Extra Credit 5

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Q16.2.11

Predict the sign of the entropy change for the following processes:

An ice cube is warmed to near its melting point.
Exhaled breath forms fog on a cold morning.
Snow melts.

Solution:

Positive \((+) \Delta S \)
1. If you increase the temperature of a solid, you're increasing the degrees of vibrational freedom, thus increasing entropy.
Negative \((-) \Delta S \)
1. The hot air condenses as it interacts with the cold morning air, creating fog. As the phase changes from gas to liquid, the degrees of translational freedom, as well as rotational and vibrational, are decreased, meaning a decrease in entropy.
Positive \((+) \Delta S \)
1. As snow melts, the phase changes from a solid to a liquid phase. Liquids have more degrees of translational, and rotational freedom than solids.

Q5.3.18

One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 × 10³ pounds).

Q10.1.12

The molecular mass of butanol, C₄H₉OH, is 74.14; that of ethylene glycol, CH₂(OH)CH₂OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference.

Solution:

The usual trend (that as molar mass increases, the boiling point will also increase) does not apply in this case. If you look at the molecular structures of both molecules, you see that ethylene glycol has two OH bonds versus butanol's one OH bond. When oxygen is bonded to a hydrogen, the large bond dipole and short dipole-dipole distance leads to a strong interaction called hydrogen bonding (applied to nitrogen, oxygen, and fluorine.) Since ethylene glycol has twice the number of hydrogen bonds, it will take more energy to break the interactions caused by them than it will for butanol, thus increasing the boiling point.

Q13.2.16

Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?

\[ SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g) \]

[SO₂Cl₂] = 0.12 M, [Cl₂] = 0.16 M and [SO₂] = 0.050 M. K_c for the reaction is 0.078.

Solution:

Step 1: Calculate Q (reaction quotient) using the values given for that point in time.

\( Q = \dfrac{[SO_2][Cl_2]}{[SO_2Cl_2]} = \dfrac{[0.05][0.26]}{[0.12]} = .067\)

Step 2: Compare Q to the given K_C value.

\( Q < K_c \)

Step 3: Interpret

Since \( Q < K_c \), the reaction will shift towards the reactants, or in the reverse direction in order to establish equilibrium.

Final Answer:

Is the system at equilibrium? No, because \( Q \neq K_c \)

In which direction will the system need to shift? In order to increase Q so as to reach the K_C value, the reaction will shift to the left so as to create more reactants, or go in the reverse direction.

Q14.1.10

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

\(\ce{NO2- + H2O ⟶ HNO2 + OH-}\)
\(\ce{HBr + H2O ⟶ H3O+ + Br-}\)
\(\ce{HS- + H2O ⟶ H2S + OH-}\)
\(\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}\)
\(\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}\)
\(\ce{[Fe(H2O)5(OH)]^2+ + [Al(H2O)6]^3+ ⟶ [Fe(H2O)6]^3+ + [Al(H2O)5(OH)]^2+}\)
\(\ce{CH3OH + H- ⟶ CH3O- + H2}\)

Solution:

Brønsted-Lowry acid: donates a proton (\(H^+)\)

Brønsted-Lowry base: accepts a proton

Conjugate Base: accepts a proton in reverse reaction

Conjugate Acid: donates proton in reverse reaction

	Brønsted-Lowry Acid	Brønsted-Lowry Base	Conjugate Acid	Conjugate Base
a.	\( H_2O \)	\( NO_2 \)	\( HNO_2 \)	\( OH^- \)
b.	\( HBr \)	\( H_2O \)	\( H_3O^+ \)	\( Br^- \)
c.	\( H_2O \)	\( HS^- \)	\( H_2S \)	\( OH^- \)
d.	\( H_2PO_4 \)	\( OH^- \)	\( H_2O \)	\( HPO_4 \)
e.	\( HCl \)	\( H_2PO_4 \)	\( H_3PO_4 \)	\( Cl^- \)
f.	\(\ce{[Al(H2O)6]^3+ }\)	\(\ce{[Fe(H2O)5(OH)]^2+ }\)	\(\ce{ [Fe(H2O)6]^3+ }\)	\( \ce{[Al(H2O)5(OH)]^2+} \)
g.	\( CH_3OH \)	\( H^- \)	\( H_2 \)	\( CH_3O^- \)