Extra Credit 12

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Q5.4.16

Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(l) and O₂(g) under standard conditions?

Answer:

$$\ce{HgO}(s)\rightleftharpoons\ce{Hg}(l)+\ce{\frac{1}{2}O2}(g) $$

$$\ce{\Delta H_{rxn}}=\ce{\Delta H_{Hg}}+\ce{\frac{1}{2}\Delta H_{O2}}-\ce{\Delta H_{HgO}} $$

$$\ce{\Delta H_{rxn}}=\ce{0}+\ce{0}-\mathrm{(-90.8\:kJ/mol)}=\mathrm{90.8\:kJ/mol}$$

$$\ce{Q}=\ce{n\Delta H_{rxn}}=\ce{1mol}\times {90.8\:kJ/mol}=\mathrm{90.8\:kJ}$$

You can find the data you need in T1 and T2.

Q10.3.21

Which contains the compounds listed correctly in order of increasing boiling points?

N₂ < CS₂ < H₂O < KCl
H₂O < N₂ < CS₂ < KCl
N₂ < KCl < CS₂ < H₂O
CS₂ < N₂ < KCl < H₂O
KCl < H₂O < CS₂ < N₂

Answer:

$$\ce {a. N2 < CS2 < H2O < KCl}$$

The boiling point depends on the intermolecular forces (IMF) of the substances. The stronger the force between molecules, the higher the boiling poinnt is. $\ce{KCl}$ is ionic solid. The IMF between molecules is ionic bonding. The IMF of $\ce{H2O}$ molecules is mainly hydrogen bonding. The IMF between $\ce{KCl}$ and $\ce{CS2}$ are both London dispersion force. Ionic bonding is stronger than the hydrogen bonding, and hydrogen bonding is stronger than the London dispersion force. And the bigger the molecule is and the more electrons molecule has, the stronger the London dispersion force is; so the IMF of $\ce{CS2}$ is stronger than the IMF of $\ce{N2}$. Therefore, we can find the strength of IMF of those substance in the following order: $\ce{N2}$ < $\ce{CS2}$ < $\ce{H2O}$ < $\ce{KCl}$. So we can find the relationship of the boiling point should be a.

Q13.3.9

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

$\ce{2H2O}(g)\rightleftharpoons\ce{2H2}(g)+\ce{O2}(g) \hspace{20px} ΔH=\ce{484\:kJ}$
$\ce{N2}(g)+\ce{3H2}(g)\rightleftharpoons\ce{2NH3}(g) \hspace{20px} ΔH=\mathrm{-92.2\:kJ}$
$\ce{2Br}(g)\rightleftharpoons\ce{Br2}(g) \hspace{20px} ΔH=\mathrm{-224\:kJ}$
$\ce{H2}(g)+\ce{I2}(s)\rightleftharpoons\ce{2HI}(g) \hspace{20px} ΔH=\ce{53\:kJ}$

Answer:

a. Increase T: more products. Decrease volume: more reactants.

b. Increase T: more reactants. Decrease volume: more products.

c. Increase T: more reactants. Decrease volume: more products.

d. Increase T: more products. Decrease volume: more reactants.

When the temperature increases, the equilibrium constant K will increase for the endothermic reaction ( $\ce\Delta H > 0$ ) and will decrease for the exothermic reaction ( $\ce\Delta H < 0$ ). The increase in K will shift the reaction to products and the decrease in K will shift the reaction to the reactants.

When the volume of the reaction vessel decreases, the equilibrium constant K will increase for the reaction that has more moles of gases in the reactants for 1 mole reaction and will decrease for the reaction that has more moles of gases in the products for 1 mole reaction. The increase in K will shift the reaction to products and the decrease in K will shift the reaction to the reactants.

Q14.3.3

Use this list of important industrial compounds (and Figure) to answer the following questions regarding: CaO, Ca(OH)₂, CH₃CO₂H, CO_2, HCl, H₂CO₃, HF, HNO₂, HNO₃, H₃PO₄, H₂SO₄, NH₃, NaOH, Na₂CO₃.

Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
List those compounds in that can behave as Brønsted-Lowry acids with strengths lying between those of H₃O⁺ and H₂O.
List those compounds in that can behave as Brønsted-Lowry bases with strengths lying between those of H₂O and OH⁻.

Answer:

a. Strong Brønsted-Lowry acids: $\ce{HCl, HNO3, H2SO4}$. Strong Brønsted-Lowry bases: $\ce{Ca(OH)2, NaOH}$

b. $\ce{H3O+, H2SO4, HCl, HNO3, H3PO4, HF, HNO2, CH3CO2H, H2CO3, CO2, H2O }$

c. $\ce{H2O, NH3, Na2CO3, CaO, Ca(OH)2, NaOH, OH-}$

Q15.1.X

Calculate the concentration of Sr²⁺ when SrF₂ starts to precipitate from a solution that is 0.0025 M in F^–.

for the endothermic reaction ( $\ce\Delta H > 0$ )

Answer:

$$\ce{K_{sp} = [Sr^2+] * [F^{-}]^2} $$

$$\ce{[Sr^2+] = \frac{K_{sp}}{[F^{-}]^2}} $$

$$\ce{[Sr^2+]} = \mathrm{\frac{4.33 * 10^{-9}}{(0.0025)^2}} $$

$$\ce{[Sr^2+]} = \mathrm{6.928 * 10^{-4} M}$$

You can find the data you need here.

Five answers for Extra Credit 15

Q14.3.5

$$\ce{[H2O] > [NH3] > [OH^{-}] > [NH4+] > [H3O+]} $$

Q15.1.X

$$\ce{[F^{-}]} = \mathrm{5.91*10^{-5}M} $$

Q16.4.8

$$ b.Hydrazine: \ce{ N2H4 (g)} ⟶ \ce{N2(g) + 2H2(g)} $$

Q5.4.20

$$ \ce{OS(s) + 2O2(g)} ⟶ \ce{OsO4(s)} \text{will produces more heat.}$$

Q10.4.3

$$\ce {ln (\frac{P1}{P2}) = \frac{\Delta H}{R} * (\frac{1}{T_2} - \frac{1}{T_1})} $$

$$\ce{ln} (\mathrm{\frac{1atm}{3.4atm}}) = \frac{\mathrm{40650J/mol}}{\mathrm{8.314J/(K\ce{*}mol)}} \ce{*} (\frac{1}{T_2} - \frac{1}{373.15K}) $$

$$\ce {T_2 = \mathrm{411.59K}}$$

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Q5.4.16

Q10.3.21

Q13.3.9

Q14.3.3

Q15.1.X