concentrations (molarity) rate constant, (k), depends on size, speed, kind of molecule, temperature, etc.

Transcription

1 #80 Notes Ch. 12, 13, 16, 17 Rates, Equilibriums, Energies Ch. 12 I. Reaction Rates NO 2(g) + CO (g) NO (g) + CO 2(g) Rate is defined in terms of the rate of disappearance of one of the reactants, but it can also be defined by the rate of appearance of one of the products. Rate = - [NO 2 ] = - [CO] = [NO] = [CO 2 ] t t t t - rate can be measured for a specific time (instantaneous rate = slope) or over a time interval (average rate) II. Integrated Rate Laws Rate = k [reactant 1] x [reactant 2] y -x,y are usually integers concentrations (molarity) rate constant, (k), depends on size, speed, kind of molecule, temperature, etc. Ex. 1) A + B C initial [A] initial [B] initial rate rate = - [B] row M M 4.0 x 10-5 t row M M 4.0 x 10-5 row M M 1.6 x 10-4 = 16 x 10-5 Compare two rows where only one concentration is changing (only A or only B ). comparing row 1 & 2: A remains the same, B is doubling and the rate is the same. (going down the rows) rate = k [A] [B] 1X 2X 2 0 equals 1, so [B] 0 comparing row 1 & 3: A is doubling, B is the same and the rate is four times bigger. (going down the rows) rate = k [A] [B] 4X 2X 2 2 equals 4, so [A] 2

2 Rate = k[a] 2 [B] 0 or Rate = k[a] 2 (2 nd order, since powers add up to 2 ) Find rate constant: (A is 2 nd order, B is zero order) Rate = k[a] x 10-5 = k (0.100 M) 2 plugging in values from row # x 10-3 = k Book does all 3 rows and averages (just do one row). Ex. 2) (CH 3 ) 3 CBr + OH - (CH 3 ) 3 COH + Br - rate = + [Br - ] [(CH 3 ) 3 CBr] 0 [OH - ] 0 t c row M 0.18 M 1.0 x 10-3 row M 0.10 M 2.0 x 10-3 row M 0.10 M 1.0 x 10-3 comparing row 1 & 3: (CH 3 ) 3 CBr is the same, (going up the row) OH - is basically doubling and the rate is the same. rate = k [(CH 3 ) 3 CBr] [OH - ] 1X 2X 2 0 equals 1, so [OH - ] 0 comparing row 2 & 3: (CH 3 ) 3 CBr is doubling, (going up the row) OH - is the same and the rate is doubling. rate = k [(CH 3 ) 3 CBr] [OH - ] 2X 2X 2 1 equals 2, so [(CH 3 ) 3 CBr] 1 rate = k [(CH 3 ) 3 CBr] 1 [OH - ] 0 = k [(CH 3 ) 3 CBr] 1 (1 st order overall) Find the rate constant: rate = k [(CH 3 ) 3 CBr] 1 (1.0 X10-3 ) = k (0.10 M) using row #1 1.0 X10-2 = k What would the rate be, if [(CH 3 ) 3 CBr] = 0.26 M & [OH - ] = 0.45 M? rate = k [(CH 3 ) 3 CBr] 1 rate = (1.0 X10-2 ) (0.26 M) rate = 2.6 X10-3

3 #81 Notes III. Reaction Mechanisms -are the series of steps or reactions necessary to achieve an overall process. *The slowest step always determines the rate law. 2 NO 2 NO 3 + NO slow NO 3 + CO NO 2 + CO 2 fast NO 2 + CO NO + CO 2 *NO 3 is an intermediate. (It s first a product, then a reactant, which cancel.) Rate = k [NO 2 ] 2 rate law from the slowest step! Ex.1a) A products (slow) rate = k [A] 1 b) 2A + B X (slow) rate = k [A] 2 [B] 1 X + B Y (fast) Y + Z products (fast) *Find the slow reaction or if only one reaction, assume it is slow! Use the slow reaction. IV. Chemical Kinetics/ Collision Model NaNO 3 + KCl NaCl + KNO 3 Double Displacement Ionic compounds in reactions do not need to rearrange their electrons, since the ions just change partners. These reactions occur quickly. Covalent compounds in reactions must change their electron clouds. The energy to do this comes from collisions! *The rate can be increased by increasing the temperature, since this increases the velocity of the particles (KE = ½ mv 2 ). The number and power of collisions increase. But the rate does not go up by as much as it should. Arrhenius proposed the existence of a threshold or activation energy that must be overcome in order to produce a reaction.

4 E a = activation energy needed for reaction to occur. E = overall energy change for the reaction, above it is (-) so exothermic. The kinetic energy of the moving molecules is changed into potential energy as the bonds are broken and formed. **Only collisions with enough energy (activation energy) will be able to form the activated complex and then the products. The reaction rate is still smaller, than the rate of collisions with enough energy to form the activated complex, because of molecular orientations. The OH - must hit from the opposite side to eject the Br -. *Catalysts: lower the E a, allowing reactions to go faster. *Inhibitors: slow the reaction.

