Physical Chemistry Study Packet Flashcards

Physical Constants and Solvent Properties

• Universal Gas Constant ($R$):

  • $R = 8.314\,L\cdot kPa/mole\cdot K$

  • $R = 0.008314\,kJ/mole\cdot K$

• Solvent Properties ($K_f$ and $K_b$):

  • Water: $K_f = 1.86\,^{\circ}C/m$; $K_b = 0.520\,^{\circ}C/m$

  • Benzene: $K_f = 5.12\,^{\circ}C/m$; $K_b = 2.53\,^{\circ}C/m$

  • Ethanol: $K_f = 1.99\,^{\circ}C/m$; $K_b = 1.22\,^{\circ}C/m$

  • Acetic acid: $K_f = 3.90\,^{\circ}C/m$; $K_b = 2.93\,^{\circ}C/m$

  • Cyclohexane: $K_f = 20.0\,^{\circ}C/m$; $K_b = 2.79\,^{\circ}C/m$

• Electrolyte van’t Hoff Factors ($i$):

  • $HCl$: $1.9$

  • $NaCl$: $1.9$

  • $MgSO_4$: $1.3$

  • $MgCl_2$: $2.7$

  • $FeCl_3$: $3.4$

• Vapor Pressure of Solvents at $25\,^{\circ}C$ ($kPa$):

  • Acetone: $30.8\,kPa$

  • Benzene: $12.7\,kPa$

  • Cyclohexane: $13.0\,kPa$

  • Ethanol: $7.87\,kPa$

  • Methanol: $16.9\,kPa$

  • Toluene: $3.79\,kPa$

Solutions and Intramolecular Forces

• Key Definitions:

  • Adhesion: Attractive forces occurring between dissimilar atoms.

  • Cohesion: Attractive forces occurring between similar atoms.

  • Viscosity: The resistance of a fluid to flow.

  • Capillary Action: Caused by the difference between adhesive and cohesive forces.

• Relationship with Increasing Intramolecular Forces:

  • Vapor Pressure: Decreases as forces increase.

  • Viscosity: Increases as forces increase.

  • Surface Tension: Increases as forces increase.

  • Boiling Point: Increases as forces increase.

  • Melting Point: Increases as forces increase.

Colligative Properties: Freezing and Boiling Point Changes

• Freezing Point Change ($\Delta T_f$) Data:

  • $0.100\,m\,HCl$: Water: $0.353$; Benzene: $0.973$; Ethanol: $0.3781$

  • $0.200\,m\,NaCl$: Water: $0.707$; Benzene: $1.95$; Ethanol: $0.756$

  • $0.300\,m\,MgSO_4$: Water: $0.725$; Benzene: $2.00$; Ethanol: $0.776$

• Boiling Point Change ($\Delta T_b$) Data:

  • $0.100\,m\,HCl$: Water: $0.0988$; Benzene: $0.481$; Ethanol: $0.232$

  • $0.200\,m\,NaCl$: Water: $0.198$; Benzene: $0.961$; Ethanol: $0.464$

  • $0.300\,m\,MgSO_4$: Water: $0.203$; Benzene: $0.987$; Ethanol: $0.476$

• Specific Calculation Examples:

  • Freezing point change for $2.00\,g$ of $MgSO_4$ in $0.300\,kg$ of ethanol: Result = $0.143$

  • Boiling point change for $2.00\,g$ of $MgCl_2$ in $0.400\,kg$ of acetic acid: Result = $0.415$

  • Freezing point change for $1.00\,g$ of $NaCl$ in $0.100\,kg$ of water: Result = $2.42$

  • Boiling point change for $1.00\,g$ of $HCl$ in $0.200\,kg$ of benzene: Result = $0.658$

  • Freezing point change for $3.00\,g$ of $FeCl_3$ in $0.300\,kg$ of cyclohexane: Result = $4.19$

  • Boiling point change for $2.00\,g$ of $MgSO_4$ in $0.300\,kg$ of ethanol: Result = $0.0879$

Vapor Pressure of Mixtures at $25\,^{\circ}C$

• Total Vapor Pressure for Specific Mixtures ($kPa$):

  • $10\,mol\%$ Acetone + $90\,mol\%$ Ethanol: $10.2\,kPa$

  • $20\,mol\%$ Benzene + Ethanol: $8.84\,kPa$

  • $30\,mol\%$ Cyclohexane + Ethanol: $9.41\,kPa$

  • $10\,mol\%$ Acetone + Methanol: $18.3\,kPa$

  • $20\,mol\%$ Benzene + Methanol: $16.1\,kPa$

  • $30\,mol\%$ Cyclohexane + Methanol: $15.7\,kPa$

  • $10\,mol\%$ Acetone + Toluene: $6.49\,kPa$

  • $20\,mol\%$ Benzene + Toluene: $5.57\,kPa$

  • $30\,mol\%$ Cyclohexane + Toluene: $6.55\,kPa$

• Theoretical Vapor Pressure of Binary Mixtures:

