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ENTHUSE
IIT CHEMISTRY
PHYSICAL CHEMISTRY
CHEMICAL KINETICS HINT & SOLUTIONS
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Ind i r a V i h ar , K ot a ( R a j .) 324 005 P h . 0744 - 2799900
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SOLUTION : CHEMICAL KINETICS
EXERCISE # S-I
1.
1 1.08 10–2
1 [NO]
Rate of reaction =
= 
4
3
4 t
(i)
= 9 × 10–4 M sec–1
1 [NH3 ]
1 [NO]

4
t
4 t
[NH3 ]
= 3.6 × 10–3 M sec–1

t
1 [H2O] 1 [NO]

6
t
4 t
[H2O]
= 5.4 × 10–3 M sec–1
t
–
(ii)
(iii)
Rate = k [A]1 [B]2 [C]0
On doubling concentration rate will increases by a factor of 8 times.
2.
(i)
(ii)
3.
Rate of reaction =
4.
Rate = k [A]2 [B2]
If volume is reduced to one third then concentration of A and B2 each will increase by a factor
of 3.
5.
6.
2 –1.1
atm min–1= 0.012 atm min–1
75
C P 1
P = CRT   P =  CRT 


t
t RT
C
0.012
= 8.33× 10–6 M sec–1

t 0.08  300
k BrO–
3

k BrO–
3
1

k Br –
2
For 1st order reaction
1  0.4  0.7
k=
sec–1
ln 


20  0.2  20
Rate = k [A]
0.7
=
 0.2
20
= 7 × 10–3 M sec–1
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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1
7.
Pi
Pf
A
+
2B

0.6
0.8
0.6–0.2
0.8–0.4
rateinitially = k × 0.6 × 0.82
ratefinally = k × 0.4 × 0.42
C + D
0
0
0.2
ratefinal 0.4  0.42

rateinitial 0.6  0.82
8.
9.
10.
[S2O82– ]
[SO2–
1 [I – ]
4 ]
–

3 t
t
t
[S2O82– ]
1 [I – ]
–
(a) –
3 t
t
–
[S2O82– ]
[I ]
–
 –3 
=3  1.5× 10–3 M sec–1
t
t
2–
[S2O8 ]
[SO2–
4 ]

(b) –
t
t
[SO2–
4 ]

 1.5× 10–3 M sec–1
t
–
[O2 ] 1 [H2O]
1 [H2O2 ]

–
t
2 t
2 t
[H2O]
[O2 ]
(a) 
 2 
t
t
36
[H2O]
[O2 ]
[H2O]
36
–1
–1  2 
 18 gmin–1
M min M min

 2 
 2 
 
32
t
t
t
32
[H2O]
[H2O2 ]
(b) 
–
t
t
36
[H2O2 ] 36
M min–1  32  34 gmin–1


t
32
For zero order reaction
[A]t = [A]0 – kt
At 10 min, [A]t = 10 – 1.2× 10–2 ×600 = 10 – 7.2 =2.8M
 A0
10 = 833.33 sec
k
1.2  102
so at 20 min = 1200 sec  [A]t =0
t100% 

C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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2
 A o –  A  t =
0.1– 0.09
= 0.1 M min–1
1
11.
k=
12.
t100% =
13.
k=
14.
[A]t = [A]0 – kt = 10 – 2 × 10–6 × 2 × 24 × 60= 9.994 M
15.
(i)
t
 Ao  6 10–6 / 0.1 10–3
107
k
= 6 × 10–9 sec.
 Ao –  At   Ao – 0.25 Ao = 0.75[A]
t
1
 A0 – 0.1 A0  0.9  A0 = 1.2 hr.
t=
k
0.75 A0
0
1  ao 
ln 

72  0.25a o 
k=
(ii)
2ln 2 ln 2

72
36
t1/2 = 36 min.
t87.5% = 3 × t50% = 3 × 36= 108 min.
(i)
k=
1  100 
= 0.022 min–1
ln 

10  80 
(ii)
k=
1
 100 
= 62.125 min–1
ln 

0.022  25 
k=
16.
17.
t1/2 =
0.693
k
t99.9% =
18.
k=
0.69
1  100 
= 10 ×
ln 

k
k  0.1 
1  500  ln 2
ln

20  250  20
t1/2 = 40 month
t=
1  100 
= 40 month
ln
k  25 
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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3
19.
1  100 
k = ln 
1  98 
20.
 At
 A o
21.
(a)
Rate
(b)
[A]t = [A]0 e–kt= 1 × e–510
× 100 = e–kt × 100 = e3.810
Rate
4
90 60
× 100
= k [A]
= 5 × 10–5 × 1= 5 × 10–5 M sec–1
–5
6060
= 0.84 M
–5
= 5 × 10 × 0.84 = 4.2 × 10–5 M sec–1
22.
Let initial concentration of A and B are 3ao & 2ao.
After three half life (t = 15 sec)
3a
3a
Concentration of A = 3o  o
2
8
2a o 2a o
Concentration of B = 2 
2
4
CA 3

CB 4
23.
r = K [NO]x [H2]y
(i)
 when [H2] is halved rate of reaction is also halved so order w.r.t. [H2] is 1
 when [NO] is doubled rate of reaction become four time so order w.r.t. [NO] is 2 so
(ii)
(iii)
total order of reaction = 2 + 1 = 3
r = K[NO]2 [H2]
4.4 × 10–4 = K = × (1.5 × 10–4)2 × 4 × 10–3
K = 0.488 × 107 sec–1 M–2
rate = 0.488 × 107 × (1.1 × 10–3)2 × 1.5 × 10–3= 0.88572 × 10–2
24.
Order of reaction = 1 +
26.
A
t = 0 P0
t = t P0–x
t= 0

ln  (t1/2 )2 / (t1/2 )1 
B
0
x
P0

ln a o1 / a o2
+

C
0
x
P0
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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2P0 = P3  P0 =
P3
2
P2 = P0 + x  x = P2 – P0= P2 –
P3
2P  P
= 2 3
2
2
P3
P0
1
1
2
 ln
K = ln
t P0 – x t P3 – P2
25.
26.
27.
Use Hit and Trial method.
Use Hit and Trial method.
Use Hit and Trial method.
28.
t1/2  A0
1– n
1–n
2
1
 
