Chemical Kinetics Practice Questions - DPP for JEE

 Page 1


1. (d) Rate constant k = 0.6 × 10
–3
 mole per second. (unit mol L
–1
S
–1
shows zero order reaction)
For a zero order reaction
[A] = [A]
0
 – kt
and [A
0
] – [A] = [B] = kt
= 0.6 × 10
–3
 × 20 × 60 = 0.72 M
2. (c)
3. (c) For a first order reaction
when t = t
½
4. (a)
....(i)
when p
c
 = 0.20 atm p
A
 is reduced to 0.40 and p
B
 = 0.40 (See
stoichiometric representation)
Rate = k[0.40] [0.40]
2
...(ii)
5. (a) From data 1 and 3, it is clear that keeping (B) const, When [A] is
doubled, rate remains unaffected. Hence rate is independent of [A].
from 1 and 4, keeping [A] constant, when [B] is doubled, rate
Page 2


1. (d) Rate constant k = 0.6 × 10
–3
 mole per second. (unit mol L
–1
S
–1
shows zero order reaction)
For a zero order reaction
[A] = [A]
0
 – kt
and [A
0
] – [A] = [B] = kt
= 0.6 × 10
–3
 × 20 × 60 = 0.72 M
2. (c)
3. (c) For a first order reaction
when t = t
½
4. (a)
....(i)
when p
c
 = 0.20 atm p
A
 is reduced to 0.40 and p
B
 = 0.40 (See
stoichiometric representation)
Rate = k[0.40] [0.40]
2
...(ii)
5. (a) From data 1 and 3, it is clear that keeping (B) const, When [A] is
doubled, rate remains unaffected. Hence rate is independent of [A].
from 1 and 4, keeping [A] constant, when [B] is doubled, rate
become 8 times. Hence
rate .
6. (a) Since initial velocity is ten times the permissible value
? A
0
 = 10A
t
1/2
 =  = 
=  = 100 days.
7. (d) Since the slow step is the rate determining step hence  if we
consider option (A) we find
Rate = 
Now if we consider option (B) we find
Rate = ...(i)
For equation,
H
2
S  H
+
 + HS
–
or  
Substituting this value in equation (i) we find
Rate = 
Thus slow step should involve 1 molecule of Cl
2
 and1 molecule of H
2
S.
hence only , mechanism (A) is consistent with the given rate equation.
8. (d) From 1
st
 and 2
nd
 sets of data - no change in rate is observed with
Page 3


1. (d) Rate constant k = 0.6 × 10
–3
 mole per second. (unit mol L
–1
S
–1
shows zero order reaction)
For a zero order reaction
[A] = [A]
0
 – kt
and [A
0
] – [A] = [B] = kt
= 0.6 × 10
–3
 × 20 × 60 = 0.72 M
2. (c)
3. (c) For a first order reaction
when t = t
½
4. (a)
....(i)
when p
c
 = 0.20 atm p
A
 is reduced to 0.40 and p
B
 = 0.40 (See
stoichiometric representation)
Rate = k[0.40] [0.40]
2
...(ii)
5. (a) From data 1 and 3, it is clear that keeping (B) const, When [A] is
doubled, rate remains unaffected. Hence rate is independent of [A].
from 1 and 4, keeping [A] constant, when [B] is doubled, rate
become 8 times. Hence
rate .
6. (a) Since initial velocity is ten times the permissible value
? A
0
 = 10A
t
1/2
 =  = 
=  = 100 days.
7. (d) Since the slow step is the rate determining step hence  if we
consider option (A) we find
Rate = 
Now if we consider option (B) we find
Rate = ...(i)
For equation,
H
2
S  H
+
 + HS
–
or  
Substituting this value in equation (i) we find
Rate = 
Thus slow step should involve 1 molecule of Cl
2
 and1 molecule of H
2
S.
hence only , mechanism (A) is consistent with the given rate equation.
8. (d) From 1
st
 and 2
nd
 sets of data - no change in rate is observed with
the change in concentration of ‘C’. So the order with respect to ‘C’
is zero.
From 1
st
 and 4
th
 sets of data
Dividing eq. (4) by eq. (1)
or 0.25 = (0.5)
x
 or (0.5)
2
 = (0.5)
x
? x = 2
The order with respect to ‘A’ is 2 from the 1
st
 and 3
rd
 sets of data
dividing eq. (1) by eq. (3)
or (0.5)
1
 = (0.5)
y 
? y = 1
The order with respect to ‘B’ is 1
So the order with respective the reactants A, B and C is 2, 1 and 0.
9. (b)
Thus energy of activation for reverse reaction depend upon whether
reaction is exothermic or endothermic
If reaction is exothermic,
 
