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Rate constant k of a reaction varies wit...

Rate constant k of a reaction varies with temperature according to the equation
`logk=" constant"-(E_(a))/(2.303).(1)/(T)`
where `E_(a)` is the energy of activation for the reaction. When a graph is plotted for log k versus `1//T`, a straight line with a slope -6670 K is obtained. Calculate the energy of activation for this reaction. State the units `(R=8.314" JK"^(-1)mol^(-1))`

Text Solution

Verified by Experts

Slope of the line `=-(E_(a))/(2.303"R")=-6670" K"`
or `" "E_(a)=2.303xx8.314 ("JK"^(-1)mol^(-1))xx6670" K"=127711.4" J mol"^(-1).`
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