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Chemical Kinetics Test - 5

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Chemical Kinetics Test - 5
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  • Question 1
    4 / -1

    Q. For zeroth order reaction,

    A → B

    [A]0 = 0.01 M, [A]t = 0.008Matter 10 min.

    Thus, half-life is

     

    Solution

     

     

    For zeroth order reaction

    When reaction is 50% completed. 

     

     

  • Question 2
    4 / -1

    A Certain Zeroth Order reaction has k = 0.025 Ms-1 for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?

    Solution

    x(product) formed after 15 s
    = 0.025 Ms-1 x 15s
    = 0.375 M
    Then, reactant left
    = 0.500 - 0.375
    = 0.125 M

  • Question 3
    4 / -1

    For Zeroth Order Reaction, variation of x with time is shown as 

    Q. At initial concenration of the reactant as 17.32 min dm-1 , half - life period is 

    Solution

  • Question 4
    4 / -1

    For the first order reaction, rate constant k is given by

    kt = logC0 - log Ct

    where, C0 is the intial concenration and Ct is the concentration at time t. Graph between log Ct and time t is

    Solution



    This represents a straight line

  • Question 5
    4 / -1

    In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M is 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

    Solution

    For first order reaction,

    time = 15 min to change its concentration from 0.8 m to 0.4 m Thus, 50% reaction.
    Thus,  t50 = 15 min



  • Question 6
    4 / -1

    For the First order reaction, concentration of the reactant after two averagelives is reduced to

    Solution

    For first order reaction,




  • Question 7
    4 / -1

    The half-life period of a first order chemical reaction is 6.93 min. The time required for the completion of 99% of the chemical reaction will be (log2 = 0.3010)

    [AIEEE 2009]

    Solution



    ∴  

    when, a = 100,

    (a-x) = 100-99 = 1, t = t99

  • Question 8
    4 / -1

    For the first order reaction,

    A → Product

    Q. The concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of the reaction when the concentration of A is 0.01 M is

    [AIEEE 2012]

    Solution

    For the first order kinetic, 



  • Question 9
    4 / -1

    For the reaction,

    A → Products,      

    Concentration of A at different time intervals are given :

    At 20 min, rate will be

    Solution





  • Question 10
    4 / -1

    For the first order reaction with, C0 as the initial concentration and C at time  t, (C0 - C) =

    Solution




  • Question 11
    4 / -1

    The radioisotope N-13 which has a half-life of 10 min is used to image organs in the body. If the injected sample has an activity of 40 μCi (microcurie), what is the activity after 30 min?

    Solution

    Disintegration follows first-order kinetics
    t = total time = 30 min
    t 50 = half-life = 10 min


  • Question 12
    4 / -1

    The following data were obtained during the first order decomposition of

    2A(g) → B(g) + C(s)

    at constant volume and at a particular temperature : 

    Thus, rate constant is

    Solution







  • Question 13
    4 / -1

    A G .M. Counter is used to study the process of first order of radioactive disintegration. In the absence of radioactive substance A, it counts 3 dps (disintegration per second). When A is placed in G.M Counter it records 23 dps at the start and 13 dps after 10 min. It records x dps after next 10 min. Half-life period of A is y min. Thus, x and y are

    Solution

    When there is no radioactive substance in G.M. Counter, it records = 3 dps. Thus, it is the zero-error in the counter and in each it is added. Thus, initial counts = 23 - 3 = 20 dps.
    Attimef (after 10min) = 13- 3 = 10dps
    Thus, half-life is 10 min = y
    Hence, after next 10 min, dps = 5
    Hence, reading in G.M. Counter = 5 + 3 = 8 dps = x

  • Question 14
    4 / -1

    Inversion of sucrose (C12H22O11) is a first order reaction and is studied by measuring angle of rotation at different interval of time

    r0 = angle of rotation at the start, rt = angle of rotation at time t
    r∞ = angle of rotation at the complete reaction

    There is 50% inversion, when

    Solution

    r0 = rotation at the start, rt = rotation at time t
    r∞ = rotation at the end then (r∞ - r0) = a and (r∞ - rt) = (a - x)

    When 50% inversion has taken place


    ∴  2 (a - x) = a
    2(r∞ - rt) = r∞ - r0
    2r∞ - 2rt = r∞ - r0
    ∴  r0 = (2rt - r∞)

  • Question 15
    4 / -1

    Following radioactive disintegration follows first order kinetics :

    Starting with 1 mol of A in a 1L flask at 300 K, pressure set up after two half-lifes is approximately

    Solution

    When B is formed from A, two a-particies (He) are formed.
    A taken initially = 1 mol

    This helium remains in closed vessel.
    Thus, pressure set up is due to helium.

