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Rate constant 'k' of a reaction varies w...

Rate constant 'k' of a reaction varies with temperature 'T' according to the equation:
`log k= logA-(E_(a))/(2.303R)((1)/(T))`
where `E_(a)` is the activation energy. When a graph is plotted for log k vs `(1)/(T)`, a straight line with a slope of -4250 K is obtained. Calculate `'E_(a)'` for the reaction. `[R=8.314JK^(-1)"mol"^(-1)]`

Text Solution

Verified by Experts

The equation
`log k=log A -(E_(a))/(2.303R) ((1)/(T))`
is of the type `y=mx +C` where m is the slope of the line.
`therefore (-E_(a))/(2.303R)= -4250 K or E_(a)=2.303R xx 4250 K`
or `E_(a)=2.303xx 8.314 JK^(-1)"mol"^(-1) xx 4250K`
`=81375"J mol"^(-1)=81.375 "kJ mol"^(-1)`
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Knowledge Check

  • Rate constant k of a reaction varies with temperature according to the equation, log k = constant - (E_a)/(2.303RT) When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

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