Physical Chemistry PDF Notes

## Physical Chemistry ### Matter * Matter exists in three states: * **Gas**: The individual molecules have little attraction for one another and are free to move about. * **Liquid**: The individual molecules are attracted to one another but can slide over each other. * **Solid**: The i...

## Physical Chemistry ### Matter * Matter exists in three states: * **Gas**: The individual molecules have little attraction for one another and are free to move about. * **Liquid**: The individual molecules are attracted to one another but can slide over each other. * **Solid**: The individual molecules are strongly attracted to one another and cannot move around. ### Units * Distance: cm, nm, Km, A°, m * Volume: cm³, mL, L = 1000mL = 1000 cm³ * Pressure: atm, torr, Pascal, N/m² * atm = 760 torr = 1.013 * 10^5 Pa * Temperature: °C, Tk, K(T) = T + 273 * Force: N, dyne * dyne: The force acting on a body of mass 1g to move it with acceleration of 1 cm/s². * dyne = (g * cm * s²) * Newtons: The force acting on a body of mass 1 kg to move it with acceleration of 1 m/s². * N = kg * m * s² * Energy (E) = Work (W) = Force * distance = F * d * J = kg * m² * s² * erg= g * cm² * s² = dyne * cm * 1J = 10^7 erg * 1 Cal = 4.184 J * Energy = 1.887 Cal * mol * day ### Concentration * Concentration (M): Moles per liter. Mol/L = M * Molecular Weight (MW): Molecular weight of a material measured in grams. * Number of moles (n): * n = W / MW ### General characteristics of Gases * **Expansibility:** Gases have limitless expansibility and will expand to fill the entire vessel. * **Compressibility:** Gases are easily compressed by applying pressure to a movable piston fitted in the container. * **Diffusibility:** Gases diffuse rapidly through each other to form a homogeneous mixture. * **Pressure:** Gases exert pressure on the walls of the container in all directions. * **Effect of heat:** When a gas confined in a vessel is heated, its pressure increases. Upon heating in a fitted vessel, the volume of the gas also increases. ### Parameters of a Gas * **Volume (V)** * **Pressure (p)** * **Temperature (T)** * **Number of moles of gas in the container (n)** * **Types of motion**: There are two types of motion inside a container: * **Rotational Motion:** Occurs during the transition of the molecule from place to place. * **Translational motion:** The transition of the molecule from its original position. ### Gas Laws * **Boyle's Law**: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved. * The relationship is inverse, meaning that as one value increases, the other decreases. * P * V = constant = k * **Charles's Law:** At constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature. If the absolute temperature is doubled, the volume is doubled * V / T = constant = k * V = V0 * (1 + t / 273) * **Gay-Lussac's Law:** At constant volume, the pressure of a given mass of gas is directly proportional to the absolute (Kelvin) temperature. * P / T = constant = k * The combined gas law is P1 * V1 / T1 = P2 * V2 / T2 and combines Boyle's law and Charles's law. * **Avogadro's Law**: At constant temperature and pressure, equal volumes contain the same number of moles. * 1 mole of any gas at STP (standard temperature and pressure) contains 6.023 * 10^23 molecules. * 1 mole of any gas at STP occupies 22.4 L (molar volume). * V = nRT ### Ideal Gas Equation * PV = nRT ### Dalton's Law * At constant temperature, the total pressure of a mixture of any gases in a definite volume is equal to the sum of the partial pressures of the individual gases, if each gas occupies the same volume alone. * P = P₁ + P₂ + .... + Pn ### Examples **Example 1**: The pressure of a container of He is 650 mmHg at 25°C. If the container is cooled to 0°C, what will the pressure be? * P₁ = 650 mmHg * T₁ = 25°C + 273 = 298 K * T₂ = 0°C + 273 = 273 K * P₂ = P₁ * T₂ / T₁ = 650 * 273 / 298 = 595.4 mmHg **Example 2:** A sample of ammonia gas has a volume of 120 mL at 5°C and 700 mmHg. What is the volume of the gas at 50°C and 1.2 atm? * V₁ = 120 mL * T₁ = 273 + 5 K = 278 K * P₁ = 700 mmHg = 0.92 atm * T₂ = 273 + 50 K = 323 K * P₂ = 1.2 atm * V₂ = V₁ * P₁ * T₂ / P₂ * T₁ = 120 * 0.92 * 323 / 1.2 * 278 = 17.84 mL **Example 3:** A sample of He gas with a mass of 18 gm occupies 1.6 liters at a particular temperature and pressure. What mass of oxygen would occupy 1.6 L at the same temperature and pressure? * No. of moles of He: 18 / 4 = 4.5 * Number of moles is equal for the same volume. Therefore, the number of moles for oxygen is 4.5 * Mass of oxygen = 4.5 * 32 = 144 g **Example 4:** A sample of gas has a volume of 2.5 L at 600 mmHg and 20°C. What is the volume of this gas at STP? * V₁ = 2.5 L * P₁ = 600 mmHg * T₁ = 20 + 273 = 293 K * P₂ = 760 mmHg * T₂ = 273 K * V₂ = V₁ * P₁ * T₂ / P₂ * T₁ = 2.5 * 600 * 273 / 