18 Chemical Kinetics

1. For the reaction 4. For the reaction H2(g) + Br2 (g) ® 2HBr (g), the
N 2O5 (g) ¾¾
® 2NO2 (g) + 1/2 O 2(g) the value experimental data suggest, rate = k[H2][Br2]1/2.
The molecularity and order of the reaction are
of rate of disappearance of N 2 O5 is given as respectively
6.25 × 10–3 mol L–1s–1. The rate of formation of
3 3 3 1
NO 2 and O2 is given respectively as : (a) 2, (b) , (c) 1, 1 (d) 1,
2 2 2 2
(a) 6.25 × 10–3 mol L–1s–1 and
5. For a chemical reaction t1/2 is 2.5 hours at room
6.25 × 10–3 mol L–1s–1 temperature. How much of the reactant will be
(b) 1.25 × 10–2 mol L–1s–1 and left after 7.5 hours if initial weight of reactant was
3.125 × 10–3 mol L–1s–1 160 g?
(c) 6.25 × 10–3 mol L–1s–1 and
3.125 × 10–3 mol L–1s–1 (a) 10 g (b) 40 g (c) 80 g (d) 20 g
(d) 1.25 × 10–2 mol L–1s–1 and 6. The rate law for a reaction between the
6.25 × 10–3 mol L–1s–1 substances A and B is given by rate = k[A]n[B]m.
2. The reactions rate N 2 (g) + 3H 2 (g) ® 2 NH3 (g) On doubling the concentration of A and halving
the concentration of B, the ratio of the new rate
was measured d[ NH 3 ] = 2 ´10 - 4 mol sec–1. to the earlier rate of the reaction will be as
dt
The rates of reactions expressed in terms of N2
(a) (m + n) (b) (n – m)
and H2 are
Rate in terms of N2. Rate in terms of H2 1
(mol L–1sec–1 ) (mol L–1sec–1 ) (c) 2(n + m) (d) ( m+ n)
–4 2
(a) 2×10 2×10–4 7. The order of a reaction, with respect to one of
(b) 3×10–4 1×10–4 the reacting component Y, is zero. It implies that:
(c) 1×10 –4 3×10–4 (a) the reaction is going on at a constant rate
(d) 2×10 –1 2×10–3 (b) the rate of reaction does not vary with
3. For the reaction A + B ¾¾® C + D. The variation temperature
of the concentration of the products is given by (c) the reaction rate is independent of the
the curve concentration of Y
(d) the rate of formation of the activated
Y complex is zero
Z 8. For a first order reaction, a plot of log (a – x)
against time is a straight line with a negative slope
Conc equal to
W -k
(a) (b) – 2.303 k
2.303
X
Time 2.303 Ea
(c) (d) -
(a) Z (b) Y (c) W (d) X k 2.303 R
 EBD_7587
208 Chemistry Objective MCQs
9. A following mechanism has been proposed for a 16. The activation energy of the reaction,
reaction A + B ® C + D + 38 kcal is 20 kcal. What would
2A + B ® D + E be the activation energy of the following reaction.
A + B ® C + D (slow) C + D ®A+ B
A + C ® E (fast) (a) 20 kcal (b) –20 kcal
(c) 18 kcal (d) 58 kcal
The rate law expression for the reaction is:
17. The half-life of a reaction is inversely proportional
(a) r = k [A]2 [B] (b) r = k [A] [B]
to the square of the initial concentration of the
(c) r = k [A]2 (d) r = k [A] [C] reactant. Then the order of the reaction is
10. The rate law for the reaction below is given by (a) 0 (b) 1 (c) 2 (d) 3
the expression k [A][B]
18. For a reaction A + 2B ® C, the amount of C
A + B ® Product formed by starting the reaction with 5 moles of
If the concentration of B is increased from 0.1 to A and 8 moles of B is
0.3 mole, keeping the value of A at 0.1 mole, the (a) 5 moles (b) 8 moles
rate constant will be: (c) 16 moles (d) 4 moles
(a) 3k (b) 9k (c) k/3 (d) k 19. The rate law for the reaction 2X + Y ® Z is
11. The half life of a radioactive element is 20 min. Rate = k[X][Y]. The correct statement with
The time interval between the stages of its 33% regard to this relation is
and 67% decay is
(a) the unit of k is s–1
(a) 40 min (b) 20 min (c) 30 min (d) 25 min
(b) the rate of the reaction is independent of
12. Which of the following reactions is not of the
[X] and [Y]
first order?
(c) for this reaction t1/2 is independent of
(a) Inversion of sucrose in the presence of acid initial concentrations of reactant
(b) Acid-catalyzed hydrolysis of ethyl acetate (d) the rate of formation of Z is half of the rate
(c) Hydrolysis of tertiary butyl halide using of disappearance of X
alkali 20. In a reaction, 2A ® products, the concentration
(d) Oxidation of I– ion by S2O82– ion of A decreases from 0.50 M to 0.38 M in
13. Rate constant k = 1.2 × 103 mol– 1 L s– 1 and 10 min. What is the rate of reaction (in M s–1)
Ea = 2.0 × 102 kJ mol– 1. When T ® ¥: during this interval?
(a) A = 2.0 × 102 kJ mol– 1 (a) 0.012 (b) 0.024
(c) 2 × 10 –3 (d) 2 × 10–4
(b) A = 1.2 × 103 mol– 1 L s– 1
(c) A = 1.2 × 103 mol L– 1 s– 1 21. A chemical reaction is catalyzed by a catalyst X.
Hence X
(d) A = 2.4 × 103 kJ mol– 1 s– 1
(a) reduces enthalpy of the reaction
14. A catalyst lowers the activation enegy of a (b) decreases rate constant of the reaction
reaction from 20 kJ mol– 1 to 10 kJ mol– 1. The (c) increases activation energy of the reaction
temperature at which the uncatalyzed reaction (d) does not affect equilibrium constant of the
will have the same rate as that of the catalyzed at reaction
27 ° C is
22. In respect of the equation k = Ae- Ea / RT in
(a) – 123 °C (b) 327 °C chemical kinetics, which one of the following
(c) 32.7 °C (d) + 23 °C statements is correct ?
15. For a reaction, the rate constant is expressed as (a) A is adsorption factor
k = Ae–40000/T. The energy of the activation is (b) Ea is energy of activation
(a) 40000 cal (b) 88000 cal (c) R is Rydberg’s constant
(c) 80000 cal (d) 8000 cal (d) k is equilibrium constant
Chemical Kinetics 209
23. The minimum energy a molecule should possess 29. Consider the two hypothetical reactions given
