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  • The rate constant of a first order reaction, <img alt="\mathrm{\mathrm{X} \rightarrow \mathrm{Y}, is \; 1.6 \times 10^{-3} \mathrm{~s}^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

The rate constant of a first order reaction, \mathrm{\mathrm{X} \rightarrow \mathrm{Y}, is \; 1.6 \times 10^{-3} \mathrm{~s}^{-1}} at 300 K. Given that the activation energy of the reaction is 28 \mathrm{~kJ} \mathrm{~mol}^{-1} and assuming Arrhenius behavior for the temperature dependence, the total time required to obtain \mathrm{90 % \; of \; \mathrm{Y} \; at \; 350 \mathrm{~K}} is in sec.

\mathrm{(Use\; \mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ).}

Option: 1

2.93 \times 10^2 \mathrm{sec}


Option: 2

4.93 \times 10^2 \mathrm{sec}


Option: 3

3.93 \times 10^2 \mathrm{sec}


Option: 4

5.93 \times 10^2 \mathrm{sec}


Answers (1)

best_answer

\begin{aligned} & \mathrm{\ln \left(k_2 / k_1\right)=\frac{E_q}{8.314}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)} \\ \\& \mathrm{\ln \left(\frac{k_2}{1.6 \times 10^3}\right)=\frac{28 \times 10^3}{8.314}\left(\frac{1}{300}-\frac{1}{300}\right)} \\ \\&\mathrm{ \left.k_2=7.86 \times 10^{-3} \mathrm{~s}\right\}} \\ \\& \mathrm{7.86 \times 10^{-3}=\frac{2.303}{t} \log \left\{\frac{100}{10}\right\}} \\ & \mathrm{t=2.927 \times 10^2 \mathrm{sec}} \end{aligned}

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