A first order rkn at 298K can be expressed by generalisation logK(ln s -1 )=17.64-(2.5*10 4 /T) then what is the activation energy of the rkn will be [K=rate constant of rkn ;T=Kelvin temp.] ?

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SPARSH JAUHARI · Answer

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

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A first order rkn at 298K can be expressed by generalisation logK(ln s -1 )=17.64-(2.5*10 4 /T) then what is the activation energy of the rkn will be [K=rate constant of rkn ;T=Kelvin temp.] ?

**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.] ?**

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A first order rkn at 298K can be expressed by generalisation logK(ln s -1 )=17.64-(2.5*10 4 /T) then what is the activation energy of the rkn will be [K=rate constant of rkn ;T=Kelvin temp.] ?

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.] ?

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**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.] ?**