A first order reaction has a specific reaction rate of 10-2 sec-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

  • Option 1)

    238.6 sec

  • Option 2)

    138.6 sec

  • Option 3)

    346.5 sec

  • Option 4)

    693.0 sec

 

Answers (1)

As discussed in:

First Order Reaction -

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

- 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}

 

 

\frac{t_{1}}{2} \:\:of \:\:the \:\:reaction=\frac{0.693}{K}

\frac{0.693}{10^{-2}}sec=69.3\:sec

A=\frac{A_{o}}{2^{t/ t\frac{1}{2}}}

A_{o}= Initial concentration

A= Final concentration

\Rightarrow 2^{t/ t\frac{1}{2}}= \frac{A_{o}}{A}= \frac{20g}{5g}=4=2^{2}

\therefore \frac{t}{t \frac{1}{2}}=2\:\:\:\:or\:\:\:\:t=2t \frac{1}{2}=138.6 sec

 

 


Option 1)

238.6 sec

This option is incorrect.

Option 2)

138.6 sec

This option is correct.

Option 3)

346.5 sec

This option is incorrect.

Option 4)

693.0 sec

This option is incorrect.

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