The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271 K and 281 K. What must the activation energy be for the rate constant to double as described?

Answer : The activation energy for the reaction is, 119.7 J

Explanation :

According to the Arrhenius equation,

or,

where,

= rate constant at 271 K

= rate constant at 281 K =

= activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

= initial temperature = 271 K

= final temperature = 281 K

Now put all the given values in this formula, we get:

Therefore, the activation energy for the reaction is, 119.7 J

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explanation:

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