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For a first order reaction, A \rightarrow P , t_{\frac{1}2} (half-
life) is 10 days. The time required for \frac{1}{4}^{th}converstion of A (in days) is :

(ln 2=0.693, ln 3=1.1)

  • Option 1)

    5

  • Option 2)

    3.2

  • Option 3)

    4.1

  • Option 4)

    2.5

 

Answers (2)

best_answer

As we have leant,

 

First Order Reaction -

The rate of the reaction is proportional to the first power of the concentration of the reactant.

- wherein

Formula:

R    \rightarrow        P

a                 0

a-x             x

rate[r]=K[R]^{1}

\frac{-d(a-x)}{dt}=K(a-x)

\frac{-dx}{dt}=K(a-x)  [differentiate rate law]

ln \:[\frac{a}{a-x}]=kt \:(Integrated rate law)

Unit of k=sec^{-1}

t_\frac{1}{2}=\frac{0.693}{k}

 

 

 \\*K = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{10} \\* K = \frac{2.303}{t}\log \frac{R_{o}}{R_{t}} \\* \frac{0.693}{10} = \frac{2.303}{t}\log \frac{R_{o}\times 4}{3R_{0}} \\*\frac{0.693}{10} = \frac{2.303}{t} \left [\log 4 - \log 3 \right ] = \frac{2.303}{t}\left[0.6020 - 0.4771 \right ] \\*\frac{0.693}{10} = \frac{2.303}{t}\times 0.1249 \\* t = \frac{2.303\times0.1249\times10}{0.693} = 4.15 days

 


Option 1)

5

Option 2)

3.2

Option 3)

4.1

Option 4)

2.5

Posted by

Aadil

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