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Bogdan [553]
3 years ago
8

A certain reaction has an activation energy of 66.41 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times fas

ter than it did at 293 K?
Chemistry
1 answer:
swat323 years ago
8 0

Answer : The temperature will be, 392.462 K

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at T_1  = K_1

K_2 = rate constant at T_2 = 3K_1

Ea = activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 293 K

T_2 = final temperature = ?

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

\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]

T_2=392.462K

Therefore, the temperature will be, 392.462 K

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