Solved Problems (Subjective) st
Problem 1:
A (half life = 12.8 h) transforms transforms to B (35%), C (19%) and D (43%) each by 1 order kinetics as shown below.
Solution:
Let the rate constants of the above processes be k 1, k 2 and k 3, respectively and the overall rate constant be k. Then
Also, k 1 = 0.38 k =
Similarly, = 67.36h
Where t1, t2 and t3 are the partial half-lives for the conversion of A and B, C and D, st respectively, by 1 order kinetics. Problem 2:
212
The half-life of Pb is 10.6 h. It undergoes decay to its daughter (radioactive) 212 element Bi of half-life 60.5 minute. Calculate the times at which daughter element will have the maximum activity?
Solution:
= 227.1 min
Problem 3:
Solution:
–6
t1/2 of a first order reaction at 27°C is 100 min. 10 % of the reactant molecules exists in the activated state at 27°C calculate ca lculate maximum rate constant of the reaction.
= –Ea/RT
k = Ae A=
= 6.93
When Problem 4:
Solution:
107 min –1
Thus A is the maximum rate constant of the reaction. 32
The half life of P is 14.3 days. Calculate the specific activity of a phosphorus 32 containing specimen having1.0 ppm P (Atomic weight of P = 31) 6 32 Activity of 1g sample is the specific activity of the sample 10 g contains 1g P –6
1g sample will contain 10 P
32
32
i.e.
atoms of P
= 10
= 1.089 10 dps Since this is activity i.e. rate of decay of 1g sample Specific activity = 1.089 1010 dps per gram = Problem 5:
Solution:
= 0.295 CI per g
The gas phase decomposition of dimethyl ether follows first order kinetics. CH3OCH3(g) CH4(g_ + + H2(g) + CO(g) The reaction is carried out in a constant volume container at 500°C. Initially only dimethyl ether is present at a pressure of 0.40 atm. The total pressure of the gaseous reaction mixture after 12 minute is 0.75 atm. a tm. Calculate half-life of the reaction. CH3OCH3(g) CH4(g_ + H2(g) + CO(g) P0 0 0 0 …t = 0 P0 – x x x x …t = 12 minute From question P0 = 0.40 atm Total pressure after 12 minute = P0 = x + x + x + x = 0.75 atm or P0 + 2x = 0.75 x = 0.175 atm At constant volume mole pressure So, = -1
= 0.0475 min
Problem 6:
Solution:
Problem 7:
–1
–1
The rate of a firs order reaction is 0.8 mol L min after 10 min and 0.1 after 30 min from the start of the reaction. Find half time o f the reaction. n Rate C , where n is the order of reaction For a first order reaction: N = 1 so
14
The half-life period of C is 5760year. An ancient piece of wood has -activity of 4.4 atm per gram carbon. A living tree has a -activity of 17.6 dpm per gram carbon. A dead coal when examined in the same Gniger-Muller instrument, it was found to have an activity of 1.1 compounds per gram carbon. Calculate how long before the tree was cut.
Solution:
= 13380.13 years 238
235
Problem 8:
The isotpe U - and U occur in nature in the ratio of 140 : 1. Assuming that at the time of earth formation, they were present in equal ratio, make an estimation of the 238 235 9 8 age of the earth. The half life period of U and U as 4.5 10 and 7.13 10 year respectively.
Solution:
In nature At the time of earth formation
If
and
238
235
be the decay constants of U and U
respectively, then
Putting the value of 9 T = 6.04 10 year
Problem 9:
and
, we get
Rate law for the following reaction; is Ester + H+ Acid + Alcohol; is [ester]1 [H+]0 What would be the effect on the rate if i) Concentration of ester is doubled ? ii) concentration of H+ ion is doubled ?
Solution:
The rate law expression in this question, suggests that concentration of acid is nothing to play with velocity. i) When concentration of ester is doubled; velocity of the reaction will become double. + ii) When concentration of H ion is doubled velocity will be unaffected.