in order to enter into a fruitful collision is known below :
as I a A ® Products, k = x mol–1 L min–1
(a) reaction energy (b) collision energy II b B ® Products, k = y min–1
(c) activation energy (d) threshold energy The half-lives of both the reactions are the same,
24. For a reaction, activation energy (Ea) = 0 and rate equal to 1 hr when molar concentration of the
constant reactant is 1.0 M in each case. If these reactions
( k ) = 3.2 ´106 s -1 at 300 K. What is the value of are started at the same time taking 1M of the
the rate constant at 310 K reactant in each case, the ratio [A]/[B] after 3 hr
will be :
(a) 3.2 ´ 10-12 s -1 (b) 3.2 ´ 106 s -1 (a) 0.5 (b) 4 (c) 1 (d) 2
(c) 6.4 ´ 1012 s -1 (d) 6.4 ´ 106 s -1 30. Consider the consecutive reactions :
1
25. In the Arrhenius plot of ln k Vs , a linear plot
T k = 2´10-5 s -1 k =8´10-6 s -1 k = 3´10-3 s -1
is obtained with a slope of –2 × 104 K. The energy A ¾¾¾¾¾¾
® B ¾¾¾¾¾¾
®.C ¾¾¾¾¾¾
®D
of activation of the reaction (in kJ mole–1) is (R k = 2 ´10 -5 s -1 k = 8´10 -6 s - 1 k = 3´10 -3 s -1
A ¾¾ ¾¾ ¾ ¾
® B ¾¾ ¾ ¾ ¾¾
® C ¾¾¾ ¾ ¾¾
®D
value is 8.3 J K–1 mol–1) The rate determining step of the reaction is :
(a) A ® B (b) C ® D
(a) 83 (b) 166 (c) 249 (d) 332 (c) B ® C (d) A ® D
26. The rate of reaction is doubled for every 10°C 31. A + 2B ® C, the rate equation for this reaction
rise in temperature. The increase in reaction rate is given as Rate = K[A][B].
as a result of temperature rise from 10°C to 100°C If the concentration of A is kept the same but
is that of B is doubled what will happen to the
rate itself ?
(a) 112 (b) 512 (c) 400 (d) 614
(a) halved (b) the same
27. Consider the reaction A ® 2 B + C, DH = – 15 (c) doubled (d) quadrupled
kcal. The energy of activation of backward 32. In the reaction of formation of sulphur trioxide
reaction is 20 kcal mol–1. In presence of catalyst
by contact process 2SO2 + O2 ƒ 2SO3 the rate
the energy of activation of forward reaction is 3
kcal mol–1. At 400 K the catalyst causes the rate of reaction was measured as
of the reaction to increase by the number of times d [O2 ]
= -2.5 ´10-4 mol L-1s -1 . The rate of
equal to dt
reaction is terms of [SO2] in mol L–1 s–1 will be:
(a) – 1.25 × 10–4 (b) – 2.50 × 10–4
(a) e3.5 (b) e2.5 –4
(c) – 3.75 × 10 (d) – 5.00 × 10–4
(c) e –2.5 (d) e 2.303
33. For the reaction, 2N2O5 ® 4NO2 + O2, the rate
28. A reaction takes place in various steps. The rate equation can be expressed in two ways
constant for first, second, third and fifth steps
d [ N 2O5 ] d [ NO 2 ]
are k1, k2, k3 and k5 respectively. The overall rate - = k [N 2 O5 ] and + = k ¢ [ N 2 O5 ]
dt dt
1/ 2
k 2 æ k1 ö k and k¢ are related as:
constant is given by k = ç ÷
k3 è k5 ø (a) k = k¢ (b) 2k = k¢
If activation energy are 40, 60, 50 and 10 kJ/mol (c) k = 2k¢ (d) k = 4k¢
respectively, the overall energy of activation (kJ/ 34. The reaction 2N2O5 2N2O4 + O2 is
mol) is : (a) bimolecular and of second order
(b) unimolecular and of first order
(a) 10 (b) 20 (c) bimolecular and of first order
(c) 25 (d) none of these (d) bimolecular and of zero order
 EBD_7587
210 Chemistry Objective MCQs
35. For the reaction system : 41. The reaction L ¾¾® M is started with 10.0 g of
2NO (g) + O2 (g) ® 2 NO2 (g) volume is L. After 30 and 90 minutes 5.0 g and 1.25 g of L
suddenly reduced to half its value by increasing respectively are left. The order of the reaction is
the pressure on it. If the reaction is of first order (a) 0 (b) 1 (c) 2 (d) 3
with respect to O2 and second order with respect 42. A first order reaction is half-completed in
to NO, the rate of reaction will 45 minutes. How long does it need for 99.9% of
(a) diminish to one-eighth of its initial value the reaction to be completed?
(b) increase to eight times of its initial value (a) 20 hours (b) 10 hours
(c) increase to four times of its initial value 1
(d) diminish to one-fourth of its initial value (c) 7 hours (d) 5 hours
2
36. Units of rate constant of first and zero order 43. A reaction which is of first order w.r.t. reactant A,
reactions in terms of molarity M unit are has a rate constant 6 min –1. If we start with [A] =
respectively 0.5 mol L–1, when would [A] reach the value of
(a) sec–1, M.sec–1 (b) sec–1, M 0.05 mol L–1
–1
(c) M.sec , sec –1 (d) M, sec–1 (a) 0.384 min (b) 0.15 min
37. The plot of concentration of the reactant Vs time (c) 3 min (d) 3.84 min
for a reaction is a straight line with a negative 44. The time taken for 90% of a first order reaction to
slope. The reaction follows a rate equation of complete is approximately
(a) 1.1 times that of half-life
(b) 2.2 times that of half-life
(a) zero order (b) first order
(c) 3.3 times that of half-life
(c) second order (d) third order
38. The differential rate law for the reaction (d) 4.4 times that of half-life
45. The rate constant of a reaction is 0.0693 min– 1.
H2 (g) + I2 (g) ® 2HI (g) is
Starting with 10 mol, the rate of the reaction after
d[ H 2 ] d[ I 2 ] d[ HI ] 10 min is
(a) - =- =-
dt dt dt (a) 0.0693 mol min– 1
d[H2 ] d[I2 ] 1 d[HI] (b) 0.0693 × 2 mol min– 1
(b) = = (c) 0.0693 × 5 mol min– 1
dt dt 2 dt
(d) 0.0693 × (5)2 mol min– 1
1 d[H 2 ] 1 d[I 2 ] d[HI]
(c) = =- 46. The plot that represents the zero order reaction
2 dt 2 dt dt is :
d[H 2 ] d[I ] d[HI]
(d) -2 = -2 2 =
dt dt dt [R] [R]
(a) (b)
39. Which of the following will react at the highest
t t
rate?

