Find the energy of activation for the reaction who rate doubles when the temperature changes from $30^{o} C to 40^{o} C$ .

5.46 k J m o l^{- 1}

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54.65 k J m o l^{- 1}

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23.5 k J m o l^{- 1}

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81.21 k J m o l^{- 1}

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Solution

The correct option is B $54.65 k J m o l^{- 1}$
Expression for rate constant at two different temperature is given by
$l o g_{10} \frac{k_{2}}{k_{2}} = \frac{E_{a}}{R \times 2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Given: $\frac{k_{2}}{k_{1}} = 2; R = 8.314 J K^{- 1} m o l^{- 1}; T_{1} = 30 + 273 = 303 K$
and $T_{2} = 40 + 273 = 313 K$ Substituting the given values in the equation,
$l o g_{10} 2 = \frac{E_{a}}{8.314 \times 2.303} [\frac{313 - 303}{313 \times 303}]$
$E_{a} = \frac{2.303 \times 8.314 \times 313 \times 303 \times 0.301}{10}$
$= 54658.46 J m o l^{- 1}$
$= 54.65 k J m o l^{- 1}$

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Similar questions

30^{o} C

40^{o} C

30^{\circ} C to 40^{\circ} C

27^{o} C

37^{o} C

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

27^{0}

37^{0}

300 K

310 K

(R = 8.314 J K^{- 1} {m o l}^{- 1} a n d log 2 = 0.301)

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