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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 = 8.3Jk^–1mol^–1) (Antilog 0.1667 = 1.468)

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1.25

1.75

1.5

2.0

answer is C.

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Detailed Solution

$\begin{array}{l} \log \frac{k_{2}}{K_{1}} = \frac{Ea}{2 .303 \times R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}] \\ = \frac{58 .3}{2 .303 \times 8 .3 \times 10^{- 3}} [\frac{5}{300 \times 305}] = 0 .1672 \\ \frac{K_{2}}{K_{1}} ≅ 1 .5 \end{array}$

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