Stephanie Yeung Per. 3 Semester 2

Properities of Solutions

Thermochemistry

Gas Laws

Percent Composition, Empirical and Molecular Formulas

q = mCΔT

Heat (enthalpy) change, ∆H: amount of heat energy released or absorbed during process

exothermic processes, -∆H: processes where energy is released as it proceeds, and surroundings become warmer

endothermic processes, +∆H: processes where energy is absorbed as it proceeds, and surroundings become cooler

∆H from ∆Hf: standard enthalpy change of reaction = sum of standard molar enthalpies of formation of products multiplied by its coefficient, n, in balanced equation, minus corresponding sum of standard molar enthalpies of formation of reactants.

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(reactants)

water phase changes

specific heat: amount of heat required to raise temperature of 1 g of substance by 1 °C

ΔT = temperature change (°C) = Tfinal - Tinitial

m = mass (g)

C = specific heat (J/g°C)

q = heat lost/gained (J)

Cwater = 4.184 J/g°C

Cice = 2.09 J/g°C

Csteam = 1.87 J/g°C

∆Hfus = 333 J/g

∆Hvap = 2240 J/g

heat of fusion: energy that must be absorbed in order to convert 1 mol of solid to liquid at its melting point

heat of vaporization: energy that must be absorbed in order to convert 1 mol of liquid to gas at its boiling point

mCΔT = -[mCΔT]

heat gained = heat lost

Joule: measures heat

calorie: heat required to raise temperature of 1 g of water by 1 °C

1 calorie = 4.184 Joules

pressure = force/unit area; caused by collisions of molecules with walls of container

Kinetic Molecular Theory

deviations from ideal behavior

1 atm = 101,325 Pa = 101.3 kPa = 760 mmHg = 760 torr

  1. Gas particles are in constant, rapid motion. They, therefore, possess kinetic energy, the energy of motion.
  1. Collisions between gas particles and between particles and the walls of the container are elastic collisions. No energy is lost in elastic collisions.
  1. Gases consist of tiny particles that are far apart relative to their size.
  1. There are no forces of attraction between gas particles.
  1. The average kinetic energy of gas particles depends on temperature. The temperature of a gas is proportional to the average kinetic energy of the gas particles.

likely to behave nearly ideally: gases at high temp and low pres, small non-polar gas molecules

likely not to to behave nearly ideally: gases at low temp and high pres, large polar gas molecules

Boyle's Law: P1V2 = P2V2

Charles's Law: V1/T1 = V2/T2

The Combined Gas Law: P1V1/T1 = P2V2/T2

Gay Lussac's Law: P1/T1 = P2/T2

T = temperature (K) K = °C + 273

V = volume (same units for both)

P = pressure (same units for both)

V = volume (same units for both)

T = temperature (K)

T = temperature (K)

V = volume (same units for both)

P = pressure (same units for both)

P = pressure (same units for both)

Avogadro's Law: V1/n1 = V2/n2

V = volume (same units for both)

n = moles

Dalton's Law of Partial Pressures: Ptotal = P1 + P2 + P3 + ...

Standard Temperature and Pressure

where P1, P2, P3 are partial pressures

Ideal Gas

PV = nRT

n = mole

R = proportionality constant, 0.08206 L⋅atm/mol⋅K

V = volume (L)

T = temperature (K)

P = pressure (atm)

m/M, where m = mass M = molar mass

at STP,

D = MP/RT

gas density = molar mass/22.4 L

P = pressure

R = gas constant

D = mass/volume

M = molar mass

T = temperature

finding empirical formula

finding molecular formula

molecular formula: true # of atoms of each element in formula of compound

empirical formula: lowest whole # ratio of atoms in compound

  1. mass to mole
  1. divide by small Divide each value of moles by the smallest of the values
  1. percent to mass Assume there is 100 g of each compound.
  1. multiply till whole If necessary, multiply each # by same integer to get all whole #s
  1. Divide the molecular mass by the empirical mass
  1. Multiply empirical formula by this # to get molecular formula
  1. Find the empirical formula mass

finding empirical formula by combustion analysis

H2O g × 2.02g/18.02g = H g

N can be NH3, N2, NO2,...