  • Methanol-Ethanol ($50.0\%$ Methanol): $12.4\,kPa$

  • Benzene-Cyclohexanol ($20.0\%$ Benzene): $12.9\,kPa$

  • Cyclohexanol-Ethanol ($30.0\%$ Cyclohexanol): $9.41\,kPa$

  • Toluene-Benzene ($60.0\%$ Toluene): $7.35\,kPa$

  • Acetone-Methanol ($10.0\%$ Acetone): $18.3\,kPa$

  • Ethanol-Acetone ($40.0\%$ Ethanol): $21.6\,kPa$

Osmotic Pressure ($\Pi$)

• Osmotic Pressure Calculations (using $R = 8.314\,kPa\cdot L/mole\cdot K$):

  • Data Set 1 ($HCl$): $P = 6070\,kPa$, Temp = $25\,^{\circ}C$, $i = 1.9$, Molarity = $1.29\,M$

  • Data Set 2 ($NaCl$): $P = 3780\,kPa$, Temp = $25\,^{\circ}C$, $i = 1.9$, Molarity = $0.803\,M$, Vol = $0.500\,L$, Mass = $23.5\,g$

  • Data Set 3 ($MgSO_4$): $P = 775\,kPa$, Temp = $50\,^{\circ}C$, $i = 1.3$, Molarity = $0.222\,M$

  • Data Set 4 ($MgCl_2$): $P = 748\,kPa$, Temp = $60\,^{\circ}C$, $i = 2.7$, Molarity = $0.100\,M$, Vol = $0.750\,L$, Mass = $7.15\,g$

  • Data Set 5 ($FeCl_3$): $P = 122\,kPa$, Temp = $15\,^{\circ}C$, $i = 3.4$, Molarity = $0.0150\,M$, Vol = $0.800\,L$, Mass = $1.95\,g$

• Specific Solutions in Water:

  • $10.0\,g$ $NaCl$ in $0.750\,L$ at $25\,^{\circ}C$: $\Pi = 1070\,kPa$

  • $10.0\,g$ $HCl$ in $0.750\,L$ at $35\,^{\circ}C$: $\Pi = 1780\,kPa$

  • $20.0\,g$ $MgSO_4$ in $0.500\,L$ at $15\,^{\circ}C$: $\Pi = 1030\,kPa$

  • $20.0\,g$ $MgCl_2$ in $0.500\,L$ at $40\,^{\circ}C$: $\Pi = 2950\,kPa$

  • $5.00\,g$ $FeCl_3$ in $1.00\,L$ at $20\,^{\circ}C$: $\Pi = 253\,kPa$

Advanced Freezing, Boiling, and Osmotic Pressure Tables

• Freezing Point Changes (Variable Molecular Weights):

  • Water: $\Delta T = 5.81$, $K_f = 1.86$, $i = 1.00$, Molality = $3.13$, Mass = $10.0\,g$, Solvent = $0.100\,kg$, MW = $32.0\,g/mole$

  • Benzene: $\Delta T = 11.1$, $K_f = 5.12$, $i = 1.00$, Molality = $2.17$, Mass = $10.0\,g$, Solvent = $0.100\,kg$, MW = $46.0\,g/mole$

  • Ethanol: $\Delta T = 2.69$, $K_f = 1.99$, $i = 1.00$, Molality = $1.35$, Mass = $20.0\,g$, Solvent = $0.200\,kg$, MW = $74.0\,g/mole$

• Boiling Point Changes (Variable Molecular Weights):

  • Acetic Acid: $\Delta T = 3.76$, $K_b = 2.93$, $i = 1.00$, Molality = $1.28$, Mass = $20.0\,g$, Solvent = $0.200\,kg$, MW = $78.0\,g/mole$

  • Cyclohexane: $\Delta T = 0.775$, $K_b = 2.79$, $i = 1.00$, Molality = $0.278$, Mass = $5.00\,g$, Solvent = $0.300\,kg$, MW = $60.0\,g/mole$

• Osmotic Pressure (Variable Molecular Weights):

  • Condition A: $\Pi = 1032$, Temp = $25.0\,^{\circ}C$, $i = 1.00$, Molarity = $0.417$, Vol = $0.750\,L$, Mass = $10.0\,g$, MW = $32.0\,g/mole$

  • Condition B: $\Pi = 742$, Temp = $35.0\,^{\circ}C$, $i = 1.00$, Molarity = $0.290$, Vol = $0.750\,L$, Mass = $10.0\,g$, MW = $46.0\,g/mole$

  • Condition C: $\Pi = 1294$, Temp = $15.0\,^{\circ}C$, $i = 1.00$, Molarity = $0.541$, Vol = $0.500\,L$, Mass = $20.0\,g$, MW = $74.0\,g/mole$

