In general, it is observed that the rate of a chemical reaction doubles with every 10° rise in temperature. If this generalisation holds for a reaction in the temperature range 298 K to 308 K, what would be the value of activation energy for this reaction ? (R` = 8.314 JK^(-1) mol^(-1)`)

Explanation

According to Arrhenius equation,
`log ""(K_2)/(k_1) = (E_a)/(k_1) = (E_a)/(2.303 R ) [(T_2-T_1])/(T_1 T_2)`
here ` T_1 = 298 K_1 =k ` (say ) `T_2 = 308 K ,K_2 = 2k`
` therefore log "" (2k)/(k ) = (E_a)/(2.303 xx 8.314 ) xx [(308 -298 )/(298 xx 308 )]`
` or E_a =(2.303 xx 8.314 xx 298 xx 308 xx log 2)/( 10 )`
` = 52903 J mol^(-1) = 52.9 KJ mol^(-1)`