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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)

best_answer

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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Aadil

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