NEET Exam  >  NEET Notes  >  Chemistry Class 12  >  NCERT Solutions: Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Page No.66 - Intext Questions

Q3.1: 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.
Ans: Average rate of reaction

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics 

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 6.67 × 10−6 M s−1

Q3.2: 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? 
Ans: 

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
= 0.005 mol L−1 min−1
= 5 × 10−3 M min−1

Page No.71 - Intext Questions

Q3.3: For a reaction, A + B → Product; the rate law is given by, r = k [ A]1/2 [B]2. What is the order of the reaction? 
Ans: 
The order of the reaction = NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.4: 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?
Ans: The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = [X]2 (1)

Let [X] = a mol L−1, then equation (1) can be written as:

Rate1 = k.(a)= 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)= 9(ka2)

Hence, the rate of formation will increase by 9 times.

Page No.78 - Intext Questions

Q3.5: A first order reaction has a rate constant 1.15 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g?
Ans: From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10−3 s−1

We know that for a 1st order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 444.38 s

= 444 s (approx)

Q3.6: Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. 
Ans: 
We know that for a 1st order reaction,
NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
It is given that t1/2 = 60 min
NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
= 0.693/60
= 0.01155 min-1
= 1.155 min-1
Or k = 1.925 x 10-4 s-1

Page No.84 - Intext Questions

Q3.7: What will be the effect of temperature on rate constant?
Ans: 
With increase in temperature, the rate of the reaction and the rate constant increases. As a generalization, the rate of the reaction (and the rate constant) becomes almost double for every ten degree rise in temperature. This is also called temperature coefficient.

Q3.8: The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Ans: 
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
Now, substituting these values in the equation:
NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.9: 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?
Ans: 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:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Exercises

Q3.1: 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) H2O(aq) + 3 I− (aq) 2 H  → 2 H2O (l) + NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical KineticsRate = k[H2O2][I]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = [CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = [C2H5Cl]
Ans: (i) Given rate = [NO]2

Therefore, order of the reaction = 2

Dimension of NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(ii) Given rate = [H2O2] [I]

Therefore, order of the reaction = 2

Dimension of NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(iii) Given rate = [CH3CHO]3/2

Therefore, order of reaction = 3/2

Dimension of NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(iv) Given rate = [C2H5Cl]

Therefore, order of the reaction = 1

Dimension of NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.2: For the reaction: 
2A + B → A2B
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 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.
Ans: The initial rate of the reaction is

Rate = [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 mol−2 L2 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  = NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics  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 = [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89 mol L−1 s−1

Q3.3: The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if = 2.5 × 10−4 mol−1 L s−1?
Ans: The decomposition of NH3 on platinum surface is represented by the following equation.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

However, it is given that the reaction is of zero order.

Therefore,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, the rate of production of N2 is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

And, the rate of production of H2 is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 7.5 × 10−4 mol L−1 s−1

Q3.4: The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = [CH3OCH3]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.,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Ans: If pressure is measured in bar and time in minutes, then

Unit of rate = bar min−1

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, unit of rate constantsNCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.5: Mention the factors that affect the rate of a chemical reaction.
Ans:
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

Q3.6: 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?
Ans: Let the concentration of the reactant be [A] = a

Rate of reaction, R = [A]2

= ka2

(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 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. NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics, then the rate of the reaction would be

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

=1/4 ka2

=1/4 R

Therefore, the rate of the reaction would be reduced to 1/4 th.

Q3.7: What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on the rate constant be represented quantitatively?
Ans: 
The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise of temperature. The effect can be represented quantitatively by Arrhenius equation,
k = Ae-Ea/RT
Where,
A = Arrhenius constant
Ea = Activation energy
R = Gas constant
T = Temperature
k = Rate Constant

Q3.8: In a pseudo first order hydrolysis of ester in water, the following results were obtained:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Ans: (i) Average rate of reaction between the time interval, 30 to 60 seconds,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 4.67 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

For t = 30 s, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 1.911 × 10−2 s−1

For t = 60 s, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 1.957 × 10−2 s−1

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics     

For t = 90 s,

= 2.075 × 10−2 s−1

Then, average rate constant,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.9: 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?
Ans: (i) The differential rate equation will be

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(ii) If the concentration of B is increased three times, then

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, the rate of reaction will increase 9 times.

