CHM 1046 FINAL REVIEW - PDF Free Download

CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART I Chapter Description 6 Thermochemistry 11 States of Matter; Liquids and Solids 12 Solutions 13 Rates of Reactions 18 Thermodynamics and Equilibrium

CH. 6 THERMOCHEMISTRY Thermochemistry: Energy changes in a system during chemical/physical changes. Internal Energy (U): The total energy of a system. To calculate the change in Internal Energy ( U): U = U final U initial Work (w): Energy transfer between a system and its surrounding. W= -P V Heat (q): Energy transfer between a system and its surrounding that results from a temperature difference. State Functions: a property of system that depends on its present state independent of pathway. (ex: Heat, Work, Temperature, Pressure, volume, Energy) First Law of Thermodynamics: Energy cannot be created nor destroyed only an exchange between a system and its surrounding. U = q + w Ch. 6 Thermochemistry

Heats of Reaction: The heat absorbed/evolved in a reaction system and its surrounding at a given temperature. q = U + P V Enthalpy of Reaction( H): the amount of heat content of a reaction at a fixed pressure and a given temperature. H = q Energy is Added H > 0 Ex: Reaction Vessel cools Energy is Removed H < 0 Ex: Reaction Vessel warms Ch. 6 Thermochemistry

Heat Capacity (C):The amount of heat needed to raise the temperature of a system by 1 C or 1K. (Units: J/ C or J/K) q = C T Specific Heat (C p or S): The amount of heat needed to raise the temperature of 1g of substance by 1 C or 1K. (Units: J/g. C or J/g.K) q = m C p T Calorimeter: used to measure evolved/absorbed heat of a reaction. Coffee-cup Calorimeter: used for reactions in solutions at constant pressure. q reaction = -q water Bomb Calorimeter: used for reactions of gases at a constant volume. q reaction = -(q water + q bomb ) Ch. 6 Thermochemistry

Hess s Law: The total enthalpy change of a reaction is the sum of the enthalpy changes for each individual step. Standard Heat/Enthalpy of formation ( H f ): The enthalpy change for the formation of 1 mole of a substance in its standard state from elements in their standard states. H rxn = n H f (products) - n H f (reactants) Ch. 6 Thermochemistry

Questions 1. Indicate if the following reactions Endothermic or Exothermic A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) + 184 KJ B) 2 NH 3 (g) + 91.8 KJ N 2 (g) + 3 H 2 (g) C) NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) H = +12.7 KJ D) 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 KJ Ch. 6 Thermochemistry

Questions 1. Indicate if the following reactions Endothermic or Exothermic A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) + 184 KJ Exothermic B) 2 NH 3 (g) + 91.8 KJ N 2 (g) + 3 H 2 (g) C) NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) H = +12.7 KJ D) 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 KJ Ch. 6 Thermochemistry

Questions 1. Indicate if the following reactions Endothermic or Exothermic A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) + 184 KJ Exothermic B) 2 NH 3 (g) + 91.8 KJ N 2 (g) + 3 H 2 (g) Endothermic C) NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) H = +12.7 KJ D) 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 KJ Ch. 6 Thermochemistry

Questions 1. Indicate if the following reactions Endothermic or Exothermic A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) + 184 KJ Exothermic B) 2 NH 3 (g) + 91.8 KJ N 2 (g) + 3 H 2 (g) Endothermic C) NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) H = +12.7 KJ Endothermic D) 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 KJ Ch. 6 Thermochemistry

Questions 1. Indicate if the following reactions Endothermic or Exothermic A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) + 184 KJ Exothermic B) 2 NH 3 (g) + 91.8 KJ N 2 (g) + 3 H 2 (g) Endothermic C) NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) H = +12.7 KJ Endothermic D) 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 KJ Exothermic Ch. 6 Thermochemistry

Questions 2. How many joules of heat must be added to 20 grams of water to increase its temperature by 10 C? Ch. 6 Thermochemistry

2. How many joules of heat must be added to 20 grams of water to increase its temperature by 10 C? Solution: Questions q = m C p T q = (20 g) (4.184 J/g. C) (10 C) Ch. 6 Thermochemistry

Questions 3. When 24 g of Al (s) at 98 C is placed in 50 g of H 2 O, the final temperature is 26.5 C. What was the initial temperature of H 2 O? (C p Al(s) = 0.903 J/g. C). Ch. 6 Thermochemistry

