The rate of a reaction triples when temperature changes from `50^@` C to `100^@ `C. Calculate the energy of activation for such a reaction. (`R = 8.314 JK^(-1) mol^(-1) log 3 = 0.4771`)

Answer

`22.012 mol^(-1)`

Explanation

According to Arrhenius equation
`log (k_2)/(k_1) = ( E_a )/(2.303 R) [(1)/(T_1)-(1)/(T_2)]`
Substiting the values , We get
`log 3= ( E_a)/( 2.303 xx 8.314 ) [(1)/( 323 ) -(1)/( 373)]`
` 0.4771 = (E_a )/( 19.147 ) xx ( 50) /( 120479 ) , 0.4771 = ( 50 E_a)/(19.147 xx 120479 )`
` E_a = ( 0.4771 xx 19.147 xx 120479 )/( 50 )`
`=22011.59 J mol^(-1) = 22.012 KJ mol^(-1)`