1.

A first order reaction is 50% complete in 30 minutes at 27^@C and in 10 minutes at 47^@C. The energy of activation of the reaction is

Answer»

`43.83 kJ mol^(-1)`
`34.84kJmol^(-1)`
`84.00kJ mol^(-1)`
`30.00kJmol^(-1)`

Solution :`k_(27^@C)= (0.693)/30 MI n^(-1)`,
`k_(47^@C)=0.693/10 mi n^(-1)`
`:. K_(47^@C)/k_(27^@C)=3`
`log. (k_(47^@C))/(k_(27^@C))= log 3 = 0.4771`
`:. 0.4771= E_a/(2.303 xx 8.314)(20/(300xx 320))`
`E_a=(0.4771 xx 2.303 xx 8.314xx 300xx320)/20`
`= 43.84 kJ`


Discussion

No Comment Found

Related InterviewSolutions