50
25
1 – n = 1 n = 0

O

[H ]
CH3–C–OC2H5 
 CH3COOH + C2H5OH
30.
t=0
t=t
t=
a
a–x
0
a  42.03 – 19.24
a – x  24.02 – 19.25
k=
31.
0
x
a
0
x
a
1 42.03 –19.24 1 22.79
=
ln
ln
75 24.02 19.25 75 4.77
VKMnO4 [H2O2 ]
1 [H2O2 ]0 1 V0
1
22.8
ln

ln
= ln
t [H2O2 ]t t Vt 600 13.8
K=
(CH3)2O(g)  CH4(g) + H2(g) + CO(g)
32.
t=0
t=t
312
312–x
0
x
0
x
0
x
Pt = 312 + 2x
P = 2x = 96  x = 48
k=
1
312
ln
390 312  48
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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2N2O5(g)  4NO2(g) + O2(g)
33.
34.
35.
t=0
Po
0
t=t
Po–x
2x
t=
0
2Po
0
x
2
Po
2
5Po
= 584.5  Po = 233.8
2
3x
Pt =Po +
= 284.5  x = 33.8
2
11
233.8
k=
ln
t 30 233.8  33.8
2A(g)  4B(g) + C(b)
t=0
Po
0
0
x
t=t
0.1–x
2x
2
3x
0.1 +
= 0.145  x = 0.03
2
1
0.1
1000 0.1
t=
=
= 47.69 sec
ln
ln
3
7.48 10
0.1  0.03 7.48 0.07
S  G + F
t=0 a
0 0
t = t a–x
x x
t= 0
a a
rt  x × rG + x × rF
.......(i)
r  a × rG + a × rF
.......(ii)
r  a( rG + rF ) and rt  x ( rG + rF )
r –rt  (a–x)( rG + rF )
K=
36.
k=
r
1
ln 
t r  rt
1 r  r0
ln
t r  rt
1 3.8 13.1 1 16.9
=
ln
ln
60 3.8 11.6 60 15.4
1 r r
t = ln  0
K r  rt
Put rt = 0
=
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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37.

A
c
t=0
0
t=t
c–x
t=
0
At
t = 6.93
38.
After 1 hr
[A]t = [A]0 e –(x 10 x)1
= [A]0 e–11x

x
A0 1  e–11x

x  10x
k1 2 10–3 1


k 2 3 10–2 15
[C]t =
39.
k1
A
t = 0 ao
t= 0
 t1/2 overall
=
 A0
11
1 e 
–11x
B 1
ao×
C 15
ao× 16
ao
 100
16
ao = 1600
a 
1
1
 1600 
ln  o  =
ln 
t=
= 86.625 min
–3
–2
k1  k 2  a t   2 10    3 10   100 
[B] =

1

16
k2

40.
0
x
x
c
c
rt = 2.5 = 20(c–x) + 30x – 40x
2.5 = 20c – 30x
… (1)
r = –5 = 30c– 40c
1
–5 = –10c  c = = 0.5
2
 x = 0.25
1
 0.5 
k=
ln 
0.93  0.5 – 0.25 
k = 0.1 min–1
t=
At
B + C

1 1

60 90
(t1/2)overall = 36 min.
41.

k1 >>> k2

[C] =  A0 [1– e– k2t ] = 20 (1 – e–0.0693×10)
= 10 M
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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ln(k 2 / k1 )
k 2 – k1
42.
tmax =
43.
H = Eaf – Eab  72 = 77 – Eab  Ea b = 5 kJ
44.
45.
46.
k  E 1 1 
ln  2   a  – 
 k1  R  T1 T2 
E  1
1 
ln (4) = a 
–
2  300 320 
Ea = 13.44 kcal
ln 1.75 =
a  1
1 

8.314  298 308 
0.693
360
k  E 1 1 
ln  2   a  – 
 k1  R  T1 T2 
k380ºC 
47.
Fraction of molecules = e–Ea/RT
H = Eaf – Eab  0.12 = Eaf – 0.02  Ea b = 0.14 kcal
49.
Increase in rate constant due to presence of catalyst = e
 Decrease in Ea due to catalyst = x = 2.25 kJ
Ea in the absence of catalyst = 6.05 + 2.25 = 8.3 kJ
E
8.31000
Slope = – a  –
= –1000
R
R
51.
Rate = k1[N2O5]
52.
Rate = k2[NOBr2] [NO]
[NOBr2 ]

Keq. =
[NO][Br2 ]


53.
x
RT
 1.718
[NOBr2] = Keq. [NO] [Br2]
Rate = k2 Keq [NO]2 [Br2]
Rate = k2 [N2O2] [H2]
[N 2O 2 ]

Keq =
[NO]2

[N2O2] = Keq [NO]2
Rate = k2 Keq. [NO]2 [H2]
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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EXERCISE # S-II
1.
Rate = k[A]2 [B]
Given : [A] = 0.2 M
[B] = 2 × 103 M
Since [B] is very large in comparison to [A]
therefore [B] will remain almost constant with the progress of reaction.

Rate = k'[A]2
where k' = k[B]
= 5 × 10–5 × 2 × 103 = 0.1

For 2nd order reaction
1
t1 = t1 
2
2
k '[A]
=
2.

