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Grade 12Physical Chemistry

A first order rkn at 298K can be expressed by generalisation logK(ln s-1)=17.64-(2.5*104/T) then what is the activation energy of the rkn will be [K=rate constant of rkn ;T=Kelvin temp.] ?

Profile image of basit ali
14 Years agoGrade 12
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1 Answer

Profile image of SPARSH  JAUHARI
ApprovedApproved Tutor Answer14 Years ago

K=Ae^(-Ea/RT)
log k= logA -(Ea)/RT
here logA=17.64
and Ea/R=2.5*10^4
Put R=8.314