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= 6.67 × 10
−6 M s−1
Question
2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L
−1
to 0.4 mol L
−1
in 10 minutes. Calculate the rate during this interval?
Solution :
= 0.005 mol L
−1 min−1
= 5 × 10
−3
M min
−1
Question
3. For a reaction, A + B → Product; the rate law is given by, r = k [A]
½
[B]
2
. What is the order of the reaction?
Solution :
The order of the reaction r = k [A]
½
[B]
2
= 1 / 2 + 2
= 2.5
= 444.38 s
= 444 s (approx)
= 52897.78 J mol
−1
= 52.9 kJ mol
−1
Question
9. The activation energy for the reaction 2HI
(
g
)
→ H
2
+ I
2(
g
)
is 209.5 kJ mol
−1
at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution :
In the given case:
E
a
= 209.5 kJ mol
−1
= 209500 J mo
l−1
T = 581 K
R = 8.314 JK
−1
mol
−1
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x = e
−Ea / RT
⇒Inx= −Ea / RT
⇒logx=−Ea / 2.303RT
⇒logx= −209500Jmol
−1
/ 2.303 × 8.314JK
−1
mol
−1
×581
=−18.8323
Now,x= Antilog (−18.8323)
=1.471×10
−19
Question
10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N
2
O
(g) Rate = k[NO]
2
(ii) H
2
O
2
(aq) + 3 I
−
(aq)
+ 2 H
+
→ 2 H
2
O (l) + I
3
-
Rate = k[H
2
O
2
][I
−
]
(iii) CH
3
CHO(g) → CH
4
(g) + CO(g) Rate = k [CH
3
CHO]
3/2
(iv) C
2
H
5
Cl(g) → C
2
H
4
(g) + HCl(g) Rate = k [C
2
H
5
Cl]
Solution :
(i) Given rate = k [NO]
2
Therefore, order of the reaction = 2
Dimension of k = Rate / [NO]
2
= mol L
-1
s
-1
/ (mol L-1)
2
= mol L
-1
s
-1
/ mol
2
L
-2
= L mol
-1
s
-1
(ii) Given rate = k [H
2
O
2
] [I
−
]
Therefore, order of the reaction = 2
Dimension of
k = Rate / [H
2
O
2
][I
-
]
= mol L
-1
s
-1
/ (mol L
-1
) (mol L
-1
)
= L mol
-1
s
-1
(iii) Given rate = k [CH
3
CHO]
3/2
Therefore, order of reaction = 3 / 2
Dimension of k = Rate / [CH3CHO]
3/2
= mol L
-1
s
-1
/ (mol L
-1
)
3/2
= mol L
-1
s
-1
/ mol
3/2
L
-3/2
= L½ mol-½ s-1
(iv) Given rate = k [C
2
H
5
Cl]
Therefore, order of the reaction = 1
Dimension of k = Rate / [C
2
H
5
Cl]
= mol L
-1
s
-1
/ mol L
-1
= s
-1
Question
11. For the reaction:
2A + B → A
2
B
the rate = k[A][B]
2
with k = 2.0 × 10
−6
mol
−2
L
2
s
−1
. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L
−1
. Calculate the rate of reaction after [A] is reduced to 0.06 mol L
−1
.
Solution :
The initial rate of the reaction is
Rate = k [A][B]
2
= (2.0 × 10
−6 mol−2
L
2
s
−1
) (0.1 mol L
−1
) (0.2 mol L
−1
)
2
= 8.0 × 10
−9
mol
−2
L
2
s
−1
When [A] is reduced from 0.1 mol L
−1
to 0.06 mol
−1
, the concentration of A reacted = (0.1 − 0.06) mol L
−1
= 0.04 mol L
−1
Therefore, concentration of B reacted 1/2 x 0.04 mol L
-1
= 0.02 mol L
−1
Then, concentration of B available, [B] = (0.2 − 0.02) mol L
−1
= 0.18 mol L
−1
After [A] is reduced to 0.06 mol L
−1
, the rate of the reaction is given by,
Rate = k [A][B]
2
= (2.0 × 10
−6
mol
−2
L
2
s
−1
) (0.06 mol L
−1
) (0.18 mol L
−1
)
2
= 3.89 mol L
−1
s
−1
Question
12. The decomposition of NH
3
on platinum surface is zero order reaction. What are the rates of production of N
2
and H
2
if k = 2.5 × 10
−4
mol
−1
L s
−1
?
Solution :
The decomposition of NH
3
on platinum surface is represented by the following equation.
= 7.5 × 10
−4
mol L
−1
s
−1
Question
13. The decomposition of dimethyl ether leads to the formation of CH
4
, H
2
and CO and the reaction rate is given by
Rate = k [CH
3
OCH
3
]
3/2
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
rate = k (P
CH3OCH3
)
3/2
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Solution :
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min
−1
Rate = k [CH
3
OCH
3
]
3/2
⇒ k =
Rate / [CH
3
OCH
3
]
3/2
Therefore, unit of rate constants(k) = bar min
−1 /
bar
3/2
= bar
-½
min
-1
Question
14. Mention the factors that affect the rate of a chemical reaction.
Solution :
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
Question 15. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled (ii) reduced to half?
