A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is again 50% complete in 5 minutes at 350 K, Calculate the activation energy if it is a first order reaction. [`R = 8.314 JK^(-1) mol^(-1) , log 4 = 0.602`]

Answer

`24.210 KJ mol^(-1)`

Explanation

`k=( 0.693 )/(t_(1//2))`
`k_1 =( 0.693 )/( 20 ) min ^(-1) , k_2 = ( 0.693 ) /(5) min""^(-1)`
According to Arrhenius equation
`log (k_2 )/(k_1) = (E_a )/( 2.303 R [(T_2 - T_1)/(T_1 T_2) ]`
` log (0.693//5 )/( 0.693 //20) = (E_a )/(2.303 xx 8.314 JK^(-1) mol^(-1)) [(350 -300 )/(350 xx 300 )]`
`log 4 = (E_a)/( 2.303 xx 8.314 ) xx (50)/( 350 xx 300 J mol^(-1))`
` or E_a = 0.602 xx 2.303 xx 8.314 xx 7 J mol^(-1)`
`= 24210 J mol^(-1) = 24.210 KJ mol^(-1)`