1.

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k log A(E_a)/(2.303R)(1/T) where E_a is the activation energy . When a graph is plotted is or logk Vs 1/T a straight line with a slope of - 400K is obtained . Calculate the activation energy .

Answer»

SOLUTION :LOG k = log A - `E_a/(2.303R)(1/T)`
y = c + mx
`m=-E_a/(2.303R)`
`E_a=-2.303Rm`
`E_a=-2.303xx8.314"J K "^(-1)"MOL"^(-1)XX(-400K)`
`E_a=76,589"J mol"^(-1)`
`E_a = 76 589 "kJ mol"^(-1)` .


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