[R]
(a) 1 mole of A and 1 mole of B in a 1-L vessel (c) (d) [R]
(b) 2 mole of A and 2 mole of B in a 2-L vessel t t
(c) 3 mole of A and 3 mole of B in a 3-L vessel 47. Which of the following reactions is not of the
(d) All would react at the same rate first order?
40. In the reaction, A + 2B ¾¾ ® 6C + 2D, If the initial
d[A] (a) Inversion of sucrose in the presence of acid
rate - at t = 0 is 2.6 × 10–2 M sec–1, what
dt (b) Acid-catalyzed hydrolysis of ethyl acetate
d [B]
will be the value of at t = 0? (c) Hydrolysis of tertiary butyl halide using
dt alkali
(a) 8.5 × 10 M sec (b) 2.5 × 10–2 M sec–1
–2 –1
(d) Oxidation of I– ion by S2O82– ion
(c) 5.2 × 10–2 M sec–1 (d) 7.5 × 10–2 M sec–1
Chemical Kinetics 211
48. A first order reaction is 50% completed in 20 (a) log k versus log T will give a straight line
minutes at 27°C and in 5 minutes at 47°C. The with a slope as –25000
energy of activation of the reaction is : (b) log k versus T will give a straight line with
slope as 25000
(c) log k versus 1/T will give a straight line with
(a) 43.85 kJ/mol (b) 55.14 kJ/mol slope as –25000/R
(c) 11.97 kJ/mol (d) 6.65 kJ/mol (d) log k versus 1/T will give a straight line
49. For the first order reaction A ® B + C is carried 55. The velocity of a reaction is doubled for every
out at 27°C. If 3.8 × 10–16 % of the reactant 10°C rise in temp. If the temp. is raised to 50°C
molecules exists in the activated state, the Ea from 0 °C the reaction velocity increases by about
(activation energy) of the reaction is: (a) 12 times (b) 16 times
(a) 12 kJ/mol (b) 831.4 kJ/mol (c) 32 times (d) 50 times
(c) 100 kJ/mol (d) 88.57 kJ/mol 56. Which of the following statements is incorrect?
50. A catalyst lowers the activation energy of a (a) Activation energy for the forward reaction
certain reaction from 83.314 to 75 kJ mol–1 at is equals to activation energy for the reverse
500 K. What will be the rate of reaction as reaction
compared to uncatalysed reaction? Assume (b) For a reversible reaction, an increase in
other things being equal. temperature increases the reaction rate for
(a) Double (b) 28 times both the forward and the backward reaction
(c) 7.38 times (d) 7.38 × 103 times (c) The larger the initial reactant concentration
51. Rate of a reaction can be expressed by Arrhenius for a second order reaction, the shorter is
equation as: k = Ae - Ea / RT its half-life.
(d) When Dt is infinitesimally small, the average
In this equation, Ea represents
rate equals the instantaneous rate
(a) the total energy of the reacting molecules
57. Consider the following statements:
at a temperature, T
(b) the fraction of molecules with energy greater
than the activation energy of the reaction
(c) the energy below which all the colliding I. Increase in concentration of reactant
molecules will react increases the rate of a zero order reaction.
(d) the energy below which colliding molecules
II. Rate constant k is equal to collision
will not react
frequency A if Ea = 0.
52. The reason for almost doubling the rate of
III. Rate constant k is equal to collision
reaction on increasing the temperature of the
frequency A if Ea = ¥.
reaction system by 10°C is
(a) The value of threshold energy increases IV. ln k Vs T is a straight line.
(b) Collision frequency increases V. In k Vs 1/T is a straight line.
(c) The fraction of the molecule having energy Correct statements are
equal to threshold energy or more increases (a) I and IV (b) II and V
(d) Activation energy decreases (c) III and IV (d) II and III
53. In Arrhenius plot, intercept is equal to 58. The time required for 10% completion of a first
-Ea order reaction at 298 K is equal to that required
(a) (b) ln A for its 25% completion at 308 K. If the pre-
R
(c) ln K (d) log10A exponential factor for the reaction is
54. A reaction rate constant is given by 3.56 × 109 s–1, the rate constant at 318 K is:

(a) 18.39 kcal mol–1 (b) 20 kcal mol–1

k = 1.2 ´ 1014 e -25000 / RTsec -1 . It means (c) 16 kcal mol–1 (d) 21.5 kcal mol–1
 EBD_7587
212 Chemistry Objective MCQs
59. For the reaction, 3A + 2B ® C + D, the differential 65. A reaction proceeds by first order, 75% of this
rate law can be written as: reaction was completed in 32 min. The time

1 d [A ] d [C] required for 50% completion is

= k [A ] [B]
n m
(a) =
3 dt dt
(a) 8 min (b) 16 min
d [A ] d [C ]
= k [A ] [B]
n m (c) 20 min (d) 24 min
(b) - =
dt dt
66. Consider the reaction : N2 (g) + 3H2 (g) ® 2 NH3 (g)
1 d [A ] d [C]
= k [A ] [B]
n m
(c) + =- d[ NH 3 ]
3 dt dt The equality relationship between and
dt
d[H 2 ]
1 d [A ] d [C] -
= k [A ] [B] is
n m
(d) - = dt
3 dt dt
60. The instantaneous rate of disappearance d[ NH 3 ] 2 d[H 2 ]
(a) + =-
of MnO4– ion in the following reaction is dt 3 dt
4.56 × 10–3 Ms–1 2MnO4– + 10I– + 16H+ ® d[ NH 3 ] 3 d[H 2 ]
2Mn2+ + 5I2 + 8H2O (b) + =-
dt 2 dt
The rate of appearance I2 is :
d[ NH 3 ] d[H 2 ]
(a) 4.56 × 10–4 Ms–1 (b) 1.14 × 10–2 Ms–1 (c) =-
dt dt
(c) 1.14 × 10–3 Ms–1 (d) 5.7 × 10–3 Ms–1
61. The rate constant of a zero order reaction is d[ NH 3 ] 1 d[H 2 ]
2.0 × 10–2 mol L–1 s–1. If the concentration of the (d) =-
dt 3 dt
reactant after 25 seconds is 0.5 M. What is the
initial concentration? 67. The rate of a reaction increases four-fold when
(a) 0.5 M (b) 1.25 M the concentration of reactant is increased 16 times.
If the rate of reaction is 4 × 10– 6 mol L– 1 s– 1 when
(c) 12.5 M (d) 1.0 M
the concentration of the reactant is 4 × 10– 4 mol
62. The reaction A ® B follows first order kinetics.
L– 1. The rate constant of the reaction will be
The time taken for 0.8 mole of A to produce 0.6
mole of B is 1 hour. What is the time taken for
conversion of 0.9 mole of A to produce 0.675 mole
(a) 2×10–4 mol1/2 L–1/2 s–1
of B?
(b) 1×10–2 s–1
(a) 2 hours (b) 1 hour (c) 2×10–4 mol–1/2 L1/2 s–1
(c) 0.5 hour (d) 0.25 hour (d) 25 mol–1 L min–1
63. The rate of a first order reaction is 1.5 × 10–2 mol L–1 68. In the reaction A ® B + C, rate constant is 0.001
min–1 at 0.5 M concentration of the reactant. The Ms–1. If we start with 1 M of A then conc. of A
half life of the reaction is and B after 10 minuter are respectively.
(a) 0.383 min (b) 23.1 min
(a) 0.5 M, 0.5 M (b) 0.6 M, 0.4 M
(c) 8.73 min (d) 7.53 min
(c) 0.4 M, 0.6 M (d) 0.6 M 0.5 M
64. For a first order reaction A ¾® B the reaction
rate at reactant concentration of 0.01 M is found 69. The half life for the virus inactivation if in the
to be 2.0 ´10-5 mol L-1 s-1. The half life period of beginning 1.5% of the virus is inactivated per
the reaction is minute is (Given: The reaction is of first order)
(a) 30 s (b) 220 s (c) 300 s (d) 347 s (a) 76 min (b) 66 min (c) 56 min (d) 46 min
Chemical Kinetics 213
70. The reaction of A2 and B2 follows the equation 75. For an exothermic reaction, the energy of
A 2 (g ) + B2 (g ) ® 2AB ( g ) activation of the reactants is
(a) equal to the energy of activation of products
The following data were observed
(b) less than the energy of activation of
Initial rate of appearance of products
[ A 2 ]0 [ B2 ]0 (c) greater than the energy of activation of
AB(g) (in Ms -1) products
0.10 0.10 2.5 ´ 10 -4 (d) sometimes greater and sometimes less than
that of the products
0.20 0.10 5 ´ 10 -4 76. When a biochemical reaction is carried out in
0.20 0.20 10 ´ 10 -4 laboratory in the absence of enzyme then rate of
reaction obtained is 10–6 times, then activation
The value of rate constant for the above reaction energy of reaction in the presence of enzyme is
is:
6
(a) 2.5 × 10–4 (b) 2.5 × 10–2 (a)
(c) 1.25 × 10 –2 (d) None of these RT
71. In th e presence of an acid, the initial (b) different from Ea obtained in laboratory
concentration of cane sugar was reduced from (c) P is required
0.20 to 0.10 M in 5 hours and from 0.2 to 0.05 M (d) can't say anything
in 10 hours. The reaction is of : 77. A chemical reaction was carried out at 300 K and
280 K. The rate constants were found to be k1
(a) Zero order (b) First order and k2 respectively. then
(c) Second order (d) Third order (a) k2 = 4k1 (b) k2 = 2k1
72. The activation energies of the forward and (c) k2 = 0.25 k1 (d) k2 = 0.5 k1
backward reactions in the case of a chemical
78. In a reaction at 27ºC, 10–3% reactant molecules
reaction are 30.5 and 45.4 kJ/mol respectively.
manage to cross over the barrier of transition
The reaction is :
state. The energy of these molecules in excess of
(a) exothermic
the average value will be (R = 2 cal K–1 mol–1) :
(b) endothermic
(c) neither exothermic nor endothermic
(d) independent of temperature
73. A homogeneous catalytic reaction takes place (a) 6.91 kcal mol–1 (b) 3.00 kcal mol–1
through the three alternative plots A, B and C (c) 4.15 kcal mol –1 (d) 5.10 kcal mol–1
shown in the given figure. Which one of the 79. The activation energy for a simple chemical
following indicates the relative ease with which reaction A ® B is Ea in forward direction. The
the reaction can take place? activation energy for reverse reaction
A (a) is always double of Ea
B
C
(b) is negative of Ea
(c) is always less than Ea
Energy

(d) can be less than or more than Ea

80. The slope in Arrhenius plot, is equal to:
Ea Ea
(a) - (b)
Reaction course 2.303 R R
(a) A > B > C (b) C > B > A
R
(c) A > C > B (d) A = B = C (c) - (d) None of these
74. A catalyst is a substance which : 2.303 Ea
(a) is always in the same phase as in the reaction 81. A radioactive isotope having a half - life period
(b) alters the equilibrium in a reaction of 3 days was received after 12 days. If 3g of the
(c) does not participate in the reaction but alters isotope is left in the container, what would be the
the rate of reaction initial mass of the isotope?
(d) participates in the reaction and provides an (a) 12g (b) 36g (c) 48g (d) 24g
easier pathway for the same
 EBD_7587
214 Chemistry Objective MCQs
82. In a chemical reaction A is converted into B. The 88. The rate constant for a first order reaction whose
rates of reaction, starting with initial half life is 480 sec, is :
concentrations of A as 2 × 10–3 M and 1 × 10–3 (a) 1.44 × 10–3 sec–1 (b) 1.44 × sec–1
M, are equal to 2.40 × 10–4 Ms–1 and 0.60 × 10–4 (c) 0.72 × 10–3 sec–1 (d) 2.88 × 10–3 sec–1
Ms–1 respectively. The order of reaction with 89. The hypothetical reaction
respect to reactant A will be
(a) 0 (b) 1.5 (c) 1 (d) 2 A 2 + B2 ¾
¾® 2AB ; follows the following
83. For a reaction A ® Products, a plot of log t1/2 mechanism A 2 ¾Fast
¾¾® A + A ,
versus log a0 is shown in the figure. If the initial Slow Fast
concentration of A is represented by a0, the order A + B 2 ¾¾¾® AB + B , A + B ¾¾ ¾® AB .
of the reaction is The order of the overall reaction is
(a) 0 (b) 1 (c) 2 (d) 3/2
90. If half-life of a substance is 5 yrs, then the total
amount of substance left after 15 years, when initial
amount is 64 grams is
log t1/2 45° (a) 16 g (b) 2 g (c) 32 g (d) 8 g
91. In a reversible reaction the energy of activation
of the forward reaction is 50 kcal. The energy of
activation for the reverse reaction will be
log a0 (a) < 50 kcal
(b) either greater than or less than 50 kcal
(a) one (b) zero (c) two (d) three
(c) 50 kcal
84. The rate constant of a reaction with a virus is (d) > 50 kcal
3.3 × 10– 4 s– 1. Time required for the virus to 92. Activation energy of a chemical reaction can be
become 75% inactivated is
determined by
(a) 35 min (b) 70 min
(c) 105 min (d) 17.5 min
(a) evaluating rate constant at standard
85. A Geigger Muller counter is used to study the
temperature
radioactive process. In the absence of radioactive
(b) evaluating velocities of reaction at two
substance A, it counts 3 disintegration per different temperatures
second (dps). At the start in the presence of A, it (c) evaluating rate constants at two different
records 23 dps; and after 10 min 13 dps, temperatures
(i) What does it count after 20 min (d) changing concentration of reactants
(ii) What is the half life of A? 93. For the exothermic reaction A + B ® C + D, DH
(a) 8 dps, 10 min (b) 5 dps, 10 min is the heat of reaction and Ea is the energy of
(c) 5 dps, 20 min (d) 5 dps, 5 min activation. The energy of activation for the
formation of A + B will be
86. The half life period for catalytic decomposition
(a) Ea (b) DH
of AB3 at 50 mm Hg is 4 hrs and at 100 mm Hg it (c) Ea + DH (d) DH – Ea
is 2 hrs. The order of reaction is 94. A reaction having equal energies of activation
(a) 1 (b) 3 (c) 2 (d) 0 for forward and reverse reaction has :
87. The rate equation for a reaction, (a) DG = 0
N2O ¾® N2 + 1/2O2 (b) DH = 0
is Rate = k[N2O]0 = k. If the initial concentration (c) DH = DG = DS = 0
of the reactant is a mol Lit–1, the half-life period (d) DS = 0
of the reaction is 95. The rate coefficient (k) for a particular reactions
a is 1.3 × 10–4 M–1 s–1 at 100°C, and 1.3 × 10–3 M–1 s–1
(a) t 1 = (b) - t 1 = ka at 150°C. What is the energy of activation (Ea)
2
2k 2
(in kJ) for this reaction? (R = molar gas constant
a k = 8.314 JK–1 mol–1)
(c) t1 = (d) t1 =
k 2
a
2 (a) 16 (b) 60 (c) 99 (d) 132
Chemical Kinetics 215
96. The reaction X ® Y is an exothermic reaction. initial value. If the rate constant for a first order
Activation energy of the reaction for X into Y is reaction is k, the t1/4 can be written as
150 kJ mol–1. Enthalpy of reaction is 135 kJ mol–1.
The activation energy for the reverse reaction,
Y ® X will be :
(a) 280 kJ mol–1(b) 285 kJ mol–1 (a) 0.75/k (b) 0.69/k (c) 0.29/k (d) 0.10/k
–1
(c) 270 kJ mol (d) 15 kJ mol–1 99. A ® B, DH = – 10kJ mol– 1, Ea(f) = 50 kJ mol– 1, then
97. The activation energy for a reaction which Ea of B ® A will be
doubles the rate when the temperature is raised (a) 40 kJ mol– 1 (b) 50kJ mol– 1
from 298 K to 308 K is (c) – 50kJ mol – 1 (d) 60 kJ mol– 1
100. For a certain reaction, rate = k × [H+]n. If pH of
(a) 59.2 kJ mol–1 (b) 39.2 kJ mol–1 reaction changes from two to one, the rate
(c) 52.9 kJ mol–1 (d) 29.5 kJ mol–1 becomes 100 times of its value at pH = 2, the
98. t1/4 can be taken as the time taken for the order of reaction is –
3 (a) 1 (b) 2 (c) 0 (d) 3
concentration of a reactant to drop to of its
4