760 * 293 = 1.84 L **Example 5:** A sample of H₂ gas has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the number of moles of gas present, and the number of molecules. * P * V = n * R * T * n = ( P * V ) / ( R * T ) = (1.5 * 8.56) / (0.082 * 273) = 0.6 moles * Numberof molecules = n * N * A = 0.6 * 6.02 * 10^23 = 3.45 * 10^23 molecules **Example 6:** A mixture of gases contains 8 g of O₂ and 14 g of N₂ gas, with a total pressure of 3 atm. What is the partial pressure of each gas? * n (O₂) = 8 / 32 = 0.25 moles * n (N₂) = 14 / 28 = 0.5 moles * P (O₂) = (n (O₂) / n (total)) * P (total) = (0.25 / 0.75) * 3 = 1 atm * P (N₂) = (n (N₂) / (n (total)) * P (total) = (0.5 / 0.75) * 3 = 2 atm ### Graham's Law of Diffusion * **At constant temperature and pressure, the rate of diffusion of a given amount of any gas is inversely proportional to the square root of its molecular weight or density** * The rate of diffusion of gas A is inversely proportional to the square root of its molecular weight: rA = 1 / √MA. * Similarly, the rate of diffusion of gas B is inversely proportional to the square root of its molecular weight: rB = 1 / √MB. * rB / rA = 6 and MA = 40 g. * Finding the molecular weight of gas B: MB = (MA / 6²) = 1.11 g ### Kinetic Molecular Theory of Gases * **Assumptions of the Theory:** * **Any gas consists of small particles called molecules.** * **These molecules move in straight lines in random motion with high velocity in all directions.** * **These molecules of gas collide with each other, and with the walls of the container. The total number of collisions on the wall represent the pressure of gas, and must be elastic.** * **The mean kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas (T).** * **There is no intermolecular force between molecules of gas.** * **The volume of molecules themselves can be neglected in comparing with the whole volume of the container.** ### Behavior of Real Gases * All gases follow ideal behavior at certain conditions: low pressure, high temperature. * This means that the ideal gas law (PV=nRT) is accurate under these conditions. * For real gases, the ideal gas law is not accurate. * PV = Z * nRT ### Compressibility Factor * The compressibility factor (Z) is a correction factor that accounts for the deviation of real gases from ideal behavior. * Z = 1 for ideal gases, and is greater than 1 for real gases. * Z = PV / nRT * The compressibility factor is calculated by comparing the value of PV obtained from the ideal gas law to the experimental value of PV. ### Van der Waals Equation * The van der Waals equation is a more realistic equation for the behavior of real gases. This equation takes into account the intermolecular forces and the volume of the molecules: * (P + a(n/V)²) * (V - nb) = nRT * The "a" coefficient accounts for the attractive forces between gas molecules. * The "b" coefficient accounts for the volume of the molecules. * At low pressures: * V >> nb * The van der Waals equation reduces to the ideal gas equation. * At moderate pressure: * V is not much larger than nb. * The van der Waals equation accounts for the attractive forces between gas molecules. ### Andrews' Isotherms * This is a graphical plot of pressure versus volume at a constant temperature for a given mass of gas. * The critical temperature is the temperature above which a gas cannot be liquefied at any pressure. * The critical pressure is the pressure at which a gas can be liquefied at the critical temperature. * These isotherms show that real gases do not always behave ideally, especially at high pressures and low temperatures. ### Kinetic Theory Proof * **Suppose** the container is a cube with length l. * **Suppose** the container contains n molecules. * **Suppose** each molecule has a mass m. * **Suppose** each molecule moves with a velocity of v. * **Kinetic Theory Equation**: PV=(1/3) * n * m * v² ### Gas Laws and Kinetic Theory * **Boyle's Law**: At constant temperature, P = (1/3) * n * m * v² / V * **Charles's Law**: At constant volume, P = (1/3) * n * m * (v² / T) * **Avogadro's Law**: At constant pressure and temperature, n * V remains constant, which means that the number of particles in a given volume will be the same. ### Dalton's Law and Kinetic Theory * The total pressure is the sum of the pressures due to each gas: * P = P₁ + P₂ + ... + Pn * This can be written as: * P = (1/3) * (n₁ * m₁ * v₁² + n₂ * m₂ * v₂² + ... + nn * mn * vn²) / V * This shows that the total pressure is the sum of the partial pressures of each gas. ### Graham's Law and Kinetic Theory * The rate of effusion of a gas is proportional to the average speed of the molecules. * The average speed of the molecules is proportional to the square root of the temperature and inversely proportional to the square root of the molecular weight. * Combining these two relationships, we get Graham's law of diffusion: r₁ / r₂ = √M₂ / √M₁ ### Bond Enthalpy * **Bond enthalpy** is the change in enthalpy when one mole of a specific kind of bond is broken in the gas phase. * **Bond enthalpies** are always **positive** quantities because energy is always required to **break** a chemical bond. * **For endothermic reactions, the sign of the enthalpy change is negative because the bonds in the products are stronger than those in the reactants.