Updated Mar.17/06
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NUMERICAL ANSWERS TO ASSIGNED TUTORIAL PROBLEM SETS FOR CHEM206
FROM KOTZ & TREICHEL'S CHEMISTRY & CHEMICAL REACTIVITY
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NOTE: these answers have NOT been verified -- NOTE #101's answer has been fixed!
NOTE: "^" means "raised to the power of..." to indicate scientific notation
Ch. Q# Ans. Units SFs Comments
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15 1 explain using collision theory (see Z Ch.12.7)
15 19 3.0x10^-4 mol/(L min) 2 rate of appearance of ammonia
15 23a Rate = k[NO]2[Br2]
23b if [Br2] is tripled, rate will triple
23c if [NO] decreases by 1/2, rate will decrease by x 1/4
15 25a In first 2 data sets: [H2] is constant; [NO] is x1/2
rate decreases by factor of 4 --> rxn = 2nd order in NO
In 2nd & 3rd data sets: [NO] constant; [H2] doubled
rate also doubles --> rxn is 1st order in H2
25b Rate = k[NO]2[H2]
25c 0.136 mol/(L s) 3 rate of formation of N2
6.32 L2/(mol2 s) 3 rate constant, k
25d 0.159 mol/(L s) 3 initial rate of reaction calculated using rate law
15 27a 1st 2 data sets: [CO] constant; [NO2] halved
rate also goes down by factor of 2 --> 1st order in NO2
Sets 1 & 3: [NO2] constant; [CO] doubles
rate doubles --> rxn is 1st order in NO2
Rate = k[CO][NO2]
27b 1.9 L/(mol h) 2 rate constant, k
27c 2.0x10^-7 mol/(L h) 2 initial rxn rate calculated using rate law
15 29a rate=k[H2PO4-][OH-]2
15 29b 4.2x106 L2/(mol2min) 2 k, calculated using initial concentrations & rate
15 29c 0.0044 mol/L 2 initial conc. H2PO4- required for rate=0.0020 M/min
15 31 0.00557 min-1 3 k, calculated using integrated 1st order rate law
15 35 0.0490 min 3 time required to drop [NO2] from 2.00 to 1.50 mol/L
15 37 580 s 2 time required to decompose 3/4 of sample; 1/4 remains
15 41 150 min 2 time required for 7.50mg --> 0.25mg; using k=0.023 min-1
15 43 4.3 mg 2 remaining after 1.0 day; using k = 0.26 d-1
15 51 0.025 min-1 2 k,= -slope of linear graph: ln[HOF] vs time (1st order)
15 55 85 kJ/mol 2 Ea -- rxn 3x faster at 310K vs 300K; use Arrenhius' eqn
15 57 270 kJ/mol 2 Ea calculated using Arrhenius' eqns for 2 T's (2 k's)
15 59a endothermic
59b yes, rxn occurs in two steps
15 61a rate = k[Cl][ICl]
61b rate = k[O][O3]
61c rate = k[NO2]2
15 63a add up 2 rxns & cancel species occurring on both sides...
gives overall rxn: NO2 + CO --> NO + CO2
63b both steps are bimolecular
63c rate = k[NO2]2
63d NO3 is an intermediate
15 67a (CH3)3CBr + H2O --> (CH3)3CBr + HBr
67b step 1, the slow step, is rate-limiting (rate-determining)
67c rate = k[(CH3)3CBr]
15 69a false: likely occurs in 1 step, but not necessarily true
69b true
69c false: raising temp. T increases value of rate const. k
69d false: T has no effect on value of Ea (just # of collisions > Ea!)
69e false: if [ ] of both reactants x2, rate will increase by x4
69f true
15 71a decrease
71b increase
71c no change
71d no change
71e no change
71f no change (note: should read "1/2-life of cyclopropane")
15 75a rate = k[CO2]
75b 0.028 s-1 2 rate constant, k
75c 24 s 2 half life, t1/2
15 77a 1st order in CH3NC: rate = k[CH3NC]
77b ln[CH3NC]t = ln[CH3NC]o - kt
77c 2x10^-4 s-1 1 k = -slope (slope estimated from graph)
77d 3x10^3 s 1 [ ] remaining after 10000s, starting with 0.0166 mol/L
initial [ ] from y-intercept of graph: [ ]o = e^intercept
15 81a 0.59 mg 2 using k = 0.18 h-1 (from 1/2-life) & integrated 1st order rate law
81b 75 h 2 using integrated 1st order rate law
15 83 add up steps to get overall: 2O3--> 3O2
Cl atoms regenerated, = catalyst. ClO = intermediate
15 85a true
85b true
85c false: as rxn proceeds, [reactant] decreases --> rate decreases
85d false: possible to have 1step mechanism for 3rd order rxn
if the rate-limiting step is termolecular
15 87 103 kJ/mol 3 Ea = -slope of graph of lnk vs. 1/T
15 89a 75.8 s 3 when 3/4 of PH3 gone, 1/4 remains; 2 half-lives passed.
89b 0.33 unitless 2 NOTE: 2SF if treat as exactly 1 min; otherwise should be 1SF...
k = 0.0183 s-1; use 1st order int. rate law to calc. [PH3]t/[PH3]o
15 90a 2NO(g) + Br2(g) --> 2BrNO(g)
15 90b 1: termolecular; 2: step1=bi, step2=bi; 3: step1=bi, step2=bi
15 90c mech 2: intermediate = Br2NO(g); mech 3: intermediate = N2O2(g)
15 90d Assuming step1 = RLS for each mech., all predicted rate eqns different.
mech1: 2nd order in NO, 1st order in Br2
mech2: 1st order in both NO & Br2
mech3: 2nd order in NO, zero order in Br2
If assume 2nd step is slow, will get different results....
15 93 slowest rxn has smallest k: rxn (d)
fastest rxn has largest k: rxn (c)
15 97a 3 steps in mechanism
97b slowest step has largest Ea; steps 1&2 have ~ equal, large Ea's
97c overall rxn is exothermic
15 99 graphs.: same net E change for rxns; DIFFERENT Ea's
catalyzed rxn has LOWER activation E
15 101 4.76 min 3 must recognize that: t(90C)/t(100C) = k(100C)/k(90C)
i.e., shorter rxn time implies faster rate constant
and time required to complete a rxn is inversely proportional to k
(...from thinking about form of integrated rate law...)
then use Arrhenius' eqns for both T's & combine to solve for Ea
15 103 132 mm Hg 3 total P after 45min (1.5 half-lives)
P(HOF)=35.4mm Hg, P(HF)=65mm Hg, P(O2) = 32mm Hg
remember: partial P's are proportional to gas concentration
so can sub P's into integrated rate laws...
and here: had first used half life to find k, = 0.0231 min-1