CHEMICAL KINETICS

1 For the reaction N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) under certain conditions of temperature and partial pressure
of the reactants, the rate of formation of NH 3 is 0 .001 kg h −1 . The rate of conversion of H 2 under the same
conditions is

(a) 1 .82  10 −4 kg / hr (b) 0.0015 kg / hr

(c) 1.52  10 kg / hr
4
(d) 1.82  10 −14 kg / hr
2 For the reaction N 2 + 3 H 2 → 2 NH 3
[ NH 3 ] −[H 2 ]
if = 2  10 − 4 mol l −1 s −1 , the value of would be
t t
(a) 1  10 −4 mol l −1 s −1 (b) 3  10 −4 mol l −1 s −1
(c) 4  10 −4 mol l −1 s −1 (d) 6  10 −4 mol l −1 s −1

3 A gaseous hypothetical chemical equation 2 A ⇌ 4 B + C is carried out in a closed vessel. The

concentration of B is found to increase by 5  10 −3 mol l −1 in 10 second. The rate of appearance of B is
(a) 5  10 −4 mol l −1 sec −1 (b) 5  10 −5 mol l −1 sec −1
(c) 6  10 −5 mol l −1 sec −1 (d) 4  10 −4 mol l −1 sec −1
4 The rate of disappearance of SO 2 in the reaction 2SO 2 + O2 → 2SO 3 is 1.28  10 −3 g / sec then the rate of formation
of SO 3 is
(a) 0.64  10 −3 g / sec (b) 0.80  10 −3 g / sec
(c) 1.28  10 −3 g / sec (d) 1.60  10 −3 g / sec

5 The velocity of the chemical reaction doubles every 10 o C rise of temperature. If the temperature is
raised by 50 o C , the velocity of the reaction increases to about
(a) 32 times (b) 16 times
(c) 20 times (d) 50 times
6 The rate of a reaction is doubled for every 10 o rise in temperature. The increase in reaction rate as a
result of temperature rise from 10 o to 100 o is
(a) 112 (b) 512
(c) 400 (d) 614
7 A first order reaction complete its 10% in 20 minutes then time required to complete its 19% is
(a) 30 minutes (b) 40 minutes
(c) 50 minutes (d) 38 minutes
(e) 45 minutes
8 The main function of a catalyst in speeding up a reaction is
(a) To increase the rate of the forward reaction
(b) To change the reaction path so as to decrease the energy of activation for the reaction
(c) To reduce the temperature at which the reaction can occur
(d) To increase the energy of the molecules of the reactants
9 The rate law for the reaction
[H + ]
Sucrose + Water ⎯⎯ ⎯→ Glucose + Fructose is given by
(a) Rate = K [sucrose] [water]
(b) Rate = K [sucrose] [water] 0
(c) Rate = K [sucrose] 0 [water]
(d) Rate = K [sucrose] 1 / 2 [water] 1 / 2
d [ A] d [B]
10 A + 2B → C + D . If − = 5  10 − 4 mol l −1 s −1 1, then − is [DPMT 2005]
dt dt

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 (a) 2 .5  10 −4 mol l −1 s −1 (b) 5 . 0  10 −4 mol l −1 s −1

(c) 2 . 5  10 − 3 mol l −1 s −1 (d) 1 . 0  10 −3 mol l −1 s −1

11 The data for the reaction A+B →C is[CBSE PMT 1994]

Exp. [A]0 [B]0 Initial rate
(1) 0.012 0.035 0.10
(2) 0.024 0.070 0.80
(3) 0.024 0.035 0.10
(4) 0.012 0.070 0.80
The rate law corresponds to the above data is
(a) Rate = k [B]3 (b) Rate = k [B]4
(c) Rate = k [ A][B]3 (d) Rate = k [ A] 2 [ B] 2

