Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper

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The document Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev is a part of the JEE Course Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper.
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1. Rate of a reaction A + B →  products, is given below as a function of different initial concentrations of A and B :         (1982 - 4 Marks)

[A] (mol/l)[B] (mol/l)Initial rate (mol/l/min)
0.010.010.005
0.020.010.010
0.010.020.005

 

Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the reaction?

Solution: 

From data (i) and (ii) it is obvious that when the concentration of B is kept constant (0.01 mol litre–1) and the concentration of A is doubled (0.01 to 0.02 mol litre–1), the rate of reaction is also doubled (0.005 to 0.010 mol litre–1 min–1). This shows that the rate of reaction varies directly as the first power of the concentration. Hence the order of reaction with respect to A is 1.

Similarly, from data (i) and (iii) it is obvious that when the concentration of A is kept constant (0.01 mol litre–1) and the concentration of B doubled (0.01 to 0.02 mol litre–1), the rate of reaction remains constant (0.005 mol litre–1 min–1). This shows that the order of reaction with respect to B is zero. Now we know that the rate of reaction, A + B ® Products, is given by
Rate r = k [A]1[B]0 ⇒ r = k [A]
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


2. Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half life of 5770 years. What is the rate constant (in years–1) for the decay? What fraction would remain after 11540 years?                 (1984 - 3 Marks)

Solution: 
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

 Half life, t1/2 = 5770 years

Let the original sample be 1 gram.

∴ After every 5770 years one-half of radioactive carbon would decay or disintegrate.

Thus, 1 g sample becomes 1⁄2 g after 5770 years and
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

∴ 25% of radioactive carbon remains after 11540 years. Rate constant, k for first order reaction,
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


3. While studying the decomposition of gaseous N2O5 it is observed that a plot of logarithm of its partial pressure versus time is linear. What kinetic parameters can be obtained from this observation? (1985 - 2 Marks)

Solution: 

 Assuming that the decomposition of N2O5 is a first order reaction, then
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Thus log P vs time graph is linear with slope = -k/2.303 if the given reaction is of first order which is in accordance with the given statement. Thus the reaction obeys first order reaction.


4.   Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev disintegrates to give   Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRevas the final product. How many alpha and beta particles are emitted during this process?          (1986 - 2 Marks)

Solution: 
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

NOTE : For emission of one a-particle, atomic mass decreases by 4 and atomic number by 2. Further for the emission of one b-particle, the atomic mass does not change but the atomic number increases by 1. So we first find the a-particles :

Decrease in atomic mass = 234 – 206 = 28
No. of a-particles emitted = 28/4 = 7

Hence, atomic number should have decrease to

90 – (7 × 2) = 76

Now, atomic number of Pb = 82,

which is more by (82 – 76) = 6

This increase is due to ionisation of β-particles.

Therefore, β-particles emitted = 6.

 

5. A first order reaction has K = 1.5 × 10–6 per second at 200ºC. If the reaction is allowed to run for 10 hours, what percentage of the initial concentration would have changed in the product? What is the half life of this reaction?                (1987 - 5 Marks)

Solution: 
For a first order reaction we know that
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Thus, 5.2% of the initial concentration has changed into product.

Again we know that

 Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


6. A first order reaction is 50% complete in 30 minutes at 27ºC and in 10 minutes at 47ºC. Calculate the reaction rate constant at 27ºC and the energy of activation of the reaction in kJ/mole.                  (1988 - 3 Marks)

Solution: 
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

7. An experiment requires minimum beta activity product at the rate of 346 beta particles per minute. The half life period of  Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev , which is a beta emitter is 66.6 hours. Find the minimum amount of  Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev required to carry out the experiment in 6.909 hours.             (1989 - 5 Marks)

Solution: 

Minimum number of b-particles required in one minute = 346

No. of b-particles required for carrying out the experiment for 6.909 × 60 minutes = 346 × 6.909 × 60 = 143431

∴ Amount of b-particles required
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


8. In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 1013 sec–1 and 98.6 kJ mol–1 respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes?                      (1990 - 3 Marks)

Solution: 
According to Arrhenius equation

 Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

Substituting the various values in the above equation, we get
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


9. The decomposition of N2O5 according to the equation :      (1991 - 6 Marks)
  2N2O5(g) → 4NO2(g) + O2(g)
  is a first order reaction. After 30 min. from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm of Hg and on complete decomposition, the total pressure is 584.5 mm of Hg. Calculate the rate constant of the reaction.

Solution: 

2N2O5(g) ® 4NO2(g) + O2(g)

2 mol of gaseous nitrogen pentoxide on complete decomposition gives 5 mol of gaseous products.
Therefore, initial pressure of N2O5 = 584.5 × 2/5
= 233.8 mm Hg.
Let x be the amount of N2O5 decomposed after 30 min.

∴ After 30 min.

 Pressure due to N2O= 233.8 – x ; Pressure due to NO2 = 2x

and pressure due to O= x/2

Total pressure after 30 min
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Hence pressure of N2O5 after 30 min.
= 233.8 – 33.8 = 200 mm Hg
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev


10. Two reactions (i) A → products, (ii) B → products, follows first order kinetics. The rate of the reaction : (i) is doubled when the temperature is raised from 300K to 310K. The half life for this reaction at 310K is 30 minutes. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction, (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300K.                    (1992 - 3 Marks)

Solution: 

Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev
Subjective Qns. of Chemical Kinetics & Nuclear Chemistry, Past year Qns. (Part - 1) - JEE Advance JEE Notes | EduRev

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