Solution :
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]
2
= ka
2
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)
2
= 4ka
2
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a , then the rate of the reaction would be
R = k(1/2a)
2
= 1/4 Ka
2
= 1/4 R
Therefore, the rate of the reaction would be reduced to
Question 16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?
Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
k=Ae
−Ea/RT
Where,
k = rate constant,
A = Frequency factor / Arrhenius factor,
R = gas constant
T = temperature
Ea = activation energy for the reaction.
Question
17. In a pseudo first order hydrolysis of ester in water, the following results were obtained:
Question
19. In a reaction between A and B, the initial rate of reaction (r
0
) was measured for different initial concentrations of A and B as given below:
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
Question
20. The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Question
21. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Question
24. The experimental data for decomposition of N
2
O
5
[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:
(ii) Time corresponding to the concentration, 1630x10
2
/ 2 mol L
-1
= 81.5 mol L
-1
is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
(iv) The given reaction is of the first order as the plot, Log[N
2
O
5
] v/s t, is a straight line. Therefore, the rate law of the reaction is
Rate = k [N
2
O
5
]
(v) From the plot, Log[N
2
O
5
] v/s t, we obtain
- k /2.303
Again, slope of the line of the plot Log[N
2
O
5
] v/s t is given by
- k / 2.303. = -0.67 / 3200
Therefore, we obtain,
- k / 2.303 = - 0.67 / 3200
⇒ k = 4.82 x 10-4 s-1
(vi) Half-life is given by,
t
½
= 0.693 / k
= 0.639 / 4.82x10
-4
s
=1.438 x 103
This value, 1438 s, is very close to the value that was obtained from the graph.
Question 25. The rate constant for a first order reaction is 60 s
−1
. How much time will it take to reduce the initial concentration of the reactant to its 1/16
th
value?
Solution :
It is known that,
Question
26. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Solution :
Therefore, 0.7814 μg of 90
Sr
will remain after 10 years.
Again,
Therefore, 0.2278 μg of 90
Sr
will remain after 60 years.
Question
27. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution :
For a first order reaction, the time required for 99% completion is
t
1
= 2.303/k Log 100/100-99
= 2.303/k Log 100
= 2x 2.303/k
For a first order reaction, the time required for 90% completion is
t
2
= 2.303/k Log 100 / 100-90
= 2.303/k Log 10
= 2.303/k
Therefore, t
1
= 2t
2
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.
Question
28. A first order reaction takes 40 min for 30% decomposition. Calculate t
1/2
.
Solution :
For a first order reaction,
t = 2.303/k Log [R] º / [R]
k = 2.303/40min Log 100 / 100-30
= 2.303/40min Log 10 / 7
= 8.918 x 10-3 min
-1
Therefore, t
1/2
of the decomposition reaction is
t1/2 = 0.693/k
= 0.693 / 8.918 x 10-3 min
= 77.7 min (approximately)
= 77.7 min (approximately)
Question
29. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
After time, t, total pressure, Pt = (P
º
- p) + p + p
⇒ Pt = (P
º
+ p)
⇒ p = P
t
- P
0
therefore, P
º
- p = P
0
- Pt - P
0
= 2P
0
− P
t
For a first order reaction,
k = 2.303/t Log P
0
/P
0
- p
= 2.303/t Log P
0
/ 2 P
0
- P
t
When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0
= 2.175 × 10
−3
s
−1
When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0
= 2.235 × 10
−3
s
−1
Hence, the average value of rate constant is
k = (2.175 × 10
- 3
+ 2.235 × 10
- 3
) / 2 s
- 1
= 2.21 × 10
−3
s
−1
Question
30. The following data were obtained during the first order thermal decomposition of SO
2
Cl
2
at a constant volume.
SO
2
Cl
2
(g) → SO
2
(g) + Cl
2
(g)
After time, t, total pressure, P
t
= (P
º
- p) + p + p
⇒ P
t
= (P
º
+ p)
⇒ p = Pt - Pº
therefore, Pº - p = Pº - Pt - Pº
= 2 P
º
- P
t
For a first order reaction,
k = 2.303/t Log P
º
/ P
º
- p
= 2.303/t Log P
º
/ 2 P
º
- Pt
When t= 100 s,
k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6
= 2.231 × 10 - 3s - 1
When Pt= 0.65 atm,
P0+ p= 0.65
⇒ p= 0.65 - P0
= 0.65 - 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is
PSOCL2
= P
0
- p
= 0.5 - 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(p
SOCL2
)
= (2.23 × 10 - 3s - 1) (0.35 atm)
= 7.8 × 10
- 4
atm s
- 1
Question
31. The rate constant for the decomposition of N
2
O
5
at various temperatures is given below:
Slope of the line,
In k= - 2.8
Therefore, k = 6.08x10-2s-1
Again when T = 50 + 273K = 323K,
1/T = 3.1 x 10-3 K
In k = - 0.5
Therefore, k = 0.607 s-1
Question
32. The rate constant for the decomposition of hydrocarbons is 2.418 × 10
−5
s
−1
at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Solution :
k = 2.418 × 10
−5
s
−1
T = 546 K
E
a
= 179.9 kJ mol
−1
= 179.9 × 10
3
J mol
−1
According to the Arrhenius equation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10
12
s
−1
(approximately)
Question
33. Consider a certain reaction A → Products with k = 2.0 × 10
−2
s
−1
. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L
−1
.
