NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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NCERT QUESTIONS - (Chemical Kinetics)

Ques 4.1: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev 

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 6.67 × 10−6 M s−1

Ques 4.2: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Ques 4.3: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

At 298 K, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

At 308 K,  NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

According to the question,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev  

From Arrhenius equation, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

To calculate at 318 K,

It is given that, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Again, from Arrhenius equation, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.4: 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.

Ques 4.5: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 444.38 s

= 444 s (approx)

Ques 4.6: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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.

Ques 4.7: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

The given equation is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 668.95 K

= 669 K (approximately)

Ques 4.8: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

We get:

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 52897.78 J mol−1

= 52.9 kJ mol−1

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

Ques 4.9: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.10: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N2O (g) Rate = k[NO]2
 (ii) H2O(aq) + 3 I− (aq) 2 H  → 2 H2O (l) + NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRevRate = 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

(ii) Given rate = [H2O2] [I]

Therefore, order of the reaction = 2

Dimension of NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Therefore, order of reaction = 3/2

Dimension of NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

(iv) Given rate = [C2H5Cl]

Therefore, order of the reaction = 1

Dimension of NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.11: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev  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

Ques 4.12: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Therefore,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, the rate of production of N2 is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

And, the rate of production of H2 is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 7.5 × 10−4 mol L−1 s−1

Ques 4.13: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, unit of rate constantsNCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.14: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev(ii)

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 8.314 J K−1 mol−1 × 28000 K

= 232792 J mol−1

= 232.792 kJ mol−1

Ques 4.15: 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?

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev, then the rate of the reaction would be

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

=1/4 ka2

=1/4 R

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

Ques 4.16: What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Ans: The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

where, k is the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

Ea is the energy of activation for the reaction

Ques 4.17: In a pseudo first order hydrolysis of ester in water, the following results were obtained:

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev(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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 4.67 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

For t = 30 s, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 1.911 × 10−2 s−1

For t = 60 s, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 1.957 × 10−2 s−1

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev     

For t = 90 s,

= 2.075 × 10−2 s−1

Then, average rate constant,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.18: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, the rate of reaction will increase 9 times.

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, the rate of reaction will increase 8 times.

 

Ques 4.19: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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.

 

Ques 4.20: The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

According to the question,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev 

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, the rate law is

Rate = [A] [B]2

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRevNCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

From experiment I, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 6.0 L2 mol−2 min−1

From experiment II, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 6.0 L2 mol−2 min−1

From experiment IV, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 6.0 L2 mol−2 min−1

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

Ques 4.21: 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:

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

 

Ans: 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

Ques 4.22: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

(ii) Half life,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev= 0.35 min (approximately)

(iii) Half life,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 0.173 years (approximately)

Ques 4.23: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

 NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

It is known that,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Ques 4.24: The experimental data for decomposition of N2ONCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev in gas phase at 318K are given below:

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

(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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

(iii)  

 NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

(v) From the plot, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRevv/s t, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Again, slope of the line of the plot NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRevv/s t is given by

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev.

Therefore, we obtain,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

(vi) Half-life is given by,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Ques 4.25: 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 1/16th value?

Ans: It is known that,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Ques 4.26: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev 

It is known that,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Again,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

Ques 4.27: For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. 

Ans: For a first order reaction, the time required for 99% completion is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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.

Ques 4.28: A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Ans: For a first order reaction,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, t1/2 of the decomposition reaction is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 77.7 min (approximately)

Ques 4.29: For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Calculate the rate constant.

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

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

After time, t, total pressure, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 2P0 − Pt

For a first order reaction,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

When t = 360 s, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 2.175 × 10−3 s−1

When t = 720 s, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 2.21 × 10−3 s−1

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

Ques 4.30: The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

After time, t, total pressure, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Therefore, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 2 P0 − Pt

For a first order reaction,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

When t = 100 s, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev= 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev)

= (2.23 × 10−3 s−1) (0.35 atm)

= 7.8 × 10−4 atm s−1

Ques 4.31: The rate constant for the decomposition of N2O5 at various temperatures is given below:

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev
Draw a graph between ln and 1/and calculate the values of and Ea.

Predict the rate constant at 30º and 50ºC.

Ans: From the given data, we obtain

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Slope of the line,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

According to Arrhenius equation,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Again,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

When NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Then, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Again, when NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Then, at NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev,

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

Ques 4.32: 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.

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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= (0.3835 − 5) 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s−1 (approximately)

Ques 4.33: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 0.135 mol L−1 (approximately)

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

Ques 4.34: 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 - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

It is given that, t1/2 = 3.00 hours

Therefore, NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

= 0.231 h−1

Then, 0.231 h−1NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

NCERT Solutions - Chemical Kinetics, Class 12, Chemistry Class 12 Notes | EduRev

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

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