The activation energy....

Asked by  | 11th Oct, 2009, 09:23: PM

Expert Answer:

The formula to consider is: log k = -(Ea/2.303RT) + log A

For the first case,

log k2 = -(Ea2/2.303RT) + log A

log k2 = (-50.14/2.303RT) + log A

log k1=(-75.2/2.303RT) + log A

Subtracting the two equations, we get

log k2 - log k1 = (-50.14/2.303RT) - (-75.2/2.303RT)

                         = 0.0043

k2/k1 = 27

Therefore, in the presence of catalyst the rate of reaction increases 27 times.

Answered by  | 13th Oct, 2009, 02:39: PM

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