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Chemical Kinetics Practice Questions - DPP for NEET

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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
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FAQs on Chemical Kinetics Practice Questions - DPP for NEET

1. What is the definition of chemical kinetics?
Ans. Chemical kinetics is the branch of chemistry that studies the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which reactions occur.
2. How can the rate of a chemical reaction be determined?
Ans. The rate of a chemical reaction can be determined by measuring the change in concentration of reactants or products over time. This can be done by monitoring the consumption of reactants or the formation of products using various experimental techniques.
3. What are the factors that affect the rate of a chemical reaction?
Ans. The rate of a chemical reaction can be influenced by several factors, including temperature, concentration of reactants, surface area, presence of a catalyst, and the nature of the reactants.
4. What is the role of a catalyst in chemical kinetics?
Ans. A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative pathway with lower activation energy. It does not undergo any permanent change in its own chemical composition during the reaction.
5. How can the rate of a chemical reaction be expressed mathematically?
Ans. The rate of a chemical reaction can be expressed mathematically using the rate equation, which relates the rate of reaction to the concentrations of reactants. This equation can be determined experimentally and can help in understanding the relationship between reactant concentrations and reaction rate.
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