KINETICS, CH 14 #10, ch 14: A 2 + AB + AC  BA 2 + A + AC BA 2 + A + AC  A 2 + BA 2 + C AB + AC  BA 2 + C CATALYST IS CONSUMED AND REGENERATED REACTIVE INTERMEDIATE IS PRODUCED AND THEN CONSUMED.

#16 SLOPE OF LINES GIVES RATES: AT 75 MIN RATE IS 4.2 * M/min. OR 7.0 * M/sec AT 250 MIN RATE IS 2.1 * M/min (3.5 * M/sec ) TIME (MIN) INTERVAL (MIN) MOLARITY CHANGE OF MOLARITY RATE (M/s) * * * * 10 -5

#20, ch 14: 2 THE C 2 H 4 DISSAPEARS AT A RATE OF 0.37 M/s, AND WATER AND CO 2 APPEAR TWICE AS FAST 2(0.37 M/s) = 0.74 M/s. - [C 2 H 4 ]/ t [CO 2 ]/2 t [H 2 O]/2 t == - [C 2 H 4 ]/ t [CO 2 ]/ t [H 2 O]/ t == INDICATES THE CO 2 AND H 2 O APPEAR IN ½ THE TIME (TWICE AS FAST) AS IT TAKES THE C 2 H 4 TO DISSAPEAR. INDICATES C 2 H 4 NEEDS TWICE THE TIME TO DISSAPEAR AS IT TAKES FOR CO 2 AND H 2 O TO APPEAR. C 2 H 4 + 3O 2  2 CO H 2 O N 2 H 4 + H 2  2 NH 3 N 2 H 4 IS CONSUMED AT AT RATE OF 63 TORR/hr, THUS NH 3 IS PRODUCES AT DOUBLE THAT RATE: 2(63 TORR/hr) = 126 TORR/hr

TRIAL[S 2 O 8 2- ][I-]INITIAL RATE(M/s) * * * *10 -5 #32 CH 14 S 2 O I -  2 SO I 3 - [S 2 O 8 2- ] INCRESES A FACTOR OF 1.5 WHILE THE RATE INCRESES BY 1.5: S 2 O 8 2- IS FIRST ORDER.

TRIAL[S 2 O 8 2- ][I-]INITIAL RATE(M/s) * * * *10 -5 #32 CH 14 S 2 O I -  2 SO I 3 - [S 2 O 8 2- ] INCRESES A FACTOR OF 2, I- BY A FACTOR OF 1.5, WHILE THE RATE INCRESES BY 3 FOLD: S 2 O 8 2- IS FIRST ORDER(KNOWN) AND I - MUST BE FIRST ORDER. 2 * 1.5 = 3 : 3 1 = 3 The exponent of 1 makes the equation true, the reactants are first order

TRIALRATE M/s DIVIDE [S 2 O 8 2- ] (M) [I-] (M) K (M -1 s -1 ) *10 -6 / x0.036 = 4.0 * *10 -6 / x0.036 = 4.0 * *10 -6 / x0.050 = 4.0 * *10 -5 / x0.072 = 3.98 *10 -3 #32-B S 2 O I -  2 SO I 3 - AVERAGE k = 4.0 * M -1 s -1 #32-C DISSAPEARS 1/3 AS RAPIDLY AS I -, THEREFORE RATE = 3 (4.0 * )(0.015 M )(0.040 M) = 7.2 * M/s = RATE OF I- DISSAPEARANCE S 2 O 8 2- RATE=k [S 2 O 8 2- ][I-], VALUES FROM TABLE SUBSTITUTED IN.

#42 CH 14 timeMol Aln (Mol A)1/Mol A Plot of ln A vs t is linear,Reaction is 1 st order in A ln(A) t 1/mol(A)

#42 CH 14-b timeMol Aln (Mol A) 1/Mol A Plot of ln A vs t is linear,Reaction is 1 st order in A ln A 42-b k = -slope k = -[-3.91 –(-2.70)]/120 = s c t1/2 = 0.693/k = 0.693/ = 68.7s

f =e -Ea/RT #48-a CH 14 a) -Ea/RT = 1.60 J/mol x mol*K = K J f =e = 1.92 * b) -Ea/RT = 1.60 J/mol x mol*K = K J f =e = 4.09 * * / 1.92 * = 2.13 times more at 510K

#54 CH 14-b ln [k 1 /k 2 ]= Ea/R [1/T 2 – 1/T 1 ] ln [0.0796/0.0815]= Ea/8.314J/mol [1/1220 – 1/1010] = Ea ( * ) / J/mol Ea = 8.313( )J/mol = * 10 3 J/mol

1) HBr + O 2  HOOBr (SLOW) 2) HOOBr + HBr  2 HOBr 3)2 HOBr + 2 HBr  2 H 2 O + 2 Br 2 4 HBr + O 2  2H 2 O + Br 2 #68 CH 14 b) GIVEN VERBALLY IN PROBLEM RATE = [HBr][O 2 ], BOTH FIRST ORDER. THIS AGGREES WITH STEP 1) WHICH MUST BE SLOW. c) HOOBr AND HOBr ARE INTERMEDIATES d) THE INTERMEDIATES MY NOT ACCUMULATE IF THEY ARE IN A FAST STEP AFTER A SLOW ONE AS HERE.

#72 CH 14 1) 2 [ N 2 O + NO  N 2 + NO 2 ] 2) 2 NO 2  2 NO + O 2 2 N 2 O  2N 2 + O 2 CATALYST INTERMEDIATE 1) 2 N 2 O + 2 NO  2 N NO 2 (SLOW) 2) 2 NO 2  2 NO + O 2 2 N 2 O  2N 2 + O 2 THE INTERMEDIATE NO 2 DOES NOT ACCUMULATE AS IT IS PRODUCED IN THE SKOW STEP AND CONSUMED IN THE FAST STEP

ln kc – ln k = [ Ea – Eac/RT] or Ea-Eac = RT ln (kc/k) Where c is with catalyst. #80 CH 14 Ea-Eac = RT ln (kc/k) Ea-Eac = J * 310K * ln (1 * 10 5 ) Ea-Eac = * 10 4 = 30. kJ Represents the ratio of rates asked for in problem.