Extra Credit 23

Q16.4.16

The standard free energy change is \(ΔG^\circ_{298}=−RT\ln K=\mathrm{4.84\: kJ/mol}\). When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and \(ΔG_{298}\) becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.

Q15.1.X

The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:

1. AgBr: [Ag⁺] = 5.7 × 10^–7 M, [Br^–] = 5.7 × 10^–7 M
2. CaCO₃: [Ca²⁺] = 5.3 × 10^–3 M, \(\ce{[CO3^2- ]}\) = 9.0 × 10^–7 M
3. PbF₂: [Pb²⁺] = 2.1 × 10^–3 M, [F^–] = 4.2 × 10^–3 M
4. Ag₂CrO₄: [Ag⁺] = 5.3 × 10^–5 M, 3.2 × 10^–3 M
5. InF₃: [In³⁺] = 2.3 × 10^–3 M, [F^–] = 7.0 × 10^–3 M

Solutions:

For the following <-> equates to left and right harpoons

(a) AgBr(s) <-> Ag+ (aq) + Br- (aq)

Ksp = [Ag+][Br-]

Ksp = [5.7x10^-7][5.7x10^-7] = 3.25x10^-13

(b) CaCO3(s) <-> Ca2+(aq) + Co3^2-(aq)

Ksp = [Ca2+][Co3^2-]

Ksp = [5.3 × 10^–3][9.0 × 10^–7] = 4.77x10^-9

(c) PbF2(s) <-> Pb2+(aq) + 2F-(aq)

Ksp = [Pb2+][F-]^2

Ksp = [2.1x10^-3][4.2x10^-3]^2 = 3.7x10^-8

(d) Ag2CrO4(s) <-> 2Ag+(aq) + CrO4^- (aq)

Ksp = [Ag+]^2[CrO4^-]

Ksp = [5.3x10^-5]^2[3.2x10^-3] = 8.9x1-^12

(e) InF3(s) <-> In3+(aq) + 3F-(aq)

Ksp = [In3+][F-]^3

Ksp = [2.3x10^-3][7.0x10^-3]^3 = 7.89x10^-28

Q14.6.1

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 × 10⁻⁵ as K_a for acetic acid.

1. What is the pH of the solution?
2. Is the solution acidic or basic?

Solutions:

"<->" equates to left and right harpoons

1) CH3CO2H+(aq) + H2O(l) <-> H3O+(aq) + CH3CO2- (aq)

Ka = ([CH3CO2-][H3O+])/[CH3CO2H+]

Ka = 1.8x10^-5

Because the problem says that the buffer solution is prepared of equal volumes of the two substances we can do the following assumption. In 1 Liter of the final solution of acetic acid and sodium acetate there is 0.5 Liters of acetic acid and 0.5 Liters of sodium acetate. Because of this assumption we must then do the following calculations.

Acetic Acid = (0.200 M) x (0.5L/1L) = 0.100 M

Sodium Acetate = (0.600 M) x (0.5L/1L) = 0.300 M

Using these newly calculated values we can now use an ICE table to obtain the equilibrium concentrations which will then allow us to obtain the pH

	CH3CO2H+	H3O+	CH3CO2-
Initial Concentration (M)	0.1	0	0.3
Change (M)	-x	x	x
Equilibrium (M)	0.1-x	x	0.3+x

Using, Ka = ([CH3CO2-][H3O+])/[CH3CO2H+], we can substitute our values from the ICE table to obtain...

1.8x10^-5 = [(x)(0.3+x)]/(0.1-x)

We can assume that the value of x will be very small and then simplify this equation to

1.8x10^-5 = [(x)(0.3)]/(0.1)

x = 6.0x10^-6 M

The assumption that we made that x is small is valid as evident by the fact that the value for x is less than 5% of 0.1 as well as less than 0.3

Using our value for x we can now solve for the pH

pH = -log(H3O+)

pH = -log (x)

pH = -log(6.0x10^-6)

pH = 5.222

2) Because the question does not implicitly tell us that the reaction takes place at a certain temperature, we can assume that it takes place at room temperature. At room temperature, a pH above 7 is basic, and below 7 is acidic. Since 5.222 is less than 7, the solution is acidic.

Q13.3.17

How can the pressure of water vapor be increased in the following equilibrium?

\[\ce{H2O}(l)\rightleftharpoons\ce{H2O}(g) \hspace{20px} ΔH=\ce{41\:kJ}\]

During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).

1. Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L.
2. How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C₃H₈: density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO₂(g) and H₂O(l)] and the furnace used to burn the LPG has the same efficiency as the gas furnace.
3. What mass of carbon dioxide is produced by combustion of the methane used to heat the house?
4. What mass of water is produced by combustion of the methane used to heat the house?
5. What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L.
6. How many kilowatt–hours (1 kWh = 3.6 × 10⁶ J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house.
7. Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?

