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For the reaction 2NOCl(g) hArr 2NO(g)+Cl...

For the reaction `2NOCl(g) hArr 2NO(g)+Cl_(2)(g)`, the equilibrium constant is `2.8xx10^(-5)` at `300 K` and `7.0xx10^(-1)` at `400 K`. What is the activation energy for the reaction?

Text Solution

Verified by Experts

Given `K_(1)=2.8xx10^(-5)` at `T_(1)=300 K`
`K_(2)=7.0xx10^(-1)` at `T_(2)=400 K`
Using expression
ln`(K)=(-Ea)/(RT)`
we get, ln`(2.8xx10^(-5))=(-Ea)/(R(300K))+ Constant …(i)`
in`(7.0xx10^(-1))=(-Ea)/(R(400K))+ Constant …(ii)`
Subtracting (i) from (ii), we get
ln`((7.0xx10^(-1))/(2.8xx10^(-5)))=(Ea)/(R)(3/(300K)-1/(400K))`
`=(Ea)/(8.314 J K^(-1) "mol"^(-1))(100/(300xx400))`
`:. Ea=(8.314)((300xx400)/(100))[2.303log((7.0xx10^(-1))/(2.8xx))]`
`=101048 J "mol"^(-1)=101.048 K J "mol"^(-1)`
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