The activation energy of a reaction is 9 kcal `"mole "^(-1) ` . The increase in the rate cnstant when its temperature is raised from 295 to 300 approximately

Options

A. `1.289` times
B. `12.89` times
C. `0.1289` times
D. 0.25

Answer

A

Explanation

`log ""(K_(2))/(K_(1)) =(E_(a). DT )/(2.303RT_(2)T_(1))`
`=(9000xx5)/(2.303xx2xx300xx295)=0.1104`
`log ""(k_(2))/(k_(1))=0.1104 therefore (K_(2))/(K_(1))=1.289`
` therefore K_(2)=K_(1)xx1.289`