12 For a reaction 2 A + B → Products, doubling the initial concentration of both the reactants increases the
rate by a factor of 8, and doubling the concentration of B alone doubles the rate. The rate law for the
reaction is
(a)  = k[ A][B]2 (b)  = k[ A]2 [B]
(c)  = k[ A][B] (d)  = k[ A]2 [B]2
13 Point out the wrong statement :
For a first order reaction
(a) Time for half-change (t1 / 2 ) is independent of initial concentration
(b) Change in the concentration unit does not change the rate constant (K )
(c) Time for half-change  rate constant = 0.693
(d) The unit of K is mole −1 min −1
14 The rates of a certain reaction (dc/dt) at different times are as follows
Time Rate (mole litre–1 sec –1 )
0 2 . 8  10 −2
10 2.78  10 −2
20 2 .81  10 −2
30 2.79  10 −2
The reaction is
(a) Zero order (b) First order
(c) Second order (d) Third order
1
15 The decomposition of N 2 O5 is a first order reaction represented by N 2 O5 → N 2 O4 + O2 . After 15
2
minutes the volume of O2 produced is 9 ml and at the end of the reaction 35 ml . The rate constant is equal
to
1 35 1 44
(a) ln (b) ln
15 44 15 26
1 44 1 35
(c) ln (d) ln
15 35 15 26
16 Arrhenius equation is
d ln K d ln K
(a) = E * / RT (b) = E * / RT 2
dT dT
d ln K d ln K
(c) = −E * / RT 2 (d) = −E * / RT
dT dT

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 17 An endothermic reaction A → B has an activation energy 15 kcal / mole and energy of reaction 5 kcal / mole .
The activation energy of the reaction B → A is
(a) 20 kcal/mole (b) 15 kcal/mole
(c) 10 kcal/mole (d) None of these
18 The rate constant, the activation energy and the arrhenius parameter of a chemical reaction at 25 o C
are 3 .0  10 −4 s −1 , 104 .4 kJ mol −1 and 6.0  10 14 s −1 respectively. The value of the rate constant as T →  is
(a) 2 .0  10 18 s −1 (b) 6.0  10 14 s −1
(c) Infinity (d) 3 .6  10 30 s −1
19 Nitrogen monoxide, NO, reacts with hydrogen, H2, according to the following
equation:
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
If the mechanism for this reaction were,
2NO(g) + H2(g) →N2(g) + H2O2(g) (fast)
H2O2(g) + H2(g) → 2H2O (g) (slow)
which of the following rate laws would we expect to obtain experimentally?
(a) Rate = k[H2O2][H2] (b) Rate = k[NO]2[H2]
[NO]2 [H 2 ]2
(c) Rate = k[NO]2[H2]2 (d) Rate = k '
[N 2 ]
20 The rate constant for the first order decomposition of a reaction is given by
1.25 104 K
log K ( S −1 ) = 14.34 − Then the activation energy of the reaction is
T
a) 239KJ / mole b) 220KJ / mole c) 190KJ / mole d) 270KJ / mole

21 For the reaction A + B → C , it is found that doubling the concentration of A increases the rate by 4
times, and doubling the concentration of B doubles the reaction rate. What is the overal order of the
reaction.

22. For a first-order reaction, the ratio of times to complete 99.9% and half of the
reaction is

23 The inversion of Cane sugar proceeds with constant half-life of 500 minutes at
pH = 5 for any concentration of sugar. However if p H = 6, the half-life changes to
50minutes. For this reaction rate equation is rate = k [sugar] m [H+]n. The value of”n”
is.
24 Half life of a reaction is 20 sec. If t2 is second half life of reaction assuming it to
t3
be zero order and t3 is third half life assuming it to be 2nd order reaction, then =?
t2
25 A first order chemical reaction. A —→P is occurring at 500 K. If the same
reaction is to be carried out in presence of a catalyst, temperature of 400 K is
required to be maintained in order to maintain the same rate of reaction. If the
catalyst lowers the activation energy by 30 kJ, then what would be the value of
activation energy of the original pathway (uncatalysed reaction)/25.

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CHEMICAL KINETICS – KEY SHEET

1 b 2 b 3 a 4 c 5 a 6 b
7 b 8 b 9 b 10 a 11 a 12 b
13 d 14 a 15 d 16 b 17 c 18 b
19 d 20 a 21 3 22 10 23 0 24 8
25 6

CHEMICAL KINETICS – SOLUTIONS

−dN 2 −1 dH 2 1 dNH 3
1 (b) = =
dt 3 dt 2 dt
dH 2 3
=  0 .001 = 0 .0015 kg hr −1 .
dt 2

2 (b) N 2 + 3H 2 ⇌ 2NH 3
−[ N 2 ] 1 [H 2 ] 1 [ NH 3 ]
=− =
t 3 t 2 t
[H 2 ] 3 [ NH 3 ] 3
 =  =  2  10 −4
t 2 t 2
= 3  10 −4 mol litre −1 sec −1

3 (a) Increase in concentration of B = 5  10 −3 mol l −1 Time = 10 sec

Increase of conc. B
Rate of appearance of B=
Time taken

5  10 −3 mol l −1
= = 5  10 −4 mol l −1 Sec −1
10 sec

4 (c) The rate of formation of SO 3 is 1 . 28  10 −3 g / sec .