Answer KEYs
1 (b) 11 (b) 21 (d) 31 (c) 41 (b) 51 (d) 61 (d) 71 (b) 81 (c) 91 (b)
2 (c) 12 (d) 22 (b) 32 (d) 42 (c) 52 (b) 62 (b) 72 (a) 82 (d) 92 (c)
3 (b) 13 (b) 23 (d) 33 (b) 43 (a) 53 (b) 63 (b) 73 (b) 83 (b) 93 (c)
4 (a) 14 (b) 24 (b) 34 (c) 44 (c) 54 (c) 64 (d) 74 (c) 84 (b) 94 (b)
5 (d) 15 (c) 25 (b) 35 (b) 45 (c) 55 (c) 65 (b) 75 (b) 85 (a) 95 (b)
6 (c) 16 (d) 26 (b) 36 (a) 46 (d) 56 (a) 66 (a) 76 (b) 86 (c) 96 (b)
7 (c) 17 (d) 27 (b) 37 (a) 47 (d) 57 (b) 67 (a) 77 (c) 87 (a) 97 (c)
8 (a) 18 (d) 28 (c) 38 (d) 48 (b) 58 (a) 68 (c) 78 (a) 88 (a) 98 (c)
9 (b) 19 (d) 29 (d) 39 (d) 49 (c) 59 (d) 69 (d) 79 (d) 89 (d) 99 (d)
10 (d) 20 (d) 30 (c) 40 (c) 50 (c) 60 (b) 70 (c) 80 (a) 90 (d) 100 (b)

1. (b) N2O5 (g) ¾¾

® 2 NO2(g) + 1/2 O2 (g) -d[ N 2 ] 1d[ H 2 ] 1 d ( NH 3 ]
=- =
d 1 d d
dt 3dt 2 dt
– [ N 2O5 ] = + [NO2 ] = 2 [O 2 ] 1
Rate of disappearance of N2 = the rate of
dt 2 dt dt
2
d formation of NH3 = 1 × 10–4
[ NO2 ] = 1.25 ´ 10 -2 mol L–1s–1 and Rate of disappearance of H2 =3/2 the rate of
dt
formation of NH3 = 3 × 10–4
d
[O2 ] = 3.125 ´ 10 -3 mol L–1s–1 3. (b) The curve Y shows the increase in
dt concentration of products with time.
2. (c) N2+3H2 2NH3; Rate is given by any of 3
4. (a) The order of reaction is and molecularity
the expressions 2
is 2.
 EBD_7587
216 Chemistry Objective MCQs
5. (d) Using the relation – Ea / RT
[A] = [A]0 (1/2)n [n = number of half-lives] 15. (c) k = Ae
T = n × t1/2 – Ea
7.5 \ = – 40000
Here, n = =3 R
2.5 \ Ea = 40000 × 2 = 80000 cal
3
æ 1ö 1
\ [A] = 160 ´ çè ÷ø = 160 ´ = 20g 16. (d) E a = 58
2 8 2

