If $60 %$ of a first order reaction was completed in $60$ minute, $50 %$ of the same reaction would be completed in approximately?

46

minute

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60

minute

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40

minute

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50

minute

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Solution

The correct option is A $46$ minute
Its a 1st Order reaction.

$k = \frac{2.303}{60} l o g \frac{100}{40}$

$k = 0.015 {m i n}^{- 1}$

$t = \frac{2.303}{0.015} l o g \frac{100}{50}$

$t = 46 m i n$

Hence, the correct option is $A$

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If 60% of a first order reaction was completed in 60 minute, 50% of the same reaction would be completed in approximately?

The correct option is A 46 minuteIts a 1st Order reaction.k=2.30360log10040k=0.015 min−1t=2.3030.015log10050t=46 minHence, the correct option is A

If $60 %$ of a first order reaction was completed in $60$ minute, $50 %$ of the same reaction would be completed in approximately?

The correct option is A $46$ minute
Its a 1st Order reaction.

$k = \frac{2.303}{60} l o g \frac{100}{40}$

$k = 0.015 {m i n}^{- 1}$

$t = \frac{2.303}{0.015} l o g \frac{100}{50}$

$t = 46 m i n$

Hence, the correct option is $A$