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 1/16th 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 t1/2.
Answer:
For a first order reaction,
Therefore, t1/2 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