NCERT Solutions for Class 12 Chemistry Part 1 Chapter 4
Chemical Kinetics Class 12
Chapter 4 Chemical Kinetics Exercise Solutions
In text : Solutions of Questions on Page Number : 98
Q1 :
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer :
Average rate of reaction
= 6.67 × 10 - 6M s - 1
Q2 :
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L-1to 0.4 mol L-1in 10 minutes. Calculate the rate during this interval?
Answer :
Average rate
= 0.005 mol L - 1 min - 1
Q3 :
For a reaction, A + B →Product; the rate law is given by, . What is the order of the reaction?
Answer :
The order of the reaction
Q4 :
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer :
The reaction X → Y follows second order kinetics. Therefore, the rate equation for this reaction will be: Rate = k[X]2 (1)
Let [X] = a mol L-1, then equation (1) can be written as: Rate1 = k .(a)2
= ka2
If the concentration of X is increased to three times, then [X] = 3a mol L-1
Now, the rate equation will be:
Rate = k (3a)2
= 9(ka2)
Q5 :
A first order reaction has a rate constant 1.15 10-3s-1. How long will 5 g of this reactant take to reduce to 3 g?
Answer :
From the question, we can write down the following information: Initial amount = 5 g
Final concentration = 3 g Rate constant = 1.15 10 - 3s - 1
We know that for a 1storder reaction,
= 444.38 s
Q6 :
Time required to decompose SO2Cl2to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer :
We know that for a 1storder reaction,
It is given that t1/2= 60 min
Q7 :
What will be the effect of temperature on rate constant?
Answer :
Where,
A is the Arrhenius factor or the frequency factor
Tis the temperature
Ris the gas constant
Q8 :
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Answer :
It is given that T1 = 298 K
∴T2 = (298 + 10) K
= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10°. Therefore, let us take the value of k1 = k and that of k2 = 2k
Also, R = 8.314 J K - 1 mol - 1
We get:
= 52897.78 J mol - 1
= 52.9 kJ mol - 1
Q9 :
The activation energy for the reaction 2HI(g) → H2 + I2(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?
Answer :
In the given case:
Ea = 209.5 kJ mol - 1 = 209500 J mol - 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:
Exercise : Solutions of Questions on Page Number : 117
Q1 :
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N2O(g) Rate = k[NO]2
(ii) H2O2 (aq) + 3 I - (aq) + 2 H+→ 2 H2O (l) + Rate = k[H2O2][I - ]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]
Answer :
(i) Given rate = k [NO]2
Therefore, order of the reaction = 2 Dimension of
(ii) Given rate = k [H2O2] [I - ] Therefore, order of the reaction = 2
Dimension of
(iii) Given rate = k [CH3CHO]3/2 Therefore, order of reaction =
Dimension of
(iv) Given rate = k [C2H5Cl] Therefore,order of the reaction = 1
Q2 :
For the reaction:
2A + B → A2B
the rate = k[A][B]2with k= 2.0 x 10-6mol-2L2s-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.
Answer :
The initial rate of the reactionis Rate = k [A][B]2
= (2.0 × 10 - 6mol - 2L2s - 1) (0.1 mol L - 1) (0.2 mol L - 1)2
= 8.0 × 10 - 9mol - 2L2s - 1
When [A] is reduced from 0.1 mol L - 1to 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 = 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 - 6mol - 2L2s - 1) (0.06 mol L - 1) (0.18 mol L - 1)2
Q3 :
The decomposition of NH3on platinum surface is zero order reaction. What are the rates of production of N2and H2if k = 2.5 x 10-4mol-1L s-1?
Answer :
Therefore,
However,it is given that the reaction is of zero order. Therefore,
And, the rate of production of H2is
Q4 :
The decomposition of dimethyl ether leads to the formation of CH4, H2and CO and the reaction rate is given by
Rate = k [CH3OCH3]3/2
If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?
Answer :
If pressure is measured in bar and time in minutes, then Unit of rate = bar min - 1
Therefore, unit of rate constants
Q5 :
Mention the factors that affect the rate of a chemical reaction.
Answer :
The factors that affect the rate of a reaction areas follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii)
Q6 :
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?
