Extra Credit 13

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Q16.4.6

Is the formation of ozone (O₃(g)) from oxygen (O₂(g)) spontaneous at room temperature under standard state conditions?

solution:

Based on the reaction formula:

$$3O_2(g)\rightarrow2O_3(g).$$

We can use the equation

$${\Delta G_{raction}=\Delta G_{f Product}-\Delta G_{fReactance}}$$

to calculate weather the reaction spontaneous.

Based on the data provided on T1 and T2, we can find the standard Gibbs free energy of O₂ and O₃. As a result, we can calculate the gibbs energy of overall reaction:

$${\Delta G_{raction}=\Delta G_{f Product}-\Delta G_{fReactance}=2*163.2kJ/mol-3*0kJ/mol=326.4kJ/mol}$$

Because the ΔG of the reaction is positive, the reaction cannot spontaneous at room temperature.

Q5.4.18

Both graphite and diamond burn.

$\ce{C}(s,\:\ce{diamond})+\ce{O2}(g)⟶\ce{CO2}(g)$

For the conversion of graphite to diamond:

$\ce{C}(s,\:\ce{graphite})⟶\ce{C}(s,\:\ce{diamond})\hspace{20px}ΔH^\circ_{298}=\mathrm{1.90\:kJ}$

Which produces more heat, the combustion of graphite or the combustion of diamond?

solution:

According to Hess's Law, we can rewrite the reaction of combustion of diamond to:

$${C_{s, diamond}+O_2(g)\rightarrow C_{s,graphite}+O_2(g)+\Delta H_{conversion}\rightarrow CO_2(g)+\Delta H_{conversion}+\Delta H_{graphite combustion}}$$

so the enthalpy of system during the combustion of diamond is more negative than the combustion of graphite, because the enthalpy change of diamond to graphite is negative. As a result, the combustion of diamond will produce more heat.

Q10.3.22

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)

solution:

Overall, we can separate the overall change to two part: the water reach the body temperature and these water absorb heat to convert to vapor.

We can calculate the enthalpy of vaporization of water from the equation

$${\Delta H_{vaporization}=\Delta H_{f H_2O(g)}-\Delta H_{f H_2O(l)}=-241.82 kJ/mol-(-285.83 kJ/mol)= 44.01 kJ/mol}$$

When we calculate the heat of convert 3.8 °C water to the vapor, we also need the heat absorbed that allow the water raise to body temperature, so we use the equation:

$$ {Q=c_{sp} * \Delta T * m_{H_2O}+ \Delta H_{vaporization} * n_{H_2O}} $$

Then we plug in the data from 15.1, T1, and T2:

Q= 4.18 J/(g * K)* (36.6 °C- 3.8°C)* 566.99 g/ 1000J/ kJ +44.01 kJ/mol * (566.99 g/ 18 g/ mol) = 1462.17 kJ

As a result, we need 1462.17 kJ to convert the water to vapor.

Q13.3.13

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

Write the expression for the equilibrium constant (K_c) for the reversible reaction \[\ce{Fe2O3}(s)+\ce{3H2}(g)\rightleftharpoons\ce{2Fe}(s)+\ce{3H2O}(g) \hspace{20px} ΔH=\mathrm{98.7\:kJ}\]
What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
What will happen to the concentration of each reactant and product at equilibrium if H₂O is removed?
What will happen to the concentration of each reactant and product at equilibrium if H₂ is added?
What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

solution:

a. $$ K_c = \left(\frac{[H_2O]^3}{[H_2]^3} \right) $$

b. Due to Le Chatelier's Principe, added Fe will shift the reaction to left, and therefore increase the concentration of H₂ and decrease the concentration of H₂O.

c. Due to Le Chatelier's Principe, added H₂O will shift the reaction to left, and therefore increase the concentration of H₂ and increase on the concentration of H₂O.

d.Due to Le Chatelier's Principe, added H₂ will shift the reaction to right, and therefore increase the concentration of H₂ and increase on the concentration of H₂O.

e.Due to Le Chatelier's Principe, added pressure will shift the reaction to the side which has less molecules , but the number of molecules in each side are equal to each other, so the decreasing volume cannot shift the reaction to either side, and therefore as the volume decrease, both concentration of H₂ and H₂O increasing.

f. As the ΔH of this reaction is positive, the reaction need to absorb heat from the surrounding. Increasing temperature will push the reaction to the left side and therefore increase the concentration of H₂O and decrease the concentration of H_2.

Q14.3.4

The odor of vinegar is due to the presence of acetic acid, CH₃CO₂H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.

solution:

[H₂O] > [CH₃CO₂H] > $\ce{[H3O+]}$ ≈ $\ce{[CH3CO2- ]}$ > [OH⁻]

Five Answer from Extra Credit 19

Q13.4.9

At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant K_P for the transformation at 60 °C?

$${H_2O(l)}⇌{H_2O(g)}$$

$\ce{K_p}= \left(\frac{P [H_2O]}{1}\right)$ = 0.196

Q14.3.25

Which of the following will increase the percent of NH₃ that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)?

addition of NaOHNaOH
addition of HClHCl
addition of NH4OH

a.addition of NaOH

Q14.4.1

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

Al(NO3)3Al(NO3)3
RbIRbI
KHCO2KHCO2
CH3NH3Br

a. Al(NO3)3 Acidic

b. RbI Neutral

c.$\ce{KHCO_2}$ Basic

d.$\ce{CH_3NH_3Br}$ Acidic

Q16.4.13

Calculate the equilibrium constant at the temperature given.

O2(g)+2F2(g)⟶2F2O(g)(T=100°C)
I2(s)+Br2(l)⟶2IBr(g)(T=0.0°C)
2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)(T=575°C)
N2O3(g)⟶NO(g)+NO2(g)(T=−10.0°C)
SnCl4(l)⟶SnCl4(g)(T=200°C)

1. Kc= 1.8*10^(-8)

2. Kc= 9.6*10^(-8)

that is too lot for calculate! I cannot continue calculate this problem. Sorry.

Q15.1.4

How do the concentrations of Pb²⁺ and S^2– change when K₂S is added to a saturated solution of PbS?

when adding the $\ce{K_2S}$ into the saturated solution of PbS, the more $\ce{S^{2-}}$ ions will shift the solvent reaction of PbS to the left side, which creat the PbS solids. As a reuslt, the concentration of $\ce{Pb^{2+}}$ ions decrease but the concentration of $\ce{S^{2-}}$ will not change.

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Q13.3.13

Q16.4.6

Q5.4.18

Q10.3.22

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Q14.3.4

Five Answer from Extra Credit 19

Q13.4.9

Q14.3.25

Q14.4.1

Q16.4.13

Q15.1.4

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