Chapter 4 Chemical Kinetics Solutions
Question - 21 : - The following results have been obtained during the kineticstudies of the reaction:
Answer - 21 : -
2A + B → C + D
| Experiment | A/ mol L−1 | B/ mol L−1 | Initial rate of formation of D/mol L−1 min−1 |
| I | 0.1 | 0.1 | 6.0 × 10−3 |
| II | 0.3 | 0.2 | 7.2 × 10−2 |
| III | 0.3 | 0.4 | 2.88 × 10−1 |
| IV | 0.4 | 0.1 | 2.40 × 10−2 |
Determinethe rate law and the rate constant for the reaction.
Answer
Let the order of the reactionwith respect to A be x andwith respect to B be y.
Therefore, rate of thereaction is given by,
Question - 22 : - The following results havebeen obtained during the kinetic studies of the reaction:
2A + B → C + D
Answer - 22 : -
| Experiment | A/ mol L−1 | B/ mol L−1 | Initial rate of formation of D/mol L−1 min−1 |
| I | 0.1 | 0.1 | 6.0 × 10−3 |
| II | 0.3 | 0.2 | 7.2 × 10−2 |
| III | 0.3 | 0.4 | 2.88 × 10−1 |
| IV | 0.4 | 0.1 | 2.40 × 10−2 |
Determine the rate law and therate constant for the reaction.
Answer
Let the order of the reactionwith respect to A be x and with respect to B be y.
Therefore, rate of the reactionis given by,
Question - 23 : - Calculate the half-life of afirst order reaction from their rate constants given below:
(i) 200 s−1 (ii) 2 min−1 (iii) 4 years−1
Answer - 23 : - (i) Half life,
=3.47 ××10 -3 s(approximately)
(ii) Halflife,= 0.35 min (approximately)
(iii) Half life,
= 0.173 years (approximately)
The half-life for radioactivedecay of 14C is 5730 years. An archaeological artifactcontaining wood had only 80% of the 14C found in a living tree.Estimate the age of the sample.
Answer - 24 : - Here,
It is known that,
= 1845 years (approximately)
Hence, the age of the sample is1845 years.
The rate constant for a firstorder reaction is 60 s−1. How much time will it take to reduce theinitial concentration of the reactant to its 1/16th value?
Answer - 25 : -
It is known that,
Hence, the required time is 4.6 ×10−2 s.
Question - 26 : - During nuclear explosion, one ofthe products is 90Sr with half-life of 28.1 years. If 1μgof 90Sr was absorbed in the bones of a newly born baby insteadof calcium,
Answer - 26 : - how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Answer
Question - 27 : - For the decomposition ofazoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
Answer - 27 : -
| t (sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
Answer
The decomposition ofazoisopropane to hexane and nitrogen at 543 K is represented by the followingequation.
After time, t, total pressure,
= 2P0 − Pt
For a first order reaction,
= 2.21 × 10−3 s−1
Question - 28 : - The following data were obtainedduring the first order thermal decomposition of SO2Cl2 ata constant volume.
Answer - 28 : - 
| Experiment | Time/s−1 | Total pressure/atm |
| 1 | 0 | 0.5 |
| 2 | 100 | 0.6 |
Calculatethe rate of the reaction when total pressure is 0.65 atm.
Answer
The thermal decomposition of SO2Cl2 ata constant volume is represented by the following equation.
= 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 totalpressure 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 atms−1
Answer - 29 : -
| T/°C | 0 | 20 | 40 | 60 | 80 |
| 
| 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and1/T and calculate the values of A and Ea.
Predictthe rate constant at 30º and 50ºC.
Answer
From the given data, we obtain
| T/°C | 0 | 20 | 40 | 60 | 80 |
| T/K | 273 | 293 | 313 | 333 | 353 |
| 
| 3.66×10−3 | 3.41×10−3 | 3.19×10−3 | 3.0×10−3 | 2.83 ×10−3 |
| 
| 0.0787 | 1.70 | 25.7 | 178 | 2140 |
| ln k | −7.147 | − 4.075 | −1.359 | −0.577 | 3.063 |
When 
Again, when 
Question - 30 : - The rate constant for thedecomposition of hydrocarbons is 2.418 × 10−5 s−1 at546 K. If the energy of activation is 179.9 kJ/mol, what will be the value ofpre-exponential factor.
Answer - 30 : -
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 Arrheniusequation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s−1 (approximately)