Extra Credit 7

Q5.3.23

Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories.

Solution:

A note before getting started: all the data applied here are collected under standard conditions.

Cereal: First, convert the unit from calories per ounce to kilojoules per ounce as required:

130 cal/ounce x 10^-3 kcal/cal x 4.184 kJ/kcal = 0.544kJ/ounce

Then, calculate the total amount of energy in the 32-ounce cereal

0.544kJ/ounce x 32.0 ounce =17.408kJ

The 32-ounce breakfast cereal costs $4.23. In a certain sense, we buy the cereal in order to obtain the energy it provides, so it is reasonable to replace the concept of "cereal" with the "energy" amount we just calculated in the last step in this case. Hence, the sentence becomes: the 17.408kJ energy costs $4.23. To calculate how much energy can each dollar buy:

17.408kJ / $4.23 = 4.12kJ/$ (keep only three effective numbers)

isooctane: To calculate the expense of each kilojoule of isooctane, we will use the formular:

\[Q=△_cH^{○}n\]

\[Q=\textrm{ energy released by combustion}\]

\[△_cH^{○}=\textrm{ heat of combustion of isooctane, obtained from NIST profile of isooctane} =-5461.3 kJ/mol\]

\[n=\textrm{number of moles of isooctane}\]

First, convert the 1L volume of isooctane to mass and then to number of moles

1L x 10³ mL/L x 0.6919g/ml = 691.9g

691.9g / 114.2285g·mol(molar mass of isooctane obtained from NIST) = 6.057 mol (keep only four effective number)

Then, calculate the total amount of energy we can obtain from combusting 6.057 mol of isooctane.

-5461.3 kJ/mol x 6.057mol = -3.307 x 10⁴ kJ

The negative sign here means that these -3.307 x 10⁴ kJ of energy is released(in contrast of absorbed) by the combustion of isooctane. Since being released is the only activity the energy in this question is going to do, we ignore the negative sign here to avoid creating confusion to later calculation. Hence,

Q=3.307 x 10⁴ kJ

Again, the same as how we dealt with the concept of "the price of the cereal" and "the price of the energy in the cereal", we will do the concept substitution here, too. Now we will say that "this -3.307 x 104 kJ energy, which comes from the 1L isooctane, costs $0.45". We can now calculate how much energy can each dollar purchase.

3.307 x 10⁴ kJ / $0.45 = 7.349 x 10⁴ kJ/$

Do a comparison between the kilojoules of energy per dollar data we just obtained,

7.349 x 10⁴ kJ/$ ＞＞4.12kJ/$

We can see that, we can obtain a great lot more kilojoules of energy from burning isooctane than eating cereal by buying one dollar of them.

(Too bad, human's digestion system cannot process isooctane, so in real-life it is still better to choose a cereal for yourself.)

Q10.3.1

Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?

Answer:

In essence, the boiling process is a process in which water molecules, held together by intermolecular forces to form a liquid phase in macro observation, obtain enough energy from heat to over come the intermolecular forces that hold them together and started leaving the league of water molecules with the energy they have obtained. The energy will be consumed exclusively due to this evaporation process.

The amount of water molecules that obtain enough energy to escape changes. As more heat is added, more molecules obtain enough energy and escape, resulting in more violently boiling water with a constant temperature.

Q13.2.21

Convert the values of K_c to values of K_P or the values of K_P to values of K_c.

1. \(\ce{Cl2}(g)+\ce{Br2}(g) \rightleftharpoons \ce{2BrCl}(g) \hspace{20px} K_c=4.7×10^{−2}\textrm{ at 25°C}\)
2. \(\ce{2SO2}(g)+\ce{O2}(g) \rightleftharpoons \ce{2SO3}(g) \hspace{20px} K_P=\textrm{48.2 at 500°C}\)
3. \(\ce{CaCl2⋅6H2O}(s) \rightleftharpoons \ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25°C}\)
4. \(\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g) \hspace{20px} K_P=\textrm{0.196 at 60°C}\)

Solutions:

In all the sub questions, we will use the following equation,

\[K_p=(RT)^{△n}K_c\]

\[K_p=\textrm{equilibrium constant of pressure}\]

\[R=\textrm{gass constant}=0.082L·atm·mol^{-1}·K^{-1}\]

\[T=\textrm{temperature in Kelvin}\]

\[△n=\textrm{moles of products - moles of reactants}\]

\[K_c=\textrm{equilibrium constant of concentration}\]

a. ∵K_p=(RT)^△nK_c ∴K_p=(0.082\L·atm·mol^-1·K^-1 x 298.15 K)^(2-1-1) x 4.7 x 10^-2 =4.7 x 10^-2

b. ∵K_p=(RT)^△nK_c ∴K_c=K_p/(RT)^△n=48.2/(0.082L·atm·mol^-1·K^-1 x 773.15K)^(2-1-2) =3.06x10³

c. ∵K_p=(RT)^△nK_c ∴K_c=K_p/(RT)^△n=5.09x10^-44/(0.082L·atm·mol^-1·K^-1 x 298.15K)^(6-0)=2.38x10^-52

d. ∵K_p=(RT)^△nK_c ∴K_c=K_p/(RT)^△n=0.196/(0.082L·atm·mol^-1·K^-1 x 333.15K)^(1-0)=7.17x10^-3

