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Otrada [13]
3 years ago
8

A certain reaction has an activation energy of 28.90 kJ / mol. At what Kelvin temperature will the reaction proceed 5.00 times f

aster than it did at 313 K?
Chemistry
1 answer:
BabaBlast [244]3 years ago
4 0

Answer: 365 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 = 1.00

K_2 = rate constant at T_2 = 5.00

Ea = activation energy for the reaction = 28.90 kJ/mol= 28900 j/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 313 K

T_2 = final temperature = ?

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

\log (\frac{5.00}{1.00})=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]

0.69=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]

T_2=365K

Therefore, 365 K is required to increase the reaction rate by 5.00 times.

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