  • Question 16
    4 / -1

    For the first order reaction Tav (average life), T50 (half-life) and T75 (time 75% reaction) are in the ratio of

    Solution




  • Question 17
    4 / -1

    For the first order reaction,


    Following equations are given :



    Select the correct equations.

    Solution

    For the first order reaction,










    ∴ (a-x) = a (1/2)y    (Equation II)

  • Question 18
    4 / -1

    Rate constant of a first order reaction (A → B) is 0.0693 min-1. If initial concentration is 1.0 M, rate after 30 min is

    Solution

    C0 = 1 M



    ∴  

  • Question 19
    4 / -1

    Following reaction is pseudo-unimolecular w.r.t. C6H5N2CI (A)

    50 mL of 1M benzene diazonium chloride (A) is taken. After 1 h, 1.226 L of N2 gas at 1 atm and 300 K is obtained. Thus, half-life of the reaction is (log 250 = 2.40) 

    Solution






    Half-life(t50)= 
     

  • Question 20
    4 / -1

    Minimum half-life of an isotope needed so that not more than 0.1 % of the nuclei undergo decay during a 3.0 h laboratory experiment is

    Solution



  • Question 21
    4 / -1

    Naturally occurring potassium consists of 0.01% at 40 K which has a half-life of 1.28 x 109 yr. Activity of 1.00 g sample of KCl is

    Solution


    Number of 40 K atoms in 1g KCI


  • Question 22
    4 / -1

    Direction (Q. Nos. 22-27) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d).

    Passage l

    Isotope of oxygen with mass number less than 16 undergoes β+ -emission. Assume an equimolar mixture of 14O and 15O, Also t50 (14O) = 71s, t50 (15O) = 124s.

    Q. Ratio of rate constants of 14O to 15O is

    Solution





  • Question 23
    4 / -1

    Passage l

    Isotope of oxygen with mass number less than 16 undergoes β+ -emission. Assume an equimolar mixture of 14O and 15O, Also t50 (14O) = 71s, t50 (15O) = 124s.

    Q. Ratio of 14O to 15O left after 1.00 h is

    Solution





  • Question 24
    4 / -1

    Passage II

    Consider acid-hydrolysis of ester which is first order reaction.

    CH3COOC2H5 + H2O  CH3COOH + C2H5OH

    Kinetics is studied by titration of the acid with NaOH at a given time.
    V0 = Volume of NaOH at start, Vt = Volume of NaOH at time t, V∞ = Volume of NaOH at the end of the complete hydrolysis.

    Q. 
    Rate constant k is given by

    Solution



    At time t = 0, volume of NaOH is due to H+ (catalyst) = V0 and remains constant
    at time t, volume of NaOH is due to H+ and CH3COOH = Vt


    At infinite time (complete reaction), volume of NaOH is due to total CH3COOH (or ester) and H+ = a + V0





    If ester is 50% hydrolysed, 



  • Question 25
    4 / -1

    Passage II

    Consider acid-hydrolysis of ester which is first order reaction.

    CH3COOC2H5 + H2O  CH3COOH + C2H5OH

    Kinetics is studied by titration of the acid with NaOH at a given time.
    V0 = Volume of NaOH at start, Vt = Volume of NaOH at time t, V∞ = Volume of NaOH at the end of the complete hydrolysis.

    Q.

    If ester is 50% hydrolysed, then 

    Solution



    At time t = 0, volume of NaOH is due to H+ (catalyst) = V0 and remains constant
    at time t, volume of NaOH is due to H+ and CH3COOH = Vt


    At infinite time (complete reaction), volume of NaOH is due to total CH3COOH (or ester) and H+ = a + V0





    If ester is 50% hydrolysed, 



  • Question 26
    4 / -1

    Passage III

    I, II and III are three vessels of equal volumes containing molecules of A as shown and in each A decomposes into products.

    Q. 
    Initial rate of decomposition of A will be in order in these three vessels as

    Solution


    Thus, ratio fo rates = 6 : 1 2 : 9 = 1 : 2 : 1 5
    For first order reaction T50 is independent of initial concentration.
    Thus, T50 will be equal to 1 :1 :1.

  • Question 27
    4 / -1

    Passage III

    I, II and III are three vessels of equal volumes containing molecules of A as shown and in each A decomposes into products.

    Q. 
    Time of 50% decomposition is in order

    Solution


    Thus, ratio fo rates = 6 : 1 2 : 9 = 1 : 2 : 1 5
    For first order reaction T50 is independent of initial concentration.
    Thus, T50 will be equal to 1 :1 :1.

  • Question 28
    4 / -1

     

    The rate of a chemical reaction tells us about

  • Question 29
    4 / -1

     

    In the rate equation, when the concentration of reactants is unity then the rate is equal to

  • Question 30
    4 / -1

    The average rate and instantaneous rate of a reaction are equal

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