Extra Credit 39

Q17.5.7

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

S17.5.7

We know the relationship between ΔG, ΔG° and the equilibrium constant K. We can obtain a relationship between E∘cell and K.

$$\triangle G^o=RT/nK$$

The equation for Nerst equation is derived as

$$E ^ocell=(\frac{RT}{nF})lnK$$

Therefore, we can see that $$E ^ocell$$ depends on Concentration from The Nernst Equation. As the cell is been used, the concentration on the reactant side decreased thus decreasing the cell potential. As the battery continues to be used the (\frac{RT}{nF})lnK$$ will move towards zero indicting an overall cell potential of zero and dead battery.

Q12.3.2

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

S12.3.2

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

For a reaction with the general equation:

$$aA+bB\rightarrow cC+dD$$

the reaction order and rate law are described as

$$rate=k[A]^m[B]^n$$

Since the concentration is doubled, [A] increased to [2A], and the rate is 4 times than before. Thus the eqauation will be

$$4rate=k[2A]^m[B]^n$$

Since there is no change in [B]

We can simplify the equation to calculate m

$$rate=k[A]^m$$

$$4rate=k[2A]^m$$

$$m=2$$

Therefore, the order for this reaction is 2.

2.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?

The similar method from the previous question.

The reaction form is

$$rate=k[A]^m[B]^n$$

As the concentration of the reactant tripled, the rate is also tripled, the equation can be written as

$$3rate=k[3A]^m[B]^n$$

Therefore, we can calculate m which equals to one.

The order of this reaction is one.

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?

S12.5.11

In order to determine the activation energy, we can use the Arrhenius equation:

$$k=Ae^{-(Ea/RT)}$$

We can further simplify this equation as

$$lnk=lnA-\frac{Ea}{RT}$$

Given:

rate increases 1.47 time

the temperature changed from 30 °C to 37 °C.

By using the formula

$$ln \frac{K1}{K2}=\frac{Ea}{R}(1/T2-1/T1)$$

$$ln(1.47)=\frac{Ea}{8.314}(1/303-1/T310)$$

$$Ea=43.74kJ/mol$$

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.

S21.4.6

(a)

This the band of stability

From this band, we can see nuclides with the proton number greater than 83 are not stable.

Below the decay pattern shows how those nuclides decay to form more stable nuclides.

From those graph, we can see that if the element is blow the belt, such as Po. It will undergo an alpha decay.

$$_{84}^{214}\textrm{Po}\rightarrow _{82}^{208}\textrm{Pb}+_{2}^{4}\textrm{He}$$

(b) If the nuclide is above the band of stability such as Th, it will undergo a beta decay

$$_{90}^{230}\textrm{Th}\rightarrow _{91}^{213}\textrm{Pa}+_{-1}^{0}\textrm{e}$$

Q20.2.10

Balance each redox reaction under the conditions indicated.

1.acidic solution$$MnO4^-(aq) + S_2O_3^{2-}(aq)\rightarrow Mn^{2+}(aq) + SO_4^{2-}(aq);$$

2.acidic solution$$Fe^{2+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Fe^{3+}(aq) + Cr^3{3+}$$

3.Basic solution $$Fe(s) + CrO_4^{2-}(aq) \rightarrow Fe_2O_3(s) + Cr_2O_3(s)$$

4.acidic solution:$$Cl_2(aq) \rightarrow ClO_3^-(aq) + Cl^-(aq)$$

5.Basic soolution: $$CO_3^{2-}(aq) + N_2H_4(aq) \rightarrow CO(g) + N_2(g)$$

S20.2.10

1.acidic solution$$MnO4^-(aq) + S_2O_3^{2-}(aq)\rightarrow Mn^{2+}(aq) + SO_4^{2-}(aq);$$

(1)Oxidatoin state change: Mn changed from 7 to 2; S changed from 2 to 6

(2) Oxidation: $$S_2O_3^{2-}(aq) \rightarrow SO _4^{2-}(aq)$$

Reduction:$$MnO_ 4^-(aq) \rightarrow Mn^{2+}(aq) $$

(3)Balance each reaction by adding H2O to first balance oxygen

Oxidation:$$5H_2O+S_2O_3^{2-}(aq) \rightarrow 2SO_ 4^{2-}(aq)$$

Reduction:$$MnO_4^-(aq) \rightarrow Mn^{2+}(aq)+4H_2O$$

(4) Add $$H^+$$ and electron to balance the equation and charge

Oxidation:$$5H_2O+S_2O_3^{2-}(aq) \rightarrow 2SO _4^{2-}(aq)+10H^++8e^-$$

Reduction:$$5e^-+8H^++MnO_4^-(aq) \rightarrow Mn^{2+}(aq)+4H_2O$$

(5)Mutilply oxidation reaction by 5 and reduction reaction by 8 to balance the charge

