1.

Rate constant K varies with temperature by equation log K("min"^(-1))=5-((2000))/T.We can conclude that (R=8.314J"mol"^(-1)K^(-1) (or) cal "mol"^(-1)K^(-1))

Answer»

Pre exponential factor A is 5
Ea is 4 K cal/mol
Pre exponential factor, A is `10^(5)`
Ea is 9.212 Kcal/mol

Solution :`K=A.e^(-Ea//RT),logK=logA-(epsilon_(a))/(2.303R)xx1/T`
`logA=5,A=10^(5),(E_(a))/(2.303RT)=2000/T,E_(a)=(2000xx2.303xx2)/1,E_(a)=9.212KCal`


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