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Solve · Q8

Q.viii. The rate constant for the first order reaction is given by log10_{10} kk = 14.34 - 1.25 ×\times 104^4 T. Calculate activation energy of the reaction. (239.3 kJ/mol)
[!NOTE]
The book prints the expression exactly as shown -- "1.25 ×\times 104^4 T", with no division -- which is dimensionally impossible and inconsistent with the printed answer: 239.3 kJ/mol requires the Arrhenius form log10_{10} kk = 14.34 - 1.25 ×\times 104^4/T. The solution uses the /T form.

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Step 1. Given log⁡10k=14.34−1.25×104T\log_{10}k=14.34-\dfrac{1.25\times10^4}{T}, compare with the Arrhenius straight-line form log⁡10k=−Ea2.303R⋅1T+log⁡10A\log_{10}k=-\dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}+\log_{10}A.

Step 2. Matching the coefficient of 1/T1/T (the slope): −Ea2.303R=−1.25×104-\dfrac{E_a}{2.303R}=-1.25\times10^4, so Ea2.303R=1.25×104\dfrac{E_a}{2.303R}=1.25\times10^4. …

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