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  • For the first-order reaction A → 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half-life of the reaction is ...... min. (Round off to the nearest integer). [ Use : ln 2 = 0.69, ln 10 = 2.3, ln 9 =0.95  <br

For the first-order reaction A → 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half-life of the reaction is ...... min. (Round off to the nearest integer).
[ Use : ln 2 = 0.69, ln 10 = 2.3, ln 9 =0.95  
Properties of logarithms: \ln x^{y}=y \ln x;
                                    \left.\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\ln \mathrm{x}-\ln \mathrm{y}\right]
(Round off to the nearest integer)
 

Answers (1)

best_answer

For the given first-order reaction,

\begin{matrix} & \mathrm{A }& \xrightarrow[]{ \ \quad \ } & \mathrm{2B} \\ t=0 & 1 & & 0\\ t=100 & (1-x) & & 2x \end{matrix}

Given, 2 x=0.2 \Rightarrow x=0.1

Now, for a first-order reaction, 

            \ln \left(\frac{a_{0}}{a_{t}}\right)=k t

\Rightarrow \quad \ln \left(\frac{1}{0.9}\right)=k(100)

\Rightarrow \quad 2.3(\log 10-\log 9)=100 k

\Rightarrow k=\frac{2.3(1-0.95)}{100}=1.15 \times 10^{-3}

\therefore\ t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.69}{1.15 \times 10^{-3}}=600\ min

Hence, the correct answer is 600

Posted by

sudhir.kumar

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