Review Chapters : General, Organic, & Biological Chemistry Janice Gorzynski Smith

Chapter 6 & 7 Concepts 2  Energy  conversions, conservation of energy  Breaking bonds requires E, forming bonds releases E  Endothermic & Exothermic Reactions  Energy diagrams, Activation Energy, heat absorbed or released  Factors affecting rates of reactions  Concentration, temperature, catalysts  Equilibrium  Equilibrium constant expressions  Le Chatlier Principle  States of matter: g, l, s & their properties  Effect of intermolecular forces on behavior  Gas laws: combined, ideal, & dalton’s law partial pressure  Intermolecular forces  London-Dispersion, Dipole-Dipole, Hydrogen-Bonding  Relative strength, importance in g, l, s behavior  Phase Changes  Navigate a heating/cooling curve  Enthalpy of phase changes

Equations & Conversions 1 cal = J 1,000 J = 1 kJ 1,000 cal = 1 kcal 1 kcal = kJ PV = nRT R= L atm mol K R = 62.4 L mm Hg mol K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 P total = P A + P B + P C Equations to memorize in orange K = [C] c [D] d [A] a [B] b = [products] [reactants]

Energy of Reactions E Reactants Products EXOTHERMIC ENDOTHERMIC Heat released Heat absorbed ENDOTHERMIC Heat + A + B  C + D Products have weaker bonds and a higher energy then Reactants. Heat is absorbed by the system. ΔE +ΔH + EXOTHERMIC A + B  C + D + heat Products have stronger bonds and a lower energy then Reactants. Heat is released by the system. ΔE - ΔH - Energy required to break bonds Energy released as bonds form ΔH EaEa EaEa Transition State

Rates of Reactions Increase the Rate of a Reaction Increase Temperature Increase Average KE of particles, so more likely to collide with enough energy to overcome E a Increase Concentration Reactants Increase the number of collisions per second Add a Catalyst Decrease E a Same likelyhood rxn will happen when particles collide, but more collisions Greater likelyhood that particles will have enough KE to react

Equilibrium & Le Chatlier’s Principle a A + b Bc C + d D equilibrium constant =K = [C] c [D] d [A] a [B] b = [products] [reactants] A + BC + D + heat reactantEq Shiftproduct increase  decrease   increase  decrease  T increase  T decrease A + B + heatC + D reactantEq Shiftproduct increase  decrease   increase  decrease  T increase  T decrease K > 1 products favoredK < 1 reactants favored K = 1 equilibrium

Intermolecular Forces London Dispersion Forces Dipole-Dipole Forces Hydrogen Bonds Ion-Dipole Forces Weakest Strongest Forces experienced by states of matter Gas < Liquids < Solids Increasing Average Kinetic Energy

Physical Properties Property of s, l, gIncreasesDecreasesExample Boiling Point increasing total intermolecular forces decreasing total intermolecular forces Water has a high boiling point because it has H-bonding, dipole, and dispersion forces. It is close to heptane (C7H16), a heavier molecule that only experiences dispersion forces. Melting Point increasing total intermolecular forces decreasing total intermolecular forces The melting point of ionic solids is extremely high compared to water which experiences all other intermolecular forces, but not ion-dipole forces. (NaCl is 1074 K and water is 273 K) Retention of V & Shape Increasing intermolecular forces and decreasing T & P Decreasing intermolecular forces, and increasing kinetic energy of particles or T & P Gases will fill the volume and shape of the container that holds them, while solids will retain their own shape and volume regardless of the container. Surface Tension with increasing intermolecular forces with decreasing intermolecular forces The molecules on the surface have less neighbors (and therefore less stabilizing intermolecular forces) and so have a higher potential energy, which the material will try to reduce with its shape (sphere): water beading. Viscosity increasing intermolecular forces and decreasing temperature decreasing intermolecular forces and increasing temperature Not just a property of liquids, also gases and solids. Amorphous solids change shape over time because of their viscosity. Vapor Pressure Decreasing intermolecular forces and increasing temperature Increasing intermolecular forces and decreasing temperature Ether has weaker intermolecular forces than water and a higher vapor pressure, so it evaporates much faster then water.

Gas Behavior Non Rigid Container: Piston balloon Rigid Container: Closed Flask P constant V increase w/ T or # of moles V constant P increase w/ T or # of moles PV = nRT P1V1P1V1 T1T1 = P2V2P2V2 T2T2 P total = P A + P B + P C

Phase Changes SOLID LIQUID GAS fusion freezing evaporation condensation deposition sublimation endothermic exothermic System absorbs energy from surrounds in the form of heat o Requires the addition of heat System releases energy into surrounds in the form of heat or light o Requires heat to be decreased

Phase Changes TEMPERATURE HEAT ADDED solid liquid gas s l l g fusion ΔH fus evaporation or vaporization ΔH vap endothermic

Chapter 8 & 9 Concepts 12  Identify the solvent and solute in a solution  Like dissolves like, predict which molecules will form solutions  Predict the effect of temperature or pressure on a solution  Perform concentration calculations & conversions  Perform dilution calculations  Predict relative changes in colligative properties between multiple solutions  Understand osmotic pressure & how your kidney’s work.  Identify an acid/base reaction, the acid, base, conjugate acid/base  Caculate K a, K b  Use K w to determine concentration of H 3 O + or OH -  Discuss how water acts as both an acid and a base  Perform titration calculations  Communicate how a buffer prevents large pH changes

