Page No 119: - Chapter 4 - Chemical Kinetics - Exercise Solutions - Ncert Solutions class 12 - Chemistry

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Question 4.13:

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

Answer:

(i) Half life, 

= 3.47

×10 ^-3 s (approximately)

(ii) Half life, 

= 0.35 min (approximately)

(iii) Half life, 

= 0.173 years (approximately)

Question 4.14:

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.

Answer:

Here,

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Question 4.15:

The experimental data for decomposition of N₂O₅

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

(ii) Find the half-life period for the reaction.(i) Plot [N₂O₅] against t.

(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 k and compare it with (ii).

Answer:

1.

(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) The given reaction is of the first order as the plot,   v/s t, is a straight line. Therefore, the rate law of the reaction is

(v) From the plot,   v/s t, we obtain

Again, slope of the line of the plot   v/s t is given by

.

Therefore, we obtain,

(vi) Half-life is given by,

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

Question 4.16:

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 1/16^th value?

Answer:

It is known that,

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

Question 4.17:

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:

Here, 

It is known that,

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

Again,

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

Question 4.18:

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

Answer:

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

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

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.

Question 4.19:

A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.

Answer:

For a first order reaction,

Therefore, t_1/2 of the decomposition reaction is

= 77.7 min (approximately)

Question 4.20:

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

Answer:Calculate the rate constant.

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

After time, t, total pressure, 

= 2P₀ − Pt

For a first order reaction,

When t = 360 s, 

= 2.175 × 10⁻³ s⁻¹

When t = 720 s, 

= 2.235 × 10⁻³ s⁻¹

Hence, the average value of rate constant is

= 2.21 × 10⁻³ s⁻¹

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

Question 4.21:

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

 

Experiment	Time/s−1	Total pressure/atm
1	0	0.5
2	100	0.6

Answer:Calculate the rate of the reaction when total pressure is 0.65 atm.

The thermal decomposition of SO₂Cl₂ at a constant volume is represented by the following equation.

After time, t, total pressure, 

Therefore, 

= 2 P₀ − Pt

For a first order reaction,

When t = 100 s, 

= 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

= 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 × 10⁻³ s⁻¹) (0.35 atm)

= 7.8 × 10⁻⁴ atm s⁻¹