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  • For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given: <img alt="\mathrm{ \ln 10 =

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given: \mathrm{ \ln 10 = 2.303 \text{ and} \log 2 = 0.3010})

Option: 1

1.12


Option: 2

2.43


Option: 3

3.32


Option: 4

33.31


Answers (1)

best_answer

First order Reaction,

\mathrm{\ln \frac{[A]}{[A]_{0}}=\text{-kt}, \mathrm{t_{1 / 2}}=\frac{\ln 2}{K}}

Now,

90 \% \text { Reaction completed}\Rightarrow [A]=\frac{10}{100}\mathrm{[A]_{0} \Rightarrow[A]=0.1[A]_{0}}

Putting this in the intgrated rate equation, we have 

\ln \left(\frac{0 . 1[A]_{0}}{[A]_{0}}\right)=\mathrm{-k t_{90 %}}

So, \operatorname{t} _{90 \%}=\frac{\ln \left(\frac{10}{1}\right)}{\text{k}}=\mathrm{x \times t_{50\%}=x \frac{\ln 2}{k}}

\Rightarrow \quad \text{x}=\frac{\ln \left(\frac{10}{1}\right)}{\ln 2}=\frac{\ln 10}{\ln 2}=3.32

Hence, the correct answer is Option (3)

Posted by

Pankaj

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