3. When 24 g of Al (s) at 98 C is placed in 50 g of H 2 O, the final temperature is 26.5 C. What was the initial temperature of H 2 O? (C p Al(s) = 0.903 J/g. C). Solution: Questions q water = - q Al (m water )(C p water )(T f - T i ) = -(m Al )(C p Al )(T f - T i ) Ch. 6 Thermochemistry

Questions 4. What is the heat capacity of a given calorimeter when 80.0 J of energy was absorbed and the temperature is increased from 312 K to 325K? Ch. 6 Thermochemistry

Questions 4. What is the heat capacity of a given calorimeter when 80.0 J of energy was absorbed and the temperature is increased from 312 K to 325K? Solution: q = C T *Rearrange equation to solve for C Ch. 6 Thermochemistry

Questions 4. What is the heat capacity of a given calorimeter when 80.0 J of energy was absorbed and the temperature is increased from 312 K to 325K? Solution: q = C T C = q/ T Ch. 6 Thermochemistry

Questions 4. What is the heat capacity of a given calorimeter when 80.0 J of energy was absorbed and the temperature is increased from 312 K to 325K? Solution: q = C T C = q/ T C = 80.0 J/ 13K Ch. 6 Thermochemistry

Questions 5. Given the reaction below, determine the H for the equations below. 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + H 2 O (g) H = -906 KJ A. 2 NH 3 (g) + 5/2 O 2 (g) 2 NO(g) + H 2 O(g) B. 12 NH 3 (g) + 15 O 2 (g) 12 NO (g) + 3 H 2 O (g) C. 2 NO (g) + 1/2 H 2 O (g) 2 NH 3 (g) + 5/2 O 2 (g) Ch. 6 Thermochemistry

Questions 6. Calculate H for : 2Al (s) + Cr 2 O 3(s) Al 2 O 3(s) + 2Cr (s) Ch. 6 Thermochemistry

Questions 6. Calculate H for : 2Al (s) + Cr 2 O 3(s) Al 2 O 3(s) + 2Cr (s) Solution: H rxn = [ Al 2 O 3(s) + 2 Cr (s) ] - [ 2 Al (s) + Cr 2 O 3(s) ] H rxn = [ 1 mol (-1675.7 kj/mol) + 2 mol (0 kj/mol) ] - [ 2 mol (0 kj/mol) + 1 mol (-1134.7 kj/mol) ] H rxn = [ -1675.7 kj ] - [ -1134.7 kj ] Ch. 6 Thermochemistry

Ch. 11 States of Matter; Liquids and Gases CH. 11 STATES OF MATTER; LIQUIDS AND GASES Comparing Gases, Liquids, and Solids: Gases: compressible fluids Liquids: Incompressible fluids Solids: Incompressible and rigid Phase Transitions: Melting: Solid Liquid Freezing: Liquid Solid Vaporization: Liquid Gas Condensation: Gas Liquid Sublimation: Solid Gas Deposition: Gas Solid Heat of Phase Transition: Heat of Fusion ( H fus ): Heat needed to melt a solid. Heat of Vaporization ( H vap ): Heat needed to vaporize a liquid.

Ch. 11 States of Matter; Liquids and Gases Phase Diagrams: Locate the three phases of water on the diagram.

Ch. 11 States of Matter; Liquids and Gases Phase Diagrams: Liquid Solid Gas

Ch. 11 States of Matter; Liquids and Gases Phase Diagrams: Triple Point: the point at which all three phases of a substance coexist at equilibrium (at 0.01 C and 0.0060 atm) Liquid Solid Gas

Ch. 11 States of Matter; Liquids and Gases Phase Diagrams: Triple Point: the point at which all three phases of a substance coexist at equilibrium (at 0.01 C and 0.0060 atm) Liquid Critical Point: a temperature and pressure at which liquid and gas coexist. Critical Temperature: the highest temperature at which a gas can condense into a liquid (only increasing the pressure). Solid Gas

Phase Diagrams: Triple Point: the point at which all three phases of a substance coexist at equilibrium (at 0.01 C and 0.0060 atm) Liquid Critical Point: a temperature and pressure at which liquid and gas coexist. Critical Temperature: the highest temperature at which a gas can condense into a liquid (only increasing the pressure). Supercritical Fluid: The fluid that exist past above a critical point. Solid Gas Properties of liquids: Surface Tension: The force per unit area that is required to expand its surface area. Viscosity: the resistance of flow, property of liquids and gases. Ch. 11 States of Matter; Liquids and Gases

