Extra Credit 20

Q10.4.5

Determine the phase changes that carbon dioxide undergoes as the pressure changes if the temperature is held at −50 °C? If the temperature is held at −40 °C? At 20 °C?

The pressure and temperature axes on this phase diagram of carbon dioxide are not drawn to constant scale in order to illustrate several important properties.

Q10.4.5 SOLUTION

\[\ce{CO2}(g)\rightleftharpoons\ce{CO2}(l)\] \[\ce{CO2}(l)\rightleftharpoons\ce{CO2}(s)\]

1. Since the temperature is being held at \(\ce{-50^{\circ}C}\), we can follow the first line (a.) up from the x-axis. At approximately 10 kPa, carbon dioxide is a gas and as the pressure increases, it remains a gas until it reaches the line between the gas and liquid phase (\(\ce{P_1}\)) at which point the carbon dioxide gas will condense into its liquid form. As the pressure continues to increase, the liquid carbon dioxide will eventually reach the line that divides the liquid and solid phase (\(\ce{P_4}\)) and the liquid carbon dioxide will turn into its solid form (dry ice).
2. For \(\ce{-40^{\circ}C}\), we can follow the second line (b.). At 10 kPa, carbon dioxide is in its gas form and when the pressure reaches \(\ce{P_2}\), the carbon dioxide will condense into its liquid form. The liquid carbon dioxide will eventually reach \(\ce{P_5}\) and become dry ice.
3. At \(\ce{20^{\circ}C}\), we will follow the third line (c.). Like the others, the carbon dioxide starts in its gas form at 10 kPa and condenses into a liquid at \(\ce{P_3}\). \(\ce{P_3}\) is a lot higher than \(\ce{P_1}\) and \(\ce{P_2}\), and this shows that at higher temperatures, it requires more pressure, and therefore more energy, to perform non-spontaneous phase changes. The liquid carbon dioxide will transform into solid carbon dioxide at \(\ce{P_6}\).

Q10.4.5 Answer

a. As you can see from picture in the solution. At −50 °C when pressure is below P1 it is gas. Above P1 and below P4 it is liquid. Above P4 it is solid.

b.As you can see from picture in the solution. At −40 °C when pressure is below P2 it is gas. Above P2 and below P5 it is liquid. Above P5 it is solid.

c.As you can see from picture in the solution. At 20 °C when pressure is below P3 it is gas. Above P3 and below P6 it is liquid. Above P6 it is solid.

Q13.3.15

Acetic acid is a weak acid that reacts with water according to this equation:

\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\]

Will any of the following increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion?

1. Addition of HCl
2. Addition of NaOH
3. Addition of NaCH₃CO₂

Q13.3.15 SOLUTION

1. Answer: No \[\ce{HCl}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{H3O+}(aq)+\ce{Cl-}(aq)\] When \(\ce{HCl}\) is added to the solution, it will react with the \(\ce{H2O}\) to form hydronium and chloride ions. Following Le Chatelier's Principle, the addition of hydronium ions in the solution would result in the reaction moving towards the left, creating more acetic acid. This is an example of the common ion effect where a substance that adds an ion already in the solution is dissolved into the solution.
2. Answer: Maybe \[\ce{NaOH}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{OH-}(aq)+\ce{Na+}(aq)+\ce{H2O}(l)\] When \(\ce{NaOH}\) is added to the solution, it creates sodium and hydroxide ions. Neither of these ions are in the original reaction so the common ion effect does not apply here. But the hydroxide ions will react with the hydronium ions that are already in the solution and create water. \[\ce{H3O+}(aq)+\ce{OH-}(aq)\rightleftharpoons2\ce{H2O}(l)\] The addition of water molecules in the solution will shift the reaction to the right and will increase the amount of acetic acid that reacts. However, this also depends on which of the reactant species is the limiting reactant because if the \(\ce{CH3CO2H}) was the limiting reactant, an increase in water molecules would not result in any changes in the acetic acid's ability to react.
3. Answer: No \[\ce{NaCH3CO2}(aq)+\ce{H2O}(l)\rightleftharpoons\ce{Na+}(aq)+\ce{CH3CO2-}(aq)+\ce{H2O}(l)\] When \(\ce{NaCH3CO2}\) is added to the solution, it dissociates into sodium and acetate ions. This is another example of the common ion effect where acetate ions are present in both the original reaction and the compound that was added. Le Chatelier's Principle dictates that with the addition of the acetate ions, the reaction will go towards the left and form more acetic acid.

