दिया है: `T_(1) = 293 K, T_(2) = 313 K, k_(2)= 4 xx k_(1)`,
`R= 8.314 xx 10^(-3) kJ K^(-1) "mol"^(-1)`
आरहीनियम समीकरण के अनुसार,
`log_(10) k_(2)/k_(1) = E_(a) / (2.303R) [1/T_(1)-1/T_(2)]`
`log_(10)(4 xx k_(1))/k_(1) = E_(a)/(2.303 xx 8.314 xx 10^(-3)) xx [1/293 01/313]`
`0.6021 = E_(a)/0.0919 xx 20/(293 xx 313)`
`E_(a) = (0.6021 xx 0.0191 xx 293 xx 313)/(20)`
`=52.73 kJ mol^(-1)`