D + 2E; is found to be first order in A; second order in B Problem 10: The reaction 2A + B + C and zero order in C. i) Give the rate law for the above reaction in the form of a differential equation. ii) What is the effect on the rate of increasing the concentration of A, B and C two times? Solution:
i) The rate law according to given information may be given as,
ii) When concentration of A, B and C are doubled then rate will be
i.e., rate becomes 8 fold, the original rate.
Problem 11: At 27°C it was observed, during a reaction of hydrogenation that the pressure of H2 gas decreases from 2 atm to 1.1 atm in 75 min. Calculate the rate of reaction –1 –1 (molarity/sec). Given (R = 0.082/litre atom K mole ) Solution:
Rate = atm/min Rate in atm/sec
atm/sec.
Answer is required in molarity per second PV = nRT
sec = molarity per second
. Rate in molarity/sec = (n/V)/sec Problem 12: In presence of an acid N-chloro acetanilide changes slowly into p-chloro acetanilide. Former substance liberated iodine from KI and not the later and hence progress of reaction can be measured by titrating iodine liberated with Na2S2O3 solution, the results obtained were as follows : Time (hours) 0 1 2 4 68 (a–x)i.e., hypo 45 32 22.5 11.3 3.7 2.9
show that reaction is uniomlecular and find out the fraction of N-chloroacetanilide decomposed after three hours. Solution:
The present reaction is —
Let us apply the kinetics of first order reaction. After 1 hours
–1
= 0.34098 hour
After 2 hours
–1 = 0.34660 hour
After 4 hours
= 0.3455 hour
–1
–1
After 6 hours
= 0.4164 hour
After 8 hours
= 0.3428 hour
–1
–1
Average value of constant = 0.3584 hours
Since, on applying first order kinetics, we get almost same values of rate constant after different time intervals, hence, the reaction is of first order. Let us see the fraction decomposed after 3 hours.
0.3584 =
[fraction decomposed] –E/RT
Problem 13: For the decomposition of dimethyl ether, A in the Arrhenious equation K = Ae has a value of 1.26 1013 and Ea value of 58.5 kcal. Calculate half life period for first order decomposition at 527°C. Solution:
Taking logarithm of Arrheniuos equation –E/RT K = Ae we get log K = log A –
… (i) 13
Given A = 1.26 10 E = 58.5 kcal T = 527 + 273 = 800 K Substituting these value in Eq. (i), we get log K = log (1.26
1013) –
= 13.1003 – 15.9799 = –2.8796 –3 –1 K = 1.3194 10 sec
sec = 525 sec Problem 14: For the reaction — 2NO(g) + H2(g) N2O(g) + H2O(g) at 900 K, the following data are obtained : Initial pressure Initial pressure Rate of NO (atm) of H2(atm) (atm min –1) 0.150 0.400 0.020 0.075 0.400 0.005 0.150 0.200 0.010 Find the rate law and the value of rate constant. Solution:
Let order with respect to NO(g) is `m’ and order with respect to H2(g) is `n’ Then, Rate = … (1) m
n
0.020 = K [0.15] [0.40] m n 0.005 = K [0.075] [0.40] m n 0.010 = K [0.15] [0.2] Dividing Eq. (2) by (3), we get
… (2) … (3) … (4)
m=2 Dividing Eq. (2) by (4), we get
n=1 Rate = Substituting the values of m, n in Eq. (2), we get 0.020 = K (0.15)2 [0.40] –2 –1 K = 2.22 atm min Problem 15: 10 gram atoms of an -active radio isotope are disintegrating in a sealed container. 3 In one hour the helium gas collected at STP is 11.2 cm . Calculate the half-life of the radio-isotope. Solution:
No. of atoms of helium is 11.2 cc at NTP =
6.02
23
10 = 3.01
20
10 atoms
Since, helium atom corresponds to -particle. Thus, Rate of disintegration = 3.01 1020 per hour. We know, Rate = Rate constant concentration in atom 20 23 3.01 10 = K 10 6.02 10 –3 –1 K = 0.05 10 hour = 13860 hours
years A
A–4
Problem 16: A carbon radio isotope ZX (half life 10 days) decays to give Z–2Y . If 1.00 gm atom of ZXA is kept in a sealed tube, how much helium will accumulate in 20 days ? 3 Express the result in cm at STP. Solution:
Initial concentration (N0) of radio-isotope is 1 gm atom. Concentration remained after 20 days may be calculated as Nt = N0 where n = n0 of half lives = 20/10 = 2 = Concentration decayed to
-particles = 1 –
gm atom. An
-particle takes 2
electron from air and from helium gas. Thus, Helium formed
gm atom
cc
cc
Problem 17: Radioactive decay is a first order process. Radioactive carbon wood sample decays with a half life of 5770 years. What is the rate constant in (years) –1 for the decay? What fraction would remain after 11540 years? Solution:
10 –4 year –1
= 1.201
–4
1.201
10
4.002 = (Remaining fraction) = 131 Problem 18: A sample of 53I , as iodide ion, was administered to a patient in a carrier consisting of 0.10 mg of stable iodide ion. After 4.00 days, 67.7% of the initial radioactivity was detected in the thyroid gland of the patient. What mass of the stable iodide ion had 131 migrated to the thyroid gland? Given T1/2 I = 8 days.
