1. DPP No. -1
Total Marks : 42 Max. Time : 42 min.
Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18]
Subjective Questions ('–1' negative marking) Q.7 to Q.12 (4 marks 4 min.) [24, 24]
Integer Type Questions ('–1' negative marking) Q.13 (4 marks 4 min.) [4, 4]
ANSWER KEY DPP No.7
1. (B) 2. (C) 3. (B) 4. (D) 5. (B)
6. (D) 7. Rate = k [A]1
[B]2
, k = 5 × 10–2
M–2
hr–1
8. k =
t
1
ln 







t
0
P
P
9. k =
t
1
ln 







t
0
0
P
–
P
2
P
10. k =
t
1
ln 









t
P
–
P
P
11. 3 12. k =
t
1
ln 










)
P
P
(
5
P
3
t
13. (i) 3. (ii) 5 (iii) 1
1. Which of the following leads to bonding?
(A) +
–
s-orbital p-orbital
(B*) + –
s-orbital p-orbital
(C)
+
+ –
–
p-orbital
p-orbital
(D)
f
u
E
u
f
y
f
[
k
re
s
al
sd
k
S
uc
a
/
kc
u
k
u
sd
si
{
ke
s
ag
S
&
(A) (B*)
(C) (D)
2. Which one of the following is the correct set with respect to molecule, hybridization and shape?
(A) BeCl2
, sp2
, linear (B) BeCl2
, sp2
, triangular planar
(C*) BCl3
, sp2
, triangular planar (D) BCl3
, sp3
, tetrahedral
f
u
E
u
f
y
f
[
k
re
s
al
sd
k
S
u
l
kl
e
w
gv
.
k
q
]l
a
d
j
.
kr
F
k
kv
k
d
`
f
rd
sl
a
n
H
k
Ze
s
al
g
hg
S
&
(A)BeCl2
,sp2
,js[kh; (B)BeCl2
,sp2
,f=kdks.kh;leryh;
(C*)BCl3
,sp2
, f=kdks.kh;leryh; (D)BCl3
,sp3
, prq"Qydh;
Sol. Cl–Be–Cl (sp, linear) ; (sp2
, triangular planar)
3. Number of natural life times (Tav
) required for a first-order reaction to achieve 99.9% level of completion is :
i
z
F
k
ed
k
s
f
Vd
hv
f
H
k
f
Ø
;
kd
sf
y
,99.9%r
dv
f
H
k
f
Ø
;
kd
si
w
.
k
Zg
k
s
u
sd
sf
y
,v
k
o
'
;
di
z
k
d
`
f
r
dv
k
;
q
d
k
y(Tav
)d
hl
a
[
;
kg
S
(A) 2.3 (B*) 6.9 (C) 9.2 (D) 0.105
Sol. We know, k =
t
303
.
2
log )
x
–
a
(
a
99.9% completion
a = 100
a – x = (100 – 99.90) = .10
DAILY PRACTICE PROBLEMS (DPP)
Subject : Physical/Inorganic Chemistry Date : DPP No. 7 Class : XII Course :

2. Then t =
k
303
.
2
log 





10
.
100
t = 2.303 × 3 × 





k
1
; t = 6.9 × tav
Sol. g
et
k
u
r
sg
S
] k =
t
303
.
2
log )
x
–
a
(
a
99.9%i
w
.
k
Z
r
kd
sf
y
,
a = 100
a – x = (100 – 99.90) = .10
r
k
s t =
k
303
.
2
log 





10
.
100
t = 2.303 × 3 × 





k
1
; t = 6.9 × tav
4. Consider the plots for the types of reaction nA  B + C
–
dt
]
A
[
d
[A]
]
A
[
1
These plots respectively correspond to the reaction orders :
nAB+Cv
f
H
k
f
Ø
;
kd
sf
o
f
H
k
U
uç
d
k
j
k
s
ad
sf
y
,f
u
E
uv
k
j
s
[
k
k
si
jf
o
p
k
jd
h
f
t
,%
–
dt
]
A
[
d
[A]
]
A
[
1
;
gv
k
j
s
[
kØ
e
'
k
%f
d
ld
k
s
f
Vd
hv
f
H
k
f
Ø
;
kl
sl
E
c
a
f
/
k
rg
S
A
(A) 0, 2, 1 (B) 0, 1, 2 (C) 1, 1, 2, (D*) 1, 0, 2
Sol. nA 
 B + C
dt
]
A
[
d

 [A] (Ist
order)
[A]t
 t (zero order)
t
]
A
[
1
 t (2nd
order)
5. Reaction A + B  C + D follows rate law, r = k[A]1/2
[B]1/2
starting with 1 M of A and B each. What is the time taken
for concentration of A become 0.1 M ?
Given k = 2.303 × 10–2
sec–1
.
(A) 10 sec (B*) 100 sec (C) 1000 sec (D) 434 sec
vfHkfØ;kA+BC+Dnjfu;er=k[A]1/2
[B]1/2
dkvuqlj.kdjrhgSA;fnArFkkBçR;sddks1MlkanzrkdslkFk vfHkfØ;k
ç
k
j
E
H
kd
ht
k
r
hg
SAd
hl
k
a
n
z
r
k0.1Mg
k
s
u
se
s
af
d
r
u
kl
e
;y
x
s
x
k
Given k = 2.303 × 10–2
sec–1
.
(A)10lSd.M (B*)100lSd.M (C)1000lSd.M (D)434lSd.M
Sol. A + B  C + D
t = 0 1 1 0 0
t = t 1 – x 1 – x x x
r = k[A]1/2
[B]1/2

dt
dx
= k (1 – x)1/2
(1 – x)1/2
.

