1.

The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = 8.314 JK^(-1) mol^(-1) and log2 = 0.301)

Answer»

`53.6 kJ mol^(-1)`
48.6 kJ `mol^(-1)`
58.5 kJ `mol^(-1)`
60.5 kJ `mol^(-1)`

SOLUTION :log `(K_(2))/(K_(1)) = (-E_(a))/(2.030R) ((1)/(T_(2)) - (1)/(T_(1)))`
`(K_(2))/(K_(1)) = 2 , T_(2) = 310 K"" T_(1) = 300` K
`implies ` log `2 = (-E_(a))/(2.303 XX 8.314 ) ((1)/(310) - (1)/(300))`
`implies E_(a) = 53598.6` J/mol = 53.6 KJ/mol


Discussion

No Comment Found

Related InterviewSolutions