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Arrhenius Equation

The rate cons...

Question

The rate constant of a reaction at $300 K$ is $5.0 \times 10^{- 4} m i n u t e^{- 1}$ . The temperature was increased by $20 K$ and the value of rate constant $^{'} K^{'}$ increased three times. Calculate the energy of activation of the reaction?
What will be the value of rate constant at $37^{\circ} C$ ? $[R = 1.987 c a l o r i . K e l v i n^{- 1} . m o l^{- 1}]$ .

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Solution

Using Arrheneius equation we get
$ln k = - \frac{E_{a}}{R T} ln (\frac{k_{1}}{k_{2}}) = \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
on substitution we get:
$ln (\frac{5 \times 10^{- 4}}{3 \times 5 \times 10^{- 4}}) = \frac{E_{a}}{1.987} (\frac{1}{320} - \frac{1}{300})$
$E_{a} = 10478.12$ cal/mol
Rate constant at $37^{\circ} C$ T=310 K
$ln k = - \frac{10478.12}{1.987 \times 310} = - 17.01 k = 4.09 \times 10^{- 8}$

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