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} $
• E. $ 41760 \, kJ \, mol^{ - 1} $

Answer 1

Answer: (A)

log $ \frac{ k_2 }{ k_1 } = \frac{ E-a }{ 2.303 \, R } \bigg [ \frac{ T_2 - T_1 }{ T_1 - T_2 } \bigg ] $
$ \frac{ k_2 }{ k_1} = 6, \, T_1 = 350 \, K, \, T_2 = 400 K$
log 6 $ = \frac{ E_a }{ 2.303 \times 8.314 } \bigg [ \frac{ 400 - 350 }{ 400 \times 350 } \bigg ] $
0.7782 = $ \frac{ E_a }{ 2.303 \times 8.314 } \times \frac{ 50 }{ 400 \times 350 } $
$ E_a = 41.721 \, J \, mol^{ - 1} $
= 41.721 kJ $ mol^{ - 1} $