Chem II 6/1

Exam One Feedback and Course Administration

• Exam One Reflections:

  • The instructor noted that the time allocated (100 minutes) seemed sufficient as many students did not take the full time (average finish time was approximately 60 to 70 minutes).

  • The instructor expressed appreciation for the ability to see students during the online exam to prevent cheating.

  • Settings for the exam will be updated so that students can view their specific point results and feedback while the material is fresh, which is recommended for cumulative final exam preparation.

• Submission Policy:

  • Students are required to submit their scratch paper pages either via the exam platform (screenshots) or by emailing the instructor immediately after submission.

• Upcoming Schedule:

  • Chapter 14 and Chapter 15 homework is due this Sunday.

  • Exam Two will cover the second half of Chapter 14, all of Chapter 15, Chapter 16, and potentially a portion of Chapter 17.

Quantifying Solution Concentration: Volume Percent and Parts per Million

• Volume Percent ($\text{v/v}\%$):

  • Definition: Primarily used for solutions where both the solute and solvent are liquids.

  • Formula: $\text{Volume \%} = \frac{\text{milliliters of solute}}{\text{milliliters of solution}} \times 100$

• Parts per Million (ppm) and Parts per Billion (ppb):

  • These units are used for very dilute solutions.

  • ppm Formula: $\text{ppm} = \frac{\text{milliliters of solute}}{\text{milliliters of solution}} \times 10^6$

  • ppb Formula: $\text{ppb} = \frac{\text{milliliters of solute}}{\text{milliliters of solution}} \times 10^9$

• Practice Problem: Volume Percent:

  • Question: How many milliliters of ethyl acetate are in $400\,mL$ of a $1.8\%\,v/v$ solution of ethyl acetate in water?

  • Setup: $1.8\% = \frac{x}{400\,mL} \times 100$

  • Solving for $x$: $\frac{1.8}{100} \times 400 = 7.2\,mL$ of ethyl acetate.

  • Solvent calculation: $400\,mL\text{ (total)} - 7.2\,mL\text{ (solute)} = 392.8\,mL\text{ (water)}$.

Mole Fraction and Mole Percent

• Mole Fraction ($\chi$):

  • Definition: The ratio of the moles of solute to the total moles of the solution.

  • Formula: $\chi = \frac{\text{moles of solute}}{\text{total moles of solution}}$

  • Characteristics: This value is unitless as moles cancel out.

• Mole Percent:

  • Formula: $\text{Mole \%} = \chi \times 100$

• Practice Problem: Concentration Conversion:

  • Given: A $0.509\,m$ (molality) aqueous sodium chloride ($NaCl$) solution.

  • Goal: Find mole fraction and mole percent.

  • Step 1 (Assumptions): Since molality is moles of solute per kilogram of solvent, assume $0.509\text{ moles of } NaCl$ in $1\,kg\text{ (1000 g) of water}$.

  • Step 2 (Convert solvent to moles): $\frac{1000\,g\,H_2O}{18.015\,g/mol} = 55.51\text{ moles of } H_2O$.

  • Step 3 (Calculate $\chi$): $\chi = \frac{0.509}{0.509 + 55.51} = 0.00909$.

  • Step 4 (Calculate %): $0.00909 \times 100 = 0.909\%$.

Introduction to Colligative Properties

• Definition: Physical properties of a solvent that change when a solute is added to form a solution. These changes depend on the concentration of the solution, not the identity of the solute particles.

• The Four Primary Colligative Properties:

  • Freezing Point Depression.

  • Boiling Point Elevation.

  • Osmotic Pressure.

  • Vapor Pressure Lowering.

Freezing Point Depression

• Concept: The freezing point of a solution is always lower than the freezing point of the pure solvent. This occurs because solute particles interfere with the ability of solvent molecules to lock into a crystalline structure, requiring more energy removal to freeze.

• Real-World Applications:

  • Salting roads in winter to lower the freezing point of water to prevent ice formation.

