For an elementary reaction the variation of rate constant (k) with temperature is given by the following equation.
$\log_{10} k = 5.4 - \frac{100}{T}$ where 'T' is temperature on Kelvin scale and k is in terms of $\sec^{- 1}$ . Identify the incorrect option.

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Fraction of activated molecules will be $e^{- 100 / T}$ at any temperature.

Activation energy for the reaction will be approx. 460.6 $cal$

Rate of reaction will vary linearly with concentration of reactant.

There is no finite temperature at which rate constant can be $4 \times 10^{6} \sec^{- 1}$

answer is B.

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Detailed Solution

$A = 10^{5.4} = 2.5 \times 10^{5} \sec^{- 1}$

$- \frac{E_{a}}{2.303 R} = - 100$

$\frac{k}{A} = e^{- \frac{E_{a}}{R T}}$

$E_{a} = 460.6$

Option A, C, and is a correct option and option B is a incorrect option for these equation
$\log_{10} k = 5.4 - \frac{100}{T}$ .

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