Keq =
1
= 50 min.
0.1 0.2
kf
= 0.16
kb
3.3 10 –4
= 0.16
kb
3.3 10 –4
= 2.0625 × 10–3
0.16
0.693
0.693


–4
kf  k b (3.310  2.0625 10–3 )

kb =

t1
2
= 389.96 sec
= 4.83 min.
3.
t=0
t = 30 min
t =  min
At
2P(g)  4Q(g) + R(g) + S(l)
Pº
0
0
0
x
Pº–x
2x
32.2 mmHg
2
Pº
0
2Pº
32.2 mmHg
2
t=
Total pressure = 2Pº +
At
Pº
+ 32.5 = 617  Pº = 233.8 mm.
2
t = 30 min
Total pressure = Pº–x + 2x +
x
+ 32.5 = 317  x = 33.8 mm.
2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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9

k=
1  Pº  1  233.8 
ln 
=
= 5.2 × 10–3 min.
ln 


t  Pº x  30  233.8  33.8 
After t = 75 min.
x = Pº(1 – e–kt) = 233.8(1– e–5.210
4.

Total pressure = Pº–x + 2x +

Rate  k 

k 20ºC
k3ºC
(a)

k  E
ln  2   a
 k1  R

ln(3) 
–3
75
) = 75.56 mmHg
x
+ 32.5 = 379.64 mmHg
2
1
time
3  64

3
64
1 1
  
 T1 T2 
Ea  1
1 


8.314  276 293 
Ea = 43.45 kJ/mole
(b)
k 
t  E 1 1
ln  2   ln  1   a   
 k1 
 t 2  R  T1 T2 
 64  43.45 103  1
1 
ln   

8.314  293 313 
 t2 

t2 = 20.47 hr.
A 
a0 M
a0–x
0
5.
B +
0
x
a0
C
0
x
a0
t=0
t = 20 min
t=
At
t=
Rotation r = a0 × 40 + a0 × (–80) = –20  a0 = 0.5
At
t = 20 min
Rotation rt = (a0 – x) × 60 + x × 40 + x × (–80) = 5


t=0
6.
t=1hr.
60 × a0 – 100 x = 5  x = 0.25
a
At
t = 20 min
x= 0
2
Half life = 20 min
a0 k1
A
a k2
0
B
&
C
k1 1

k2 9
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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
k2 = 9 k1
= 9 × 1.3 × 10–5 sec–1
After t = 1hr.
x = [A]0 1– e–(k1 K2 )t

7.


= 0.374 a0
k2
[C]
9

x 
 0.374
10
[A] k1  k 2
(a 0  0.347a 0 )
(a 0 – x)
At
T = 43.3°C
3163
+ 12 = 28
(273  43.3)
k = 1022 M–1 min–1
For 2nd order reaction
1
1
t1 

= 10–19 min–1
22
2
a 0  k 0.00110
log k =
8.
For reaction (i)
A  product
k300 1
0.693
 & k310 
= 0.0231 min–1
k310 2
30

k  E
ln  2   a
 k1  R
ln  2 
1 1
  
 T1 T2 
Ea  1
1 

8.314  300 310 
Ea = 53.59 kJ/mole
For reaction (ii)
k 
(ii) 310K

  k(i) 
k  E
ln  2   a
 k1  R
310K
× 2 = 0.0231 × 2 & Ea(ii) 
Ea(i) 53.59

kJ / mole
2
2
1 1
  
 T1 T2 
 0.0231 2 
53.59  1
1 
ln 




k1

 2  8.314  300 310 
k = 0.0327 min–1
9.
Rate = k3[CoCl] [Cl2]

…(1)
k2
[CoCl]

(from 2nd reaction equilibrium)
k –2 [Cl][Co]
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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11
k2
[Cl] [Co]
k –2

[CoCl] =

k1 [Cl]2

k –1 [Cl2 ]

[Cl] =

From (1), (2) & (3)
Rate
k1
[Cl2 ]
k –1
= k3
1mole k1
A
k2
0
10.
t=
…(3)
k2
k1

 [Co]  [Cl2 ]
k –2
k –1
= k3 
t=0
…(2)
3
k2
k1

 [Co]  [Cl2 ] 2
k –2
k –1
B
k1
1mole
k1  k 2
C
k2
1mole
k1  k 2

11.
Eaoverall 
k1
k2
Ea1 
Ea
k1  k 2
k1  k 2 2
k
2k
E 
 2E  ...
k  2k  ...
k  2k  ...
2
2 2
2E
8
(2n 1)
E 
 2E  ... =

E  ... =
=
E
n(n  1)
n(n 1)
n(n  1) n(n 1)
3
Eaoverall 
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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EXERCISE # O-I
3.
4.
d[N2 ]
1 d[H 2 ] 1 d[NH3 ]
= 
=
dt
3 dt
2 dt
k1'
k ''
k
k
= 1 = 1 =
2
2
1
2
d[A]
d[B] d[C] 1 d[D]
= 
=
=
4dt
dt
dt
2 dt
5.

6.
Rate = k [A]1 [B]0
7.
Rate = slope =
8.
Rate = k [A]1 [B]2
Initial Rate
= 1 × 10–2 = k × 1 × 12
K = 10–2
 Final Rate = kc [A]1 [B]1
= 10–2 × 0.5 × 0.52
= 1.25 × 10–3 M sec–1
9.

10
= 0.5 M sec–1
20
1 d[A] 1 d[B]
=
y dt
x dt
d[A] x d[B]
=

y dt
dt
d[A] 
x
d[B] 
log  
= log  log 


y
 dt 
 dt 
x

log
= 0.3
y

x
=2
y
10.
‘k’ depends only upon temperature and catalyst.
12.
[A]t = [A]0 – kt
[A]0 = [A]t + kt
= 0.05 + 0.2 ×
30
60
= 0.15 M
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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13.
For zero order reaction
Rate = k
14.
[B]t = 2 kt
= 2 × 10–3 × 100
= 0.2 M
15.
t1 =
 A0
2k
log( t 1 ) = log [A]0 – log (2k)
2
2
18.
Rate = k [A]
19.
Rate = 4 × 10–3 × 0.02
= 8 × 10–5 M sec–1
20.
k=
1  A0
ln
t  A t
1  1 
n
20  0.25 
= 0.06931 min–1
=
21.
k=
0.693 0.693

