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Page No 119: - Chapter 4 Chemical Kinetics Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


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, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7ad0f420.gif

= 3.47

×10 -3 s (approximately)

(ii) Half life, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_753ea805.gif

= 0.35 min (approximately)

(iii) Half life, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7f8f9078.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6348/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4451f518.gif

= 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,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m37f6ea30.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c431e67.gif

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6350/NS_14-11-08_Utpal_12_Chemistry_4_30_html_26185dc1.gif

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Question 4.15:

The experimental data for decomposition of N2O5

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m7508b7ed.gif

in gas phase at 318K are given below:

 

t(s)

0

400

800

1200

1600

2000

2400

2800

3200

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cc8818c.gif

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 and compare it with (ii).

Answer:

1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_me506e51.jpg

(ii) Time corresponding to the concentration, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_7c1699df.gif  is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

 

t(s)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_27d7e51c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6d9ac63a.gif

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, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif  v/s t, is a straight line. Therefore, the rate law of the reaction is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_47ff23ff.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_40b6aaf6.gif

(v) From the plot, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif  v/s t, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_720ab5a7.gif

Again, slope of the line of the plot https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_2c7892c1.gif  v/s t is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m30f2689c.gif .

Therefore, we obtain,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m29d84705.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2841e4e4.gif

(vi) Half-life is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6372/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3ee340dc.gif

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 1/16th value?

Answer:

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6351/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m774930f5.gif

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, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m4f924778.gif

It is known that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m25b3f4d2.gif

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

Again,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6353/NS_14-11-08_Utpal_12_Chemistry_4_30_html_16b7b6cb.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6354/NS_14-11-08_Utpal_12_Chemistry_4_30_html_b7c8ada.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6354/NS_14-11-08_Utpal_12_Chemistry_4_30_html_1367300f.gif

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 t1/2.

Answer:

For a first order reaction,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5cd18479.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5fff9d0b.gif

Therefore, t1/2 of the decomposition reaction is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6356/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m624d0e08.gif

= 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m5e306863.gif

After time, t, total pressure, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_670c95fe.gif

= 2P0 − Pt

For a first order reaction,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m510c488d.gif

When t = 360 s, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_5f71fbd9.gif

= 2.175 × 10−3 s−1

When t = 720 s, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_9f83889.gif

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6358/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3e8a9bbb.gif

= 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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_3572ff23.gif

 

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_6f5f1162.gif

After time, t, total pressure, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_35bdefa5.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_73a3ad92.gif

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m2b66381a.gif

= 2 P0 − Pt

For a first order reaction,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_m3f1e8ec2.gif

When t = 100 s, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_110029b4.gif

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_68f4cf67.gif = 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(https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/263/6359/NS_14-11-08_Utpal_12_Chemistry_4_30_html_68f4cf67.gif )

= (2.23 × 10−3 s−1) (0.35 atm)

= 7.8 × 10−4 atm s−1

Also Read : Page-No-120:-Chapter-4-Chemical-Kinetics-Exercise-Solutions-class-12-ncert-solutions-Chemistry

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