The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to double as described?

Answer : The activation energy for the reaction is, 51.9 kJ

Explanation :

According to the Arrhenius equation,

or,

where,

= rate constant at 295 K

= rate constant at 305 K =

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

= initial temperature = 295 K

= final temperature = 305 K

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

Therefore, the activation energy for the reaction is, 51.9 kJ

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.

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