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JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Rate law and Rate Constant

  • question_answer
    In a first order reaction the concentration of reactant decreases from \[800\,mol/d{{m}^{3}}\] to \[50\,mol/d{{m}^{3}}\] is \[2\times {{10}^{2}}\sec \]. The rate constant of reaction in \[{{\sec }^{-1}}\] is [IIT-JEE (Screening) 2003]

    A)                 \[2\times {{10}^{4}}\]   

    B)                 \[3.45\times {{10}^{-5}}\]

    C)                 \[1.386\times {{10}^{-2}}\]         

    D)                 \[2\times {{10}^{-4}}\]

    Correct Answer: C

    Solution :

               \[k=\frac{2.303}{t}{{\log }_{10}}\frac{a}{a-x}\]; \[t=2\times {{10}^{2}},\,\,a=800,\,\,a-x=50\]                    \[k=\frac{2.303}{2\times {{10}^{2}}}{{\log }_{10}}\frac{800}{50}=\frac{2.303}{2\times {{10}^{2}}}{{\log }_{10}}16\]                       \[=\frac{2.303}{2\times {{10}^{2}}}{{\log }_{10}}{{2}^{4}}=\frac{2.303}{2\times {{10}^{4}}}\times 4\times 0.301\]                          \[=1.38\times {{10}^{-2}}\,\,{{s}^{-1}}\]


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