## Q – The rate constant for a reaction was measured as a function of temperature. A plot of lnK versus T ( in kelvin ) is linear and has a slope of -1.075 x 10 ^{4} K. Find the activation energy ?

### Solution-

*Arrhenius equation , *

*K = Ae *^{-Ea/RT}

^{-Ea/RT}

*ln K = ln A + ln e ^{-Ea/RT}*

* = [-Ea/R]1/T + ln A*

*we know, *

*y = m . x + c*

*slope ‘m’ = – Ea/R*

*R = 8.314 J mole ^{-1} K^{-1}*

*slope = -1.075 x 10 ^{4}*

* – Ea = -1.075 x 10 ^{4} x 8.314*

*Ea = 8.94 x 10 ^{4}*

*Activation energy = 8.94 x 10*^{4}

^{4}