Q13.3.15 Answer

a. no, will not increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion.

b. no, will not increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion.

c. Yes, will increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion.

Q14.3.7

Explain why the ionization constant, K_a, for HI is larger than the ionization constant for HF.

Q14.3.7 Answer

The ionization constant, K_a, for HI is larger than the ionization constant for HF. HI is a strong acid and HF is a weak acid. Strong acid always have larger K_a than weak acid, so the ionization constant,K_a, for HI is larger than the ionization constant for HF.

Q15.1.X

Iron concentrations greater than 5.4 × 10^–6 M in water used for laundry purposes can cause staining. What [OH^–] is required to reduce [Fe²⁺] to this level by precipitation of Fe(OH)₂?

Q15.1.X Answer

[OH^–]= 2.065× 10^–10M

Q16.4.10

Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.

1. \(\ce{Cl2}(g)+\ce{Br2}(g)⟶\ce{2BrCl}(g) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.7×10^{−2}\)
2. \(\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} \mathrm{T=500\:°C} \hspace{20px} K_p=48.2\)
3. \(\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} \mathrm{T=60\:°C} \hspace{20px} K_p=\mathrm{0.196\: atm}\)
4. \(\ce{CoO}(s)+\ce{CO}(g)\rightleftharpoons\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} \mathrm{T=550\:°C} \hspace{20px} K_p=4.90×10^2\))
5. \(\ce{CH3NH2}(aq)+\ce{H2O}(l)\rightarrow\ce{CH3NH3+}(aq)+\ce{OH-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=4.4×10^{−4}\)
6. \(\ce{PbI2}(s)⟶\ce{Pb^2+}(aq)+\ce{2I-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=8.7×10^{−9}\)

Q16.4.10 Answer

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(298K)ln(\(4.7×10^{−2}\))

ΔG°=7575.44J=7.58kJ

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(773K)ln(\(48.2\))

ΔG°=-24905.85J=-24.91kJ

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(333K)ln(\(0.196\))

ΔG°=4511.76J=4.51kJ

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(823K)ln(\(4.90×10^{2}\))

ΔG°=-42384.74J=-42.38kJ

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(298K)ln(\(4.4×10^{−4}\))

ΔG°=19148.5J=19.15kJ

ΔG° =-RTln(K) ΔG° =-(8.314J/(K × Mol)(298K)ln(\(8.7×10^{−9}\))

ΔG°=45983.59J=45.98kJ

ANSWERS (PS 4)

Q5.3.16 ANSWER

q = 15.6 kJ

Q13.2.14 ANSWER

1. \(Q=1.0×10^{-3}\) The reaction will go towards the products (right).
2. \(Q=2.5×10^3\) The reaction will go towards the products (right).
3. \(Q=2\) The reaction will go towards the reactants (left).
4. \(Q=\infty\) The reaction will go towards the reactants (left).
5. \(Q=0\) The reaction will go towards the products (right).
6. \(Q=50\) The reaction will go towards the reactants (left).

Q14.1.8 ANSWER

1. conjugate acid = \(\ce{H3S+}\) conjugate base = \(\ce{HS-}\)
2. conjugate acid = \(\ce{H3PO4}\) conjugate base = \(\ce{HPO4^2-}\)
3. conjugate acid = \(\ce{PH4+}\) conjugate base = \(\ce{PH2-}\)
4. conjugate acid = \(\ce{H2S}\) conjugate base = \(\ce{S^2-}\)
5. conjugate acid = \(\ce{H2SO3}\) conjugate base = \(\ce{SO3^2-}\)
6. conjugate acid = \(\ce{H4O2^2+}\) conjugate base = \(\ce{H2O2}\)
7. conjugate acid = \(\ce{H5N2+}\) conjugate base = \(\ce{H3N2-}\)
8. conjugate acid = \(\ce{CH3OH2+}\) conjugate base = \(\ce{CH3O-}\)