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

342 kJ

m o l^{- 1}

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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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Solution

The correct option is C 34.7 kJ $m o l^{- 1}$
We know that Rate $\propto K$ where k = rate constant which varies with temperature.
So $\frac{k_{f}}{k_{i}} = 2$
Now we can simply apply,
log $\frac{k_{f}}{k_{i}} = \frac{E a}{2.303 \times R} (\frac{1}{T_{i}} - \frac{1}{T_{f}})$
which gives Ea=34673 J/mol $≃ 34.7 K J / m o l$

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Similar questions

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)

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:

Q. What is the activation energy for a reaction if its rate doubles when the temperature is raised from

20^{0} C

35^{0} C

? (R = 8.314 J

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

)

Q. What is the activation energy for a reaction if its rate doubles when the temperature is raised from

20^{\circ} C

35^{\circ} C

? (

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

)

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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 mol−1K−1)

The correct option is C 34.7 kJ mol−1We know that Rate ∝K where k = rate constant which varies with temperature. So kfki=2Now we can simply apply, logkfki=Ea2.303×R(1Ti−1Tf)which gives Ea=34673 J/mol ≃34.7KJ/mol

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