The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction ( R = 8.314 JK - 1 mol - 1 )

Question

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction ( R = 8.314 JK - 1 mol - 1 ) (A)  41.7 KJ mol - 1  (B)  4.17 kJ mol - 1  (C)  417 kJ mol - 1  (D)  0.417 kJ mol - 1

Tardigrade Admin · Accepted Answer

log ( k2 / k1 ) = ( E-a / 2.303 R ) [ ( T2 - T1 / T1 - T2 ) ] ( k2 / k1) = 6, T1 = 350 K, T2 = 400 K log 6 = ( Ea / 2.303 × 8.314 ) [ ( 400 - 350 / 400 × 350 ) ] 0.7782 = ( Ea / 2.303 × 8.314 ) × ( 50 / 400 × 350 ) Ea = 41.721 J mol - 1 = 41.721 kJ mol - 1

Q. The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction
$(R = 8.314 J K^{- 1} m o l^{- 1})$

$41.7 K J m o l^{- 1}$

$4.17 k J m o l^{- 1}$

$417 k J m o l^{- 1}$

$0.417 k J m o l^{- 1}$

$41760 k J m o l^{- 1}$

Q. The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction (R=8.314JK−1mol−1)

41.7KJmol−1

4.17kJmol−1

417kJmol−1

0.417kJmol−1

41760kJmol−1