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Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    Time required to decompose\[S{{O}_{2}}C{{l}_{2}}\]to half of its initial amount is 60 min. If the decomposition is a first order reaction, what is the rate constant of the reaction?

    A)  \[1.925\times {{10}^{-4}}s\]

    B)  \[2.092\times {{10}^{-4}}{{s}^{-1}}\]

    C)  \[1.925\times {{10}^{-4}}{{s}^{-1}}\]

    D)  \[2.925\times {{10}^{-4}}{{s}^{-1}}\]

    Correct Answer: C

    Solution :

     \[{{t}_{1/2}}=60\text{ }min=60\times 60\text{ }s\] \[k=\frac{0.693}{{{t}_{1/2}}}\] \[=\frac{0.693}{60\times 60\,s}\] \[=1.925\times {{10}^{-4}}{{s}^{-1}}\]


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