^{0}\rm{\beta}\) (positron) emission:

\[_{1}^{1}\textrm{p} \to _{0}^{1}\textrm{n} + _{+1}^{0}\rm{\beta}\]

b.) If the heavy nuclides are above the band of stability, that means they are neutron rich. To stabilize, they need to decay in a way to increase their neutron-to-proton ratio. Two ways they can do this is through \(_{-1}^{0}\rm{\beta}\) emission:

\[_{0}^{1}\textrm{n} \to _{1}^{1}\textrm{p} + _{-1}^{0}\rm{\beta}\]

or spontaneous fission, if the nuclei is very massive:

\[\text{ex. } _{98}^{254}\textrm{Cf} \to _{46}^{118}\textrm{Pd} + _{52}^{132}\textrm{Te}\ + 4 _{0}^{1}\textrm{n}\]

Q20.2.10

Balance each redox reaction under the conditions indicated.

a.) MnO₄⁻_(aq) + S₂O₃²⁻_(aq) → Mn²⁺_(aq) + SO₄²⁻_(aq); acidic solution

b.) Fe²⁺_(aq) + Cr₂O₇²⁻_(aq) → Fe³⁺_(aq) + Cr³⁺_(aq); acidic solution

c.) Fe_(s) + CrO₄²⁻_(aq) → Fe₂O_3(s) + Cr₂O_3(s); basic solution

d.) Cl_2(aq) → ClO₃⁻_(aq) + Cl⁻_(aq); acidic solution

e.) CO₃²⁻_(aq) + N₂H_4(aq) → CO_(g) + N_2(g); basic solution

A20.2.10

To balance a redox reactions, 1.) split up the reaction into the reduction and oxidation half-reactions, 2.) balance the elements other than O and H, 3.) balance O by adding H₂O, 4.) balance H by adding H⁺, 5.) balance the charge by adding \(e^-\)s, 6.) add up the two half-reactions and multiply by the necessary coefficients to cancel out the \(e^-\)s, 6.5) neutralize all H⁺ by adding OH^- and balancing on both sides.

A. 1.) determine the reduction and oxidation reactions:

(Reduction): MnO₄^- (aq) + 8H⁺ (aq) + 5e^- → Mn²⁺(aq) + 4H₂O (l)

(Oxidation) : S₂O₃²⁻(aq) + 5H₂O (l) → 2SO₄²⁻(aq) + 10H⁺(aq)+ 4e^-

Balance the electrons:

(MnO₄^- (aq) + 8H⁺ (aq) + 5e^- → Mn²⁺(aq) + 4H₂O (l))x 4

(S₂O₃²⁻(aq) + 5H₂O (l) → 2SO₄²⁻(aq) + 10H⁺(aq)+ 4e^-)x5

Completely balanced reaction: 4MnO₄^- (aq) + 5S₂O₃²⁻(aq) + 9H₂O (l) → 10SO₄²⁻(aq) + 4Mn²⁺(aq) + 18H⁺(aq)

B. (Oxidation) Fe²⁺_(aq)→ Fe³⁺ + e^-

(Reduction) 14H₃O⁺+Cr₂O₇²⁻+6e^- → 2Cr³⁺_(aq)+ 21H₂O

balance the electrons:

6Fe²⁺_(aq)→ 6Fe³⁺ + 6e^-

14H₃O⁺+Cr₂O₇²⁻+6e^- → 2Cr³⁺_(aq)+ 21H₂O

Cancel everything out:

6Fe²⁺_(aq)+14H₃O⁺+Cr₂O₇²⁻→ 2Cr³⁺_(aq)+ 21H₂O + 6Fe³⁺

C.

(Reduction): CrO₄²⁻(aq) → Cr₂O₃(s)

2CrO₄²⁻(aq) + 10H⁺(aq) + 10 OH^-(aq) → Cr₂O₃(s) + 5H₂O (l) + 10 OH^-(aq)

2CrO₄²⁻(aq) + 5H₂O (l) + 6e^- → Cr₂O₃(s) + 10 OH^-(aq)

(Oxidation): Fe(s) → Fe₂O₃(s)

2Fe(s) + 3H₂O(l) + 6OH^-(aq)→ Fe₂O₃(s) + 6H⁺(aq) + 6OH^-(aq)