CO2 g × 12.01g/44.01g = C g

subtract total mass by mass of other elements to get O

solutions are homogenous (uniform composition) mixtures

calculations of solution concentration

grams per liter: g of solute/ L of solution

parts per million: (g of solute/ g of solution) × (1 ⋅ 10^6)

percent composition: % mass = (g of solute/ g of solution) × 100

molarity: M = mol of solute/ L of solution

in dilute water-based solution, 1 ml = 1 g

unsaturated solution: less than max amount of solute dissolved

supersaturated solution: more than max amount of solute dissolved

saturated solution: max amount of solute dissolved

solvent: dissolving substance

solute: dissolved substance in solution

Chemical Kinetics

Factors affecting rate

Collision Model

  1. collisions must have sufficient energy to produce reaction
  1. colliding particles must be correctly oriented to one another in order to produce reaction

increased surface area: increases rate. particles collide more

increased concentration: usually increases rate. particles collide more

increased temp: always increases rate. particles collide more

presence of catalysts: lower reaction rate by providing alternate pathways

rate expression

rate law: describes way in which reactant concentration affects reaction rate

rate = -∆[reactant]/∆t

rate = ∆[product]/∆t

coefficient in balanced equation is multiplied to ∆t

reaction rate: rate at which chemical reaction takes place; measured by rate of formation of product or rate of disappearance of reactants

rate = k[reactant]^n

[] = M of reactant

n = order of reactant

k = rate constant

Acids and Bases

Chemical Equilibrium

Le Chatelier's Principle: when system at equilibrium is placed under stress, system will undergo change in such way to relieve that stress

remove concentration from one side of the reaction: system shifts to that side

increase temp of exothermic forward reaction: system shifts to reactants

add more concentration to one side of the reaction: system shifts to the other side

increase temp of endothermic forward reaction: system shifts to products

increase pressure of reaction: system shifts to side with fewer moles in balanced equation

equilibrium expression

Keq = [product]/[reactant]

raise each substance'a concentration to power equal to substance's coefficient in balanced equation

Keq = equilibrium constant

[] = M of substance

concentration of any solid or pure liquid that takes part in reaction is left out because these concentrations never change

equilibrium expression involving pressure

reaction quotient

Kp = Pproduct/Preactant

where Pproduct,Preactant are equilibrium partial pressures

Draw a # line from least to greatest. Plot approximate values for Q and K the # line. Queen to King. Q → K. The direction of the arrow determines the direction the system shifts until equilibrium is achieved.

Q = [product]/[reactant]

Bronsted-Lowry

Arrhenius

conjugate acid: forms when base gains proton (H+)

conjugate base: forms when acid loses proton (H+)

base: accepts proton (H+)

acid: donates proton (H+)

base: produce hydroxide ions (OH-) in aqueous solutions

acid: produce hydronium ions (H3O+) in aqueous solutions

neutralization: acid and base react to produce salt and water

pH and pOH calculations

at 25 °C, Kw = [H3O+][OH-] = (1 × 10^-7)(1 × 10^-7) = 1 × 10^-14

[OH-] = 10^-pOH

[H3O+] = 10^-pH

pH + pOH = 14

pOH = -log[OH-]

pH = -log[H3O+]

weak acids and bases

strong acids and bases

pH directly from concentration of acid

ICE table, solve for x, which is also [H3O+] or [OH-]

pOH directly from concentration of base

Classifying salts as acidic, basic, or neutral

strong acid + weak base: acidic

weak acid + strong base: basic

strong acid + strong base: neutral

weak acid + weak base: which Ka/Kb is greater

Strong Acids: HCl, HBr, HI, HCIO4, HClO4, HNO3, H2SO4

Strong Bases: KiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

5% Rule: If Ka/Kb is 1/1000 smaller than [salt], remove the change in x for [salt]

1 atm

0°C, 273 K

ideal gas: molar volume = 22.4 L