Oxidation:$$25H_2O+5S_2O_3^{2-}(aq) \rightarrow 10SO _4^{2-}(aq)+50H^++40e^-$$

Reduction:$$40e^-+64H^++8MnO_4^-(aq) \rightarrow 8Mn^{2+}(aq)+32H_2O$$

(6) add two equation together

$$14H^++5S_2O_3^{2-}+8MnO_4^-\rightarrow 10SO_4^{2-}+7H_2O+8Mn^{2+}$$

2. acidic solution$$Fe^{2+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Fe^{3+}(aq) + Cr^3{3+}$$

(1)Oxidatoin state change: Fe changed from 2 to 3;Cr changed from 6 to 3

(2)Oxidation: $$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) $$

Reduction:$$Cr_2O_7^{2-}(aq)\rightarrow Cr^{3+}$$

(3)Balance each reaction by adding H2O to first balance oxygen

Oxidation: $$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) $$

Reduction:$$Cr_2O_7^{2-}(aq)\rightarrow Cr^{3+}+7H_2O$$

(4) Add $$H^+$$ and electron to balance the equation and charge

Oxidation: $$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)+e^- $$

Reduction:$$6e^-+14H^++Cr_2O_7^{2-}(aq)\rightarrow Cr^3+7H_2O$$

(5)Mutilply oxidation reaction by 6 to balance the charge

Oxidation: $$6Fe^{2+}(aq) \rightarrow 6Fe^{3+}(aq)+6e^- $$

(6) add two equation together

$$14H^++Cr_2O_7^{2-}+6Fe^{2+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_2O$$

3.Basic solution $$Fe(s) + CrO_4^{2-}(aq) \rightarrow Fe_2O_3(s) + Cr_2O_3(s)$$

(1)Oxidatoin state change: Fe changed from 0 to -3;Cr changed from -6 to -3

(2)Oxidation:$$CrO_4^{2-}(aq) \rightarrow Cr_2O_3(s)$$

Reduction:$$Fe(s) \rightarrow Fe_2O_3(s) $$

(3)Balance each reaction by adding H2O to first balance oxygen

Oxidation:$$2CrO_4^{2-}(aq) \rightarrow Cr_2O_3(s)+5H_2O$$

Reduction:$$3H_2O+Fe(s) \rightarrow Fe_2O_3(s) $$

(4) Add $$H^+$$ and electron to balance the equation and charge

Oxidation:$$6e^-+10H^++2CrO_4^{2-}(aq) \rightarrow Cr_2O_3(s)+5H_2O$$

Reduction:$$3H_2O+Fe(s) \rightarrow Fe_2O_3(s) +6H^+$+6e^-$$

(5)add two eqaution together

$$4H^++2CrO4^{2-}+2Fe\rightarrow CrO_3+2H_2O+Fe_2O_3$$

(6)Since this reaction is basic, we need to add hydroxide to balance the hydrogen ion.

