If the rate constants of a reaction at 500K and 700K are `0.002 s^(-1) and 0.06 s^(-1)` respectively, the value of `K^(-1)` activation energy is `(R = 8.314 J mol^(-1) K^(-1), log 3 = 0.477)`

Options

A. 49.49 kJ `mol^(-1)`
B. 98.98 kJ `mo^(-1)`
C. 24.75 kJ `mol^(-1)`
D. 12.37 kJ `mol^(-1)`

Answer

A

Explanation

`log((k_2)/(k_1))= (Ea)/(2.303 R) ((T_2 - T_1)/(T_1 T_2))`
`log((0.06)/(0.0020)) = (Ea)/(2.303 xx 8.314)((700 - 500)/(700 xx 500))`
`log(30) = (Ea)/(19.147) ((200)/(350000))`
`1.477 = (Ea.)/(19.147) ((200)/(3500))`
`(therefore log (30) = log (3) + log (10 = 0.477+1 = 1.477)`
`therefore Ea = 1.477 xx 19.147 xx 1750 = 49.49` kJ `mol^(-1)`