Chapter 111 atm*101,325 Pa760 mmHg*760 torr*1.01325 bar14.7 psi1. Describe the properties of gases1) A sample of gas assumes both the shape and volume of its container2) Gases are compressible.3) The densities of gases are much smaller than those of liquids and solids; and the density of a gaseous substance is highly variable depending on temperature and pressure. (g/L)4) Gases form homogeneous mixtures with one another in any proportion.

2. Use the assumptions of kinetic molecular theory to explain the behavior of gases1) There is a lot of empty space between the gasparticles compared to the size of the particles.2) The attraction (or repulsion) between particlesis negligible.3) The particles of the gas are constantly movingrandomly, in straight paths. When they collide in perfectly elastic collisions.4) The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature.

3. Understand the relationship between effusion and molecular mass Graham's law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass:

Effusion is the escape of gas molecules from a container to a region of vacuum. Lighter gases effuse and diffuse more rapidly than heavier gases

4. Define the standard temperature and pressure (STP) Standard Temperature and Pressure1 atm and 273.15 K (0C) Volume of 1 mole at STP = 22.4 L

5. List factors that may cause a gas to deviate from ideal behavior No attractions between gas molecules Gas molecule volume is not significant. At high pressure molecules are close together and individual volume becomes significant At low temperatures molecules are moving slower and any intermolecular forces become significant Molar volume of a real gas is larger than predicted by the ideal gas law at high pressures Molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures

6. Use the relationships among P, V, T, and n in calculations Boyles Law

Charless Law

Avogadros Law

7. Use the ideal gas law to determine density or molar mass Density, d = g/L

Molar mass

P = atm, V = L, n = moles, R = 0.08206 L*atm/K*mol, T = K

8. Perform gas mixture calculations using Daltons Law of Partial Pressures and mole fractions Dalton's law of partial pressures states that the total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture:

The mole fraction (i) of a component of a mixture is the number of moles of the component divided by the total number of moles in the mixture:

The mole fraction of a mixture component is always less than 1. The sum of mole fractions for all components of a mixture is always 1. Mole fraction is dimensionless.

9. Perform stoichiometric calculations for reactions with gaseous reactants and/or products. Reaction occurring at constant temperature and pressure, and involves only gases, coefficients in balanced chemical equation apply to units of volume, as well as numbers of molecules or moles. Balanced chemical equation and ideal gas equation can be used to determine volumes of gaseous reactants and/or products in a reaction.

Chapter 12Attractive forces are greater in:Larger molecules > smaller moleculesPolar molecules > nonpolar molecules (roughly equal size)Hydrogen bonds > cant form hydrogen bonds (same size)Ion-ion > other types of intermolecular forces

1. Describe the physical properties of liquids (surface tension, viscosity, vapor pressure, boiling point) and relate these to the types of attractive forces experienced by the sample of matter Surface tension is net pull inward on molecules at the surface of a liquid. Cohesion is attractive forces between molecules within a substance. Adhesion is attractive forces between molecules in substance and their container. Cohesion > adhesion convex meniscus. Adhesion > cohesion concave meniscus. Strong intermolecular forces have higer surface tension. Decreases with temp. Viscosity is resistance to flow. Stong intermolecular forces have higher viscosities. Decreases with temp. Vapor pressure measure how easily molecules escape to the vapor phase. Volatile substance has high vapor pressure. Weaker attractive forces, faster rate of evaporation. Increases with temp.Surface tension, viscosity, and vapor pressure are all temperature dependent. Boiling point is the temperature at which its vapor pressure equals the external atmospheric pressure Normal boiling point is temperature when vapor pressure is equal to 1 atm. Stronger intermolecular forces higher boiling point.

2. Predict the physical properties of various solids (melting point, vapor pressure, and amorphous vs. crystalline structures) 1. Melting point of solid is temperature which energies of particles break free from fixed positions and flow past one another. Strong intermolecular forces melt at higher temperatures.2. Vapor pressure magnitude of solids much lower than corresponding liquid.3. Amorphous solids if solid forms under certain extraordinary conditions, may not be sufficient time for molecules to move into positions of regular crystal.4. Crystalline solid possess rigid and long range order; atoms occupy specific positions.