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 K t + 10 rt + 10
5 (a) = =2; For an increase of temperature to 50 o C , i.e. 5 times, the rate increases by 25 times,
Kt rt
i.e. 32
k t + 10 rt + 10
6 (b) = =2
Kt rt

For an increase of temperature to 90 o C i.e. 9 times, the rate increases by 29 times i.e. 512.
2 . 303 90
7 (b) In first phase, K= log .....(i)
20 100

In second phase K = 2.303 log 81 .....(ii)

t 100

From eq. (i) 2 .303 log 90 = 2 .303 log 81

20 100 t 100
20 (log 81 − log 100 )
t=
(log 90 − log 100 )
20 (1.908 − 2) 20  (−0.092 )
= = = 40 minutes
(1.954 − 2) (−0.046 )
8 (b) Catalyst decrease energy of activation.
9 (b) It is a pseudo-unimolecular reaction.
10 (a) A + 2 B → C + D
−d [ A]
= 5  10 − 4
dt
1 d[B] 5  10 −4
− = = 2 .5  10 −4 mol −1 sec −1
2 dt 2
11 coceptual
12 (b) 2 A + B → Products
According to question : Rate of reaction of ‘A’  [B] as increase in rate is double when [B] is
doubled.
Rate of reaction  [A] [B] as increase in rate is 8 times when concentration of both reactant is
doubled. It means that order of reaction is 3 and overall rate reaction should be r = K[ A]2 [B]
13 (d) Unit of K for Ist order reaction is Time −1 .
14 (a) The concentration of reactant does not change with time for zero order reaction (unit of K suggests
zero order) since reactant is in excess.
1  a  1  35  1  35 
15 (d) K= log e  = log e   = log e  
t  a − x  15  35 − 9  15  26 
16 conceptual
17 conceptual
18 (b) T2 = T (say), T = 25 o C = 298 K,
Ea = 104 .4 kJ mol −1 = 104 .4  10 3 J mol −1
K1 = 3  10 −4 , K 2 = ,
K2 Ea 1 1
log =  − 
K1 2 .303 R  T1 T2 
K2 104 .4  10 3 J mol −1
log =
3  10 − 4 2.303  (8.314 J k −1 mol −1 )
 1 1 1
 298 K − T  As T → , T → 0
 
K2 104 .4  10 3 J mol −1
 log −4
=
3  10 2.303  8 .314  298
K2 K2
log = 18 .297 , = 1 .98  10 18 or
3  10 − 4 3  10 − 4
K 2 = (1 .98  10 18 )  (3  10 −4 ) = 6  10 14 s −1

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19 2NO(g) + H2(g) N2(g) + H2O2(g) (fast)
H2O2(g) + H2(g) → 2H2O (g) (slow)
Rate = k[H2O2][H2]
[N 2 ][H 2 O 2 ]
K=
[NO]2 [H 2 ]
[NO]2 [H 2 ]2
Rate = Kk
[N 2 ]

.
20 a
Ea
109k = 109 A −
2.303RT
Ea 1.25 104
=
2.303RT T
 Ea = 2.303  8.314 1.25 104
Ea = 239kj / mole
21 A+B →C
On doubling the concentration of A rate of reaction increases by four times. Rate  [A]2
However on doubling the concentration of B , rate of reaction increases two times. Rate  [B ]
Thus, overall order of reaction = 2 + 1= 3
22 10

23 Half life 500 for all concentrations of sugar.

Half life is independent of the conc. of the sugar. So reaction is Ist order
W.R.T. sugar
For H+ conc.
n −1
t1  a2 
= 
t2  a1 
n −1
500  10 −6 
= 
50  10 −5 
n = 0
24
For zero order reaction t 1/2  [a]
1
For second order reaction t1/2 
[a]

[a] = initial conc.

So, t2 = 10 sec and t3 = 80
So, t3/t2 = 8
Ea c
25 log kuncatalysed = logA –
500 RT

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 Ea uc
logkcat = log A –
400 RT
Ea uc Ea c
=
5 4
4Eauc = 5 Eac
Eauc – Eac = 30
5
Ea c − Ea c = 30
4
Eac = 120 kJ
Eauc = 150 kJ
150  25 = 6.

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