CH 8 Equations & Conversions Molarity = moles of solute (mol) V of solution (L) M 1 V 1 = M 2 V 2

CH 9 Equations & Conversions K a = [H 3 O + ][ ] A − [HA] K b = [OH - ][BH + ] [ B] K w = [H 3 O + ][OH − ] = 1.0 x 10 −14 pH = -log[H 3 O + ] Acidic solution: pH 1 x 10 −7 Basic solution: pH > 7  [H 3 O + ] < 1 x 10 −7 Neutral solution: pH = 7  [H 3 O + ] = 1 x 10 −7

Solutions, Solubility, & Concentration 1.The solute is the substance present in a lesser amount. 2.The solvent is the substance present in a larger amount. Solubility is the amount of solute that dissolves in a given amount of solvent. REMEMBER: LIKE DISSOLVES LIKE.  In aqueous or liquid phase solutions solubility increases with increasing temperature  Gases dissolved in liquids increase solubility with decreasing temperature and increasing pressure Communicate how much of a solute is dissolved in a solvent using concentration:  % w/v  % v/v  % mass / mass  ppm  Molarity initial valuesfinal values M 1 V 1 = M 2 V 2 Dilution: Adding more solvent to the initial solution. The number of moles solute DOES NOT CHANGE.

Colligative Properties Colligative properties are properties of a solution that depend on the concentration of the solute but not its identity.  One mole of any nonvolatile solute raises the boiling point of 1 kg of H 2 O the same amount, 0.51 o C.  One mole of any nonvolatile solute lowers the freezing point of 1 kg of H 2 O by the same amount,1.86 o C. Apply pressure to reverse osmosis. This is how our kidneys filter blood Reverse Osmosis

Acids / Bases A Brønsted–Lowry acid is a proton (H + ) donor. A Brønsted–Lowry base is a proton (H + ) acceptor. HA+ B A − HB+B+ + gain of H + acidbaseconjugate base conjugate acid loss of H + HOH add H + HOH H + conjugate acid H 2 O as an acid HO − Conjugate base remove H + Strong: Weak: K w = [H 3 O + ][OH − ] H 2 O as a base

Acid / Base Equilibrium & pH H 3 O + (aq) + (aq)HA(g) + H 2 O(l)A − K a = [H 3 O + ][ ] A − [HA] pH = -log[H 3 O + ] OH - (aq) + BH + (aq)B (g) + H 2 O(l) K b = [OH - ][BH + ] [ B] Base dissociation constant acid dissociation constant Low pH (0 ~ 7) [H 3 O + ] high Acidic Conditions High pH (7 ~ 14) [H 3 O + ] low Basic Conditions

Common Acid / Base Reactions Neutralization reaction: An acid-base reaction that produces a salt and water. H + (aq) + OH − (aq) H—OH(l) H + (aq) + HCO 3 − (aq) H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) A bicarbonate base, HCO 3 −, reacts with one H + to form carbonic acid, H 2 CO 3. A carbonate base, CO 3 2–, reacts with two H + to form carbonic acid, H 2 CO 3. 2 H + (aq) + CO 3 2– (aq) H 2 O(l) + CO 2 (g) H 2 CO 3 (aq)

Titration AH + B  A- + BH + Acid + Base  Conjugate Base + Conjugate Acid Moles of base Moles of base Volume of acid Volume of acid mole–mole conversion factor mole–mole conversion factor M (mol/L) conversion factor M (mol/L) conversion factor Moles of acid Moles of acid Volume of base M (mol/L) conversion factor [1] [2] [3]

Buffers [ ] A − [HA] = [H 3 O + ] xKaKa pH of buffer = -log[H 3 O + ] where

Chapter 10 Concepts 22  Interpret Atomic number and mass number  Know radioactive particles: alpha, beta, positron, gamma  Write & solve radioactive decay equations  Determine the number of half lives that pass in a given amount of time.  Familiar with measurements of the amount of radioactivity  Familiar with measurements of radiation absorbed  Understand how radioisotopes are used in medicine

Atomic Symbols & Nuclear Particles 12 6 C mass number (A) atomic number (Z) the number of protons + the number of neutrons mass number (A) number of neutrons 6 12 – 6 = 6 atomic number (Z)= the number of protons = alpha particle:  or 4 2 He beta particle: β or 0 −1 e positron: β + or 0 +1 e gamma ray:  number of protons

Nuclear Equations & Half Life original nucleus new nucleus + radiation emitted 4 2 He 0 −1 e 0 +1 e  radiation emitted = The half-life (t 1/2 ) of a radioactive isotope is the time it takes for one-half of the sample to decay.

Radioactivity Radioisotopes can be injected or ingested to determine if an organ is functioning properly or to detect the presence of a tumor. amount of radioactivity 1 Ci = 3.7 x Bq. radiation absorbed The rad—radiation absorbed dose The rem—radiation equivalent for man