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PHYSICAL CHEMISTRY TARGET : PRE-MEDICAL 2025
CHEMICAL KINETICS
1. In gaseous reaction it is important for the
understanding of the upper atmosphere H2O and
O react bimolecularly to form two OH radicals. (1)
∆H for this reaction is 72 kJ at 500 K and Ea is
Reaction coordinate
77 kJ, then Ea for the bimolecular recombination
of two OH radicals to form H2O and O is :-
(1) 3 kJ (2) 4 kJ (3) 5 kJ (4) 7 kJ
2. Half lives and initial concentration of a first order
(2)
and zero order reactions are same. Then the
ratio of the initial rates of the first order reaction
Reaction coordinate
to that of zero order reaction is :-

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(1) 1/0.693
(2) 2 × 0.693

P.E.
(3) 2/0.693 (3)
A+B
(4) 6.93 AB+I A+P
3. In the following first order reactions IAB
Reaction coordinate
K1 K2
(A)  → Product, (B)  → Product,
the ratio K1/K2 if 90% of (A) has been reacted in
time ‘t’ while 99% of (B) has been reacted in
P.E.

(4)
time 2t is :- A+B
IAB A+P
(1) 1 AB+I
Reaction coordinate
(2) 2
6. The reaction A(g) + 2B(g) → C(g) + D(g) is an
(3) 1/2
elementary process. In an experiment, the initial
 1 − 2 log 3  partial pressure of A & B are PA = 0.6 and
(4) 2  
 2 − log11 − 2 log 3  PB = 0.8 atm. Calculate the ratio of rate of
4. For a reaction : A → Product reaction relative to initial rate when PC becomes
0.2 atm.
d[A]
Rate law is – = K[A]0 1 1 1
dt (1) (2) (3) (4) 2
4 10 6
The concentration of A left after time t when
7. In the following reaction, rate constant is
t = 1/K is :-
1.2 × 10–2 M s–1 A → B. What is concentration
[A]0 [A]0 1
(1) (2) [A]0e (3) (4) of B after 10 min., if we start with 10 M of A.
e e2 [A]0
(1) 2.8 M (2) 7.2 M (3) 8.2 M (4) 2.7 M
5. The following mechanism has been proposed for
8. Two substances A (t1/2=5 min) and B (t1/2=15 min)
the exothermic catalyzed complex reaction. follow first order kinetics are taken in such a way

 fast
 IAB 
A + B  → AB +  k1
→ IP + A k2 that initially [A] = 4[B]. Calculate the time after
which the concentration of both the substance
If k1 is much smaller than k2. The most suitable
will be equal.
qualitative plot of potential energy (P.E.) versus
(1) 15 min (2) 20 min
reaction coordinate for the above reaction.
(3) 10 min (4) 25 min

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TARGET : PRE-MEDICAL 2025 PHYSICAL CHEMISTRY
9. A reaction takes place in three steps with 13. Consider a chemical reaction involving
individual rate constant and activation energy as compounds A and B, which is found to be first
follows- order in A and second order in B. At what rate
Rate cons tant Activation energy will the reaction occur in experiment 2 ?
–1
Step 1 k1 E a1 = 180kJ / mol Experiment Rate (Ms ) Initial [A] Initial [B]
Step 2 k2 E a2 = 80kJ / mol 1 0.10 1.0 M 0.20 M
Step 3 k3 E a3 = 50kJ / mol 2 ? 2.0 M 0.60 M
–1
2/3
(1) 1.8 M s
k k  (2) 0.20 M s–1
If overall rate constant, (k) =  1 2  , then
 k3  (3) 1.2 M s
–1

overall activation energy of the reaction will be: (4) 0.36 M s–1
(1) 140 kJ/mol 14. In a reaction carried out at 400 K, 0.0001% of
(2) 150 kJ/mol

