What is the activation energy of a reaction if its rate doubles when temperatures is raised from $20^{o}$ C to $35^{o}$ C?(R $= 8.314$ J $K^{- 1}$ $m o l^{- 1}$ )

34.7

m o l^{- 1}

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15.1

m o l^{- 1}

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342

m o l^{- 1}

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269

m o l^{- 1}

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Solution

The correct option is A $34.7$ kJ $m o l^{- 1}$
Let us consider the Arrhenius equation,

We have

$log \frac{K_{2}}{K_{1}} = - \frac{E_{a}}{2.303 R} [\frac{T_{1} - T_{2}}{T_{1} T_{2}}]$

Given that
Initial temperature $T_{1} = 20 + 273 = 293 K$
Final temperature $T_{2} = 35 + 273 = 308 K$

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

Since, the rate becomes double on raising the temperature,
$∴ r_{2} = 2 r_{1}$

$\frac{r_{2}}{r_{1}} = 2$

Hence the rate constant, $K \propto r$

Since $\frac{K_{2}}{K_{1}} = 2$

putting the values

$log 2 = - \frac{E_{a}}{2.303 \times 8.314} [\frac{293 - 308}{293 \times 308}]$

$E_{a} = 34673.48 J m o l^{- 1}$

$E_{a} = 34.7 K J m o l^{- 1}$

Hence, the correct option is $A$

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What is the activation energy of a reaction if its rate doubles when temperatures is raised from 20oC to 35oC?(R=8.314 JK−1 mol−1)

What is the activation energy of a reaction if its rate doubles when temperatures is raised from $20^{o}$ C to $35^{o}$ C?(R $= 8.314$ J $K^{- 1}$ $m o l^{- 1}$ )