The rate constant for a first order reaction becomes six times when the temperature is raised from $350 K$ to $400 K$ . Calculate the activation energy for the reaction. $[R = 8.314 J K^{- 1} m o l^{- 1}]$

4.17 k J m o l^{- 1}

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41.7 k J m o l^{- 1}

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417.0 k J m o l^{- 1}

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4170 k J m o l^{- 1}

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Solution

The correct option is A $41.7 k J m o l^{- 1}$
$K = A e^{\frac{- E_{a}}{R T}}$
$K_{2} = 6 K_{1}$

$\frac{K_{2}}{K_{1}} = e^{\frac{E_{a}}{R} \times (\frac{1}{T_{1}} - \frac{1}{T_{2}})}$

$6 = e^{\frac{E_{a}}{R} \times (\frac{1}{350} - \frac{1}{400})}$

$E_{a} = 41.7 k J m o l^{- 1}$

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Similar questions

5

350

400

= 8.314

K^{- 1} m o l^{- 1}

(E_{a})

[R = 8.314 J K^{- 1} m o l^{- 1}, l o g 4 = 0.6021]

20^{o} C

50^{o} C

(R = 8.314 J K^{- 1} {m o l}^{- 1})

30^{0} C

50^{0} C

J m o l^{- 1} K^{- 1}

27^{o} C

37^{o} C

[R = 8.314 J K^{- 1} m o l e]

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