`log K_(2) - log K_(1) = (E_(a))/(2.303 R)[(1)/(T_(1))-(1)/(T_(2))]`
`log k_(2) = log k_(1) + (E_(a))/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
`k_(1)= 1.60 xx 10^(-5) s^(-1) ,E_(a) = 209 kJ // mol`
`=209 xx10^(3) J mol^(-1) k_(2) = ? ,T _(1)= 600 k`
` T_(2) = 700 k , R = 8.314 J mol^(-1) k^(-1)`
मान रखने पर
`log k_(2) = log (1.60 xx 10^(-5) ) + (209 xx 10^(3))/(2.303 xx 8.314) [(700-600)/(600 xx 700)]`
`log K_(2) = - 5 log 10 + log 1.60 + (209 xx 10^(3))/(19.1471) [ (100)/(42000)]`
`log k_(2) = (-5 + 0.204 ) + 10.9154 xx 10^(3) (2.3809 xx 10^(-4))`
`log k_(2) = - 4.7959 + 2.5958`
`=- 2.197`
`k_(2)= “ Antilog “ (-2.1971)`
`k_(2)`= Antilog `(var(3).8029)`
`=6.352 xx 10^(-3) s^(-1)`