Ea1 20
6. (c) Rate1 = k [A]n [B]m; Rate2 = k [2A]n [½B]m
Ea2

Energy
Rate2 k[2A]n [½B]m A+B
\ =
Rate1 k[A]n [B]m 38

= [2]n [½]m = 2n.2–m C+D

= 2n–m Reaction Co-ordinate
7. (c) Let us consider a reaction, 1
17. (d) t1/2 µ
x X + y Y ¾¾ ® aA+bB a2 1
x
rate = [X] [Y] y We know that t1/2 µ n -1
a
It is given that order of reaction w.r.t. i.e. n = 3
rd
Thus reaction is of 3 order.
component Y is zero.
Hence, rate = [X]x 18. (d) A + 2B —® C
i.e., rate becomes independent of the 1 mole of A reacts with 2 moles of B to give
concentration of Y. 1 mol of C.
2.303 a \ 5 moles of A would react with 10 moles of
8. (a) t= log B to give 5 moles of C.
k a-x
But, only 8 moles of B are available
2.303 2.303
or t = log a - log(a - x) \ B acts as a limiting reagent.
k k 2 moles of B gives 1 mole of C
-k
log (a – x) = t + log a \ 8 moles of B will give 1/2 × 8 = 4 moles of C.
2.303 19. (d) The given reaction is : 2X + Y —® Z
9. (b) From slow reaction d[X] d[Z]
Rate = k [A] [B] – =
2dt dt
10. (d) Rate constant is independent of concentration. \ Rate of formation of Z is half of the rate
11. (b) Change in 67% to 33% is almost half the of disappearance of X.
concentration change.
20. (d) Rate of reaction = d[A]
Half dt
67% ¾¾¾¾ ®
change 33.5(»33%)
Given, [A]initial = 0.50 M
So time interval between the stages of its [A]final = 0.38 M
33% and 67% decay is same as t1/2 = 20 min. dt = 10 min = 600 sec
12. (d) It is a second order reaction, first order both d[A] = 0.12
w.r.t S2O82– and I–. 0.12
\ r = k[S2O82–][I–] Rate = = 2 ´ 10 –4 M s –1.
600
All other options are of first order reaction. 21. (d) A catalyst affects equally both forward and
13. (b) k = Ae– Ea/RT backward reactions, therefore it does not
If T ¾¾ ® ¥, k = A affect equilibrium constant of reaction.

E'a E'a 10 20 22. (b) In equation k = Ae - Ea / RT ; A = Frequency

14. (b) = = = T factor
T1 T2 300 2
k = velocity constant, R = gas constant and
\ T2 = 600 K = 327°C Ea = energy of activation
 Chemical Kinetics 217
23. (d) The definition of threshold energy. dSO2 dO
24. (b) When Ea = 0, rate constant is independent \- = -2 ´ 2
of temperature. dt dt
25. (b) k = Ae a
–E / RT = – 2 × 2.5 × 10–4
= – 5 × 10–4 mol L–1 s–1
lnk = ln A – Ea /RT
33. (b) Rate of disappearance of reactant = Rate of
For ln k Vs 1/T
appearance of products
ln A = intercept
– Ea/R = slope = –2 × 104 K 1 d [ N 2 O5 ] 1 d [ NO2 ]
\ Ea = 8.3 × 2 × 104 J mol–1 - =
2 dt 4 dt
= 16.6 × 104 J mol–1 or 166 kJ mol–1
26. (b) As the rate of reaction get doubled for every 1 1
k [ N 2 O 5 ] = k ¢ [ N 2 O5 ]
10°C rise in temperature. Hence the increase 2 4
in reaction rate as a result of temperature
rise from 10°C to 100°C is equal to = 29 = 512 k k¢
=
27. (b) Ea (f) – Ea (b) = DH = – 15 k cal 2 4
Þ Ea (f) = – 15 + 20 = 5 k cal \ k ¢ = 2k
Ea - Ea (catalyst) 34. (c) It is bimolecular first order reaction since
k (catalyst)
=e RT =e =e Rate µ [N2O5]
k 3 35. (b) r = k [O2][NO]2. When the volume is reduced
(5 -3) ´10
2.5 to 1/2, the conc. will double
=e =
= e 2 ´ 400 =e
\ New rate = k [2O2][2 NO]2 = 8 k [O2][NO]2
The new rate increases to eight times of its
28. (c) k = A .e - Ea /(RT) initial.
\ Effective overall energy of activation 36. (a) For a zero order reaction.
1 1 rate =k[A]º i.e., rate = k
Ea = Ea (2) - E a (3) + Ea (1) - Ea (5)
2 2 hence unit of k = M.sec–1
1 1 For a first order reaction.
= 60 - 50 + ´ 40 - ´ 10 = 25 kJ/mol
2 2 rate = k [A]
29. (d) Units of k indicate that reaction I is of k = M.sec–1/M = sec–1
second order and reaction II is first order. 37. (a) Plots of conc. [A] Vs time, t
For I reaction, t1/2 µ 1/a,
first t1/2 = 1 hr, second t1/2 = 2 hr Zero First
1hr 2 hr order order
[ A] = 1M ¾¾¾
® 0.5M ¾¾¾
® 0.25M
log [A]

[ A]
1hr 1hr 1hr
[B] = 1M ¾¾¾
® 0.5M ¾¾¾
®0.25M ¾¾¾
® 0.125M t t
1hr 1hr 1hr
= 1M ¾¾¾ ® 0.5M ¾¾¾ ® 0.25M ¾¾¾ ® 0.125M
First
[ A] 0.25M order
= =2 1 1
[ B] 0.125M [ A] Second [ A ]2 Third
30. (c) The slowest step determines the rate. order order
t t t
31. (c) Rate = k [A][B] = R
R' = k [A][2B] 1 d [ HI]
38. (d) rate of appearance of HI =
R k[A][B] k[A][B] 2 dt
= =
R ¢ k[A][2 B] 2k[A][B]
-d ëé H 2 ûù
Þ 2R = R' i.e., rate become doubles. rate of disappearance of H2 =
32. (d) From rate law dt

1 dSO2 dO 1 dSO3 -d [ I 2 ]
- =- 2 = rate of disappearance of I2 =
2 dt dt 2 dt dt
 EBD_7587
218 Chemistry Objective MCQs

-d [ H 2 ] d [I 2 ] 1 d [HI] 0.693 0.693

hence =- = 45. (c) t1/2 = = = 10 min
dt dt 2 dt k 0.0693
Reactant after 10 min = 5 mol
2d [H 2 ]2d [I2 ] d [HI] æ dx ö
or – =- =
dt dt dt Rate çè ÷ø = k[A] = 0.0693 × 5 mol min– 1
dt
39. (d) Since all have same concentration of 46. (d) For a zero order, rate of reaction does not
reactants, all would react at same time. change with time.
40. (c) Rate of reaction 47. (d) It is a second order reaction, first order both
Rate of disappearance / appearance w.r.t S2O82– and I–.
= \ r = k[S2O82–][I–]
Stoichiometric coefficient
All other options are of first order reaction.
d [A ] 1 d [B] 0.693 0.693
=- = = 2.6 ´10-2 48. (b) k1 (300 ) = ; k 2 (320 ) =
dt 2 dt 20 5
1 d [B] k 2 (320 )
=
Ea é1 1 ù
=- In ê - ú
2 dt k1 (300 ) R ë T1 T2 û
d [B] 2.303 RT1T2 k
5.2 ´10-2 = - Ea = log 2
dt ( T2 - T1 ) k1
41. (b) After every 30 minutes the amount is
2.303 ´ 8.314
1 = ´ 300 ´ 320 log 4
reduced to therefore t1/2 is 30 minutes. 20 ´ 1000
2
In 90 minutes the amount is reduced to = 55.14 kJ/mol
1 1 E
i.e. n . Here n = 3. True for 1st order - a 10 -16
8 2 49. (c) e RT = 3.8 ´ -
reaction. 100
0.693 Ea
42. (c) k= further - = ln 3.8 ´10 -18
45 RT
2.303 100 Ea = 100 kJ/mol
t= ´ 45 log
0.693 100 - 99.9 -E /RT
k 2 Ae a2 (E - E ) /RT
2.303 ´ 45 ´ 3 1 50. (c) = - E
= e a1 a2
= = 7 hours . k1 Ae a1 /RT
0.693 2
k Ea - Ea
2.303 a 2.303log 2 = 1
43. (a) t= log
k a-x k1 RT

