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Rate Constant

The rate cons...

Question

The rate constant of a first order reaction at $25^{o} C$ is ${0.24 s}^{- 1}$ . If the energy of activation of the reaction is ${88 k J m o l}^{- 1}$ , at what temperature would this reaction have rate constant of ${4 \times 10}^{- 2} s^{- 1}$ (283.7 K) ?

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Solution

$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g \frac{6}{1} = \frac{88000}{2.303 \times 8.314} (\frac{1}{T_{1}} - \frac{1}{298})$

$T_{1} = 283.7 K$

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104 k J m o l^{- 1}

25^{o} C

3.7 \times 10^{- 5} s^{- 1}

30^{o} C

k_{1} =

T_{1}

k_{2} =

T_{2}

E_{a} =

(E_{a})

2 \times 10^{- 2} s^{- 1}

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8 \times 10^{- 2} s^{- 1}

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0.58 s^{- 1}

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