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Arrhenius Equation

Rate of a fir...

Question

Rate of a first order reaction becomes 4 times when temperature (in K) is doubled (assuming the initial concetration same for both cases) having almost constant activation energy of 1.386 kcal. Find the value of $T_{1}$ (in K)?

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Solution

$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$\Rightarrow l n 4 = \frac{1386}{2} (\frac{1}{T_{1}} - \frac{1}{2 T_{1}})$
$\Rightarrow T_{1} = \frac{1386}{0.693 \times 4 \times 2} = 250 K$

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