  • Biological antifreeze: Frogs and turtles flood their cells with sugars to prevent their bodies from freezing at zero degrees Celsius.

• Equation: $\Delta T_f = i \times m \times K_f$

  • $\Delta T_f$: Change in freezing point ($T_{solvent} - T_{solution}$).

  • $i$: Van\'t Hoff factor.

  • $m$: Molality ($mol/kg$).

  • $K_f$: Freezing point depression constant (specific to the solvent).

• The Van\'t Hoff Factor ($i$):

  • Used to account for the dissociation of ionic compounds into multiple particles.

  • Covalent Solutes (non-metals): Do not dissociate; $i = 1$.

  • Ionic Solutes (metals and non-metals): $i > 1$ (calculated by counting ions in the chemical formula).

    • $NaCl \rightarrow Na^+ + Cl^- \implies i = 2$

    • $MgCl_2 \rightarrow Mg^{2+} + 2Cl^- \implies i = 3$

    • $Li_3PO_4 \rightarrow 3Li^+ + PO_4^{3-} \implies i = 4$

    • $Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^- → i = 3$

    • $Mg_3N_2 \rightarrow 3Mg^{2+} + 2N^{3-} → i = 5$

• Final Step Calculation: To find the new freezing point: $T_{new} = T_{original} - \Delta T_f$.

• Sample Calculation: Find the freezing point of a $2.6\,m$ aqueous sucrose ($C_{12}H_{22}O_{11}$) solution ($K_f = 1.86\,^∘C/m$).

  • $i = 1$ (sucrose is covalent).

  • $\Delta T_f = 1 \times 2.6 \times 1.86 = 4.836\,^∘C$.

  • $T_{new} = 0\,^∘C - 4.836\,^∘C = -4.8\,^∘C$.

Boiling Point Elevation

• Concept: The boiling point of a solution is higher than that of the pure solvent because solute-solvent interactions (like ion-dipole forces) make it harder for molecules to escape into the gas phase.

• Real-World Applications:

  • Antifreeze in radiators (ethylene glycol in water): Prevents engine overheating by raising the boiling point of the coolant above $100\,^∘C$.

  • Adding salt to pasta water: Increases the boiling temperature to cook food faster (oil does not have this effect because it does not form a solution).

• Equation: $\Delta T_b = i \times m \times K_b$

  • $K_b$ is the boiling point elevation constant.

• Final Step Calculation: To find the new boiling point: $T_{new} = T_{original} + \Delta T_b$.

Osmotic Pressure

• Osmosis: The flow of solvent through a semi-permeable membrane from a region of low solute concentration to high solute concentration. This is driven by entropy (the natural tendency toward disorder/dispersion).

• Osmotic Pressure Equation: $\Pi = iMRT$

  • $\Pi$: Osmotic pressure (measured in atmospheres, $\text{atm}$; not the number $3.14$).

  • $i$: Van\'t Hoff factor.

  • $M$: Molarity ($mol/L$).

  • $R$: Ideal Gas Constant ($0.08206\,L\cdot atm / mol\cdot K$).

  • $T$: Temperature in Kelvin ($^∘C + 273.15$).

• Practice Problem: Calculate the osmotic pressure of seawater at $20\,^∘C$ ($3.4\,g$ of $NaCl$ per Liter).

  • $i = 2$ ($NaCl$).

  • $M = \frac{3.4\,g / 58.44\,g/mol}{1\,L} = 0.0582\,M$.

  • $T = 20 + 273.15 = 293.15\,K$.

  • $\Pi = 2 \times 0.0582 \times 0.08206 \times 293.15 = 2.8\,atm$.

• Applications: Biological systems (red blood cells), water purification, and hydraulic systems.

Vapor Pressure Lowering (Raoult\'s Law)

• Concept: Solutions have a lower vapor pressure than pure solvents because solute-solvent interactions tether solvent molecules in the liquid phase and solute particles reduce the surface area available for evaporation.