= 10–2 sec–1
t1
69.3
2
Rate = k × [A] = 10–2 × 0.1 = 10–3 M sec–1
22.
1
Fraction of reactant left after n half lift =  
2

40 min is 2 half life

at  1  1


a o  2  4
n
2
24.
t1 =
2
26.
0.693
= 600 sec.
1.155 103
k1 [D]2
1 

k=
ln 
 = 0.0255
k 2 [P]
 0.6 
1
0.6 
t=
n 
= 20 min.
0.0225  0.36 
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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27.
29.
1
1
+kt

at ao
1 d[NO2 ] d[O2 ]
=
= 1.5 × 10–4
2 dt
dt
d[NO2 ]

= 3 × 10–4 = 3 × 10–3 [NO2]2

dt
[NO2] = 0.316 M

1
100  2ln 2
ln 
 = 100
100
 100  75 
1
200 
t=
ln 
 c = 100 min.
k
 200  150 
30.
k=
31.
k=
33.
1
100 
ln 
 = 0.0805
20
 100  80 
0.693
t1 =
= 8.66 min.
2
0.0805
A  nB
A0
0
A0– x
nx
At intersection point [A] = [B]  A0– x =nx  x =
 [B] =nx =
34.
A
n 1
nA
n 1
For nth order reaction
1
c
t 1 × n1  n1
2
ao
ao
log ( t 1 ) = log c – (n – 1) = log ao
2

Slope = –(n – 1) = – 1

n=2
37.
Rate = k [A]t = k [A]0 e–kt
38.
| Slope | = k
=
2
2  2.303
× 2.303 min–1=
sec–1 = 7.7 × 10–3 sec–1
10
10  60
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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39.
t1 =
2
0.693
min
8 105
= 8.6625 × 103 min.
40.
Concentration of A remain constant because it is present in large amount so order w.r.t. A will
be zero in overall order or reaction.
41.
t1/2 
1
 [A]01
[A]0
On comparing with
t1/2  [A]10n
1 – n = –1  n = 2
42.
r = K [A]x[B]y
from exp. 1 and 2 on increasing [A]0 2 times rate becomes 8 time, so x = 3
from exp. 2 and 4 on increasing [B]0 2 times rate does not change, so y = 0
r = K[A]3[B]0
43.
r = k[A]x[B]y
from exp 1 Q 2 x = 1
from exp 2 Q 3 y = 0
r = k[A][ B]0
ln 2
t1/2 =
k
6.93×10–6 = k× 0.01
k = 6.93×10–4
0.693
t1/2 =
= 1000 sec
6.93 104
t = 50 min = 3000 sec
[A] 0.5 1
 M
[A]t = 3 0 
2
8 16
1
r = 6.93×10–4×
M sec–1
16
44.
t1/2  P01n
1n
950  250 

235  500 
950
1
log
= (1 – n)log
235
2
0.6 = –(1–n)×0.3
1–n = –2  n = 3
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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(CH3)2.CHN=NCH(CH3)2(g)  N2(g) + C6H10(g)
45.
t=0
Po
0
0
t=t
Po–x
x
x
Pt = Po + x  x = Pt – Po
Po – x = Po – (Pt – Po) = 2Po – Pt
k=
P
Po
1
1
ln o = ln
t Po  x t 2Po  Pt
H2O2(aq)  H2O() +
46.
t=0
C
0
t=t
C–x
x
t=
0
C
1
O2(g)
2
0
x
2
C
2
C
x
 50 and  5
2
2
k=
1
C
1 50
=
ln
ln
20 C  x 20 45
A  2B + C
47.
t=0
Po
0
0
t=t
Po–x
2x
x
2Po
Po
t= 0
3Po = 270  Po = 90
Pt = P0 + 2x = 176 2x = 80  x = 43
(PA)t = Po – x = 90 – 43 = 47
48.
49.
NH4NO2(aq)  N2(g) + 2H2O()
t=0
C
0
0
t=t
C–x
x
2x
t=
0
C
2C
C  70 and x  40
1 70
k = ln
t 30
A  A + C
1 r r
1 100  0
1
K = ln  0 =
=
ln
ln 2 = 0.0693 min–1
t r  rt 10 100  50 10
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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50.
K=
1 r  r0
ln
t r  rt
1 10  40
ln
10 10 15
1 50
ln
10 25
0.0693 sec
=
51.
[B] k1

[C] k2
52.
2
A
0.3

2B
2
2C
1
[B] k1 1


[C] k2 2
Total moles = 0.5 + 2 + 1 = 3.5
53.
r = Z11.e–Ea/RT.P
54.
The min extra amount of energy given to reactant molecule so that they can cross energy
barrier.
EIn
PE
55.
P
R
Reaction co-ordination
56.
Molecules with sufficient amount of energy and proper orientation
58.
k = A e–Ea/RT
59.
k = A if
T.
60.
k = A if
T.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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61.
 k  Ea  1 1 
n  2  
  
 k1  R  T1 T2 
62.
 k  Ea  1 1 
n  2  
  
 k1  R  T1 T2 
63.
Fraction of molecules = e–Ea/RT =
64.
3.8  1016
100
–Ea/RT
% of activated molecules= e
×100
65.
The reaction which has more steep curve will be more temp sensitive.
66.
n k = n A –
68.
Slope = 
70.
Rate - k [A] [B]
71.
Rate = k3 [Q]2 [P]
Ea  1 
R  T 
Ea
8.3 1000
= –1000

