VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    For a first order reaction at \[27{}^\circ C\], the ratio of time required for 75% completion to 25% completion of reaction is

    A)  \[{{H}_{3}}C-C{{H}_{2}}-C\equiv CH\]  

    B)  \[{{H}_{2}}C=CH-C\equiv CH\]

    C)  \[HC\equiv C-C{{H}_{2}}-C\equiv CH\]  

    D)  \[NaOH\]

    Correct Answer: C

    Solution :

     For a first order reaction, \[t=\frac{2.303}{\lambda }{{\log }_{10}}\frac{a}{a-x}\] Let initial amount of reactant is 100. \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\log \frac{100}{100-75}}{\log \frac{100}{100-25}}\]  [\[\because \]\[\lambda \]remains constant] \[=\frac{\log \frac{100}{25}}{\log \frac{100}{75}}=\frac{\log 4}{\log 4/3}\] \[=\frac{log4}{\log 4-\log 3}\] \[=\frac{0.6020}{0.1249}=4.81\]


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