$$4OH^-+4H^++2CrO4^{2-}+2Fe\rightarrow CrO_3+2H_2O+Fe_2O_3+4OH^-$$

(7) Overall balanced equation

$$2H_2O+2CrO4^{2-}+2Fe\rightarrow CrO_3+Fe_2O_3+4OH^-$$

4. acidic solution:$$Cl_2(aq) \rightarrow ClO_3^-(aq) + Cl^-(aq)$$

(1). oxidation state : Cl changed from 0 to 5 and -1

(2) Oxidation: $$Cl(aq) \rightarrow ClO_3^-(aq) $$

Reduction:$$Cl(aq) \rightarrow Cl^-(aq)$$

(3)Balance each reaction by adding H2O to first balance oxygen

Oxidation: $$3H_2O+Cl(aq) \rightarrow ClO_3^-(aq) $$

(4) Add $$H^+$$ and electron to balance the equation and charge

Oxidation: $$3H_2O+Cl(aq) \rightarrow ClO_3^-(aq)+6H^++5e^- $$

Reduction:$$Cl(aq)+e^- \rightarrow Cl^-(aq)$$

(5)Mutilply reduction reaction by 5 to balance the chargere

Reduction:$$5Cl(aq)+5e^- \rightarrow 5Cl^-(aq)$$

(6) add two equation together

$$3H_2O+6Cl\rightarrow 5Cl^-+ClO_3^-+6H^+$$

5. Basic soolution: $$CO_3^{2-}(aq) + N_2H_4(aq) \rightarrow CO(g) + N_2(g)$$

(1)Oxidation state: C changed from -4 to -2; N changed from 2 to 0

(2)Oxidation:$$CO_3^{2-}(aq) \rightarrow CO(g) $$

Reduction:$$N_2H_4(aq) \rightarrow N_2(g)$$

(3)Balance each reaction by adding H2O to first balance oxygen

Oxidation:$$CO_3^{2-}(aq) \rightarrow CO(g) +2H_2O$$

(4) Add $$H^+$$ and electron to balance the equation and charge

Oxidation:$$2e^-+4H^++CO_3^{2-}(aq) \rightarrow CO(g) +2H_2O$$

Reduction:$$N_2H_4(aq) \rightarrow N_2(g)+4H^+$+4e^-$$

(5)Mutilply oxidation reaction by 2 to balance the charge

Oxidation:$$4e^-+8H^++2CO_3^{2-}(aq) \rightarrow 2CO(g) +4H_2O$$

(6) add two equation together

$$4H^++N_2H_4+2CO_3^{2-}\rightarrow 2CO+N_2+4H_2O$$

(7)Since this reaction is basic, we need to add hydroxide to balance the hydrogen ion.

$$4OH^-+4H^++N_2H_4+2CO_3^{2-}\rightarrow 2CO+N_2+4H_2O+4OH^-$$

(8)The overall balanced equation

$$N_2H_4+2CO_3^{2-}\rightarrow 2CO+N_2+4OH^-$$

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.

S20.5.5

Yes, I agree. The irreversible cells are those can't be recharged such as the dry cell. Those cells have an electrolyte in an acidic water-based paste. The reversible cell is those can be recharged such as nickel–cadmium (NiCad) battery which does not require the electrodes to be in separate compartments. Therefore, the study of electrochemistry can be useful to determine the reversibility and irreversibility. Those methods also can be further applied to different types of cells.

Q24.6.1

Describe crystal field theory in terms of its

1. assumptions regarding metal–ligand interactions.

2. weaknesses and strengths compared with valence bond theory.

S24.6.1

1.The Crystal Field Splitting described the energy difference between two sets of d orbitals in a metal atom when ligands are present. With the

Spectrochemical Series, we can see the order of their abilities to split the d orbital energy levels.

Since Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases(Libretext). The side toward I^- has a weak field and the CO side has a strong field.

2.weaknesses and strengths compared with valence bond theory.

The valence bond theory is Aufbau principle, which states that electrons are most stable when they filled each orbital with one electron before further filling it.

Crystal field theory is a model that describes the electronic structure of transition metal compounds, all of which can be considered coordination complexes.

Comparison:

The valence bond theory states the bond between the metal and the ligand is covalent; Crystal field theory believes the bond is ionic.

The Crystal field theory also provides the satisfactory explanation for the color of transition metal complexes such as spectral properties but it overlooks the attractive forces the d-electrons of the metal ion and nuclear charge on the ligand atom.

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.)

S14.7.9

(a) There is an isoelectric point (pI) where the enzyme generally has a minimum solubility in aqueous solutions. In a similar manner to the effect on enzymes, the charge and charge distribution on the substrate(s), product(s) and coenzymes (where applicable) will also be affected by pH changes. Increasing hydrogen ion concentration will, additionally, increase the successful competition of hydrogen ions for any metal cationic binding sites on the enzyme, reducing the bound metal cation concentration.(Effect of PH)

The graph shows how relation between PH value and enzyme activity

(b) The pH of a solution can have several effects on the activity of enzymes. Since the enzymes are made of amino, pH can have an effect on the state of ionization of acidic or basic amino acids. Due to the side chains of acidic amino acid which contains carboxyl functional groups and the amine functional groups for basic amino acid.

Reference

1.OpenStax CNX. N.p., n.d. Web. 11 June 2017.

2.Zumdahl, Steven, and Donald DeCoste. Chemical Principle. 8th ed. N.p.: n.p., 2015. Print.

3."23.5 Bonding in Coordination Compounds: Crystal Field Theory - Ms. Smith." Google Sites. N.p., n.d. Web. 11 June 2017.

4." Crystal Field Theory Versus Valence Bond Theory Engineering Essay." UKEssays. N.p., n.d. Web. 11 June 2017.

5. PH and Enzymes. N.p., n.d. Web. 11 June 2017.

6." Effect of PH and Ionic Strength." Effect of PH and Ionic Strength on Enzyme Activity. N.p., n.d. Web. 11 June 2017.