** * **There are more bonds in the reactants than in the products in endothermic reactions.** * **The bond enthalpy increases as the order of the bond increases (single, double, triple).** * **When a bond involves two different atoms, the enthalpy is an approximation because it must be averaged over two different species.** * **To calculate the enthalpy change, we use the bond enthalpies.** ### Hess's Law * **Hess's Law states that the enthalpy change for a reaction is the same whether the reaction takes place in one step or several steps.** * **This is because enthalpy is a state function, meaning that it only depends on the initial and final states of the system and not on the pathway taken.** ### Heat Capacity * The amount of heat required to increase the temperature of the substance by 1°C. * C = (q / ΔT) * **Specific heat capacity (c):** The amount of heat required to rise the temperature of 1 gram of a substance by 1°C (1°K). * c = (q/(m * ΔT)) * **Molar specific heat capacity (Cm):** The amount of heat required to raise the temperature of 1 mole of a substance by 1°C (1°K) * Cm = (q / (n * ΔT)) ### Heat of Combustion * The change in the heat content of the system when 1 gram molecule (1 mole) of an organic compound is completely burnt in the presence of oxygen. ### Heat of Hydrogenation * The change in the heat contained in the system when 1 gram molecule (1 mole) of unsaturated hydrocarbon is hydrogenated with hydrogen gas. ### Heat of Formation * The change in heat contained in the system when 1 gram molecule (1 mole) of a compound is formed from its elements. * The heat of formation of elements is 0. ### Heat of Neutralization * The change in heat contained in the system when 1 g-equivalent of an acid is neutralized with 1 gram-equivalent of a base in an aqueous solution. ### Measuring Heat of Reaction * **Calorimeter:** calorimeter is a device used to measure the heat flow of a reaction, especially for exothermic reactions. * **Bomb Calorimeter:** A bomb calorimeter is used for reactions involving gases at high temperature. The bomb is a heavy metal vessel that is surrounded by water. * q (reaction) = -C (cal) * ΔT ### Chemical Equilibrium * **Reversible Reactions:** * AB + C ↔ CB + A * **Dynamic (equilibrium)**: The rates of the forward and reverse reactions are equal. * The concentrations of reactants and products remain constant. * The system appears to be static, but it is actually dynamic. * **Factors that affect the equilibrium state:** * **Temperature:** If the temperature is increased, the equilibrium shifts in the direction that absorbs heat (endothermic). * **Pressure:** If the pressure is increased, the equilibrium shifts in the direction that produces fewer moles of gas. * **Concentration:** If the concentration of a reactant is increased, the equilibrium shifts in the direction that consumes that reactant. * **Equilibrium Constant (K):** * Kc = ( [C]^c * [D]^d ) / ( [A]^a * [B]^b ) * Kp = ( P(C)^c * P(D)^d ) / ( P(A)^a * P(B)^b ) * For ideal gases: Kp = Kc * (RT)^Δn. ### Law of Mass Action * **At constant temperature, the rates of chemical reaction is directly proportional to the product of the molar concentrations (active masses) of the reacting substances.** * This is also known as the **rate law**. * **Equilibrium constant:** Kc = ( [C]^c * [D]^d ) / ( [A]^a * [B]^b ) * **Partial Pressure** for gases: The partial pressure of a gas is the pressure it would exert if it were the only gas present in the container. ### Heterogeneous Equilibrium * **Heterogeneous equilibrium** involves reactants and products in different phases (solid, liquid, gas). * **The partial pressure of a pure solid or liquid is constant and does not affect the value of Kp.** ### How to Find Equilibrium Concentration * If K is known, we can calculate the equilibrium concentration using the equilibrium expression. * The equilibrium concentration is determined by evaluating the equilibrium constant (Kc or Kp). ### Applications * **Haber Process:** This process is used for the industrial synthesis of ammonia. The reaction is: N₂ + 3H₂ ↔ 2NH₃. * **Ostwald Process:** This process is used for the industrial synthesis of nitric acid. The reaction is: 4NH₃ + 5O₂ ↔ 4NO + 6H₂O. This is just a summary of the main points in the document. There were many other important topics discussed in the document.