(iii) When the concentrations of both A and B are doubled,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, the rate of reaction will increase 8 times.

Q3.10: 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:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

What is the order of the reaction with respect to A and B?
Ans: Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Dividing equation (i) by (ii), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 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.

Q3.11: The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Determine the rate law and the rate constant for the reaction.
Ans: Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

According to the question,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics 

Dividing equation (iv) by (i), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, the rate law is

Rate = [A] [B]2

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical KineticsNCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

From experiment I, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 6.0 L2 mol−2 min−1

From experiment II, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 6.0 L2 mol−2 min−1

From experiment IV, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 6.0 L2 mol−2 min−1

Therefore, rate constant, k = 6.0 L2 mol−2 min−1

Q3.12: 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:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical KineticsAns: 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 = [A]1 [B]0

⇒ Rate = [A]

From experiment I, we obtain

2.0 × 10−2 mol L−1 min−1 = k (0.1 mol L−1)

k = 0.2 min−1

From experiment II, we obtain

4.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.2 mol L−1

From experiment III, we obtain

Rate = 0.2 min−1 × 0.4 mol L−1

= 0.08 mol L−1 min−1

From experiment IV, we obtain

2.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.1 mol L−1

Q3.13: 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

Ans: (i) Half life,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

=0.693 / 200s-1  =  3.47 x 10-3 s (approximately)

(ii) Half life,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics= 0.35 min (approximately)

(iii) Half life,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 0.173 years (approximately)

Q3.14: 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.
Ans: Here, 

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

 NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

It is known that,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Q3.15: The experimental data for decomposition of N2ONCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics in gas phase at 318K are given below:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(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) Calculate the rate constant.  
(vi) Calculate the half-life period from and compare it with (ii).

Ans: (i)

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(ii) Time corresponding to the concentration NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics , is the half life. From the graph, the half life is obtained as 1450 s.

(iii)  

 NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(iv) The given reaction is of the first order as the plot, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kineticsv/s t, is a straight line. Therefore, the rate law of the reaction is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(v) From the plot, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kineticsv/s t, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Again, slope of the line of the plot NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kineticsv/s t is given by

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics.

Therefore, we obtain,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

(vi) Half-life is given by,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

This value, 1438 s, is very close to the value that was obtained from the graph.

Q3.16: 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 1/16th value?
Ans: It is known that,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Hence, the required time is 4.6 × 10−2 s.


Q3.17: During nuclear explosion, one of the products is 90 Sr with half-life of 28.1 years. If 1μg of 90 Sr 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.

Ans: Here,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics 

It is known that,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, 0.7814 μg of 90 Sr will remain after 10 years.

Again,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, 0.2278 μg of 90 Sr will remain after 60 years.


Q3.18: For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. 
Ans: For a first order reaction, the time required for 99% completion is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

For a first order reaction, the time required for 90% completion is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, t1 = 2t2

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.

Q3.19: A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans: For a first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, t1/2 of the decomposition reaction is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 77.7 min (approximately)

Q3.20: For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Calculate the rate constant.
Ans: The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

After time, t, total pressure, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2P0 − Pt

For a first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

When t = 360 s, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2.175 × 10−3 s−1

When t = 720 s, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2.21 × 10−3 s−1

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Q3.21: The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Calculate the rate of the reaction when total pressure is 0.65 atm.
Ans: The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

After time, t, total pressure, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Therefore, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2 P0 − Pt

For a first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

When t = 100 s, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2.231 × 10−3 s−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

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics= 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(NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics)

= (2.23 × 10−3 s−1) (0.35 atm)

= 7.8 × 10−4 atm s−1

Q3.22: The rate constant for the decomposition of N2O5 at various temperatures is given below:

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical KineticsDraw a graph between ln and 1/and calculate the values of and EaPredict the rate constant at 30º and 50ºC.
Ans: From the given data, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Slope of the line,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

According to Arrhenius equation,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Again,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

When NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Then, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Again, when NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Then, at NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.23: 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.

