Chemistry 1B Notes (7)

Topic: Lecture 7 - Some Review Problems, Polymers, Structure and Phase Stability

Problem Statement:
- Historical context: The average amount of water in the atmosphere was estimated to be $3.75 imes 10^{13}$ gallons when the average surface temperature was 60 °F (around the year 1800).
- Current context: The average surface temperature is rising toward 65 °F.
- The problem asks about the change in humidity in the atmosphere, specifically the additional volume of water that will accumulate due to the temperature increase.
Key Concept:
- The change in atmospheric water is related to the change in vapor pressure of water, which is influenced by temperature. The water in the atmosphere is derived from the vapor coming from ice and water on the ground.

Temperature Conversion Steps:
- Convert 60 °F to Kelvin:
  $60 - 32 \cdot \frac{5}{9} + 273.15 = 288.706 K$
- Convert 65 °F to Kelvin:
  $65 - 32 \cdot \frac{5}{9} + 273.15 = 291.483 K$

Description of Concept:
- An increase in temperature leads to an increase in vapor pressure. Consequently, this results in a higher capacity for the atmosphere to hold water.
Mathematical Expression:
- The relationship between vapor pressures at different temperatures is expressed by:
 $ext{ln} \left( \frac{p1}{p2} \right) = - \frac{\Delta H{vap}}{R} \left( \frac{1}{T1} - \frac{1}{T_2} \right)$
- Where:
 - $p_1$ = vapor pressure at the higher temperature
 - $p_2$ = vapor pressure at the lower temperature
 - $\Delta H_{vap} = 40.63 \times 10^3 J/mol$
 - $R = 8.31447 J/(K \, mol)$
 - $T1$ and $T2$ = temperatures in Kelvin

Substituting Values:
- Insert values into the equation:
 $\text{ln} \left( \frac{p1}{p2} \right) = - \frac{40.63 \times 10^3}{8.31447} \left( \frac{1}{291.483} - \frac{1}{288.706} \right)$
- Result:
 $\text{ln} \left( \frac{p1}{p2} \right) = +0.161257$
Final Calculation:
- Therefore:
 $\frac{p1}{p2} = e^{0.161257} = 1.175$

Understanding the Result:
- This denotes that $p1$ (vapor pressure at 65 °F) is approximately 17.5% larger than $p2$ (vapor pressure at 60 °F).
Calculating Additional Water:
- Since the additional water in the atmosphere is directly proportional to the vapor pressure increase, calculate additional water volume:
  $0.175 \times 3.75 \times 10^{13} \text{ gal} = 6.56 \times 10^{12} \text{ gal}$
- Approximate result: 7 million billion extra gallons of water.
Consequences of Increased Water:
- The influx of additional water can lead to extreme weather conditions, including wider temperature swings and increased potential for flooding.

Understanding Dimethyl Ether:
- Chemical formula: H₃C-O-CH₃
- Task: Draw the Lewis structure, indicating all lone pairs of electrons and formal charges.
Hydrogen Bonding Potential:
- Query: Which of the following molecules can hydrogen bond with dimethyl ether?
- a) HCOOH (formic acid)
- b) H₃C-O-H (methanol)
- c) H₃C-S-H
- d) HF (hydrogen fluoride)
- e) H₂O (water)
- f) CO₂
- g) PH₃
- h) CH₃-NH₂ (dimethyl amine)

Criteria for Hydrogen Bonding:
- A hydrogen atom must be bonded with either N, O, or F (the hydrogen bond donor).
- The partner atom must possess a lone pair of electrons and must also be N, O, or F (the hydrogen bond acceptor).
Observation for Dimethyl Ether:
- Dimethyl ether contains lone pairs on O but lacks hydrogen atoms bonded to N, O, or F.
Identified Candidates for Hydrogen Bonding:
- Possible candidates:
- HCOOH, H₃C-O-H, HF, H₂O, and CH₃-NH₂ can form hydrogen bonds with dimethyl ether.
- Exclusions:
- H₃C-S-H, CO₂, and PH₃ cannot form hydrogen bonds with dimethyl ether since they do not have the requisite hydrogen atoms bonded to N, O, or F.

Concept: Porous nickel, referred to as metallic wood.
Process Description:
- Step 1: Initially, plastic nanospheres are placed in pure water and allowed to evaporate, leading to a nearly perfect packing arrangement.
- Step 2: Electrochemistry is then applied to deposit nickel metal within the voids left by the evaporated plastic.
- Step 3: Finally, the plastic is dissolved, leaving behind a porous structure of nickel.