Solution:
We know,
70.7% of initial activity is present. Given that 67.7% activity is migrated to thyroid gland. Thus, weight of I – migrated to thyroid gland may be calculated as 100 = 95.75% i.e, 0.1
= 0.09575 mg
Solved Problems (Objective) Problem 1:
The rate constant of a reaction: 3P, is for the rate of consumption of A at 27°C. The initial rate of formation of P when concentration of A is 1.5 2/3 M,is: –3 –1 –1 –3 –1 –1 (a) 1.67 10 mol L min (b) 2.5 10 mole L min –3 —1 –1 (c) 2.5 10 mole L min (d) None of these
Solution:
The unit of K indicates that it is a 1st order reaction
= (b) Problem 2:
A catalyst lowers Ea of a reaction by 25%. The temperature at which rate of uncatalysed reaction will be equal to that of the catalysed reaction at 27°C under the same condition of concentration of reactant is equal to (a) 127°C (b) –48°C (c) 27°C (d) None of these
Solution:
, cat Let k at T1 is equal to k at T2, then
(a) Problem 3:
If a reaction A + B ! C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy 70 kJ/mol, the activation energy for the reverse reaction is (a) 30 kJ/mol (c) 70 kJ/mol
(b) 40kJ/mol (d) 100 kJ/mol
Solution:
By seeing the curve, activation energy for backward reaction = 100 kJ (d)
Problem 4:
st
The progress of a 1 order reaction: 3B carried out at 27°C is shown below graphically. Half time of the reaction at 27° is nearly (a) 20 min (b) 40 min (c) 49 min (d) None of these
Solution: a 0 t=0 a–x 3x t=t From curve: a – x – 3x when t = 20 min a = 4x
Upon solving: t1/2 – 48.94 min (c)
49 min
Problem 5:
When a reaction is carried out taking 4M concentration of the reactant, the half-time of the reaction is 20 min. When the same reaction is carried out taking 1 M concentration of the reactant, in concentration of the reactant is reduced to 0.25 M in 4 hours. Hence order of reaction is (a) one (b) two (c) half (d) zero
Solution:
For a second order reaction:
(b) Problem 6:
A catalyst lowers the activation energy of a reaction from 20 kJ mole –1 to –1 10 kJ mole . The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27 C is (a) –123 C (b) 327 C (c) 327 C (d) + 23 C °
°
°
°
°
Solution: °
T2 = 600 K = 327 C (b)
"
Problem 7:
Solution:
Another first order reaction B is 75% complete in 40 min at the same temperature. If these two reactions are carried out simultaneously at 27°C taking equal initial concentration of the reactants A and B in each reaction, the ratio of concentration of to that of B after 60 minute will be (a) 1:2 (b) 1:1 (c) 1:3 (d) 2:1 For the reaction:
For the reaction:
(d) Problem 8:
–6
In a first order reaction 10 % of the reactant molecule are able to cross-over the potential energy barrier at 27°C. The Ea (kJmol –1) of the reaction will be nearly (a) 100 (b) 45 (c) 50 (d) 20
Solution:
= fraction of the reactant molecules crossing over the barrier at temperature T So,
(b) Problem 9:
Solution:
The incorrect statement among the following is: (a) Order of reaction is an experimental property (b) Order of reaction is a fixed property i.e. independent of experimental conditions of the reaction. (c) Moleculartiy is a theoretical property and it concerns with mechanism. (d) Order of reaction may be any number integral, fractional and zero. The reaction
