CBSE Class 12-science Answered
The activation energy....
Asked by | 11 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 | 13 Oct, 2009, 02:39: PM
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