5 #82 Notes Ch. 13 Chemical Equilibrium I. Equilibrium Equations N 2(g) + 3H 2(g) 2NH 3(g) at equilibrium: Forward Rate = Reverse Rate k f [N 2 ] 1 [H 2 ] 3 = k r [NH 3 ] 2 k f = [NH 3 ] 2 s k r [N 2 ] 1 [H 2 ] 3 K = [NH 3 ] 2 s products [N 2 ] 1 [H 2 ] 3 Molarity reactants *Exception: Pure solids and pure liquids are not included ( aq and g are included). {H 2 O (l) not included, H 2 O (g) included} K p is for gases only! 2 K p = P NH3 d 3 P N2 P H2 K p = K(RT) n **P in atmospheres R = L atm mol K n = mols products mols reactants Ex. 1) Write the K equation and is K = K p? a) Mg (s) + 2HCl (aq) MgCl 2(aq) + H 2(g) **(Solids and liquids do not count in either equation.) K = [MgCl 2 ] 1 [H 2 ] 1 n=2 mols product 2 mols reactant K p = K (RT) [HCl] 2 K p = K (RT) 0 K p = K b) Fe 2 O 3(s) + 3 CO (g) 2 Fe (s) + 3 CO 2(g) 3 mols product 3 mols reactant K = [CO 2 ] 3 K p = K (RT) [CO] 3 K p = K (RT) 0 so Kp = K c) 4 Fe (s) + 3 O 2(g) 2 Fe 2 O 3(s) 0 mols product 3mols reactant K = 1 K p = K (RT) [O 2 ] 3 K p = K (RT) -3 so K p K

6 Ex.2) For N 2(g) + 3 H 2(g) 2 NH 3(g) : What is the concentration of NH 3 at equilibrium, if at equilibrium there is mol N 2 and mol H 2 in 2.00 L? K = K = [NH 3 ] 2 M N2 = mol = M M H2 = mol = M [N 2 ] [H 2 ] L 2.00 L = [NH 3 ] 2 (0.0201M) 1 ( M) X10-7 = [NH 3 ] X10-4 M = [NH 3 ]

7 #83 Notes II. Le Chatelier s Principle If a system is in equilibrium and a condition is changed, then the system will shift toward restoring the equilibrium. A) Concentration N H 2 2 NH 3 Ex. 1) increase N 2 : increases collisions, so it speeds the reaction going to the right (products) shifts right (too much N 2, so it will shift to the other side (products)) Ex. 2) increase NH 3 : shifts left (too much NH 3, shifts to the other side (reactants)) Ex. 3) decrease H 2 : shifts left (too little H 2, so it must shift toward H 2 to increase it (reactants)) B) Pressure (only affects gases) Ex. 1) increase P, by decreasing volume (if decreasing volume, the reaction will shift toward the side that is the most compact (smallest volume)) N H 2 2 NH 3 4 mol 2 mol x 22.4 L x 22.4 L 88 L 44 L smallest volume (= least number of mols) the right side is more compact shifts right (if increase volume, shifts to the left, larger volume side) Ex. 2) increase P, by adding Ne, no change (Ne is not in the reaction, other gases will not change the volume, since gases are point masses (ideal)) C) Temperature Ex. 1) Girls + Boys Dancing + heat (exothermic, heat produced, ΔH = (-) energy on product side) Increase Temperature, Add heat, less dancing, shifts left Ex. 2) Decrease T, for exothermic reaction (ΔH = (-), energy on product side) A + B C + energy decrease T, too little energy, shifts right Ex. 3) Increase T for endothermic reaction (ΔH = (+),energy on reactant side) energy + A C increase T, too much energy, shifts right

8 Ch. 16 Energies I. H, G, S H = Enthalpy (heat energy) (+) H = endothermic (-) H = exothermic G = Gibb s Free Energy (chemical potential energy) (+) G = nonspontaneous (-) G = spontaneous S = Entropy (energy of disorder) (+) S = increasing (-) S = decreasing ΔH reaction = Σ n ΔH products(final) Σ n ΔH reactants(initial) ΔG reaction = Σ n ΔG products(final) Σ n ΔG reactants(initial) ΔS reaction = Σ n ΔS products(final) Σ n ΔS reactants(initial) see p. A-21 Ex. 1a) Calculate ΔG reaction for: 2 HF (g) H 2(g) + F 2(g) Is it spontaneous? ΔG reaction = Σ n ΔG products(final) Σ n ΔG reactants(initial) ΔG reaction = [(1 mol H 2 ) ( 0 kj/mol) + (1 mol F 2 ) ( 0 kj/mol) ] [( 2 mol HF) ( kj/mol)] ΔG reaction = kj = kj nonspontaneous Ex. 1b) Calculate ΔS reaction ΔS reaction = Σ n ΔS products(final) Σ n ΔS reactants(initial) ΔS reaction = [(1 mol H 2 ) ( J/(mol K)) + (1 mol F 2 ) ( J/(mol K)) ] [( 2 mol HF) ( J/(mol K))] ΔS reaction = J/K J/K J/K = J/K decreasing (+) H = endothermic (-) H = exothermic