Ans: k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1

According to the Arrhenius equation,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= (0.3835 − 5) 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s−1 (approximately)

Q3.24: Consider a certain reaction A → Products with = 2.0 × 10−2 s−1. Calculate the concentration of remaining after 100 s if the initial concentration of is 1.0 mol L−1.
Ans: 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, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 0.135 mol L−1 (approximately)

Hence, the remaining concentration of A is 0.135 mol L−1.


Q3.25: Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Ans: For a first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

It is given that, t1/2 = 3.00 hours

Therefore, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 0.231 h−1

Then, 0.231 h−1NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Q3.26: The decomposition of hydrocarbon follows the equation = (4.5 × 1011 s−1) e−28000 K/T Calculate Ea.
Ans: The given equation is

= (4.5 × 1011 s−1) e−28000 K/T (i)

Arrhenius equation is given by,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics(ii)

From equation (i) and (ii), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 8.314 J K−1 mol−1 × 28000 K

= 232792 J mol−1

= 232.792 kJ mol−1

Q3.27: The rate constant for the first order decomposition of H2O2 is given by the following equation:

log = 14.34 − 1.25 × 10K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Ans: Arrhenius equation is given by,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

The given equation is

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

From equation (i) and (ii), we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol1 (approximately)

= 239.34 kJ mol−1

Also, when t1/2 = 256 minutes,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 2.707 × 10−3 min−1

= 4.51 × 10−5 s−1

It is also given that, log k = 14.34 − 1.25 × 104 K/T

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 668.95 K

= 669 K (approximately)

Q3.28: The decomposition of A into product has value of as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would be 1.5 × 104 s−1?
Ans: From Arrhenius equation, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

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

Then,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

= 297 K

= 24°C

Hence, k would be 1.5 × 104 s−1 at 24°C.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Q3.29: 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 is 4 × 1010 s−1. Calculate at 318 K and Ea.
Ans: For a first order reaction,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

At 298 K, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

At 308 K,  NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

According to the question,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics  

From Arrhenius equation, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

To calculate at 318 K,

It is given that, NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Again, from Arrhenius equation, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Q3.30: 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.
Ans: From Arrhenius equation, we obtain

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

Hence, the required energy of activation is 52.86 kJmol−1.

The document NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics is a part of the NEET Course Chemistry Class 12.
All you need of NEET at this link: NEET
135 videos|347 docs|182 tests

Up next

FAQs on NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

1. What is chemical kinetics?
Ans. Chemical kinetics is the study of the rates at which chemical reactions occur and the factors that influence these rates.
2. How can we determine the rate of a chemical reaction?
Ans. The rate of a chemical reaction can be determined by measuring the change in concentration of reactants or products over time.
3. What are the factors that can affect the rate of a chemical reaction?
Ans. Factors that can affect the rate of a chemical reaction include temperature, concentration of reactants, presence of catalysts, and surface area of reactants.
4. Why is it important to study chemical kinetics?
Ans. Studying chemical kinetics helps us understand the mechanisms of chemical reactions, optimize reaction conditions for desired products, and improve the efficiency of industrial processes.
5. How can we express the rate of a chemical reaction mathematically?
Ans. The rate of a chemical reaction can be expressed mathematically using rate laws, which relate the rate of reaction to the concentrations of reactants.
135 videos|347 docs|182 tests
Download as PDF

Up next

Explore Courses for NEET exam

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

pdf

,

Previous Year Questions with Solutions

,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

,

Free

,

MCQs

,

practice quizzes

,

Viva Questions

,

video lectures

,

ppt

,

Important questions

,

Sample Paper

,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

,

Objective type Questions

,

study material

,

Extra Questions

,

past year papers

,

Exam

,

mock tests for examination

,

NCERT Solutions Class 12 Chemistry Chapter 3 - Chemical Kinetics

,

shortcuts and tricks

,

Semester Notes

,

Summary

;