Solutions:

(a) Amount of heat provided by Methane can be found using the following equation

q = (3500 kWh) x [(3.6x10^6 J)/(1 kWH)] x (1/.89) = 1.42x10^7 kJ

Using the heat of combustion of methane, we can find the number of moles of methane required in the following step

# mols methane = 1.42x10^7 kJ/890.8 kJ/mol = 15940 mol

Using the molar mass of methane (16.04 g/mol), we can convert 15940 mol into 255700 g.

Using the density of methane (0.681 g/L) and the mass of methane (255700 g) we can find the volume through the equation, Volume = Mass/Density

Volume = 255700 g/ 0.681 g/L

Volume = 375500 L

The next step converts L into cubic feet

375500 L x (1 cubic foot/28.316 L) = 13260 cubic feet

(b) Using the heat of combustion of propane and heat produced by the combustion of gas calculated earlier, we can find the number of moles of propane required in the following step

# mols propane = 1.42x10^7 kJ/2219 kJ/mol =6399 mol

Using the molar mass of propane (44.1 g/mol) we can convert 6399 mol into 28200 g.

Using the density of propane (0.5318 g/mL) and the mass of methane (28200 g) we can find the volume through the equation, Volume = Mass/Density

Volume = 28200 g/ 0.5318 g/mL

Volume = 53.07 L

The next step converts L into gallons

53.07 L x (1 gallon/ 0.37852 L) = 141.2 gallons

(c) Using the reaction of the combustion of methane, it is evident that the number of moles of methane is equivalent to the number of moles of carbon dioxide used in this reaction. As a result, we can use the calculated value of 15940 mol and multiply it by the molar mass of CO2 (44.01) to obtain the mass of CO2 produced in the combustion of methane.

15940 mol x 44.01 g/mol = 701.52 kg

(d) Using the reaction of the combustion of methane, it is evident that the number of moles of water is two times as much as the number of moles of methane in this reaction. As a result, we can use the calculated value of 15940 mol and double it. Then we can multiply it by the molar mass of H2O (18.02) to obtain the mass of H2O produced in the combustion of methane.

15940 mol x 2 x 18.02 g/mol = 574.32 kg

(e) Using the reaction of the combustion of methane, it is evident that the number of moles of O2 is two times as much as the number of moles of methane in this reaction. As a result, we can use the calculated value of 15940 mol and double it. Then we can multiply it by the molar mass of O2 (32.0) to obtain the mass of O2 used in the combustion of methane.

15940 mol x 2 x 32.0 g/mol = 1020160 g

Since only 23% of air is oxygen we must then divide 1020160 by 0.23 to obtain 4435478.3 g

Then using the density of air (1.22 g/L) and the mass of air (4435478.3 g) we can find the volume through the equation, Volume = Mass/Density

Volume = 4435478.3 g/ 1.22 g/L = 3635637.92 L

(f) Because electricity is 100% efficient, the heat required will be the same 3500 kWh

(g) (40% efficient) q = 3500 kWh/0.4 = 8750 kWh

Mass of coal required: 8750 kWh/2.26 kWh/lb = 3872 lb

3872 lb x 0.454 kg/lb = 1756.34 kg

Q11.5.13

Calculate the molality of each of the following solutions:

1. 583 g of H₂SO₄ in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 × 10² g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, C₁₈H₂₁NO₃, in 125.5 g of ethanol, C₂H₅OH
4. 25 g of I₂ in 125 g of ethanol, C₂H₅OH

Solutions:

(a) To find the moles of H2SO4, divide the mass(583 g) by the molar mass (98.079 g/mol) to obtain 5.944 mol. The mass of the solvent (water) is 1.50 kg

Molality = mols of solute/mass of solvent

Molality = 5.944/150

Molality = 3.96m

(b) To find the moles of NaCl, divide the mass (0.86 g) by the molar mass (58.4 g/mol) to obtain 0.0147 mol. The mass of the solvent (water) is .1 kg

Molality = mols of solute/mass of solvent

Molality = 0.0147/.1

Molality = 0.147m

(c) To find the moles of codeine, divide the mass (46.85 g) by the molar mass (299.364 g/mol) to obtain 0.156 mol. The mass of the solvent (ethanol) is .1255 kg

Molality = mols of solute/mass of solvent

Molality = 0.156/.1255

Molality = 1.247m

(d) To find the moles of I2, divide the mass (25 g) by the molar mass (253.8089 g/mol) to obtain 0.09849 mol. The mass of the solvent (ethanol) is .125 kg

Molality = mols of solute/mass of solvent

Molality = 0.09849/.125

Molality = 0.7880m