What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{0} C$ to $35^{0} C$ ? (R = 8.314 J $m o l^{- 1} K^{- 1}$ )

269 kJ

m o l^{- 1}

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34.7 kJ

m o l^{- 1}

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15.1 kJ

m o l^{- 1}

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342 kJ

m o l^{- 1}

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Solution

The correct option is C 34.7 kJ $m o l^{- 1}$
$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Substituting the values

$l o g 2$ = $\frac{E_{a}}{2.303 \times 8.314} (\frac{1}{293} - \frac{1}{308})$

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

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Similar questions

The rate of a reaction doubles when its temperature changes form 300 K to 310 K. Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

Q. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

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

(IIT-JEE-2013)

Q. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20C to 35C ? (R = 8.314 J

m o l^{- 1}

K^{- 1}

)

For the reaction $A + B \to C + D Δ H = - 20 k J m o l^{- 1}$ . The activation energy of the forward reaction is $85 k J m o l^{- 1}$ . The activation energy for backward reaction is

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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200C to 350C? (R = 8.314 J mol−1K−1)

The correct option is C 34.7 kJ mol−1logk2k1 = Ea2.303×R(1T1−1T2)Substituting the valueslog 2 = Ea2.303×8.314(1293−1308)Ea=34.7 kJ mol−1

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{0} C$ to $35^{0} C$ ? (R = 8.314 J $m o l^{- 1} K^{- 1}$ )

The correct option is C 34.7 kJ $m o l^{- 1}$
$l o g \frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Substituting the values

$l o g 2$ = $\frac{E_{a}}{2.303 \times 8.314} (\frac{1}{293} - \frac{1}{308})$

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