Week 3 Lecture 5

Attendance and Responsibility

• Students are responsible for staying informed about class material regardless of attendance.

  • If absent, students should find out what was covered, including assignments and announcements.

  • Proactive communication with the instructor is encouraged for any missed information.

Upcoming Test and Expectations

• Reminder that this is the third week of the semester.

  • A test will be scheduled soon, and it is crucial for students to prepare adequately.

Ethanol Phase Changes and Energy Calculations

• State and properties of ethanol:

  • Melting point, boiling point, heat of fusion, heat of vaporization, specific heat capacities are key focus areas.

• Energy calculations during phase changes:

  • Determine energy changes required during solid to liquid and gas phases.

  • Break down calculations into four stages:

Stage 1: Heating Solid Ethanol

• Initial Conditions:

  • Starting temperature: -125°C to -114°C.

• Energy Calculation:

  • Formula: $Q<em>1 = m c</em>s \Delta t$

  • Where:

    • m = mass (46 g)

    • c_s = specific heat of solid ethanol (0.97 J/g°C)

    • $\Delta t = -114 - (-125) = 11$°C.

• Substitute into formula:

  • $Q_1 = 46 imes 0.97 imes 11$

Stage 2: Melting of Ethanol

• Phase Transition: Solid to liquid (melting point).

  • Total moles of ethanol involve using moles =grams/molar mass = $\frac{46}{46} = 1$ mole.

• Energy Calculation for Fusion:

  • Heat of fusion = 5.2 kJ/mol.

• Energy required:

  • $Q_2 = 1 ext{ mole} imes 5.2 ext{ kJ/mol} = 5.2 ext{ kJ}$.

Stage 3: Heating Liquid Ethanol

• Conditions: Heat from -114°C to a higher temperature (assumed 78°C).

• Energy Calculation:

  • Formula: $Q<em>3 = m c</em>l \Delta t$

  • Where:

    • c_l = specific heat of liquid ethanol (2.3 J/g°C)

    • $\Delta t = 78 - (-114) = 192$°C.

• Substitute into formula:

  • $Q_3 = 46 imes 2.3 imes 192$

Stage 4: Vaporization of Ethanol

• Phase Transition: Liquid to gas.

• Energy Calculation for Vaporization:

  • Heat of vaporization = 38.56 kJ/mol.

• Energy required:

  • $Q_4 = 1 ext{ mole} imes 38.56 ext{ kJ/mol} = 38.56 ext{ kJ}$.

Total Energy Required

• To find the total energy required for the entire process, sum all energies:

  • $Q<em>{total} = Q</em>1 + Q<em>2 + Q</em>3 + Q_4$.

• Ensure proper unit conversions, especially between joules and kilojoules when necessary.

Vapor Pressure Concepts

Effect of Solutes on Vapor Pressure

• Vapor Pressure of Solutions:

  • The pressure above a solution is lower than that of the pure solvent due to solute addition.

• General formula:

  • $P<em>{solution} = X</em>{solvent} imes P^{ ext{°}}_{solvent}$

  • Where:

    • $X_{solvent}$ = mole fraction of the solvent.

• Example Calculation:

  • 90 grams of water (approximately 5 moles) with 18 grams of sugar.

  • Pressure calculation at 25°C leading to approximate values obtained from standard tables.

• As more solute is added, the vapor pressure continues to decrease.

Comparison of Different Solutions

• Analyze three solutions:

  • 0.1M sodium chloride (A)

  • 0.1M calcium chloride (B)

  • 0.1M sugar (C)

• Determine which has the highest and lowest vapor pressure based on particle contribution:

  • Sodium chloride disassociates into 2 ions, calcium chloride into 3 ions, and sugar provides 1.

Colligative Properties

Boiling Point Elevation and Freezing Point Depression

• Boiling Point Elevation (BPE): The application of solute increases the boiling point.

• Freezing Point Depression (FPD): The addition of solute lowers the freezing point.

• Equations:

  • Boiling point elevation: $\Delta T<em>b = K</em>b imes m$

  • Freezing point depression: $\Delta T<em>f = K</em>f imes m$

  • Where m = molality, Kb and Kf depend on the solvent.

• Example Problem: Sodium chloride's impact on freezing point versus a non-dissociating solute like sugar.

Osmotic Pressure

• Definition: The pressure needed to prevent the flow of solvent into a solution through a semipermeable membrane.

• Relation to Solute Concentration:

  • Higher concentrations of solute result in higher osmotic pressure.

  • Equation for osmotic pressure: $\Pi = CRT$

  • Where C = molarity, R = gas constant, T = temperature in Kelvin.

• Practical Application: Osmotic pressure is particularly relevant in biological systems and the functioning of cells in varying solute environments.

Problem Solving for Lab Assignments

• Example of determining molecular weight from depression points in freezing point experiments with lauric acid and benzoic acid.

  • Monitor temperature changes to assess points of melting and freezing, collect data and apply to given equations to solve for molecular weights.

• Reminder: Regular practice with assignments and prior material will assist in grasping these concepts.