3. or
dt
dx
= k (1 – x).
 t =
k
1
ln 





x
–
1
1
; t = 2
–
10
303
.
2
303
.
2

log 





1
.
0
1
= 100 sec.
6. At high temperature (504ºC) dimethyl ether decomposes as per the reaction
(CH3)2 O (g) 
 CH4(g) + H2(g) + CO(g)
time (sec.) : 0 400 1200 
ptotal (mm) : 312 412 560 935
Determine the half life of the reaction during this run. (ln 





523
623
= 0.175, ln 





375
623
= 0.5076)
m
P
pr
k
i(504ºC)i
jM
k
b
e
s
f
F
k
yb
Z
F
k
jf
u
E
uv
f
H
k
f
Ø
;
kd
sv
u
q
l
k
jf
o
;
k
s
f
t
rg
k
s
r
kg
S
A
(CH3)2 O (g) 
 CH4(g) + H2(g) + CO(g)
l
e
;(l
S
d
.
M
-e
s
a
) : 0 400 1200 
pdqy
(mm) : 312 412 560 935
bliz;ksxdsnkSjkuvfHkfØ;kdhv)Z&vk;qKkrdjksA(ln 





523
623
=0.175,ln 





375
623
=0.5076)
(A)1311sec¼lSd.M½ (B)1411sec¼lSd.M½ (C)1511sec¼lSd.M½ (D*)1611sec¼lSd.M½
Sol. (CH3
)2
O (g) — CH4
(g) + H2
(g) + CO(g)
t=0 a 0 0 0 a  P
º
= 312 mm
t=400 sec a–x x x x a+2x  p
400
= 412mn
t=1200 sec a – y y y y a+2y  p
1200
= 560 mm
t= 0 a a a 3a P

= 935 mm
Kt = ln 







t
o
C
C
= ln 





x
–
a
a
 Kt = ln 









t
o
p
–
p
p
–
p
(t = 400sec) K1
=
400
1
ln 





412
–
935
312
–
935
=
400
1
ln 





523
623
sec–1
= 0.175 ×
400
1
(t = 1200sec) K2
=
1200
1
ln 





560
–
935
312
–
935
=
1200
1
ln 





375
623
sec–1
= 0.5076 ×
1200
1
2
1
t =
2
k
k
2
ln
2
1 
= )
k
k
(
2
ln
2
2
1  =
2
1638
1584 
= 1611 sec.
7. A reaction between A and B is represented stoichiometrically by A + B  C. Observations on the rate of this
reaction are obtained in three separate experiments as follows :
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
initial concentrations final concentration
[A]0
, M [B]0
, M duration of experiment t, hr [A]f
, M
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
(1) 0.1000 1.0 0.5 0.0975
(2) 0.1000 2.0 0.5 0.0900
(3) 0.0500 1.0 2.0 0.0450
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
What is the order of the reaction with respect to each reactant and what is the value of the rate constant?
Ar
F
kkBd
se
/
;v
fH
k
f
Ø;
kd
k
sj
l
l
eh
d
j.
k
f
er
h}
k
j
kA+BCl
sç
n
f'
k
Zrf
d;
kt
k
rkg
S]blv
fH
k
fØ
;
kd
sf
y
,n
jd
kç
s
{
k.
k
r
h
ui
`
F
k
dç
;
k
s
x
k
s
ae
s
af
u
E
uç
d
k
jç
k
I
rd
j
r
sg
S
a
A
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
ç
k
j
f
E
H
k
dl
k
U
n
z
r
k v
f
U
r
el
k
U
n
z
r
k
[A]0
, M [B]0
, M ç;ksxdsnkSjku t,hr [A]f
, M
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
(1) 0.1000 1.0 0.5 0.0975
(2) 0.1000 2.0 0.5 0.0900

4. (3) 0.0500 1.0 2.0 0.0450
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
ç
R
;
s
s
dv
f
H
k
d
k
j
dd
sl
k
i
s
{
kv
f
H
k
f
Ø
;
kd
hd
k
s
V
hD
;
kg
S
a
]r
F
k
kn
jf
u
;
r
k
a
dd
ke
k
uD
;
kg
S
a
A
Ans. Rate = k [A]1
[B]2
, k = 5 × 10–2
M–2
hr–1
Sol. Initial conc. duration of experiment (hr) final conc. [A]f
Rate
[A]0
,M [B]0
, M
0.1000 1.0 0.5 0.0975 5 × 10–3
M hr–1
0.1000 2.0 0.5 0.0900 20 × 10–3
M hr–1
0.0500 1.0 2.0 0.0450 2.5 × 10–3
M hr–1
A + B — C
Rate = –
dt
]
A
[
d
= –
dt
]
B
[
d
=
dt
]
C
[
d
Rate = K[A]a
[B]b
R1
= K [0.1]a
[1]b
= 5 × 10–3
M hr–1
R2
= K [0.1]a
[2]b
= 20 × 10–3
M hr–1
2
1
R
R
=
b
2
1






=
4
1
 b = 2
R3
= K [0.05]a
[1]2
= 2.5 × 10–3
M hr–1
R1
= K [0.1]a
[1]2
= 5 × 10–3
M hr–1
1
3
R
R
=
a
2
1