2Fe(s) + 6OH^-(aq)→ Fe₂O₃(s) + 3H₂O (aq) +6e^-

Completely balanced reaction: 2CrO₄²⁻(aq) + 2Fe(s) + 2H₂O (l) → Fe₂O₃(s) + Cr₂O₃(s) + 4OH^-(aq)

D. (Reduction): Cl₂(aq) → Cl⁻(aq)

(Cl₂(aq) + 2e^-→ 2Cl⁻(aq)) x 5

(Oxidation): Cl₂(aq) → ClO₃⁻(aq)

Cl₂(aq) + 6H₂O (l) → 2ClO₃⁻(aq) + 12H⁺(aq) + 10e^-

Completely balanced reaction: 6Cl₂(aq) + 6H₂O(l) → 2ClO₃⁻(aq) + 10Cl⁻(aq) + 12H⁺(aq)

E.

(Reduction): 4H⁺+ CO₃²⁻_(aq) + 2e^- → CO_(g)+2H₂O

(Oxidation): N₂H_4(aq)→ 4H⁺+ N_2(g)+ 4e^-

balance electrons:

8H⁺+ 2CO₃²⁻_(aq) + 4e^- → 2CO_(g)+4H₂O

N₂H_4(aq)→ 4H⁺+ N_2(g)+ 4e^-

since electrons are balanced, we may proceed to cancel them out :

2CO₃²⁻_(aq) +N₂H₄(aq)+ 8H⁺→ 4H⁺ + N_2(g)+ 2CO_(g)+4H₂O

2CO₃²⁻_(aq) +N₂H₄(aq)+ 4H⁺→ N_2(g)+ 2CO_(g)+4H₂O

since this is a basic solution, add OH- to both sides:

2CO₃²⁻_(aq) +N₂H₄(aq)+ 4H₂O → N_2(g)+ 2CO_(g)+4H₂O + 4OH^-

(sorry for the this particular problem (20.2.10) , I did not have enough time to completely edit perfectly (i.e. A through E of this problem) -Matthew Sam)

Q20.5.5

State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.

A20.5.5

I agree with this statement because through the Big Triangle of Chemistry, E^º and E can be related to G and K through the equations:

\[E =E^{\circ }- \frac{RT}{nF}Q\]

\[E^{\circ }=\left ( \frac{RT}{nF} \right )lnK\]

It can be inferred from these equations the spontaneity of an equation because a positive E value corresponds to a negative G value which means it is spontaneous. Similarly, a negative E values corresponds to a positive G values which means the reaction is not spontaneous. The relationship between E and G can be found through this equation:

\[G^{\circ }=-nFE^{\circ }_{cell}\]

\[G=-nFE_{cell}\]

(Matthew Sam had to answer all of this question)

Q24.6.1

Describe crystal field theory in terms of its

a.) assumptions regarding metal–ligand interactions.

b.) weaknesses and strengths compared with valence bond theory.

A24.6.1

a.) One of the assumptions made by CFT is that ligands are simple, dimensionless point charges and that their only important part is their electron lone pair. Another assumption is that the metal-ligand interactions are purely electrostatic and that covalent interactions don't play a part.

b.) Weaknesses: Only \(d\) electrons are taken into account in CFT and the role of the other \(s\) and \(d\) orbitals are ignored, thus painting an incomplete picture. VBT takes into account the sum total of all the lower orbitals and their interactions. Also, CFT doesn't take into account orbitals of the ligands, unlike VBT where the orbitals of both the ligand and the central atom is accounted for.

Strengths: CFT explains the variety of colors found among metal complexes with through the concept of crystal field splitting energies. The CFT qualitatively describes the strength of the metal ligand bonds. It describes the breaking of the orbital degeneracy in transition metals due to the presence of ligands. (Matthew Sam had to answer the last part of this question)

Q14.7.9

Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to a.) effects at the catalytic center and b.) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.)

A14.7.9

a.) Amino acid residues in the active catalytic center perform optimally in a narrow pH range due to their specific pKa values, which affects their protonation state.

b.) The effects elsewhere in the enzyme could change the catalytic activity since the side chains of the amino acids could interact with the catalytic sites in different pH conditions. For example, basic, versus acidic amino acids have different functional groups. Basic ones have an amine functional group, whereas acidic one shave a carboxyl functional group. The difference in the hydrogen ion concentration may greatly affect the interaction the amino acid may have with the catalytic site due to the different conformation it can take due to the different functional groups. (Matthew Sam had to answer this part of the question)