If reaction is endothermic
 
10. (b) For a first order reaction
t
1/2
 = = 0.5 × 10
–3
s
–1
11. (c)
Page 4


1. (d) Rate constant k = 0.6 × 10
–3
 mole per second. (unit mol L
–1
S
–1
shows zero order reaction)
For a zero order reaction
[A] = [A]
0
 – kt
and [A
0
] – [A] = [B] = kt
= 0.6 × 10
–3
 × 20 × 60 = 0.72 M
2. (c)
3. (c) For a first order reaction
when t = t
½
4. (a)
....(i)
when p
c
 = 0.20 atm p
A
 is reduced to 0.40 and p
B
 = 0.40 (See
stoichiometric representation)
Rate = k[0.40] [0.40]
2
...(ii)
5. (a) From data 1 and 3, it is clear that keeping (B) const, When [A] is
doubled, rate remains unaffected. Hence rate is independent of [A].
from 1 and 4, keeping [A] constant, when [B] is doubled, rate
become 8 times. Hence
rate .
6. (a) Since initial velocity is ten times the permissible value
? A
0
 = 10A
t
1/2
 =  = 
=  = 100 days.
7. (d) Since the slow step is the rate determining step hence  if we
consider option (A) we find
Rate = 
Now if we consider option (B) we find
Rate = ...(i)
For equation,
H
2
S  H
+
 + HS
–
or  
Substituting this value in equation (i) we find
Rate = 
Thus slow step should involve 1 molecule of Cl
2
 and1 molecule of H
2
S.
hence only , mechanism (A) is consistent with the given rate equation.
8. (d) From 1
st
 and 2
nd
 sets of data - no change in rate is observed with
the change in concentration of ‘C’. So the order with respect to ‘C’
is zero.
From 1
st
 and 4
th
 sets of data
Dividing eq. (4) by eq. (1)
or 0.25 = (0.5)
x
 or (0.5)
2
 = (0.5)
x
? x = 2
The order with respect to ‘A’ is 2 from the 1
st
 and 3
rd
 sets of data
dividing eq. (1) by eq. (3)
or (0.5)
1
 = (0.5)
y 
? y = 1
The order with respect to ‘B’ is 1
So the order with respective the reactants A, B and C is 2, 1 and 0.
9. (b)
Thus energy of activation for reverse reaction depend upon whether
reaction is exothermic or endothermic
If reaction is exothermic,
 
If reaction is endothermic
 
10. (b) For a first order reaction
t
1/2
 = = 0.5 × 10
–3
s
–1
11. (c)
12. (d) Overall order = sum of orders w.r.t each reactant.
Let the order be x and y for G and H respectively
Q  For (1)  and (3), the rate is doubled when conc. of G is doubled
keeping that of H constant i.e.,  ? x = 1
From (2) and (3),  y = 2
?  Overall order is 3.
13. (c) Rate
1
 = k [A]
n
 [B]
m
; Rate
2
 = k [2A]
n
 [½B]
m
= [2]
n
 [½]
m
 = 2
n
.2
–m
 = 2
n–m
14. (d)
when k
1
 and k
2
 are equal at any temperature T, we have
Page 5


1. (d) Rate constant k = 0.6 × 10
–3
 mole per second. (unit mol L
–1
S
–1
shows zero order reaction)
For a zero order reaction
[A] = [A]
0
 – kt
and [A
0
] – [A] = [B] = kt
= 0.6 × 10
–3
 × 20 × 60 = 0.72 M
2. (c)
3. (c) For a first order reaction
when t = t
½
4. (a)
....(i)
when p
c
 = 0.20 atm p
A
 is reduced to 0.40 and p
B
 = 0.40 (See
stoichiometric representation)
Rate = k[0.40] [0.40]
2
...(ii)
5. (a) From data 1 and 3, it is clear that keeping (B) const, When [A] is
doubled, rate remains unaffected. Hence rate is independent of [A].
from 1 and 4, keeping [A] constant, when [B] is doubled, rate
become 8 times. Hence
rate .
6. (a) Since initial velocity is ten times the permissible value
? A
0
 = 10A
t
1/2
 =  = 
=  = 100 days.
7. (d) Since the slow step is the rate determining step hence  if we
consider option (A) we find
Rate = 
Now if we consider option (B) we find
Rate = ...(i)
For equation,
H
2
S  H
+
 + HS
–
or  
Substituting this value in equation (i) we find
Rate = 
Thus slow step should involve 1 molecule of Cl
2
 and1 molecule of H
2
S.
hence only , mechanism (A) is consistent with the given rate equation.
8. (d) From 1
st
 and 2
nd
 sets of data - no change in rate is observed with
the change in concentration of ‘C’. So the order with respect to ‘C’
is zero.
From 1
st
 and 4
th
 sets of data
Dividing eq. (4) by eq. (1)
or 0.25 = (0.5)
x
 or (0.5)
2
 = (0.5)
x
? x = 2
The order with respect to ‘A’ is 2 from the 1
st
 and 3
rd
 sets of data
dividing eq. (1) by eq. (3)
or (0.5)
1
 = (0.5)
y 
? y = 1
The order with respect to ‘B’ is 1
So the order with respective the reactants A, B and C is 2, 1 and 0.
9. (b)
Thus energy of activation for reverse reaction depend upon whether
reaction is exothermic or endothermic
If reaction is exothermic,
 
If reaction is endothermic
 
10. (b) For a first order reaction
t
1/2
 = = 0.5 × 10
–3
s
–1
11. (c)
12. (d) Overall order = sum of orders w.r.t each reactant.
Let the order be x and y for G and H respectively
Q  For (1)  and (3), the rate is doubled when conc. of G is doubled
keeping that of H constant i.e.,  ? x = 1
From (2) and (3),  y = 2
?  Overall order is 3.
13. (c) Rate
1
 = k [A]
n
 [B]
m
; Rate
2
 = k [2A]
n
 [½B]
m
= [2]
n
 [½]
m
 = 2
n
.2
–m
 = 2
n–m
14. (d)
when k
1
 and k
2
 are equal at any temperature T, we have
15. (a) Arrhenius equation is given by
k = 
Taking log on both sides, we get
log k = log A – 
Arrhenius plot a graph between log k and  whose slope is .
16. (a) When concentration of A is doubled, rate is doubled. Hence order
with respect to A is one.
When concentrations of both A and B are doubled, rate increases by 8
times hence order with respect to B is 2.
? rate = k [A]
1
 [B]
2
17. (c) Third order