1.

What is the activation energy for a reaction if the rate is doubled when the temperature is raised from 20^(@) C to 35^(@)C (R=8.314 J mol^(-1)K^(-1))

Answer»

15.1 kJ `MOL^(-1)`
`342 kJ mol^(-1)`
`269 kJmol^(-1)`
`34.7 kjmol^(-1)`

Solution :d) `k_(2)/k_(1) = 2, T_(1)=20+273=293 K`,
`T_(2)=35 + 273=308`K
`k=8.314 JK^(-1)mol^(-1)`
`log k_(2)/k_(1) =E_(a)/(2.303 ) [1/T_(1)-1/T_(2)]`
`log2= (E_(a))/(2.303 xx 8.314(JK^(-1)mol^(-1))[1/(293K)-1/(308K)]`
`0.3010= (E_(a) xx 15)/(2.303 xx 8.314 (Jmol^(-1)) xx 293 xx 308)`
`E_(a) = (0.3010 xx 2.303 xx 8.314 x 293 xx 308)/(15) (Jmol^(-1))`
`=34.673 J mol^(-1) = 34.673 kJ mol^(-1)`


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