·astreat
4 -ofherjetocts"cont
comparing expts 182, comparing expts 102,
a 10x10-K-2.16x10= (49 when [I] doubles and [OR] & [OU] remains constant,
12)
9 F 2
rate= oxoxioa doubles.
r a te of first order
a = reaction is word. [I] and a 1.
1
connsexptoLOO =
fos)" comparing expts 183,
4
= (4)" when [OCT] increases by 4 times, with II & IOH] remaining constant,
b=
the rate of reaction increases by 4 times.
comparing expts 18 4.
:rate of reaction is first order w.r.t. [OUT. b =
2.40 x 10-4-2.16x10-4
1.20 x 10-4-1.14x10-4" (EtC
4
=Ec comparing expts 204
10st4 = c when [Or] doubles with II and [0U] remaining constant,
10-4-1.14x10-4
120 x 10-4-1.08 x 10-4 1. 20 x
c = -
1 initial rate of reaction of expt.2= 10 initial rate of reaction of expt. 4I 10
rate=1IT
hot==
=6x10-7
of
IOH-]
Or rate of reaction
rate of reaction halved.
rate of reactionic -
order w.r.t IOMT and c =-1.
LSEOU
:rate:
men [IT = 1.20 x 104, [OU=16x10-4, [0nT =200 and rate of reaction = 6x10-7,
x 10-4]
6. 00x1077=1<[1.20x10-4351.16
[2]
K:
IT*)
Nooxo
=
62. 5
C) No, the reaction occurs by a multi-step mechanism as the experimental rate law does not match the balanced equation in part (a)
(d) UroU) + octiaal = n0U199) or 199) fast step, reversible
I-199) + 10C 1991 > H0E (ag) + (7199) slow, rate -
determining step
401 (ag) + or 199) + 0l (ag) + HroCl fast
overall equation: [+OC -> C70]
rate= K2 [I] [HOCT
rate forward K, L0U]
= rate reverse 1
=
[HOCIIORT
K,COCK =
K_CHOUTIOH-
rate= K2 FI-JTHOC KTOCIT
Ky[HOU] [0H-]
[oftheveK
:Etis =
2 At 20, 100 mol of X, 1500 since reaction is first order w.r.t. X. entASt=-1t + entAjo
At 5 = 1h,0.500 mol of X
↓
↑
temp to 32°
At t =3h,0.200 mol of X
when t = (hs
ent=00] k(1+enIT] = -
This study source was downloaded by 100000899606396 from CourseHero.com on 06-26-2025 04:32:14 GMT -05:00
en(0.500] =-K+en1
1 = InCO)=ench) temperature 298.154,K= when
https://www.coursehero.com/file/175660629/BS1012-chem-kinetics-tutorial-worked-solutionspdf/
= McaT
4 -ofherjetocts"cont
comparing expts 182, comparing expts 102,
a 10x10-K-2.16x10= (49 when [I] doubles and [OR] & [OU] remains constant,
12)
9 F 2
rate= oxoxioa doubles.
r a te of first order
a = reaction is word. [I] and a 1.
1
connsexptoLOO =
fos)" comparing expts 183,
4
= (4)" when [OCT] increases by 4 times, with II & IOH] remaining constant,
b=
the rate of reaction increases by 4 times.
comparing expts 18 4.
:rate of reaction is first order w.r.t. [OUT. b =
2.40 x 10-4-2.16x10-4
1.20 x 10-4-1.14x10-4" (EtC
4
=Ec comparing expts 204
10st4 = c when [Or] doubles with II and [0U] remaining constant,
10-4-1.14x10-4
120 x 10-4-1.08 x 10-4 1. 20 x
c = -
1 initial rate of reaction of expt.2= 10 initial rate of reaction of expt. 4I 10
rate=1IT
hot==
=6x10-7
of
IOH-]
Or rate of reaction
rate of reaction halved.
rate of reactionic -
order w.r.t IOMT and c =-1.
LSEOU
:rate:
men [IT = 1.20 x 104, [OU=16x10-4, [0nT =200 and rate of reaction = 6x10-7,
x 10-4]
6. 00x1077=1<[1.20x10-4351.16
[2]
K:
IT*)
Nooxo
=
62. 5
C) No, the reaction occurs by a multi-step mechanism as the experimental rate law does not match the balanced equation in part (a)
(d) UroU) + octiaal = n0U199) or 199) fast step, reversible
I-199) + 10C 1991 > H0E (ag) + (7199) slow, rate -
determining step
401 (ag) + or 199) + 0l (ag) + HroCl fast
overall equation: [+OC -> C70]
rate= K2 [I] [HOCT
rate forward K, L0U]
= rate reverse 1
=
[HOCIIORT
K,COCK =
K_CHOUTIOH-
rate= K2 FI-JTHOC KTOCIT
Ky[HOU] [0H-]
[oftheveK
:Etis =
2 At 20, 100 mol of X, 1500 since reaction is first order w.r.t. X. entASt=-1t + entAjo
At 5 = 1h,0.500 mol of X
↓
↑
temp to 32°
At t =3h,0.200 mol of X
when t = (hs
ent=00] k(1+enIT] = -
This study source was downloaded by 100000899606396 from CourseHero.com on 06-26-2025 04:32:14 GMT -05:00
en(0.500] =-K+en1
1 = InCO)=ench) temperature 298.154,K= when
https://www.coursehero.com/file/175660629/BS1012-chem-kinetics-tutorial-worked-solutionspdf/
= McaT
