The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 `JK^(-1) mol^(-1)`)

Answer

`55.33 KJ mol^(-1)`

Explanation

According to Arrhenius equation
`log (k_2)/(k_1) = (E_a)/(2.303 R) [ (1)/(T_1)- (1)/(T_2)]`
here ` T_1 = 300 K , T_2 = 320 K`
` log (k_2)/(k_1) = log 4=0.6020 `
substituting the values we get
`log 4= ( E_a)/( 2.303 xx 8.314 ) [(1)/(300 ) - (1)/(320)]`
` or 0.6020 = (E_a)/(17.147 ) xx [(320-300 )/( 300 xx 320 )]`
` or 0.6020 = (E_a )/( 19.147 ) xx (20 )/( 300 xx 320 )`
` or E_a = ( 0.6020 xx 19.147 xx 300 xx 300 )/(20 )`
`= 55327.2 J mol^(-1) = 55.33 KJ mol^(-1)`