Answer :
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2
= ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
= 4ka2
= 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. , then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to
Q7 :
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Answer :
The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.
where, kis the rate constant,
A is the Arrhenius factor or the frequency factor, R is the gas constant,
Tis the temperature, and
Q8 :
In a pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s |
0 |
30 |
60 |
90 |
[Ester]mol L- 1 |
0.55 |
0.31 |
0.17 |
0.085 |
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer :
(i)
= 4.67 × 10 - 3mol L - 1s - 1
(ii)
For t= 30 s,
= 1.911 × 10 - 2s - 1
For t= 60 s,
= 1.957 × 10 - 2s - 1
For t= 90 s,
= 2.075 × 10 - 2s - 1
Q9 :
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer :
(i)
(ii) If the concentration of B is increased three times, then
Therefore, the rate of reaction will increase 9 times.
(iii)
Therefore, the rate of reaction will increase 8 times.
Q10 :
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
A/ mol L- 1 |
0.20 |
0.20 |
0.40 |
B/ mol L- 1 |
0.30 |
0.10 |
0.05 |
r0/ mol L- 1 s - 1 |
5.07 × 10- 5 |
5.07 × 10- 5 |
1.43 × 10- 4 |
What is the order of the reaction with respect to A and B?
Answer :
Let the order of the reaction with respect to A be xand with respect to B be y. Therefore,
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Q11 :
Experiment A/ mol L- 1 B/ mol
L- 1 Initial rate
of formation
of D/mol L- 1 min - 1 I 0.1 0.1 6.0 × 10- 3
The following results have been obtained during the kinetic
studies of the reaction: 2A + B → C +
D
II |
0.3 |
0.2 |
7.2 × 10- 2 |
III |
0.3 |
0.4 |
2.88 × 10- 1 |
IV |
0.4 |
0.1 |
2.40 × 10- 2 |
Determine the rate law and the rate constant for the reaction.
Answer :
Let the order of the reaction with respect to A be xand with respect to B be y. Therefore, rate of the reaction is given by,
Dividing equation (iv) by (i), we obtain
Dividing equation (iii) by (ii), we obtain
Therefore, the rate law is Rate = k [A] [B]2
= 6.0 L2mol - 2min - 1
= 6.0 L2mol - 2min - 1
= 6.0 L2mol - 2min - 1
= 6.0 L2mol - 2min - 1
Q12 :
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:
Experiment |
A/ mol L- 1 |
B/ mol L- 1 |
Initial rate/mol L- 1 min - 1 |
I |
0.1 |
0.1 |
2.0 × 10- 2 |
II |
-- |
0.2 |
4.0 × 10- 2 |
III |
0.4 |
0.4 |
-- |
IV |
-- |
0.2 |
2.0 × 10- 2 |
Answer :
The given reaction is of the first order with respect to A and of zero order with respect to B. Therefore, the rate of the reaction is given by,
Rate = k [A]1[B]0
⇒ Rate = k [A]
From experiment I, we obtain
2.0 x 10-2mol L-1min-1= k (0.1 mol L-1)
⇒ k= 0.2 min-1
From experiment II, we obtain 4.0 x 10-2mol L-1min-1= 0.2 min-1[A]
⇒ [A] = 0.2 mol L-1
From experiment III, we obtain Rate = 0.2 min-1 x 0.4 mol L-1
= 0.08 mol L-1min-1
From experiment IV, we obtain 2.0 x 10-2mol L-1min-1= 0.2 min-1[A]
Q13 :
Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s-1 (ii) 2 min-1 (iii) 4 years-1
Answer :
(i)
= 3.47 ×10 -3 s (approximately)
(ii)
= 0.35 min (approximately)
(iii)
= 0.173 years (approximately)
Q14 :
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
Answer :
It is known that,
= 1845 years (approximately)
Q15 :
in gas phase at 318K are given below:
t(s) |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
|
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N2O5] and t.
(iv) What is the rate law? (v)
(ii) Time corresponding to the concentration, is the half life. From the graph, the half life is obtained as 1450 s.
(iii)
t(s) |
|
|
0 |
1.63 |
- 1.79 |
400 |
1.36 |
- 1.87 |
800 |
1.14 |
- 1.94 |
1200 |
0.93 |
- 2.03 |
1600 |
0.78 |
- 2.11 |
2000 |
0.64 |
- 2.19 |
2400 |
0.53 |
- 2.28 |
2800 |
0.43 |
- 2.37 |
3200 |
0.35 |
- 2.46 |
(iv)
(v) From the plot, v/s t, we obtain
Again, slope of the line of the plot v/s t is given by .
Q16 :
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 1/16thvalue?
Answer :
It is known that,
Q17 :
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.