Q14.1.14

Is the self ionization of water endothermic or exothermic? The ionization constant for water (K_w) is \(2.9 \times 10^{-14}\) at 40 °C and \(9.6 \times 10^{-14}\) at 60 °C.

\[\ce{H2O}(l)\rightleftharpoons\ce{H3O+}(aq)+\ce{OH-}(aq)\]

Solution:

Calculate the heat of reaction of the self ionization of water under 40℃ first.

40℃:

We can first recall the equation that links chemical equilibrium with thermodynamic explanations:

△G_rxn = △G^○+ RTlnQ_rxn

△G_rxn= change in Gibbs free energy due to the reaction

△G^○= change in Gibbs free energy of pure reactants going into pure products under equilibrium under standard condition

R= ideal gas constant= 8.314 J/mol·K = 0.082 L·atm/mol·K

T= temperature in Kelvin

Q_rxn= reaction quotient

We will then modify it since we are solving for the change in Gibbs free energy under equilibrium condition.

∵The solution has reached equilibrium,

∴△G_rxn=0

∴ 0=△G^○+ RTlnK

∴△G^○= -RTlnK

We will then apply this eqation to calculate the change in Gibbs free energy when the self ionization of water has reached equilibrium at 40℃ (313.15K)

△G^○= -RTlnK_w= -8.314J/mol·K x 313.15K x ln2.9x10^-14=8.12x10⁴J/mol

We will be using this equation for the following steps:

△G= △H- T△S

△G= change in Gibbs free energy

△H= change in enthalpy

△S= change in entropy

T= temperature in Kelvin

However, we will need to obtain the change in entropy of this reaction first. We extend the Hess law to:

△S_rxn= Σ△S_products - Σ△S_reactants

Also, we will use the data of entropy of formation for H₃O⁺(H⁺), OH^- and H₂O from this table to calculate the entropy of products and reactants.

Now we can continue calculating for the entropy of the reaction.

△S_rxn= Σ△S_products - Σ△S_reactants = [△S(H₃O⁺)+△S(OH^-)]-△S(H₂O)=(0-10.8J/mol·K)-70.0J/mol·K=-80.8J/mol·K

Then, from △G= △H- T△S we can obtain that:

△H=△G-T△S

So:

△H_rxn=△G_rxn-T△S_rxn=8.12x10⁴ J/mol - 313.15K x (-80.8 J/mol·K) = 1.07x10⁵J/mol

A positive result indicates that this reaction is endothermic.

Q15.1.X

Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products).

1. TlCl
2. BaF₂
3. Ag₂CrO₄
4. CaC₂O₄•H₂O
5. the mineral anglesite, PbSO₄

Answers:

Assume that all the reactions take place in standard condition.

a. First, write the dissolution equation of TlCl:

\[\ce{TlCl}(s)\rightleftharpoons\ce{Tl+}(aq)+\ce{Cl-}(aq)\]

The solubility product of this reaction should be:

K_sp=[Tl⁺][Cl^-]= 1.86 × 10⁻⁴

Assume the concentration of Tl⁺ to be S, that of Cl^- will also equal to S since whenever 1 mol of TlCl is dissociated it will produces 1 mol of Tl⁺ and 1 mol of Cl^-.

Hence, we can obtain that:

S²= 1.86 × 10⁻⁴

S= 1.36 x 10^-2

∴ Under saturation, the concentration of Tl⁺ in the solution is 1.36 x 10^-2 M, that of Cl^- is 1.36 x 10^-2 M.

b. The same process as that mentioned in a, we will write the dissolution equation of BaF₂ first.

\[\ce{BaF2}(s)\rightleftharpoons Ba^{2+} (aq)+ 2F^{-} (aq)\]

The solubility product of this reaction should be:

K_sp= [Ba²⁺][F^-]²= 1.84 × 10⁻⁷

Assume the concentration of Ba²⁺ to be S, then that of F^- would be 2S since whenever 1 mol of BaF₂ dissociates it will produce 1 mol of Ba²⁺ and 2 mol of F^-.

