We have,
`A=4.0xx10^(13)sec^(-1), E_(a)=98.6 xx10^(3)J//mol,`
`t_(1//2)=10xx60 sec, K=?`
As, `K=Ae^(-E_(a)//RT)`
Therefore, `K=4.0xx10^(13)e^(-[(98.6xx10^(3))//(8.314xxT)])`
Or `log_(e) (0.693//600)=log_(e)(4xx10^(13))`
`-[(98.6xx10^(3))(8.314xxT)]`
Or, `log_(10)(0.693//600)=log_(10)(4xx10^(13))-[(98.6xx10^(3))//(8.314xx2.303xxT)]`
`T=311.35K`