=
2
1
 a = 1
R1
= K [0.1] [1]2
= 5 × 10–3
M hr–1
K =
1
.
0
10
5 3
–

= 5 × 10–2
M–2
hr–1
8. Let there be as first-order reaction of the type, A 
 B + C. Let us assume that all the three species are gases.
We are required to calculate the value of rate constant based on the following data.
0 T 
P0 Pt –
Time
Partial pressure of A
e
k
u
kf
dA 
 B+C,
dç
F
k
ed
k
s
f
Vç
d
k
jv
f
H
k
f
Ø
;
kg
S
a
]v
cg
e;
gt
k
u
r
sg
S
af
dr
h
u
k
s
aç
t
k
f
rx
S
l
h
;g
S
a
Ag
e
s
af
u
E
uv
k
W
d
M
k
si
j
v
k/
kk
fj
rnjf
u;
rk
add
kek
ui
fj
df
yrd
ju
kvk
o'
;dg
Sa
A
Ans. k =
t
1
ln 







t
0
P
P
Sol. A 
 B + C
t=0 a 0 0 a  P0
t=t a-x x x (a–x)  Pt
Kt = ln 





 x
a
a
 k =
t
1
ln 







t
0
P
P
9. Let there be a first order reaction, A 
 B + C. Let us assume all three are gases. We are required to calculate
the value of rate constant based on the following data
0 t 
P0 Pt –
Time
Total pressure
Calculate the expression of rate constant.
e
k
u
kf
dA 
 B+Cç
F
k
ed
k
s
V
hd
hv
f
H
k
f
Ø
;
kg
S
a
Av
c;
ge
k
u
r
sg
Sf
dr
h
u
k
s
aç
t
k
f
rx
S
l
h
;g
S
a
]g
e
s
af
u
E
uv
k
W
d
M
k
s
ai
jv
k
/
k
k
f
j
r
njfu;rkaddkekuifjdfyrdjukvko';dgSaA

5. njfu;arkddkO;tadKkrdhft,A
Ans. k =
t
1
ln 







t
0
0
P
–
P
2
P
Sol. A 
 B + C
t=0 a 0 0 a  P0
t=t a-x x x (a+x)  Pt
 x  (Pt
– P0
)
Kt = ln 





 x
a
a
 k =
t
1
ln 








 )
P
P
(
P
P
0
t
0
0
 k =
t
1
ln 







t
0
0
P
–
P
2
P
10. A(g) 
 B(g) + C(g)
Calculate the expression of rate constant.
A(g) 
 B(g) + C(g)
njfu;arkddkO;tadKkrdhft,A
Ans. k =
t
1
ln 









t
P
–
P
P
Sol. A 
 B + C
t=0 a 0 0
t=t a-x x x 2x  Pt
 x 
2
Pt
t=  0 a a 2a  P   a 
2
P
Kt = ln 





 x
a
a
 k =
t
1
ln 









t
P
–
P
P
11. For the reaction, 2NO + H2

 N2
O + H2
O the value of -dp/dt was found to be 1.50 Torr s–1
for a pressure of 359
Torr of NO and 0.25 Torr s-1
for a pressure of 152 Torr, the pressure of H2
being constant. On the other hand, when
the pressure of NO was kept constant, –dp/dt was 1.60 Torr s–1
for a hydrogen pressure of 289 Torr and 0.79
Torr s–1
for a pressure of 147 Torr. Determine the order of the reaction.
vfHkfØ;k 2NO+H2

 N2
O+H2
Ods fy, H2
ds fu;r nkc ij -dp/dtds eku NOds 359Vksj nkc ij 1.50Vksj
lSd.M–1
,oaNOds152Vksjnkcij0.25VksjlSd.M–1
ik;sx;stcfd;fnNOdkfu;rnkcfy;ktkrkrks –dp/dtdseku
H2
ds289Vk
sji
j1.60Vks
jlSd
.M
–1
,o
aH2
ds147Vksjij0.79Vks
jlsd
.M
–1
ik;
sx;sAblvfH
kfØ;
kdhdksf
VKk
rdj
ksA
Ans. 3
Sol. rate = K [NO]a
[H2
]b
a
2
1
152
359
25
.
0
5
.
1
r
r







 
1
6
= (2.36)a
 a = 2
b
4
3
147
289
79
.
0
6
.
1
r
r







  2 = (1.966)b
 b = 1
order of the reaction = 1 + 2 = 3

6. 12. Decomposition of N2
O5
follows first-order kinetics, 2N2
O5
(g) 
 4NO2
(g) + O2
(g). If pressure of the system at
time, t and  are Pt
and 
P respectively, find the expression of rate contant (k).
N2
O5
dkfo;kstuizFkedksfVcyxfrdhdkvuqlj.kdjrkgS,2N2
O5
(g) 
 4NO2
(g)+O2
(g);fnle;trFkk  ijfudk;
dsnk
cØe
'k%Pt
rF
kk 
P gk
s]rksnjfu;
rkad(k)dkO;
atdKk
rd
hft
,A
Ans. k =
t
1
ln 










)
P
P
(
5
P
3
t
Sol. 2N2
O5
(g) 
 4NO2
(g) + O2
(g)
t = 0 a 0 0
t = t a – x 2x
2
x







2
x
3
a  Pt
t=  0 2a
2
a
2
a
5
 P
kt = ln 





 x
a
a
a 
5
2
P
& x 
3
2






 
P
3
2
Pt
k =
t
1
ln





















 


