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  • For a certain first order reaction of the recant is left after <img alt="560 \mathrm{~s}" src="https://entrance

For a certain first order reaction 36 \% of the recant is left after 560 \mathrm{~s}. The rate constant of the reaction will be

Option: 1

\mathrm{3 \times 10^{-6} \mathrm{~s}^{-1}}


Option: 2

\mathrm{5 \times 10^{-7} s^{-1}}


Option: 3

\mathrm{9 \times 10^{-4} \mathrm{~s}^{-1}}


Option: 4

\mathrm{8 \times 10^{-4} \mathrm{~s}^{-1}}


Answers (1)

best_answer

for first order reaction

\mathrm{ k \times t=\ln \frac{A_0}{A_t} }\, \, \, \, ----(a)

\mathrm{\text { here } A_0=100 \text { { Initial concentration\} }}

\mathrm{A t=36 \quad\{\text { conc of reactant after } t\}}
By eq (a)
 \mathrm{ \begin{aligned} & k \times 560=\ln \frac{100}{64} \\\\ & k \times 560=2.303\left(\log 10^2-\log 2^6\right) \\\\ & k=\frac{2.303}{560}(2-6 \log 2) \\\\ & k=\frac{2.303}{560}[2-(6 \times 0.3)] \end{aligned} }

\mathrm{\begin{aligned} & k=\frac{2.303}{560} \times 0.2 \\\\ & k=8 \times 10^{-4} \mathrm{~s}^{-1} \end{aligned}}

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Anam Khan

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