Question:medium

For a first order reaction, rate constants at \(50^\circ\mathrm{C}\) and \(100^\circ\mathrm{C}\) are \(1.5\times10^{-5}\,\mathrm{s^{-1}}\) and \(4.5\times10^{-5}\,\mathrm{s^{-1}}\), respectively. What is the approximate activation energy of the reaction (in \(\mathrm{kJ\,mol^{-1}}\))? \[ (\log3=0.48,\;R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}) \]

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For two temperatures, \[ \boxed{ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac1{T_1}-\frac1{T_2} \right). } \] Always convert temperatures into Kelvin before substitution.
Updated On: Jul 18, 2026
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The Correct Option is A

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