Solution :
k = 2.0 × 10
−2
s
−1
T = 100 s
[A]
o
= 1.0 moL
−1
Since the unit of k is s
−1
, the given reaction is a first order reaction.
Therefore, k = 2.303/t Log [A]º / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s Log 1.0 / [A]
⇒2.0 × 110-2 s-1 = 2.303/100s ( - Log [A] )
⇒ - Log [A] = - (2.0 x 10-2 x 100) / 2.303
⇒ [A] = antilog [- (2.0 x 10-2 x 100) / 2.303]
= 0.135 mol L
−1 (approximately)
Hence, the remaining concentration of A is 0.135 mol L
−1.
Question
34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t
1/2
= 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Solution :
For a first order reaction,
k = 2.303/t Log [R]º / [R]
It is given that, t1/2 = 3.00 hours
Therefore, k = 0.693 / t1/2
= 0.693 / 3 h-1
= 0.231 h - 1
Then, 0.231 h - 1 = 2.303 / 8h Log [R]º / [R]
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Question
35. The decomposition of hydrocarbon follows the equation k = (4.5 × 10
11
s
−1
) e
−28000
K/T Calculate E
a
.
Solution :
The given equation is
k = (4.5 × 10
11
s
−1
) e
−28000
K/T (i)
Arrhenius equation is given by,
k= Ae
-Ea/RT
(ii)
From equation (i) and (ii), we obtain
Ea / RT = 28000K / T
⇒ Ea = R x 28000K
= 8.314 J K
−1
mol
−1
× 28000 K
= 232792 J mol
−1
= 232.792 kJ mol
−1
Question
36. The rate constant for the first order decomposition of H
2
O
2
is given by the following equation:
log k = 14.34 − 1.25 × 10
4 K/T
Calculate E
a
for this reaction and at what temperature will its half-period be 256 minutes?
Solution :
Arrhenius equation is given by,
k= Ae -Ea/RT
⇒In k = In A - Ea/RT
⇒In k = Log A - Ea/RT
⇒ Log k = Log A - Ea/2.303RT (i)
The given equation is
Log k = 14.34 - 1.25 104 K/T (ii)
From equation (i) and (ii), we obtain
Ea/2.303RT = 1.25 104 K/T
= 1.25 × 10
4
K × 2.303 × 8.314 J K
−1
mol
−1
= 239339.3 J mol
−1
(approximately)
= 239.34 kJ mol
−1
Also, when t
1/2
= 256 minutes,
k = 0.693 / t
1/2
= 0.693 / 256
= 2.707 × 10
- 3
min
- 1
= 4.51 × 10 - 5s
- 1
= 2.707 × 10
−3
min
−1
= 4.51 × 10
−5
s
−1
It is also given that, log k = 14.34 − 1.25 × 10
4
K/T
= 668.95 K
= 669 K (approximately)
Question
37. The decomposition of A into product has value of k as 4.5 × 10
3
s
−1
at 10°C and energy of activation 60 kJ mol
−1
. At what temperature would k be 1.5 × 10
4
s
−1
?
Solution :
From Arrhenius equation, we obtain
log k2/k1 = Ea / 2.303 R (T
2
- T
1
) / T
1
T
2
Also, k
1
= 4.5 × 10
3 s−1
T
1
= 273 + 10 = 283 K
k
2
= 1.5 × 10
4
s
−1
E
a
= 60 kJ mol
−1
= 6.0 × 10
4
J mol
−1
Then,
= 297 K
= 24°C
Hence, k would be 1.5 × 10
4
s
−1
at 24°C.
Question
38. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10
10
s
−1
. Calculate k at 318 K and E
a
.
Solution :
For a first order reaction,
t = 2.303 / k log a / a - x
At 298 K,
t = 2.303
/
k
log
100
/
90
= 0.1054 /
k
At 308 K,
t' = 2.303
/
k'
log
100
/
75
= 2.2877 /
k'
According to the question,
t = t'
⇒
0.1054 /
k =
2.2877 /
k'
⇒ k'
/
k = 2.7296
From Arrhenius equation, we obtain
To calculate k at 318 K,
It is given that, A = 4 x 1010 s-1, T = 318K
Again, from Arrhenius equation, we obtain
Therefore, k = Antilog (-1.9855)
= 1.034 x 10-2 s -1
Question
39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution :
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol
−1.