Answer :
Therefore, 0.7814 μg of 90Sr will remain after 10 years. Again,
Q18 :
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer :
For a first order reaction, the time required for 90% completion is
Therefore, t1= 2t2
Q19 :
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Answer :
= 77.7 min (approximately)
Q20 :
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
t (sec) |
P(mm of Hg) |
0 |
35.0 |
360 |
54.0 |
720 |
63.0 |
Calculate the rate constant.
Answer :
= 2P0 - Pt
When t = 360 s,
= 2.175 × 10 - 3 s - 1
When t = 720 s,
= 2.235 × 10 - 3 s - 1
= 2.21 × 10 - 3 s - 1
Q21 :
Experiment |
Time/s- 1 |
Total pressure/atm |
1 |
0 |
0.5 |
2 |
100 |
0.6 |
Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer :
The thermal decomposition of SO2Cl2at a constant volume is represented by the following equation.
Therefore,
= 2 P0 - Pt
When t= 100 s,
= 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 SOCl2is
= P0 - 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( )
= (2.23 × 10 - 3s - 1) (0.35 atm)
Q22 :
The rate constant for the decomposition of N2O5 at various temperatures is given below:
T/°C |
0 |
20 |
40 |
60 |
80 |
|
0.0787 |
1.70 |
25.7 |
178 |
2140 |
Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30 º and 50 ºC.
Answer :
From the given data, we obtain
T/°C |
0 |
20 |
40 |
60 |
80 |
T/K |
273 |
293 |
313 |
333 |
353 |
|
3.66×10- 3 |
3.41×10 - 3 |
3.19×10- 3 |
3.0×10 - 3 |
2.83 ×10 - 3 |
|
0.0787 |
1.70 |
25.7 |
178 |
2140 |
ln k |
- 7.147 |
- 4.075 |
- 1.359 |
- 0.577 |
3.063 |
According to Arrhenius equation,
When ,
Again, when ,
Q23 :
The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Answer :
k= 2.418 × 10 - 5s - 1
T= 546 K
Ea= 179.9 kJ mol - 1= 179.9 × 103J mol - 1
= (0.3835 - 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
Q24 :
Consider a certain reaction A → Products with k = 2.0 x 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.
Answer :
k= 2.0 × 10 - 2s - 1
T= 100 s
[A]o= 1.0 moL - 1
Sincethe unit of kis s - 1, the given reaction is a first order reaction. Therefore,
= 0.135 mol L - 1(approximately)
Q25 :
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, witht1/2 =
3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Answer :
It is given that, t1/2= 3.00 hours Therefore,
= 0.231 h - 1
Then, 0.231 h - 1
Q26 :
The decomposition of hydrocarbon follows the equation
k = (4.5 x 1011 s-1) e-28000 K/T
Calculate Ea.
Answer :
The given equation is
k = (4.5 × 1011 s - 1) e - 28000 K/T (i)
Arrhenius equation is given by,
(ii)
= 8.314 J K - 1mol - 1× 28000 K
= 232792 J mol - 1
Q27 :
The rate constant for the first order decomposition of H2O2is given by the following equation: log k = 14.34 - 1.25 x 104 K/T
Calculate Eafor this reaction and at what temperature will its half-period be 256 minutes?
Answer :
From equation (i) and (ii), we obtain
= 1.25 × 104K × 2.303 × 8.314 J K - 1mol - 1
= 239339.3 J mol - 1 (approximately)
= 239.34 kJ mol - 1
= 2.707 × 10 - 3min - 1
= 4.51 × 10 - 5s - 1
It is also given that, log k= 14.34 - 1.25 × 104K/T
= 668.95 K
Q28 :
The decomposition of A into product has value of k as 4.5 x 103 s-1 at 10°C and energy of activation 60 kJ mol-
1. At what temperature would k be 1.5 x 104 s-1?
Answer :
Also, k1 = 4.5 × 103 s - 1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s - 1
Ea = 60 kJ mol - 1 = 6.0 × 104 J mol - 1
= 297 K
= 24°C
Hence, k would be 1.5 × 104 s - 1 at 24°C.
Q29 :
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 x 1010 s-1. Calculate k at 318 K and Ea.
Answer :
At 298 K,
According to the question,
From Arrhenius equation, we obtain
To calculate k at 318 K,
Q30 :
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.
Answer :
From Arrhenius equation, we obtain
Hence, the required energy of activation is 52.86 kJmol - 1.