chapter 17 part 3. intermolecular forces vs. intramolecular forces

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Chapter 17 Part 3

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Page 1: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Chapter 17

Part 3

Page 2: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Intermolecular forces vs. Intermolecular forces vs. Intramolecular forcesIntramolecular forces

Page 3: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

What happens when the water What happens when the water evaporates?evaporates?

• Intermolecular attractions (van der Waals forces) are broken.

• Intramolecular attractions (bonds) are not broken.

Page 4: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

alcohol vs. water

How do the vapor pressures, rates of evaporation, and van der Waals forces compare?

How would the boiling point of alcohol compare to water?

Page 5: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Names Ethanol, Ethyl alcohol, rubbing alcohol

Molecular formula C2H6O Boiling point 78.4°C

Why is it called rubbing alcohol?

Page 6: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Substances with weaker Van der Waals Forces

1. Are easier to evaporate

2. Have higher vapor pressure

3. Be more volatile

4. Have lower boiling points

Page 7: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Equilibrium

• A state of equilibrium is the most stable state for a reversible system.

Page 8: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

LeChatlier’s Principle

• If a stress is placed on a system in equilibrium the system will tend to readjust so that the stress is reduced.

• 3 Stresses are changing the:– Concentration– Temperature– Pressure

• LeChatlier’s Principle = “Do the opposite”

Page 9: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

I smell gasoline

C8H18(l) ↔ C8H18(g)

Page 10: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

C8H18(l) ↔ C8H18(g)

• Can is sealed

• Contents exert certain pressure due to “vapor pressure”

• Can is opened

• Pressure decreases

• Reaction shifts to increase the pressure

• More liquid goes into the vapor state

Page 11: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

2NO2(g) ↔N2O4(g)brownbrown colorlesscolorless

How does applying LeChatlier’s Principle explain that this reaction is exothermic?

Rule: An increase in temperature will always shift a reaction in the endothermic direction.

What other rule could we use?

Note: The rule for temperature applies “do the opposite” already. So don’t do the opposite of the rule.

Page 12: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

H2O(l) H2O(g)

Water Vapor Pressure Table Temperature Pressure (°C) (mmHg)

Temperature Pressure (°C) (mmHg)

Temperature Pressure (°C) (mmHg)

0.0 4.6 5.0 6.5 10.0 9.2 12.5 10.9 15.0 12.8 15.5 13.2 16.0 13.6 16.5 14.1 17.0 14.5 17.5 15.0 18.0 15.5 18.5 16.0 19.9 16.5

19.5 17.0 20.0 17.5 20.5 18.1 21.0 18.6 21.5 19.2 22.0 19.8 22.5 20.4 23.0 21.1 23.5 21.7 24.0 22.4 24.5 23.1 25.0 23.8 26.0 25.2

27.0 26.7 28.0 28.3 29.0 30.0 30.0 31.8 35.0 42.2 40.0 55.3 50.0 92.5 60.0 149.4 70.0 233.7 80.0 355.1 90.0 525.8 95.0 633.9 100.0 760.0

Is the reaction above exothermic or endothermic?

Page 13: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Predict the effect of the following changes on the reaction:

2SO3(g) ↔ 2SO2 (g) + O2 (g) ∆H = +197.78 kJ

(a) Increasing the temperature of the reaction.

Page 14: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Predict the effect of the following changes on the reaction:

2SO3(g) ↔ 2SO2 (g) + O2 (g) ∆H = +197.78 kJ

(b) Increasing the pressure on the reaction.

Page 15: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Predict the effect of the following changes on the reaction:

2SO3(g) ↔ 2SO2 (g) + O2 (g) ∆H = +197.78 kJ

(c) Adding more O2.

Page 16: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Predict the effect of the following changes on the reaction:

2SO3(g) ↔ 2SO2 (g) + O2 (g) ∆H = +197.78 kJ

(d) Removing O2.

Page 17: Chapter 17 Part 3. Intermolecular forces vs. Intramolecular forces

Homework

Worksheet 1 Chapter 17.