Overview 4-8 An unorthodox chemistry student decides to set out on a round-the-world trip using a balloon filled with hydrogen. Barely remembering his physics lessons, he figures out from his weight that he ll need 5.03 kg of hydrogen to get airborne. The most convenient source of hydrogen available to him from household items is described by the following reaction: Al + NaOH + H 2 O NaAl(OH) 4 + H 2 [unbalanced] He knows that this is a redox reaction, and that he has 12.0 kg of aluminum (the combined weight of his utensils and those of his roommates) and a 15 kg bag of NaOH. Too busy to do the calculation, he throws everything in a big bucket with water and hopes for the best. Overview 4-8 (a) Is there enough aluminum to get the required amount of H 2? Is there enough NaOH? (b) Will there be any utensils (Al) left after the reaction is complete? (c) The weather that day is: P = 1 atm, T = 22 o C. How much volume will the produced hydrogen occupy (in pure form)? (d) If the weather changes to STP, what would the volume of H 2 be? (e) If he collects the hydrogen right over the aqueous reaction solution at 22 o C and 1 atm barometric pressure, what is the partial pressure of H 2? (P H2O = 19.8 torr at 22 o C) (f) What is the standard enthalpy change for the reaction (H o f (H 2 O) = -285.8 kj/mol; Ho f (NaOH) = -311.8 kj/mol; H o f(naal(oh) 4 ) = -2027.0 kj/mol) (g) What is the amount of heat released when the reaction is complete (assume the standard molar!h)?
Overview 4-8 (h) If 4500 L of water was used, and the starting temperature was 22 o C, what is the final temperature of the aqueous solution (c = 4.18 J/g. K)? in class we found that the amount of heat energy (q p, or!h, which is the same) produced in the reaction, that is, when 375 mmol of NaAl(OH) 4 was produced, is 375 mol X -858.7 kj/mol (that is, per mole of NaAl(OH) 4 produced) = -322000 kj. If this heat is used to warm 4500 L, that is 4500 kg, of water, the temperature will raise by c = q/m.!t;!t = q/m. c;!t = 322000000J/(4500000gx4.18J/g. K) = 17.1 K. Notice the sign change for heat, that is because the heat is consumed when water is heated. (i) What is the amount of work produced by the expanding hydrogen at the end of the reaction? work = w = -P!V; our P has been 1 atm, we determined in class that the volume of H2 produced at the end (at 295 K) is 13623.5 L. since there is no hydrogen at the beginning of the reaction,!v = V final - V initial = 13623.5 L; w = -1atm X 13623.5 L = -13623.5 atm. L. If you re given the conversion factor (1 J = 9.87.10-3 atm. L), the answer in joules is -13623.5/0.00987 = -1343.8 kj. (j) What is the!e?!e = w + q (= -P!V +!H) = -1343.8kJ -322000 kj = -323343 kj
Acid-Base Titrations Practice Problem 50.00 ml of an HCl solution of unknown concentration required 33.32 ml of 0.1524 M standard NaOH solution to reach the endpoint. What is the concentration of the HCl solution? HCl + NaOH " NaCl + H 2 O n(hcl) = n(naoh) = 0.03332 L # 0.1524 mol/l = 0.005078 mol 0.005078 mol 0.05000L = 0.1016 mol/l Oxidation-Reduction Red + Ox " Products e electrons reducing agent: loses electrons oxidizing agent: gains electrons
Oxidation-Reduction + $+ Cl Cl H H H Cl $% non-polar covalent bond: electrons distributed evenly polar covalent bond: electrons distributed unevenly OXIDATION STATE: A MODEL, compounds treated as if electrons were transferred completely, not actually shared. +1-1 Cl H Oxidation-Reduction redox reactions are characterized by change in oxidation state H 2 + Cl 2 " 2HCl reducing agent: loses electrons oxidizing agent: gains electrons +1-1 Cl H
Oxidation Number (State): General For an atom in its elemental form equals 0 (H 2, N 2 ) For monoatomic ion equals the ion charge (Ag +, S 2- ) The sum of O.N. for the atoms in a compound is zero (HCl) The sum of O.N. for the atoms in a polyatomic ion equals the ion charge ClO +1-2 Oxidation Number (State): General Rules for assigning an oxidation number 1. For Group 1A +1 in all compounds 2. For Group 2A +2 in all compounds 3. For hydrogen +1 in combination with nonmetals -1 in combination with metal and B 4. For fluorine -1 in all compounds 5. For oxygen -2 in all compounds except F and peroxides 6. For Group 7A (halogens) -1 except with O and halogens lower in the group