P
5
2
P
3
2
P
5
2
P
5
2
t
=
t
1
ln 










)
P
P
(
5
P
3
t
13.
Integer Answer Type
This section contains 3 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.
i
w
.
k
k
±
dm
Ù
k
ji
z
d
k
j
b
l[
k
.
Me
s
a3i
z
'
ug
S
a
Ai
z
R
;
s
di
z
'
ud
km
Ù
k
j0l
s9r
di
w
.
k
k
±
dg
S
A
(i) The number of lone pairs of electrons present on the central atom in IBr2
–
is
IBr2
–
d
sd
s
U
n
z
h
;i
j
e
k
.
k
qi
jm
i
f
L
F
k
rb
y
s
D
V
ª
k
s
u
k
s
ad
s,
d
k
d
h;
q
X
e
k
s
ad
hl
a
[
;
kg
S&
Ans.
So answer is 3.
(ii) How many B—O—B bond are present in the structure of borax ?
c
k
s
j
s
D
ld
hl
a
j
p
u
ke
s
af
d
r
u
sB—O—Bc
a
/
km
i
f
L
F
k
rg
S
Ans. 5
(iii) The number of S–S bond(s) in dithionic acid is/are
M
k
b
ZF
k
k
;
k
s
f
u
dv
E
ye
s
aS–Sc
a
/
k
k
s
ad
hl
a
[
;
kg
S
@
g
S
a
Ans. 1

7. DPP No. -2
Total Marks : 46 Max. Time : 51 min.
Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [18, 18]
Multiple choice objective ('–1' negative marking) Q.6 (5 marks 4 min.) [5, 4]
Subjective Questions ('–1' negative marking) Q.7 to Q.11 (4 marks 5 min.) [20, 25]
Match the Following (no negative marking) Q.12 (8 marks 8 min.) [8, 8]
Integer Type Questions ('–1' negative marking) Q.13 (3 marks 3 min.) [3, 3]
ANSWER KEY DPP No.8
1. (A) 2. (C) 3. (D) 4. (B) 5. (B)
6. (A) 7. 4 8. k =
t
1
ln 







t
0
V
V
9. k =
t
1
ln 









t
0
V
–
V
V
–
V
10. k =
t
1
ln 









t
0
V
–
V
V
–
V
11. k =
t
1
ln 









t
0
r
–
r
r
–
r
12. (A) p, q ; (B) p, q, r ; (C) p, q, r, s ; (D) p, q, r, s 13. (i) 8 (ii) 3 (iii) 3
1. The decomposition of N2
O5
in chloroform wasfollowed by measuring the volume of O2
gasevolved : 2N2
O5
(CCI4
) 
2N2
O4
(CCI4
) + O2
(g). The maximum volume of O2
gas obtained was 100 cm3
. In 500 minutes, 90 cm3
of O2
were
evolved. The first order rate constant (in min–1
) for the disappearance of N2
O5
is :
DyksjksQkeZesaN2
O5
dsfo;kstu 2N2
O5
(CCI4
)2N2
O4
(CCI4
)+O2
(g)dksfu"dkflrO2
dsvk;rudsekiuesaç;qDrfd;k
t
k
r
kg
SO2
x
S
ld
kv
f
/
k
d
r
ev
k
;
r
u100cm3
ç
k
I
rg
k
s
r
kg
Sr
F
k
k500f
e
u
Ve
s
a90cm3
O2
f
u
"
d
k
f
l
rg
k
s
r
hg
S
AN2
O5
d
sf
o
y
q
I
rg
k
s
u
s
d
sf
y
,ç
F
k
ed
k
s
f
Vd
ko
s
xf
u
;
r
k
a
d(f
e
u
V
–1
e
s
a
)g
k
s
x
k
A
(A*)
500
303
.
2
(B)
90
100
log
500
303
.
2
(C)
100
90
log
500
303
.
2
(D)
500
10
100

Sol. k t = In 







t
O
C
C
2 N2
O5

 2N2
O4
+ O2
t=0 200 cm3
0 0
t=t 20 cm3
180cm3
90 cm3
t=  0 200cm3
100 cm3
Initial volume of N2
O5
= 200 cm3
.
because Max. volume of O2
= 100 cm3
.
 K × 500 = In 





20
200
 k =
500
10
In
=
500
303
.
2
.
2. The reaction A(g) + 2B(g)  C(g) is an elementary reaction. In an experiment involving this reaction, the initial
partial pressures of A and B are PA
= 0.40 atm and PB
= 1.0 atm respectively. When pressure of
C becomes 0.3 atm in the reaction the rate of the reaction relative to the initial rate is :
(A)
12
1
(B)
50
1
(C*)
25
1
(D) none of these
vfHk
fØ;kA(g)+2B(g)C(g),dljy¼
,dyin½v
fHkfØ
;kg
Saft
ls,dç;ks
xes
aç;q
Drfd
;kt
krkg
S]ArF
kkBdkçk
jfEH
kd
nkcØ
e'k%PA
=0.40ok;
qe.My
h;rF
kkPB
=1.0ok;
qe.Myh
;gSAtcPC
=0.3ok;
qe.Myh
;gkstkrkgSAç
kjfEHk
dnjdslki
s{k
v
f
H
k
f
Ø
;
kd
hn
jg
k
s
x
h
(A)
12
1
(B)
50
1
(C*)
25
1
(D)m
i
j
k
s
D
re
s
al
sd
k
s
b
Zu
g
h
a
DAILY PRACTICE PROBLEMS (DPP)
Subject : Physical/Inorganic Chemistry Date : DPP No. 8 Class : XII Course :