9 #84 Notes II. Entropy (S) -is the disorder in a system. In nature systems get more messed up (gain entropy). Ex. 1) Is entropy increasing or decreasing? a) melting Al Al (s) Al (l) More ordered More messy (lower entropy) (higher entropy) ΔS = (+) increasing b) sublimating I 2 I 2(s) I 2(g) More ordered More messy (lower entropy) (higher entropy) ΔS = (+) increasing c) 2 CO (g) + O 2(g) 2 CO 2(g) all gases, so look at amounts 3mols 2 mols Three objects can get more disorganized than 2 objects. High entropy Low entropy ΔS = (-) decreasing d) Zn (s) + 2 HCl (aq) ZnCl 2(aq) + H 2(g) aqueous to aqueous is no change Solid Gas More messy, higher entropy ΔS = (+) increasing Ex. 2) Which has more entropy? a) C 6 H 12 O 6(s) or H 2 O (s) C 6 H 12 O 6(s) is more complex b) He at 0 o C or He at 200 o C 200 o C is a higher temperature, more motion c) He at 1 atm or He at 2 atm 1 atm has lower pressure, less organized ** The higher the temperature and the lower the pressure, the higher the entropy! STP = 273 K & 1 atm

10 III. Relationship between H, G, & S ΔG = ΔH TΔS ΔG in kj/mol, ΔH in kj/mol, T in Kelvin, ΔS in J/(mol K) **so change to kj/(mol K) Ex.1) Will a reaction be spontaneous at 25 o C, if ΔH = 542 kj/mol and ΔS = -14 J/(mol K)? -14 J 1 kj = kj/(mol K) mol K 1 X10 3 J ΔG = ΔH TΔS = (542 kj/mol) [(298 K) ( kj/(mol K))] ΔG = 542 kj/mol kj/mol = 546 kj/mol (+) so nonspontaneous

11 #85 Notes IV. Free Energy and Equilibrium No Element! G = G o + RT ln Q p G is at non-equilibrium conditions G o is at equilibrium, calculated by G reaction = n G products - n G reactants R = 8.31 J/mol K Q p is K p at non-equilibrium conditions Remember: K p = K (RT) n At equilibrium G = 0, so 0 = G o + RT ln K p G o = -RT ln K p Ex. 1) Find G for N 2(g) + 3 H 2(g) 2 NH 3(g) at 25 o C, if P N2 = atm, P H2 = atm, and P NH3 = atm. Appendix 4: G o reaction = n G products - n G reactants G o = [ (2 mol NH 3 ) ( kj/mol) ] ( -16 on paper) [ (1 mol N 2 ) (0 kj/mol) + (3 mol H 2 ) (0 kj/mol) ] G o = kj 2 Q p = P NH3 P 1 3 N2 P H2 = (0.012 atm) 2 = 2.88 X10 6 (0.050 atm) 1 ( atm) 3 G = G o + RT ln Q p G = (-32.9 kj 1 X10 3 J ) + (8.31 J/mol K) (298 K) ln (2.88 X10 6 ) 1 kj (14.87) G = J J = 3932 = 3.9 X10 3 J or 3.9 kj Start #86 Notes Ch.17 Electrochemistry I. Galvanic Cells Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Zn (s) Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e - Cu (s) oxidizing reducing

12 Oxidized Reduced Anode (-) Cathode (+) The salt bridge adds ions to keep the solutions balanced: (-) ions are added to (+) ions are added to balance the arriving replace the departing Zn 2+ (+) ions. Cu 2+ (+) ions. ** or cells separated by porous disk Cell potential = electromotive force (emf) = ε cell Reduction Potentials Table: Zn 2+ (aq) + 2 e - Zn (s) v Cu 2+ (aq) + 2 e - Cu (s) 0.34 v Zn (s) Zn 2+ (aq) + 2 e v Cu 2+ (aq) + 2 e - Cu (s) 0.34 v Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) 1.10 v (+) is spontaneous

13 #86 Notes I. continued ** If given 2 reduction potentials, make the reaction spontaneous. (flip one of them) Ex.1a) Find the cell potential for: Al e - Al v Ni e - Ni v Reaction must be spontaneous! Which should we flip? 2X Al Al e v **Don t multiply voltages! 3X Ni e - Ni v 3 Ni Al 2 Al Ni 1.419v (spontaneous) b) Find cell potential and sketch the galvanic cell. Ag + + e - Ag v Co e - Co v 2X Ag + + e - Ag v Co Co e v 2 Ag + + Co 2Ag + Co v Reducing Oxidizing Cathode (+) Anode (-) The salt bridge adds ions to keep the solutions balanced: (+) ions are added to (-) ions are added to replace the departing balance the arriving Ag + (+) ions. Co 2+ (+) ions. Read Effect of Temperature on Solubility Lab