nd
is a first order at low gas and 2 order at high gas pressure. (b) Problem 10: The conc. vs time at curve of a reaction is an given below. The order of reaction is (a) zero (b) one (c) two (d) half
Solution:
The curve shows that
Thus,
i.e. order = 2 (c)
Problem 11: The reaction A(g) + 2B(g) C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is (a) 1/48 (b) 1/24 (c) 9/16 (d)1/6 Solution:
After reaction A + 2B 0.6 – 0.2 0.8 – 0.4 0.4 0.4
C + 0.2 0.2
D 0.2 0.2
(d) Problem 12: Thermal decomposition of a compound is of first order. If 50% of a sample of the compound is decomposed in 120 minute show long will it take for 90% of the compound to decompose? (a) 399 min (b) 410 min (c) 250 min (d) 120 min Solution:
…(1)
=
…(2)
Equating (1) and (2)
t = 399 minutes (a) Problem 13: If concentration are measured in mole/litre and time in minutes, the unit for the rate rd constant of a 3 order reaction are –1 –1 2 –2 –1 (a) mol lit min (b) lit mol min
–1
–1
–1
(c) lit mol min Solution:
1–n
(d) min –1
K = [conc.] min rd For 3 order reaction 1–3 –1 = [mole/litre] min 2 –2 –1 = lit .mole min (b)
Problem 14: What is the activation energy for the decomposition of N2O5 as
N2O5
2NO2 +
O2 –5
If the values of the rate constants are 3.45 # 10 and 6.9 respectively. (a) 102 # 102 kJ (b) 488.5 kJ (c) 112 kJ (d) 112.5 kJ
#
–3
10
at 27°C and 67°C
Solution:
Ea = 112.5 kJ (d) 2AB are as given below: Problem 15: The kinetic data for the reaction: 2A + B2 [A] [B2] Rate –1 –1 –1 –1 mol L molL molL min –3 0.5 1.0 2.5 10 –3 1.0 1.0 5.1 10 0.5 2.0 1 10 –2 Hence the order of reaction w.r.t A and B2 an, respectively (a) 1 and 2 (b) 2 and 1 (c) 1 and 1 (d) 2 and 2 Solution: Data suggest that when the conc. of A is doubled keeping the conc. of B2 constant, the rate is just doubled and when conc. of B2 is doubled keeping the conc. of A constant, the rate increases 4 times i.e. Rate [A], when [B2] is constant and 2 Rate [B2] , when [A] is constant Hence order w.r.t. A and B2 are 1 and 2, respectively
Problem 16: For the first order reaction: A B, the conc. vs time curve is as given below: Hence, the rate of constant of the reaction is as follows: –1 (a) 0.0231 min –1 (b) 0.033 min –1 (c) 0.0693 min (d) None of these
Solution:
When t = 30 min, CA = CB, so t1/2 = 80 minutes –1
= 0.0231 min (a)
238
Problem 17: Lead remains always present in uranium ore due to radio active decay of U (t1/2 = 9 238 10 years) to lead – 2906 (the enol product). In an uranium mineral, U to 206 Pb mole rate was found to be 1:3. Hence age of the mineral is: 9 9 (a) 2.25 10 years (b) 1.5 10 years 9 9 (c) 9.0 10 years (d) 1.25 10 years
Solution:
238
206
U : Pb :: 1:3 238 th Massof U remains (1/4 ) of original compound N =
n=2
age T =
(c) 235 Problem 18: During the fission of U , energy of the order of 180 MeV is generated per nucleus 235 fissioned. The amount of energy released by the fission of 0.235g of U is: 7 25 (a) 1.73 10 kJ (b) 1.08 10 kJ 16 7 (c) 1.73 10 kJ (d) 1.08 10 kJ
Solution:
235
No. of nuclei of U
undergoing fission = =
Energy released = = (a)
=
MCQ for practice Level 1 1.
