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(D) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_a}{RT}$.Converting to base $10$,we get $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 R

Vedclass · Accepted Answer

$36.8$

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(D) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_a}{RT}$.Converting to base $10$,we get $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$.Comparing this with the given equation $\log_{10} k = -2000 \left(\frac{1}{T}ight) + 6$,we identify the slope as $-\frac{E_a}{2.303 R} = -2000$.Thus,$E_a = 2000 	imes 2.303 	imes R$.Using $R = 8 \, J \, mol^{-1} K^{-1}$ and the approximation $\ln x \approx 2.3 \log_{10} x$ provided in the question:$E_a = 2000 	imes 2.3 	imes 8 = 36800 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$,we get $E_a = \frac{36800}{1000} = 36.8 \, kJ \, mol^{-1}$.

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