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JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Mock Test - Chemical Kinetics

  • question_answer
    For the reaction, \[P\text{ }+\text{ }Q+\text{ }R\to S\] Experimental data for the measured initial rates in given below.
    Expt. (P) (Q) (R) Initial rate (in \[m{{s}^{-1}}\])
    1. 0.2M 0.5M 0.4M \[8.0\times {{10}^{-5}}\]
    2. 0.4M 0.5M 0.4M \[3.2\times {{10}^{-4}}\]
    3. 0.4M 2.0M 0.4M \[1.28\times {{10}^{-3}}\]
    4. 0.1M 0.2M 1.6M \[4.0\times {{10}^{-8}}\]
    The order of the reaction with respect of P, Q and R respectively is

    A) 4, 1, 3  

    B) 2, 2, 1

    C) 1, 2, 1  

    D) 2, 1, 1

    Correct Answer: D

    Solution :

    [d] By rate law, Rate \[=k{{[P]}^{x}}{{[Q]}^{y}}{{[R]}^{z}}\]
    x=order w.r.t P
    y=order w.r.t. Q
    z=order w.r.t. R
    \[8.0\times {{10}^{-5}}=k{{(0.2)}^{x}}{{(0.5)}^{y}}{{(0.4)}^{z}}\]
    \[3.2\times {{10}^{-4}}=k{{(0.4)}^{x}}{{(0.5)}^{y}}{{(0.4)}^{z}}\]
    \[1.28\times {{10}^{-3}}=k{{(0.4)}^{x}}{{(2.0)}^{y}}{{(0.4)}^{z}}\]
    \[4.0\times {{10}^{-5}}=k{{(0.1)}^{x}}{{(0.25)}^{y}}{{(1.6)}^{z}}\]
    From (1) and (2), \[\frac{3.2\times {{10}^{-4}}}{8.0\times {{10}^{-5}}}={{(2)}^{x}}\]
    \[(4)={{(2)}^{x}}\Rightarrow x=2\]
    From (2) and (3), \[\frac{1.28\times {{10}^{-3}}}{3.2\times {{10}^{-4}}}={{(4)}^{y}}\]
    \[(4)={{(4)}^{y}}\Rightarrow y=1\]
    From (1) and (4),
    \[\frac{8\times {{10}^{-5}}}{4\times {{10}^{-5}}}={{(2)}^{x}}{{(2)}^{y}}{{\left( \frac{1}{4} \right)}^{z}}=(4)(2){{\left( \frac{1}{4} \right)}^{z}}\]
    \[\Rightarrow z=1\]
    Thus, order w.r.t. P=2
    w.r.t. Q=1
    w.r.t. R=1


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