3. Describe the basic types of crystalline solids (ionic, covalent, molecular, metallic)

4. Identify phase changes and their associated enthalpies

The molar heat of vaporization (Hvap) is the amount of heat required to vaporize 1 mole of a substance at its boiling point.The molar heat of fusion ( Hfus) is the amount of heat required to melt 1 mole of a substance at its melting point.The molar heat of sublimation (Hsub) is equal to the sum of the molar heats of fusion and vaporization: Hsub = Hfus + Hvap.

5. Interpret heating/cooling curves6. Calculate heat(q) changes with temperature and/or phase changes

7. Interpret a phase diagramTriple point is combination of pressure and temperature where 3 phases exist in equilibrium.Critical temperature (Tc) is the temperature above which a gas cannot be liquefied by applying pressure.Critical pressure (Pc) is the pressure necessary to liquefy a gas at its critical temperature.A substance above its critical temperature and pressure is a supercritical fluid has properties of both gas and liquid.Supercooling is the process of rapidly lowering a liquid's temperature below its freezing point.

Chapter 131. Describe types of solutions, including all phases

Saturated solutions contain the maximum possible amount of dissolved solute. Unsaturated solutions contain less than the maximum possible amount of solute. Supersaturated solutions contain more solute than specified by the solubility.

2. Describe changes in entropy and enthalpy when solutions are formedEnthalpy changes

The overall solution-formation process is exothermic () when the energy given off in step 3 is greater than the sum of energy required for steps 1 and 2. The overall process is endothermic () when the energy given off in step 3 is less than the total required for steps 1 and 2.

3. Describe how environmental factors (T and P) affect the solubility of a gas in a liquid1. Increasing the temperature decreasesthe solubility of most gases in water1. Increasing the pressure increases the solubility of gases in water

4. Use Henrys law to calculate solubility of a gas in a solution

c = molar concentration (mol/L), P = pressure (atm)

5. Explain and calculate different physical properties between solutions and pure solvent (colligative properties)Colligative properties depend on the number of solute particles in solution but not on the nature of the solute particles. vapor pressure loweringnonvolatile solute is dissolved in liquid, vapor pressure exerted by liquid decreasesSolvent in a solution will always exert a lower vapor pressure than the pure solvent.If both components are volatile

boiling point elevation elevationNon volatile solute lowers the vapor pressure of a solution, so increases the boiling point of the solution relative to the pure liquid

Tb = boiling-point elevation, Tb = boiling point of solution, Tb = boiling point of pure solvent

m = molality of solution (mol/kg), Kb = molal boiling point elevation constant (C/m)

freezing point depression

Tf = freezing point depression, Tf = freezing point of solution, Tf = freezing point of pure solvent

m = molality of solution (mol/kg), Kf = molal freezing point depression constant (C/m)

osmotic pressureosmotic pressure of a solution is pressure require to stop osmosis.

= osmotic pressure (atm), M = molarity (mol/L), R = gas constant (0.08206 Latm/molK), T = absolute temperature (K)

Chapter 141. Distinguish between a spontaneous and a nonspontaneous process

2. Define entropy (S) and Gibbs free energy (G) Gibbs free energy (G) is energy available to do work.

G has units of energy just as H and TS. Free energy is a state function.

Entropy of a system is a measure of how spread out or how dispersed the systems energy is

3. Identify trends in standard entropy values For a given substance, the standard entropy is greater in the liquid phase than in the solid phase. For a given substance, the standard entropy is greater in the gas phase than in the liquid phase. For two monatomic species, the one with the larger molar mass has the greater standard entropy. For two substances in the same phase, and with similar molar masses, the substance with the more complex molecular structure has the greater standard entropy. In cases where an element exists in two or more allotropic forms, the form in which the atoms are more mobile has the greater entropy.

