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Chemical Kinetics

    • Question 1
    CBSEENCH12006140
    Wired Faculty App

    For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M is 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

    Solution
    Given that 
    Initial concentration, [R1] = 0.03
    Final concentration, [R2] = 0.02
    Time taken ∆t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )
    The formula of average rate of change 
    rav =-∆R∆t =∆[P]∆t

    (i) Average rate
                         = (0.03 - 0.02) M25 × 60 sec= 0.01 M25×60 s = 6.66 M s-1
    (ii) Average rate
                            = (0.03-0.02) M25 min =  0.01 M25= 0.0004 Ms-1.
    Question 2
    CBSEENCH12006141
    Wired Faculty App

    In a reaction 2A → Products, the concentration of A decreases from 0.5 mol to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval.

    Solution
    Given that 
    Initial concentration [A1] =0.5
    Final concentration [A2] =0.4
    Time is  = 10 min

    Rate of reaction = Rate of disappearance of A.
    Rate of reaction = -12∆[A]∆t
                             = - 12(0.4-0.5) mol L-110 minute= 0.1 mol L-15 minutes= 0.005 mol litre-1 min-1.
    Question 3
    CBSEENCH12006142
    Wired Faculty App

    For a reaction, A + B → Product; the rate law is given by, r = k[ A]1/2 [B]2. What is the order of reaction?

    Solution
    The order of the reaction is sum of the powers on concentration.
    So that sum will 

    r = k[A]    1/2[B]2

    Order of reaction = 12+2 = 2.5.
    Question 4
    CBSEENCH12006143
    Wired Faculty App

    The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

    Solution
    Let the reaction is X →Y

    This reaction follows second order kinetics.
    So that, the rate equation for this reaction will 
    Rate, R = k[X]2 .............(1)
    Let initial concentration is x mol L−1,
    Plug the value in equation (1)
    Rate, R1 = k .(a)2
    = ka2
    Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
    Plug the value in equation (1) we get 
    Rate, R2 = k (3a)2
    = 9ka2
    We have already get that R1 = ka2 plus this value we get
    R2 = 9 R1 
    So that, the rate of formation will increase by 9 times.
    Rate = k[A]2
    If concentration of X is increased to three times,
    Rate = k[3A]2
    or Rate = 9 k A2
    Thus, rate will increase 9 times.

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