2.
3.
4.
st
-
For the 1 order reaction A(g) 2B(g) + C(s) t1/2 = 24 minute the reaction is carried out taking certain mass of A enclosed in a vessel in which it exerts a pressure of 400 mm Hg. The pressure of the reaction mixture after exp iry of 48 minutes will be (a) 700mm (b) 600mm (c) 800mm (d) 1000mm The ratio of the rate constant of a reaction at any temperature T to the rate constant is equal to (a) Energy of activation of the reaction (b) Fraction of molecules in the activated state (c) Average life of the reaction (d) Pre-exponential factor in the Arrhenius equation The rate constant of a reaction: at 27°C is at 27°C and at this temperature per cent of the reactant molecules are able to cross-over the P.E. barrier. The maximum rate constant of the reaction is: (a) (b) –1 –1 (c) 0.2 s (d) 20s For a first order reaction: A B, whose concentration vs. time curve is as shown in the figure. The rate constant is equal to (a) 41.58 h –1 –1 (b) 4.158 s (c) –1 (d) 6.93 min
5.
The rate of reaction: is doubled when concentration of A is increased 4-fold. If half-time of the reaction with 0.2 M concentration of A is 25 minutes then that for 0.8 M concentration of A the half time of the reaction will be (a) 12.50 min (b) 50 min (c) 5 min (d) 625 min
6.
The incorrect statement amongst the following about order of reaction is: (a) It is an experimental property (b) It concerned with kinetics (c) It may be any number including zero (d) It is a fixed property which does not change in any case.
7.
8.
If of a reaction be positive and k 1 and k 2 be the rate constants of forward reaction and backward reaction, respectively, at temperature t°C and be the respective rate constants at (t + 10)°C then (a)
(b)
(c)
(d) None of these
For which of the following reactions, the average life will be the minimum? (a) (b) (c) (d) PCl5
9.
PCl3 + Cl 2
A decomposes following two parallel first order mechanisms
the rate of disappearance of A i.e. (a) (c) 10.
taking 1 M concentration of A is equal to (b) (d) None of these
Among equilibrium constant (K), rate constant and Ea the one or more than one that is (are) not affected by catalyst is (one) (a) Rate constant and (b) K and Ea (c) and K (d) Only K
Answers
1.
a
2.
b
3.
d
4.
c
5.
b
6.
d
7.
c
8.
b
9.
a
10.
c
Level 2 1.
The decomposition of nitrogen pentoxide can be represented as N2O5(g)
2NO2(g) +
O2(g)
The rate of the reaction can be expressed as (a) (b) (c) (d) 2.
The instantaneous rate of disappearance of the Mn –3
ion in the following reaction is 4.56
–1
10 Ms 2Mn
–
+
+ 10I + 16H
2+
2Mn + 5I2 + 8H2O
The rate of appearance of I2 is (a) 1.14 10 –3 M s –1 (c) 4.56 10 –4 M s –1
(b) 5.7 10 –3 M s –1 (d) 1.14 10 –2 M s –1
3.
The rate constant for the reaction 2N2O5 $ ! 4NO2 + O2 –5 –1 –5 –1 –1 is 3 " 10 s . If the rate at a given time is 2.40 " 10 mol L s , then concentration of N2O5 at that time is: (a) 1.4 (b) 1.2 (c) 0.04 (d) 0.8
4.
The rate expression for the reaction A(g) + B(g)
C(g) is rate =
. What changes
in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight ? (a) CA 2; CB 2 (b) CA 2; CB 4 (c) CA 1; CB 4 (d) CA 4; CB 1 5.
