Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer
    The reaction \[A\xrightarrow{{}}B\] follows first order kinetics. The time taken for \[0.8\text{ }mole\]of A to produce 0.6 mole of B is\[1\text{ }h\]. What is the time taken for conversion of \[0.9\text{ }mole\]of A to produce  \[0.675\text{ }mole\]of B?

    A)  \[1h\]

    B)  \[0.5\text{ }h\]

    C)  \[0.25\text{ }h\]                             

    D)  \[2\text{ }h\]

    Correct Answer: A

    Solution :

    Kinetics for first order reaction is \[k=\frac{2.303}{t}\log \frac{a}{(a-x)}\]                 \[\Rightarrow \]               \[k=\frac{2.303}{1}\log \frac{0.8}{0.2}\]                 \[\Rightarrow \]               \[k=\frac{2.303}{1}\log 4\]                 \[\therefore \]  \[k=2.303\times 0.6020\]              ?..(i) Again, \[\Rightarrow \]               \[k=\frac{2.303}{t}\log \frac{0.9}{0.225}\] \[\Rightarrow \]               \[k=\frac{2.303}{t}\log 4\] \[\therefore \]  \[k=\frac{2.303}{t}\times 0.6020\]            ??(ii) From Eqs (i) and (ii) \[\Rightarrow \]               \[2.303\times 0.6020=\frac{2.303}{t}\times 0.6020\] \[\therefore \]  \[t=1h\]


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