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The activation energy of a reaction is 58.3 kJ/mole. The ratio of the rate constants at 305K and 300K is about $(R = 8.3 J K^{- 1} m o l^{- 1}) (A n t i \log 0.1.468)$

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1.25

1.5

1.75

2.0

answer is B.

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Detailed Solution

$L o g (\frac{k_{2}}{k_{1}}) = \frac{E a}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} \times T_{2}}]$

$L o g (\frac{K_{2}}{k_{1}}) = \frac{58.3}{2.303 \times 8.314 \times 10^{- 3}} [\frac{305 - 300}{300 \times 305}]$

$\frac{k_{2}}{k_{1}} = 1.5$

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