R
8.3
 Keq =
k1 [Q]2

k 2 [P]
 [Q]2 =
k1
[P]
k2
 Rate =
k1
k3 [P]2
k2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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19
EXERCISE # O-II
1.
(A) Order of reaction is experimentally decided.
(B) r = K[A]°  r = K
(C) Zero order reaction does not depends on conc. of reactant
(D) Zero order reaction is always complex reaction
2.
3.
1
d[A]
= k[A]2
dt
On solving
2
1/2
k = [A1/2
0 A ]
t

d[C]
= k[A]1 [B]–1
dt
[A] 1


[B] 2
1


4.
d[C]
A
= k [A]   = 2k
dt
2
[C] = 2kt
3A
P°

2B
+
2C
t=0
0
0
2
2
t = 20 min
P° – x
x
x
3
3
2
2
t=
0
P°
P°
3
3
2P 2
4
At t =  : Total pressure=
+ P = P = 4  P° = 3
3
3
3
2
2
At t = 20 min : Total pressure = P° – x + x + x = 3.5  x = 1.5
3
3
o
P

At t = 20 min.
x=
2

t1/2 = 20 min.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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5.
For 2nd order reaction
1
1
+ kt

[A]t [A]0
Curve between
6.
1
1
vs t is straight line with slope k and intercept
[A]t
[A]0
B 1
2 M
3
A
t=0
2M
t=
0
C 2
2 M
3
At t = :
2
4
× (–72) + × (42)
3
3
=8
0.693 10
Hal life =
min.

k1  k 2 3
Rotation =
7.
k = A eEa /RT
At
T :
k=A
When Ea = 0
:
k=A
8.
(A)
(B)
(C)
(D)
Unit of A = Unit of k1
Because molecularity can’t be zero.
For complex reaction molecularity is not defined.
k increases with increase in temperature.
9.
(A)
(B)
(C)
For elementary reaction order must be equal to molecularity but reverse is not valid.
Same as (A)
Order can be found experimentally.
d CH3Br  k1[CH 4 ][Br2 ]
=
[HBr]
dt
1 k2
[Br2 ]
10.
(B)
If
[HBr]  0
d CH3Br 
= k1 [CH4] [Br2]
dt
(C)
If
[Br2]  0
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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d
[CH3Br]
dt
=
=
k1 [CH4 ][Br2 ]
[HBr]
1 k2
[Br2 ]
k1 [CH4 ][Br2 ]2
k [CH 4 ][Br2 ]2
= 1
[Br2 ]  k 2 [HBr]
k 2 [HBr]
11.
Because for simple reaction order is equal to molecularity and molecularity of a reaction on not
be fractional.
12.
For reaction of the order one or grater rate depends upon concentration of reactant and as the
concentration of reactant decreases its rate also decreases so it take infinite time for completion.
13.
For endothermic reaction :
Ea f
Eab
1
P.E.
2
Threshold energy
H
3
Progress of reaction
14.
15.
16.
Presence of catalyst charges the mechanism of reaction.
df
= k(1–f)
dt
1  1 

k = ln 
t  1– f 
ln(1–f) = –kt
f = 1–e–kt
t=0
17.
t=t1/2
k1=k
2mole
A
0
k2=2k
Total mole = 1 =
18.
mole of B =
2B
k
2
1 mole  mole
k  2k
3
2C
2k
4
1 mole  mole
k  2k
3
2
2
2 4
 = 3 mole
3 3
2
mole
3
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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
t75%
= 100 min. = 2 t1/2

t1/2


k


t90%
= 50 min.
0.693
=
50
50
1  a 
= n  0  
× n 10= 166.66 min.
k  0.1a 0  0.693

20.




Since reaction is occuring at constant pressure it means volume of the contains is variable.
0.693

t1/2 =
= 50 min
1.386 102
19.
A(g)

3B(g)
+
2C(g)
t=0
2 mole
0
0
2
t = 100 min.
=0.5 mole
3×1.5
2×1.5 mole
22

Total moles = 0.5 + 4.5 + 3 = 8 mole
V2
n

= 2
at constant pressure.
V1
n1

V2 = 12.5 ×

[B] =
21.
tmax
=
22.

k1 >>> k2
8
= 50 t
2
4.5
= 0.09 M
50
n(k 2 / k1 )
(k 2  k1 )
Option (C) will be correct.
2B
23.
A
3C
[A]0
= [A]t +
[B] [C]

2
3
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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23
Y
24.
X
Z
[X]t
[Y]t
[Z]
25.
= a0 × e(k1 k2 )t
k1
=
a0[1 e(k1 k2 )t ]
k1  k 2
=
k2
a0[1 e(k1 k2 )t ]
k1  k 2
k1 = 3
CH4 + CO2
CH3 COOH
k2 = 4
CH2CO + H2O
Fraction of CH3 COOH reacting as per reaction (i)
k1
3
3
=


k1  k 2 3  4 7
26.
k1
3B
k2
2C
A
dB
= 3k1 [A]
dt
dC
= 2k2 [A]
dt
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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EXERCISE # JEE MAINS
1.
2.
Let Rate = k[A]m [B]n
On putting given data
0.045 = k (0.5)m (0.05)n
0.090 = k (0.10)m (0.05)n
0.720 = k (0.20)m (0.10)n
On solving
m = 1, n = 2
…(1)
…(2)
…(3)
K1
K2
A 
 B 
C
Net, rate of formation of B =
d[B]
= K1[A] – k2[B]
dt
d[B]
=0
dt
K1[A] – K2[B] = 0
K
[B] = 1 [A]
K2
As,




3.
For zero order reaction
Ct = Co –kt
Ct /[R]
time
For Ist order reaction
Ct = Coe–kt
lnCt = lnCo – kt
lnCt/[lnR]
time
4.
Activation energy to from C is 5kJ/mole greater than that of to from D
(Ea)C = 15 kJ/mole
(Ea)D = 10 kJ/mole
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5.
N0
 e– t
N
before 1 Hour
N0
Will decrease
N
N
dN
N
N0
2
  –5dt
1
1

 5t
N N0
6.
N0
 1  5t N0
N
N
After 1Hr. 0 Will increase.
N
k
Ea  1 1 
In 2 

k1 R  T1 T2 
Ea  1
1 
1



=


4 
 2.5 10  8.314  600 800 
In 
Ea = 166 KJ
7.
xA  yB
1 d[A]
1 d[B]
–
=+
y dt
x dt
d[A] x d[B]
=
y dt
dt
Taking log both the side,
–