Intermolecular Forces: Forces of attraction between Molecules. Strongest to Weakest: Intermolecular Forces (IMF) 1) H-Bonding: (POLAR) any molecule that have a bond between a H covalently bonded to a very electronegative atom (N/O/F) is capable of forming an intermolecular H-bonding with each other. (EX. H 2 O, CH 3 OH, NH 3 ) 2) Dipole-Dipole: (POLAR) the attractive force between molecules when polar molecules are present, negative end of the molecule is aligned near the positive end of another. 3) London/Dispersion: (NON-POLAR) The attractive force between molecules that results fro the small instantaneous dipoles due to the varying position of electrons. Increase with increasing molecular mass. Decrease in molecules of the same molecular formula but with more branching (less surface area or more compact). Ch. 11 States of Matter; Liquids and Gases

Ch. 11 States of Matter; Liquids and Gases

Ch. 11 States of Matter; Liquids and Gases Questions 1. Which of the following molecules have a permanent dipole moment? a. H 2 O b. CO 2 c. CH 4 d. N 2 e. CO f. NH 3

Ch. 11 States of Matter; Liquids and Gases Questions 2. Ethanol C 2 H 5 OH and methyl ether CH 3 OCH 3 have the same molar mass. Which has a higher boiling point?

CH. 12 SOLUTIONS Solute (present in smaller amount) + Solvent (present in higher amount) = SOLUTION Solute: the constituent of a solution that is present in smaller amount. Solvent: the constituent of a solution that is present in greater amount. Ch. 12 Solutions

Solubility increases with temperature (mostly) Except with Gases: Solubility of gases decrease at higher temperature. Henry s Law: Solubility of gases increase with increasing partial pressure of the gas above the solution. S = k H P Ch. 12 Solutions

Colligative Properties Colligative Properties: dependent on the concertation (M) of ions/solutes in the solution. Molarity: Other Units of Concentration: Parts Per million (ppm): mg/l = g solute/g solution X 10 6 Mass Percentage of Solute: Molality: Parts Per billion (ppb): µg/l = g solute/g solution X 10 9 Parts Per trillion (ppt): ng/l = g solute/g solution X 10 12 Mole Fraction: same as Vapor Pressure= (vapor pressure of pure solvent) (vapor pressure of solution) Raoult s Law: Ch. 12 Solutions

Colligative Properties Cont d Boiling Point Elevation ( T b )= Boiling Point of Solution Boiling Point of Pure Solvent T b = i K b m Freezing Point Depression ( T f )= Freezing Point of Pure Solvent Freezing Point of Solution T f = i K f m Osmosis: movement of solvent through a semipermeable membrane until solute concentration on both sides of the membrane is equal. Osmotic Pressure (π ): Pressure required to stop osmosis in a solution. π = M R T (M: Molar Mass, R: Gas Constant=0.0821 L.atm/K.mol., T: Temperature in Kelvin) Ch. 12 Solutions

Questions 1. How many grams of CuSO 4 * 5H 2 O will be needed to make 75.0 ml of 0.500 M solution? Ch. 12 Solutions

Questions 1. How many grams of CuSO 4 * 5H2O will be needed to make 75.0 ml of 0.500 M solution? Answer: 0.075 L solution x 0.500 molcuso 4 5H 2 O 1 L Solution x 249.6 gcuso 4 5H 2 O 1 molcuso 4 5H 2 O = Ch. 12 Solutions

Questions 2. What is the molality of a solution made by dissolving 16.6 g of sodium bromide (NaBr) in 1330 g of ethanol (C 2 H 5 OH)? Ch. 12 Solutions

Questions 2. What is the molality of a solution made by dissolving 16.6 g of sodium bromide (NaBr) in 1330 g of ethanol (C 2 H 5 OH)? Answer: 16.6 g NaBr x 1 mol NaBr 102.89 g NaBr = 0.161 mol NaBr m = 0.161 mol NaBr 1.330 Kg C 2 H 5 OH = Ch. 12 Solutions

Questions 2. What is the mole fraction of oxygen gas in a mixture which contains 66.8 g of O 2, 44.1 g of N 2, and 21.5 g of H 2? Ch. 12 Solutions

Questions 2. What is the mole fraction of oxygen gas in a mixture which contains 66.8 g of O 2, 44.1 g of N 2, and 21.5 g of H 2? Answer: 66.8 g O 2 x 1 mol O 2 32 g O 2 = 2.0875 mol O 2 44.1 g N 2 x 1 mol N 2 28 g N 2 = 1.575 mol N 2 21.5 g H 2 x 1 mol H 2 2 g H 2 = 10.75 mol H 2 Total number of moles = 2.0875 mol O 2 + 1.575 mol N 2 + 10.75 mol H 2 = 14.4125 moles x = 2.0875 mol O 2 14.4125 moles = Ch. 12 Solutions

