Question

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

Answer 1

Explanation:

Solution:

log k2k1=E−a2.303R[T2−T1T1−T2]k2k1=E−a2.303R[T2−T1T1−T2]

k2k1=6,T1=350K,T2=400Kk2k1=6,T1=350K,T2=400K

log 6 =Ea2.303×8.314[400−350400×350]=Ea2.303×8.314[400−350400×350]

0.7782 = Ea2.303×8.314×50400×350Ea2.303×8.314×50400×350

Ea=41.721Jmol−1Ea=41.721Jmol−1

= 41.721 kJ mol−1