8.
 d[B]  log  x 
 –d[A] 
log 
= log 
 

+
 dt 
 dt 
 y
x
log   = 0.3010 = log2


 y
x 2
  A is C2H4 & B is C4H8
y 1
2N2O5  4NO2(g) + O2(g)
t = 0 3M
t = 30 2.75 M
[N2O5 ] 0.25
=
t
30
1 – (N2 O5 )
1 [NO2 ]
×
= ×
t
t
2
4
[NO2 ]
0.25
–2
–1
–1
=
 2 = 1.667 × 10 mol L min
t
30
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9.
r = k [A]x [B]y
When double the conc. of A. rate is also doubled, So order w.r.t. A is 1. But by changing the
conc. of B rate does not change so order w.r.t. B is zero.
R = k[A]1
6.93 × 10–3 = k × 0.1
K = 6.93 × 10–2
kA
 kr
2
kA = 2 × 3.93 × 10–2
t1 
2
0.693
= 5 min
2  6.93 10–2
10.
2A + B  product
R.O.R = k[A]x [B]y
0.3 = k[A]x [B]y
....(1)
x
y
2.4 = k[2A] [2B]
....(2)
x
y
0.6 = k[2A] [B]
....(3)
Solving the above equations, use get,
x = 1 and y = 2
i.e.
order w.r.t A is 1
and order w.r.t B is 2
11.
k = Ae–Ea/RT
Graph(I): Higher is the value of Ea, lower will be rate constant for the reaction.
Graph(II): On increasing temperature rate constant always is increases.
12.
Rate of formation of A = 2 k1[A2]
Rate of disappearance of A = k–1[A]2
 Net rate of formation of A = 2 k1[A2] – 2k–1[A]2
13.
K = Ae–Ea/RT
lnK = lnA -
Ea
RT
y = C – mx
slope = –Ea = –y
Ea = y
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14.
For zero order :
Co
k=
2 t 1
2

k=
Now,

15.
0.2
1
 k=
2 6
60
Co – Ct
=t
k
0.5 – 0.2
t=
× 60 = 18 hrs
1
For zero order reaction t 1 =
2
16.
 ln k = lnA –
Ea
R
[A]o
5
=
= 50 years
2K
2  0.05
1
 
T
–Ea
= –4606
R
k  E 1 1 
ln  2   a  – 
 k1  R  T1 T2 

17.
k
1
1 
2.303 log  2–5   4606 
–
 400 500 
 10 
 k2 = 10–4
1
N2O5(g)  2NO2(g) + O2(g)
2
t = 0 50
0
0
50 – x
t=t
t= 0


2x
100
P =  Po 
x
2
50
3x 
– Po = 87.5 – 50 1.5x = 37.5 x = 25
2 
If Po = 50 and x = 25, t = 50 min is t1/2
At t = 100 min
Pt = Po +
3x
= 50 + 1.5 × 37.5= 106.25 mm hg
2
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18.
19.
20.
ln 2
ln 2
K=
days–1
K
10
1
1
1
4
t = ln
  ln = 4.1 days
K 3/ 4 K
3
t1/2 =
Ao t Ao


2
4
Same % is reacted is in same time interval. So it is 1st orde
t
A 

r = K[acetaldehyde]n
1 = K×[363 × 0.95]n
0.5 = K ×[365 × 0.67]n
– (1)
– (2)
n
n
n
 0.95 
2= 
 (2) 2  = 1  n = 2

2
 0.67 
21.
K1 = A.e–a2/RT
K1 = A.e–a1/RT
a1 – a2 = 10 kJ/mol
K 2 e Ea 2 /RT
(Ea1 Ea2 )

 Ea1 /RT = e
K1 e
K 
10 103
ln  2  =
=4
8.314

300
K
 1
22.
K 
(E a)A  1 1 
ln  2  

R  T1 T2 
 K1  A
(Ea)A  1
1 


 300 310 
R
(Ea)A
10
ln2 =

R
300  310
(Ea)A
10
2(E a) B
=

R
R
300  310
ln 2 =
.......(1)
 1
1

 
 300 T2 
T2 = 304.92
23.
Ea  1
1 


R  300 310 
2  0.693 8.314  300  310
a =
kJ / mol = 107.2 kJ/mol
10 1000
ln4 =
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24.
In 50 minutes, concentration of H2O2 becomes
1
of initial.
4


2 × t1/2 = 50 minutes
t1/2 = 25 minutes
.693

k=
per minute
25
.693
× 0.05 = 1.386 × 10–3
rH2O2 =
25
2H2O2  2H2O + O2
1
rO2   rH2O2 ;
rO2 = 0.693 × 10–3
2
rO2 = 6.93 × 10–4 mol/minute × litre
25.
Ki < Kii so Step-(i) is RDS
26.
r1 = K[A][B]
......(1)
r2 = K[A]×3[B]
......(2)
r2 = 3r1
More than 3 molecules can’t colloid simultaneously.
27.
2N2O5(g)  4NO2(g) + O2(g)
28.
t = 0 50
t = 30 50 – 2x
0
4x
0
x


87.5 = 50 + 3x


3x = 37.5


PN2O5 after 30 min = 50 – 25 = 25


t1/2 = 30 min.

x = 12.5
Hence after 60 min, (two half lives), PN2O5 remaining =


Hence decrease in PN2O5 = 50 – 12.5 = 37.5 torr.