8. Sol. A(g) + 2B(g)  C(g)
t = 0 0.4 atm 1 atm 0 atm
t = t (0.4 –0.3)atm (1 – 0.6)atm 0.3 atm
Since reaction is elementary.
So, Rate of reaction w.r.t. A & B will be of order equal to stoichiometric coefficient
Rate = K [A] [B]2
Rate(Initial)
= K [0.4] [1]2
Rate(aftar t = t)
= K [0.1][0.4]2
)
o
t
(
)
t
t
(
R
R


=
]
1
[
]
4
.
0
[
K
]
4
.
0
[
]
1
.
0
[
K 2
=
25
1
3. A reaction is catalysed by H+
ion;and in the rate law the dependence of rate is of first order with respect to the
concentration of H+
ions, in presence of HA rate constant is 2 × 10–3
min–1
and in presence of HB rate constant is
1 × 10–3
min–1
. HA and HB (both strong acids) have relative strength as :
,
dv
f
H
k
f
Ø
;
kH+
v
k
;
u}
k
j
km
R
i
z
s
f
j
rd
ht
k
r
hg
Sr
F
k
kb
l
e
s
ao
s
xf
u
;
eH+
v
k
;
ud
hl
k
U
n
z
r
kd
sl
k
i
s
{
ki
z
F
k
ed
k
s
f
Vd
hv
f
H
k
f
Ø
;
kg
S
A
HAd
hm
if
L
F
k
f
re
san
jf
u
;
rk
a
d2×10–3
min–1
g
Sr
Fk
kHBd
hm
i
f
LF
k
f
re
s
an
jf
u;
r
k
a
d1×10–3
min–1
g
SrcHAr
F
kkHB(n
k
s
u
k
s
a
i
z
c
yv
E
yg
S
)v
k
i
s
f
{
k
dl
k
e
F
;
Zg
S:
(A) 0.5 (B) 0.002 (C) 0.001 (D*) 2
Sol. We know
Rate = k [conc.]
Given Rxn catalysed by HA and HB
Rate constant kA
= k1
[H+
]A
kB
= k1
[H+
]B
Then relative strength of acidsA and B is
B
A
k
k
=
B
A
]
H
[
]
H
[


1
2
=
B
A
]
H
[
]
H
[


= strength of
]
B
Acid
[
]
A
Acid
[
Sol. g
et
k
u
rsg
Sf
d
nj=k[lkUnzrk]
f
n
;
kx
;
kg
Sf
dv
f
H
k
f
Ø
;
kHAr
F
k
kHBl
sm
R
i
z
s
f
j
rg
S
A
n
jf
u
;
r
k
a
d kA
= k1
[H+
]A
kB
= k1
[H+
]B
v
cv
E
y
k
s
aAr
F
k
kBd
hv
k
i
s
f
{
k
dl
k
e
F
;
Z
B
A
k
k
=
B
A
]
H
[
]
H
[


1
2
=
B
A
]
H
[
]
H
[


= ]
B
[
]
A
[
vEy
vEy
dhlkeF;ZrkA
4. The gaseous decomposition reaction, A(g) 
 2B(g) + C(g) is observed to first order over the excess of liquid
water at 25ºC. It is found that after 10 minutes the total pressure of system is 188 torr and after very long time it is
388 torr. The rate constant of the reaction (in hr–1
) is : [Given : vapour pressure of H2
O at 25º is 28 torr (ln 2 = 0.7, ln
3 = 1.1, ln 10 = 2.3)]
(A) 0.02 (B*) 1.2 (C) 0.2 (D) none of these.
25ºCijn
zotydsvk
f/
kD;es
axSl
h;fo
;ks
tuvf
Hkf
Ø;
kA(g) 
 2B(g)+C(g)dsf
y,çF
kedk
sfVç
sf{
krdhtk
rhgS
A;g
ik;kx;kgSfd 10feuVi'pkr~ra=kdkdqynkc188torr rFkkdkQhyEcsle;i'pkr~;g388torrgSArksvfH
kfØ;kdsfy,
njfu;rkad(hr–1
esa)gS%[fn;kx;kgS: 25ºCijH2
O dkok"inkc28torrgSA
[(ln 2 = 0.7, ln 3 = 1.1, ln 10 = 2.3)]
(A) 0.02 (B*) 1.2 (C) 0.2 (D)m
i
j
k
s
D
re
s
al
sd
k
s
b
Zu
g
h
a
A
Sol. A(g) 
 2B(g) + C(g)
Let initial pressure P0
0 0

9. After 10 min. (P0
– x) 2x x
After long time )
t
( 
 0 2P0
P0
as per given (P0
– x) + 2x + x + vaour pressure of H2
O = 188
P0
+ 2x = 160 and 3P0
+ 28 = 388
so, P0
= 120 and x = 20 torr
k = 