  • Condition D: $\Pi = 1335$, Temp = $40.0\,^{\circ}C$, $i = 1.00$, Molarity = $0.513$, Vol = $0.500\,L$, Mass = $20.0\,g$, MW = $78.0\,g/mole$

Clausius-Clapeyron Equation Experiments

• Vapor Pressure Calculations:

  • If $P = 12.5\,torr$ at $T = 25\,^{\circ}C$ and $\Delta H_{vap} = 23\,kJ/mole$, then $P$ at $65\,^{\circ}C$ is $37.5\,torr$

  • If $P = 21.6\,torr$ at $T = 36\,^{\circ}C$ and $\Delta H_{vap} = 40.5\,kJ/mole$, then $P$ at $76\,^{\circ}C$ is $131.6\,torr$

  • If $P = 33.8\,torr$ at $T = 44.6\,^{\circ}C$ and $\Delta H_{vap} = 4.47\,kJ/mole$, then $P$ at $71\,^{\circ}C$ is $38.5\,torr$

  • If $P_1 = 3.4\,torr$ at $11.6\,^{\circ}C$ and $P_2 = 28.8\,torr$ at $54.9\,^{\circ}C$, then $\Delta H_{vap} = 38.4\,kJ/mole$

  • If $P_1 = 21.7\,torr$ at $14.8\,^{\circ}C$ and $P_2 = 22.1\,torr$ at $15.3\,^{\circ}C$, then $\Delta H_{vap} = 25.2\,kJ/mole$

  • If $P_1 = 6.50\,torr$ at $37.6\,^{\circ}C$ and $P_2 = 12.5\,torr$ at $48.6\,^{\circ}C$, then $\Delta H_{vap} = 49.4\,kJ/mole$

• Clausius-Clapeyron Multi-Property Table:

  • Liquid 1: $P_1 = 12.30$, $P_2 = 10.20$, $T_1 = 25.0\,^{\circ}C$, $T_2 = 20.0\,^{\circ}C$, $\Delta H_{vap} = 27.2\,kJ/mole$

  • Liquid 2: $P_1 = 0.1495$, $P_2 = 0.0235$, $T_1 = 125.0\,^{\circ}C$, $T_2 = 53.5\,^{\circ}C$, $\Delta H_{vap} = 27.2\,kJ/mole$

  • Liquid 3: $P_1 = 1.589$, $P_2 = 0.236$, $T_1 = 13.3\,^{\circ}C$, $T_2 = 103.9\,^{\circ}C$, $\Delta H_{vap} = 1.31\,kJ/mole$

  • Liquid 4: $P_1 = 12.6$, $P_2 = 3.28$, $T_1 = 823\,K$, $T_2 = 469\,K$, $\Delta H_{vap} = 25.7\,kJ/mole$

  • Liquid 5: $P_1 = 56.31$, $P_2 = 285$, $T_1 = 189\,K$, $T_2 = 22.7\,K$, $\Delta H_{vap} = 20.36\,kJ/mole$. Note: There is a discrepancy in the Liquid 5 work output provided in the packet (listed as $P_1 = 56.31$ and $\Delta H = 20.36$).

• Formula Derivations Used in Work:

  • Solving for $\Delta H$: $\Delta H = R \times \ln\left(\frac{P_1}{P_2}\right) \times \frac{1}{\frac{1}{T_2} - \frac{1}{T_1}}$

  • Solving for $T_2$: T_2 = \frac{1}{\frac{R \times \ln\frac{P_1}{P_2}}{\Delta H} + \frac{1}{T_1}}

  • Solving for $P_2$: $P_2 = P_1 \times e^{\left(\frac{\Delta H}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right)}$

Phase Diagrams and Transformations

• Carbon Dioxide ($CO_2$) Path:

  • Initial Condition: Heated isobarically at $70\,atm$ from $-100\,^{\circ}C$ to room temperature.

  • Expansion: Expanded isothermally to $1\,atm$.

• Water Pathway:

  • Compression: Compressed from $22,100\,kPa$ to $70\,kPa$ at $0\,^{\circ}C$.

  • Initial State: Water starts as a liquid.

  • Transition 1: It transitions to a solid.

  • Heating Stage: At $70\,kPa$, the substance is heated to $200\,^{\circ}C$.

  • Transition 2: The solid water melts to a liquid.

  • Transition 3: The liquid vaporizes into water vapor.

• Copper Oxide Pathway:

  • Heating Stage: Heated at $1\,torr$ from $600\,^{\circ}C$ to $1300\,^{\circ}C$.

  • Initial State: Starts as $CuO$.

  • Transition 1: Transitions to $Cu_2O$.

  • Transition 2: Ends as liquid $Cu_2O$ at the end of heating.

  • Decompression: Isothermally decompressed to $10^{-4}\,torr$.

  • Final Result: The $Cu_2O$ forms liquid copper.