2.303 0.5 (83.314 - 75) ´ 103

= log = 0.384 min . = =2
6 0.05 8.314 ´ 500
2
2.303 100 log k 2 = = 0.868
44. (c) t90% = log (I) 2.303
k 100 - 90 Taking Antilog
2.303 100 k2 = 7.38
t50% = log (II)
k 100 - 50 51. (d) In Arrhenius equation k = Ae - Ea / RT , Ea is
t log10 the energy of activation, which is required
Dividing 90% =
t50% log 2 by the colliding molecules to react resulting
in the formation of products.
\ t90% = 3.3t50% 52. (b) When the temperature is increased, energy
in form of heat is supplied which increases
Chemical Kinetics 219
the kinetic energy of the reacting molecules. 2.303 ´ 8.314 ´ 298 ´ 308
this will increase the number of collisions = log 2.73
and ultimately the rate of reaction will be 308 - 298
enhanced. Ea = 76.62 kJ mol–1 = 18.39 kcal mol–1
53. (b) According to Arrhenius equation,
59. (d) For the reaction
k = Ae. - Ea / RT
where k = rate constant 3A + 2B ¾¾ ® C+D
Rate of disappearance of A = Rate of
E
or ln k = lnA - a appearance of C reaction
RT
A = frequency factor 1 d [ A] d [C ]
= k [ A] [ B ]
n m
Ea = Energy of activation. =- =
3 dt dt
54. (c) k = 1.2 ´ 1014 e -25000 / RT sec -1 or
dMnO-4
14 25000 1 60. (b) Given - = 4.56 × 10–3 Ms–1
log k = log 1.2 × 10 - . dt
R T
From the reaction given,
logk 1 dMnO 4 – 4.56 ´ 10 -3
- = Ms -1
2 dt 2
1/T
Equation of straight line
1 d MnO4- 1 d I2
25000 - =
slope = - 2 dt 5 dt
R
55. (c) There are 5 tens hence (2)5 = 32.
5 dMnO-4 dI2
56. (a) Ea (F.R.) ¹ Ea (B.R.) Ea can be calculated. \ - =
2 dt dt
57. (b) According to Arrhenius equation, On substituting the given value
k = Ae–Ea/RT
\ when Ea = 0, k = A dI2 4.56 ´ 10-3 ´ 5
\ = = 1.14 × 10–2 M/s
Also ln k ns 1/T is a straight line with slope dt 2
= –Ea/R. 61. (d) For a zero order reaction
\ Statements (ii) and (v) are correct.
a-x
58. (a) Let the initial concentration (A) = 100 Rate constant = k =
Final concentration at 298 K = 100 – 10 = 90 t
Final concentration at 308 K = 100 – 25 = 75 a - 0.5
2 × 10–2 =
Substituting the values in the 1st order rate 25
reaction a – 0.5 = 0.5
2.303 100 a = 1.0 M
t= log ...(i) 62. (b) A ® B For a first order reaction
k298 90
Given a = 0.8 mol, (a – x) = 0.8 – 0.6 = 0.2
2.303 100
t= log ...(ii) 2.303 0.8
k308 75 k= log or k = 2.303 log 4
1 0.2
k308 again a = 0.9, a – x = 0.9 – 0.675 = 0.225
From (i) and (ii) = 2.73
k 208 2.303 0.9
k= log
Substituting the value in the following t 0.225
relation
2.303
2.303 R ´ T1 ´ T2 k 2.303log 4 = log 4
Ea = log 2 t
T2 - T1 k1 Hence t = 1 hour
 EBD_7587
220 Chemistry Objective MCQs
63. (b) For a first order reaction, A ® Products
4 ×10 – 6 4 ×10 – 6
r = =
r = k[A] or k = (4 ´ 10 – 4 )1/ 2 2 ´ 10 – 2
[A]
= 2 × 10– 4 mol1/2 L– 1/2 s– 1
1.5 ´ 10-2 68. (c) [A ]t = [A ] - kt = 1 - 0.001´10 ´ 60
Þk = = 3 × 10–2
0.5 = 0.4 M

Further, t1/ 2 =
0.693
=
0.693
= 23.1 [B]t = 0.001´ 10 ´ 60 = 0.6 M
k 3 ´ 10-2 69. (d) For the first order reaction for small finite
64. (d) Given [A] = 0.01 M change
Rate = 2.0 × 10–5 mol L–1 s–1
For a first order reaction
1 D[A] D[ A] /[ A]
k1 = Þ = 1.5% min– 1
Rate = k[A] [A] Dt Dt
= 0.015 min– 1
2.0 ´ 10 -5
k= = 2 × 10–3 0.693
[0.01] t1/2 = = 46.2 min » 46 min
0.693 0.015 min –1
t1/2 = -3 = 347 sec 70. (c) Order w.r.t. A = 1; order w.r.t. B = 1
2 ´ 10
65. (b) Given: 75% reaction gets completed in 32 1 d ( AB)
Rate = = k r [ A ][ B ] ;
min 2 dt
2.303 a
Thus, k = log 1
t (a - x ) ´ (2.5 ´104 ) = k r ( 0.1)( 0.1)
2
2.303 100 kr = 1.25 × 10–2
= log
32 (100 - 75) 1 0.2 1 0.2
71. (b) k1 = In ; k 2 = In ; k1 = k 2 .
2.303 5 0.1 10 0.05
= log 4 = 0.0433 min–1
32 72. (a) Exothermic because of activation energy
Now we can use this value of k to get the Eb > E¦
value of time required for 50% completion 73. (b) More activations energy, slow reaction rate.
of reaction
74. (c) A catalyst does not take part in chemical
2.303 a 2.303 100 reaction but it alters the rate of reaction.
t= log = log
k (a - x ) 0.0433 50 75. (b) Activation energy of reactant is less than
2.303 the energy of activation of products.
= log 2 = 16 min 76. (b) The presence of enzyme (catalyst) increases
0.0433
the speed of reaction by lowering the energy
66. (a) If we write rate of reaction in terms of
barrier, i.e., a new path is followed with lower
concentration of NH3 and H2,then
activation energy.
1 d[ NH 3 ] 1 d[H 2 ]
Rate of reaction = =- ET
2 dt 3 dt
E'T
d[ NH 3 ] 2 d[H 2 ]
So, =-
dt 3 dt Ea
Products
Energy