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the total number of molecules are in activated
(3) 130 kJ/mol state. The energy of activation of the reaction is
(4) 120 kJ/mol (1) zero
10. A first order reaction is 50% completed in (2) 7.37 k cal/mol
20 minutes at 27°C and in 5 min at 47°C. The (3) 9.212 k cal/mol
energy of activation of the reaction is
(4) 11.05 k cal/mol
(1) 43.85 kJ/mol 15. Two first-order reactions have half-lives in the
(2) 55.14 kJ/mol ratio 3 : 2. Calculate the ratio of time intervals
(3) 11.97 kJ/mol t1 : t2 , t1 is the time period for 25% completion
(4) 6.65 kJ/mol of the first reaction and t2 for 75% completion of
11. Under the same reaction conditions, initial the second reaction :-
–3
concentration is 1.386 mol dm of a substance (1) 0.311 : 1 (2) 0.420 : 1
becomes half in 40 seconds and 20 seconds (3) 0.273 : 1 (4) 0.119 : 1
through first order and zero order kinetics, 16. A certain reaction obeys the rate equation (in the
k 
respectively. Ratio  l  of the rate constants for integrated form) C(1-n) - C(1-n)  = (n – 1) kt where
 k0 
0

n is the order of the reaction; C0 the initial

first order (k1) and zero order (k0) of the reactions
concentration and C, the concentration after
is
–1 3 time t. What is the unit of k for n = 3 :-
(1) 0.5 mol dm
–1
(1) sec
(2) 1.0 mol dm–3 –1 –1
–1 3
(2) litre mol sec
(3) 2.0 mol dm
–3
(3) mol litre–1 sec–1
(4) 0.1 mol dm
(4) litre2 mol–2 sec–1
12. The rate constant of a first order reaction at
17. The half-life of a first order reaction is one hour.
37°C is 10–3 min–1. The temperature coefficient
During what time interval will the concentration
of this reaction is 2. What is the rate constant
7
–1
(in min ) at 17°C for this reaction ? be reduced to of its initial value :-
–3
8
(1) 10
(1) 6.32 minutes (2) 17.4 minutes
(2) 5 × 10–4
–3
(3) 23.2 minutes (4) 11.6 minutes
(3) 2.5 × 10
(4) 2.5 × 10–4

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PHYSICAL CHEMISTRY TARGET : PRE-MEDICAL 2025
18. The time elapsed between 33% and 67% 21. A gaseous reaction, 2A(g) → B(g) + 5C(g) shows
completion of a first order reaction is 30 minutes increase in pressure from 80 mm to
What is the time needed for 25% completion :- 100 mm in 5 minutes. The rate of
(1) 15.5 min (2) 12.3 min disappearance of A is :-
(3) 18.5 min (4) 16.5 min (1) 4 mm min
–1
(2) 8 mm min
–1

19. Kinetics of the reaction, A(g) → 2B(g) + C(g) is (3) 16 mm min

–1
(4) 2 mm min
–1

followed by measuring the total pressure at 22. On introducing a catalyst at 500K, the rate
different times. It is given that : constant of a first order reaction increases 2.718
Initial pressure of A = 0.5 atm. times. If the activation energy in the presence of
Total pressure of reaction mixture after 2 hours –1
a catalyst is 4.15 kJ mol . Then what will be Ea
= 0.7 atm. in absence of catalyst (value of e = 2.718) :-
–3 –1
Rate constant of the reaction = 1 × 10 s (1) 4.15 kJ (2) 2.08 kJ
What is the rate of reaction when the total (3) 2.718 kJ (4) 8.3 kJ

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pressure is 0.7 atm ?
–4 –1
23. For a first order reaction A → products, the
(1) 2.0 × 10 atm s
–4 –1
concentration of [A] is reduced from 2M to
(2) 4.0 × 10 atm s
0.125 M in one hour, the t1/2 of this reaction
(3) 5.0 × 10–4 atm s–1
(in min) is :-
(4) 7.0 × 10–4 atm s–1
(1) 30 (2) 45 (3) 15 (4) None
20. For N2 + 3H2 → 2NH3 + 22 kcal, Ea
(energy of activation) is 70 kcal. Hence, Ea of
2NH3 → N2 + 3H2 is :-
(1) 92 kcal (2) 70 kcal
(3) – 70 kcal (4) – 92 kcal

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 3 2 1 1 1 3 2 1 1 2 1 4 1 4 1
Que. 16 17 18 19 20 21 22 23
Ans. 4 4 2 2 1 4 4 3

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