Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer The half-life period of a first order process is 1.6 min. It will be 90% complete in

    A) 5.3 min                

    B) 10.6 min

    C) 43.3mm         

    D)        99.7 min

    Correct Answer: A

    Solution :

    For a first order reaction                 \[k=\frac{2.303}{t}\log \frac{a}{a-x}\] At half-time,                 \[x=\frac{a}{2}\] \[\therefore \]  \[k=\frac{2.303}{1.6}\times \log \frac{a}{a-\frac{a}{2}}\]                    \[=\frac{2.303}{1.6}\times \log 2\]                            \[=\frac{2.303}{1.6}\times 0.3010\]                    \[=0.4332\] For 90% completion,                 \[x=0.9a\] \[\therefore \]  \[k=\frac{2.303}{t}\log \frac{a}{a-0.9a}\]      \[0.4332=\frac{2.303}{t}\log \frac{a}{0.1a}\]                    \[=\frac{2.303}{t}\log 10\]                    \[=\frac{2.303}{t}\times 1\]                 \[t=\frac{2.303}{0.4332}\]                   \[=5.3\,\,\min \]


You need to login to perform this action.
You will be redirected in 3 sec spinner