…Somebody at some point made this. Ehh it’s ok. Semester 2 Review …Somebody at some point made this. Ehh it’s ok.

Stoich

Stoichiometry Vocab Limiting vs. Excess Reagent Theoretical vs. Actual Yield % yield = (Actual yield / Theoretical yield) x 100

Thermo.

Vocab Thermochemistry Potential Energy Kinetic Energy Energy Heat Temperature Calorimetry Enthalpy Entropy Hess’s Law Specific Heat Capacity Endothermic Reactions Exothermic Reactions

Specific Heat Calculations q = (m)(c)(ΔT) heat = mass • specific heat • change in temp ∆𝑇= 𝑇 𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑖𝑛𝑖𝑡𝑖𝑎𝑙

Application of Hess’s Law… Reg. Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: SO3 (g) + H2O (g)  H2SO4 (l) H2SO4 (l)  H2S (g) + 2 O2 (g) ΔH = 235.5 kJ H2O (g)  H2O (l) ΔH = -44 kJ H2S (g) + 2 O2 (g)  SO3 (g) + H2O (l) ΔH = -207 kJ Change in Enthalpy = -72.5kJ

Application of Hess’s Law… Honors Calculate the enthalpy of the following chemical reaction: CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g) Given: C(s) + O2(g) ---> CO2(g) ΔH = -393.5 kJ/mol S(s) + O2(g) ---> SO2(g) ΔH = -296.8 kJ/mol C(s) + 2S(s) ---> CS2(ℓ) ΔH = +87.9 kJ/mol

Calculations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Enthalpies of formation: CH4: -74.86 kJ/mol O2: 0.0 kJ/mol CO2: -393.5 kJ/mol H2O: -241.8 kJ/mol

Calculations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Enthalpies of formation: CH4: -74.86 kJ/mol X 1 mol = -74.86 kJ O2: 0.0 kJ/mol X 2 mol = 0.0 kJ CO2: -393.5 kJ/mol X 1 mol = -393.5 kJ H2O: -241.8 kJ/mol X 2 mol = -483.6 kJ

Calculations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Enthalpies of formation: ΔH = (-393.5 kJ + -483.6 kJ) – (-74.86 kJ + 0.0 kJ) ΔH = -802.24kJ

Molarity

Vocab: Be a part of the solution! Solute: The particles that become dissolved. Solvent: the dissolving medium (what the solute dissolves in). Aqueous Solution: solution where the solvent is water. Concentration: a measure of the amount of solute that is dissolved in a given quantity of solvent.

Solution making 101 To reach a specific concentration of solution, we must carefully calculate and measure out both the solute and the solvent. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒= 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 Volume must be in Liters!

Calculations with Dilutions M1V1 = M2V2 (Molarity)(Volume) = (Molarity)(Volume)

Kinetics and Equilibrium

Vrooom! Collision Theory: Particles must collide with proper orientation and enough energy to react. Factors that affect reactions rates: Concentration Surface area Particle size Temperature Catalysts

Law of Mass Action Chemical Equilibrium: a state in which the forward and reverse reactions balance each other because they take place at equal rates. For the reaction: 𝑗𝐴+𝑘𝐵 ↔𝑙𝐶+𝑚𝐷 Where K is the Equilibrium Constant, and is unitless. Ignore pure solids and pure liquids with equilibrium.

Le Chatelier’s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Possible changes: Concentration Pressure Volume Inert Gas Temperature

Nuclear Chemistry

Radiation Radioisotopes: Isotopes with unstable nuclei. Three types of radiation: Alpha particles Beta particles Gamma rays

Nuclear Equations

Fission vs. Fusion Nuclear Fission: the splitting of a nucleus into fragments. This is accompanied by a very large release of energy. Nuclear Fusion: the combination of atomic nuclei. This releases more energy than fission and uses materials that are easier to come by. The activation energy is around 5,000,000K.

Acids and Bases

Not so Mathy Bronsted-Lowry acid vs. Bronsted-Lowry base Conjugate acid and base Amphoterism Strong acids HCl, HBr, HI, HClO4, H2SO4, HNO3 Strong bases Group I and II hydroxides

pH and pOH Calculations

Gas Laws

Not so Mathy What causes pressure? What can we use to measure pressure? Relationships between pressure, volume, temperature and moles of gas.

Gas Laws Boyle’s Law PV = PV Charles’s Law V/T = V/T Gay-Lussac’s Law P/T = P/T Combined Gas Law PV/T = PV/T Ideal Gas Law PV = nRT Dalton’s Law of Partial Pressures Total pressure = Sum of partial n1/nT = P1/PT