67. (a) Rate µ Ea

Concentration 1

= k Concentration
Rate Reactants + catalyst
k= 1/2
(Concentration) Progress of reaction
Chemical Kinetics 221
Here ET is the threshold energy. 82. (d) A B
Ea and Ea1 is energy of activation of reaction Initial concentration Rate of reaction
in absence an d presence of catalyst 2 × 10–3 M 2.40 × 10–4 Ms–1
respectively. 1 × 10–3 M 0.60 × 10–4 Ms–1
77. (c) The rate constant doubles for 10º C rise in rate of reaction
temperature. r = k[A]x
For 20º C rise, the rate constant will be 4 where x = order of reaction
times
hence
\ k1 = 4k2 or k2 = 0.25 k1
2.40 × 10–4 = k [2 × 10–3]x ......(i)
78. (a) e - Ea / RT = 10 -3 % = 10 -5 ; 0.60 × 10–4 = k [1 × 10–3]x ......(ii)
Ea = 2.303 × 2 × 300 × 5 cal On dividing eqn.(i) from eqn. (ii) we get
= 6.91 k cal mol–1 4 = (2)x
79. (d) The activation energy of reverse reaction \ x=2
will depend upon whether the forward i.e. order of reaction = 2
reaction is exothermic or endothermic. 83. (b) Plot given is for zero order reaction.
As DH = E a (forward reaction) – E a
0.693
(backward reaction) 84. (b) t1/2 = = 2100 s = 35 min
k
For exothermic reaction
t75% = 2t1/2 = 2 × 35 = 70 min
DH = –ve
85. (a) In the absence of A, 3 dps is zero error, hence
\ –DH = Ea(f) – Ea(b)
or Ea(f) = Ea(b) – DH Initial count = 23 – 3 = 20 dps
\ Ea(f) < Ea(b) After 10 min = 13 – 3 = 10 dps
for endothermic reaction After 20 min = 5 dps
DH = + ve (recorded = 5 + 3 = 8 dps)
\ DH = Ea(f) – Ea(b) or Ea(f) = DH + Ea(b) (50% fall in 10 min, T50 = 10 min)
\ Ea(f) > Ea(b). 1
86. (c) t1/ 2 µ where n is the order of
80. (a) Arrhenius equation is given by ( p ) n -1
k = Ae - Ea / RT reaction
Taking log on both sides, we get n -1 n -1
Ea
2 æ 50 ö 1 æ 1ö
log k = log A – =ç ÷ or =ç ÷
4 è 100 ø 2 è 2ø
2.303 RT
Arrhenius plot a graph between log k and n – 1 = 1; n = 2
1 - Ea 87. (a) For a zero order reaction
whose slope is .
T 2.303R a
t1/2 =
81. (c) Given t1/2 = 3 2k
Total time T = 12 0.693 0.693
12 88. (a) k= = = 1.44 ´ 10 -3 s-1
No. of half lives (n) = =4 t1/2 480
3
n
89. (d) ¾® 2AB ;
A 2 + B2 ¾
æ1ö N 4
ç ÷ = \ æç 1 ö÷ = 3 A2 ¾
¾® A + A ( Fast );
è2ø No è2ø N
3 1 A + B2 ¾
¾® AB + B (Slow )
=
N 16 Rate law = k[A][B 2 ] put value of [A] from
N = 48 g Ist reaction since A is intermediate
k[A 2 ] = A
 EBD_7587
222 Chemistry Objective MCQs

\ Rate law equation = K k[A 2 ][B2 ] Ea é 1 1 ù

1= -
1 3 2.303 ´ 8.314 ë 373 423 úû
ê
\ Order = +1 =
2 2 Ea = 60 kJ / mol
90. (d) t1/2 = 5 years, T = 15 years hence total num- 96. (b) X ¾® Y ; DH = –135 kJ/mol,
Ea = 150 kJ/mol
ber of half life periods = 15 = 3
5 For an exothermic reaction
64 Ea(F.R.) = DH + E¢a(B.R.)
\ Amount left = = 8g 150 = – 135 + E¢a(B.R.)
( 2) 3
E¢a(B.R.) = 285 kJ/mol
91. (b) DH = Ea (f) - Ea (b) 97. (c) Activation energy can be calculated from
Thus energy of activation for reverse the equation.
reaction depend upon whether reaction is
log K 2 – Ea æ 1 1 ö
exothermic or endothermic. = ç – ÷
log K1 2.303 R è T2 T1 ø
If reaction is exothermic,
DH = - ve , Ea (b) > Ea (f)
log K 2
If reaction is endothermic, Given = 2 T2 = 308 K; T1 = 298 K
log K1
DH = + ve Ea (b) < Ea (f)
92. (c) We know that the activation energy of – Ea æ 1 1 ö
\ log 2 = ç – ÷
chemical reaction is given by formula 2.303 ´ 8.314 è 308 298 ø

Ea é T2 - T1 ù Ea = 52.9 kJ mol–1
k2
= = ê ú , where k1 is the
k1 2.303R ë T1T2 û 2.303 1 2.303 4
98. (c) t1/ 4 = log = log
k 3/ 4 k 3
rate constant at temperature T1 and k2 is
the rate constant at temperature T2 and Ea 2.303 2.303
is the activation energy. Therefore activation = (log 4 - log 3) = (2log 2 - log 3)
k k
energy of chemical reaction is determined
2.303 0.29
by evaluating rate constant at two different = (2 ´ 0.301 - 0.4771) =
k k
temperatures.
93. (c) For the exothermic reaction the energy of 99. ® B, DH = – 10 kJ mol– 1
(d) A ¾¾
products is always less than the reactants.
It is an exothermic reaction.
If E a is the energy of activation for the
Ea(b) = Ea(f) – (DH)
forward reaction, the energy of activation
for backward reaction is Ea + DH = 50 – (– 10) = 60 kJ
100. (b) pH = 2 ; r1 = k × (10–2)n {Q [H+] = 10–pH}
94. (b) DH = Ea (f ) - Ea (b) = 0
pH = 1 ; r2 = k × (10–1)n
95. (b) According to Arrhenius equation
Given r2 = 100 r1
k2 Ea æ 1 1 ö
log = - æ 10-1 ö
n
k1 2.303R çè T1 T2 ÷ø Þ ç ÷ = 100
ç 10-2 ÷
è ø
1.3 ´ 10 -3 Ea é 1 1 ù 10n = 100
log = -
1.3 ´ 10 -4 2.303 ´ 8.314 êë 373 423 úû \ n=2