PNO2 = 2 × 37.5 = 75 torr

37.5
= 18.75 torr
2
Ptotal = 12.5 + 75 + 18.75= 106.25 torr.
50
= 12.5 torr.
4
PO2 =

C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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29.
H = Eaf – Eab
 –40 = 2x – 3x
 Eaf = 80 kJ/mol

Eab = 120 kJ/mol
30.
1.2 × 10–3 = K (0.1)x (0.1)y
1.2 × 10–3 = K (0.1)x (0.2)y
2.4 × 10–3 = K (0.2)x (0.1)y
R = K [A]1 [B]0
31.
log
K2
E a  1 1 

  
K1 2.030R  T2 T1 
K2
 2 ; T2 = 310 K and T1 = 300 K
K1
 log2 
Ea
1 
 1


2.303  8.134  310 300 
 Ea  53598.6J / mol = 53.6 KJ/mol
Ans is (1)
32.
[A]0
2n
n = 2 so t75% = 40 min
[A]t =
t75% = 2×
ln 2
= 40
K
ln 2
min–1
20
ln 2
r = K[A] =
×0.01
20
= 3.47×10–4 M/min
K=
33.
rt = ri (T.C.)
T/10
= ri×(2)50/10 = 32 ri
34.
K1 = A1eEa1 /RT
K2 = A2eEa2 / RT
K1
A (E  E )/ RT
= 1 e a2 a1
A2
K2
K1 = K 2A  e
Ea1 / RT
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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35.
From mechanism [A]
r = K[Cl2][H2S]
From mechanism [B]
r = K[Cl2][HS]
Keq =
[H  ][HS ]
[H 2S]
r = K1[Cl2][H2S][H+]–1
36.
A  product
For zero order reaction
1
t1/2  n 1
a = initial concentration of reactant
a
t1/2  a
1
2
(t1/ 2 )1 a1


;
(t1/ 2 )2 a 2 (t1/ 2 )2 0.50
0.5
t1/2 =
= 0.25 h.
2
1 100
ln
K
1
ln 2 1
k=
 min–1
6.93 10
t99% = 10ln100 = 20ln10
= 46.06 min
37.
t99% =
38.
–2
39.
HR = E – Eb = 180 – 200 = – 20 kJ mol–1
The correct answer for this question should be –20kJ mol–1. But no option given is correct.
Hence we can ignore sign and select option
ln2
K=
days-1
30
40.
t=
d[A]
 d[B]

dt
2 dt
1
30
ln 10 =
×ln10 = 100 days
K
ln 2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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A1
X 
Z X  0n1
41.
A
Z
42.

NO(g) + Br2(g) 
 NOBr2(g)
NOBr2(g) + NO(g)  2NOBr(g) [rate determining step]
Rate of the reaction (r) = k [NOBr2] [NO]
where [NOBr2] = Kc [NO] [Br2]
r = k. KC [NO] [Br2] [NO]
r = k' [NO]2 [Br2].
The order of the reaction with respect to NO(g) = 2.
43.
Generally, molecularity of simple reactions is equal to the sum of the number of molecules of
reactants involved in the balanced stoichiometric equation. Thus, a reaction involving two
different reactants can never be unimolecular.
44.
H = a.f – a,b > 0
a,f > a,b
48.
2 A + B  C, rate = k [A] [B]
The value of k (velocity constant) is always independent of the concentration of reactant and it
is a function of temperature only.
51.
In Arrhenius equation, k = Ae–Ea/RT
k = rate constant, A = frequency factor
T = temperature, R = gas constant, Ea = energy of activation.
This equation can be used for calculation of energy of activation.
r = K[NO]2 [O2]1
52.
1
Pf = 2Pi
V
Rate1 = k [A]n [B]m
On doubling the concentration of A and halving the concentration of B
Rate2 = k [2A]n [B/2]m
P
53.
m
Ratio between new and earlier rate
54.
k[2A]n [B/2]m
1
 2n     2n  m
n
m
k [A] [B]
2
H2 + I2  2HI
When 1 mole of H2 and 1 mole of I2 reacts, 2 moles of HI are formed in the same time interval.
[H2 ] [I2 ] 1 [HI]
Thus the rate may be expressed as


t
t
2 t
The negative sign signifies a decrease in concentration of the reactant with increase of time.
H2 + I2  2HI
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55.
Order is the sum of the power of the concentrations terms in rate law expression. R = [A].[B]2
Thus, order of reaction = 1 + 2 = 3.
1n
56.
 mol 
k= 

 L 
59.
Surface catalysed reaction are zero order reaction
time1 where n = order
,
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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EXERCISE # JEE ADVANCED
1.
From the given data rate of reaction is obtained as :
r = k[A]1[B]0[C]1
From experiment–1 : r = k[0.2][0.1] = 6 × 10–5
So,
k = 3 × 10–3 s–1

r = 3 × 10–3 × 0.15 × 0.15 = 6.75 × 10–5 mol dm–3s–1
2.
Rate = –
1 d(N2O5 )
= k[N2O5]1
2 dt
d(N2O5 )
= 2k[N2O5]1 = k1[N2O5]1
–
dt
k1 = 2k = 2 × 5 × 10–4 = 10–3sec–1

2N2O5(g) 
 2N2O4(g) + O2(g)
t=0
1
0
0
P
t
1–P
P
2
P
Ptotal after time ' t ' 1  1.45 atm
2
P =0.9 atm
1 P
k’ =  n
t
Pt
1
1
1
n

n 10
10–3 =
3
y 10
0.1 y 103
y = n10 = 2.303
3.
(A,D)
A(g)  2B(g) + C(g)
t = 0 P0
0
0
t = t P0 – x
2x
x
Total pressure at t time Pt = P0 – x + 2x + x = P0 + 2x =
Pt – P0
2

x=

P0 – x = P0 – 

 Pt – P0  3P0  Pt
=
2
 2 


1 
P0
  kt = ln  2P0 
t = ln



k  3P0 – Pt 
 3P0  Pt 

2
ln(3P0 – Pt) = ln(2P0) – kt
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At
t  t1 ;
3