CH. 13 RATES OF REACTIONS Reaction Rate: Molar concentration of products increase per unit time and/or Molar concentration decrease of reactants per unit time. N2O4( g ) 2NO2( g ) Rate of decomposition of N2O4 = [N2O4] t Rate of formation of NO2 = [NO2] t Reaction Rates may be affected (direct relationship) by: 1. Temperature (increasing T usually increases rate) 2. Concentration of Catalyst 3. Concentration of Reactants (remember for rate law we only use concentrations of reactants) 4. Surface are of Catalyst or solid Reactant Ch. 13 Reaction of Rates

CH. 13 RATES OF REACTIONS Rate Law: used to find rate of reactions based on concentrations of reactants. For Example, The Rate Law For the following reaction is CO2 (g) + CF4 (g) 2COF2 (g) Rate = k [CO2] [CF4] (K is the rate constant, varies with T and its units depend on the rate law) Units here: k = Rate [CO2] [CF4] = mol/(l.s) ( mol L )(mol L *The rate-determining step in a reaction mechanism is the slowest step. mol/(l.s) x = mol ) L = L 2 mol.s Reaction Order: is used to classify the reaction. It is used in respect to each reactant written/effecting the rate law. It equals to the exponent of the concentrations in the rate law. For Example, Using the reaction above: This reaction is first order with respect to CO2 and in first order with respect to CF4. The reaction is second order overall. Ch. 13 Reaction of Rates

Questions 1. The reaction between NO (g) and Cl 2(g) has been studied at 50 C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g) NOCl (g) Experiment [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 1 0.250 0.250 1.43 x 10-6 2 0.250 0.500 2.86 x 10-6 3 0.500 0.500 11.4 x 10-6 a) What is the Rate law Equation for this reaction? b) What is the order of reaction with respect to NO (g)? c) What is the order of reaction with respect to Cl 2(g)? d) The over all order of reaction? e) What is the value of the rate constant K, at 50 C? f) What is the rate of formation of NOCl (g) When the [NO] and [Cl 2 ] is 0.220 for both? g) What happens to the rate when we double the concentration of NO (g)? h) What happens to the rate if the concentration of Cl 2(g) is tripled? Ch. 13 Reaction of Rates

1. The reaction between NO (g) and Cl 2(g) has been studied at 50 C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g) NOCl (g) a) What is the Rate law Equation for this reaction? For NO (g) : Rate 3 Rate 2 = ( [NO (g)]3 [NO (g)]2 11.4 x 10 6 2.86 x 10 6 = ( [0.500] [0.250] )x 4 = ( 2 ) x ) x Questions Experiment [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 1 0.250 0.250 1.43 x 10-6 2 0.250 0.500 2.86 x 10-6 3 0.500 0.500 11.4 x 10-6 X = 2 Ch. 13 Reaction of Rates

Questions 1. The reaction between NO (g) and Cl 2(g) has been studied at 50 C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g) NOCl (g) Experiment [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 1 0.250 0.250 1.43 x 10-6 2 0.250 0.500 2.86 x 10-6 3 0.500 0.500 11.4 x 10-6 a) What is the Rate law Equation for this reaction? Rate = k [NO (g) ] 2 [Cl 2(g) ] b) What is the order of reaction with respect to NO (g)? Second order with respect to NO (g) c) What is the order of reaction with respect to Cl 2(g)? First order with respect to Cl 2(g) d) The over all order of reaction? The reaction is third order overall Ch. 13 Reaction of Rates

Questions 1. The reaction between NO (g) and Cl 2(g) has been studied at 50 C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g) NOCl (g) Experiment [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 1 0.250 0.250 1.43 x 10-6 2 0.250 0.500 2.86 x 10-6 3 0.500 0.500 11.4 x 10-6 g) What happens to the rate when we double the concentration of NO (g)? Since the reaction is in 2nd order with respect to NO (g) Then if concentration is doubled, the rate would h) What happens to the rate if the concentration of Cl 2(g) is tripled? Since the reaction is in 1st order with respect to Cl 2(g) Then if concentration is tripled, the rate would Ch. 13 Reaction of Rates

Ch. 18 Thermodynamics And Equilibrium CH. 18 THERMODYNAMICS AND EQUILIBRIUM (at a given T, S > q T )