• Nonvolatile Solutes (Solids):

  • Raoult\'s Law: $P_{solution} = \chi_{solvent} \times P^0_{solvent}$

    • $P_{solution}$: Vapor pressure of the solution.

    • $\chi_{solvent}$: Mole fraction of the solvent ($\frac{mol_{solvent}}{mol_{total}}$).

    • $P^0_{solvent}$: Vapor pressure of the original pure solvent.

• Practice Problem (Solid Solute): $5\,g$ sugar in $25\,g$ water at room temperature ($P^0 = 23.8\,torr$).

  • $n_{water} = 1.388\,mol$; $n_{sucrose} = 0.0146\,mol$.

  • $\chi_{solvent} = \frac{1.388}{1.388 + 0.0146}$.

  • $P_{soln} = \chi_{solvent} \times 23.8 = 23.5\,torr$.

• Volatile Solutes (Liquids):

  • If both solute and solvent are liquids, both contribute to vapor pressure.

  • Formula: $P_{total} = P_A + P_B = (\chi_A P^0_A) + (\chi_B P^0_B)$.

• Practice Problem (Volatile Mixture): $25\%\,w/w$ benzene in toluene mixture.

  • Assume $25\,g$ benzene and $75\,g$ toluene.

  • Calculate vapor pressure for benzene ($P_{benzene} = 26.58\,torr$) and toluene ($P_{toluene} = 20.39\,torr$).

  • Total Vapor Pressure: $26.58 + 20.39 = 47\,torr$.

Colloids

• Definition: Solutions containing larger particles ($5$ to $1000\,nm$) that are evenly distributed but do not settle out.

• Properties: They scatter light (Tyndall effect) and do not appear transparent.

• Examples: Smoke, fog.

Chapter 15: Chemical Kinetics

• Kinetics Defined: The study of reaction speeds (rates), activation energy (energy to start), and reaction mechanisms (step-by-step processes).

• Reaction Rate: The speed at which reactants are consumed or products are formed.

  • Fast reactions: Electron jumps ($10^{-15}\,s$), explosions ($s$).

  • Slow reactions: Rusting (years), rock formation (millions of years).

• Quantifying Rates:

  • Formula: $\text{Rate} = \frac{\Delta [Concentration]}{\Delta t}$

  • Brackets ($[…]$) denote molarity ($M$), which is the standard unit used in kinetics.

• Reaction Progress and Equilibrium:

  • As a reaction proceeds, reactants decrease (downward slope) and products increase (upward slope).

  • Eventually, concentrations stabilize as they reach chemical equilibrium. This often happens because reactants become too dilute to encounter each other.

• Relative Rates and Stoichiometry:

  • Rates are relative to coefficients in a balanced equation ($aA + bB \rightarrow cC + dD$).

  • General Formula: $\text{Rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = \frac{1}{c} \frac{\Delta [C]}{\Delta t} = \frac{1}{d} \frac{\Delta [D]}{\Delta t}$

  • Reactants are always negative in the raw calculation, but rate is reported as a positive value by convention.

  • Example: $2O_3 \rightarrow 3O_2$

    • $\text{Rate} = -\frac{1}{2} \frac{\Delta [O_3]}{\Delta t} = \frac{1}{3} \frac{\Delta [O_2]}{\Delta t}$.

• Calculation Example: Between $0$ and $10$ seconds, a reactant A decreases from $1.0\,M$ to $0.85\,M$.

  • $\text{Rate} = -\frac{0.85 - 1.0}{10 - 0} = 0.015\,M/s$.

Questions & Discussion

• Question: For the scratch work, do we have to submit it on the platform and also email it?

• Answer: The instructor stated it is either/or. Students can upload screenshots to the exam site or email them immediately after the exam.

• Question: Why is the calculation using $0.85\,M$ instead of $0.3\,M$ in the final kinetics practice problem?

• Answer: The instructor clarified that the calculation was specifically for the $0$ to $10$ second time interval. The $0.3\,M$ value was the concentration at $80$ seconds, which was outside the requested timeframe.