PA =
P0
3
1  P  ln 3
= constant
t 1  ln  0  =
k  P0 / 3 
k
3
Value of k doesn't depends on initial concentration of reactant.
4.
(A) Ea is independent on stearic factor
(B) k = P.Z.e–Ea /RT
k = Ae–Ea /RT
According to collision theory, P value is generally less than unity but for some reactions P is
greater than one and for such reactions, observed rate is greater than rate predicted from
Arrhenius equation. For a reaction with P value greater than 1, implies that the experimentally
determined value of frequency factor (A) is higher than that predicted by Arrhenius equations
and such reactions proceeds rapidly without the use of a catalyst.  (A) and (C).
5.
(A) High activation energy means lower value of k and slow reaction.
(B) On increasing temperature energy of particles increases hence greater number of collisions
occurs whose energy exceeds the activation energy.
K = Ae–Ea /RT
E
dk
 k a2
dT
RT
E
dk
 a2
dT RT
(C) Rate of increase of k with temperature is higher when Ea has a large value
(D) A =Frequency factor = No of collisions per unit time per unit volume.
6.
Initially on increasing temperature rate of reaction will increase, so % yield will also increase
with time. But at equilibrium % yield at high temperature (T2 ) would be less than at T1 as
reaction is exothermic .
7.
8H+ + MnO4 + [Fe(H2O)2 (C2O4)2]2–  Mn2+ + Fe3+ + 4CO2 + 6H2O
rate of change of [H ]
=8
rate of change of [MnO4 ]
8.
M  N
r = K [M]x
as [M] is doubled, rate increases by a factor of 8.
i.e.
8 r = K [2M]x

8 = (2)x
x=3
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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10.
 C 
Kt1/8 = In  O  = In 8
 CO / 8 
 CO 
Kt1/10 = In 
 = In 10
C
/
10
 O

then
11.

12.
t1/8
In 8
log 8
10 =
10 =
×10 = 9
log 10
In 10
t1/10
Ct = C0e–Kt
1
t1/2 
,
K on increasing T.
K
After eight half lives,
C
C = 8o
2
C
Co – 8o
2 100 = 99.6%
% completion =
C0
K=
C0  C 1  0.75 0.25
=
=
=5
t
0.05
0.05
0.75  0.40 0.35
=
=5
0.07
0.07
So, reaction must be of zero order.
K=
14.
From Arrhenius equation
K = Ae–Ea/RT
Ea
nk = nA –
RT
2.303 log K = 2.303 log A –
Ea
RT
Ea
1
×
+ log A
2.303R T
1
log K = – (2000)
+6
T
log K =
....... (A)
........(B)
On comparing equation (A) and (B)
Ea
= –2000
2.303R
Ea = 2.303 × 8.314 × 2000 = 38.29 kJ and log A = 6A = 106
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15.
aG + bH  products
Rate = k[G]x [H]y
R = K [G]0x [H]0y (Let initial conc. are [G]0 & [H]0)
8R = K[2G]0x [2H]0y = K2x.2yR
so
2x + y = 8
 x + y = 3
16.
Order of reaction can have any value zero, positive ,negative or any fractional value
2X(g)  3Y(g) + 2Z(g)
17.
800
–
–
800 – 2x
3x
2x
Pt = (800 + 3x)
from given data in time 100 min the partial pressure of X decreases from 800 to 400 so t 1/2 100
min. Also in next 100 min Px decreases from 400 to 200 so again t1/2 = 100 min. Since half life
is independent of initial concentration so reaction must Ist order with respect to X.
t=0
t=t
Rate constant K =
n2
= 6.93 × 10–3 min–1.
t1/ 2
Time taken for 75% completion = 2 × t1/2 = 200 min.
Now when Px = 700 = 800 – 2x so x = 50 mm of Hg
so total pressure = 800 + 3x = 950 mm of Hg
18.
k=
1
0.1
1
0.1
1
=
= ln 4
ln
ln
40 0.025 40 0.025 40
r = k[X] =
1
ln 4  0.01=3.47 × 10–4 M/minute
40
1
800
1
=
ln
ln 4 = 1.386× 10–4
4
4
2  10
50 2  10
19.
k=
20.
For a chemical reaction,
N2(g) + 3H2(g)  2NH3(g)
Concentration of N2 and H2 are decreased so negative sign is represented in expression, while
concentration of NH3 is increased. So +ve sign is represented in expression. Also, from
stoichiometry it can be said that the time during which 1 mole of N2 will be consumed in the
same time 3 moles of H2 is consumed and 2 moles of NH3 is formed.
22.
r = k [A]= k [A]0e–kt
0.04 = k [A]0 e k 10 (1)
0.03 = k [A]0 e k 20 (2)
From equation (1) and (2)
k=
1 4
.693
ln t1/2 =
10 3
k
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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38
23.
r = k [A]  2.4 × 10–5 = 3 × 10–5 × [A]  [A] = 0.8 M
24.
According to arrhenius equation K = Ae–Ea /RT
Let Ea of the reaction in absence of catalyst = x kJ mol–1
Therefore Ea of the reaction in presence of catalyst = x – 20 kJ mol–1
The Arrhenius equations in the two conditions can thus be written as
K = Ae
–
x
R 500
....(i)
K = Ae
–
x  20
R 400
...(ii)
Dividing equation (i) by (ii), we get
–e
26.
x
500R
–
=e
x 20
R400

x
x  20
=
500
400
or
x = 100 kJ mol–1
r = k [A]= k [A]0e–kt
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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39
17.
(a) From the rate law expression, R0 = k[A0]a[B0]b and from the table it is clear that :
(i) when the concentration of [A0] is doubled, keeping [B0] constant (see readings 1 and 2), the
rate also doubles i.e. rate is directly proportional to [A0] or a = 1.
(ii) when the concentration of [B0] is reduced, keeping [A0] constant (see readings 1 and 3), the
rate remains constant i.e., rate is independent of [B0] or b = 0. Thus, rate equation becomes
R0 = k[A0].
R0
0.05
(b) k =
=
= 0.5 sec–1.
[A0 ] 0.10
18.
For Ist order Rxn
1
[A] = [A0]  
 2
1
50 = 800  
 2
n=4
Then, Half life =
Then, K =
n
n  No. of Half life.
n
2  104
= 5 × 103 sec.
4
0.6932
= 1.38 × 10–4 sec–1.
3
5  10
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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