Calculate the half-life of the first-order reaction C2H4O(g)
CH4(g) + CO(g)
if the initial pressure of C2H4O(g) is 80 mm and the total pressure at the end of 20 minutes is 120 mm. (a) 40 min (b) 120 min (c) 20 min (d) 80 min 6.
7.
Rate constant of a first order reaction is 0.0693 min –1. If we start with 20 mol L –1, it is –1 reduced to 2.5 mol L in: (a) 10 min (b) 20 min (c) 30 min (d) 40 min The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/litre and concentration at some instant is 0.01 mol/litre, then t = (a) 10 minutes (b) 30 minutes (c) 20 minutes (d) 40 minutes
8.
In the Arrhenius equation k = A exp (–EA/RT), the rate constant (a) Decreases with increasing activation energy and increases with temperature (b) Increases with decrease of activation energy and increase of temperature (c) Decreases with decrease of activation energy and increase of temperature (d) Increases with decrease of activation energy and decreasing temperature
9.
Rate of formation of SO3 in the following reaction 2SO2 + O2 $ ! 2SO3 is 100 g min –1. Hence, rate of disappearance of O2 is: –1 –1 (a) 50 g min (b) 100 g min –1 –1 (c) 200 g min (d) 40g min
10.
For a reaction A(g) + 2B(g) $ ! C(g) + D(g)
Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, relative to the initial value is: (a)
(b)
(c)
(d)
11.
In the first order reaction the concentration of reactant decreases from 2 M to 0.50 M in 20 minutes. The value of specific rate is (a) 69.32 (b) 6.932 (c) 0.6932 (d) 0.06932
12.
The rate of a gaseous reaction is given by the expression K[A] [B] . If the volume of reaction vessel is suddenly reduced to one half of the initial volume. The reaction rate relative to the original rate will be (a) 1/24 (b) 1/32 (c) 32 (d) 24
2
3
13.
The half life period t1/2 is independent of initial concentration of reactant when the order of reaction is (a) Negative (b) 0 (c) 1 (d) Fractional
14.
A reaction X2 + Y2 $ ! 2XY occurs by the following mechanism X2 X + Y2
15.
X+X
… (slow)
XY + Y
… (fast)
X+Y XY The order of the overall reactions (a) 2
(b) 1
(c) 1
(d) zero
The graph between log K and
… (fast)
–1
[K is rate constant (s ) and T the
temp. (K)] is a straight line with OX = 5 and
–1
= tan
.
Hence Ea will be
(a) 2.303
2 cal
(b)
(c) 2 cal
(d) None of these
16.
For a chemical reaction A B, the rate of reaction increases by a factor of 1.837 when the concentration of A is increased by 1.5 times. The order of reaction with respect to A is (a) 1 (b) 1.5 (c) 2 (d) –1
17.
Rate of the chemical reaction : nA products, is doubled when the concentration of A is increased four times. If the half time of the reaction at any temperature is 16 min. then time required for 75% of the reaction to complete is (a) 24.0 min. (b) 27.3 min. (c) 48 min. (d) 49.4 min.
18.
The mechanism of the reaction A + 2B + C D is Step – 1 A + B X (Fast) equilibrium Step-2
X+C
Y (Slow)
Step-3 Y+B D (Fast) which rate law is correct (a) R = K[C] (c) R = K[A][B][C]
2
(b) R = K[A][B ][C] (d) R = K[D]
19.
For a first order reaction the ratio of t0.75 to t0.50 would be (a) 4 : 3 (b) 3 : 2 (c) 2 : 1 (d) 1 : 2
20.
Two substances A(t1/2 = 5 min.) and B(t1/2 = 15 min.) are taken in such away that initially [A]0 = 4[B]0. The time after which both the concentrations will be equal is (a) 5 min. (b) 15 min. (c) 20 min. (d) Concentrations can never be equal
Answers 1.
(d)
2.
(d)
3.
(d)
4.
(b)
5.
(c)
6.
(c)
7.
(b)
8.
(b)
9.
(d)
10.
(a)
11.
(d)
12.
(c)
13.
(c)
14.
(b)
15.
(c)
16.
(b)
17.
(b)
18.
(c)
19.
(c)
20.
(b)