 x
P
P
ln
t
1
0
0
 10
1
100
120
ln
10
1







× (ln 4 + ln 3 – ln 10)
= 0.02 min–1
= 1.2 hr–1
5. The hydrolysis of cane sugar was studied using an optical polarimeter and the following readings were taken:
time (min.) : 0 84 min 
observed rotation (degrees) : 50 20 –10
When was the mixture optically inactive? (log 2 = 0.3, log 3 = 0.48)
x
U
u
sd
h'
k
d
Z
j
kd
st
y
&
v
i
?
k
V
ud
sv
/
;
;
ui
z
d
k
'
k
h
;i
k
s
y
s
j
h
e
h
V
jd
sm
i
;
k
s
x}
k
j
kf
d
;
kt
k
r
kg
S
]v
k
S
jf
u
E
ui
k
B
~
;
k
a
df
y
;
sx
;
sg
S
A
le; (feuV): 0 84 min 
i
z
s
f
{
k
r?
k
w
.
k
Z
u(f
M
x
z
h
): 50 20 –10
feJ.kizdk'kh;vlfØ;dcgksrkgSA(log2=0.3,log3=0.48)
(A) 118 min (B*) 218 min (C) 318 min (D) 418 min
Sol. Kt = ln 











t
0
r
r
r
r
K =
84
1
ln 









20
10
50
10
=
84
1
ln (2)
Solution is optically inactive i.e. optical rotation is zero.






2
ln
84
1
t = ln 









0
10
50
10
t = 84
2
ln
6
ln
= 218 min.
6. For a certain reaction A 
 products, the t½ as a function of [A]0 is given as below :
[A]0 (M) : 0.1 0.025
t1/2 (min.) : 100 50
Which of the following is/are true ?
(A*) The order is
2
1
(B*) t½ would be min
10
100 for [A]0 = 1 M
(C) The order is 1 (D) t½ would be 100 min for [A]0 = 1 M
,dfuf'prvfHkfØ;kA 
 mRikn]dsfy;st½ dks[A]0 dsQyuds:iesafuEuçdkjlsfn;kx;kgSA
[A]0 (M) : 0.1 0.025
t1/2 (min.) : 100 50
f
u
E
ue
s
al
sd
k
S
u
l
kl
R
;g
S
A
(A*)
2
1
d
k
s
f
Vg
S
A (B*) [A]0 =1Mdsfy;s t½, min
10
100 gksxkA
(C)ç
F
k
ed
k
s
f
Vg
S
A (D)[A]0 = 1 Mijt½, 100min gksxkA
Sol. A  Products.
t1/2
= [conc.]1 – n
n  order of reaction.
50
100
=
n
–
1
025
.
0
1
.
0







10. 2 = 2 – 2n
 2n = 1.
 n =
2
1
.
order of reaction is
2
1
.
2
/
1
t
100
=
2
/
1
1
1
.
0






 t1/2
= min
10
100 for [A0
] = 1M.
7. The plot of log (V – V) versus t (where V is the volume of nitrogen
collected under constant temperature and pressure conditions)
for the decomposition of C6H5N2Cl is given at 50°C with an amount
of C6H5N2Cl equivalent to 58.3 cc N2.
5 10 15 20 25 30
1.0
1.75
time(min.)
log(V – V)

Calculate the rate constant for the reaction in hr–1 expressing
your answer in a single significant digit.
log(V –V)dktdslkis{kvkjs[k(tgk¡fu;rrkionkcifjfLFkfre
sa
ukbVªkstudkvk;ruV fy;kx;kgS½C6H5N2Cldsfo;kstudsfy,50°C
rkifn;kx;kgSAftlijC6H5N2Cldhek=kk58.3ccN2dslekugSA
n
jf
u
;
r
k
a
dd
hx
.
k
u
k?
k
.
V
s
–1e
s
ad
j
k
som
Ù
k
jd
s
o
y,
dv
a
de
s
an
k
s
A
5 10 15 20 25 30
1.0
1.75
time(min.)
log(V – V)

Ans. 4
Sol. Slope =
0
25
)
75
.
1
1
(


= – 0.03
k = – 2.303 (slope) min–1
= – 2.303 (– 0.03) min–1 = 0.06909 min–1 = 4.14 hr–1  4 hr–1
8. Now, let us assume a first-order reaction, A 
 B + C, such that A, B and C, such that A, B and C are in
solution. At time zero, a small amount of the solution is taken, cooled (to stop the reaction from proceeding) and
titrated with a suitable reagent. Let us assume that the reagent reacts only with A and not with B and C. The same
process is repeated at time t.
0 t
V0 Vt
Time
Volume of reagent
Calculate the expression of rate constant.
v
cg
e
u
se
k
u
kf
df
u
E
uv
f
H
k
f
Ø
;
kç
F
k
ed
k
s
f
Vd
hg
S
a
]
A 
 B+C,blç
dkjdhv
fHkfØ;kesaA,BrFkkCfoy;
uesag
SsA'kw
U;le;ijfoy
;udhd
eek=kkyhtkr
hgSa]ftls
B
.
M
+
k¼
v
f
H
k
f
Ø
;
kd
k
sv
k
s
jv
k
x
sj
k
s
d
u
sd
sf
y
,
½r
F
k
k,
dm
i
;
q
D
rv
f
H
k
d
e
Z
dd
sl
k
F
kv
u
q
e
k
f
i
rf
d
;
kt
k
r
kg
S
a
Av
cg
e;
ge
k
u
r
sg
S
a
fdvfH
kde
ZddsoyAdslk
FkfØ;
kdj
rkgSa
]BrF
kkCdslk
Fku
gha
Al
ekuçØedht le
;i
ji
quj
ko`
frd
ht
krhgS
aA
njfu;arkddkO;tadKkrdhft,A
Ans. k =
t
1
ln 