Balancing Redox Reactions +1-2 +1 +1 +3 +1-2 K 2 Cr 2 O 7 + HI " KI + CrI 3 + I 2 + H 2 O +6-1 -1-1 0 Step 1. Assign oxidation numbers Step 2. Find and the oxidant and reducing agent Step 3. Determine the number of electrons: (a) gained by the oxidant (3), and (b) lost by the reducing agent (1) Step 4. (the key step) Assign coefficients to match the number of electrons lost and gained. Step 5. Complete balancing by inspection (by guessing) K 2 Cr 2 O 7 + 14HI " 2KI + 2CrI 3 + 3I 2 + 7H 2 O Ideal Gas Laws PV = nrt PV nt = R
Ideal Gas Laws PV nt = R R = 8.31 J mol -1 K -1 universal gas constant R = 0.0821 atm L mol -1 K -1 fixed n and T Ideal Gas Laws PV nt = R fixed n and P fixed P and T PV = constant Boyle s law V T = constant Charles law V n = constant Avogadro s law
Gas at Standard Conditions Standard conditions are standard temperature and pressure STP: 0 o C and 1 atm (760 torr) Volume of 1 mol of gas is 22.4 L Ideal Gas molecules have mass, but no volume no attraction or repulsion between molecules move in straight lines between collisions no loss of energy during collisions, only transfer
Partial Pressure of Gases P total = P 1 + P 2 + P 3 + partial pressure is proportional to mole fraction that is, P 1 n = 1 P total n total P 1 = n 1 n total # P total Heat and Work: Two Forms of Energy Heat (q) Work (w) as thermal energy as all other (mechanical, electrical etc)!e = q + w
Energy is a State Function!E is independent of how the change takes place 1. C + O 2 CO + O 2 CO 2 2. overall!e is the same! C + O 2 CO2 Enthalpy (H): is Heat at Constant P pressure is equal in the initial and final states w = -P!V initial state final state Enthalpy: H = E + PV
Enthalpy is a State Function!H is independent of how the change takes place 1. C + O 2 CO + O 2 CO 2 2. overall!h is the same! C + O 2 CO2 Exothermic and Endothermic Processes new definitions for processes - exothermic: heat is released endothermic: heat is consumed energy diagram for exothermic process: A B H A!H < 0 B Time
Types of Enthalpy Heat of combustion (!H comb ) CH 4 + 2O 2 " CO 2 + 2H 2 O!H comb = -891 kj/mol Heat of formation (!H f ) 2K + Br 2 " 2KBr!H =!H f < 0 Heat of fusion (!H fus ) NaCl(s) " NaCl(l)!H =!H fus Heat of vaporization (!H vap ) C 6 H 6 (l) " C 6 H 6 (g)!h =!H vap Specific Heat Capacity, c c (J/g K) = q mass #!T specific heat capacity: the quantity of heat required to change the temperature of 1 gram of substance by 1 K
Stoichiometry of Thermochemical Equations sign magnitude 2H 2 O " 2H 2 + O 2 2 moles 2H 2 + O 2 " 2H 2 O 2 moles 1 mole 2 moles 2 moles 1 mole!h = 572 kj!h = -572 kj 1 H 2 + O 2 " H 2 O!H = -286 kj 2 1 mole 1 mole 0.5 moles Hess s Law Hess s law of heat summation: the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps A + B " C + D C + D " E + F A + B " E + F!H = a!h = b!h overall = a+b
Standard Heats of Reaction (!H rxn ) o!h at standard states is!h o standard states: gases: 1 atm and the gas behaving ideally aqueous solution: 1 M concentration pure substance: most stable form at 1 atm, 25 o C usually!h o rxn and!h f o!h o rxn = &m!ho f(products) - &n!ho f(reactants) m,n - stoichiometric coefficients
Atomic Structure orbitals are populated by electrons according to certain rules Additional quantum number!! spin quantum number (m s ) spin -1/2 spin +1/2 Atomic Structure orbitals are populated by electrons according to certain rules Name Symbol Permitted Values Property Principal n 1, 2, 3, etc orbital size (energy) Angular Momentum l 0 to n-1 orbital shape Magnetic m l -l to 0 to +l orbital orientation Spin m s +1/2 or 1/2 direction of e - spin
Atomic Structure orbitals are populated by electrons according to certain rules exclusion principle: no two electrons in the same atom can have the same four quantum numbers Atomic Structure exclusion principle: no two electrons in the same atom can have the same four quantum numbers an orbital electrons
Atomic Structure exclusion principle: no two electrons in the same atom can have the same four quantum numbers empty an orbital one electron electrons only three options! two electron
Energy Profile of Sublevels E N E R G Y 2s 2p 3s 3p 3d 4s 4p 1s What did I miss?