General Chemistry Midterm Exam I: Study Notes and Worked Examples

Gases and Gas Laws

Calculating Moles of a Gas Using the Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and amount of a gas. It is expressed as:

• Equation: $PV = nRT$

• P: Pressure (in kPa or atm)

• V: Volume (in liters)

• n: Number of moles

• R: Universal gas constant ($R = 8.314$ J/mol·K)

• T: Temperature (in Kelvin)

Example: To find the number of moles of NO(g) in a 275.0 mL container at 32.5°C and 75 kPa:

• Convert volume to liters: $V = 0.275$ L

• Convert temperature to Kelvin: $T = 32.5 + 273.15 = 305.65$ K

• Plug values into the equation and solve for n:

$n = \frac{PV}{RT} = \frac{75 \times 0.275}{8.314 \times 305.65}$ *Additional info: This calculation demonstrates the use of the ideal gas law for determining the amount of gas present under non-standard conditions.*

Partial Pressures in Gas Mixtures (Dalton’s Law)

In a mixture of gases, each gas exerts a pressure independently of the others. The partial pressure of a gas is the pressure it would exert if it occupied the container alone.

• Dalton’s Law of Partial Pressures: $P_{total} = P_1 + P_2 + ...$

• For each gas: $P_i = \frac{n_iRT}{V}$

Example: For 2.0 mol H2 and 1.0 mol N2 in a 22.4 L vessel at 298 K:

• Calculate $P_{H_2}$ and $P_{N_2}$ separately using the ideal gas law.

Intermolecular Forces

Types of Intermolecular Interactions

Intermolecular forces are attractions between molecules that affect physical properties like boiling and melting points.

• Hydrogen Bonding: Strong attraction between H and N, O, or F (e.g., H2O, NH2OH)

• Dipole-Dipole: Attraction between polar molecules (e.g., NH2OH)

• London Dispersion Forces: Present in all molecules, especially nonpolar ones (e.g., CBr4)

Examples:

• NH2OH: Hydrogen bonding, dipole-dipole, dispersion

• CBr4: Dispersion only

• H2O: Hydrogen bonding, dipole-dipole, dispersion

Thermochemistry

Calculating Heat for Temperature Change

The heat required to change the temperature of a substance is given by:

• Equation: $q = mc\Delta T$

• q: Heat (in Joules)

• m: Mass (in kg)

• c: Specific heat capacity (J/kg·K)

• \Delta T: Change in temperature (K or °C)

Example: To heat 300.0 g (0.300 kg) of water from 20.0°C to 100.0°C:

• $\Delta T = 100.0 - 20.0 = 80.0$°C

• $q = 0.300 \times 4184 \times 80.0$

Enthalpy of Reaction and Formation

The enthalpy of formation ($\Delta H_f$) is the heat change when one mole of a compound forms from its elements. The reaction enthalpy change ($\Delta H_{rxn}$) can be calculated using enthalpies of formation:

• Equation: $\Delta H_{rxn} = \sum \Delta H_f^{\circ} (products) - \sum \Delta H_f^{\circ} (reactants)$

Example: Calculating the reaction enthalpy for burning TNT using its $\Delta H_f$ value.

Spontaneity and Thermodynamics

Criteria for Spontaneity

A chemical reaction is spontaneous at constant pressure and temperature if the Gibbs free energy change ($\Delta G$) is negative:

• Equation: $\Delta G = \Delta H - T\Delta S$

• If $\Delta G < 0$, the reaction is spontaneous.

Temperature for Spontaneity

To estimate the temperature at which a reaction becomes spontaneous:

• Set $\Delta G = 0$ and solve for $T$:

$T = \frac{\Delta H}{\Delta S}$

Phase Changes and Vapor Pressure

Clapeyron-Clausius Equation

The Clapeyron-Clausius equation relates vapor pressure and temperature for phase changes:

$\ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$

• Used to estimate boiling points and vapor pressures at different temperatures.

Solutions and Solubility

Solvent Choice and Solubility Principles

Solubility depends on the principle "like dissolves like":

• Ionic and polar substances dissolve best in polar solvents (e.g., water).

• Nonpolar substances dissolve best in nonpolar solvents (e.g., benzene).

Examples:

• KCl: Water (polar)

• CCl4: Benzene (nonpolar)

• CH3COOH: Water (polar, but also some solubility in benzene due to nonpolar tail)

Molality Calculation

Molality (m) is the number of moles of solute per kilogram of solvent:

$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

Example: Dissolving 25.0 g NaCl in 500.0 g water:

• Find moles of NaCl: $\text{moles} = \frac{25.0}{58.44}$

• Convert solvent to kg: $0.500$ kg

• Calculate molality.

Bonus: Balancing Equations and Thermodynamic Feasibility

Balancing Chemical Equations

Balancing equations ensures the same number of each atom on both sides. For reactions involving rocket fuels, balance the equation before calculating thermodynamic properties.

Assessing Rocket Fuel Feasibility

To determine if a fuel is suitable, check if its oxidation is spontaneous at high temperature (e.g., 2000 K) using $\Delta G$ values. If $\Delta G < 0$, the reaction is thermodynamically favorable.

Reference: Periodic Table of the Elements

The periodic table organizes elements by atomic number and properties. It is essential for identifying element symbols, atomic masses, and chemical families.