The values of the rate constant of a reaction were
determined at several temperatures. A plot
logk Vs 1/T gave a straight line with slope of -2.1x10 K. What is the activation energy of the?
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Answer:
According to Arrhenius equation
logk=logA−
2.303
E
a
⋅
T
1
Plot of log k vs
T
1
is a straight line.
Slope = −
2.303R
E
a
Intercept = log A
solution
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