t
0
V
V
Sol. A 
 B + C
t=0 a 0 0 a  V0
t=t a-x x x (a–x)  Vt

11. Kt = ln 





 x
a
a
 k =
t
1
ln 







t
0
V
V
9. A(soln.) 
 B(soln.) + C(soln.)
Reagent reacts with all species A, B and C.
0 t 
V
Vt
V0
Time
Volume of reagent
The n factor of A, B and C with the reagent and that of the reagent with A, B and C are not known. Calculate the
expression of rate constant.
A¼foy;u½ 
 B¼foy;u½+C¼foy;u½
v
f
H
k
d
e
Z
dl
H
k
hç
t
k
f
r
;
k
s
a
A,Br
F
k
kCd
sl
k
F
kf
Ø
;
kd
j
r
sg
S
a
A
v
fH
kd
eZ
ddsl
kF
kA,Br
Fk
kCdsnx
q.
kk
ad rF
kkA,Br
Fk
kCd
slk
Fkm
lvf
Hk
de
Z
ddsK
kru
ghg
Sa
A
njfu;rkaddkO;tadKkrdhft,A
Ans. k =
t
1
ln 









t
0
V
–
V
V
–
V
Sol. A(soln.) 
 B(soln.) + C(soln.)
t=0 a 0 0
t=t a-x x x
t=  0 a a
Let n-factor of A, B & C are n1
, n2
& n3
& that of reagent is n & molarity of reagent is M
At t= 0,
meq. of A= meq. of reagent
n1
× a = M × n × V0
....... (i)
At t= t,
meq. of reagent = meq. of A + meq of B + meq of C
MVt
n = (a – x)n1
+ xn2
+ xn3
....... (ii)
At t=  ,
meq. of reagent = meq of B + meq of C
MV  n = an2
+ an3
....... (iii)
From (iii) & (i)
(n2
+ n3
) =
a
n
MV
& n1
=
a
n
MV0
 MVt
n = (a – x)
a
n
MV0
+ x
a
n
MV
 










 

t
0
V
V
V
V
x
a
a
Kt = ln
x
a
a

 K =
t
1
ln 











t
0
V
V
V
V
10. A(soln.) 

 

.)
ln
so
(
D
B(soln.) + C(soln.)
D is catalyst present in the solution whose n-factor is not known. Catalyst have a constant concentration through-
out the reaction.

12. 0 t 
V
Vt
V0
Time
Volume of reagent
Calculate the expression of rate constant.
A¼foy;u½ 

 

.)
ln
so
(
D
B¼foy;u½+C¼foy;u½
Df
o
y
;
ud
sm
i
f
L
F
k
rm
R
ç
s
j
dg
S
a]f
t
l
d
knx
q
.
k
k
a
dK
k
ru
g
hg
S
a
Am
R
ç
s
j
dd
hi
w
j
hv
f
H
k
f
Ø
;
kd
sn
k
S
j
k
ul
k
U
n
z
r
kf
u
;
rj
[
k
r
kg
S
a
A
njfu;rkaddkO;tadKkrdhft,A
Ans. k =
t
1
ln 









t
0
V
–
V
V
–
V
Sol. Same as of question 9.
11. Now, let us assume that A, B and C are optically active compounds, which rotate the plane polarized light in the
clockwise or anticlockwise direction.
A(soln.) 
 B(soln.) + C(soln.)
0 t 
r
rt
r0
Time
Total rotation in degrees
Calculate the expression of rate constant.
v
ce
k
u
kf
d
A,Br
F
k
kCç
d
k
'
k
h
;:
il
sl
f
Ø
;;
k
S
f
x
dg
S
a
]t
k
sf
dl
e
r
y/
k
z
q
o
.
kç
d
k
'
kd
hv
k
S
jn
f
{
k
.
k
k
o
r
Zv
F
k
o
ko
k
e
k
o
r
Zf
n
'
k
ke
s
a
?
k
q
.
k
Z
u
d
j
k
r
s
g
S
a
A
A¼foy;u½ 
 B¼foy;u½+ C¼foy;u½
njO;atdfu;arkddkO;atdKkrdhft,A
Ans. k =
t
1
ln 









t
0
r
–
r
r
–
r
Sol. A(soln.) 
 B(soln.) + C(soln.)
t=0 a 0 0
t=t a-x x x
t=  0 a a
Let ,  &  are specific rotation of A, B & C respectively.
At t=0,
r0
= a ........... (i)
At t=t,
rt
= (a–x) + x+ x ........... (ii)
At t=  ,
r  = a+ a ........... (iii)
From (i) & (iii)
=
a
r0
& (+ ) =
a
r
Putting in (ii)
rt
= (a – x)
a
r0
+
a
r
x 
t
0
r
r
r
r
x
a
a



 