Ch. 18 Thermodynamics And Equilibrium Entropy (S) Entropy is a measure of the disorder of a system ΔS > 0 More Disorder ΔS < 0 More Order S > q/t (spontaneous Process) S = q/t (Equilibrium Process) In order of decreasing Entropy: Gas > Aqueous > Liquid > Solid Entropy increases if : a molecule brakes into two or more smaller molecules Moles of gases increase Solid Liquid or gas Liquid Gas Standard Change in Entropy, ΔS⁰: ΔS⁰ = n S⁰ (products) m S⁰ (reactants)

Gibbs Free Energy (G) Free Energy determines the Spontaneity of the reaction at a given T and P ΔG > 0 Nonspontaneous (larger than 10 KJ) ΔG < 0 Spontaneous (more negative than -10 KJ) ΔG⁰ = ΔH⁰ - T ΔS⁰ *Small negative/positive value of ΔG, means that at equilibrium there will be significant amounts of both reactants and products. Standard Gibbs free energy of formation, ΔG⁰ f : ΔG⁰ = n G⁰ f (products) m G⁰ f (reactants) ΔG⁰ and thermodynamic equilibrium constant, K: ΔG = ΔG⁰ + RT ln Q At equilibrium, free energy no longer changes, so ΔG = 0, then ΔG⁰ = - RT ln Q *Spontaneous Change: A change that occurs without external change. If a forward reaction is spontaneous, then the reverse is non-spontaneous. Ch. 18 Thermodynamics And Equilibrium

Ch. 18 Thermodynamics And Equilibrium Questions 1. Indicate if ΔS⁰ is positive or negative for the following reactions: A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) B) 2 NH 3 (g) N 2 (g) + 3 H 2 (g) C) 2 H 2 (g) + O 2 (g) 2 H 2 O (l)

Ch. 18 Thermodynamics And Equilibrium Questions 1. Indicate if ΔS⁰ is positive or negative for the following reactions: A) 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) 8 moles of gas 4 moles of gas so B) 2 NH 3 (g) N 2 (g) + 3 H 2 (g) C) 2 H 2 (g) + O 2 (g) 2 H 2 O (l)

Ch. 18 Thermodynamics And Equilibrium Questions 2. Calculate ΔS⁰ for the following reaction then calculate ΔG⁰ at 25 ⁰C and determine if the reaction is Spontaneous or nonspontaneous: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + H 2 O (g) H = -906 KJ Answer: Prediction: 9 moles of gas 5 moles of gas (ΔS⁰ is Negative??) Let s Calculate to find out the value of ΔS⁰ ΔS⁰ = n S⁰ (products) m S⁰ (reactants) 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + H 2 O (g) S⁰ = 4 x 193 5 x 205 4 x 211 189 ΔS⁰ = [(4 x 211) + 189] [(4 x 193) + (5 x 205)] J/K ΔS⁰ = [844 + 189] [772 + 1025] J/K ΔS⁰ = [1033] [1797] J/K

Ch. 18 Thermodynamics And Equilibrium Questions 2. Calculate ΔS⁰ for the following reaction then calculate ΔG⁰ at 25 ⁰C and determine if the reaction is Spontaneous or nonspontaneous: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + H 2 O (g) H = -906 KJ Answer: Now let s Calculate ΔG⁰ using the given T= 25 ⁰C = 298 K and given H = -906 KJ And the calculated ΔS⁰ = -764 J/K

Ch. 18 Thermodynamics And Equilibrium Questions 2. Calculate ΔS⁰ for the following reaction then calculate ΔG⁰ at 25 ⁰C and determine if the reaction is Spontaneous or nonspontaneous: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + H 2 O (g) H = -906 KJ Answer: Now let s Calculate ΔG⁰ using the given T= 25 ⁰C = 298 K and given H = -906 KJ And the calculated ΔS⁰ = -764 J/K = -0.764 KJ/K ΔG⁰ = ΔH⁰ - T ΔS⁰ ΔG⁰ = -906 KJ 298 K (-0.764 KJ/K)

Bibliography Ebbing, D. D., & Gammon, S. D. (2013). General Chemistry. Cengage Learning. General Chemistry Course Information. (2014). Retrieved from Dr. Sapna Gupta: http://drsapnag.manusadventures.com/index.php/general-chemistry Epp, E. (1999, July 19). Thermochemistry. Retrieved from Erik's Page: http://eppe.tripod.com/thermchm.htm