Hence, we can obtain that,

S·(2S)²= 1.84 × 10⁻⁷

S= 3.58 x 10^-3

∴Under saturation, the concentration of Ba²⁺ is 3.58 x 10^-3 M, and the concentration of F^- is 2 x 3.58 x 10^-3 =7.17 x 10^-3 M

c. The dissolution equation of Ag₂CrO₄ is:

\[\ce{Ag2CrO4}(s)\rightleftharpoons\ce{2Ag+}(aq)+CrO_4 ^{2-}(aq)\]

The solubility product of this reaction should be

K_sp= [Ag⁺]²[CrO₄^2-]= 1.12 × 10⁻¹²

Assume the concentration of CrO₄^2- to be S, then that of Ag⁺ will be 2S since whenever 1 mol of Ag₂CrO₄ dissolves it forms 2 mol of Ag⁺ and 1 mol of CrO₄^2- .

Hence, we can obtain that

S·(2S)²= 1.12 × 10⁻¹²

S= 6.54 x 10^-5

∴ Under saturation, the concentration of Ag⁺ is 6.54 x 10^-5 x 2=1.31x10^-4 M, and that of CrO₄^2- is 6.54 x 10^-5 M

d. The equation of dissociation of CaC₂O₄•H₂O is:

\[\ce{CaC2O4•H2O}(s)\rightleftharpoons Ca^{2+} (aq)+C_2 O_4 ^{2-} (aq)+\ce{H2O}(l)\]

The solubility product of this reaction is:

K_sp= [Ca²⁺][C₂O₄^2-]= 2.3 x 10^-9 (this data is not available in the table provided. This data is found on this site.)

Assume the concentration of Ca²⁺ is S, then that of C₂O₄^2- will also be S since whenever 1 mol of CaC₂O₄•H₂O is dissociated it forms 1 mol of Ca²⁺ and 1 mol of C₂O₄^2- .

Hence, we can obtain that

S·S=2.3 x 10^-9

S= 4.8 x 10^-5

∴Under saturation, the concentration of Ca²⁺ is 4.8 x 10^-5 M, and that of C₂O₄^2- is also 4.8 x 10^-5 M

e. The equation of dissociation of PbSO₄ is:

\[\ce{PbSO4}(s)\rightleftharpoons Pb^{2+}(aq)+SO_4 ^{2-} (aq)\]

The solubility product of this reaction is:

K_sp=[Pb²⁺][SO₄^2-]=1.7 x 10^-8 (this data is also obtained from the site mentioned in d.)

Assume the concentration of Pb²⁺ is S, then that of SO₄^2- would also be S since whenever 1 mol of PbSO₄ dissociates it produces 1 mol Pb²⁺ and 1 mol SO₄^2- .

S·S= 1.7 x 10^-8

S= 1.3 x 10^-4

∴Under saturation, the concentration of Pb²⁺ is 1.3 x 10^-4 M, that of SO₄^2- is also 1.3 x 10^-4 M.

Answers to Zehao's EC questions

5.3.16. Q=15.6 kJ

10.1.10. a. Because the type of interatomic/intermolecular forces between these two substances are different. Ne is dominated by London dispersion forces, while HF is dominated by dipole-dipole forces. These two interatomic/intermolecular forces have different strength, so they need absorb different amount of heat to overcome the intermolecular forces between particles and start boiling.

b. As the atomic mass increases, the substance's boiling point increases. However, as the molecular mass increases in the row from HF to HI, the boiling point decreases greatly first and then started to increases. For atomic gases, as the atomic mass increases, the electronic cloud around the nucleus becomes larger, hence increase the polarizability and hence increase the London dispersion forces. More heat would needed to overcome these forces, resulting in a higher boiling temperature.

The same reason work for that, when comparing their boiling point, HI ＞ HBr ＞HCl. HF has a even higher boiling point due to the stronger hydrogen bonding caused by the high elecronegativity of fluorine.

13.2.14.

a. Q=1.03 x 10^-3 reaction shift toward products.

b. Q=2.5 x 10³ reaction shift toward products.

c. Q=2 reaction shift toward reactants.

d. Q is not calculable. Reaction shift toward reactants.

e. Q=0 Reaction shift toward products.

f. Q=50 Reaction shift toward reactants.

14.1.8.

a. Conjugate base: HS^-

b. Conjugate base: HPO₄^2-

c. Conjugate acid: PH₄⁺

d. Conjugate base: S^2- Conjugate acid: H₂S

e. Conjugate acid: H₂SO₃ Conjugate base:SO₃^2-

f. Conjugate base: H₂O₂

g. Conjugate base: N₂H₃^-

h. Conjugate base: CH₃O^-

15.1.X. 1.9 x 10^-24M (K_sp=1.9 x 10^-33)