4. Calculate the standard entropy change (S) for a given reaction

5. Predict the sign of S for a given process and use the sign to indicate whether the system has undergone an increase or decrease in entropyIncrease in entropy: Melting Vaporization or sublimation Temperature increase Reaction resulting in a greater number of gas molecules

Sfor: Decomposition of CaCO3(s) to give CaO(s) and CO2(g) is positive Heating bromine vapor from 45C to 80C is positive Condensation of water vapor on a cold surface is negative Reaction of NH3(g) and HCl(g) to give NH4Cl(s) is negative Dissolution of sugar in water is positive

6. State in words or symbols the second law and third law of thermodynamics Second law of thermodynamics, the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. For a process to be spontaneous as written (in the forward direction), Sunivmust be positive. Therefore, the system may undergo adecreasein entropy, as long as the surroundings undergoes a largerincreasein entropy, and vice versa. A process for which Sunivis negativeis not spontaneous as written.

Third law of thermodynamics the entropy of a pure crystalline solid is 0 at absolute zero. The significance of the third law of thermodynamics is that it enables us to determine experimentally the absoluteentropies of substances. Starting with the knowledge that the entropy of a pure crystalline substance is zero at 0 K, we can measure the increase in entropy of the substance as it is heated.

7. Calculate G from temperature, H, and S

8. Use the sign of G to indicate if a process is spontaneous

9. Predict the sign of G given the signs of H and S at high and low temperatures

10. Calculate the standard free energy change (G) of a given reaction

11. Calculate the temperature at which a process becomes spontaneous H = 199.5 kJ/mol and S = 476 J/K mol

Chapter 151. Write the equilibrium constant (K) expression for a given reaction

2. Calculate K given equilibrium concentrations of reactants and products

3. Predict the relative amounts of reactants and products at equilibrium given the equilibrium constant (K)1) Large Kc: The reaction will go essentially to completion and the equilibrium mixture will consist predominately of products. Greater than 1*1022) Small Kc: The reaction will not occur to any significant degree, and the equilibrium mixture will consist predominantly of reactant. Smaller than 1*10-23) Kc between 1*10-2 and 1*102: The reaction will proceed to a significant degree but will not go to completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.

4. Calculate the equilibrium concentration of reactants or products given initial concentrations.

5. Differentiate between heterogeneous and homogeneous equilibria Homogenous equilibrium reactants and products exist in same phase either gaseous or aqueous. Heterogeneous equilibrium reversible reaction where species are not all in the same phase. Only aqueous and gaseous species in equilibrium expressions.

6. Calculate the reaction quotient (Q) and predict the direction of a reaction given initial concentrations of reactants and products and the equilibrium constant (K).Predictions are made based on comparisons between Qc and Kc. There are three possibilities:

Q < K The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds in the forward direction. Q = K The initial concentrations are equilibrium concentrations. The system is at equilibrium. Q > K The ratio of initial concentrations of products to reactants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.

7. Calculate Gand G of a reaction at a specified temperature givenQorK.

R gas constant 8.314 J/K*mol or 8.314*10-3 kJ/K*molT absolute temperature in K

8. Use an ICE table to determine equilibrium, initial or final concentrations of reactants and products

9. Predict the shift of a reaction using Le Chtelier's principle given a change in one of the following: removal or addition of reactant or product, change in volume or pressure, and temperature change.

Decreased volume (pressure increase) shifts to direction with fewer moles of gasIncreased volume (pressure decrease) shifts to direction with more moles of gas

Increasing temperature adding heat, decreasing temp removing heat.Exothermic heat is productEndothermic heat is reactantEquilibrium shifts away from added heat

Chapter 161. Write base/conjugate acid and acid/conjugate base pairs

2. Define through words and examples Brnsted acids/bases and Lewis acids/bases Bronsted-Lowry acid are H+ donators. When acid donates proton, what remains of acid is conjugate base. Bronsted-Lowry base are H+ acceptors. When base accepts proton, new protonated species is conjugate acid. Amphoteric substance can be both acid and base. 3. Determine the relative strength of acids based on their composition and structure Strength of acid is measured by tendency to dissociate or ionize. Ionization/dissociation factors:a) Strength of H-X bondb) Polarity of H-X bond

Hydrohalic acid strength biggest factor is bond strength.HF