13. Kt = ln
x
a
a

 k =
t
1
ln 









t
0
r
–
r
r
–
r
12. Match the pair of species given in column-I with the identical characteristic(s)/type of hybridisation given in
column-II :
Column - Column -
(Pair of species) (Identical Property in pairs of species)
(A) PCl3
F2
and PCl2
F3
(p) Hybridisation of central atom
(B) BF3
and BCl3
(q) Shape of molecule/ion
(C) CO2
and CN2
–2
(r)  (dipole moment)
(D) C6
H6
and B3
N3
H6
(s) Total number of electrons
L
r
E
H
k
-Ie
s
an
hx
;
h
si
z
t
k
f
r
;
k
s
a¼
L
i
h
'
k
h
t
½d
s;
q
X
e
k
s
ad
k
sm
u
d
sl
g
hx
q
.
k
k
s
a
@
i
z
d
k
jd
sl
a
d
j
.
kt
k
sL
r
E
H
k
-IIe
s
af
n
;
sx
;
sg
Sd
sl
k
F
kl
q
e
s
f
y
rd
h
f
t
,
A
dkWye& dkWye&
¼
L
i
h
'
k
h
td
k;
q
X
e
½ ¼
L
i
h
'
k
h
td
s;
q
X
ee
s
al
e
k
ux
q
.
k
½
(A) PCl3
F2
rFkkPCl2
F3
(p) d
s
U
n
z
h
;i
j
e
k
.
k
qd
kl
a
Ø
e
.
k
(B) BF3
rFkkBCl3
(q) v
.
k
q
@
v
k
;
ud
hv
k
d
`
f
r
(C) CO2
rFkkCN2
–2
(r) ¼
f
}
/
k
z
q
o
v
k
?
k
w
.
k
Z
½
(D) C6
H6
rFkkB3
N3
H6
(s) b
y
s
D
V
ª
k
W
ud
hd
q
yl
a
[
;
k
Ans. (A) p, q ; (B) p, q, r ; (C) p, q, r, s ; (D) p, q, r, s
Sol. (A) PCl3
F2
& PCl2
F3
Hybridisation sp3
d sp3
d
shape TBP TBP
Dipole
moment
Total no. of
electrons 84 76
(B) BF3
BCl3
Hybridisation sp2
sp2
Shape Trigonal planar Trigonal planar
Dipole moment
Total no. of
electrons 32 56
(C) CO2
CN2
2–
Hybridisation sp sp
Shape linear linear
 O = C = O –
N = C = N–
 = 0  = 0
Total no. of
electrons 22 22
13.
Integer Answer Type
This section contains 3 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.
i
w
.
k
k
±
dm
Ù
k
ji
z
d
k
j
b
l[
k
.
Me
s
a3i
z
'
ug
S
a
Ai
z
R
;
s
di
z
'
ud
km
Ù
k
j0l
s9r
di
w
.
k
k
±
dg
S
A
(i) Find the sum of average oxidation number of S in H2
SO5
(peroxy monosulphuric acid) and Na2
S2
O3
(sodium
thiosulphate).

14. H2
SO5
(ijvkWDlheksukslY¶;wfjd vEy) rFkk Na2
S2
O3
(lksMh;e Fkk;kslYQsV) esa S ds
vkSlr vkWDlhdj.k vadks dk ;ksx Kkr dhft;sA
Ans. 8
Sol. H2
SO5
and Na2
S2
O3
(+6) (+2)
(ii) The equilibrium N2
(g) + 3H2
(g) 2NH3
(g) is established in a closed container by initially taking only NH3
. Find
out the number of moles of H2
present in 102 g of the equilibrium mixture if the molar massof the equilibrium mixture
is observed to be
4
51
g/mole.
,
dc
a
ni
k
=
ke
s
ai
z
k
j
E
H
ke
s
ad
s
o
y
]NH3
d
k
sy
s
d
j
]l
k
E
;N2
(g)+3H2
(g) 2NH3
(g)d
k
sL
F
k
k
f
i
rf
d
;
kx
;
k
Al
k
E
;f
e
J
.
kd
s102
ge
s
am
i
f
L
F
k
rH2
d
se
k
s
y
k
s
ad
hl
a
[
;
kK
k
rd
h
f
t
,;
f
nf
e
J
.
kd
ke
k
s
y
jn
z
O
;
e
k
u
4
51
g/molei
z
s
f
{
k
rf
d
;
kx
;
k
A
Ans. 3
Sol. Mobs
=


 )
1
n
(
1
MTh

4
51
=


 )
1
2
(
1
17
  =
3
1
Mass of eq. mixture = Initial mass = 102 g
l
kE
;f
eJ
.
kd
kn
zO
;e
k
u=i
zk
j
f
EH
k
dn
zO
;
e
ku=102g
 i
zk
j
E
Hkes
afy
;
sx
;
s 3
NH
n =
17
102
=6moles
N2
(g) + 3H2
(g) 2NH3
(g)
t = 0 6moles
eq. 6(/2) 6(3/2) 6(1 – )
 2
H
n present = 6 




 
2
3
= 9 = 9 ×
3
1
= 3 moles
(iii) Let Z be the compressibility factor for a real gas at critical conditions and Vander Waal constant be is dependent as
b = 







C
C
P
T
m
R
, then find the product (Z × m).
e
k
u
kd
h
]Ø
a
k
f
r
di
f
j
f
L
F
k
f
ri
j
],
do
k
L
r
f
o
dx
S
ld
sf
y
,
]l
E
i
h
M
~
;
r
kx
q
.
k
k
a
dZg
Sr
F
k
ko
k
.
M
j
o
k
yf
u
;
r
k
a
db,f
u
E
uç
d
k
jl
sf
u
H
k
Z
j
djrk gS]b= 







C
C
P
T
m
R
,rc(Z×m)dkxq.kuQyKkrdhft;s
Ans. 3
Sol. For a real gas at critical point, Z =
C
C
C
RT
V
P
=
8
3
Also, b = 







C
C
P
T
8
R
 m = 8
 Product (Z × m) =
8
3
× 8 = 3.
g
y
- ØkafrdfcUnqij,dokLrfodxSldsfy,]Z=
C
C
C
RT
V
P
=
8
3
rFkk,b= 







C
C
P
T
8
R
 m = 8
 (Z × m)dk xq.kuQy=
8
3
× 8= 3.